9 Wave phenomena

Contents

91 ndash Simple harmonic motion

92 ndash Single-slit diffraction

93 ndash Interference

Essential Ideas

The solution of the harmonic oscillator can be framed around the variation of kinetic and potential energy in the system

Single-slit diffraction occurs when a wave is incident upon a slit of approximately the same size as the wavelength

Interference patterns from multiple slits and thin films produce accurately repeatable patterns copy IBO 2014

PHYSICS 2015indb 283 150514 337 PM

Chapter 9

284

AH

L

91 Simple harmonic motion

Essential idea The solution of the harmonic oscillator can be framed around the variation of kinetic and potential energy in the system

Understandingsbull ThedefiningequationofSHMbull Energychanges

Recall from chapter 4 that if the acceleration a of a system is

bull directlyproportionaltoitsdisplacementxfromitsequilibriumposition

andbull directedtowardstheequilibriumposition

thenthesystemwillexecuteSHM

ThisistheformaldefinitionofSHM

Weexpressedthisdefinitionmathematicallyas

a = ndash const times x

The negative sign indicated that the acceleration was directedtowardstheequilibriumpositionMathematicalanalysis shows that the constant is in fact equal to ω2whereω is theangular frequency(definedabove)of thesystemHenceequation45becomes

a=ndashω2 x

If a system is performing SHM then to produce theaccelerationaforcemustbeactingonthesysteminthedirectionoftheaccelerationFromourdefinitionofSHMthe magnitude of the force F is given by

F = ndash kx

where k is a constant and the negative sign indicates that theforce isdirectedtowardstheequilibriumpositionofthesystem(DonotconfusethisconstantkwiththespringconstantHoweverwhendealingwiththeoscillationsofamassontheendofaspringkwillbethespringconstant)

To understand the solutions of the SHM equation letus consider the oscillations of a mass suspended from a vertically supported spring We shall consider the mass of the spring to be negligible and for the extension x to obey the rule F = kx F (= mg) is the force causing the extension Figure901(a)showsthespringandasuspendedweightofmass m inequilibriumInFigure901(b) theweighthasbeen pulled down a further extension x0

spring

weight of mass m x0

x0

P x

unbalanced force F on weight = -kx

(a) (b)

equilibrium x = 0

unstretched length

e ke

mg

Figure 901 SHM of a mass suspended by a spring

InFigure901(a)theequilibriumextensionofthespringis e and the net force on the weight is mg ndash ke = 0

InFigure408(b)iftheweightisheldinpositionatx=x0 and thenreleasedwhentheweightmovestopositionPadistance xfromtheequilibriumpositionx=0thenetforceon the weight is mg ndash ke ndash kxClearlythentheunbalancedforce on the weight is ndashkx When the weight reaches a point distance xabovetheequilibriumpositionthecompressionforce in the spring provides the unbalanced force towards theequilibriumpositionoftheweight

The acceleration of the weight is given by Newtonrsquos second law

F = ndashkx = ma

ie

This is of the form a = -ω2 x where ω = that is the weight will execute SHM with a frequency

The displacement of the weight x is given by

ω

ThisistheparticularsolutionoftheSHMequationfortheoscillation of a weight on the end of a spring This system

NATURE OF SCIENCEInsights The equation for simple harmonic motion (SHM) can be solved analytically and numerically Physicists use such solutions to help them to visualize the behaviour of the oscillator The use of the equations is very powerful as any oscillation can be described in terms of a combination of harmonic oscillators Numerical modelling of oscillators is important in the design of electrical circuits (111)

copy IBO 2014

PHYSICS 2015indb 284 150514 337 PM

Wave Phenomena

285

AH

L

is often referred to as a harmonic oscillator

Alternativelyiftheweightwasatitsequilibriumpositionwhent=0thenthetrigonometricexpressionwouldbe

x = x0sinωt

The velocity v of the weight at any instant can be found byfinding thegradient of the displacement-time graph at that instant The displacement time graph is a cosine function and the gradient of a cosine function is a negative sine function The gradient of

x = x0cosωt is in fact T 2T00

00

velo

city

dis

pla

cem

ent

t

so

ω ω ω

where v0isthemaximumandminimumvelocityequalinmagnitude to ωx0

Studentsfamiliarwithcalculuswillrecognisethevelocityv as

( )0d d cos sind d oxv x t x tt t

ω ω ω= = = minus Similarly

( ) 2 20 0

d d sin cosd dva x t x t xt t

ω ω ω ω ω= = minus = minus = minus

whichofcourseisjustthedefiningequationofSHM

However we have to bear in mind that ωt varies between 0 and 2π where cosωt is negative for ωt for

and sinωt is negative for ωt in the range π to

2π This effectively means that when the displacement fromequilibriumispositivethevelocityisnegativeandsodirectedtowardsequilibriumWhenthedisplacementfromequilibriumisnegativethevelocityispositiveandsodirectedawayfromequilibrium

ThesketchgraphinFigure902showsthevariationwithtime t of the displacement x and the corresponding variation with time t of the velocity v This clearly demonstrates the relation between the sign of the velocity and sign of the displacement

T 2T00

00

velo

city

disp

lace

men

t

t

x0

ndashx0

v0

ndashv0

Figure 902 Displacement-time and velocity-time graphs

We can also see how the velocity v changes with displacement x

We can express sinωt in terms of cosωt using the trigonometric relation

2 2sin cos 1θ θ+ =

From which it can be seen that

Replacing θ with ωt we have

20 1 cosv x tω ω= minus minus

Remembering that and putting x0 inside the squarerootgives

v=ndashωradic_______

( x0 2 ndash x2 )

Bearing in mind that v can be positive or negative wemust write

v=plusmnωradic_______

( x0 2 ndash x2 )

The velocity is zero when the displacement is a maximum and is a maximum when the displacement is zero

The graph in Figure 903 shows the variation with x of the velocity v for a system oscillating with a period of 1 secondandwithanamplitudeof5cmThegraphshowsthe variation over a time of any one period of oscillation

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

-5 -4 -3 -2 -1 1 2 3 4 5

v c

m s

-1

x cm

Figure 903 Velocity-displacement graph

PHYSICS 2015indb 285 150514 337 PM

Chapter 9

286

AH

L

Boundary ConditionsThe two solutions to the general SHM equation are

and Which solution applies to aparticularsystemdependsasmentionedaboveontheboundary conditions for that system For systems such astheharmonicoscillatorandthesimplependulumtheboundary condition that gives the solution is that the displacement x = x0 when t = 0 For some other systems it might turn out that x = 0 when t =0 This will lead to the solution From a practical pointofviewthetwosolutionsareessentiallythesamefor example when timing the oscillations of a simple pendulumyoumightdecidetostartthetimingwhenthependulumbobpasses through theequilibriumpositionIneffectthetwosolutionsdifferinphaseby

The table in Figure 904 summarises the solutions we have forSHM

x = x 0 cosωt x = x 0 sinωt

ν = minus ν 0 sinωt ν = ν 0 cosωt

ν = ndash ωx 0 sinωt ν = ndash ωx 0 cosωt

ν = plusmn ω radic______

x0 2 ndash x2 ν = plusmn ω radic

______ x0

2 ndash x2

Figure 904 Common equations

We should mention that since the general solution to the SHMequation is there are infactthreesolutionstotheequationThisdemonstratesa fundamental property of second order differential equationsthatoneofthesolutionstotheequationisthesum of all the other solutions This is the mathematical basis of the so-called principle of superposition

SHM is a very good example in which to apply the NewtonianmethoddiscussedinChapter2ieiftheforcesthatactonasystemareknownthenthefuturebehaviourofthesystemcanbepredictedHerewehaveasituationinwhich the force is given by ndashkx From Newtonrsquos second law therefore ndashkx = mawherem is the mass of the system and a istheaccelerationofthesystemHowevertheaccelerationis not constant For those of you who have a mathematical benttherelationbetweentheforceandtheaccelerationis written as

2dd

xkx mt

minus =2

This is what is called a ldquosecond orderdifferentialequationrdquoThesolutionoftheequationgives x as a function of tTheactualsolutionisoftheSHMequationis

x = Pcosωt + Qsinωt where P and Q are constants and ω is theangularfrequencyofthesystemandisequalto

Whether a particular solution involves the sine function or thecosinefunctiondependsontheso-calledlsquoboundaryconditionsrsquo If for example x = x0 (the amplitude)whent =0thenthesolutionisx = x0cosωt

The beauty of thismathematical approach is that oncethe general equation has been solved the solution forall systemsexecutingSHM isknownAll thathas tobeshowntoknowifasystemwillexecuteSHMisthattheaccelerationofthesystemisgivenbyEquation45ortheforceisgivenbyequation47Thephysicalquantitiesthatω will depend on is determined by the particular system Forexampleforaweightofmassm oscillating at the end of a vertically supported spring whose spring constant is k then ω = or from equation 44 For a simple pendulum ω = where l is the length of the pendulum and g is the acceleration of free fall such

that T=2πradic_

l _ g

Examples

ThegraphinFigure905showsthevariationwithtime t of the displacement x ofasystemexecutingSHM(SomequestionpartswerealreadydoneintheSLchapter4)

-10

-8

-6

-4

-2

0

2

4

6

8

10

05 1 15 2 25 3 35t s

xcm

Figure 905 Displacement ndash time graph for SHM

Use the graph to determine the

(i) periodofoscillation(ii) amplitudeofoscillation(iii) maximumspeed(iv) speedatt = 13 s(v) maximumacceleration

Solutions

(i) 20 s (time for one cycle)(ii) 80 cm(iii) Using gives

(remember that ) = 25 cm s-1

PHYSICS 2015indb 286 150514 337 PM

Wave Phenomena

287

AH

L

(iv) v = minusv0sinωt = minus25sin (13π) To find the value of the sine function we have to convert the 13π into degrees (remember ω and hence ωt is measured in radians)

1 deg therefore 13π = 13 times 180 = 234deg therefore v1 = minus25sin (234deg) = +20 cm s ndash1

Or we can solve using ωradic_______

( x0 2 ndash x2 )

from the graph at t = 13 s x = ndash 48 cm

therefore v = π times = 20 cm s-1

(v) Using = π2 times 80 = 79 m s-2

Energy changesWe must now look at the energy changes involved in SHMTodosowewillagainconcentrateontheharmonicoscillator The mass is stationary at x = +x0 (maximumextension)andalsoatx = ndashx0 (maximumcompression)At these two positions the energy of the system is all potential energy and is in fact the elastic potential energy stored in the spring This is the total energy of the system ET and clearly

Equation412

ThatisthatforanysystemperformingSHMtheenergyofthesystemisproportionaltothesquareoftheamplitudeThis is an important result and one that we shall return to when we discuss wave motion

At x =0thespringisatitsequilibriumextensionandthemagnitude of the velocity v of the oscillating mass is a maximum v0TheenergyisallkineticandagainisequaltoET We can see that this is indeed the case as the expression for the maximum kinetic energy Emax in terms of v0 is

Clearly ET and Emax areequalsuchthat

ET = 1 __ 2 kx0 2 = Emax = 1 __ 2 mv0

2

From which

v0 2 = k __ m x0

2(asv0ge0)

Therefore

v0 = radic__

k __ m x0 = ωx0

whichtiesinwiththevelocitybeingequaltothegradientofthedisplacement-timegraph(see415)Asthesystemoscillates there is a continual interchange between kinetic energy and potential energy such that the loss in kinetic energyequalsthegaininpotentialenergyandET = EK + EP

Remembering that wehavethat

Clearly thepotentialenergyEP at any displacement x is given by

At any displacement xthekineticenergyEK is EK = 1__2mv2

Hencerememberingthat

v_______( x0

2 ndash x2 ) wehave

EK = 1__2m 2 ( x0

2 ndash x2 )

Although we have derived these equations for a harmonicoscillatortheyarevalidforanysystemoscillatingwithSHMThe sketch graph in Figure 906 shows the variation with displacement x of EK and EP for one period

displacement

ener

gy potentialkinetic

Figure 906 Energy and displacement

Example

The amplitude of oscillation of a mass suspended by a vertical spring is 80 cm The spring constant of the spring is74Nmndash1Determine

(a) thetotalenergyoftheoscillator

(b) thepotentialandthekineticenergyoftheoscillator at a displacement of 48 cm from equilibrium

PHYSICS 2015indb 287 150514 337 PM

Chapter 9

288

AH

L

Solution

(a) Spring constant k = 74 N m-1 and

x0 = 80 times 10-2 m

ET = frac12 kx02

= frac12 times 74 times 64 times 10-4 = 024 J

(b) At x = 48 cm

Therefore EP = frac12 times 74 times (48 times 10-2)2

= 0085 times 10-4 J

EK = ET minus EP = 024 minus 0085 = 016 J

EXERCISE 911 Which graph shows the relationship between the

acceleration a and the displacement x from the equilibriumpositionofanobjectundergoingsimple harmonic motion

Acceleration

Displacement

Acceleration

Displacement

A B

Acceleration

Displacement

Acceleration

Displacement

C D

2 AnobjectisundergoingsimpleharmonicmotionaboutafixedpointPandthemagnitudeofitsdisplacementfromPisx Which one of the following statements is correct

Magnitudeoftheresultant force

Directionoftheresultant force

A Proportionaltox AwayfrompointP

B Inversely proportional to x AwayfrompointP

C Proportionaltox TowardspointP

D Inversely proportional to x TowardspointP

3 Theangularspeedoftheldquominuterdquohandofananalogue watch is

A π1800rads-1 C π30rads-1

B π60rads-1 D 120rads-1

4 Which of the following is true of the magnitude of the acceleration of the object that is undergoing simple harmonic motion

A It is greatest at the midpoint of the motionB It is greatest at the end points of the motionC It is uniform throughout the motionD Itisgreatestatthemidpointsandthe

endpoints

5 Aparticleoscillateswithsimpleharmonicmotionwith a period T

At time t=0theparticlehasitsmaximumdisplacement Which graph The variation with time t of the kinetic energy of the particle is shown in which diagram below

A

00

Ek

T t

D

00

Ek

T t

C

00

Ek

T t

B

0

Ek

t0 T

6 Figure907belowshowshowthedisplacementofthe magnet of mass 03 kg hanging from a spring varies with time for two oscillations

0 02 04 06 08 10

0

2

-2

-4

4

Disp

lace

men

t (cm

)

Time (s)

Figure 907

PHYSICS 2015indb 288 150514 337 PM

Wave Phenomena

289

AH

L

Using information from this graph calculate the

(a) valueofthespringconstant(b) maximumkineticenergyofthemagnet

7 AsystemisoscillatingwithSHMasdescribedbythe graph in the Figure 908 below

Disp

lace

men

t (cm

)

005 1 15 2 25 3 35

2

-2

-6

-4

4

6

Time (s)

Figure 908

(a) Usethegraphtodeterminethe

(i) periodofoscillation(ii) amplitudeofoscillation(iii) maximumspeed(iv) speedatt=13s(v) maximumacceleration

(b) Statetwovaluesoftforwhenthemagnitudeofthevelocity is a maximum and two values of t for when the magnitude of the acceleration is a maximum

9 Figure 909 shows the relationship that exists between the acceleration and the displacement from the equilibriumpositionforaharmonicoscillator

Acce

lera

tion

(m-2

)

Displacement (cm)0

-4000

-2000

2000

-05 050

Figure 909

(a) Stateandexplaintworeasonswhythegraphopposite indicates that the object is executing simple harmonic motion

(b) Determinethefrequencyofoscillation

92 Single-slit diffraction

Essential idea Single-slitdiffractionoccurswhenawaveis incident upon a slit of approximately the same size as the wavelength

Understandingsbull Thenatureofsingle-slitdiffraction

The nature of single-slit diffractionSingle slit diffraction intensity distribution

When plane wavefronts pass through a small aperture they spread out as discussed in chapter 4 This is an example ofthephenomenoncalleddiffractionLightwavesarenoexception to this and ways for observing the diffraction of light have also been discussed previously

Howeverwhenwelookatthediffractionpatternproducedbylightweobserveafringepatternthatisonthescreenthere is a bright central maximum with ldquosecondaryrdquomaxima either side of it There are also regions where there is no illumination and these minima separate the maxima If we were to actually plot how the intensity of illumination varies along the screen then we would obtain a graph similar to that as in Figure 910

intensity

distance along screen

Figure 910 Intensity distribution for single-slit diffraction

We would get the same intensity distribution if we were to plot the intensity against the angle of diffraction θ(Seenextsection)

NATURE OF SCIENCEDevelopment of theories When light passes through an aperture the summation of all parts of the wave leads to an intensity pattern that is far removed from the geometrical shadow that simple theory predicts (19)

copy IBO 2014

PHYSICS 2015indb 289 150514 337 PM

Chapter 9

290

AH

L

This intensity pattern arises from the fact that each point ontheslitactsinaccordancewithHuygenrsquosprincipleasa source of secondary wavefronts It is the interference between these secondary wavefronts that produces the typical diffraction pattern

Obtaining an expression for the intensity distribution is mathematically a little tricky and it is not something that we are going to attempt here However we can deducea useful relationship from a simple argument In this argument we deal with a phenomenon called Fraunhofer diffractionthatisthelightsourceandthescreenareaninfinitedistanceawayformtheslitThiscanbeachievedwith the set up shown in Figure 911

lens 1 lens 2

single slit

source

screen

Figure 911 Apparatus for viewing Fraunhofer diffraction

The source is placed at the principal focus of lens 1 and the screen isplacedat theprincipal focusof lens2Lens1 ensures that parallel wavefronts fall on the single slit and lens2ensuresthattheparallelraysarebroughttoafocuson the screen The same effect can be achieved using a laser and placing the screen some distance from the slit If the light and screenarenot an infinitedistance from the slitthen we are dealing with a phenomenon called Fresnel diffraction and such diffraction is very difficult to analyse mathematically To obtain a good idea of how the single slit patterncomesaboutweconsiderthediagramFigure912

Xθ2

Yλ

d

P

b

f

screen

θ1

Figure 912 Single slit diffraction

In particular we consider the light from one edge of the slittothepointPwherethispointis justonewavelengthfurther from the lower edge of the slit than it is from the

upper edge The secondary wavefront from the upper edge willtraveladistanceλ2furtherthanasecondarywavefrontfrom a point at the centre of the slitHencewhen thesewavefronts arrive at P theywill be out of phase andwillinterfere destructively The wavefronts from the next point below the upper edge will similarly interfere destructively with the wavefront from the next point below the centre of the slit In this way we can pair the sources across the whole width of the slit If the screen is a long way from the slit then the angles θ1 and θ2becomenearlyequal (If the screen is at infinity then they are equal and the two lines PX and XY are at right angles to each other)FromFigure1116weseethereforethattherewillbeaminimumatPif

λ b θ1 sin= where b is the width of the slit

Howeverbothanglesareverysmallequaltoθ saywhereθ is the angle of diffraction

Soitcanbewritten θ λb---

=

This actually gives us the half-angular width of the central maximum We can calculate the actual width of the maximum along the screen if we know the focal length of the lens focussing the light onto the screen If this is f then we have that

θ df---

= Suchthat d fλb-----

=

To obtain the position of the next maximum in the pattern we note that the path difference is 3

2---λ We therefore divide

theslitintothreeequalpartstwoofwhichwillproducewavefronts that will cancel and the other producing wavefronts that reinforce The intensity of the second maximum is therefore much less than the intensity of thecentralmaximumof theorderof005or120of theoriginalintensity(Muchlessthanonethirdinfactsincethewavefrontsthatreinforcewillhavedifferingphases)

We can also see now how diffraction effects become more and more noticeable the narrower the slit becomes If light of wavelength 430 nm was to pass through a slit of width say10cmandfallonascreen30mawaythenthehalfangular width of the central maximum would be 013 microm

d = f λ

___ b = 3 times 430 times 10 ndash9 ____________ 01 = 013 μm

There will be lots of maxima of nearly the same intensity and the maxima will be packed very closely together (Thefirstminimumoccursatadistanceof012micromfromthe centre of the central maximum and the next occurs effectivelyatadistanceof024microm)Weeffectivelyobservethe geometric pattern Refer to Example

We also see now how for diffraction effects to be noticeable the wavelength must be of the order of the slit width The width of the pattern increases in proportion to the wavelength and decreases inversely with the width of the

PHYSICS 2015indb 290 150514 337 PM

Wave Phenomena

291

AH

L

93 Interference

Essential idea Interference patterns from multiple slits andthinfilmsproduceaccuratelyrepeatablepatterns

Understandingsbull Youngrsquosdouble-slitexperimentbull Modulationoftwo-slitinterferencepatternbyone-slit

diffraction effectbull Multipleslitanddiffractiongratinginterferencepatternsbull Thinfilminterference

Youngrsquos double slit experimentPlease refer to interference in chapter 4 However wewill reiterate the condition for the interference of waves from two sources to be observed The two sources must be coherentthatistheymusthavethesamephaseorthephase difference between them must remain constant

Also to reinforce topics concerning the principle ofsuperpositionandpathdifferenceandphasedifferenceincluded in Figure 913 is another example of two source interference

We can obtain evidence for the wave nature of sound by showing that sound produces an interference pattern Figure 913 shows the set up for demonstrating this

CRO

speaker

speaker

signalgenerator

microphoneX

Y

Figure 913 Interference using sound waves

The two speakers are connected to the same output of the signal generator andplaced about 50 cmapartThesignalgeneratorfrequencyissettoabout600Hzandthe

slit If the slit width is much greater than the wavelength then the width of the central maxima is very small

Insummary

for destructive interference and minima

dsinθ=nλ wheren=123hellip

for constructive interference and maxima

dsinθ=(n+frac12)λ wheren=123hellip

Note that waves from point sources along the slit arrive in phase at the centre of the fringe pattern and constructively interferetoproducethecentralmaximumcalledthezero ordermaximum(n=0)

Example

LightfromalaserisusedtoformasingleslitdiffractionpatternThe width of the slit is 010 mm and the screen is placed 30 m from the slit The width of the central maximum is measured as26cmCalculatethewavelengthofthelaserlight

Solution

Since the screen is a long way from the slit we can use the small angle approximation such that the f is equal to 30 m

The half-width of the central maximum is 13 cm so we have

λ13 10 2ndashtimes( ) 10 10 4ndashtimes( )times

30------------------------------------------------------------------=

To give λ = 430 nm

This example demonstrates why the image of a point source formed by a thin converging lens will always have a finite width

Exercise 921 Aparallelbeamoflightofwavelength500nmis

incidentonaslitofwidth025mmThelightisbroughttofocusonascreenplaced150mfromthe slit Calculate the angular width and the linear width of the central diffraction maximum

2 Lightfromalaserisusedtoformasingleslitdiffraction pattern on a screen The width of the slit is 010 mm and the screen is 30 m from the slit The width of the central diffraction maximum is 26cmCalculatethewavelengthofthelaserlight

3 Determinetheanglethatyouwouldexpecttofindconstructiveinterference(forn=0n=1andn=2) fromasingleslitofwidth20μmandlightofwavelength 600 nm

NATURE OF SCIENCECuriosity Observed patterns of iridescence in animals such as the shimmer of peacock feathers led scientists to develop the theory of thin film interference (15)

Serendipity The first laboratory production of thin films was accidental (15)

copy IBO 2014

PHYSICS 2015indb 291 150514 337 PM

Chapter 9

292

AH

L

microphoneismovedalongthelineXYthatisaboutonemetre from the speakers As the microphone is moved along XY the trace on the cathode ray oscilloscope isseen to go through a series of maxima and minima corresponding to points of constructive and destructive interference of the sound waves

Aninterestinginvestigationistofindhowtheseparationof the points of maximum interference depends on the frequency of the source and also the distance apart ofthe speakers

In the demonstration of the interference between two soundsourcesdescribedaboveifweweretomoveoneofthe speakers from side to side or backwards and forwards the sound emitted from the two speakers would no longer be in phase No permanent points of constructive or destructive interference will now be located since the phase difference between the waves from the two sources is nolongerconstantiethesourcesarenolongercoherent

Light from an incandescent source is emitted with acompletely random phase Although the light from two separate sources will interfere because of the randomlychanging phase no permanent points of constructive or destructive interference will be observed This is why a singleslit isneededintheYoungrsquosdoubleslitexperimentByactingasapointsourceitessentiallybecomesacoherentlight source The light emitted from a laser is also very nearly coherent and this is why it is so easy to demonstrate optical interference and diffraction with a laser

Youngrsquos double slit experiment is one of the great classic experiments of physics and did much to reinforce the wave theoryoflightThomasYoungcarriedouttheexperimentin about 1830

It is essentially the demonstration with the ripple tank and the sound experiment previously described butusing light The essential features of the experiment are shown in Figure 914

screendouble slit

single slit

coloured lter

S 1

S2

Figure 914 Youngrsquos double slit experiment

YoungallowedsunlighttofallontoanarrowsingleslitA few centimetres from the single slit he placed a double slit The slits are very narrow and separated by a distance equal to about fourteen slitwidthsA screen is placed

about ametre from thedouble slitsYoungobservedapatternofmulticolouredldquofringesrdquo in the screenWhenhe placed a coloured filter between the single slit anddouble slit he obtained a pattern that consisted of bright coloured fringes separated by darkness

The single slit essentially ensures that the light falling on the double slit is coherent The two slits then act as the two speakers in the sound experiment or the two dippers in the ripple tank The light waves from each slit interfere and produce the interference pattern on the screen Without thefilterapatternisformedforeachwavelengthpresentinthesunlightHencethemulticolouredfringepattern

You can demonstrate optical interference for yourselfSmoking a small piece of glass and then drawing twoparallel lines on it can make a double slit If you then look throughthedoubleslitatasingletungstenfilamentlampyouwillseethefringepatternByplacingfiltersbetweenthe lamps and the slits you will see the monochromatic fringe pattern

YoucanalsoseetheeffectsofopticalinterferencebylookingatnetcurtainsEachlsquoholersquointhenetactsasapointsourceand the light from all these separate sources interferes and producesquiteacomplicatedinterferencepattern

A laser can also be used to demonstrate optical interference Sincethelightfromthelaseriscoherentitisveryeasytodemonstrate interference Just point the laser at a screen and place a double slit in the path of the laser beam

LetusnowlookattheYoungrsquosdoubleslitexperimentinmore detail The geometry of the situation is shown in Figure915

S1

S2 X

P

Q

D

y

d

screen

θ

θrsquo

Figure 915 The geometry of Youngrsquos double slit experiment

S1andS2 are the two narrow slits that we shall regard as twocoherentmonochromaticpointsourcesThedistancefrom the sources to the screen is D and the distance between the slits is d

The waves from the two sources will be in phase at Q and therewill be abright fringehereWewish tofindtheconditionfortheretobeabrightfringeatPdistancey from Q

PHYSICS 2015indb 292 150514 337 PM

Wave Phenomena

293

AH

L

We note that that D (asymp1metre)isverymuchgreaterthaneither y or d(asymp fewmillimetres)Thismeansthatbothθ and θʹ areverysmallanglesandforintentsandpurposesequal

From the diagram we have that

θ = Dy

And

2S Xd

θʹ =

(Remembertheanglesareverysmall)

Sinceθ asymp θʹ then

2S XyD d

=

ButS2X is the path difference between the waves as they

traveltoPWehavethereforethat

path difference = Dyd

TherewillthereforebeabrightfringeatPif

Dyd

= nλ

Supposethatthereisabrightfringeaty = y1 corresponding

to n = n1 then

Ddy1 = n

1λ

If the next bright fringe occurs at y = y2 this will correspond

to n = n1+1Hence

Ddy2 =(n

1+1)λ

This means that the spacing between the fringes y2 ndash y

1 is

given by

y2 ndash y

1 = d

Dλ

Youngactuallyusethisexpressiontomeasurethewavelengthof the light he used and it is a method still used today

We see for instance that if in a given set up we move the slits closer together then the spacing between the fringes will get greaterEffectivelyourinterferencepatternspreadsoutthatis there will be fewer fringes in a given distance We can also increase the fringe spacing by increasing the distance betweentheslitsandthescreenYouwillalsonotethatfora

givensetupusinglightofdifferentwavelengthsthenldquoredrdquofringeswillspacefurtherapartthanldquobluerdquofringes

In this analysis we have assumed that the slits act as point sources and as such the fringes will be uniformly spaced andofequalintensityAmorethoroughanalysisshouldtakeintoaccountthefinitewidthoftheslits

ReturningtoFigure915weseethatwecanwritethepathdifferenceS2X as

S2X = dtan θʹ

But since θʹ is a small angle the sine and tangent will be nearlyequalsothat

S2X = dsinθʹ

And since θʹ asymp θ then

S2X = dsinθ

The condition therefore for a bright fringe to be found at a point of the screen can therefore be written as

dsinθ = nλ

Insummary

for constructive interference and minima

d sinθ = nλ wheren=123hellip

for destructive interference and maxima

d sinθ = ( n + frac12 )λ wheren=123hellip

Figure 916 shows the intensity distribution of the fringes on the screen when the separation of the slits is large comparedtotheirwidthThefringesareofequalintensityandofequalseparation

Intensity

distance along screen

Figure 916 The intensity distribution of the fringes

It isworthnoting that if the slits are close together theintensity of the fringes is modulated by the intensity distribution of the diffraction pattern of one of the slits

PHYSICS 2015indb 293 150514 337 PM

Chapter 9

294

AH

L

Example

Light of wavelength 500nm is incident on two smallparallel slits separated by 10mmDetermine the anglewherethefirstmaximumisformed

If after passing through the slits the light is brought toa focusona screen15m fromtheslitscalculate theobserved fringe spacing on the screen

Solution

Using the small angle approximation we have

3

7

10105minus

minustimes==

dλθ

= 5 times 10-4 rad

The fringe spacing is given by

3

7

1010551

minus

minustimestimes==

dDy λ

= 075 mm

Modulation of two-slit interference pattern by one-slit diffraction effect

In practice the intensity pattern of the maxima as shown in Figure 917 is not constant but fluctuates while decreasingsymmetrically on either side of the central maximum as shown inFigure917Theintensitypatternisacombinationofboththe single-slit diffractionenvelope and the double slit pattern That is the amplitude of the two-slit interference pattern is modulated(ieadjustedto)byasingleslitdiffractionenvelope

2λD

___ λD

D

___λd

d

___

Double slitnite width

Figure 917 Double slit pattern

ForagivenslitseparationdwavelengthoflightandfixedslittoscreendistancethevariationinintensitydependsonthewidthoftheslitDAlthoughincreasingthewidthoftheslitsincreasestheintensityoflightinthefringesitalsomakesthem less sharp and they become blurred As the slit width is widenedthefringesgraduallydisappearbecausenumerouspoint sources along the widened slit give rise to their own dark and bright fringes that then overlap with each other

Exercise 931 InFigure913thedistancebetweenthespeakers

is050mandthedistancebetweenthelineofthespeakersandthescreenis20mAsthemicrophoneismovedalongthelineXYthedistance between successive points of maximum soundintensityis030mThefrequencyofthesoundwavesis44times103HzCalculateavalueforthe speed of sound

2 Laserlightofwavelength610nmfallsontwoslits10x10-5apartDeterminetheseparationofthefirstordermaximumformedonascreen20maway

3 Twoparallelslits012mmapartareilluminatedwith red light with a wavelength of 600 nm An interferencepatternfallsonascreen15maway

Calculate

(a) thedistancefromthecentralmaximumtothefirstbrightfringe

(b) thedistancetotheseconddarkline

Multiple slit and diffraction grating interference patterns

If we examine the interference pattern produced when monochromatic light passes through a different number of slits we notice that as the number of slits increases the number of observed fringes decreases the spacingbetween them increases and the individual fringes become much sharper We can get some idea of how this comes about by looking at the way light behaves when a parallel beam passes through a large number of slits The diagram for this is shown in Figure 918

θθd

1

2

3

telescope

Figure 918 A parallel beam passing through several slits

The slits are very small so that they can be considered to act as point sources They are also very close together such that d issmall(10ndash6 m)EachslitbecomesasourceofcircularwavefrontsandthewavesfromeachslitwillinterfereLetus

PHYSICS 2015indb 294 150514 337 PM

Wave Phenomena

295

AH

L

consider the light that leaves the slit at an angle θ as shown Thepathdifferencebetweenwave1andwave2isdsinθ and ifthisisequaltoanintegralnumberofwavelengthsthenthe two waves will interfere constructively in this direction Similarly wave 2 will interfere constructively with wave3 at this angle andwave 3with 4 etc across thewholegratingHenceifwelookatthelightthroughatelescopethatisbringittoafocusthenwhenthetelescopemakesanangle θ to the grating a bright fringe will be observed The condition for observing a bright fringe is therefore

dsinθ = mλ

Supposeweuselightofwavelength500nmandsupposethat d = 16 times 10ndash6 m

Obviously we will see a bright fringe in the straight on position θ=0(thezero order)

The next position will be when m =1(thefirst order)andsubstitutionintheaboveequationgivesθ = 18deg

The next position will be when m=2(thesecond order)and this give θ = 38deg

For m=3sinθ is greater than 1 so with this set up we only obtain5fringesonezeroorderandtwoeithersideofthezero order

The calculation shows that the separation of the orders is relatively large At any angles other than 18deg or 38deg the light leaving the slits interferes destructively We can see that the fringes will be sharp since if we move just a small angle away from 18deg the light from the slits will interfere destructively

An array of narrow slits is usually made by cutting narrow transparent lines very close together into the emulsion on aphotographicplate(typically200linespermillimetre)Suchanarrangementiscalledadiffraction grating

The diffraction grating is of great use in examining the spectral characteristics of light sources

All elements have their own characteristic spectrum An element can be made to emit light either by heating it until it is incandescent or by causing an electric discharge through itwhen it is in a gaseous state Yourschool probably has some discharge tubes and diffraction gratingsIfithasthenlookattheglowingdischargetubethroughadiffractiongratingIfforexampletheelementthat you are looking has three distinct wavelengths then each wavelength will be diffracted by a different amount inaccordancewiththeequationdsinθ = nλ

Alsoiflaserlightisshonethroughagratingontoascreenyou will see just how sharp and spaced out are the maxima By measuring the line spacing and the distance of the screen fromthelaserthewavelengthofthelasercanbemeasured

If your school has a set of multiple slits say from a single slit toeight slits then it isalsoaworthwhileexercise toexamine how the diffraction pattern changes when laser light is shone through increasing numbers of slits

Consider a grating of NparallelequidistantslitsofwidthaseparatedbyanopaqueregionofwidthD We know that light diffracted from one slit will be superimposed and interfere with light diffracted from all the other slits The intensityprofile of the resultant wave will have a shape determined by

bull theintensityoflightincidentontheslitsbull thewavelengthofthislightbull theslitwidthbull theslitseparation

Typical intensity profiles formed when a plane wavemonochromaticlightatnormalincidenceononetwoandsix slits are shown in Figure 919 In this example the slit widthisonequarterofthegratingspacing

Six slits sharperfringes

d = 4a

I α N 2

(N ndash 2)= 4 subsidiary maxima

order0 4321

I = 36

Two slits

order0

single-slitdiraction

envelope

d = 4a4th principal maximum missing

principa maxima

I = 4 (nominal units)

4321

intensity

order

ISS

One slit

0 1

single-slitdiraction

pattern

N = number of slitsslit width a

Figure 919 Intensity profiles

PHYSICS 2015indb 295 150514 337 PM

Chapter 9

296

AH

L

1 For Nge2thefringepatterncontainsprincipalmaximathat are modulated by the diffraction envelope

2 ForNgt2thefringepatterncontains(Nndash2)subsidiary maxima

3 Where the grating spacing d is four time the slit width a(4xa =d)thenthefourthprincipalmaximumismissing from the fringe pattern Where the grating spacingdisthreetimetheslitwidtha(3xa =d)thenthe third principal maximum is missing from the fringepatternandsoon

4 Increasing N increases the absolute intensities of the diffraction pattern (I = N2)

If white light is shone through a grating then the central image will be white but for the other orders each will be spread out intoacontinuousspectrumcomposedofaninfinitenumberof adjacent images of the slit formed by the wavelength of the different wavelengths present in the white light At any given point in the continuous spectrum the light will be very nearly monochromatic because of the narrowness of the images of the slit formed by the grating This is in contrast to the doubleslitwhereifwhitelightisusedtheimagesarebroadand the spectral colours are not separated

In summary as is the case of interference at a doubleslit there is a path difference between the rays fromadjacent slits of d sinθ Where there is a whole number of wavelengthsconstructiveinterferenceoccurs

For constructive interference and minima

d sinθ = mλ wheren=123hellip

When there is a path difference between adjacent rays of halfawavelengthdestructiveinterferenceoccurs

For and maxima

d sinθ = ( m + frac12 )λ wheren=123hellip

Exercise 941 Lightfromalaserisshonethroughadiffraction

grating on to a screen The screen is a distance of 20mfromthelaserThedistancebetweenthecentraldiffractionmaximumandthefirstprincipalmaximum formed on the screen is 094 m The diffractiongratinghas680linespermmEstimatethe wavelength of the light emitted by the laser

2 Monochromaticlightfromalaserisnormallyincident on a six-slit grating The slit spacing is three times the slit width An interference patternisformedonascreen2montheotherside of the grating

(a) Determineandexplaintheorderoftheprincipal maxima missing from the fringe pattern

(b) Howmanysubsidiarymaximaaretherebetween the principal maxima

(c) Howmanyprincipalmaximaarethere

(i) Betweenthetwofirstorderminimain the diffraction envelope

(ii) Betweenthefirstandsecondorderminima in the diffraction envelope

(d) Drawthepatternformedbythebrightfringesin(c)

(e) Sketchtheintensityprofileofthefringepattern

Thin film interferenceYoumightwellbe familiarwith the colouredpatternoffringes that can be seen when light is reflected off thesurfaceofwateruponwhichathinoilfilmhasbeenspiltorfromlightreflectedfrombubblesWecanseehowthesepatternsarisebylookingatFigure920

Extendedlight source air

A

B

C

12

3

4

Oil lm

Water

Figure 920 Reflection from an oil film

Consider light from an extended source incident on a thinfilmWealsoconsiderawave fromonepointofthe source whose direction is represented by the ray shownSomeofthislightwillbereflectedatAandsometransmitted through the filmwhere somewill again bereflectedatB(somewillalsobetransmittedintotheair)SomeofthelightreflectedatBwillthenbetransmittedatCandsomereflectedandsoonIfweconsiderjustrays1and2thenthesewillbenotbeinphasewhenthearebrought to a focus by the eye since they have travelled

PHYSICS 2015indb 296 150514 337 PM

Wave Phenomena

297

AH

L

different distances If the path difference between them is an integral number of half wavelengths then we might expect the twowaves to be out of phaseHowever ray1 undergoes a phase change of π on reflection but ray2 does not since it is reflected at a boundary betweena more dense and less dense medium (See chapter 4)Henceifthepathdifferenceisanintegralnumberofhalf-wavelengthsrays1and2willreinforceieray1and2areinphaseHowever rays357etcwillbeoutofphasewithrays246etcbutsinceray2ismoreintensethanray3andray4moreintensethanray5thesepairswillnot cancel out so there will be a maximum of intensity

Ifthepathdifferenceissuchthatwave1and2areoutofphasesincewave1ismoreintensethanwave2theywillnotcompletelyannulHoweveritcanbeshownthattheintensitiesofwaves2345hellipaddtoequaltheintensityofwave1Sincewaves345hellipareinphasewithwave2there will be complete cancellation

The path difference will be determined by the angle at which ray1 is incident andalsoon the thickness (andtheactualrefractiveindexaswell)ofthefilmSincethesourceisanextendedsourcethelightwillreachtheeyefrom many different angles and so a system of bright and darkfringeswillbeseenYouwillonlyseefringesifthefilmisverythin(exceptifviewedatnormalincidence)sinceincreasingthethicknessofthefilmwillcausethereflected rays to get so far apart that they will not becollected by the pupil of the eye

From the argument above to find the conditions forconstructive and destructive interferenceweneedonlyfindthepathdifferencebetweenray1andray2Figure921showsthe geometry of the situation

A BE

F

C

D

1 2

φrsquo φrsquo

φ

φ

d

Figure 921 The geometry of interference

Thefilmisofthicknessd and refractive index n and the light has wavelength λ If the line BF is perpendicular to ray 1 then the optical path difference (opd) between ray 1 andray2whenbroughttoafocusis

opd = n(AC+CB)ndashAF

We have to multiple by the refractive index for the path travelledbythelightinthefilmsincethelighttravelsmoreslowly in thefilm If the light travels say adistancex in a material of refractive index n then in the time that it takes to travelthisdistancethelightwouldtraveladistancenx in air

IfthelineCEisatrightanglestoray2thenweseethat

AF = nBE

FromthediagramAC=CDsowecanwrite

opd = n(CD+CB)ndashnBE=nDE

Alsofromthediagramweseethatwhereφ is the angle of refraction

DE=2cosφ

From which opd=2ndcosφ

Bearinginmindthechangeinphaseofray1onreflectionwe have therefore that the condition for constructive interference is

2nd φcos m 12---+

λ m 1 2 hellip = =

Andfordestructiveinterference 2ndcosφ = mλ

Eachfringecorrespondstoaparticularopd for a particular value of the integer m and for any fringe the value of the angle φ is fixedThismeans that itwill be in the formof an arc of a circle with the centre of the circle at the point where the perpendicular drawn from the eye meets the surface of the film Such fringes are called fringesofequal inclinationSince theeyehasa smallaperturethesefringesunlessviewedatneartonormalincidence (φ = 0) will only be observed if the film is very thinThisisbecauseasthethicknessofthefilmincreasesthereflected rayswill get further and further apart and sovery few will enter the eye

Ifwhitelightisshoneontothefilmthenwecanseewhywe get multi-coloured fringes since a series of maxima and minima will be formed for each wavelength present in the white light However when viewed at normalincidenceitispossiblethatonlylightofonecolourwillundergoconstructiveinterferenceandthefilmwilltakeon this colour

PHYSICS 2015indb 297 150514 337 PM

Chapter 9

298

AH

L

Thickness of oil films

The exercise to follow will help explain this use

Non-reflecting films

A very important but simple application of thin filminterferenceisintheproductionofnon-reflectingsurfaces

Athinfilmofthicknessd and refractive index n1 is coated onto glass of refractive index n where n1 lt n Light ofwavelength λ that is reflected at normal incidence willundergo destructive interference if 12

2n d λ

= thatis

14d

nλ

=

(remember that there will now no phase change at the glass interface ie we have a rare to dense reflection)

The use of such films can greatly reduce the loss oflightby reflection at the various surfacesof a systemoflenses or prisms Optical parts of high quality systemsare usually all coatedwithnon-reflecting films in orderto reduce stray reflections The films are usually madebyevaporatingcalciumormagnesiumfluorideonto thesurfacesinvacuumorbychemicaltreatmentwithacidsthat leave a thin layer of silica on the surface of the glass ThecoatedsurfaceshaveapurplishhuebyreflectedlightThis is because the condition for destructive interference fromaparticularfilmthicknesscanonlybeobtainedforone wavelength The wavelength chosen is one that has a value corresponding to light near the middle of the visible spectrumThismeansthatreflectionofredandvioletlightis greater combining to give the purple colour Because of the factor cosφatanglesotherthannormalincidencethepath difference will change butnotsignificantlyuntilsayabout30deg(egcos25deg=090)

It should be borne in mind that no light is actually lost by anon-reflectingfilmthedecreaseofreflectedintensityiscompensated by increase of transmitted intensity

Non-reflecting films can be painted onto aircraft tosuppressreflectionofradarThethicknessofthefilmis determined by

4nd λ

= where λ is the wavelength of

the radar waves and ntherefractiveindexofthefilmat this wavelength

Example

A plane-parallel glass plate of thickness 20 mm isilluminated with light from an extended source The refractiveindexoftheglassis15andthewavelengthofthelight is 600 nm Calculate how many fringes are formed

Solution

We assume that the fringes are formed by light incident at all angles from normal to grazing incidence

At normal incidence we have 2nd = mλ

From which

m 2 15 2 10 3ndashtimestimestimes6 10 7ndashtimes

------------------------------------------- 10 000= =

At grazing incidence the angle of refraction φ is in fact the critical angle

Therefore φφ arcsin 115------- 42 deg= =

ie cosφ = 075

At grazing incidence 2ndcosφ = m λ

From which (and using cos(sinndash1(115) = 075)

m 2 15 2 10 3ndash 075timestimestimestimes6 10 7ndashtimes

------------------------------------------------------------ 7500= =

The total number of fringes seen is equal to m ndash m = 2500

Exercise 95When viewed from above the colour of an oil film onwater appears red in colour Use the data below to estimate theminimumthicknessoftheoilfilm

average wavelength of red light = 630 nm

refractiveindexofoilforredlight=15

refractive index of water for red light = 13

PHYSICS 2015indb 298 150514 337 PM

Wave Phenomena

299

AH

L

94 Resolution

Essential idea Resolution places an absolute limit on the extent to which an optical or other system can separate images of objects

Understandings

bull Thesizeofadiffractingaperturebull Theresolutionofsimplemonochromatic

two-source systems

The size of a diffracting apertureOur discussion concerning diffraction so far has been for rectangular slits When light from a point source enters a smallcircularapertureitdoesnotproduceabrightdotasanimagebutacirculardiscknownasAiryrsquosdiscsurroundedbyfainterconcentriccircularringsasshowninFigure922

Laser

Figure 922 Diffraction at a circular aperture

Diffractionatanapertureisofgreatimportancebecausetheeye and many optical instruments have circular apertures

What is the half-angular width of the central maximum of the diffraction formed by a circular aperture This is not easy to calculate since it involves some advanced mathematics The problemwasfirstsolvedbytheEnglishAstronomerRoyalGeorgeAiryin1835whoshowedthatforcircularapertures

θ 122 λb--------------= where b is the diameter of the aperture

Example 2

Inthefollowingdiagramparallellightfromadistantpointsource(suchasastar)isbroughttofocusonthescreenSbyaconverginglens(the lens is shown as a vertical arrow)

Thefocal length(distance fromlens toscreen) is25cmand the diameter of the lens is 30 cm The wavelength of thelightfromthestaris560nmCalculatethediameterofthe diameter of the image on the screen

Solution

The lens actually acts as a circular aperture of diameter 30 cm The half angular width of central maximum of the diffraction pattern that it forms on the screen is

75

2

122 122 56 10 23 1030 10b

λθminus

minusminus

times times= = = times

times rad

The diameter of the central maxima is therefore

25 times 10-2 times 23 times 10-5

= 57 times 10-6 m

Although this is small it is still finite and is effectively the image of the star as the intensity of the secondary maxima are small compared to that of the central maximum

The resolution of simple monochromatic two-source systemsThe astronomers tell us that many of the stars that we observe with the naked eye are in fact binary stars That iswhatwe see as a single star actually consistsof twostarsinorbitaboutacommoncentreFurthermoretheastronomerstellusthatifweusealdquogoodrdquotelescopethenwewillactuallyseethetwostarsThatiswewillresolve the single point source into its two component parts Sowhatisitthatdetermineswhetherornotweseethetwo stars as a single point source ie what determines whether or not two sources can be resolved It canrsquot just bethatthetelescopemagnifiesthestarssinceiftheyareacting as point sources magnifying them is not going to make a great deal of difference

Ineachofoureyesthereisanaperturethepupilthroughwhich the light enters This light is then focused by the eye lens onto the retina But we have seen that when light passes through an aperture it is diffracted and so when we look at

NATURE OF SCIENCEImproved technology The Rayleigh criterion is the limit of resolution Continuing advancement in technology such as large diameter dishes or lenses or the use of smaller wavelength lasers pushes the limits of what we can resolve (18)

copy IBO 2014

PHYSICS 2015indb 299 150514 337 PM

Chapter 9

300

AH

L

apointsourceadiffractionpatternwillbeformedontheretina If we look at two point sources then two diffraction patterns will be formed on the retina and these patterns will overlap The width of our pupil and the wavelength of the light emitted by the sources will determine the amount by which they overlap But the degree of overlap will also depend on the angular separation of the two point sources WecanseethisfromFigures923

pupilretina

S1

S2

P2

P1

θ

Figure 923

LightfromthesourceS1 enters the eye and is diffracted by the pupil such that the central maximum of the diffraction patternisformedontheretinaatP1SimilarlylightfromS2producesamaximumatP2 If the two central maxima are well separated then there is a fair chance that we will see the two sources as separate sources If they overlap then we will not be able to distinguish one source from another From the diagram we see as the sources are movedclosertotheeyethentheangleθ increases and so does the separation of the central maxima

Figures924925926and927showsthedifferentdiffractionpatternsandtheintensitydistributionthatmightresultonthe retina as a result of light from two point sources

S1 S21 Well resolved

Figure 924 Very well resolved

S1 S22 Well resolved

Figure 925 Well resolved

S1 S23 Just resolved Rayleigh criterion Minimum of S2 coincides with maximum peak of S1

Figure 926 Just resolved

4 Not resolved

Figure 927 Not resolved

We have suggested that if the central maxima of the two diffraction patterns are reasonably separated then we should be able to resolve two point sources In the late 19th century Lord Rayleigh suggested by how much they should be separated in order for the two sources to be just resolved Ifthecentralmaximumofonediffractionpattern coincides with the firstminima of the otherdiffraction pattern then the two sources will just beresolved This is known as the Rayleigh Criterion

Figure926showsjustthissituationThetwosourcesarejust resolved according to the Rayleigh criterion since the peak of the central maximum of one diffraction pattern coincideswiththefirstminimumoftheotherdiffractionpattern This means that the angular separation of the peaks of the two central maxima formed by each source is just the half angular width of one central maximum ie

θ λb---

=

where b is the width of the slit through which the light fromthesourcespassesHoweverweseefromFigure1117that θ is the angle that the two sources subtend at the slit Henceweconcludethattwosourceswillberesolvedbyaslit if the angle that they subtend at the slit is greater than or equaltoλfraslb

PHYSICS 2015indb 300 150514 337 PM

Wave Phenomena

301

AH

L

Sofarwehavebeenassumingthattheeyeisarectangularslit whereas clearly it is a circular aperture and so we must use the formula

θ 122 λb--------------=

As mentioned above the angle θ is sometimes called the resolving power but should more accurately be called the minimum angle of resolution (θmin)

Clearly the smaller θ the greater will be the resolving power

The diffraction grating is a useful for differentiating closely spacedlinesinemissionspectraLikeaprismspectrometerit can disperse a spectrum into its components but it is better suited because it has higher resolution than the prism spectrometer

If λ1 and λ2 are two nearly equal wavelengths that canbarelybedistinguishedtheresolvanceorresolvingpowerofthegratingisdefinedas

R = λ _______ (λ2 ndash λ 1) = λ ___ Δλ

Where λ asymp λ1 asymp λ2

Therefore thediffractiongratingwithahighresolvancewill be better suited in determining small differences in wavelength

If the diffraction grating has Nlinesbeingilluminateditcan be shown that the resolving power of the mth order diffraction is given by

R = mN

Obviouslytheresolvancebecomesgreaterwiththeordernumber m and with a greater number of illuminated slits

Example

A benchmark for the resolving power of a grating is the sodium doublet in the sodium emission spectrum Two yellow lines in itsspectrumhavewavelengthsof58900nmand58959nm

(a) Determinetheresolvanceofagratingifthegivenwavelengths are to be distinguished from each other

(b) Howmanylinesinthegratingmustbeilluminated in order to resolve these lines in the second order spectrum

Solution

(a) R = λ ___ Δλ = 58900nm __________________ (58900ndash58959)nm

= 589nm _______ 059nm = 10 times 103

(b) N = R __ m = 10 times 10 3 ________ 2 =50times102 lines

It has been seen that diffraction effectively limits the resolving power of optical systems This includes such systems as the eye telescopes and microscopes Theresolving power of these systems is dealt with in the next section This section looks at links between technology and resolving power when looking at the very distant and when looking at the very small

Radio telescopes

The average diameter of the pupil of the human eye is about 25 mm This means that the eye will justresolve two point sources emitting light of wavelength 500 nm if their angular separation at the eye is 7

43

50 10122 24 1025 10

θminus

minusminus

times= times = times

times rad

If the eye were to be able to detect radio waves of wavelength 015m then tohave the same resolvingpower thepupilwould have to have a diameter of about 600 m Clearly this is nonsense but it does illustrate a problem facingastronomers who wish to view very distant objects such as quasarsandgalaxies thatemitradiowavesConventionalradio telescopes consist of a large dish typically 25 min diameter Even with such a large diameter the radiowavelength resolving power of the telescope is much less thantheopticalresolvingpowerofthehumaneyeLetuslook at an example

Example

The Galaxy Cygnus A can be resolved optically as an elliptically shaped galaxy However it is also a strongemitterof radiowavesofwavelength015mTheGalaxyisestimatedtobe50times1024mfromEarthUseofaradiotelescope shows that the radio emission is from two sources separated by a distance of 30 times 1021m Estimate thediameterofthedishrequiredtojustresolvethesources

Solution

The angle θ that the sources resolve at the telescope is given by

θ 21

424

30 10 60 1050 10

θ minustimes= = times

times rad

and 4

122 122 015 300060 10

d λθ minus

times= = =

timesm = 30 km

A radio telescope dish of this size would be impossible tomake letalonesupportThisshowsthatasingledishtype radio telescope cannot be used to resolve the sources and yet they were resolved To get round the problemastronomers use two radio telescopes separated by a large

PHYSICS 2015indb 301 150514 337 PM

Chapter 9

302

AH

L

distance The telescopes view the same objects at the same time and the signals that each receive from the objects are simultaneously superimposed The result of the superposition of the two signals is a two-slit interference pattern The pattern has much narrower fringe spacing than that of the diffraction pattern produced by either telescope on its own hence producing a much higherresolvingpowerWhentelescopesareusedlikethistheyare called a stellar interferometer

InSocorroinNewMexicothereisastellarinterferometerthat consists of 27 parabolic dishes each of diameter25 m arranged in a Y-shape that covers an area of 570km2Thisisaso-calledVeryLargeArray(VLA)Evenhigher resolution can be obtained by using an array of radio telescopes in observatories thousands of kilometres apart A system that uses this so-called technique oflsquovery-long-baseline interferometryrsquo (VLBI) is known asalsquovery-long-baselinearrayrsquo(VLBA)WithVLBAaradiowavelength resolving power can be achieved that is 100 timesgreaterthanthebestopticaltelescopesEvenhigherresolving power can be achieved by using a telescope that isinasatelliteorbitingEarthJapanrsquosInstituteofSpaceandAstronautical Science (ISAS) launched sucha system inFebruary 1997TheNationalAstronomicalObservatoryof Japan the National Science Foundationrsquos NationalRadio Astronomy Observatory (NRAO) the CanadianSpaceAgency theAustraliaTelescopeNationalFacilitythe European VLBI Network and the Joint Instituteback this project forVeryLongBaseline Interferometryin EuropeThis project is a very good example of howInternationalismcanoperateinPhysics

Electron microscope

Telescopes are used to look at very distant objects that are very large but because of their distance fromus appearverysmallMicroscopesontheotherhandareusedtolookat objects that are close to us but are physically very small Aswehave seen justmagnifyingobjects that ismakingthemappearlargerisnotsufficientonitsowntogaindetailabouttheobjectfordetailhighresolutionisneeded

Figure928isaschematicofhowanoptical microscope is usedtoviewanobjectandFigure929isaschematicofatransmission electron microscope(TEM)

Optical lens system

bright light source

glass slide containing specimen (object)

eye

Figure 928 The principle of a light microscope

magnetic lens system

electron source

thin wafer of material

ScreenCCD

Figure 929 The principle of an electron microscope

In the optical microscope the resolving power isdetermined by the particular lens system employed and the wavelength λ of the light used For example twopoints in the sample separated by a distance d will just be resolved if

2d mλ=

where m is a property of the lens system know as the numerical aperture In practice the largest value of m obtainable is about 16 Hence if the microscope slideis illuminatedwith light ofwavelength 480 nm a goodmicroscope will resolve two points separated by a distance dasymp15times10-7masymp015micromPointscloser together thanthiswillnotberesolvedHoweverthisisgoodenoughtodistinguishsomevirusessuchastheEbolavirus

Clearlythesmallerλ the higher the resolving power and this is where the electron microscope comes to the fore The electron microscope makes use of the wave nature of electrons(see1315)IntheTEMelectronspassthrougha wafer thin sample and are then focused by a magnetic field onto a fluorescent screen orCCD (charge coupleddevicesee142)Electronsusedinanelectronmicroscopehavewavelengthstypicallyofabout5times10-12mHoweverthe numerical aperture of electron microscopes is considerablysmaller thanthatofanopticalmicroscopetypicallyabout002NonethelessthismeansthataTEMcanresolvetwopointsthatareabout025nmapartThisresolving power is certainly high enough to make out the shape of large molecules

Another type of electron microscope uses a techniqueby which electrons are scattered from the surface of the sample The scattered electrons and then focused as in theTEM to form an image of the surfaceThese so-calledscanningelectronmicroscopes(SEM)havealowerresolving power than TEMrsquos but give very good threedimensional images

The eye

We saw in the last section that the resolving power of the human eye is about 2 times 10-4 rad Suppose that youare looking at car headlights on a dark night and the car

PHYSICS 2015indb 302 150514 337 PM

Wave Phenomena

303

AH

L

is a distance D away If the separation of the headlight is say 15 m then the headlights will subtend an angle θ 15

Dθ =

at your eye Assuming an average wavelength of

500 nm your eye will resolve the headlights into twoseparatesourcesifthisangleequals2times10-4 rad This gives D=75kmInotherwordsifthecarisapproachingyouona straight road then you will be able to distinguish the two headlightsasseparatesourceswhenthecaris75kmawayfrom you Actually because of the structure of the retina and optical defects the resolving power of the average eye is about 6 times 10-4 rad This means that the car is more likely tobe25kmawaybeforeyouresolvetheheadlights

Astronomical telescope

Letusreturntotheexampleofthebinarystarsdiscussedatthe beginning of this section on resolution The stars Kruger A and B form a binary system The average separation of the stars is 14 times 1012mandtheiraveragedistancefromEarthis12times1017mWhenviewedthroughatelescopeonEarththe system will therefore subtend an angle

12

17

14 1012 10

θtimes

=times

= 12times10-5 rad at the objective lens of the telescope Assuming that the average wavelength of the light emitted bythestarsis500nmthenifthetelescopeistoresolvethesystem into two separate images it must have a minimum diameter Dwhere12times10-5 =

7122 500 10D

minustimes times

This gives D=0050m which is about 5 cm So thisparticular system is easily resolved with a small astronomical telescope

Exercise 961 ItissuggestedthatusingtheISASVLBAitwould

bepossibletoldquoseerdquoagrainofriceatadistanceof5000kmEstimatetheresolvingpoweroftheVLBA

2 Thedistancefromtheeyelenstotheretinais20mm The light receptors in the central part of the retinaareabout5times10-6apartDeterminewhetherthe spacing of the receptors will allow for the eye to resolve the headlights in the above discussion whentheyare25kmfromtheeye

3 ThediameterofPlutois23times106 m and its averagedistancefromEarthis60times1012 m EstimatetheminimumdiameteroftheobjectiveofatelescopethatwillenablePlutotobeseenasadisc as opposed to a point source

95 Doppler effect

Essential ideaTheDopplereffectdescribesthephenomenonofwavelengthfrequencyshiftwhenrelativemotionoccurs

Understandingsbull TheDopplereffectforsoundwavesandlightwaves

The Doppler effect for sound waves and light waves

Consider two observers A and B at rest with respect to asoundsourcethatemitsasoundofconstantfrequencyf Clearly both observers will hear a sound of the same frequencyHoweversupposethatthesourcenowmovesat constant speed towards A A will now hear a sound of frequencyfA that is greater than f and B will hear a sound of frequency fB that is less than f This phenomenon is known as the Doppler Effect orDopplerPrinciple after C J Doppler(1803-1853)

The same effect arises for an observer who is either moving towards or away from a stationary source

Figure 930 shows the waves spreading out from a stationary source that emits a sound of constant frequency f The observersAandBhearasoundofthesamefrequency

A B

wavefront

S

Figure 930 Sound waves from a stationary source

Suppose now that the source moves towards A withconstant speed vFigure9310(a)showsasnapshotofthenew wave pattern

NATURE OF SCIENCETechnology Although originally based on physical observations of the pitch of fast moving sources of sound the Doppler effect has an important role in many different areas such as evidence for the expansion of the universe and generating images used in weather reports and in medicine (55)

copy IBO 2014

PHYSICS 2015indb 303 150514 337 PM

Chapter 9

304

AH

L

sourceA B

smallerwavelength

largerwavelength

Figure 931 (a) Sound waves from a moving source

The wavefronts are now crowded together in the direction of travel of the source and stretched out in the opposite direction This is why now the two observers will now hear notesofdifferentfrequenciesHowmuchthewavesbunchtogether and how much they stretch out will depend on the speed vEssentially A

A

cfλ

= and BB

cfλ

= where λA lt λB and v is the speed of sound

IfthesourceisstationaryandAismovingtowardsitthenthe waves from the source incident on A will be bunched up If A is moving away from the stationary source then the waves from the source incident on A will be stretched out

Christian Doppler(1803ndash1853)actuallyappliedtheprinciple(incorrectlyasithappens)totryandexplainthecolourofstars However the Doppler effect does apply to light aswellastosoundIfalightsourceemitsalightoffrequencyf then if it is moving away from an observer the observer willmeasurethelightemittedashavingalowerfrequencythan fSincethesensationofcolourvisionisrelatedtothefrequencyoflight(bluelightisofahigherfrequencythanred light) light emitted by objects moving way from anobserver is often referred to as being red-shifted whereas if the object is moving toward the observer it is referred to as blue-shiftedThisideaisusedinOptionE(Chapter16)

We do not need to consider here the situations where either the source or the observer is accelerating In a situation for example where an accelerating source is approaching astationaryobserverthentheobserverwillhearasoundof ever increasing frequency This sometimes leads toconfusion in describing what is heard when a source approaches passes and then recedes from a stationaryobserver Suppose for example that you are standingon a station platform and a train sounding its whistle is approaching at constant speed What will you hear as the train approaches and then passes through the station As the train approaches you will hear a sound of constant pitch but increasing loudness The pitch of the sound will be greater than if the train were stationary As the train passes through the station you will hear the pitch change at the momentthetrainpassesyoutoasoundagainofconstantpitch The pitch of this sound will be lower than the sound of the approaching train and its intensity will decrease as the train recedes from you What you do not hear is a sound of increasing pitch and then decreasing pitch

The Doppler equations for sound

Although you will not be expected in an IB examination to derivetheequationsassociatedwithaspectsoftheDopplereffectyouwillbeexpectedtoapplythemForcompletenesstherefore thederivationoftheequationsassociatedwiththeDopplereffectasoutlinedaboveisgivenhere

S primeS O

vs

Figure 931 (b)

InFigure931(b)theobserverOisatrestwithrespecttoasourceofsoundSismovingwithconstanspeedvs directly towards O The source is emitting a note of constant frequencyf and the speed of the emitted sound is v

S shows the position of the source ∆t later When the sourceisatresttheninatime∆t the observer will receive f∆t waves and these waves will occupy a distance v∆t ie

v f

λ =f∆tv∆t =

(Because of the motion of the source this number of waves will now occupy a distance (v∆t ndash vs∆t)The lsquonewrsquowavelength is therefore

v ndash vs

fλrsquo =

f∆tv∆t ndash vs∆t

=

If fisthefrequencyheardbyOthen

vfλ

= or

vf

λ = = sv vfminus

From which

s

vf fv v

=minus

Dividingthroughbyv gives

Equation1111

1 sf f v

v=

minus

If the source is moving away from the observer then we have

s

vf fv v

=minus

1

1 sf vv

=+

Equation112

PHYSICS 2015indb 304 150514 337 PM

Wave Phenomena

305

AH

L

We now consider the case where the source is stationary and the observerismoving towardsthesource with speed v0 In this situation the speed of the sound waves as measured by the observer will be v0 + v We therefore have that

v0 + v = ffraslλ = vfλ

times

From which

o1 vf fv

= +

Equation113

If the observerismovingawayfromthesource then

o1 vf fv

= minus

Fromequation(113)wehavethat

o o1 v vf f f f f fv v

∆ = minus = + minus =

Equation114

Thevelocitiesthatwerefertointheaboveequationsarethevelocities with respect to the medium in which the waves fromthesourcetravelHoweverwhenwearedealingwithalight source it is the relative velocity between the source and the observer that we must consider The reason for this is that lightisuniqueintherespectthatthespeedofthelightwavesdoes not depend on the speed of the source All observers irrespective of their speed or the speed of the source will measure the same velocity for the speed of light This is oneof thecornerstonesof theSpecialTheoryofRelativitywhichisdiscussedinmoredetailinOptionH(Chapter18)When applying the Doppler effect to light we aremainlyconcerned with the motion of the source We look here only at the situation where the speed of the source v is much smaller than the speed of light cinfreespace(v ltlt c)Underthesecircumstanceswhenthesourceismovingtowardstheobserverequation111becomes

vf f f fc

minus = ∆ = Equation115

andwhenthesourceismovingawayfromtheobserverequation112becomes vf f f f

cminus = ∆ = minus

Provided that v ltlt c these same equations apply for astationary source and moving observer

We look at the following example and exercise

Example

Asourceemitsasoundoffrequency440HzItmovesinastraight line towards a stationary observer with a speed of 30ms-1Theobserverhearsasoundoffrequency484HzCalculate the speed of sound in air

Solution

We use equation 111 and substitute f= 484 Hz f = 440 Hz

and vs = 30 m s-1

therefore 1

484 440 301v

=minus

such that 30 4401

484vminus = to

give v = 330 m s-1

Example

A particular radio signal from a galaxy is measured as havingafrequencyof139times109HzThesamesignalfromasourceinalaboratoryhasafrequencyof142times109Hz

Suggestwhy thegalaxy ismovingaway fromEarthandcalculateitsrecessionspeed(iethespeedwithwhichitismovingawayfromEarth)

Solution

The fact that the frequency from the moving source is less than that when it is stationary indicates that it is moving away from the stationary observer ie Earth

Using vf fc

∆ = we have

8 96

9

3 10 (142 139) 10 634 10142 10

c fvf∆ times times minus times

= = = timestimes

m s-1

It is usual when dealing with the Doppler effect of light to express speeds as a fraction of c So in this instance we have v = 0021 c

Using the Doppler effectWe have seen in the above example and exercise how the Dopplereffectmaybeusedtomeasuretherecessionspeedof distant galaxies The effect is also used to measure speed inothersituationsHerewewilllookatthegeneralprincipleinvolved in using the Doppler effect to measure speedFigure932showsasource(thetransmitter)thatemitseithersound or em waves of constant frequency f The waves

PHYSICS 2015indb 305 150514 337 PM

Chapter 9

306

AH

L

fromthesourceare incidentonareflectorthat ismovingtowards the transmitter with speed vThereflectedwavesare detected by the receiver placed alongside the transmitter

reector

v

transmitter

receiver

f f

f f

Figure 932 Using the Doppler effect to measure speed

We shall consider the situation where v ltlt c where c is the speed of the waves from the transmitter

For the reflector receiving waves from the transmitterit is effectively an observer moving towards a stationary sourceFromequation(114) itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation116

For the receiver receivingwaves from the reflector it iseffectively a stationary observer receiving waves from a movingsourceFromequation(115)itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation117

Ifweaddequations(116)and(117)wegetthatthetotalDopplershiftatthereceiver∆f is

v vf f f f fc c

minus = ∆ = +

But 1 vf fc

= +

hence

1 v v vf f fc c c

∆ = + +

But since v ltlt cwe can ignore the term2

2

vc when we

expandthebracketintheaboveequation

Therefore we have2vf fc

∆ = Equation118

If v asymp c then we must use the full Doppler equationsHowever for em radiationwewill alwaysonly considersituations in which v ltlt c

Example

Thespeedofsoundinbloodis1500times103 m s-1 Ultrasound offrequency100MHzisreflectedfrombloodflowinginan arteryThe frequencyof the reflectedwaves receivedbackatthetransmitteris105MHzEstimatethespeedofthebloodflowintheartery

Solution

Using equation (118) we have

6 63

2005 10 1015 10

vtimes = times

timesto give v asymp 36 m s-1 (We have assumed that the ultrasound is incident at right angles to the blood flow)

Exercise 971 Judy is standing on the platform of a station A

high speed train is approaching the station in a straight line at constant speed and is sounding its whistleAsthetrainpassesbyJudythefrequencyof the sound emitted by the whistle as heard by Judychangesfrom640Hzto430HzDetermine

(a) thespeedofthetrain

(b) thefrequencyofthesoundemittedbythewhistle as heard by a person on the train (Speedofsound=330ms-1)

2 AgalaxyismovingawayfromEarthwithaspeedof00500cThewavelengthofaparticularspectral line in light emitted by atomic hydrogen inalaboratoryis656times10-7 m Calculate the valueofthewavelengthofthislinemeasuredinthelaboratoryinlightemittedfromasourceofatomic hydrogen in the galaxy

PHYSICS 2015indb 306 150514 337 PM

10 Fields

Contents

101 ndash Describing fields

102 ndash Fields at work

Essential Ideas

Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Similar approaches can be taken in analysing electrical and gravitational potential problems copy IBO 2014

PHYSICS 2015indb 307 150514 337 PM

Chapter 10

308

AH

L

101 Describing fields

Essential idea Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Understandingsbull Gravitationalfieldsbull Electrostaticfieldsbull Electricpotentialandgravitationalpotentialbull Fieldlinesbull Equipotentialsurfaces

Gravitational potentialWehaveseenthatifweliftanobjectofmassm to a height h abovethesurfaceoftheEarththenitsgainingravitationalpotential energy is mghHoweverthisisbynomeansthefullstoryForastartwenowknowthatgvarieswithh and also the expression really gives a difference in potentialenergybetweenthevaluethattheobjecthasattheEarthrsquossurfaceandthevaluethatithasatheighthSowhatwereallyneedisazeropointCanwefindapointwherethepotentialenergyiszeroandusethispointfromwhichtomeasure changes in potential energy

The point that is chosen is in fact infinity At infinity the gravitationalfieldstrengthofanyobjectwillbezeroSoletusseeifwecandeduceanexpressionforthegaininpotentialenergyofanobjectwhenitisldquoliftedrdquofromthesurfaceoftheEarthtoinfinityThisineffectmeansfindingtheworknecessarytoperformthistask

δr

r r δr+

r infin=

mg

AB

Me

Figure 1001 Gravitational forces

Inthediagramweconsidertheworknecessarytomovethe particle of mass m a distance δr in thegravitationalfield of the Earth

The force on the particle at A is FG Mem

r2----------------=

IftheparticleismovedtoB then since δr isverysmallwe can assume that the field remains constant over thedistance ABTheworkδW done againstthegravitationalfieldoftheEarthinmovingthedistanceABis

δWGMem

r2----------------δrndash=

(rememberthatworkdoneagainstaforceisnegative)

TofindthetotalworkdoneW in going from the surface oftheEarthtoinfinitywehavetoaddalltheselittlebitsofworkThisisdonemathematicallybyusingintegralcalculus

WG Me m

r2----------------ndash

rdR

infin

int G Me m 1

r2----- rd

R

infin

intndash G Me m 1r---ndash

R

infinndash= = =

GMe m 0 1R---ndash

ndashndash=

G MemR----------------ndash=

HencewehavewhereRistheradiusoftheEarththattheworkdonebythegravitationalfieldinmovinganobjectofmass m from R(surfaceoftheEarth)toinfinityisgivenby

WGMe m

R---------------- ndash=

We can generalise the result by calculating the worknecessary per unit mass to take a small mass fromthe surface of the Earth to infinity This we call thegravitational potential Vie

Wewouldgetexactlythesameresultifwecalculatedtheworkdonebythefieldtobringthepointmassfrominfinityto the surface of Earth In this respect the formal definition ofgravitationalpotentialatapointinagravitationalfieldis therefore defined as the work done per unit mass in bringing a point mass from infinity to that point

Clearly then the gravitational potential at any point inthe Earthrsquos field distance r from the centre of the Earth (providingr gt R)is

VGMe

r------------ ndash=

Thepotentialisthereforeameasureoftheamountofworkthathastobedonetomoveparticlesbetweenpointsinagravitationalfieldanditsunit is theJkgndash1 We also note thatthepotentialisnegativesothatthepotentialenergy

NATURE OF SCIENCEParadigm shift The move from direct observable actions being responsible for influence on an object to acceptance of a fieldrsquos ldquoaction at a distancerdquo required a paradigm shift in the world of science (23)

copy IBO 2014

PHYSICS 2015indb 308 150514 337 PM

Fields

309

AH

L

aswemoveawayfromtheEarthrsquossurfaceincreasesuntilitreachesthevalueofzeroatinfinity

IfthegravitationalfieldisduetoapointmassofmassmthenwehavethesameexpressionasaboveexceptthatMe is replaced by m andmustalsoexclude thevalueof thepotential at the point mass itself ie at r = 0

WecanexpressthegravitationalpotentialduetotheEarth(orduetoanysphericalmass)intermsofthegravitationalfield strength at its surface

AtthesurfaceoftheEarthwehave

g 0ndash Re GMRe

---------ndash= e

Sothat

g 0Re2 GM= e

Hence at a distance r from the centre of the Earth the gravitationalpotentialVcanbewrittenas

VGMe

r------------ndashg 0 Re

2

r------------ndash= =

The potential at the surface of the Earth

(r = Re)istherefore-g0Re

It is interesting to see how the expression for thegravitational potential ties in with the expression mgh ThepotentialatthesurfaceoftheEarthis-g0Re (see the exampleabove)andataheighthwillbe-g0g 0 Re h+( )ndash ifweassume that g0doesnotchangeover thedistanceh The differenceinpotentialbetweenthesurfaceandtheheighth is therefore g0hSotheworkneededtoraiseanobjectof mass m to a height h is mgh ie m timesdifference ingravitationalpotential

This we have referred to as the gain in gravitationalpotentialenergy(see235)

However this expression can be extended to any twopoints inanygravitationalfieldsuch that ifanobjectofmass mmovesbetween twopointswhosepotentialsareV1 and V2 respectively then the change in gravitationalpotentialenergyoftheobjectism(V1 ndash V2)

Gravitational potential gradient

Let us consider now a region in space where thegravitationalfieldisconstantInFigure912thetwopointsAandBareseparatedbythedistance∆x

A B

direction of uniform gravitational eld of strength I

∆x

Figure 1002 The gravitational potential gradient

The gravitational field is of strength I and is in the direction shownThe gravitational potential at A is V andatBisV +∆V

Theworkdone is takingapointmassmfromAtoB isF∆x = mI∆x

Howeverbydefinitionthisworkisalsoequalto-m∆V

Therefore mI∆x=-m∆V

or VIx

∆= minus

∆Effectivelythissaysthatthemagnitudeofthegravitationalfield strength is equal to the negative gradient of thepotential If I is constant then V is a linear function of x and Iisequaltothenegativegradientofthestraightlinegraph formed by plotting V against x If I is not constant (asusuallythecase)thenthemagnitudeofI at any point in the field can be found by find the gradient of the V-x graph at that point An example of such a calculation can befoundinSection929

For those of you who do HL maths the relationship between field and potential is seen to follow from the expression for the potential of a point mass viz

mV Gr

= minus

2

ddV mG Ir r

minus = + =

Potential due to one or more point masses

Gravitationalpotentialisascalarquantitysocalculatingthepotential due to more than one point mass is a matter of simpleadditionSoforexamplethepotentialV due to the Moon and Earth and a distance x fromthecentreofEarthonastraightlinebetweenthemisgivenbytheexpression

E MM MV Gx r x

= minus + minus

whereME=massofEarthMM= mass of Moon and r = distancebetweencentreofEarthandMoon

PHYSICS 2015indb 309 150514 337 PM

Chapter 10

310

AH

L

Equipotentials and field linesIfthegravitationalpotentialhasthesamevalueatallpointson a surface the surface is said to be an equipotential surfaceSoforexampleifweimagineasphericalshellaboutEarthwhosecentrecoincideswiththecentreofEarththisshellwillbeanequipotentialsurfaceClearlyifwerepresentthegravitationalfieldstrengthbyfieldlinessincethelinesldquoradiaterdquooutfromthecentreofEarththentheselineswillbeatrightanglestothesurfaceIfthefieldlineswerenotnormaltotheequipotentialsurfacethentherewouldbeacomponentof the field parallel to the surfaceThis wouldmean thatpointsonthesurfacewouldbeatdifferentpotentialsandsoitwouldnolongerbeanequipotentialsurfaceThisofcourseholdstrueforanyequipotentialsurface

Figure 1003 shows the field lines and equipotentials fortwopointmassesm

m m

Figure 1003 Equipotentials for two point masses

It is worth noting that we would get exactly the samepattern ifwewere to replace thepointmasseswith twoequalpointcharges(See935)

Escape speedThe potential at the surface of Earth is which meansthattheenergyrequiredtotakeaparticleofmass mfromthesurfacetoinfinityisequalto

ButwhatdoesitactuallymeantotakesomethingtoinfinityWhentheparticleisonthesurfaceoftheEarthwecanthinkofitassittingatthebottomofaldquopotentialwellrdquoasinfigure1004

particle

surface of Earth

innity

GMR---------

FIgure 1004 A potential well

Theldquodepthrdquoofthewellis and if the particle gains an amount of kinetic energy equal to wherem is its massthenitwillhavejustenoughenergytoldquoliftrdquoitoutofthewell

In reality it doesnrsquot actually go to infinity it justmeansthat the particle is effectively free of the gravitationalattractionof theEarthWe say that ithas ldquoescapedrdquo theEarthrsquosgravitationalpullWemeetthisideainconnectionwithmolecular forcesTwomolecules in a solidwill sitat their equilibrium position the separation where therepulsiveforceisequaltotheattractiveforceIfwesupplyjust enough energy to increase the separation of themolecules such that they are an infinite distance apart then intermolecularforcesnolongeraffectthemoleculesandthesolidwillhavebecomealiquidThereisnoincreaseinthekineticenergyofthemoleculesandsothesolidmeltsat constant temperature

Wecancalculatetheescapespeedofanobjectveryeasilyby equating the kinetic energy to the potential energysuch that

12---mvesc ape

2 GMe mRe

----------------=

vesc aperArr2GMe

Re--------------- 2g0 Re = =

Substituting for g0 and Regivesavalueforvescape of about 11kmsndash1 from the surface of the Earth

Youwillnotethattheescapespeeddoesnotdependonthemassoftheobjectsincebothkineticenergyandpotentialenergy are proportional to the mass

IntheoryifyouwanttogetarockettothemoonitcanbedonewithoutreachingtheescapespeedHoweverthiswouldnecessitate anenormousamountof fuel and it islikely that therocketplus fuelwouldbesoheavythat itwouldnevergetoffthegroundItismuchmorepracticalto accelerate the rocket to the escape speed fromEarthorbitandthenintheoryjustlaunchittotheMoon

Example

Use the followingdata todetermine thepotential at thesurface of Mars and the magnitude of the acceleration of free fall

massofMars =64times1023kg

radiusofMars =34times106 m

Determinealsothegravitationalfieldstrengthatadistanceof 68 times 106mabovethesurfaceofMars

PHYSICS 2015indb 310 150514 337 PM

Fields

311

AH

L

Solution

= minus = minus times times = times23

11 76

64 1067 10 ndash13 10

34 10MV GR

minus timestimes

Nkg-1

But V = -g0R

Therefore 7

0 6

13 10 3834 10

VgR

times= minus = =

timesm s-2

To determine the field strength gh at 68 times 106 m above the surface we use the fact that 0 2

Mg GR

= such that GM = g0R2

Therefore 2 2

0h 2 2 2

h h

38 (34) 042(102)

g RGMgR R

times= = = = m s-2

(the distance from the centre is 34 times 106 + 68 times 106 = 102 times 106 m)

Exercise 1011 Thegraphbelowshowshowthegravitational

potentialoutsideoftheEarthvarieswithdistancefrom the centre

0 10 20 30 40 50 60 70

ndash1

ndash2

ndash3

ndash4

ndash5

ndash6

ndash7

r m times 106

V

Jkgndash1

times 10

7

(a) Usethegraphtodeterminethegainingravitationalpotentialenergyofasatelliteofmass200kgasitmovesfromthesurfaceof the Earth to a height of 30 times 107mabovetheEarthrsquossurface

(b) Calculatetheenergyrequiredtotakeit to infinity

(c) DeterminetheslopeofthegraphatthesurfaceoftheEarthmCommenton youranswer

Electric potential energy

The concept of electric potential energy was developedwith gravitational potential energy in mind Just as anobjectnearthesurfaceoftheEarthhaspotentialenergybecause of its gravitational interaction with the Earthsotoothereiselectricalpotentialenergyassociatedwithinteracting charges

Letusfirstlookatacaseoftwopositivepointchargeseachof 1μC that are initially bound together by a thread in a vacuuminspacewithadistancebetweenthemof10cmasshowninFigure1005Whenthethreadiscutthepointchargesinitiallyatrestwouldmoveinoppositedirectionsmovingwithvelocitiesv1 and v2 along the direction of the electrostatic force of repulsion

BEFORE AFTER

v v21

Figure 1005 Interaction of two positive particles

The electric potential energybetweentwopointchargescan be found by simply adding up the energy associated witheachpairofpointchargesForapairofinteractingchargestheelectricpotentialenergyisgivenby

∆U = ∆Ep + ∆Ek = ∆W = ∆Fr = kqQ

____ r2 times r = kqQ

____ r

Becausenoexternalforceisactingonthesystemtheenergyand momentum must be conserved Initially Ek = 0 and Ep =kqQr=9times109 times 1 times 10-1201m=009JWhentheyareagreatdistance fromeachotherEpwillbenegligibleThefinalenergywillbeequaltofrac12mv1

2+frac12mv22=009J

Momentum isalsoconservedand thevelocitieswouldbethe same magnitude but in opposite directions

Electric potential energyismoreoftendefinedintermsofapointchargemovinginanelectricfieldas

lsquothe electricpotential energybetweenany twopoints inan electric field is defined as negativeoftheworkdone by anelectricfieldinmovingapoint electric chargebetweentwolocationsintheelectricfieldrsquo

ΔU = ΔEp = -ΔW = -Fd = qEx

ΔU = ΔEp +ΔEk = ΔW = ΔFr = kqQ r2 x r = kqQ

wherexisthedistancemovedalong(oroppositeto)thedirection of the electric field

Electric potential energyismeasuredinjoule(J)Justasworkisascalarquantitysotooelectricalpotentialenergyis a scalar quantity The negative of the work done by an electric field in moving a unit electric charge between two points is independent of the path taken Inphysicswesaytheelectricfieldisalsquoconservativersquo field

PHYSICS 2015indb 311 150514 337 PM

Chapter 10

312

AH

L

Supposeanexternalforcesuchasyourhandmovesasmallpositive point test charge in the direction of a uniformelectric fieldAs it ismoving itmust be gaining kineticenergyIfthisoccursthentheelectricpotentialenergyofthe unit charge is changing

InFigure1006apointcharge+qismovedbetweenpointsAandBthroughadistancex in a uniform electric field

B

A

x

+ q

Figure 1006 Movement of a positive point charge in a uniform field

InordertomoveapositivepointchargefrompointAtopointBanexternal forcemustbeapplied to thechargeequaltoqE (F = qE)

Since the force is applied through a distance xthennegativeworkhas to bedone tomove the charge because energyis gainedmeaning there is an increaseelectric potential energybetweenthetwopointsRememberthattheworkdoneisequivalenttotheenergygainedorlostinmovingthe charge through the electric field The concept of electric potential energy is only meaningful as the electric field whichgeneratestheforceinquestionisconservative

W F xtimes E q x times= =

θ x cos θ x

Figure 1007 Charge moved at an angle to the field

If a chargemoves at an angle θ to an electric field thecomponent of the displacement parallel to the electric fieldisusedasshowninFigure918

W F x Eq x θcostimes= =

Theelectricpotentialenergyisstoredintheelectricfieldand theelectricfieldwill return theenergy to thepointcharge when required so as not to violate the Law ofconservationofenergy

Electric potential

The electric potential at a point in an electric field is defined as being the work done per unit charge in bringing a small positive point charge from infinity to that point

ΔV = Vinfin ndash Vf = ndash W ___ q

Ifwedesignatethepotentialenergytobezeroatinfinitythen it follows thatelectricpotentialmustalsobezeroat infinity and the electric potential at any point in an electricfieldwillbe

ΔV = ndash W ___ q

NowsupposeweapplyanexternalforcetoasmallpositivetestchargeasitismovedtowardsanisolatedpositivechargeTheexternalforcemustdoworkonthepositivetestchargetomoveittowardstheisolatedpositivechargeandtheworkmustbepositivewhiletheworkdonebytheelectricfieldmustthereforebenegativeSotheelectricpotentialatthatpointmustbepositiveaccordingtotheaboveequationIfanegativeisolatedchargeisusedtheelectricpotentialatapointonthepositivetestchargewouldbenegativePositivepoint chargesof their ownaccordmove fromaplaceofhighelectricpotential toaplaceof lowelectricpotentialNegative point charges move the other way from lowpotentialtohighpotentialInmovingfrompointAtopointBinthediagramthepositivecharge+qismovingfromalowelectricpotentialtoahighelectricpotential

In thedefinitiongiven the termldquoworkperunitchargerdquohas significance If the test charge is +16 times 10-19Cwherethe charge has a potential energy of 32 times 10-17Jthenthepotentialwouldbe32times10-17J+16times10-19C=200JC-1 Now if the charge was doubled the potential wouldbecome 64times10-17 J However the potential per unitchargewouldbethesame

Electric potential is a scalar quantityandithasunitsJC-1 orvoltswhere1voltequalsone joulepercoloumbThevoltallowsustoadoptaunitfortheelectricfieldintermsofthevolt

PreviouslytheunitfortheelectricfieldwasNCndash1

W = qV and F = qEsoW ___ V = F __ E

E = FV ___ W = FV ___ Fm V mndash1

That is the units of the electric field E can also beexpressed as V mndash1

The work done per unit charge in moving a point charge between two points in an electric field is again independent of the path taken

PHYSICS 2015indb 312 150514 337 PM

Fields

313

AH

L

Electric potential due to a point charge

Letus takeapoint rmetres froma chargedobjectThepotentialatthispointcanbecalculatedusingthefollowing

W = ndashFr = ndashqV and F = ndash q1 q2 _____ 4πε0r

2

Therefore

W = ndash q1 q2 _____ 4πε0r

2 times r = ndash q1 q2 _____ 4πε0r

= ndashq1 times q2 _____ 4πε0r

= ndashq1V

That is V q

4πε0r--------------- =

Orsimply

V kqr------

=

Example

Determinehowmuchworkisdonebytheelectricfieldofpointcharge150μCwhenachargeof200μCismovedfrom infinity to apoint 0400m from thepoint charge(Assumenoaccelerationofthecharges)

Solution

The work done by the electric field is W = -qV = -14πε0 times q times (Q rinfin - Q r0400)

W = (- 200 times 10-6 C times 900 times 109 NmC-2 times 150 times 10-6 C) divide 0400 m = - 0675 J

An external force would have to do +0675 J of work

Electric field strength and electric potential gradient

Letus lookbackatFigure1006Supposeagain that thecharge+qismovedasmalldistancebyaforceFfromAtoBsothattheforcecanbeconsideredconstantTheworkdoneisgivenby

∆W F x∆times=

TheforceFandtheelectricfieldEareoppositelydirectedandweknowthat

F = -qE and ΔW = q ΔV

Thereforetheworkdonecanbegivenas

q ΔV = -q E Δx

Therefore E V∆x∆-------

ndash=

The rate of change of potential ΔV at a point with respect to distance Δx in the direction in which the change is maximum is called the potential gradient We say that the electric field = - the potential gradient and the units are Vm-

1 From the equation we can see that in a graph of electric potential versus distance the gradient of the straight line equals the electric field strength

In reality if a charged particle enters a uniform electric field it will be accelerated uniformly by the field and its kinetic energy will increase This is why we had to assume no acceleration in the last worked example

Ek∆ 12---

mv2 q E xsdot sdot q Vx---

xsdot sdot q Vsdot= = = =

Example

Determine how far apart two parallel plates must besituated so that a potential difference of 150 times 102 V produces an electric field strength of 100 times 103 NC-1

Solution

Using E V∆x------- xhArrndash V∆

E-------15 102 Vtimes

100 103 N C 1ndashtimes------------------------------------------= = =

= 150 times 10-1

The plates are 150 times 10-1 m apart

The electric field and the electric potential at a point due to an evenly distributed charge +q on a sphere can be represented graphically as in Figure 1008

r

r

V

E

r0

r0

r0

Charge of +Q evenly distributed over surface

E 14πε0------------ Q

r2----- r r0gt=

V 14πε0------------Q

r----r r0gt=

Figure 1008 Electric field and potential due to a charged sphere

PHYSICS 2015indb 313 150514 337 PM

Chapter 10

314

AH

L

When the sphere becomes charged we know that thechargedistributes itselfevenlyoverthesurfaceThereforeeverypartofthematerialoftheconductorisatthesamepotential As the electric potential at a point is defined as beingnumericallyequaltotheworkdoneinbringingaunitpositivechargefrominfinitytothatpointithasaconstantvalueineverypartofthematerialoftheconductor

Since the potential is the same at all points on the conducting surfacethen∆V∆xiszeroButE=ndash∆V∆xThereforethe electric field inside the conductor is zero There is no electric field inside the conductor

SomefurtherobservationsofthegraphsinFigure1009are

bull OutsidethespherethegraphsobeytherelationshipsgivenasE α 1 r2 and V α 1 r

bull Atthesurfacer = r0Thereforetheelectricfieldandpotentialhavetheminimumvalueforr at this point and this infers a maximum field and potential

bull Insidethespheretheelectricfieldiszerobull Insidethespherenoworkisdonetomoveacharge

fromapointinsidetothesurfaceThereforethereisno potential difference and the potential is the same asitiswhenr = r0

Similargraphscanbedrawnfortheelectricfieldintensityandthe electric potential as a function of distance from conducting parallelplatesandsurfacesandthesearegiveninFigure1009

Potential plot E eld plotE eld

++++

ndashndashndashndash

xx x

+

x x

ndashx x

+ ndashx

x

+ +

x

x

Figure 1009 Electric field and electric potential at a distance from a charged surface

Potential due to one or more point charges

The potential due to one point charge can be determined byusingtheequationformula

V = kq r

Example 1

Determine the electric potential at a point 20 times 10-1 m fromthecentreofan isolatedconducting spherewithapointchargeof40pCinair

Solution

Using the formula

V = kq r we have

V90 10 9times( ) 40 10 12ndashtimes( )times

20 10 1ndashtimes( )------------------------------------------------------------------ 018 V= =

the potential at the point is 180 times 10-1 V

The potential due to a number of point charges can be determined by adding up the potentials due to individual point charges because the electric potential at any point outside a conducting sphere will be the same as if all the charge was concentrated at its centre

Example 2

Threepointchargesofareplacedattheverticesofaright-angledtriangleasshowninthediagrambelowDeterminetheabsolutepotentialat the+20μCchargeduetothetwoothercharges

- 6μC

+3μC +2μC

4 m

3 m

Solution

The electric potential of the +2 μC charge due to the ndash 6 μC charge is

V = (9 times 109 Nm2C-2 times -6 times 10-6 C) divide (radic 32 + 42) m = - 108 times 104 V

PHYSICS 2015indb 314 150514 338 PM

Fields

315

AH

L

Insummarywecanconcludethat

bull No work is done to move a charge along an equipotential

bull Equipotentials are always perpendicular to the electric lines of force

Figure1011and1012showsomeequipotential lines fortwo oppositely charged and identically positive spheresseparated by a distance

+ve ndashve

equipotential lines

Figure 1011 Equipotential lines between two opposite charges

equipotential lines

Figure 1012 Equipotential lines between two charges which are the same

+

ndash ndash ndash ndash

+ + +

40 V

30 V

20 V

10 V

50 V

Figure 1013 Equipotential lines between charged parallel plates

Figure1013showstheequipotentiallinesbetweenchargedparallel plates Throughout this chapter the similarities and differences between gravitational fields and electric fieldshavebeendiscussedTherelationshipsthatexistsbetweengravitationalandelectricquantitiesandtheeffectsofpointmassesandchargesissummarisedinTable1014

The electric potential of the +2 μC charge due to the +3 μC charge is

V = (9 times 109 Nm2C-2 times 3 times 10-6 C) divide 3m = 9 times 103 V

The net absolute potential is the sum of the 2 potentials- 108 times 104 V + 9 times 103 V =- 18 times 103 V

The absolute potential at the point is - 18 times 103 V

Equipotential surfacesRegions in space where the electric potential of a chargedistributionhasaconstantvaluearecalledequipotentials The placeswherethepotentialisconstantinthreedimensionsarecalled equipotential surfacesandwheretheyareconstantintwodimensionstheyarecalledequipotential lines

TheyareinsomewaysanalogoustothecontourlinesontopographicmapsInthiscasethegravitationalpotentialenergy is constant as amassmoves around the contourlines because the mass remains at the same elevationabovetheEarthrsquossurfaceThegravitationalfieldstrengthacts in a direction perpendicular to a contour line

Similarlybecausetheelectricpotentialonanequipotentiallinehas thesamevalueanelectric forcecandonoworkwhena testchargemovesonanequipotentialThereforethe electric field cannot have a component along anequipotentialandthusitmustbeeverywhereperpendiculartotheequipotentialsurfaceorequipotentiallineThisfactmakesiteasytoplotequipotentialsifthelinesofforceorlinesofelectricfluxofanelectricfieldareknown

Forexamplethereareaseriesofequipotentiallinesbetweentwoparallelplateconductorsthatareperpendiculartotheelectric fieldTherewill be a series of concentric circles(eachcirclefurtherapartthanthepreviousone)thatmapouttheequipotentialsaroundanisolatedpositivesphereThe lines of force and some equipotential lines for anisolatedpositivesphereareshowninFigure1010

Lines ofequipotential

Figure 1010 Equipotentials around an isolated positive sphere

PHYSICS 2015indb 315 150514 338 PM

Chapter 10

316

AH

L

Gravitational quantity Electrical quantityQ

uant

ities

g V∆x∆-------ndash= E V∆

x∆-------ndash=

Poin

t mas

ses a

nd ch

arge

s

V Gmr----

ndash= V 14πε0------------q

r---=

g G m

r2-----ndash= E 14πε0------------ q

r2-----=

F Gm1 m2

r2--------------= F 1

4πε0------------

q1q2

r2-----------=

Figure 1014 Formulas

Example

Deduce the electric potential on the surface of a gold nucleus that has a radius of 62 fm

Solution

Using the formula

V = kq r and knowing the atomic number of gold is 79 We will assume the nucleus is spherical and it behaves as if it were a point charge at its centre (relative to outside points)

V = 90 times 109 Nm2C-2 times 79 times 16 times 10-19 C divide 62 times 10-15 m = 18 times 107 V

The potential at the point is 18 MV

Example

Deduce the ionisation energy in electron-volts of theelectron in the hydrogen atom if the electron is in its ground state and it is in a circular orbit at a distance of 53 times 10-11 m from the proton

Solution

This problem is an energy coulombic circular motion question based on Bohrrsquos model of the atom (not the accepted quantum mechanics model) The ionisation energy is the energy required to remove the electron from the ground state to infinity The electron travels in a circular orbit and therefore has a centripetal acceleration The ionisation energy will counteract the coulombic force and the movement of the electron will be in the opposite direction to the centripetal force

Total energy = Ek electron + Ep due to the proton-electron interaction

ΣF = kqQ r2 = mv2 r and as such mv2 = = kqQ r

Therefore Ek electron = frac12 kqQ r

Ep due to the proton-electron interaction = - kqQ r

Total energy = frac12 kqQ r + - kqQ r = -frac12 kqQ r

= - 90 times 109 Nm2C-2 times (16 times 10-19 C)2 divide 53 times 10-11 m = -217 times 10-18 J

= -217 times 10-18 J divide 16 times 10-19 = -136 eV

The ionisation energy is 136 eV

Exercise 1021 ApointchargePisplacedmidwaybetweentwo

identicalnegativechargesWhichoneofthefollowingiscorrectwithregardstoelectricfieldandelectricpotentialatpointP

Electric field Electric potentialA non-zero zeroB zero non-zeroC non-zero non-zeroD zero zero

2 Twopositivechargedspheresaretiedtogetherinavacuumsomewhereinspacewheretherearenoexternal forces A has a mass of 25 g and a charge of20μCandBhasamassof15gandachargeof30μCThedistancebetweenthemis40cmTheyarethenreleasedasshowninthediagram

BEFORE AFTER

v1 v2

A B

(a) Determinetheirinitialelectricpotentialenergy in the before situation

(b) DeterminethespeedofsphereBafterrelease

3 Thediagrambelowrepresentstwoequipotentiallines in separated by a distance of 5 cm in a uniform electric field

+ + + + + + + +

ndash ndash ndash ndash ndash ndash ndash ndash

40 V

20 V 5 cm

PHYSICS 2015indb 316 150514 338 PM

Fields

317

AH

L

Determine the strength of the electric field

4 Thisquestionisabouttheelectricfieldduetoacharged sphere and the motion of electrons in that fieldThediagrambelowshowsanisolatedmetalsphereinavacuumthatcarriesanegativeelectriccharge of 60 μC

ndash

(a) Onthediagramdrawtheconventionalwayto represent the electric field pattern due to the charged sphere and lines to represent threeequipotentialsurfacesintheregionoutside the sphere

(b) Explainhowthelinesrepresentingtheequipotentialsurfacesthatyouhavesketchedindicatethatthestrengthoftheelectricfieldisdecreasingwithdistancefrom the centre of the sphere

(c) Theelectricfieldstrengthatthesurfaceofthe sphere and at points outside the sphere can be determined by assuming that the sphere acts as a point charge of magnitude 60 μC at its centre The radius of the sphere is 25 times 10ndash2 m Deduce that the magnitude of the field strength at the surface of the sphere is 86 times 107 Vmndash1

An electron is initially at rest on the surface of the sphere

(d) (i) Describethepathfollowedbytheelectronasitleavesthesurfaceofthesphere

(ii) Calculatetheinitialaccelerationofthe electron

5 Determinetheamountofworkthatisdoneinmovingachargeof100nCthroughapotentialdifferenceof150times102 V

6 Three identical 20 μC conducting spheres are placedatthecornersofanequilateraltriangleofsides 25 cm The triangle has one apex C pointing upthepageand2baseanglesAandBDeterminetheabsolutepotentialatB

7 Determinehowfaraparttwoparallelplatesmustbesituatedsothatapotentialdifferenceof250 times 102 V produces an electric field strength of 200 times 103 NC-1

8 Thegapbetweentwoparallelplatesis10times10-3mandthereisapotentialdifferenceof10times104 V betweentheplatesCalculate

i theworkdonebyanelectroninmovingfrom one plate to the other

ii thespeedwithwhichtheelectronreachesthe second plate if released from rest

iii theelectricfieldintensitybetweentheplates

9 Anelectronguninapicturetubeisacceleratedby a potential 25 times 103VDeterminethekineticenergygainedbytheelectroninelectron-volts

10 Determine the electric potential 20 times 10-2 m from achargeof-10times10-5 C

11 Determinetheelectricpotentialatapointmid-waybetweenachargeofndash20pCandanotherof+5pConthelinejoiningtheircentresifthecharges are 10 cm apart

12 During a thunderstorm the electric potential differencebetweenacloudandthegroundis10 times 109 V Determine the magnitude of the change in electric potential energy of an electron thatmovesbetweenthesepointsinelectron-volts

13 A charge of 15 μC is placed in a uniform electric fieldoftwooppositelychargedparallelplateswithamagnitudeof14times103 NC-1

(a) Determinetheworkthatmustbedoneagainstthefieldtomovethepointchargeadistance of 55 cm

(b) Calculatethepotentialdifferencebetweenthe final and initial positions of the charge

(c) Determinethepotentialdifferencebetweenthe plates if their separation distance is 15 cm

14 Duringaflashoflightningthepotentialdifferencebetweenacloudandthegroundwas12 times 109 V and the amount of transferred charge was32C

(a) Determinethechangeinenergyofthetransferred charge

(b) Iftheenergyreleasedwasallusedtoacceleratea1tonnecardeduceitsfinalspeed

(c) Iftheenergyreleasedcouldbeusedtomelticeat0degCdeducetheamountoficethatcould be melted

PHYSICS 2015indb 317 150514 338 PM

Chapter 10

318

AH

L

15 SupposethatwhenanelectronmovedfromAtoBin the diagram along an electric field line that the electric field does 36 times 10-19Jofworkonit

Determinethedifferencesinelectricpotential

(a) VB ndash VA(b) VC ndash VA(c) VC ndash VB

16 DeterminethepotentialatpointPthatislocatedatthecentreofthesquareasshowninthediagrambelow

- 6μC

+3μC +2μC

1m

1m

5μC

P

102 Fields at work

Essential idea Similar approaches can be taken inanalysingelectricalandgravitationalpotentialproblems

Understandingsbull Potentialandpotentialenergybull Potentialgradientbull Potentialdifferencebull Escapespeedbull Orbitalmotionorbitalspeedandorbitalenergybull Forcesandinverse-squarelawbehaviour

Orbital motion orbital speed and orbital energy

Although orbital motion may be circular elliptical orparabolic this sub-topic only dealswith circular orbitsThis sub-topic is not fundamentally new physics butan application that synthesizes ideas from gravitationcircularmotiondynamicsandenergy

TheMoonorbitstheEarthandinthissenseitisoftenreferredtoasasatelliteoftheEarthBefore1957itwastheonlyEarthsatelliteHoweverin1957theRussianslaunchedthefirstmanmadesatelliteSputnik1SincethisdatemanymoresatelliteshavebeenlaunchedandtherearenowliterallythousandsofthemorbitingtheEarthSomeareusedtomonitortheweathersome used to enable people to find accurately their position on thesurfaceoftheEarthmanyareusedincommunicationsandnodoubtsomeareusedtospyonothercountriesFigure1015showshowinprincipleasatellitecanbeputintoorbit

Theperson(whosesizeisgreatlyexaggeratedwithrespectto Earth) standing on the surface on the Earth throwssome stonesThe greater the speed with which a stoneis thrown the further it will land from her The pathsfollowedbythethrownstonesareparabolasByastretchof the imaginationwecanvisualisea situation inwhichastoneis thrownwithsuchaspeedthatbecauseof thecurvatureoftheEarthitwillnotlandonthesurfaceoftheEarthbutgointoldquoorbitrdquo(Path4onFigure1015)

NATURE OF SCIENCECommunication of scientific explanations The ability to apply field theory to the unobservable (charges) and the massively scaled (motion of satellites) required scientists to develop new ways to investigate analyse and report findings to a general public used to scientific discoveries based on tangible and discernible evidence (51)

copy IBO 2014

PHYSICS 2015indb 318 150514 338 PM

Fields

319

AH

L

1

2

34Earth

Figure 1015 Throwing a stone into orbit

TheforcethatcausesthestonestofollowaparabolicpathandtofalltoEarthisgravityandsimilarlytheforcethatkeeps the stone in orbit is gravity For circular motionto occur we have seen that a force must act at rightanglestothevelocityofanobjectthatistheremustbeacentripetalforceHenceinthesituationwedescribeherethe centripetal force for circular orbital motion about the EarthisprovidedbygravitationalattractionoftheEarth

We can calculate the speedwithwhich a stonemustbethrowninordertoputitintoorbitjustabovethesurfaceof the Earth

If the stone has mass m and speed v thenwehave fromNewtonrsquos2ndlaw

mv2

RE---------- G

MEm

RE2------------=

whereRE is the radius of the Earth and ME is the mass of the Earth

Bearinginmindthatg 0 GME

RE2--------= then

v g RE 10 64 106timestimes 8 103times= = =

Thatisthestonemustbethrownat8times103m sndash1

Clearly we are not going to get a satellite into orbit soclosetothesurfaceoftheEarthMovingatthisspeedthefrictionduetoairresistancewouldmeltthesatellitebeforeithadtravelledacoupleofkilometresInrealitythereforeasatelliteisputintoorbitabouttheEarthbysendingitattachedtoarocketbeyondtheEarthrsquosatmosphereandthengivingitacomponentofvelocityperpendiculartoaradialvectorfromtheEarthSeeFigure1016

Earth Satellite carried by rocket to hereSatellite orbit

Tangential component of velocity

Figure 1016 Getting a satellite into orbit

Keplerrsquos third law(This work of Kepler and Newtonrsquos synthesis of the work is an excellent example of the scientific method and makes for a good TOK discussion)

In 1627 Johannes Kepler (1571-1630)publishedhis lawsof planetary motion The laws are empirical in natureandwere deduced from the observations of theDanishastronomer Tycho de Brahe (1546-1601)The third lawgives a relationship between the radius of orbit R of a planet and its period TofrevolutionabouttheSunThelawisexpressedmathematicallyas

2

3 constantTR

=

We shall nowuseNewtonrsquos LawofGravitation to showhowitisthattheplanetsmoveinaccordancewithKeplerrsquosthirdlaw

InessenceNewtonwasabletousehislawofgravitytopredict the motion of the planets since all he had to do was factor theF givenby this law intohis second law F = ma to find their accelerations and hence their future positions

InFigure1017theEarthisshownorbitingtheSunandthedistancebetweentheircentresisR

RSun

Earth

F es

F se

Figure 1017 Planets move according to Keplerrsquos third law

Fes is the force that the Earth exerts on the Sun and Fse is the force that the Sun exerts on the Earth The forces are equalandoppositeandtheSunandtheEarthwillactuallyorbitaboutacommoncentreHoweversince theSun issoverymuchmoremassivethantheEarththiscommoncentrewillbeclosetothecentreoftheSunandsowecanregard the Earth as orbiting about the centre of the Sun Theotherthingthatweshallassumeisthatwecanignorethe forces that the other planets exert on the Earth (This wouldnotbeawisethingtodoifyouwereplanningtosend a space ship to theMoon for example)We shallalso assume that we have followed Newtonrsquos exampleandindeedprovedthataspherewillactasapointmasssituated at the centre of the sphere

Kepler had postulated that the orbits of the planets areellipticalbut since theeccentricityof theEarthrsquosorbit issmallweshallassumeacircularorbit

PHYSICS 2015indb 319 150514 338 PM

Chapter 10

320

AH

L

TheaccelerationoftheEarthtowardstheSunisa = Rω2

whereω ω 2πT------=

Hencea R 2π

T------ times

2 4π2 RT 2------------- = =

But the acceleration is given by Newtonrsquos Second Law F = mawhereFisnowgivenbytheLawofGravitationSo in this situation

F maGMs Me

R2-------------------= = butwealsohavethat

a a R 2πT

------ times

2 4π2 RT 2------------- = = and m = Me so that

G MsMe

R2------------------- Me4π2R

T2-------------times

GMs

4π2-----------rArr R3

T2------= =

Butthequantity

isaconstantthathasthesamevalueforeachoftheplanetssowehaveforalltheplanetsnotjustEarththat

R3

T2------ k =

wherekisaconstantWhichisofcourseKeplerrsquosthirdlaw

ThisisindeedanamazingbreakthroughItisdifficulttorefutethe idea that all particles attract each other in accordance withtheLawofGravitationwhenthelawisabletoaccountfortheobservedmotionoftheplanetsabouttheSun

The gravitational effects of the planets upon each othershould produce perturbations in their orbits Such is the predictivepoweroftheUniversalGravitationalLawthatit enabled physicists to compute these perturbations The telescopehadbeeninventedin1608andbythemiddleofthe 18th Century had reached a degree of perfection in design that enabled astronomers to actually measure the orbital perturbations of the planets Their measurements werealways inagreementwith thepredictionsmadebyNewtonrsquoslawHoweverin1781anewplanetUranuswasdiscovered and the orbit of this planet did not fit withthe orbit predicted by Universal Gravitation Such wasthe physicistrsquos faith in theNewtonianmethod that theysuspectedthatthediscrepancywasduetothepresenceofayetundetectedplanetUsingtheLawofGravitationtheFrenchastronomerJLeverrier and the English astronomer J C Adamswereable tocalculate justhowmassive thisnewplanetmustbeandalsowhereitshouldbeIn1846the planetNeptunewas discovered justwhere they hadpredicted Ina similarwaydiscrepancies in theorbitofNeptune ledto thepredictionandsubsequentdiscoveryin1930oftheplanetPlutoNewtonrsquosLawofGravitation

had passed the ultimate test of any theory it is not only abletoexplainexistingdatabutalsotomakepredictions

Satellite energyWhenasatelliteisinorbitaboutaplanetitwillhavebothkineticenergyandgravitationalpotentialenergySupposewe consider a satellite ofmass m that is in an orbit of radius r about a planet of mass M

Thegravitationalpotentialduetotheplanetatdistancer from its centre is

ThegravitationalpotentialenergyofthesatelliteVsat

is therefore GMe m

r----------------ndash

ThatisVsat GMem

r----------------ndash=

Thegravitationalfieldstrengthatthesurfaceoftheplanetisgivenby

g 0GMe

Re2------------=

Hencewecanwrite

Vsat g0 Re

2

rndash=

m

ThekineticenergyofthesatelliteKsatisequaltofrac12mv2wherev is its orbital speed

Byequatingthegravitationalforceactingonthesatellitetoitscentripetalaccelerationwehave

G Mem

r2---------------- mv2

r---------- mv2hArr

GMemr----------------= =

Fromwhich12---

mv2 12---

GMemr----------------times=

= 0 e

2g R

r

2m

Which is actually quite interesting since it shows thatirrespective of the orbital radius the KE is numericallyequal to half the PE Also the total energy E tot K sat Vsat+ 1

2---G Me m

r----------------times G Me m

r----------------ndash + 1

2---G Me m

r----------------ndash= = =of the satelliteisalwaysnegativesince

E tot K sat Vsat+ 12---

G Me mr----------------times

G Me mr----------------ndash

+ 12---

G Me mr----------------ndash= = =

PHYSICS 2015indb 320 150514 338 PM

Fields

321

AH

L

The energies of an orbiting satellite as a function of radial distancefromthecentreofaplanetareshownplottedinFigure1018

12

100806

0402

0ndash02

ndash04

ndash06ndash08ndash10

ndash12

2 4 6 8 10 12

ener

gya

rbita

ry u

nits

distance timesR

kinetic energytotal energy

potential energy

Figure 1018 Energy of an orbiting satellite as a function of distance from the centre of a planet

WeightlessnessSuppose that you are inanelevator(lift)thatisdescendingat constant speedandyou let goof abook thatyouareholdinginyourhandThebookwillfalltothefloorwithaccelerationequaltotheaccelerationduetogravityIfthecablethatsupportstheelevatorweretosnap(asituationthatItrustwillneverhappentoanyofyou)andyounowletgothebookthatyouareholdinginyourotherhandthis book will not fall to the floor - it will stay exactlyin linewithyourhandThis isbecause thebook isnowfallingwith the sameaccelerationas theelevatorandassuch the book cannot ldquocatchrdquo up with the floor of theelevatorFurthermoreifyouhappenedtobestandingonasetofbathroomscalesthescaleswouldnowreadzero-youwouldbeapparentlyweightlessItisthisideaoffreefallthatexplainstheapparentweightlessnessofastronautsin an orbiting satellite These astronauts are in free fall inthesensethattheyareacceleratingtowardsthecentre of the Earth

It is actually possible to define theweight of a body inseveral different waysWe can define it for example asthegravitationalforceexertedonthebodybyaspecifiedobject such as the EarthThiswe have seen thatwe doinlotsofsituationswherewedefinetheweightasbeingequaltomgIfweusethisdefinitionthenanobjectinfreefallcannotbydefinitionbeweightlesssinceitisstillinagravitationalfieldHoweverifwedefinetheweightofanobjectintermsofaldquoweighingrdquoprocesssuchasthereadingonasetofbathroomscaleswhichineffectmeasuresthecontact force between the object and the scales thenclearlyobjectsinfreefallareweightlessOnenowhasto

askthequestionwhetherornotitispossibleForexampletomeasurethegravitationalforceactingonanastronautin orbit about the Earth

WecanalsodefineweightintermsofthenetgravitationalforceactingonabodyduetoseveraldifferentobjectsForexample for an object out in space its weight could bedefined in terms of the resultant of the forces exerted on it bytheSuntheMoontheEarthandalltheotherplanetsin the Solar System If this resultant is zero at a particular pointthenthebodyisweightlessatthispoint

Inviewofthevariousdefinitionsofweightthatareavailabletousitisimportantthatwhenweusethewordldquoweightrdquoweareawareofthecontextinwhichitisbeingused

Example

Calculate the height above the surface of the Earth atwhichageo-stationarysatelliteorbits

Solution

A geo-stationary satellite is one that orbits the Earth in such a way that it is stationary with respect to a point on the surface of the Earth This means that its orbital period must be the same as the time for the Earth to spin once on its axis ie 24 hours

From Keplerrsquos third law we have G Ms

4π2----------- R3

T2------=

That is

RMe

mh

using the fact that the force of attraction between the satellite and the Earth is given by

FG Mem

R2----------------=

and that F = ma

where m is the mass of the satellite and a 4π2R

T2-------------=

we have

GMem

R2---------------- m 4π2R

T2-------------times

GMe

4π2------------rArr R3

T2------= =

PHYSICS 2015indb 321 150514 338 PM

Chapter 10

322

AH

L

Now the mass of the Earth is 60 times 1024 kg and the period T measured in seconds is given by T = 86400 s

So substitution gives R = 42 times 106 m

The radius of the Earth is 64 times 106 m so that the orbital height h is about 36 times 107 m

Example

Calculatetheminimumenergyrequiredtoputasatelliteofmass500kgintoanorbitthatisasaheightequaltotheEarthrsquosradiusabovethesurfaceoftheEarth

Solution

We have seen that when dealing with gravitational fields and potential it is useful to remember that

g 0G M

Re2---------= or g0 Re

2 GM=

The gravitational potential at the surface of the Earth is

g 0ndash Re GMRe

---------ndash=

The gravitational potential at a distance R

from the centre of the Earth is

=

The difference in potential between the surface of the Earth and a point distance R from the centre is therefore

V∆ g 0Re 1ReR------ndash

=

If R 2Re= then V∆g0 Re

2------------=

This means that the work required to ldquoliftrdquo the satellite into orbit is g0Rm where m is the mass of the satellite This is equal to

10 32 106 500timestimestimes = 16000 MJ

However the satellite must also have kinetic energy in order to orbit Earth This will be equal to

g 0 mRe2

2R-----------------g 0mRe

2

4----------------- 8000 MJ= =

The minimum energy required is therefore

24000 MJ

Exercise 1031 The speed needed to put a satellite in orbit does

not depend on

A the radius of the orbitB theshapeoftheorbitC thevalueofgattheorbitD the mass of the satellite

2 EstimatethespeedofanEarthsatellitewhoseorbitis400kmabovetheEarthrsquossurfaceAlsodetermine the period of the orbit

3 Calculatethespeedofa200kgsatelliteorbitingthe Earth at a height of 70 times 106 m

Assume that g = 82 m sndash2 for this orbit

4 TheradiioftwosatellitesXandYorbitingthe Earth are 2r and 8rwherer is the radius of the Earth Calculate the ratio of the periods of revolutionofXtoY

5 A satellite of mass mkgissentfromEarthrsquossurfaceinto an orbit of radius 5RwhereR is the radius of theEarthWritedownanexpressionfor

(a) thepotentialenergyofthesatelliteinorbit

(b) thekineticenergyofthesatelliteinorbit

(c) theminimumworkrequiredtosendthesatellitefromrestattheEarthrsquossurfaceintoits orbit

6 A satellite in an orbit of 10rfallsbacktoEarth(radius r)afteramalfunctionDeterminethespeedwithwhichitwillhittheEarthrsquossurface

7 The radius of the moon is frac14 that of the Earth AssumingEarthandtheMoontohavethesamedensitycomparetheaccelerationsoffreefallatthe surface of Earth to that at the surface of the Moon

8 UsethefollowingdatatodeterminethegravitationalfieldstrengthatthesurfaceoftheMoon and hence determine the escape speed from the surface of the Moon

Mass of the Moon = 73 times 1022kg

RadiusoftheMoon=17times106 m

PHYSICS 2015indb 322 150514 338 PM

Chapter 9

284

AH

L

91 Simple harmonic motion

Essential idea The solution of the harmonic oscillator can be framed around the variation of kinetic and potential energy in the system

Understandingsbull ThedefiningequationofSHMbull Energychanges

Recall from chapter 4 that if the acceleration a of a system is

bull directlyproportionaltoitsdisplacementxfromitsequilibriumposition

andbull directedtowardstheequilibriumposition

thenthesystemwillexecuteSHM

ThisistheformaldefinitionofSHM

Weexpressedthisdefinitionmathematicallyas

a = ndash const times x

The negative sign indicated that the acceleration was directedtowardstheequilibriumpositionMathematicalanalysis shows that the constant is in fact equal to ω2whereω is theangular frequency(definedabove)of thesystemHenceequation45becomes

a=ndashω2 x

If a system is performing SHM then to produce theaccelerationaforcemustbeactingonthesysteminthedirectionoftheaccelerationFromourdefinitionofSHMthe magnitude of the force F is given by

F = ndash kx

where k is a constant and the negative sign indicates that theforce isdirectedtowardstheequilibriumpositionofthesystem(DonotconfusethisconstantkwiththespringconstantHoweverwhendealingwiththeoscillationsofamassontheendofaspringkwillbethespringconstant)

To understand the solutions of the SHM equation letus consider the oscillations of a mass suspended from a vertically supported spring We shall consider the mass of the spring to be negligible and for the extension x to obey the rule F = kx F (= mg) is the force causing the extension Figure901(a)showsthespringandasuspendedweightofmass m inequilibriumInFigure901(b) theweighthasbeen pulled down a further extension x0

spring

weight of mass m x0

x0

P x

unbalanced force F on weight = -kx

(a) (b)

equilibrium x = 0

unstretched length

e ke

mg

Figure 901 SHM of a mass suspended by a spring

InFigure901(a)theequilibriumextensionofthespringis e and the net force on the weight is mg ndash ke = 0

InFigure408(b)iftheweightisheldinpositionatx=x0 and thenreleasedwhentheweightmovestopositionPadistance xfromtheequilibriumpositionx=0thenetforceon the weight is mg ndash ke ndash kxClearlythentheunbalancedforce on the weight is ndashkx When the weight reaches a point distance xabovetheequilibriumpositionthecompressionforce in the spring provides the unbalanced force towards theequilibriumpositionoftheweight

The acceleration of the weight is given by Newtonrsquos second law

F = ndashkx = ma

ie

This is of the form a = -ω2 x where ω = that is the weight will execute SHM with a frequency

The displacement of the weight x is given by

ω

ThisistheparticularsolutionoftheSHMequationfortheoscillation of a weight on the end of a spring This system

NATURE OF SCIENCEInsights The equation for simple harmonic motion (SHM) can be solved analytically and numerically Physicists use such solutions to help them to visualize the behaviour of the oscillator The use of the equations is very powerful as any oscillation can be described in terms of a combination of harmonic oscillators Numerical modelling of oscillators is important in the design of electrical circuits (111)

copy IBO 2014

PHYSICS 2015indb 284 150514 337 PM

Wave Phenomena

285

AH

L

is often referred to as a harmonic oscillator

Alternativelyiftheweightwasatitsequilibriumpositionwhent=0thenthetrigonometricexpressionwouldbe

x = x0sinωt

The velocity v of the weight at any instant can be found byfinding thegradient of the displacement-time graph at that instant The displacement time graph is a cosine function and the gradient of a cosine function is a negative sine function The gradient of

x = x0cosωt is in fact T 2T00

00

velo

city

dis

pla

cem

ent

t

so

ω ω ω

where v0isthemaximumandminimumvelocityequalinmagnitude to ωx0

Studentsfamiliarwithcalculuswillrecognisethevelocityv as

( )0d d cos sind d oxv x t x tt t

ω ω ω= = = minus Similarly

( ) 2 20 0

d d sin cosd dva x t x t xt t

ω ω ω ω ω= = minus = minus = minus

whichofcourseisjustthedefiningequationofSHM

However we have to bear in mind that ωt varies between 0 and 2π where cosωt is negative for ωt for

and sinωt is negative for ωt in the range π to

2π This effectively means that when the displacement fromequilibriumispositivethevelocityisnegativeandsodirectedtowardsequilibriumWhenthedisplacementfromequilibriumisnegativethevelocityispositiveandsodirectedawayfromequilibrium

ThesketchgraphinFigure902showsthevariationwithtime t of the displacement x and the corresponding variation with time t of the velocity v This clearly demonstrates the relation between the sign of the velocity and sign of the displacement

T 2T00

00

velo

city

disp

lace

men

t

t

x0

ndashx0

v0

ndashv0

Figure 902 Displacement-time and velocity-time graphs

We can also see how the velocity v changes with displacement x

We can express sinωt in terms of cosωt using the trigonometric relation

2 2sin cos 1θ θ+ =

From which it can be seen that

Replacing θ with ωt we have

20 1 cosv x tω ω= minus minus

Remembering that and putting x0 inside the squarerootgives

v=ndashωradic_______

( x0 2 ndash x2 )

Bearing in mind that v can be positive or negative wemust write

v=plusmnωradic_______

( x0 2 ndash x2 )

The velocity is zero when the displacement is a maximum and is a maximum when the displacement is zero

The graph in Figure 903 shows the variation with x of the velocity v for a system oscillating with a period of 1 secondandwithanamplitudeof5cmThegraphshowsthe variation over a time of any one period of oscillation

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

-5 -4 -3 -2 -1 1 2 3 4 5

v c

m s

-1

x cm

Figure 903 Velocity-displacement graph

PHYSICS 2015indb 285 150514 337 PM

Chapter 9

286

AH

L

Boundary ConditionsThe two solutions to the general SHM equation are

and Which solution applies to aparticularsystemdependsasmentionedaboveontheboundary conditions for that system For systems such astheharmonicoscillatorandthesimplependulumtheboundary condition that gives the solution is that the displacement x = x0 when t = 0 For some other systems it might turn out that x = 0 when t =0 This will lead to the solution From a practical pointofviewthetwosolutionsareessentiallythesamefor example when timing the oscillations of a simple pendulumyoumightdecidetostartthetimingwhenthependulumbobpasses through theequilibriumpositionIneffectthetwosolutionsdifferinphaseby

The table in Figure 904 summarises the solutions we have forSHM

x = x 0 cosωt x = x 0 sinωt

ν = minus ν 0 sinωt ν = ν 0 cosωt

ν = ndash ωx 0 sinωt ν = ndash ωx 0 cosωt

ν = plusmn ω radic______

x0 2 ndash x2 ν = plusmn ω radic

______ x0

2 ndash x2

Figure 904 Common equations

We should mention that since the general solution to the SHMequation is there are infactthreesolutionstotheequationThisdemonstratesa fundamental property of second order differential equationsthatoneofthesolutionstotheequationisthesum of all the other solutions This is the mathematical basis of the so-called principle of superposition

SHM is a very good example in which to apply the NewtonianmethoddiscussedinChapter2ieiftheforcesthatactonasystemareknownthenthefuturebehaviourofthesystemcanbepredictedHerewehaveasituationinwhich the force is given by ndashkx From Newtonrsquos second law therefore ndashkx = mawherem is the mass of the system and a istheaccelerationofthesystemHowevertheaccelerationis not constant For those of you who have a mathematical benttherelationbetweentheforceandtheaccelerationis written as

2dd

xkx mt

minus =2

This is what is called a ldquosecond orderdifferentialequationrdquoThesolutionoftheequationgives x as a function of tTheactualsolutionisoftheSHMequationis

x = Pcosωt + Qsinωt where P and Q are constants and ω is theangularfrequencyofthesystemandisequalto

Whether a particular solution involves the sine function or thecosinefunctiondependsontheso-calledlsquoboundaryconditionsrsquo If for example x = x0 (the amplitude)whent =0thenthesolutionisx = x0cosωt

The beauty of thismathematical approach is that oncethe general equation has been solved the solution forall systemsexecutingSHM isknownAll thathas tobeshowntoknowifasystemwillexecuteSHMisthattheaccelerationofthesystemisgivenbyEquation45ortheforceisgivenbyequation47Thephysicalquantitiesthatω will depend on is determined by the particular system Forexampleforaweightofmassm oscillating at the end of a vertically supported spring whose spring constant is k then ω = or from equation 44 For a simple pendulum ω = where l is the length of the pendulum and g is the acceleration of free fall such

that T=2πradic_

l _ g

Examples

ThegraphinFigure905showsthevariationwithtime t of the displacement x ofasystemexecutingSHM(SomequestionpartswerealreadydoneintheSLchapter4)

-10

-8

-6

-4

-2

0

2

4

6

8

10

05 1 15 2 25 3 35t s

xcm

Figure 905 Displacement ndash time graph for SHM

Use the graph to determine the

(i) periodofoscillation(ii) amplitudeofoscillation(iii) maximumspeed(iv) speedatt = 13 s(v) maximumacceleration

Solutions

(i) 20 s (time for one cycle)(ii) 80 cm(iii) Using gives

(remember that ) = 25 cm s-1

PHYSICS 2015indb 286 150514 337 PM

Wave Phenomena

287

AH

L

(iv) v = minusv0sinωt = minus25sin (13π) To find the value of the sine function we have to convert the 13π into degrees (remember ω and hence ωt is measured in radians)

1 deg therefore 13π = 13 times 180 = 234deg therefore v1 = minus25sin (234deg) = +20 cm s ndash1

Or we can solve using ωradic_______

( x0 2 ndash x2 )

from the graph at t = 13 s x = ndash 48 cm

therefore v = π times = 20 cm s-1

(v) Using = π2 times 80 = 79 m s-2

Energy changesWe must now look at the energy changes involved in SHMTodosowewillagainconcentrateontheharmonicoscillator The mass is stationary at x = +x0 (maximumextension)andalsoatx = ndashx0 (maximumcompression)At these two positions the energy of the system is all potential energy and is in fact the elastic potential energy stored in the spring This is the total energy of the system ET and clearly

Equation412

ThatisthatforanysystemperformingSHMtheenergyofthesystemisproportionaltothesquareoftheamplitudeThis is an important result and one that we shall return to when we discuss wave motion

At x =0thespringisatitsequilibriumextensionandthemagnitude of the velocity v of the oscillating mass is a maximum v0TheenergyisallkineticandagainisequaltoET We can see that this is indeed the case as the expression for the maximum kinetic energy Emax in terms of v0 is

Clearly ET and Emax areequalsuchthat

ET = 1 __ 2 kx0 2 = Emax = 1 __ 2 mv0

2

From which

v0 2 = k __ m x0

2(asv0ge0)

Therefore

v0 = radic__

k __ m x0 = ωx0

whichtiesinwiththevelocitybeingequaltothegradientofthedisplacement-timegraph(see415)Asthesystemoscillates there is a continual interchange between kinetic energy and potential energy such that the loss in kinetic energyequalsthegaininpotentialenergyandET = EK + EP

Remembering that wehavethat

Clearly thepotentialenergyEP at any displacement x is given by

At any displacement xthekineticenergyEK is EK = 1__2mv2

Hencerememberingthat

v_______( x0

2 ndash x2 ) wehave

EK = 1__2m 2 ( x0

2 ndash x2 )

Although we have derived these equations for a harmonicoscillatortheyarevalidforanysystemoscillatingwithSHMThe sketch graph in Figure 906 shows the variation with displacement x of EK and EP for one period

displacement

ener

gy potentialkinetic

Figure 906 Energy and displacement

Example

The amplitude of oscillation of a mass suspended by a vertical spring is 80 cm The spring constant of the spring is74Nmndash1Determine

(a) thetotalenergyoftheoscillator

(b) thepotentialandthekineticenergyoftheoscillator at a displacement of 48 cm from equilibrium

PHYSICS 2015indb 287 150514 337 PM

Chapter 9

288

AH

L

Solution

(a) Spring constant k = 74 N m-1 and

x0 = 80 times 10-2 m

ET = frac12 kx02

= frac12 times 74 times 64 times 10-4 = 024 J

(b) At x = 48 cm

Therefore EP = frac12 times 74 times (48 times 10-2)2

= 0085 times 10-4 J

EK = ET minus EP = 024 minus 0085 = 016 J

EXERCISE 911 Which graph shows the relationship between the

acceleration a and the displacement x from the equilibriumpositionofanobjectundergoingsimple harmonic motion

Acceleration

Displacement

Acceleration

Displacement

A B

Acceleration

Displacement

Acceleration

Displacement

C D

2 AnobjectisundergoingsimpleharmonicmotionaboutafixedpointPandthemagnitudeofitsdisplacementfromPisx Which one of the following statements is correct

Magnitudeoftheresultant force

Directionoftheresultant force

A Proportionaltox AwayfrompointP

B Inversely proportional to x AwayfrompointP

C Proportionaltox TowardspointP

D Inversely proportional to x TowardspointP

3 Theangularspeedoftheldquominuterdquohandofananalogue watch is

A π1800rads-1 C π30rads-1

B π60rads-1 D 120rads-1

4 Which of the following is true of the magnitude of the acceleration of the object that is undergoing simple harmonic motion

A It is greatest at the midpoint of the motionB It is greatest at the end points of the motionC It is uniform throughout the motionD Itisgreatestatthemidpointsandthe

endpoints

5 Aparticleoscillateswithsimpleharmonicmotionwith a period T

At time t=0theparticlehasitsmaximumdisplacement Which graph The variation with time t of the kinetic energy of the particle is shown in which diagram below

A

00

Ek

T t

D

00

Ek

T t

C

00

Ek

T t

B

0

Ek

t0 T

6 Figure907belowshowshowthedisplacementofthe magnet of mass 03 kg hanging from a spring varies with time for two oscillations

0 02 04 06 08 10

0

2

-2

-4

4

Disp

lace

men

t (cm

)

Time (s)

Figure 907

PHYSICS 2015indb 288 150514 337 PM

Wave Phenomena

289

AH

L

Using information from this graph calculate the

(a) valueofthespringconstant(b) maximumkineticenergyofthemagnet

7 AsystemisoscillatingwithSHMasdescribedbythe graph in the Figure 908 below

Disp

lace

men

t (cm

)

005 1 15 2 25 3 35

2

-2

-6

-4

4

6

Time (s)

Figure 908

(a) Usethegraphtodeterminethe

(i) periodofoscillation(ii) amplitudeofoscillation(iii) maximumspeed(iv) speedatt=13s(v) maximumacceleration

(b) Statetwovaluesoftforwhenthemagnitudeofthevelocity is a maximum and two values of t for when the magnitude of the acceleration is a maximum

9 Figure 909 shows the relationship that exists between the acceleration and the displacement from the equilibriumpositionforaharmonicoscillator

Acce

lera

tion

(m-2

)

Displacement (cm)0

-4000

-2000

2000

-05 050

Figure 909

(a) Stateandexplaintworeasonswhythegraphopposite indicates that the object is executing simple harmonic motion

(b) Determinethefrequencyofoscillation

92 Single-slit diffraction

Essential idea Single-slitdiffractionoccurswhenawaveis incident upon a slit of approximately the same size as the wavelength

Understandingsbull Thenatureofsingle-slitdiffraction

The nature of single-slit diffractionSingle slit diffraction intensity distribution

When plane wavefronts pass through a small aperture they spread out as discussed in chapter 4 This is an example ofthephenomenoncalleddiffractionLightwavesarenoexception to this and ways for observing the diffraction of light have also been discussed previously

Howeverwhenwelookatthediffractionpatternproducedbylightweobserveafringepatternthatisonthescreenthere is a bright central maximum with ldquosecondaryrdquomaxima either side of it There are also regions where there is no illumination and these minima separate the maxima If we were to actually plot how the intensity of illumination varies along the screen then we would obtain a graph similar to that as in Figure 910

intensity

distance along screen

Figure 910 Intensity distribution for single-slit diffraction

We would get the same intensity distribution if we were to plot the intensity against the angle of diffraction θ(Seenextsection)

NATURE OF SCIENCEDevelopment of theories When light passes through an aperture the summation of all parts of the wave leads to an intensity pattern that is far removed from the geometrical shadow that simple theory predicts (19)

copy IBO 2014

PHYSICS 2015indb 289 150514 337 PM

Chapter 9

290

AH

L

This intensity pattern arises from the fact that each point ontheslitactsinaccordancewithHuygenrsquosprincipleasa source of secondary wavefronts It is the interference between these secondary wavefronts that produces the typical diffraction pattern

Obtaining an expression for the intensity distribution is mathematically a little tricky and it is not something that we are going to attempt here However we can deducea useful relationship from a simple argument In this argument we deal with a phenomenon called Fraunhofer diffractionthatisthelightsourceandthescreenareaninfinitedistanceawayformtheslitThiscanbeachievedwith the set up shown in Figure 911

lens 1 lens 2

single slit

source

screen

Figure 911 Apparatus for viewing Fraunhofer diffraction

The source is placed at the principal focus of lens 1 and the screen isplacedat theprincipal focusof lens2Lens1 ensures that parallel wavefronts fall on the single slit and lens2ensuresthattheparallelraysarebroughttoafocuson the screen The same effect can be achieved using a laser and placing the screen some distance from the slit If the light and screenarenot an infinitedistance from the slitthen we are dealing with a phenomenon called Fresnel diffraction and such diffraction is very difficult to analyse mathematically To obtain a good idea of how the single slit patterncomesaboutweconsiderthediagramFigure912

Xθ2

Yλ

d

P

b

f

screen

θ1

Figure 912 Single slit diffraction

In particular we consider the light from one edge of the slittothepointPwherethispointis justonewavelengthfurther from the lower edge of the slit than it is from the

upper edge The secondary wavefront from the upper edge willtraveladistanceλ2furtherthanasecondarywavefrontfrom a point at the centre of the slitHencewhen thesewavefronts arrive at P theywill be out of phase andwillinterfere destructively The wavefronts from the next point below the upper edge will similarly interfere destructively with the wavefront from the next point below the centre of the slit In this way we can pair the sources across the whole width of the slit If the screen is a long way from the slit then the angles θ1 and θ2becomenearlyequal (If the screen is at infinity then they are equal and the two lines PX and XY are at right angles to each other)FromFigure1116weseethereforethattherewillbeaminimumatPif

λ b θ1 sin= where b is the width of the slit

Howeverbothanglesareverysmallequaltoθ saywhereθ is the angle of diffraction

Soitcanbewritten θ λb---

=

This actually gives us the half-angular width of the central maximum We can calculate the actual width of the maximum along the screen if we know the focal length of the lens focussing the light onto the screen If this is f then we have that

θ df---

= Suchthat d fλb-----

=

To obtain the position of the next maximum in the pattern we note that the path difference is 3

2---λ We therefore divide

theslitintothreeequalpartstwoofwhichwillproducewavefronts that will cancel and the other producing wavefronts that reinforce The intensity of the second maximum is therefore much less than the intensity of thecentralmaximumof theorderof005or120of theoriginalintensity(Muchlessthanonethirdinfactsincethewavefrontsthatreinforcewillhavedifferingphases)

We can also see now how diffraction effects become more and more noticeable the narrower the slit becomes If light of wavelength 430 nm was to pass through a slit of width say10cmandfallonascreen30mawaythenthehalfangular width of the central maximum would be 013 microm

d = f λ

___ b = 3 times 430 times 10 ndash9 ____________ 01 = 013 μm

There will be lots of maxima of nearly the same intensity and the maxima will be packed very closely together (Thefirstminimumoccursatadistanceof012micromfromthe centre of the central maximum and the next occurs effectivelyatadistanceof024microm)Weeffectivelyobservethe geometric pattern Refer to Example

We also see now how for diffraction effects to be noticeable the wavelength must be of the order of the slit width The width of the pattern increases in proportion to the wavelength and decreases inversely with the width of the

PHYSICS 2015indb 290 150514 337 PM

Wave Phenomena

291

AH

L

93 Interference

Essential idea Interference patterns from multiple slits andthinfilmsproduceaccuratelyrepeatablepatterns

Understandingsbull Youngrsquosdouble-slitexperimentbull Modulationoftwo-slitinterferencepatternbyone-slit

diffraction effectbull Multipleslitanddiffractiongratinginterferencepatternsbull Thinfilminterference

Youngrsquos double slit experimentPlease refer to interference in chapter 4 However wewill reiterate the condition for the interference of waves from two sources to be observed The two sources must be coherentthatistheymusthavethesamephaseorthephase difference between them must remain constant

Also to reinforce topics concerning the principle ofsuperpositionandpathdifferenceandphasedifferenceincluded in Figure 913 is another example of two source interference

We can obtain evidence for the wave nature of sound by showing that sound produces an interference pattern Figure 913 shows the set up for demonstrating this

CRO

speaker

speaker

signalgenerator

microphoneX

Y

Figure 913 Interference using sound waves

The two speakers are connected to the same output of the signal generator andplaced about 50 cmapartThesignalgeneratorfrequencyissettoabout600Hzandthe

slit If the slit width is much greater than the wavelength then the width of the central maxima is very small

Insummary

for destructive interference and minima

dsinθ=nλ wheren=123hellip

for constructive interference and maxima

dsinθ=(n+frac12)λ wheren=123hellip

Note that waves from point sources along the slit arrive in phase at the centre of the fringe pattern and constructively interferetoproducethecentralmaximumcalledthezero ordermaximum(n=0)

Example

LightfromalaserisusedtoformasingleslitdiffractionpatternThe width of the slit is 010 mm and the screen is placed 30 m from the slit The width of the central maximum is measured as26cmCalculatethewavelengthofthelaserlight

Solution

Since the screen is a long way from the slit we can use the small angle approximation such that the f is equal to 30 m

The half-width of the central maximum is 13 cm so we have

λ13 10 2ndashtimes( ) 10 10 4ndashtimes( )times

30------------------------------------------------------------------=

To give λ = 430 nm

This example demonstrates why the image of a point source formed by a thin converging lens will always have a finite width

Exercise 921 Aparallelbeamoflightofwavelength500nmis

incidentonaslitofwidth025mmThelightisbroughttofocusonascreenplaced150mfromthe slit Calculate the angular width and the linear width of the central diffraction maximum

2 Lightfromalaserisusedtoformasingleslitdiffraction pattern on a screen The width of the slit is 010 mm and the screen is 30 m from the slit The width of the central diffraction maximum is 26cmCalculatethewavelengthofthelaserlight

3 Determinetheanglethatyouwouldexpecttofindconstructiveinterference(forn=0n=1andn=2) fromasingleslitofwidth20μmandlightofwavelength 600 nm

NATURE OF SCIENCECuriosity Observed patterns of iridescence in animals such as the shimmer of peacock feathers led scientists to develop the theory of thin film interference (15)

Serendipity The first laboratory production of thin films was accidental (15)

copy IBO 2014

PHYSICS 2015indb 291 150514 337 PM

Chapter 9

292

AH

L

microphoneismovedalongthelineXYthatisaboutonemetre from the speakers As the microphone is moved along XY the trace on the cathode ray oscilloscope isseen to go through a series of maxima and minima corresponding to points of constructive and destructive interference of the sound waves

Aninterestinginvestigationistofindhowtheseparationof the points of maximum interference depends on the frequency of the source and also the distance apart ofthe speakers

In the demonstration of the interference between two soundsourcesdescribedaboveifweweretomoveoneofthe speakers from side to side or backwards and forwards the sound emitted from the two speakers would no longer be in phase No permanent points of constructive or destructive interference will now be located since the phase difference between the waves from the two sources is nolongerconstantiethesourcesarenolongercoherent

Light from an incandescent source is emitted with acompletely random phase Although the light from two separate sources will interfere because of the randomlychanging phase no permanent points of constructive or destructive interference will be observed This is why a singleslit isneededintheYoungrsquosdoubleslitexperimentByactingasapointsourceitessentiallybecomesacoherentlight source The light emitted from a laser is also very nearly coherent and this is why it is so easy to demonstrate optical interference and diffraction with a laser

Youngrsquos double slit experiment is one of the great classic experiments of physics and did much to reinforce the wave theoryoflightThomasYoungcarriedouttheexperimentin about 1830

It is essentially the demonstration with the ripple tank and the sound experiment previously described butusing light The essential features of the experiment are shown in Figure 914

screendouble slit

single slit

coloured lter

S 1

S2

Figure 914 Youngrsquos double slit experiment

YoungallowedsunlighttofallontoanarrowsingleslitA few centimetres from the single slit he placed a double slit The slits are very narrow and separated by a distance equal to about fourteen slitwidthsA screen is placed

about ametre from thedouble slitsYoungobservedapatternofmulticolouredldquofringesrdquo in the screenWhenhe placed a coloured filter between the single slit anddouble slit he obtained a pattern that consisted of bright coloured fringes separated by darkness

The single slit essentially ensures that the light falling on the double slit is coherent The two slits then act as the two speakers in the sound experiment or the two dippers in the ripple tank The light waves from each slit interfere and produce the interference pattern on the screen Without thefilterapatternisformedforeachwavelengthpresentinthesunlightHencethemulticolouredfringepattern

You can demonstrate optical interference for yourselfSmoking a small piece of glass and then drawing twoparallel lines on it can make a double slit If you then look throughthedoubleslitatasingletungstenfilamentlampyouwillseethefringepatternByplacingfiltersbetweenthe lamps and the slits you will see the monochromatic fringe pattern

YoucanalsoseetheeffectsofopticalinterferencebylookingatnetcurtainsEachlsquoholersquointhenetactsasapointsourceand the light from all these separate sources interferes and producesquiteacomplicatedinterferencepattern

A laser can also be used to demonstrate optical interference Sincethelightfromthelaseriscoherentitisveryeasytodemonstrate interference Just point the laser at a screen and place a double slit in the path of the laser beam

LetusnowlookattheYoungrsquosdoubleslitexperimentinmore detail The geometry of the situation is shown in Figure915

S1

S2 X

P

Q

D

y

d

screen

θ

θrsquo

Figure 915 The geometry of Youngrsquos double slit experiment

S1andS2 are the two narrow slits that we shall regard as twocoherentmonochromaticpointsourcesThedistancefrom the sources to the screen is D and the distance between the slits is d

The waves from the two sources will be in phase at Q and therewill be abright fringehereWewish tofindtheconditionfortheretobeabrightfringeatPdistancey from Q

PHYSICS 2015indb 292 150514 337 PM

Wave Phenomena

293

AH

L

We note that that D (asymp1metre)isverymuchgreaterthaneither y or d(asymp fewmillimetres)Thismeansthatbothθ and θʹ areverysmallanglesandforintentsandpurposesequal

From the diagram we have that

θ = Dy

And

2S Xd

θʹ =

(Remembertheanglesareverysmall)

Sinceθ asymp θʹ then

2S XyD d

=

ButS2X is the path difference between the waves as they

traveltoPWehavethereforethat

path difference = Dyd

TherewillthereforebeabrightfringeatPif

Dyd

= nλ

Supposethatthereisabrightfringeaty = y1 corresponding

to n = n1 then

Ddy1 = n

1λ

If the next bright fringe occurs at y = y2 this will correspond

to n = n1+1Hence

Ddy2 =(n

1+1)λ

This means that the spacing between the fringes y2 ndash y

1 is

given by

y2 ndash y

1 = d

Dλ

Youngactuallyusethisexpressiontomeasurethewavelengthof the light he used and it is a method still used today

We see for instance that if in a given set up we move the slits closer together then the spacing between the fringes will get greaterEffectivelyourinterferencepatternspreadsoutthatis there will be fewer fringes in a given distance We can also increase the fringe spacing by increasing the distance betweentheslitsandthescreenYouwillalsonotethatfora

givensetupusinglightofdifferentwavelengthsthenldquoredrdquofringeswillspacefurtherapartthanldquobluerdquofringes

In this analysis we have assumed that the slits act as point sources and as such the fringes will be uniformly spaced andofequalintensityAmorethoroughanalysisshouldtakeintoaccountthefinitewidthoftheslits

ReturningtoFigure915weseethatwecanwritethepathdifferenceS2X as

S2X = dtan θʹ

But since θʹ is a small angle the sine and tangent will be nearlyequalsothat

S2X = dsinθʹ

And since θʹ asymp θ then

S2X = dsinθ

The condition therefore for a bright fringe to be found at a point of the screen can therefore be written as

dsinθ = nλ

Insummary

for constructive interference and minima

d sinθ = nλ wheren=123hellip

for destructive interference and maxima

d sinθ = ( n + frac12 )λ wheren=123hellip

Figure 916 shows the intensity distribution of the fringes on the screen when the separation of the slits is large comparedtotheirwidthThefringesareofequalintensityandofequalseparation

Intensity

distance along screen

Figure 916 The intensity distribution of the fringes

It isworthnoting that if the slits are close together theintensity of the fringes is modulated by the intensity distribution of the diffraction pattern of one of the slits

PHYSICS 2015indb 293 150514 337 PM

Chapter 9

294

AH

L

Example

Light of wavelength 500nm is incident on two smallparallel slits separated by 10mmDetermine the anglewherethefirstmaximumisformed

If after passing through the slits the light is brought toa focusona screen15m fromtheslitscalculate theobserved fringe spacing on the screen

Solution

Using the small angle approximation we have

3

7

10105minus

minustimes==

dλθ

= 5 times 10-4 rad

The fringe spacing is given by

3

7

1010551

minus

minustimestimes==

dDy λ

= 075 mm

Modulation of two-slit interference pattern by one-slit diffraction effect

In practice the intensity pattern of the maxima as shown in Figure 917 is not constant but fluctuates while decreasingsymmetrically on either side of the central maximum as shown inFigure917Theintensitypatternisacombinationofboththe single-slit diffractionenvelope and the double slit pattern That is the amplitude of the two-slit interference pattern is modulated(ieadjustedto)byasingleslitdiffractionenvelope

2λD

___ λD

D

___λd

d

___

Double slitnite width

Figure 917 Double slit pattern

ForagivenslitseparationdwavelengthoflightandfixedslittoscreendistancethevariationinintensitydependsonthewidthoftheslitDAlthoughincreasingthewidthoftheslitsincreasestheintensityoflightinthefringesitalsomakesthem less sharp and they become blurred As the slit width is widenedthefringesgraduallydisappearbecausenumerouspoint sources along the widened slit give rise to their own dark and bright fringes that then overlap with each other

Exercise 931 InFigure913thedistancebetweenthespeakers

is050mandthedistancebetweenthelineofthespeakersandthescreenis20mAsthemicrophoneismovedalongthelineXYthedistance between successive points of maximum soundintensityis030mThefrequencyofthesoundwavesis44times103HzCalculateavalueforthe speed of sound

2 Laserlightofwavelength610nmfallsontwoslits10x10-5apartDeterminetheseparationofthefirstordermaximumformedonascreen20maway

3 Twoparallelslits012mmapartareilluminatedwith red light with a wavelength of 600 nm An interferencepatternfallsonascreen15maway

Calculate

(a) thedistancefromthecentralmaximumtothefirstbrightfringe

(b) thedistancetotheseconddarkline

Multiple slit and diffraction grating interference patterns

If we examine the interference pattern produced when monochromatic light passes through a different number of slits we notice that as the number of slits increases the number of observed fringes decreases the spacingbetween them increases and the individual fringes become much sharper We can get some idea of how this comes about by looking at the way light behaves when a parallel beam passes through a large number of slits The diagram for this is shown in Figure 918

θθd

1

2

3

telescope

Figure 918 A parallel beam passing through several slits

The slits are very small so that they can be considered to act as point sources They are also very close together such that d issmall(10ndash6 m)EachslitbecomesasourceofcircularwavefrontsandthewavesfromeachslitwillinterfereLetus

PHYSICS 2015indb 294 150514 337 PM

Wave Phenomena

295

AH

L

consider the light that leaves the slit at an angle θ as shown Thepathdifferencebetweenwave1andwave2isdsinθ and ifthisisequaltoanintegralnumberofwavelengthsthenthe two waves will interfere constructively in this direction Similarly wave 2 will interfere constructively with wave3 at this angle andwave 3with 4 etc across thewholegratingHenceifwelookatthelightthroughatelescopethatisbringittoafocusthenwhenthetelescopemakesanangle θ to the grating a bright fringe will be observed The condition for observing a bright fringe is therefore

dsinθ = mλ

Supposeweuselightofwavelength500nmandsupposethat d = 16 times 10ndash6 m

Obviously we will see a bright fringe in the straight on position θ=0(thezero order)

The next position will be when m =1(thefirst order)andsubstitutionintheaboveequationgivesθ = 18deg

The next position will be when m=2(thesecond order)and this give θ = 38deg

For m=3sinθ is greater than 1 so with this set up we only obtain5fringesonezeroorderandtwoeithersideofthezero order

The calculation shows that the separation of the orders is relatively large At any angles other than 18deg or 38deg the light leaving the slits interferes destructively We can see that the fringes will be sharp since if we move just a small angle away from 18deg the light from the slits will interfere destructively

An array of narrow slits is usually made by cutting narrow transparent lines very close together into the emulsion on aphotographicplate(typically200linespermillimetre)Suchanarrangementiscalledadiffraction grating

The diffraction grating is of great use in examining the spectral characteristics of light sources

All elements have their own characteristic spectrum An element can be made to emit light either by heating it until it is incandescent or by causing an electric discharge through itwhen it is in a gaseous state Yourschool probably has some discharge tubes and diffraction gratingsIfithasthenlookattheglowingdischargetubethroughadiffractiongratingIfforexampletheelementthat you are looking has three distinct wavelengths then each wavelength will be diffracted by a different amount inaccordancewiththeequationdsinθ = nλ

Alsoiflaserlightisshonethroughagratingontoascreenyou will see just how sharp and spaced out are the maxima By measuring the line spacing and the distance of the screen fromthelaserthewavelengthofthelasercanbemeasured

If your school has a set of multiple slits say from a single slit toeight slits then it isalsoaworthwhileexercise toexamine how the diffraction pattern changes when laser light is shone through increasing numbers of slits

Consider a grating of NparallelequidistantslitsofwidthaseparatedbyanopaqueregionofwidthD We know that light diffracted from one slit will be superimposed and interfere with light diffracted from all the other slits The intensityprofile of the resultant wave will have a shape determined by

bull theintensityoflightincidentontheslitsbull thewavelengthofthislightbull theslitwidthbull theslitseparation

Typical intensity profiles formed when a plane wavemonochromaticlightatnormalincidenceononetwoandsix slits are shown in Figure 919 In this example the slit widthisonequarterofthegratingspacing

Six slits sharperfringes

d = 4a

I α N 2

(N ndash 2)= 4 subsidiary maxima

order0 4321

I = 36

Two slits

order0

single-slitdiraction

envelope

d = 4a4th principal maximum missing

principa maxima

I = 4 (nominal units)

4321

intensity

order

ISS

One slit

0 1

single-slitdiraction

pattern

N = number of slitsslit width a

Figure 919 Intensity profiles

PHYSICS 2015indb 295 150514 337 PM

Chapter 9

296

AH

L

1 For Nge2thefringepatterncontainsprincipalmaximathat are modulated by the diffraction envelope

2 ForNgt2thefringepatterncontains(Nndash2)subsidiary maxima

3 Where the grating spacing d is four time the slit width a(4xa =d)thenthefourthprincipalmaximumismissing from the fringe pattern Where the grating spacingdisthreetimetheslitwidtha(3xa =d)thenthe third principal maximum is missing from the fringepatternandsoon

4 Increasing N increases the absolute intensities of the diffraction pattern (I = N2)

If white light is shone through a grating then the central image will be white but for the other orders each will be spread out intoacontinuousspectrumcomposedofaninfinitenumberof adjacent images of the slit formed by the wavelength of the different wavelengths present in the white light At any given point in the continuous spectrum the light will be very nearly monochromatic because of the narrowness of the images of the slit formed by the grating This is in contrast to the doubleslitwhereifwhitelightisusedtheimagesarebroadand the spectral colours are not separated

In summary as is the case of interference at a doubleslit there is a path difference between the rays fromadjacent slits of d sinθ Where there is a whole number of wavelengthsconstructiveinterferenceoccurs

For constructive interference and minima

d sinθ = mλ wheren=123hellip

When there is a path difference between adjacent rays of halfawavelengthdestructiveinterferenceoccurs

For and maxima

d sinθ = ( m + frac12 )λ wheren=123hellip

Exercise 941 Lightfromalaserisshonethroughadiffraction

grating on to a screen The screen is a distance of 20mfromthelaserThedistancebetweenthecentraldiffractionmaximumandthefirstprincipalmaximum formed on the screen is 094 m The diffractiongratinghas680linespermmEstimatethe wavelength of the light emitted by the laser

2 Monochromaticlightfromalaserisnormallyincident on a six-slit grating The slit spacing is three times the slit width An interference patternisformedonascreen2montheotherside of the grating

(a) Determineandexplaintheorderoftheprincipal maxima missing from the fringe pattern

(b) Howmanysubsidiarymaximaaretherebetween the principal maxima

(c) Howmanyprincipalmaximaarethere

(i) Betweenthetwofirstorderminimain the diffraction envelope

(ii) Betweenthefirstandsecondorderminima in the diffraction envelope

(d) Drawthepatternformedbythebrightfringesin(c)

(e) Sketchtheintensityprofileofthefringepattern

Thin film interferenceYoumightwellbe familiarwith the colouredpatternoffringes that can be seen when light is reflected off thesurfaceofwateruponwhichathinoilfilmhasbeenspiltorfromlightreflectedfrombubblesWecanseehowthesepatternsarisebylookingatFigure920

Extendedlight source air

A

B

C

12

3

4

Oil lm

Water

Figure 920 Reflection from an oil film

Consider light from an extended source incident on a thinfilmWealsoconsiderawave fromonepointofthe source whose direction is represented by the ray shownSomeofthislightwillbereflectedatAandsometransmitted through the filmwhere somewill again bereflectedatB(somewillalsobetransmittedintotheair)SomeofthelightreflectedatBwillthenbetransmittedatCandsomereflectedandsoonIfweconsiderjustrays1and2thenthesewillbenotbeinphasewhenthearebrought to a focus by the eye since they have travelled

PHYSICS 2015indb 296 150514 337 PM

Wave Phenomena

297

AH

L

different distances If the path difference between them is an integral number of half wavelengths then we might expect the twowaves to be out of phaseHowever ray1 undergoes a phase change of π on reflection but ray2 does not since it is reflected at a boundary betweena more dense and less dense medium (See chapter 4)Henceifthepathdifferenceisanintegralnumberofhalf-wavelengthsrays1and2willreinforceieray1and2areinphaseHowever rays357etcwillbeoutofphasewithrays246etcbutsinceray2ismoreintensethanray3andray4moreintensethanray5thesepairswillnot cancel out so there will be a maximum of intensity

Ifthepathdifferenceissuchthatwave1and2areoutofphasesincewave1ismoreintensethanwave2theywillnotcompletelyannulHoweveritcanbeshownthattheintensitiesofwaves2345hellipaddtoequaltheintensityofwave1Sincewaves345hellipareinphasewithwave2there will be complete cancellation

The path difference will be determined by the angle at which ray1 is incident andalsoon the thickness (andtheactualrefractiveindexaswell)ofthefilmSincethesourceisanextendedsourcethelightwillreachtheeyefrom many different angles and so a system of bright and darkfringeswillbeseenYouwillonlyseefringesifthefilmisverythin(exceptifviewedatnormalincidence)sinceincreasingthethicknessofthefilmwillcausethereflected rays to get so far apart that they will not becollected by the pupil of the eye

From the argument above to find the conditions forconstructive and destructive interferenceweneedonlyfindthepathdifferencebetweenray1andray2Figure921showsthe geometry of the situation

A BE

F

C

D

1 2

φrsquo φrsquo

φ

φ

d

Figure 921 The geometry of interference

Thefilmisofthicknessd and refractive index n and the light has wavelength λ If the line BF is perpendicular to ray 1 then the optical path difference (opd) between ray 1 andray2whenbroughttoafocusis

opd = n(AC+CB)ndashAF

We have to multiple by the refractive index for the path travelledbythelightinthefilmsincethelighttravelsmoreslowly in thefilm If the light travels say adistancex in a material of refractive index n then in the time that it takes to travelthisdistancethelightwouldtraveladistancenx in air

IfthelineCEisatrightanglestoray2thenweseethat

AF = nBE

FromthediagramAC=CDsowecanwrite

opd = n(CD+CB)ndashnBE=nDE

Alsofromthediagramweseethatwhereφ is the angle of refraction

DE=2cosφ

From which opd=2ndcosφ

Bearinginmindthechangeinphaseofray1onreflectionwe have therefore that the condition for constructive interference is

2nd φcos m 12---+

λ m 1 2 hellip = =

Andfordestructiveinterference 2ndcosφ = mλ

Eachfringecorrespondstoaparticularopd for a particular value of the integer m and for any fringe the value of the angle φ is fixedThismeans that itwill be in the formof an arc of a circle with the centre of the circle at the point where the perpendicular drawn from the eye meets the surface of the film Such fringes are called fringesofequal inclinationSince theeyehasa smallaperturethesefringesunlessviewedatneartonormalincidence (φ = 0) will only be observed if the film is very thinThisisbecauseasthethicknessofthefilmincreasesthereflected rayswill get further and further apart and sovery few will enter the eye

Ifwhitelightisshoneontothefilmthenwecanseewhywe get multi-coloured fringes since a series of maxima and minima will be formed for each wavelength present in the white light However when viewed at normalincidenceitispossiblethatonlylightofonecolourwillundergoconstructiveinterferenceandthefilmwilltakeon this colour

PHYSICS 2015indb 297 150514 337 PM

Chapter 9

298

AH

L

Thickness of oil films

The exercise to follow will help explain this use

Non-reflecting films

A very important but simple application of thin filminterferenceisintheproductionofnon-reflectingsurfaces

Athinfilmofthicknessd and refractive index n1 is coated onto glass of refractive index n where n1 lt n Light ofwavelength λ that is reflected at normal incidence willundergo destructive interference if 12

2n d λ

= thatis

14d

nλ

=

(remember that there will now no phase change at the glass interface ie we have a rare to dense reflection)

The use of such films can greatly reduce the loss oflightby reflection at the various surfacesof a systemoflenses or prisms Optical parts of high quality systemsare usually all coatedwithnon-reflecting films in orderto reduce stray reflections The films are usually madebyevaporatingcalciumormagnesiumfluorideonto thesurfacesinvacuumorbychemicaltreatmentwithacidsthat leave a thin layer of silica on the surface of the glass ThecoatedsurfaceshaveapurplishhuebyreflectedlightThis is because the condition for destructive interference fromaparticularfilmthicknesscanonlybeobtainedforone wavelength The wavelength chosen is one that has a value corresponding to light near the middle of the visible spectrumThismeansthatreflectionofredandvioletlightis greater combining to give the purple colour Because of the factor cosφatanglesotherthannormalincidencethepath difference will change butnotsignificantlyuntilsayabout30deg(egcos25deg=090)

It should be borne in mind that no light is actually lost by anon-reflectingfilmthedecreaseofreflectedintensityiscompensated by increase of transmitted intensity

Non-reflecting films can be painted onto aircraft tosuppressreflectionofradarThethicknessofthefilmis determined by

4nd λ

= where λ is the wavelength of

the radar waves and ntherefractiveindexofthefilmat this wavelength

Example

A plane-parallel glass plate of thickness 20 mm isilluminated with light from an extended source The refractiveindexoftheglassis15andthewavelengthofthelight is 600 nm Calculate how many fringes are formed

Solution

We assume that the fringes are formed by light incident at all angles from normal to grazing incidence

At normal incidence we have 2nd = mλ

From which

m 2 15 2 10 3ndashtimestimestimes6 10 7ndashtimes

------------------------------------------- 10 000= =

At grazing incidence the angle of refraction φ is in fact the critical angle

Therefore φφ arcsin 115------- 42 deg= =

ie cosφ = 075

At grazing incidence 2ndcosφ = m λ

From which (and using cos(sinndash1(115) = 075)

m 2 15 2 10 3ndash 075timestimestimestimes6 10 7ndashtimes

------------------------------------------------------------ 7500= =

The total number of fringes seen is equal to m ndash m = 2500

Exercise 95When viewed from above the colour of an oil film onwater appears red in colour Use the data below to estimate theminimumthicknessoftheoilfilm

average wavelength of red light = 630 nm

refractiveindexofoilforredlight=15

refractive index of water for red light = 13

PHYSICS 2015indb 298 150514 337 PM

Wave Phenomena

299

AH

L

94 Resolution

Essential idea Resolution places an absolute limit on the extent to which an optical or other system can separate images of objects

Understandings

bull Thesizeofadiffractingaperturebull Theresolutionofsimplemonochromatic

two-source systems

The size of a diffracting apertureOur discussion concerning diffraction so far has been for rectangular slits When light from a point source enters a smallcircularapertureitdoesnotproduceabrightdotasanimagebutacirculardiscknownasAiryrsquosdiscsurroundedbyfainterconcentriccircularringsasshowninFigure922

Laser

Figure 922 Diffraction at a circular aperture

Diffractionatanapertureisofgreatimportancebecausetheeye and many optical instruments have circular apertures

What is the half-angular width of the central maximum of the diffraction formed by a circular aperture This is not easy to calculate since it involves some advanced mathematics The problemwasfirstsolvedbytheEnglishAstronomerRoyalGeorgeAiryin1835whoshowedthatforcircularapertures

θ 122 λb--------------= where b is the diameter of the aperture

Example 2

Inthefollowingdiagramparallellightfromadistantpointsource(suchasastar)isbroughttofocusonthescreenSbyaconverginglens(the lens is shown as a vertical arrow)

Thefocal length(distance fromlens toscreen) is25cmand the diameter of the lens is 30 cm The wavelength of thelightfromthestaris560nmCalculatethediameterofthe diameter of the image on the screen

Solution

The lens actually acts as a circular aperture of diameter 30 cm The half angular width of central maximum of the diffraction pattern that it forms on the screen is

75

2

122 122 56 10 23 1030 10b

λθminus

minusminus

times times= = = times

times rad

The diameter of the central maxima is therefore

25 times 10-2 times 23 times 10-5

= 57 times 10-6 m

Although this is small it is still finite and is effectively the image of the star as the intensity of the secondary maxima are small compared to that of the central maximum

The resolution of simple monochromatic two-source systemsThe astronomers tell us that many of the stars that we observe with the naked eye are in fact binary stars That iswhatwe see as a single star actually consistsof twostarsinorbitaboutacommoncentreFurthermoretheastronomerstellusthatifweusealdquogoodrdquotelescopethenwewillactuallyseethetwostarsThatiswewillresolve the single point source into its two component parts Sowhatisitthatdetermineswhetherornotweseethetwo stars as a single point source ie what determines whether or not two sources can be resolved It canrsquot just bethatthetelescopemagnifiesthestarssinceiftheyareacting as point sources magnifying them is not going to make a great deal of difference

Ineachofoureyesthereisanaperturethepupilthroughwhich the light enters This light is then focused by the eye lens onto the retina But we have seen that when light passes through an aperture it is diffracted and so when we look at

NATURE OF SCIENCEImproved technology The Rayleigh criterion is the limit of resolution Continuing advancement in technology such as large diameter dishes or lenses or the use of smaller wavelength lasers pushes the limits of what we can resolve (18)

copy IBO 2014

PHYSICS 2015indb 299 150514 337 PM

Chapter 9

300

AH

L

apointsourceadiffractionpatternwillbeformedontheretina If we look at two point sources then two diffraction patterns will be formed on the retina and these patterns will overlap The width of our pupil and the wavelength of the light emitted by the sources will determine the amount by which they overlap But the degree of overlap will also depend on the angular separation of the two point sources WecanseethisfromFigures923

pupilretina

S1

S2

P2

P1

θ

Figure 923

LightfromthesourceS1 enters the eye and is diffracted by the pupil such that the central maximum of the diffraction patternisformedontheretinaatP1SimilarlylightfromS2producesamaximumatP2 If the two central maxima are well separated then there is a fair chance that we will see the two sources as separate sources If they overlap then we will not be able to distinguish one source from another From the diagram we see as the sources are movedclosertotheeyethentheangleθ increases and so does the separation of the central maxima

Figures924925926and927showsthedifferentdiffractionpatternsandtheintensitydistributionthatmightresultonthe retina as a result of light from two point sources

S1 S21 Well resolved

Figure 924 Very well resolved

S1 S22 Well resolved

Figure 925 Well resolved

S1 S23 Just resolved Rayleigh criterion Minimum of S2 coincides with maximum peak of S1

Figure 926 Just resolved

4 Not resolved

Figure 927 Not resolved

We have suggested that if the central maxima of the two diffraction patterns are reasonably separated then we should be able to resolve two point sources In the late 19th century Lord Rayleigh suggested by how much they should be separated in order for the two sources to be just resolved Ifthecentralmaximumofonediffractionpattern coincides with the firstminima of the otherdiffraction pattern then the two sources will just beresolved This is known as the Rayleigh Criterion

Figure926showsjustthissituationThetwosourcesarejust resolved according to the Rayleigh criterion since the peak of the central maximum of one diffraction pattern coincideswiththefirstminimumoftheotherdiffractionpattern This means that the angular separation of the peaks of the two central maxima formed by each source is just the half angular width of one central maximum ie

θ λb---

=

where b is the width of the slit through which the light fromthesourcespassesHoweverweseefromFigure1117that θ is the angle that the two sources subtend at the slit Henceweconcludethattwosourceswillberesolvedbyaslit if the angle that they subtend at the slit is greater than or equaltoλfraslb

PHYSICS 2015indb 300 150514 337 PM

Wave Phenomena

301

AH

L

Sofarwehavebeenassumingthattheeyeisarectangularslit whereas clearly it is a circular aperture and so we must use the formula

θ 122 λb--------------=

As mentioned above the angle θ is sometimes called the resolving power but should more accurately be called the minimum angle of resolution (θmin)

Clearly the smaller θ the greater will be the resolving power

The diffraction grating is a useful for differentiating closely spacedlinesinemissionspectraLikeaprismspectrometerit can disperse a spectrum into its components but it is better suited because it has higher resolution than the prism spectrometer

If λ1 and λ2 are two nearly equal wavelengths that canbarelybedistinguishedtheresolvanceorresolvingpowerofthegratingisdefinedas

R = λ _______ (λ2 ndash λ 1) = λ ___ Δλ

Where λ asymp λ1 asymp λ2

Therefore thediffractiongratingwithahighresolvancewill be better suited in determining small differences in wavelength

If the diffraction grating has Nlinesbeingilluminateditcan be shown that the resolving power of the mth order diffraction is given by

R = mN

Obviouslytheresolvancebecomesgreaterwiththeordernumber m and with a greater number of illuminated slits

Example

A benchmark for the resolving power of a grating is the sodium doublet in the sodium emission spectrum Two yellow lines in itsspectrumhavewavelengthsof58900nmand58959nm

(a) Determinetheresolvanceofagratingifthegivenwavelengths are to be distinguished from each other

(b) Howmanylinesinthegratingmustbeilluminated in order to resolve these lines in the second order spectrum

Solution

(a) R = λ ___ Δλ = 58900nm __________________ (58900ndash58959)nm

= 589nm _______ 059nm = 10 times 103

(b) N = R __ m = 10 times 10 3 ________ 2 =50times102 lines

It has been seen that diffraction effectively limits the resolving power of optical systems This includes such systems as the eye telescopes and microscopes Theresolving power of these systems is dealt with in the next section This section looks at links between technology and resolving power when looking at the very distant and when looking at the very small

Radio telescopes

The average diameter of the pupil of the human eye is about 25 mm This means that the eye will justresolve two point sources emitting light of wavelength 500 nm if their angular separation at the eye is 7

43

50 10122 24 1025 10

θminus

minusminus

times= times = times

times rad

If the eye were to be able to detect radio waves of wavelength 015m then tohave the same resolvingpower thepupilwould have to have a diameter of about 600 m Clearly this is nonsense but it does illustrate a problem facingastronomers who wish to view very distant objects such as quasarsandgalaxies thatemitradiowavesConventionalradio telescopes consist of a large dish typically 25 min diameter Even with such a large diameter the radiowavelength resolving power of the telescope is much less thantheopticalresolvingpowerofthehumaneyeLetuslook at an example

Example

The Galaxy Cygnus A can be resolved optically as an elliptically shaped galaxy However it is also a strongemitterof radiowavesofwavelength015mTheGalaxyisestimatedtobe50times1024mfromEarthUseofaradiotelescope shows that the radio emission is from two sources separated by a distance of 30 times 1021m Estimate thediameterofthedishrequiredtojustresolvethesources

Solution

The angle θ that the sources resolve at the telescope is given by

θ 21

424

30 10 60 1050 10

θ minustimes= = times

times rad

and 4

122 122 015 300060 10

d λθ minus

times= = =

timesm = 30 km

A radio telescope dish of this size would be impossible tomake letalonesupportThisshowsthatasingledishtype radio telescope cannot be used to resolve the sources and yet they were resolved To get round the problemastronomers use two radio telescopes separated by a large

PHYSICS 2015indb 301 150514 337 PM

Chapter 9

302

AH

L

distance The telescopes view the same objects at the same time and the signals that each receive from the objects are simultaneously superimposed The result of the superposition of the two signals is a two-slit interference pattern The pattern has much narrower fringe spacing than that of the diffraction pattern produced by either telescope on its own hence producing a much higherresolvingpowerWhentelescopesareusedlikethistheyare called a stellar interferometer

InSocorroinNewMexicothereisastellarinterferometerthat consists of 27 parabolic dishes each of diameter25 m arranged in a Y-shape that covers an area of 570km2Thisisaso-calledVeryLargeArray(VLA)Evenhigher resolution can be obtained by using an array of radio telescopes in observatories thousands of kilometres apart A system that uses this so-called technique oflsquovery-long-baseline interferometryrsquo (VLBI) is known asalsquovery-long-baselinearrayrsquo(VLBA)WithVLBAaradiowavelength resolving power can be achieved that is 100 timesgreaterthanthebestopticaltelescopesEvenhigherresolving power can be achieved by using a telescope that isinasatelliteorbitingEarthJapanrsquosInstituteofSpaceandAstronautical Science (ISAS) launched sucha system inFebruary 1997TheNationalAstronomicalObservatoryof Japan the National Science Foundationrsquos NationalRadio Astronomy Observatory (NRAO) the CanadianSpaceAgency theAustraliaTelescopeNationalFacilitythe European VLBI Network and the Joint Instituteback this project forVeryLongBaseline Interferometryin EuropeThis project is a very good example of howInternationalismcanoperateinPhysics

Electron microscope

Telescopes are used to look at very distant objects that are very large but because of their distance fromus appearverysmallMicroscopesontheotherhandareusedtolookat objects that are close to us but are physically very small Aswehave seen justmagnifyingobjects that ismakingthemappearlargerisnotsufficientonitsowntogaindetailabouttheobjectfordetailhighresolutionisneeded

Figure928isaschematicofhowanoptical microscope is usedtoviewanobjectandFigure929isaschematicofatransmission electron microscope(TEM)

Optical lens system

bright light source

glass slide containing specimen (object)

eye

Figure 928 The principle of a light microscope

magnetic lens system

electron source

thin wafer of material

ScreenCCD

Figure 929 The principle of an electron microscope

In the optical microscope the resolving power isdetermined by the particular lens system employed and the wavelength λ of the light used For example twopoints in the sample separated by a distance d will just be resolved if

2d mλ=

where m is a property of the lens system know as the numerical aperture In practice the largest value of m obtainable is about 16 Hence if the microscope slideis illuminatedwith light ofwavelength 480 nm a goodmicroscope will resolve two points separated by a distance dasymp15times10-7masymp015micromPointscloser together thanthiswillnotberesolvedHoweverthisisgoodenoughtodistinguishsomevirusessuchastheEbolavirus

Clearlythesmallerλ the higher the resolving power and this is where the electron microscope comes to the fore The electron microscope makes use of the wave nature of electrons(see1315)IntheTEMelectronspassthrougha wafer thin sample and are then focused by a magnetic field onto a fluorescent screen orCCD (charge coupleddevicesee142)Electronsusedinanelectronmicroscopehavewavelengthstypicallyofabout5times10-12mHoweverthe numerical aperture of electron microscopes is considerablysmaller thanthatofanopticalmicroscopetypicallyabout002NonethelessthismeansthataTEMcanresolvetwopointsthatareabout025nmapartThisresolving power is certainly high enough to make out the shape of large molecules

Another type of electron microscope uses a techniqueby which electrons are scattered from the surface of the sample The scattered electrons and then focused as in theTEM to form an image of the surfaceThese so-calledscanningelectronmicroscopes(SEM)havealowerresolving power than TEMrsquos but give very good threedimensional images

The eye

We saw in the last section that the resolving power of the human eye is about 2 times 10-4 rad Suppose that youare looking at car headlights on a dark night and the car

PHYSICS 2015indb 302 150514 337 PM

Wave Phenomena

303

AH

L

is a distance D away If the separation of the headlight is say 15 m then the headlights will subtend an angle θ 15

Dθ =

at your eye Assuming an average wavelength of

500 nm your eye will resolve the headlights into twoseparatesourcesifthisangleequals2times10-4 rad This gives D=75kmInotherwordsifthecarisapproachingyouona straight road then you will be able to distinguish the two headlightsasseparatesourceswhenthecaris75kmawayfrom you Actually because of the structure of the retina and optical defects the resolving power of the average eye is about 6 times 10-4 rad This means that the car is more likely tobe25kmawaybeforeyouresolvetheheadlights

Astronomical telescope

Letusreturntotheexampleofthebinarystarsdiscussedatthe beginning of this section on resolution The stars Kruger A and B form a binary system The average separation of the stars is 14 times 1012mandtheiraveragedistancefromEarthis12times1017mWhenviewedthroughatelescopeonEarththe system will therefore subtend an angle

12

17

14 1012 10

θtimes

=times

= 12times10-5 rad at the objective lens of the telescope Assuming that the average wavelength of the light emitted bythestarsis500nmthenifthetelescopeistoresolvethesystem into two separate images it must have a minimum diameter Dwhere12times10-5 =

7122 500 10D

minustimes times

This gives D=0050m which is about 5 cm So thisparticular system is easily resolved with a small astronomical telescope

Exercise 961 ItissuggestedthatusingtheISASVLBAitwould

bepossibletoldquoseerdquoagrainofriceatadistanceof5000kmEstimatetheresolvingpoweroftheVLBA

2 Thedistancefromtheeyelenstotheretinais20mm The light receptors in the central part of the retinaareabout5times10-6apartDeterminewhetherthe spacing of the receptors will allow for the eye to resolve the headlights in the above discussion whentheyare25kmfromtheeye

3 ThediameterofPlutois23times106 m and its averagedistancefromEarthis60times1012 m EstimatetheminimumdiameteroftheobjectiveofatelescopethatwillenablePlutotobeseenasadisc as opposed to a point source

95 Doppler effect

Essential ideaTheDopplereffectdescribesthephenomenonofwavelengthfrequencyshiftwhenrelativemotionoccurs

Understandingsbull TheDopplereffectforsoundwavesandlightwaves

The Doppler effect for sound waves and light waves

Consider two observers A and B at rest with respect to asoundsourcethatemitsasoundofconstantfrequencyf Clearly both observers will hear a sound of the same frequencyHoweversupposethatthesourcenowmovesat constant speed towards A A will now hear a sound of frequencyfA that is greater than f and B will hear a sound of frequency fB that is less than f This phenomenon is known as the Doppler Effect orDopplerPrinciple after C J Doppler(1803-1853)

The same effect arises for an observer who is either moving towards or away from a stationary source

Figure 930 shows the waves spreading out from a stationary source that emits a sound of constant frequency f The observersAandBhearasoundofthesamefrequency

A B

wavefront

S

Figure 930 Sound waves from a stationary source

Suppose now that the source moves towards A withconstant speed vFigure9310(a)showsasnapshotofthenew wave pattern

NATURE OF SCIENCETechnology Although originally based on physical observations of the pitch of fast moving sources of sound the Doppler effect has an important role in many different areas such as evidence for the expansion of the universe and generating images used in weather reports and in medicine (55)

copy IBO 2014

PHYSICS 2015indb 303 150514 337 PM

Chapter 9

304

AH

L

sourceA B

smallerwavelength

largerwavelength

Figure 931 (a) Sound waves from a moving source

The wavefronts are now crowded together in the direction of travel of the source and stretched out in the opposite direction This is why now the two observers will now hear notesofdifferentfrequenciesHowmuchthewavesbunchtogether and how much they stretch out will depend on the speed vEssentially A

A

cfλ

= and BB

cfλ

= where λA lt λB and v is the speed of sound

IfthesourceisstationaryandAismovingtowardsitthenthe waves from the source incident on A will be bunched up If A is moving away from the stationary source then the waves from the source incident on A will be stretched out

Christian Doppler(1803ndash1853)actuallyappliedtheprinciple(incorrectlyasithappens)totryandexplainthecolourofstars However the Doppler effect does apply to light aswellastosoundIfalightsourceemitsalightoffrequencyf then if it is moving away from an observer the observer willmeasurethelightemittedashavingalowerfrequencythan fSincethesensationofcolourvisionisrelatedtothefrequencyoflight(bluelightisofahigherfrequencythanred light) light emitted by objects moving way from anobserver is often referred to as being red-shifted whereas if the object is moving toward the observer it is referred to as blue-shiftedThisideaisusedinOptionE(Chapter16)

We do not need to consider here the situations where either the source or the observer is accelerating In a situation for example where an accelerating source is approaching astationaryobserverthentheobserverwillhearasoundof ever increasing frequency This sometimes leads toconfusion in describing what is heard when a source approaches passes and then recedes from a stationaryobserver Suppose for example that you are standingon a station platform and a train sounding its whistle is approaching at constant speed What will you hear as the train approaches and then passes through the station As the train approaches you will hear a sound of constant pitch but increasing loudness The pitch of the sound will be greater than if the train were stationary As the train passes through the station you will hear the pitch change at the momentthetrainpassesyoutoasoundagainofconstantpitch The pitch of this sound will be lower than the sound of the approaching train and its intensity will decrease as the train recedes from you What you do not hear is a sound of increasing pitch and then decreasing pitch

The Doppler equations for sound

Although you will not be expected in an IB examination to derivetheequationsassociatedwithaspectsoftheDopplereffectyouwillbeexpectedtoapplythemForcompletenesstherefore thederivationoftheequationsassociatedwiththeDopplereffectasoutlinedaboveisgivenhere

S primeS O

vs

Figure 931 (b)

InFigure931(b)theobserverOisatrestwithrespecttoasourceofsoundSismovingwithconstanspeedvs directly towards O The source is emitting a note of constant frequencyf and the speed of the emitted sound is v

S shows the position of the source ∆t later When the sourceisatresttheninatime∆t the observer will receive f∆t waves and these waves will occupy a distance v∆t ie

v f

λ =f∆tv∆t =

(Because of the motion of the source this number of waves will now occupy a distance (v∆t ndash vs∆t)The lsquonewrsquowavelength is therefore

v ndash vs

fλrsquo =

f∆tv∆t ndash vs∆t

=

If fisthefrequencyheardbyOthen

vfλ

= or

vf

λ = = sv vfminus

From which

s

vf fv v

=minus

Dividingthroughbyv gives

Equation1111

1 sf f v

v=

minus

If the source is moving away from the observer then we have

s

vf fv v

=minus

1

1 sf vv

=+

Equation112

PHYSICS 2015indb 304 150514 337 PM

Wave Phenomena

305

AH

L

We now consider the case where the source is stationary and the observerismoving towardsthesource with speed v0 In this situation the speed of the sound waves as measured by the observer will be v0 + v We therefore have that

v0 + v = ffraslλ = vfλ

times

From which

o1 vf fv

= +

Equation113

If the observerismovingawayfromthesource then

o1 vf fv

= minus

Fromequation(113)wehavethat

o o1 v vf f f f f fv v

∆ = minus = + minus =

Equation114

Thevelocitiesthatwerefertointheaboveequationsarethevelocities with respect to the medium in which the waves fromthesourcetravelHoweverwhenwearedealingwithalight source it is the relative velocity between the source and the observer that we must consider The reason for this is that lightisuniqueintherespectthatthespeedofthelightwavesdoes not depend on the speed of the source All observers irrespective of their speed or the speed of the source will measure the same velocity for the speed of light This is oneof thecornerstonesof theSpecialTheoryofRelativitywhichisdiscussedinmoredetailinOptionH(Chapter18)When applying the Doppler effect to light we aremainlyconcerned with the motion of the source We look here only at the situation where the speed of the source v is much smaller than the speed of light cinfreespace(v ltlt c)Underthesecircumstanceswhenthesourceismovingtowardstheobserverequation111becomes

vf f f fc

minus = ∆ = Equation115

andwhenthesourceismovingawayfromtheobserverequation112becomes vf f f f

cminus = ∆ = minus

Provided that v ltlt c these same equations apply for astationary source and moving observer

We look at the following example and exercise

Example

Asourceemitsasoundoffrequency440HzItmovesinastraight line towards a stationary observer with a speed of 30ms-1Theobserverhearsasoundoffrequency484HzCalculate the speed of sound in air

Solution

We use equation 111 and substitute f= 484 Hz f = 440 Hz

and vs = 30 m s-1

therefore 1

484 440 301v

=minus

such that 30 4401

484vminus = to

give v = 330 m s-1

Example

A particular radio signal from a galaxy is measured as havingafrequencyof139times109HzThesamesignalfromasourceinalaboratoryhasafrequencyof142times109Hz

Suggestwhy thegalaxy ismovingaway fromEarthandcalculateitsrecessionspeed(iethespeedwithwhichitismovingawayfromEarth)

Solution

The fact that the frequency from the moving source is less than that when it is stationary indicates that it is moving away from the stationary observer ie Earth

Using vf fc

∆ = we have

8 96

9

3 10 (142 139) 10 634 10142 10

c fvf∆ times times minus times

= = = timestimes

m s-1

It is usual when dealing with the Doppler effect of light to express speeds as a fraction of c So in this instance we have v = 0021 c

Using the Doppler effectWe have seen in the above example and exercise how the Dopplereffectmaybeusedtomeasuretherecessionspeedof distant galaxies The effect is also used to measure speed inothersituationsHerewewilllookatthegeneralprincipleinvolved in using the Doppler effect to measure speedFigure932showsasource(thetransmitter)thatemitseithersound or em waves of constant frequency f The waves

PHYSICS 2015indb 305 150514 337 PM

Chapter 9

306

AH

L

fromthesourceare incidentonareflectorthat ismovingtowards the transmitter with speed vThereflectedwavesare detected by the receiver placed alongside the transmitter

reector

v

transmitter

receiver

f f

f f

Figure 932 Using the Doppler effect to measure speed

We shall consider the situation where v ltlt c where c is the speed of the waves from the transmitter

For the reflector receiving waves from the transmitterit is effectively an observer moving towards a stationary sourceFromequation(114) itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation116

For the receiver receivingwaves from the reflector it iseffectively a stationary observer receiving waves from a movingsourceFromequation(115)itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation117

Ifweaddequations(116)and(117)wegetthatthetotalDopplershiftatthereceiver∆f is

v vf f f f fc c

minus = ∆ = +

But 1 vf fc

= +

hence

1 v v vf f fc c c

∆ = + +

But since v ltlt cwe can ignore the term2

2

vc when we

expandthebracketintheaboveequation

Therefore we have2vf fc

∆ = Equation118

If v asymp c then we must use the full Doppler equationsHowever for em radiationwewill alwaysonly considersituations in which v ltlt c

Example

Thespeedofsoundinbloodis1500times103 m s-1 Ultrasound offrequency100MHzisreflectedfrombloodflowinginan arteryThe frequencyof the reflectedwaves receivedbackatthetransmitteris105MHzEstimatethespeedofthebloodflowintheartery

Solution

Using equation (118) we have

6 63

2005 10 1015 10

vtimes = times

timesto give v asymp 36 m s-1 (We have assumed that the ultrasound is incident at right angles to the blood flow)

Exercise 971 Judy is standing on the platform of a station A

high speed train is approaching the station in a straight line at constant speed and is sounding its whistleAsthetrainpassesbyJudythefrequencyof the sound emitted by the whistle as heard by Judychangesfrom640Hzto430HzDetermine

(a) thespeedofthetrain

(b) thefrequencyofthesoundemittedbythewhistle as heard by a person on the train (Speedofsound=330ms-1)

2 AgalaxyismovingawayfromEarthwithaspeedof00500cThewavelengthofaparticularspectral line in light emitted by atomic hydrogen inalaboratoryis656times10-7 m Calculate the valueofthewavelengthofthislinemeasuredinthelaboratoryinlightemittedfromasourceofatomic hydrogen in the galaxy

PHYSICS 2015indb 306 150514 337 PM

10 Fields

Contents

101 ndash Describing fields

102 ndash Fields at work

Essential Ideas

Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Similar approaches can be taken in analysing electrical and gravitational potential problems copy IBO 2014

PHYSICS 2015indb 307 150514 337 PM

Chapter 10

308

AH

L

101 Describing fields

Essential idea Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Understandingsbull Gravitationalfieldsbull Electrostaticfieldsbull Electricpotentialandgravitationalpotentialbull Fieldlinesbull Equipotentialsurfaces

Gravitational potentialWehaveseenthatifweliftanobjectofmassm to a height h abovethesurfaceoftheEarththenitsgainingravitationalpotential energy is mghHoweverthisisbynomeansthefullstoryForastartwenowknowthatgvarieswithh and also the expression really gives a difference in potentialenergybetweenthevaluethattheobjecthasattheEarthrsquossurfaceandthevaluethatithasatheighthSowhatwereallyneedisazeropointCanwefindapointwherethepotentialenergyiszeroandusethispointfromwhichtomeasure changes in potential energy

The point that is chosen is in fact infinity At infinity the gravitationalfieldstrengthofanyobjectwillbezeroSoletusseeifwecandeduceanexpressionforthegaininpotentialenergyofanobjectwhenitisldquoliftedrdquofromthesurfaceoftheEarthtoinfinityThisineffectmeansfindingtheworknecessarytoperformthistask

δr

r r δr+

r infin=

mg

AB

Me

Figure 1001 Gravitational forces

Inthediagramweconsidertheworknecessarytomovethe particle of mass m a distance δr in thegravitationalfield of the Earth

The force on the particle at A is FG Mem

r2----------------=

IftheparticleismovedtoB then since δr isverysmallwe can assume that the field remains constant over thedistance ABTheworkδW done againstthegravitationalfieldoftheEarthinmovingthedistanceABis

δWGMem

r2----------------δrndash=

(rememberthatworkdoneagainstaforceisnegative)

TofindthetotalworkdoneW in going from the surface oftheEarthtoinfinitywehavetoaddalltheselittlebitsofworkThisisdonemathematicallybyusingintegralcalculus

WG Me m

r2----------------ndash

rdR

infin

int G Me m 1

r2----- rd

R

infin

intndash G Me m 1r---ndash

R

infinndash= = =

GMe m 0 1R---ndash

ndashndash=

G MemR----------------ndash=

HencewehavewhereRistheradiusoftheEarththattheworkdonebythegravitationalfieldinmovinganobjectofmass m from R(surfaceoftheEarth)toinfinityisgivenby

WGMe m

R---------------- ndash=

We can generalise the result by calculating the worknecessary per unit mass to take a small mass fromthe surface of the Earth to infinity This we call thegravitational potential Vie

Wewouldgetexactlythesameresultifwecalculatedtheworkdonebythefieldtobringthepointmassfrominfinityto the surface of Earth In this respect the formal definition ofgravitationalpotentialatapointinagravitationalfieldis therefore defined as the work done per unit mass in bringing a point mass from infinity to that point

Clearly then the gravitational potential at any point inthe Earthrsquos field distance r from the centre of the Earth (providingr gt R)is

VGMe

r------------ ndash=

Thepotentialisthereforeameasureoftheamountofworkthathastobedonetomoveparticlesbetweenpointsinagravitationalfieldanditsunit is theJkgndash1 We also note thatthepotentialisnegativesothatthepotentialenergy

NATURE OF SCIENCEParadigm shift The move from direct observable actions being responsible for influence on an object to acceptance of a fieldrsquos ldquoaction at a distancerdquo required a paradigm shift in the world of science (23)

copy IBO 2014

PHYSICS 2015indb 308 150514 337 PM

Fields

309

AH

L

aswemoveawayfromtheEarthrsquossurfaceincreasesuntilitreachesthevalueofzeroatinfinity

IfthegravitationalfieldisduetoapointmassofmassmthenwehavethesameexpressionasaboveexceptthatMe is replaced by m andmustalsoexclude thevalueof thepotential at the point mass itself ie at r = 0

WecanexpressthegravitationalpotentialduetotheEarth(orduetoanysphericalmass)intermsofthegravitationalfield strength at its surface

AtthesurfaceoftheEarthwehave

g 0ndash Re GMRe

---------ndash= e

Sothat

g 0Re2 GM= e

Hence at a distance r from the centre of the Earth the gravitationalpotentialVcanbewrittenas

VGMe

r------------ndashg 0 Re

2

r------------ndash= =

The potential at the surface of the Earth

(r = Re)istherefore-g0Re

It is interesting to see how the expression for thegravitational potential ties in with the expression mgh ThepotentialatthesurfaceoftheEarthis-g0Re (see the exampleabove)andataheighthwillbe-g0g 0 Re h+( )ndash ifweassume that g0doesnotchangeover thedistanceh The differenceinpotentialbetweenthesurfaceandtheheighth is therefore g0hSotheworkneededtoraiseanobjectof mass m to a height h is mgh ie m timesdifference ingravitationalpotential

This we have referred to as the gain in gravitationalpotentialenergy(see235)

However this expression can be extended to any twopoints inanygravitationalfieldsuch that ifanobjectofmass mmovesbetween twopointswhosepotentialsareV1 and V2 respectively then the change in gravitationalpotentialenergyoftheobjectism(V1 ndash V2)

Gravitational potential gradient

Let us consider now a region in space where thegravitationalfieldisconstantInFigure912thetwopointsAandBareseparatedbythedistance∆x

A B

direction of uniform gravitational eld of strength I

∆x

Figure 1002 The gravitational potential gradient

The gravitational field is of strength I and is in the direction shownThe gravitational potential at A is V andatBisV +∆V

Theworkdone is takingapointmassmfromAtoB isF∆x = mI∆x

Howeverbydefinitionthisworkisalsoequalto-m∆V

Therefore mI∆x=-m∆V

or VIx

∆= minus

∆Effectivelythissaysthatthemagnitudeofthegravitationalfield strength is equal to the negative gradient of thepotential If I is constant then V is a linear function of x and Iisequaltothenegativegradientofthestraightlinegraph formed by plotting V against x If I is not constant (asusuallythecase)thenthemagnitudeofI at any point in the field can be found by find the gradient of the V-x graph at that point An example of such a calculation can befoundinSection929

For those of you who do HL maths the relationship between field and potential is seen to follow from the expression for the potential of a point mass viz

mV Gr

= minus

2

ddV mG Ir r

minus = + =

Potential due to one or more point masses

Gravitationalpotentialisascalarquantitysocalculatingthepotential due to more than one point mass is a matter of simpleadditionSoforexamplethepotentialV due to the Moon and Earth and a distance x fromthecentreofEarthonastraightlinebetweenthemisgivenbytheexpression

E MM MV Gx r x

= minus + minus

whereME=massofEarthMM= mass of Moon and r = distancebetweencentreofEarthandMoon

PHYSICS 2015indb 309 150514 337 PM

Chapter 10

310

AH

L

Equipotentials and field linesIfthegravitationalpotentialhasthesamevalueatallpointson a surface the surface is said to be an equipotential surfaceSoforexampleifweimagineasphericalshellaboutEarthwhosecentrecoincideswiththecentreofEarththisshellwillbeanequipotentialsurfaceClearlyifwerepresentthegravitationalfieldstrengthbyfieldlinessincethelinesldquoradiaterdquooutfromthecentreofEarththentheselineswillbeatrightanglestothesurfaceIfthefieldlineswerenotnormaltotheequipotentialsurfacethentherewouldbeacomponentof the field parallel to the surfaceThis wouldmean thatpointsonthesurfacewouldbeatdifferentpotentialsandsoitwouldnolongerbeanequipotentialsurfaceThisofcourseholdstrueforanyequipotentialsurface

Figure 1003 shows the field lines and equipotentials fortwopointmassesm

m m

Figure 1003 Equipotentials for two point masses

It is worth noting that we would get exactly the samepattern ifwewere to replace thepointmasseswith twoequalpointcharges(See935)

Escape speedThe potential at the surface of Earth is which meansthattheenergyrequiredtotakeaparticleofmass mfromthesurfacetoinfinityisequalto

ButwhatdoesitactuallymeantotakesomethingtoinfinityWhentheparticleisonthesurfaceoftheEarthwecanthinkofitassittingatthebottomofaldquopotentialwellrdquoasinfigure1004

particle

surface of Earth

innity

GMR---------

FIgure 1004 A potential well

Theldquodepthrdquoofthewellis and if the particle gains an amount of kinetic energy equal to wherem is its massthenitwillhavejustenoughenergytoldquoliftrdquoitoutofthewell

In reality it doesnrsquot actually go to infinity it justmeansthat the particle is effectively free of the gravitationalattractionof theEarthWe say that ithas ldquoescapedrdquo theEarthrsquosgravitationalpullWemeetthisideainconnectionwithmolecular forcesTwomolecules in a solidwill sitat their equilibrium position the separation where therepulsiveforceisequaltotheattractiveforceIfwesupplyjust enough energy to increase the separation of themolecules such that they are an infinite distance apart then intermolecularforcesnolongeraffectthemoleculesandthesolidwillhavebecomealiquidThereisnoincreaseinthekineticenergyofthemoleculesandsothesolidmeltsat constant temperature

Wecancalculatetheescapespeedofanobjectveryeasilyby equating the kinetic energy to the potential energysuch that

12---mvesc ape

2 GMe mRe

----------------=

vesc aperArr2GMe

Re--------------- 2g0 Re = =

Substituting for g0 and Regivesavalueforvescape of about 11kmsndash1 from the surface of the Earth

Youwillnotethattheescapespeeddoesnotdependonthemassoftheobjectsincebothkineticenergyandpotentialenergy are proportional to the mass

IntheoryifyouwanttogetarockettothemoonitcanbedonewithoutreachingtheescapespeedHoweverthiswouldnecessitate anenormousamountof fuel and it islikely that therocketplus fuelwouldbesoheavythat itwouldnevergetoffthegroundItismuchmorepracticalto accelerate the rocket to the escape speed fromEarthorbitandthenintheoryjustlaunchittotheMoon

Example

Use the followingdata todetermine thepotential at thesurface of Mars and the magnitude of the acceleration of free fall

massofMars =64times1023kg

radiusofMars =34times106 m

Determinealsothegravitationalfieldstrengthatadistanceof 68 times 106mabovethesurfaceofMars

PHYSICS 2015indb 310 150514 337 PM

Fields

311

AH

L

Solution

= minus = minus times times = times23

11 76

64 1067 10 ndash13 10

34 10MV GR

minus timestimes

Nkg-1

But V = -g0R

Therefore 7

0 6

13 10 3834 10

VgR

times= minus = =

timesm s-2

To determine the field strength gh at 68 times 106 m above the surface we use the fact that 0 2

Mg GR

= such that GM = g0R2

Therefore 2 2

0h 2 2 2

h h

38 (34) 042(102)

g RGMgR R

times= = = = m s-2

(the distance from the centre is 34 times 106 + 68 times 106 = 102 times 106 m)

Exercise 1011 Thegraphbelowshowshowthegravitational

potentialoutsideoftheEarthvarieswithdistancefrom the centre

0 10 20 30 40 50 60 70

ndash1

ndash2

ndash3

ndash4

ndash5

ndash6

ndash7

r m times 106

V

Jkgndash1

times 10

7

(a) Usethegraphtodeterminethegainingravitationalpotentialenergyofasatelliteofmass200kgasitmovesfromthesurfaceof the Earth to a height of 30 times 107mabovetheEarthrsquossurface

(b) Calculatetheenergyrequiredtotakeit to infinity

(c) DeterminetheslopeofthegraphatthesurfaceoftheEarthmCommenton youranswer

Electric potential energy

The concept of electric potential energy was developedwith gravitational potential energy in mind Just as anobjectnearthesurfaceoftheEarthhaspotentialenergybecause of its gravitational interaction with the Earthsotoothereiselectricalpotentialenergyassociatedwithinteracting charges

Letusfirstlookatacaseoftwopositivepointchargeseachof 1μC that are initially bound together by a thread in a vacuuminspacewithadistancebetweenthemof10cmasshowninFigure1005Whenthethreadiscutthepointchargesinitiallyatrestwouldmoveinoppositedirectionsmovingwithvelocitiesv1 and v2 along the direction of the electrostatic force of repulsion

BEFORE AFTER

v v21

Figure 1005 Interaction of two positive particles

The electric potential energybetweentwopointchargescan be found by simply adding up the energy associated witheachpairofpointchargesForapairofinteractingchargestheelectricpotentialenergyisgivenby

∆U = ∆Ep + ∆Ek = ∆W = ∆Fr = kqQ

____ r2 times r = kqQ

____ r

Becausenoexternalforceisactingonthesystemtheenergyand momentum must be conserved Initially Ek = 0 and Ep =kqQr=9times109 times 1 times 10-1201m=009JWhentheyareagreatdistance fromeachotherEpwillbenegligibleThefinalenergywillbeequaltofrac12mv1

2+frac12mv22=009J

Momentum isalsoconservedand thevelocitieswouldbethe same magnitude but in opposite directions

Electric potential energyismoreoftendefinedintermsofapointchargemovinginanelectricfieldas

lsquothe electricpotential energybetweenany twopoints inan electric field is defined as negativeoftheworkdone by anelectricfieldinmovingapoint electric chargebetweentwolocationsintheelectricfieldrsquo

ΔU = ΔEp = -ΔW = -Fd = qEx

ΔU = ΔEp +ΔEk = ΔW = ΔFr = kqQ r2 x r = kqQ

wherexisthedistancemovedalong(oroppositeto)thedirection of the electric field

Electric potential energyismeasuredinjoule(J)Justasworkisascalarquantitysotooelectricalpotentialenergyis a scalar quantity The negative of the work done by an electric field in moving a unit electric charge between two points is independent of the path taken Inphysicswesaytheelectricfieldisalsquoconservativersquo field

PHYSICS 2015indb 311 150514 337 PM

Chapter 10

312

AH

L

Supposeanexternalforcesuchasyourhandmovesasmallpositive point test charge in the direction of a uniformelectric fieldAs it ismoving itmust be gaining kineticenergyIfthisoccursthentheelectricpotentialenergyofthe unit charge is changing

InFigure1006apointcharge+qismovedbetweenpointsAandBthroughadistancex in a uniform electric field

B

A

x

+ q

Figure 1006 Movement of a positive point charge in a uniform field

InordertomoveapositivepointchargefrompointAtopointBanexternal forcemustbeapplied to thechargeequaltoqE (F = qE)

Since the force is applied through a distance xthennegativeworkhas to bedone tomove the charge because energyis gainedmeaning there is an increaseelectric potential energybetweenthetwopointsRememberthattheworkdoneisequivalenttotheenergygainedorlostinmovingthe charge through the electric field The concept of electric potential energy is only meaningful as the electric field whichgeneratestheforceinquestionisconservative

W F xtimes E q x times= =

θ x cos θ x

Figure 1007 Charge moved at an angle to the field

If a chargemoves at an angle θ to an electric field thecomponent of the displacement parallel to the electric fieldisusedasshowninFigure918

W F x Eq x θcostimes= =

Theelectricpotentialenergyisstoredintheelectricfieldand theelectricfieldwill return theenergy to thepointcharge when required so as not to violate the Law ofconservationofenergy

Electric potential

The electric potential at a point in an electric field is defined as being the work done per unit charge in bringing a small positive point charge from infinity to that point

ΔV = Vinfin ndash Vf = ndash W ___ q

Ifwedesignatethepotentialenergytobezeroatinfinitythen it follows thatelectricpotentialmustalsobezeroat infinity and the electric potential at any point in an electricfieldwillbe

ΔV = ndash W ___ q

NowsupposeweapplyanexternalforcetoasmallpositivetestchargeasitismovedtowardsanisolatedpositivechargeTheexternalforcemustdoworkonthepositivetestchargetomoveittowardstheisolatedpositivechargeandtheworkmustbepositivewhiletheworkdonebytheelectricfieldmustthereforebenegativeSotheelectricpotentialatthatpointmustbepositiveaccordingtotheaboveequationIfanegativeisolatedchargeisusedtheelectricpotentialatapointonthepositivetestchargewouldbenegativePositivepoint chargesof their ownaccordmove fromaplaceofhighelectricpotential toaplaceof lowelectricpotentialNegative point charges move the other way from lowpotentialtohighpotentialInmovingfrompointAtopointBinthediagramthepositivecharge+qismovingfromalowelectricpotentialtoahighelectricpotential

In thedefinitiongiven the termldquoworkperunitchargerdquohas significance If the test charge is +16 times 10-19Cwherethe charge has a potential energy of 32 times 10-17Jthenthepotentialwouldbe32times10-17J+16times10-19C=200JC-1 Now if the charge was doubled the potential wouldbecome 64times10-17 J However the potential per unitchargewouldbethesame

Electric potential is a scalar quantityandithasunitsJC-1 orvoltswhere1voltequalsone joulepercoloumbThevoltallowsustoadoptaunitfortheelectricfieldintermsofthevolt

PreviouslytheunitfortheelectricfieldwasNCndash1

W = qV and F = qEsoW ___ V = F __ E

E = FV ___ W = FV ___ Fm V mndash1

That is the units of the electric field E can also beexpressed as V mndash1

The work done per unit charge in moving a point charge between two points in an electric field is again independent of the path taken

PHYSICS 2015indb 312 150514 337 PM

Fields

313

AH

L

Electric potential due to a point charge

Letus takeapoint rmetres froma chargedobjectThepotentialatthispointcanbecalculatedusingthefollowing

W = ndashFr = ndashqV and F = ndash q1 q2 _____ 4πε0r

2

Therefore

W = ndash q1 q2 _____ 4πε0r

2 times r = ndash q1 q2 _____ 4πε0r

= ndashq1 times q2 _____ 4πε0r

= ndashq1V

That is V q

4πε0r--------------- =

Orsimply

V kqr------

=

Example

Determinehowmuchworkisdonebytheelectricfieldofpointcharge150μCwhenachargeof200μCismovedfrom infinity to apoint 0400m from thepoint charge(Assumenoaccelerationofthecharges)

Solution

The work done by the electric field is W = -qV = -14πε0 times q times (Q rinfin - Q r0400)

W = (- 200 times 10-6 C times 900 times 109 NmC-2 times 150 times 10-6 C) divide 0400 m = - 0675 J

An external force would have to do +0675 J of work

Electric field strength and electric potential gradient

Letus lookbackatFigure1006Supposeagain that thecharge+qismovedasmalldistancebyaforceFfromAtoBsothattheforcecanbeconsideredconstantTheworkdoneisgivenby

∆W F x∆times=

TheforceFandtheelectricfieldEareoppositelydirectedandweknowthat

F = -qE and ΔW = q ΔV

Thereforetheworkdonecanbegivenas

q ΔV = -q E Δx

Therefore E V∆x∆-------

ndash=

The rate of change of potential ΔV at a point with respect to distance Δx in the direction in which the change is maximum is called the potential gradient We say that the electric field = - the potential gradient and the units are Vm-

1 From the equation we can see that in a graph of electric potential versus distance the gradient of the straight line equals the electric field strength

In reality if a charged particle enters a uniform electric field it will be accelerated uniformly by the field and its kinetic energy will increase This is why we had to assume no acceleration in the last worked example

Ek∆ 12---

mv2 q E xsdot sdot q Vx---

xsdot sdot q Vsdot= = = =

Example

Determine how far apart two parallel plates must besituated so that a potential difference of 150 times 102 V produces an electric field strength of 100 times 103 NC-1

Solution

Using E V∆x------- xhArrndash V∆

E-------15 102 Vtimes

100 103 N C 1ndashtimes------------------------------------------= = =

= 150 times 10-1

The plates are 150 times 10-1 m apart

The electric field and the electric potential at a point due to an evenly distributed charge +q on a sphere can be represented graphically as in Figure 1008

r

r

V

E

r0

r0

r0

Charge of +Q evenly distributed over surface

E 14πε0------------ Q

r2----- r r0gt=

V 14πε0------------Q

r----r r0gt=

Figure 1008 Electric field and potential due to a charged sphere

PHYSICS 2015indb 313 150514 337 PM

Chapter 10

314

AH

L

When the sphere becomes charged we know that thechargedistributes itselfevenlyoverthesurfaceThereforeeverypartofthematerialoftheconductorisatthesamepotential As the electric potential at a point is defined as beingnumericallyequaltotheworkdoneinbringingaunitpositivechargefrominfinitytothatpointithasaconstantvalueineverypartofthematerialoftheconductor

Since the potential is the same at all points on the conducting surfacethen∆V∆xiszeroButE=ndash∆V∆xThereforethe electric field inside the conductor is zero There is no electric field inside the conductor

SomefurtherobservationsofthegraphsinFigure1009are

bull OutsidethespherethegraphsobeytherelationshipsgivenasE α 1 r2 and V α 1 r

bull Atthesurfacer = r0Thereforetheelectricfieldandpotentialhavetheminimumvalueforr at this point and this infers a maximum field and potential

bull Insidethespheretheelectricfieldiszerobull Insidethespherenoworkisdonetomoveacharge

fromapointinsidetothesurfaceThereforethereisno potential difference and the potential is the same asitiswhenr = r0

Similargraphscanbedrawnfortheelectricfieldintensityandthe electric potential as a function of distance from conducting parallelplatesandsurfacesandthesearegiveninFigure1009

Potential plot E eld plotE eld

++++

ndashndashndashndash

xx x

+

x x

ndashx x

+ ndashx

x

+ +

x

x

Figure 1009 Electric field and electric potential at a distance from a charged surface

Potential due to one or more point charges

The potential due to one point charge can be determined byusingtheequationformula

V = kq r

Example 1

Determine the electric potential at a point 20 times 10-1 m fromthecentreofan isolatedconducting spherewithapointchargeof40pCinair

Solution

Using the formula

V = kq r we have

V90 10 9times( ) 40 10 12ndashtimes( )times

20 10 1ndashtimes( )------------------------------------------------------------------ 018 V= =

the potential at the point is 180 times 10-1 V

The potential due to a number of point charges can be determined by adding up the potentials due to individual point charges because the electric potential at any point outside a conducting sphere will be the same as if all the charge was concentrated at its centre

Example 2

Threepointchargesofareplacedattheverticesofaright-angledtriangleasshowninthediagrambelowDeterminetheabsolutepotentialat the+20μCchargeduetothetwoothercharges

- 6μC

+3μC +2μC

4 m

3 m

Solution

The electric potential of the +2 μC charge due to the ndash 6 μC charge is

V = (9 times 109 Nm2C-2 times -6 times 10-6 C) divide (radic 32 + 42) m = - 108 times 104 V

PHYSICS 2015indb 314 150514 338 PM

Fields

315

AH

L

Insummarywecanconcludethat

bull No work is done to move a charge along an equipotential

bull Equipotentials are always perpendicular to the electric lines of force

Figure1011and1012showsomeequipotential lines fortwo oppositely charged and identically positive spheresseparated by a distance

+ve ndashve

equipotential lines

Figure 1011 Equipotential lines between two opposite charges

equipotential lines

Figure 1012 Equipotential lines between two charges which are the same

+

ndash ndash ndash ndash

+ + +

40 V

30 V

20 V

10 V

50 V

Figure 1013 Equipotential lines between charged parallel plates

Figure1013showstheequipotentiallinesbetweenchargedparallel plates Throughout this chapter the similarities and differences between gravitational fields and electric fieldshavebeendiscussedTherelationshipsthatexistsbetweengravitationalandelectricquantitiesandtheeffectsofpointmassesandchargesissummarisedinTable1014

The electric potential of the +2 μC charge due to the +3 μC charge is

V = (9 times 109 Nm2C-2 times 3 times 10-6 C) divide 3m = 9 times 103 V

The net absolute potential is the sum of the 2 potentials- 108 times 104 V + 9 times 103 V =- 18 times 103 V

The absolute potential at the point is - 18 times 103 V

Equipotential surfacesRegions in space where the electric potential of a chargedistributionhasaconstantvaluearecalledequipotentials The placeswherethepotentialisconstantinthreedimensionsarecalled equipotential surfacesandwheretheyareconstantintwodimensionstheyarecalledequipotential lines

TheyareinsomewaysanalogoustothecontourlinesontopographicmapsInthiscasethegravitationalpotentialenergy is constant as amassmoves around the contourlines because the mass remains at the same elevationabovetheEarthrsquossurfaceThegravitationalfieldstrengthacts in a direction perpendicular to a contour line

Similarlybecausetheelectricpotentialonanequipotentiallinehas thesamevalueanelectric forcecandonoworkwhena testchargemovesonanequipotentialThereforethe electric field cannot have a component along anequipotentialandthusitmustbeeverywhereperpendiculartotheequipotentialsurfaceorequipotentiallineThisfactmakesiteasytoplotequipotentialsifthelinesofforceorlinesofelectricfluxofanelectricfieldareknown

Forexamplethereareaseriesofequipotentiallinesbetweentwoparallelplateconductorsthatareperpendiculartotheelectric fieldTherewill be a series of concentric circles(eachcirclefurtherapartthanthepreviousone)thatmapouttheequipotentialsaroundanisolatedpositivesphereThe lines of force and some equipotential lines for anisolatedpositivesphereareshowninFigure1010

Lines ofequipotential

Figure 1010 Equipotentials around an isolated positive sphere

PHYSICS 2015indb 315 150514 338 PM

Chapter 10

316

AH

L

Gravitational quantity Electrical quantityQ

uant

ities

g V∆x∆-------ndash= E V∆

x∆-------ndash=

Poin

t mas

ses a

nd ch

arge

s

V Gmr----

ndash= V 14πε0------------q

r---=

g G m

r2-----ndash= E 14πε0------------ q

r2-----=

F Gm1 m2

r2--------------= F 1

4πε0------------

q1q2

r2-----------=

Figure 1014 Formulas

Example

Deduce the electric potential on the surface of a gold nucleus that has a radius of 62 fm

Solution

Using the formula

V = kq r and knowing the atomic number of gold is 79 We will assume the nucleus is spherical and it behaves as if it were a point charge at its centre (relative to outside points)

V = 90 times 109 Nm2C-2 times 79 times 16 times 10-19 C divide 62 times 10-15 m = 18 times 107 V

The potential at the point is 18 MV

Example

Deduce the ionisation energy in electron-volts of theelectron in the hydrogen atom if the electron is in its ground state and it is in a circular orbit at a distance of 53 times 10-11 m from the proton

Solution

This problem is an energy coulombic circular motion question based on Bohrrsquos model of the atom (not the accepted quantum mechanics model) The ionisation energy is the energy required to remove the electron from the ground state to infinity The electron travels in a circular orbit and therefore has a centripetal acceleration The ionisation energy will counteract the coulombic force and the movement of the electron will be in the opposite direction to the centripetal force

Total energy = Ek electron + Ep due to the proton-electron interaction

ΣF = kqQ r2 = mv2 r and as such mv2 = = kqQ r

Therefore Ek electron = frac12 kqQ r

Ep due to the proton-electron interaction = - kqQ r

Total energy = frac12 kqQ r + - kqQ r = -frac12 kqQ r

= - 90 times 109 Nm2C-2 times (16 times 10-19 C)2 divide 53 times 10-11 m = -217 times 10-18 J

= -217 times 10-18 J divide 16 times 10-19 = -136 eV

The ionisation energy is 136 eV

Exercise 1021 ApointchargePisplacedmidwaybetweentwo

identicalnegativechargesWhichoneofthefollowingiscorrectwithregardstoelectricfieldandelectricpotentialatpointP

Electric field Electric potentialA non-zero zeroB zero non-zeroC non-zero non-zeroD zero zero

2 Twopositivechargedspheresaretiedtogetherinavacuumsomewhereinspacewheretherearenoexternal forces A has a mass of 25 g and a charge of20μCandBhasamassof15gandachargeof30μCThedistancebetweenthemis40cmTheyarethenreleasedasshowninthediagram

BEFORE AFTER

v1 v2

A B

(a) Determinetheirinitialelectricpotentialenergy in the before situation

(b) DeterminethespeedofsphereBafterrelease

3 Thediagrambelowrepresentstwoequipotentiallines in separated by a distance of 5 cm in a uniform electric field

+ + + + + + + +

ndash ndash ndash ndash ndash ndash ndash ndash

40 V

20 V 5 cm

PHYSICS 2015indb 316 150514 338 PM

Fields

317

AH

L

Determine the strength of the electric field

4 Thisquestionisabouttheelectricfieldduetoacharged sphere and the motion of electrons in that fieldThediagrambelowshowsanisolatedmetalsphereinavacuumthatcarriesanegativeelectriccharge of 60 μC

ndash

(a) Onthediagramdrawtheconventionalwayto represent the electric field pattern due to the charged sphere and lines to represent threeequipotentialsurfacesintheregionoutside the sphere

(b) Explainhowthelinesrepresentingtheequipotentialsurfacesthatyouhavesketchedindicatethatthestrengthoftheelectricfieldisdecreasingwithdistancefrom the centre of the sphere

(c) Theelectricfieldstrengthatthesurfaceofthe sphere and at points outside the sphere can be determined by assuming that the sphere acts as a point charge of magnitude 60 μC at its centre The radius of the sphere is 25 times 10ndash2 m Deduce that the magnitude of the field strength at the surface of the sphere is 86 times 107 Vmndash1

An electron is initially at rest on the surface of the sphere

(d) (i) Describethepathfollowedbytheelectronasitleavesthesurfaceofthesphere

(ii) Calculatetheinitialaccelerationofthe electron

5 Determinetheamountofworkthatisdoneinmovingachargeof100nCthroughapotentialdifferenceof150times102 V

6 Three identical 20 μC conducting spheres are placedatthecornersofanequilateraltriangleofsides 25 cm The triangle has one apex C pointing upthepageand2baseanglesAandBDeterminetheabsolutepotentialatB

7 Determinehowfaraparttwoparallelplatesmustbesituatedsothatapotentialdifferenceof250 times 102 V produces an electric field strength of 200 times 103 NC-1

8 Thegapbetweentwoparallelplatesis10times10-3mandthereisapotentialdifferenceof10times104 V betweentheplatesCalculate

i theworkdonebyanelectroninmovingfrom one plate to the other

ii thespeedwithwhichtheelectronreachesthe second plate if released from rest

iii theelectricfieldintensitybetweentheplates

9 Anelectronguninapicturetubeisacceleratedby a potential 25 times 103VDeterminethekineticenergygainedbytheelectroninelectron-volts

10 Determine the electric potential 20 times 10-2 m from achargeof-10times10-5 C

11 Determinetheelectricpotentialatapointmid-waybetweenachargeofndash20pCandanotherof+5pConthelinejoiningtheircentresifthecharges are 10 cm apart

12 During a thunderstorm the electric potential differencebetweenacloudandthegroundis10 times 109 V Determine the magnitude of the change in electric potential energy of an electron thatmovesbetweenthesepointsinelectron-volts

13 A charge of 15 μC is placed in a uniform electric fieldoftwooppositelychargedparallelplateswithamagnitudeof14times103 NC-1

(a) Determinetheworkthatmustbedoneagainstthefieldtomovethepointchargeadistance of 55 cm

(b) Calculatethepotentialdifferencebetweenthe final and initial positions of the charge

(c) Determinethepotentialdifferencebetweenthe plates if their separation distance is 15 cm

14 Duringaflashoflightningthepotentialdifferencebetweenacloudandthegroundwas12 times 109 V and the amount of transferred charge was32C

(a) Determinethechangeinenergyofthetransferred charge

(b) Iftheenergyreleasedwasallusedtoacceleratea1tonnecardeduceitsfinalspeed

(c) Iftheenergyreleasedcouldbeusedtomelticeat0degCdeducetheamountoficethatcould be melted

PHYSICS 2015indb 317 150514 338 PM

Chapter 10

318

AH

L

15 SupposethatwhenanelectronmovedfromAtoBin the diagram along an electric field line that the electric field does 36 times 10-19Jofworkonit

Determinethedifferencesinelectricpotential

(a) VB ndash VA(b) VC ndash VA(c) VC ndash VB

16 DeterminethepotentialatpointPthatislocatedatthecentreofthesquareasshowninthediagrambelow

- 6μC

+3μC +2μC

1m

1m

5μC

P

102 Fields at work

Essential idea Similar approaches can be taken inanalysingelectricalandgravitationalpotentialproblems

Understandingsbull Potentialandpotentialenergybull Potentialgradientbull Potentialdifferencebull Escapespeedbull Orbitalmotionorbitalspeedandorbitalenergybull Forcesandinverse-squarelawbehaviour

Orbital motion orbital speed and orbital energy

Although orbital motion may be circular elliptical orparabolic this sub-topic only dealswith circular orbitsThis sub-topic is not fundamentally new physics butan application that synthesizes ideas from gravitationcircularmotiondynamicsandenergy

TheMoonorbitstheEarthandinthissenseitisoftenreferredtoasasatelliteoftheEarthBefore1957itwastheonlyEarthsatelliteHoweverin1957theRussianslaunchedthefirstmanmadesatelliteSputnik1SincethisdatemanymoresatelliteshavebeenlaunchedandtherearenowliterallythousandsofthemorbitingtheEarthSomeareusedtomonitortheweathersome used to enable people to find accurately their position on thesurfaceoftheEarthmanyareusedincommunicationsandnodoubtsomeareusedtospyonothercountriesFigure1015showshowinprincipleasatellitecanbeputintoorbit

Theperson(whosesizeisgreatlyexaggeratedwithrespectto Earth) standing on the surface on the Earth throwssome stonesThe greater the speed with which a stoneis thrown the further it will land from her The pathsfollowedbythethrownstonesareparabolasByastretchof the imaginationwecanvisualisea situation inwhichastoneis thrownwithsuchaspeedthatbecauseof thecurvatureoftheEarthitwillnotlandonthesurfaceoftheEarthbutgointoldquoorbitrdquo(Path4onFigure1015)

NATURE OF SCIENCECommunication of scientific explanations The ability to apply field theory to the unobservable (charges) and the massively scaled (motion of satellites) required scientists to develop new ways to investigate analyse and report findings to a general public used to scientific discoveries based on tangible and discernible evidence (51)

copy IBO 2014

PHYSICS 2015indb 318 150514 338 PM

Fields

319

AH

L

1

2

34Earth

Figure 1015 Throwing a stone into orbit

TheforcethatcausesthestonestofollowaparabolicpathandtofalltoEarthisgravityandsimilarlytheforcethatkeeps the stone in orbit is gravity For circular motionto occur we have seen that a force must act at rightanglestothevelocityofanobjectthatistheremustbeacentripetalforceHenceinthesituationwedescribeherethe centripetal force for circular orbital motion about the EarthisprovidedbygravitationalattractionoftheEarth

We can calculate the speedwithwhich a stonemustbethrowninordertoputitintoorbitjustabovethesurfaceof the Earth

If the stone has mass m and speed v thenwehave fromNewtonrsquos2ndlaw

mv2

RE---------- G

MEm

RE2------------=

whereRE is the radius of the Earth and ME is the mass of the Earth

Bearinginmindthatg 0 GME

RE2--------= then

v g RE 10 64 106timestimes 8 103times= = =

Thatisthestonemustbethrownat8times103m sndash1

Clearly we are not going to get a satellite into orbit soclosetothesurfaceoftheEarthMovingatthisspeedthefrictionduetoairresistancewouldmeltthesatellitebeforeithadtravelledacoupleofkilometresInrealitythereforeasatelliteisputintoorbitabouttheEarthbysendingitattachedtoarocketbeyondtheEarthrsquosatmosphereandthengivingitacomponentofvelocityperpendiculartoaradialvectorfromtheEarthSeeFigure1016

Earth Satellite carried by rocket to hereSatellite orbit

Tangential component of velocity

Figure 1016 Getting a satellite into orbit

Keplerrsquos third law(This work of Kepler and Newtonrsquos synthesis of the work is an excellent example of the scientific method and makes for a good TOK discussion)

In 1627 Johannes Kepler (1571-1630)publishedhis lawsof planetary motion The laws are empirical in natureandwere deduced from the observations of theDanishastronomer Tycho de Brahe (1546-1601)The third lawgives a relationship between the radius of orbit R of a planet and its period TofrevolutionabouttheSunThelawisexpressedmathematicallyas

2

3 constantTR

=

We shall nowuseNewtonrsquos LawofGravitation to showhowitisthattheplanetsmoveinaccordancewithKeplerrsquosthirdlaw

InessenceNewtonwasabletousehislawofgravitytopredict the motion of the planets since all he had to do was factor theF givenby this law intohis second law F = ma to find their accelerations and hence their future positions

InFigure1017theEarthisshownorbitingtheSunandthedistancebetweentheircentresisR

RSun

Earth

F es

F se

Figure 1017 Planets move according to Keplerrsquos third law

Fes is the force that the Earth exerts on the Sun and Fse is the force that the Sun exerts on the Earth The forces are equalandoppositeandtheSunandtheEarthwillactuallyorbitaboutacommoncentreHoweversince theSun issoverymuchmoremassivethantheEarththiscommoncentrewillbeclosetothecentreoftheSunandsowecanregard the Earth as orbiting about the centre of the Sun Theotherthingthatweshallassumeisthatwecanignorethe forces that the other planets exert on the Earth (This wouldnotbeawisethingtodoifyouwereplanningtosend a space ship to theMoon for example)We shallalso assume that we have followed Newtonrsquos exampleandindeedprovedthataspherewillactasapointmasssituated at the centre of the sphere

Kepler had postulated that the orbits of the planets areellipticalbut since theeccentricityof theEarthrsquosorbit issmallweshallassumeacircularorbit

PHYSICS 2015indb 319 150514 338 PM

Chapter 10

320

AH

L

TheaccelerationoftheEarthtowardstheSunisa = Rω2

whereω ω 2πT------=

Hencea R 2π

T------ times

2 4π2 RT 2------------- = =

But the acceleration is given by Newtonrsquos Second Law F = mawhereFisnowgivenbytheLawofGravitationSo in this situation

F maGMs Me

R2-------------------= = butwealsohavethat

a a R 2πT

------ times

2 4π2 RT 2------------- = = and m = Me so that

G MsMe

R2------------------- Me4π2R

T2-------------times

GMs

4π2-----------rArr R3

T2------= =

Butthequantity

isaconstantthathasthesamevalueforeachoftheplanetssowehaveforalltheplanetsnotjustEarththat

R3

T2------ k =

wherekisaconstantWhichisofcourseKeplerrsquosthirdlaw

ThisisindeedanamazingbreakthroughItisdifficulttorefutethe idea that all particles attract each other in accordance withtheLawofGravitationwhenthelawisabletoaccountfortheobservedmotionoftheplanetsabouttheSun

The gravitational effects of the planets upon each othershould produce perturbations in their orbits Such is the predictivepoweroftheUniversalGravitationalLawthatit enabled physicists to compute these perturbations The telescopehadbeeninventedin1608andbythemiddleofthe 18th Century had reached a degree of perfection in design that enabled astronomers to actually measure the orbital perturbations of the planets Their measurements werealways inagreementwith thepredictionsmadebyNewtonrsquoslawHoweverin1781anewplanetUranuswasdiscovered and the orbit of this planet did not fit withthe orbit predicted by Universal Gravitation Such wasthe physicistrsquos faith in theNewtonianmethod that theysuspectedthatthediscrepancywasduetothepresenceofayetundetectedplanetUsingtheLawofGravitationtheFrenchastronomerJLeverrier and the English astronomer J C Adamswereable tocalculate justhowmassive thisnewplanetmustbeandalsowhereitshouldbeIn1846the planetNeptunewas discovered justwhere they hadpredicted Ina similarwaydiscrepancies in theorbitofNeptune ledto thepredictionandsubsequentdiscoveryin1930oftheplanetPlutoNewtonrsquosLawofGravitation

had passed the ultimate test of any theory it is not only abletoexplainexistingdatabutalsotomakepredictions

Satellite energyWhenasatelliteisinorbitaboutaplanetitwillhavebothkineticenergyandgravitationalpotentialenergySupposewe consider a satellite ofmass m that is in an orbit of radius r about a planet of mass M

Thegravitationalpotentialduetotheplanetatdistancer from its centre is

ThegravitationalpotentialenergyofthesatelliteVsat

is therefore GMe m

r----------------ndash

ThatisVsat GMem

r----------------ndash=

Thegravitationalfieldstrengthatthesurfaceoftheplanetisgivenby

g 0GMe

Re2------------=

Hencewecanwrite

Vsat g0 Re

2

rndash=

m

ThekineticenergyofthesatelliteKsatisequaltofrac12mv2wherev is its orbital speed

Byequatingthegravitationalforceactingonthesatellitetoitscentripetalaccelerationwehave

G Mem

r2---------------- mv2

r---------- mv2hArr

GMemr----------------= =

Fromwhich12---

mv2 12---

GMemr----------------times=

= 0 e

2g R

r

2m

Which is actually quite interesting since it shows thatirrespective of the orbital radius the KE is numericallyequal to half the PE Also the total energy E tot K sat Vsat+ 1

2---G Me m

r----------------times G Me m

r----------------ndash + 1

2---G Me m

r----------------ndash= = =of the satelliteisalwaysnegativesince

E tot K sat Vsat+ 12---

G Me mr----------------times

G Me mr----------------ndash

+ 12---

G Me mr----------------ndash= = =

PHYSICS 2015indb 320 150514 338 PM

Fields

321

AH

L

The energies of an orbiting satellite as a function of radial distancefromthecentreofaplanetareshownplottedinFigure1018

12

100806

0402

0ndash02

ndash04

ndash06ndash08ndash10

ndash12

2 4 6 8 10 12

ener

gya

rbita

ry u

nits

distance timesR

kinetic energytotal energy

potential energy

Figure 1018 Energy of an orbiting satellite as a function of distance from the centre of a planet

WeightlessnessSuppose that you are inanelevator(lift)thatisdescendingat constant speedandyou let goof abook thatyouareholdinginyourhandThebookwillfalltothefloorwithaccelerationequaltotheaccelerationduetogravityIfthecablethatsupportstheelevatorweretosnap(asituationthatItrustwillneverhappentoanyofyou)andyounowletgothebookthatyouareholdinginyourotherhandthis book will not fall to the floor - it will stay exactlyin linewithyourhandThis isbecause thebook isnowfallingwith the sameaccelerationas theelevatorandassuch the book cannot ldquocatchrdquo up with the floor of theelevatorFurthermoreifyouhappenedtobestandingonasetofbathroomscalesthescaleswouldnowreadzero-youwouldbeapparentlyweightlessItisthisideaoffreefallthatexplainstheapparentweightlessnessofastronautsin an orbiting satellite These astronauts are in free fall inthesensethattheyareacceleratingtowardsthecentre of the Earth

It is actually possible to define theweight of a body inseveral different waysWe can define it for example asthegravitationalforceexertedonthebodybyaspecifiedobject such as the EarthThiswe have seen thatwe doinlotsofsituationswherewedefinetheweightasbeingequaltomgIfweusethisdefinitionthenanobjectinfreefallcannotbydefinitionbeweightlesssinceitisstillinagravitationalfieldHoweverifwedefinetheweightofanobjectintermsofaldquoweighingrdquoprocesssuchasthereadingonasetofbathroomscaleswhichineffectmeasuresthecontact force between the object and the scales thenclearlyobjectsinfreefallareweightlessOnenowhasto

askthequestionwhetherornotitispossibleForexampletomeasurethegravitationalforceactingonanastronautin orbit about the Earth

WecanalsodefineweightintermsofthenetgravitationalforceactingonabodyduetoseveraldifferentobjectsForexample for an object out in space its weight could bedefined in terms of the resultant of the forces exerted on it bytheSuntheMoontheEarthandalltheotherplanetsin the Solar System If this resultant is zero at a particular pointthenthebodyisweightlessatthispoint

Inviewofthevariousdefinitionsofweightthatareavailabletousitisimportantthatwhenweusethewordldquoweightrdquoweareawareofthecontextinwhichitisbeingused

Example

Calculate the height above the surface of the Earth atwhichageo-stationarysatelliteorbits

Solution

A geo-stationary satellite is one that orbits the Earth in such a way that it is stationary with respect to a point on the surface of the Earth This means that its orbital period must be the same as the time for the Earth to spin once on its axis ie 24 hours

From Keplerrsquos third law we have G Ms

4π2----------- R3

T2------=

That is

RMe

mh

using the fact that the force of attraction between the satellite and the Earth is given by

FG Mem

R2----------------=

and that F = ma

where m is the mass of the satellite and a 4π2R

T2-------------=

we have

GMem

R2---------------- m 4π2R

T2-------------times

GMe

4π2------------rArr R3

T2------= =

PHYSICS 2015indb 321 150514 338 PM

Chapter 10

322

AH

L

Now the mass of the Earth is 60 times 1024 kg and the period T measured in seconds is given by T = 86400 s

So substitution gives R = 42 times 106 m

The radius of the Earth is 64 times 106 m so that the orbital height h is about 36 times 107 m

Example

Calculatetheminimumenergyrequiredtoputasatelliteofmass500kgintoanorbitthatisasaheightequaltotheEarthrsquosradiusabovethesurfaceoftheEarth

Solution

We have seen that when dealing with gravitational fields and potential it is useful to remember that

g 0G M

Re2---------= or g0 Re

2 GM=

The gravitational potential at the surface of the Earth is

g 0ndash Re GMRe

---------ndash=

The gravitational potential at a distance R

from the centre of the Earth is

=

The difference in potential between the surface of the Earth and a point distance R from the centre is therefore

V∆ g 0Re 1ReR------ndash

=

If R 2Re= then V∆g0 Re

2------------=

This means that the work required to ldquoliftrdquo the satellite into orbit is g0Rm where m is the mass of the satellite This is equal to

10 32 106 500timestimestimes = 16000 MJ

However the satellite must also have kinetic energy in order to orbit Earth This will be equal to

g 0 mRe2

2R-----------------g 0mRe

2

4----------------- 8000 MJ= =

The minimum energy required is therefore

24000 MJ

Exercise 1031 The speed needed to put a satellite in orbit does

not depend on

A the radius of the orbitB theshapeoftheorbitC thevalueofgattheorbitD the mass of the satellite

2 EstimatethespeedofanEarthsatellitewhoseorbitis400kmabovetheEarthrsquossurfaceAlsodetermine the period of the orbit

3 Calculatethespeedofa200kgsatelliteorbitingthe Earth at a height of 70 times 106 m

Assume that g = 82 m sndash2 for this orbit

4 TheradiioftwosatellitesXandYorbitingthe Earth are 2r and 8rwherer is the radius of the Earth Calculate the ratio of the periods of revolutionofXtoY

5 A satellite of mass mkgissentfromEarthrsquossurfaceinto an orbit of radius 5RwhereR is the radius of theEarthWritedownanexpressionfor

(a) thepotentialenergyofthesatelliteinorbit

(b) thekineticenergyofthesatelliteinorbit

(c) theminimumworkrequiredtosendthesatellitefromrestattheEarthrsquossurfaceintoits orbit

6 A satellite in an orbit of 10rfallsbacktoEarth(radius r)afteramalfunctionDeterminethespeedwithwhichitwillhittheEarthrsquossurface

7 The radius of the moon is frac14 that of the Earth AssumingEarthandtheMoontohavethesamedensitycomparetheaccelerationsoffreefallatthe surface of Earth to that at the surface of the Moon

8 UsethefollowingdatatodeterminethegravitationalfieldstrengthatthesurfaceoftheMoon and hence determine the escape speed from the surface of the Moon

Mass of the Moon = 73 times 1022kg

RadiusoftheMoon=17times106 m

PHYSICS 2015indb 322 150514 338 PM

Wave Phenomena

285

AH

L

is often referred to as a harmonic oscillator

Alternativelyiftheweightwasatitsequilibriumpositionwhent=0thenthetrigonometricexpressionwouldbe

x = x0sinωt

The velocity v of the weight at any instant can be found byfinding thegradient of the displacement-time graph at that instant The displacement time graph is a cosine function and the gradient of a cosine function is a negative sine function The gradient of

x = x0cosωt is in fact T 2T00

00

velo

city

dis

pla

cem

ent

t

so

ω ω ω

where v0isthemaximumandminimumvelocityequalinmagnitude to ωx0

Studentsfamiliarwithcalculuswillrecognisethevelocityv as

( )0d d cos sind d oxv x t x tt t

ω ω ω= = = minus Similarly

( ) 2 20 0

d d sin cosd dva x t x t xt t

ω ω ω ω ω= = minus = minus = minus

whichofcourseisjustthedefiningequationofSHM

However we have to bear in mind that ωt varies between 0 and 2π where cosωt is negative for ωt for

and sinωt is negative for ωt in the range π to

2π This effectively means that when the displacement fromequilibriumispositivethevelocityisnegativeandsodirectedtowardsequilibriumWhenthedisplacementfromequilibriumisnegativethevelocityispositiveandsodirectedawayfromequilibrium

ThesketchgraphinFigure902showsthevariationwithtime t of the displacement x and the corresponding variation with time t of the velocity v This clearly demonstrates the relation between the sign of the velocity and sign of the displacement

T 2T00

00

velo

city

disp

lace

men

t

t

x0

ndashx0

v0

ndashv0

Figure 902 Displacement-time and velocity-time graphs

We can also see how the velocity v changes with displacement x

We can express sinωt in terms of cosωt using the trigonometric relation

2 2sin cos 1θ θ+ =

From which it can be seen that

Replacing θ with ωt we have

20 1 cosv x tω ω= minus minus

Remembering that and putting x0 inside the squarerootgives

v=ndashωradic_______

( x0 2 ndash x2 )

Bearing in mind that v can be positive or negative wemust write

v=plusmnωradic_______

( x0 2 ndash x2 )

The velocity is zero when the displacement is a maximum and is a maximum when the displacement is zero

The graph in Figure 903 shows the variation with x of the velocity v for a system oscillating with a period of 1 secondandwithanamplitudeof5cmThegraphshowsthe variation over a time of any one period of oscillation

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

-5 -4 -3 -2 -1 1 2 3 4 5

v c

m s

-1

x cm

Figure 903 Velocity-displacement graph

PHYSICS 2015indb 285 150514 337 PM

Chapter 9

286

AH

L

Boundary ConditionsThe two solutions to the general SHM equation are

and Which solution applies to aparticularsystemdependsasmentionedaboveontheboundary conditions for that system For systems such astheharmonicoscillatorandthesimplependulumtheboundary condition that gives the solution is that the displacement x = x0 when t = 0 For some other systems it might turn out that x = 0 when t =0 This will lead to the solution From a practical pointofviewthetwosolutionsareessentiallythesamefor example when timing the oscillations of a simple pendulumyoumightdecidetostartthetimingwhenthependulumbobpasses through theequilibriumpositionIneffectthetwosolutionsdifferinphaseby

The table in Figure 904 summarises the solutions we have forSHM

x = x 0 cosωt x = x 0 sinωt

ν = minus ν 0 sinωt ν = ν 0 cosωt

ν = ndash ωx 0 sinωt ν = ndash ωx 0 cosωt

ν = plusmn ω radic______

x0 2 ndash x2 ν = plusmn ω radic

______ x0

2 ndash x2

Figure 904 Common equations

We should mention that since the general solution to the SHMequation is there are infactthreesolutionstotheequationThisdemonstratesa fundamental property of second order differential equationsthatoneofthesolutionstotheequationisthesum of all the other solutions This is the mathematical basis of the so-called principle of superposition

SHM is a very good example in which to apply the NewtonianmethoddiscussedinChapter2ieiftheforcesthatactonasystemareknownthenthefuturebehaviourofthesystemcanbepredictedHerewehaveasituationinwhich the force is given by ndashkx From Newtonrsquos second law therefore ndashkx = mawherem is the mass of the system and a istheaccelerationofthesystemHowevertheaccelerationis not constant For those of you who have a mathematical benttherelationbetweentheforceandtheaccelerationis written as

2dd

xkx mt

minus =2

This is what is called a ldquosecond orderdifferentialequationrdquoThesolutionoftheequationgives x as a function of tTheactualsolutionisoftheSHMequationis

x = Pcosωt + Qsinωt where P and Q are constants and ω is theangularfrequencyofthesystemandisequalto

Whether a particular solution involves the sine function or thecosinefunctiondependsontheso-calledlsquoboundaryconditionsrsquo If for example x = x0 (the amplitude)whent =0thenthesolutionisx = x0cosωt

The beauty of thismathematical approach is that oncethe general equation has been solved the solution forall systemsexecutingSHM isknownAll thathas tobeshowntoknowifasystemwillexecuteSHMisthattheaccelerationofthesystemisgivenbyEquation45ortheforceisgivenbyequation47Thephysicalquantitiesthatω will depend on is determined by the particular system Forexampleforaweightofmassm oscillating at the end of a vertically supported spring whose spring constant is k then ω = or from equation 44 For a simple pendulum ω = where l is the length of the pendulum and g is the acceleration of free fall such

that T=2πradic_

l _ g

Examples

ThegraphinFigure905showsthevariationwithtime t of the displacement x ofasystemexecutingSHM(SomequestionpartswerealreadydoneintheSLchapter4)

-10

-8

-6

-4

-2

0

2

4

6

8

10

05 1 15 2 25 3 35t s

xcm

Figure 905 Displacement ndash time graph for SHM

Use the graph to determine the

(i) periodofoscillation(ii) amplitudeofoscillation(iii) maximumspeed(iv) speedatt = 13 s(v) maximumacceleration

Solutions

(i) 20 s (time for one cycle)(ii) 80 cm(iii) Using gives

(remember that ) = 25 cm s-1

PHYSICS 2015indb 286 150514 337 PM

Wave Phenomena

287

AH

L

(iv) v = minusv0sinωt = minus25sin (13π) To find the value of the sine function we have to convert the 13π into degrees (remember ω and hence ωt is measured in radians)

1 deg therefore 13π = 13 times 180 = 234deg therefore v1 = minus25sin (234deg) = +20 cm s ndash1

Or we can solve using ωradic_______

( x0 2 ndash x2 )

from the graph at t = 13 s x = ndash 48 cm

therefore v = π times = 20 cm s-1

(v) Using = π2 times 80 = 79 m s-2

Energy changesWe must now look at the energy changes involved in SHMTodosowewillagainconcentrateontheharmonicoscillator The mass is stationary at x = +x0 (maximumextension)andalsoatx = ndashx0 (maximumcompression)At these two positions the energy of the system is all potential energy and is in fact the elastic potential energy stored in the spring This is the total energy of the system ET and clearly

Equation412

ThatisthatforanysystemperformingSHMtheenergyofthesystemisproportionaltothesquareoftheamplitudeThis is an important result and one that we shall return to when we discuss wave motion

At x =0thespringisatitsequilibriumextensionandthemagnitude of the velocity v of the oscillating mass is a maximum v0TheenergyisallkineticandagainisequaltoET We can see that this is indeed the case as the expression for the maximum kinetic energy Emax in terms of v0 is

Clearly ET and Emax areequalsuchthat

ET = 1 __ 2 kx0 2 = Emax = 1 __ 2 mv0

2

From which

v0 2 = k __ m x0

2(asv0ge0)

Therefore

v0 = radic__

k __ m x0 = ωx0

whichtiesinwiththevelocitybeingequaltothegradientofthedisplacement-timegraph(see415)Asthesystemoscillates there is a continual interchange between kinetic energy and potential energy such that the loss in kinetic energyequalsthegaininpotentialenergyandET = EK + EP

Remembering that wehavethat

Clearly thepotentialenergyEP at any displacement x is given by

At any displacement xthekineticenergyEK is EK = 1__2mv2

Hencerememberingthat

v_______( x0

2 ndash x2 ) wehave

EK = 1__2m 2 ( x0

2 ndash x2 )

Although we have derived these equations for a harmonicoscillatortheyarevalidforanysystemoscillatingwithSHMThe sketch graph in Figure 906 shows the variation with displacement x of EK and EP for one period

displacement

ener

gy potentialkinetic

Figure 906 Energy and displacement

Example

The amplitude of oscillation of a mass suspended by a vertical spring is 80 cm The spring constant of the spring is74Nmndash1Determine

(a) thetotalenergyoftheoscillator

(b) thepotentialandthekineticenergyoftheoscillator at a displacement of 48 cm from equilibrium

PHYSICS 2015indb 287 150514 337 PM

Chapter 9

288

AH

L

Solution

(a) Spring constant k = 74 N m-1 and

x0 = 80 times 10-2 m

ET = frac12 kx02

= frac12 times 74 times 64 times 10-4 = 024 J

(b) At x = 48 cm

Therefore EP = frac12 times 74 times (48 times 10-2)2

= 0085 times 10-4 J

EK = ET minus EP = 024 minus 0085 = 016 J

EXERCISE 911 Which graph shows the relationship between the

acceleration a and the displacement x from the equilibriumpositionofanobjectundergoingsimple harmonic motion

Acceleration

Displacement

Acceleration

Displacement

A B

Acceleration

Displacement

Acceleration

Displacement

C D

2 AnobjectisundergoingsimpleharmonicmotionaboutafixedpointPandthemagnitudeofitsdisplacementfromPisx Which one of the following statements is correct

Magnitudeoftheresultant force

Directionoftheresultant force

A Proportionaltox AwayfrompointP

B Inversely proportional to x AwayfrompointP

C Proportionaltox TowardspointP

D Inversely proportional to x TowardspointP

3 Theangularspeedoftheldquominuterdquohandofananalogue watch is

A π1800rads-1 C π30rads-1

B π60rads-1 D 120rads-1

4 Which of the following is true of the magnitude of the acceleration of the object that is undergoing simple harmonic motion

A It is greatest at the midpoint of the motionB It is greatest at the end points of the motionC It is uniform throughout the motionD Itisgreatestatthemidpointsandthe

endpoints

5 Aparticleoscillateswithsimpleharmonicmotionwith a period T

At time t=0theparticlehasitsmaximumdisplacement Which graph The variation with time t of the kinetic energy of the particle is shown in which diagram below

A

00

Ek

T t

D

00

Ek

T t

C

00

Ek

T t

B

0

Ek

t0 T

6 Figure907belowshowshowthedisplacementofthe magnet of mass 03 kg hanging from a spring varies with time for two oscillations

0 02 04 06 08 10

0

2

-2

-4

4

Disp

lace

men

t (cm

)

Time (s)

Figure 907

PHYSICS 2015indb 288 150514 337 PM

Wave Phenomena

289

AH

L

Using information from this graph calculate the

(a) valueofthespringconstant(b) maximumkineticenergyofthemagnet

7 AsystemisoscillatingwithSHMasdescribedbythe graph in the Figure 908 below

Disp

lace

men

t (cm

)

005 1 15 2 25 3 35

2

-2

-6

-4

4

6

Time (s)

Figure 908

(a) Usethegraphtodeterminethe

(i) periodofoscillation(ii) amplitudeofoscillation(iii) maximumspeed(iv) speedatt=13s(v) maximumacceleration

(b) Statetwovaluesoftforwhenthemagnitudeofthevelocity is a maximum and two values of t for when the magnitude of the acceleration is a maximum

9 Figure 909 shows the relationship that exists between the acceleration and the displacement from the equilibriumpositionforaharmonicoscillator

Acce

lera

tion

(m-2

)

Displacement (cm)0

-4000

-2000

2000

-05 050

Figure 909

(a) Stateandexplaintworeasonswhythegraphopposite indicates that the object is executing simple harmonic motion

(b) Determinethefrequencyofoscillation

92 Single-slit diffraction

Essential idea Single-slitdiffractionoccurswhenawaveis incident upon a slit of approximately the same size as the wavelength

Understandingsbull Thenatureofsingle-slitdiffraction

The nature of single-slit diffractionSingle slit diffraction intensity distribution

When plane wavefronts pass through a small aperture they spread out as discussed in chapter 4 This is an example ofthephenomenoncalleddiffractionLightwavesarenoexception to this and ways for observing the diffraction of light have also been discussed previously

Howeverwhenwelookatthediffractionpatternproducedbylightweobserveafringepatternthatisonthescreenthere is a bright central maximum with ldquosecondaryrdquomaxima either side of it There are also regions where there is no illumination and these minima separate the maxima If we were to actually plot how the intensity of illumination varies along the screen then we would obtain a graph similar to that as in Figure 910

intensity

distance along screen

Figure 910 Intensity distribution for single-slit diffraction

We would get the same intensity distribution if we were to plot the intensity against the angle of diffraction θ(Seenextsection)

NATURE OF SCIENCEDevelopment of theories When light passes through an aperture the summation of all parts of the wave leads to an intensity pattern that is far removed from the geometrical shadow that simple theory predicts (19)

copy IBO 2014

PHYSICS 2015indb 289 150514 337 PM

Chapter 9

290

AH

L

This intensity pattern arises from the fact that each point ontheslitactsinaccordancewithHuygenrsquosprincipleasa source of secondary wavefronts It is the interference between these secondary wavefronts that produces the typical diffraction pattern

Obtaining an expression for the intensity distribution is mathematically a little tricky and it is not something that we are going to attempt here However we can deducea useful relationship from a simple argument In this argument we deal with a phenomenon called Fraunhofer diffractionthatisthelightsourceandthescreenareaninfinitedistanceawayformtheslitThiscanbeachievedwith the set up shown in Figure 911

lens 1 lens 2

single slit

source

screen

Figure 911 Apparatus for viewing Fraunhofer diffraction

The source is placed at the principal focus of lens 1 and the screen isplacedat theprincipal focusof lens2Lens1 ensures that parallel wavefronts fall on the single slit and lens2ensuresthattheparallelraysarebroughttoafocuson the screen The same effect can be achieved using a laser and placing the screen some distance from the slit If the light and screenarenot an infinitedistance from the slitthen we are dealing with a phenomenon called Fresnel diffraction and such diffraction is very difficult to analyse mathematically To obtain a good idea of how the single slit patterncomesaboutweconsiderthediagramFigure912

Xθ2

Yλ

d

P

b

f

screen

θ1

Figure 912 Single slit diffraction

In particular we consider the light from one edge of the slittothepointPwherethispointis justonewavelengthfurther from the lower edge of the slit than it is from the

upper edge The secondary wavefront from the upper edge willtraveladistanceλ2furtherthanasecondarywavefrontfrom a point at the centre of the slitHencewhen thesewavefronts arrive at P theywill be out of phase andwillinterfere destructively The wavefronts from the next point below the upper edge will similarly interfere destructively with the wavefront from the next point below the centre of the slit In this way we can pair the sources across the whole width of the slit If the screen is a long way from the slit then the angles θ1 and θ2becomenearlyequal (If the screen is at infinity then they are equal and the two lines PX and XY are at right angles to each other)FromFigure1116weseethereforethattherewillbeaminimumatPif

λ b θ1 sin= where b is the width of the slit

Howeverbothanglesareverysmallequaltoθ saywhereθ is the angle of diffraction

Soitcanbewritten θ λb---

=

This actually gives us the half-angular width of the central maximum We can calculate the actual width of the maximum along the screen if we know the focal length of the lens focussing the light onto the screen If this is f then we have that

θ df---

= Suchthat d fλb-----

=

To obtain the position of the next maximum in the pattern we note that the path difference is 3

2---λ We therefore divide

theslitintothreeequalpartstwoofwhichwillproducewavefronts that will cancel and the other producing wavefronts that reinforce The intensity of the second maximum is therefore much less than the intensity of thecentralmaximumof theorderof005or120of theoriginalintensity(Muchlessthanonethirdinfactsincethewavefrontsthatreinforcewillhavedifferingphases)

We can also see now how diffraction effects become more and more noticeable the narrower the slit becomes If light of wavelength 430 nm was to pass through a slit of width say10cmandfallonascreen30mawaythenthehalfangular width of the central maximum would be 013 microm

d = f λ

___ b = 3 times 430 times 10 ndash9 ____________ 01 = 013 μm

There will be lots of maxima of nearly the same intensity and the maxima will be packed very closely together (Thefirstminimumoccursatadistanceof012micromfromthe centre of the central maximum and the next occurs effectivelyatadistanceof024microm)Weeffectivelyobservethe geometric pattern Refer to Example

We also see now how for diffraction effects to be noticeable the wavelength must be of the order of the slit width The width of the pattern increases in proportion to the wavelength and decreases inversely with the width of the

PHYSICS 2015indb 290 150514 337 PM

Wave Phenomena

291

AH

L

93 Interference

Essential idea Interference patterns from multiple slits andthinfilmsproduceaccuratelyrepeatablepatterns

Understandingsbull Youngrsquosdouble-slitexperimentbull Modulationoftwo-slitinterferencepatternbyone-slit

diffraction effectbull Multipleslitanddiffractiongratinginterferencepatternsbull Thinfilminterference

Youngrsquos double slit experimentPlease refer to interference in chapter 4 However wewill reiterate the condition for the interference of waves from two sources to be observed The two sources must be coherentthatistheymusthavethesamephaseorthephase difference between them must remain constant

Also to reinforce topics concerning the principle ofsuperpositionandpathdifferenceandphasedifferenceincluded in Figure 913 is another example of two source interference

We can obtain evidence for the wave nature of sound by showing that sound produces an interference pattern Figure 913 shows the set up for demonstrating this

CRO

speaker

speaker

signalgenerator

microphoneX

Y

Figure 913 Interference using sound waves

The two speakers are connected to the same output of the signal generator andplaced about 50 cmapartThesignalgeneratorfrequencyissettoabout600Hzandthe

slit If the slit width is much greater than the wavelength then the width of the central maxima is very small

Insummary

for destructive interference and minima

dsinθ=nλ wheren=123hellip

for constructive interference and maxima

dsinθ=(n+frac12)λ wheren=123hellip

Note that waves from point sources along the slit arrive in phase at the centre of the fringe pattern and constructively interferetoproducethecentralmaximumcalledthezero ordermaximum(n=0)

Example

LightfromalaserisusedtoformasingleslitdiffractionpatternThe width of the slit is 010 mm and the screen is placed 30 m from the slit The width of the central maximum is measured as26cmCalculatethewavelengthofthelaserlight

Solution

Since the screen is a long way from the slit we can use the small angle approximation such that the f is equal to 30 m

The half-width of the central maximum is 13 cm so we have

λ13 10 2ndashtimes( ) 10 10 4ndashtimes( )times

30------------------------------------------------------------------=

To give λ = 430 nm

This example demonstrates why the image of a point source formed by a thin converging lens will always have a finite width

Exercise 921 Aparallelbeamoflightofwavelength500nmis

incidentonaslitofwidth025mmThelightisbroughttofocusonascreenplaced150mfromthe slit Calculate the angular width and the linear width of the central diffraction maximum

2 Lightfromalaserisusedtoformasingleslitdiffraction pattern on a screen The width of the slit is 010 mm and the screen is 30 m from the slit The width of the central diffraction maximum is 26cmCalculatethewavelengthofthelaserlight

3 Determinetheanglethatyouwouldexpecttofindconstructiveinterference(forn=0n=1andn=2) fromasingleslitofwidth20μmandlightofwavelength 600 nm

NATURE OF SCIENCECuriosity Observed patterns of iridescence in animals such as the shimmer of peacock feathers led scientists to develop the theory of thin film interference (15)

Serendipity The first laboratory production of thin films was accidental (15)

copy IBO 2014

PHYSICS 2015indb 291 150514 337 PM

Chapter 9

292

AH

L

microphoneismovedalongthelineXYthatisaboutonemetre from the speakers As the microphone is moved along XY the trace on the cathode ray oscilloscope isseen to go through a series of maxima and minima corresponding to points of constructive and destructive interference of the sound waves

Aninterestinginvestigationistofindhowtheseparationof the points of maximum interference depends on the frequency of the source and also the distance apart ofthe speakers

In the demonstration of the interference between two soundsourcesdescribedaboveifweweretomoveoneofthe speakers from side to side or backwards and forwards the sound emitted from the two speakers would no longer be in phase No permanent points of constructive or destructive interference will now be located since the phase difference between the waves from the two sources is nolongerconstantiethesourcesarenolongercoherent

Light from an incandescent source is emitted with acompletely random phase Although the light from two separate sources will interfere because of the randomlychanging phase no permanent points of constructive or destructive interference will be observed This is why a singleslit isneededintheYoungrsquosdoubleslitexperimentByactingasapointsourceitessentiallybecomesacoherentlight source The light emitted from a laser is also very nearly coherent and this is why it is so easy to demonstrate optical interference and diffraction with a laser

Youngrsquos double slit experiment is one of the great classic experiments of physics and did much to reinforce the wave theoryoflightThomasYoungcarriedouttheexperimentin about 1830

It is essentially the demonstration with the ripple tank and the sound experiment previously described butusing light The essential features of the experiment are shown in Figure 914

screendouble slit

single slit

coloured lter

S 1

S2

Figure 914 Youngrsquos double slit experiment

YoungallowedsunlighttofallontoanarrowsingleslitA few centimetres from the single slit he placed a double slit The slits are very narrow and separated by a distance equal to about fourteen slitwidthsA screen is placed

about ametre from thedouble slitsYoungobservedapatternofmulticolouredldquofringesrdquo in the screenWhenhe placed a coloured filter between the single slit anddouble slit he obtained a pattern that consisted of bright coloured fringes separated by darkness

The single slit essentially ensures that the light falling on the double slit is coherent The two slits then act as the two speakers in the sound experiment or the two dippers in the ripple tank The light waves from each slit interfere and produce the interference pattern on the screen Without thefilterapatternisformedforeachwavelengthpresentinthesunlightHencethemulticolouredfringepattern

You can demonstrate optical interference for yourselfSmoking a small piece of glass and then drawing twoparallel lines on it can make a double slit If you then look throughthedoubleslitatasingletungstenfilamentlampyouwillseethefringepatternByplacingfiltersbetweenthe lamps and the slits you will see the monochromatic fringe pattern

YoucanalsoseetheeffectsofopticalinterferencebylookingatnetcurtainsEachlsquoholersquointhenetactsasapointsourceand the light from all these separate sources interferes and producesquiteacomplicatedinterferencepattern

A laser can also be used to demonstrate optical interference Sincethelightfromthelaseriscoherentitisveryeasytodemonstrate interference Just point the laser at a screen and place a double slit in the path of the laser beam

LetusnowlookattheYoungrsquosdoubleslitexperimentinmore detail The geometry of the situation is shown in Figure915

S1

S2 X

P

Q

D

y

d

screen

θ

θrsquo

Figure 915 The geometry of Youngrsquos double slit experiment

S1andS2 are the two narrow slits that we shall regard as twocoherentmonochromaticpointsourcesThedistancefrom the sources to the screen is D and the distance between the slits is d

The waves from the two sources will be in phase at Q and therewill be abright fringehereWewish tofindtheconditionfortheretobeabrightfringeatPdistancey from Q

PHYSICS 2015indb 292 150514 337 PM

Wave Phenomena

293

AH

L

We note that that D (asymp1metre)isverymuchgreaterthaneither y or d(asymp fewmillimetres)Thismeansthatbothθ and θʹ areverysmallanglesandforintentsandpurposesequal

From the diagram we have that

θ = Dy

And

2S Xd

θʹ =

(Remembertheanglesareverysmall)

Sinceθ asymp θʹ then

2S XyD d

=

ButS2X is the path difference between the waves as they

traveltoPWehavethereforethat

path difference = Dyd

TherewillthereforebeabrightfringeatPif

Dyd

= nλ

Supposethatthereisabrightfringeaty = y1 corresponding

to n = n1 then

Ddy1 = n

1λ

If the next bright fringe occurs at y = y2 this will correspond

to n = n1+1Hence

Ddy2 =(n

1+1)λ

This means that the spacing between the fringes y2 ndash y

1 is

given by

y2 ndash y

1 = d

Dλ

Youngactuallyusethisexpressiontomeasurethewavelengthof the light he used and it is a method still used today

We see for instance that if in a given set up we move the slits closer together then the spacing between the fringes will get greaterEffectivelyourinterferencepatternspreadsoutthatis there will be fewer fringes in a given distance We can also increase the fringe spacing by increasing the distance betweentheslitsandthescreenYouwillalsonotethatfora

givensetupusinglightofdifferentwavelengthsthenldquoredrdquofringeswillspacefurtherapartthanldquobluerdquofringes

In this analysis we have assumed that the slits act as point sources and as such the fringes will be uniformly spaced andofequalintensityAmorethoroughanalysisshouldtakeintoaccountthefinitewidthoftheslits

ReturningtoFigure915weseethatwecanwritethepathdifferenceS2X as

S2X = dtan θʹ

But since θʹ is a small angle the sine and tangent will be nearlyequalsothat

S2X = dsinθʹ

And since θʹ asymp θ then

S2X = dsinθ

The condition therefore for a bright fringe to be found at a point of the screen can therefore be written as

dsinθ = nλ

Insummary

for constructive interference and minima

d sinθ = nλ wheren=123hellip

for destructive interference and maxima

d sinθ = ( n + frac12 )λ wheren=123hellip

Figure 916 shows the intensity distribution of the fringes on the screen when the separation of the slits is large comparedtotheirwidthThefringesareofequalintensityandofequalseparation

Intensity

distance along screen

Figure 916 The intensity distribution of the fringes

It isworthnoting that if the slits are close together theintensity of the fringes is modulated by the intensity distribution of the diffraction pattern of one of the slits

PHYSICS 2015indb 293 150514 337 PM

Chapter 9

294

AH

L

Example

Light of wavelength 500nm is incident on two smallparallel slits separated by 10mmDetermine the anglewherethefirstmaximumisformed

If after passing through the slits the light is brought toa focusona screen15m fromtheslitscalculate theobserved fringe spacing on the screen

Solution

Using the small angle approximation we have

3

7

10105minus

minustimes==

dλθ

= 5 times 10-4 rad

The fringe spacing is given by

3

7

1010551

minus

minustimestimes==

dDy λ

= 075 mm

Modulation of two-slit interference pattern by one-slit diffraction effect

In practice the intensity pattern of the maxima as shown in Figure 917 is not constant but fluctuates while decreasingsymmetrically on either side of the central maximum as shown inFigure917Theintensitypatternisacombinationofboththe single-slit diffractionenvelope and the double slit pattern That is the amplitude of the two-slit interference pattern is modulated(ieadjustedto)byasingleslitdiffractionenvelope

2λD

___ λD

D

___λd

d

___

Double slitnite width

Figure 917 Double slit pattern

ForagivenslitseparationdwavelengthoflightandfixedslittoscreendistancethevariationinintensitydependsonthewidthoftheslitDAlthoughincreasingthewidthoftheslitsincreasestheintensityoflightinthefringesitalsomakesthem less sharp and they become blurred As the slit width is widenedthefringesgraduallydisappearbecausenumerouspoint sources along the widened slit give rise to their own dark and bright fringes that then overlap with each other

Exercise 931 InFigure913thedistancebetweenthespeakers

is050mandthedistancebetweenthelineofthespeakersandthescreenis20mAsthemicrophoneismovedalongthelineXYthedistance between successive points of maximum soundintensityis030mThefrequencyofthesoundwavesis44times103HzCalculateavalueforthe speed of sound

2 Laserlightofwavelength610nmfallsontwoslits10x10-5apartDeterminetheseparationofthefirstordermaximumformedonascreen20maway

3 Twoparallelslits012mmapartareilluminatedwith red light with a wavelength of 600 nm An interferencepatternfallsonascreen15maway

Calculate

(a) thedistancefromthecentralmaximumtothefirstbrightfringe

(b) thedistancetotheseconddarkline

Multiple slit and diffraction grating interference patterns

If we examine the interference pattern produced when monochromatic light passes through a different number of slits we notice that as the number of slits increases the number of observed fringes decreases the spacingbetween them increases and the individual fringes become much sharper We can get some idea of how this comes about by looking at the way light behaves when a parallel beam passes through a large number of slits The diagram for this is shown in Figure 918

θθd

1

2

3

telescope

Figure 918 A parallel beam passing through several slits

The slits are very small so that they can be considered to act as point sources They are also very close together such that d issmall(10ndash6 m)EachslitbecomesasourceofcircularwavefrontsandthewavesfromeachslitwillinterfereLetus

PHYSICS 2015indb 294 150514 337 PM

Wave Phenomena

295

AH

L

consider the light that leaves the slit at an angle θ as shown Thepathdifferencebetweenwave1andwave2isdsinθ and ifthisisequaltoanintegralnumberofwavelengthsthenthe two waves will interfere constructively in this direction Similarly wave 2 will interfere constructively with wave3 at this angle andwave 3with 4 etc across thewholegratingHenceifwelookatthelightthroughatelescopethatisbringittoafocusthenwhenthetelescopemakesanangle θ to the grating a bright fringe will be observed The condition for observing a bright fringe is therefore

dsinθ = mλ

Supposeweuselightofwavelength500nmandsupposethat d = 16 times 10ndash6 m

Obviously we will see a bright fringe in the straight on position θ=0(thezero order)

The next position will be when m =1(thefirst order)andsubstitutionintheaboveequationgivesθ = 18deg

The next position will be when m=2(thesecond order)and this give θ = 38deg

For m=3sinθ is greater than 1 so with this set up we only obtain5fringesonezeroorderandtwoeithersideofthezero order

The calculation shows that the separation of the orders is relatively large At any angles other than 18deg or 38deg the light leaving the slits interferes destructively We can see that the fringes will be sharp since if we move just a small angle away from 18deg the light from the slits will interfere destructively

An array of narrow slits is usually made by cutting narrow transparent lines very close together into the emulsion on aphotographicplate(typically200linespermillimetre)Suchanarrangementiscalledadiffraction grating

The diffraction grating is of great use in examining the spectral characteristics of light sources

All elements have their own characteristic spectrum An element can be made to emit light either by heating it until it is incandescent or by causing an electric discharge through itwhen it is in a gaseous state Yourschool probably has some discharge tubes and diffraction gratingsIfithasthenlookattheglowingdischargetubethroughadiffractiongratingIfforexampletheelementthat you are looking has three distinct wavelengths then each wavelength will be diffracted by a different amount inaccordancewiththeequationdsinθ = nλ

Alsoiflaserlightisshonethroughagratingontoascreenyou will see just how sharp and spaced out are the maxima By measuring the line spacing and the distance of the screen fromthelaserthewavelengthofthelasercanbemeasured

If your school has a set of multiple slits say from a single slit toeight slits then it isalsoaworthwhileexercise toexamine how the diffraction pattern changes when laser light is shone through increasing numbers of slits

Consider a grating of NparallelequidistantslitsofwidthaseparatedbyanopaqueregionofwidthD We know that light diffracted from one slit will be superimposed and interfere with light diffracted from all the other slits The intensityprofile of the resultant wave will have a shape determined by

bull theintensityoflightincidentontheslitsbull thewavelengthofthislightbull theslitwidthbull theslitseparation

Typical intensity profiles formed when a plane wavemonochromaticlightatnormalincidenceononetwoandsix slits are shown in Figure 919 In this example the slit widthisonequarterofthegratingspacing

Six slits sharperfringes

d = 4a

I α N 2

(N ndash 2)= 4 subsidiary maxima

order0 4321

I = 36

Two slits

order0

single-slitdiraction

envelope

d = 4a4th principal maximum missing

principa maxima

I = 4 (nominal units)

4321

intensity

order

ISS

One slit

0 1

single-slitdiraction

pattern

N = number of slitsslit width a

Figure 919 Intensity profiles

PHYSICS 2015indb 295 150514 337 PM

Chapter 9

296

AH

L

1 For Nge2thefringepatterncontainsprincipalmaximathat are modulated by the diffraction envelope

2 ForNgt2thefringepatterncontains(Nndash2)subsidiary maxima

3 Where the grating spacing d is four time the slit width a(4xa =d)thenthefourthprincipalmaximumismissing from the fringe pattern Where the grating spacingdisthreetimetheslitwidtha(3xa =d)thenthe third principal maximum is missing from the fringepatternandsoon

4 Increasing N increases the absolute intensities of the diffraction pattern (I = N2)

If white light is shone through a grating then the central image will be white but for the other orders each will be spread out intoacontinuousspectrumcomposedofaninfinitenumberof adjacent images of the slit formed by the wavelength of the different wavelengths present in the white light At any given point in the continuous spectrum the light will be very nearly monochromatic because of the narrowness of the images of the slit formed by the grating This is in contrast to the doubleslitwhereifwhitelightisusedtheimagesarebroadand the spectral colours are not separated

In summary as is the case of interference at a doubleslit there is a path difference between the rays fromadjacent slits of d sinθ Where there is a whole number of wavelengthsconstructiveinterferenceoccurs

For constructive interference and minima

d sinθ = mλ wheren=123hellip

When there is a path difference between adjacent rays of halfawavelengthdestructiveinterferenceoccurs

For and maxima

d sinθ = ( m + frac12 )λ wheren=123hellip

Exercise 941 Lightfromalaserisshonethroughadiffraction

grating on to a screen The screen is a distance of 20mfromthelaserThedistancebetweenthecentraldiffractionmaximumandthefirstprincipalmaximum formed on the screen is 094 m The diffractiongratinghas680linespermmEstimatethe wavelength of the light emitted by the laser

2 Monochromaticlightfromalaserisnormallyincident on a six-slit grating The slit spacing is three times the slit width An interference patternisformedonascreen2montheotherside of the grating

(a) Determineandexplaintheorderoftheprincipal maxima missing from the fringe pattern

(b) Howmanysubsidiarymaximaaretherebetween the principal maxima

(c) Howmanyprincipalmaximaarethere

(i) Betweenthetwofirstorderminimain the diffraction envelope

(ii) Betweenthefirstandsecondorderminima in the diffraction envelope

(d) Drawthepatternformedbythebrightfringesin(c)

(e) Sketchtheintensityprofileofthefringepattern

Thin film interferenceYoumightwellbe familiarwith the colouredpatternoffringes that can be seen when light is reflected off thesurfaceofwateruponwhichathinoilfilmhasbeenspiltorfromlightreflectedfrombubblesWecanseehowthesepatternsarisebylookingatFigure920

Extendedlight source air

A

B

C

12

3

4

Oil lm

Water

Figure 920 Reflection from an oil film

Consider light from an extended source incident on a thinfilmWealsoconsiderawave fromonepointofthe source whose direction is represented by the ray shownSomeofthislightwillbereflectedatAandsometransmitted through the filmwhere somewill again bereflectedatB(somewillalsobetransmittedintotheair)SomeofthelightreflectedatBwillthenbetransmittedatCandsomereflectedandsoonIfweconsiderjustrays1and2thenthesewillbenotbeinphasewhenthearebrought to a focus by the eye since they have travelled

PHYSICS 2015indb 296 150514 337 PM

Wave Phenomena

297

AH

L

different distances If the path difference between them is an integral number of half wavelengths then we might expect the twowaves to be out of phaseHowever ray1 undergoes a phase change of π on reflection but ray2 does not since it is reflected at a boundary betweena more dense and less dense medium (See chapter 4)Henceifthepathdifferenceisanintegralnumberofhalf-wavelengthsrays1and2willreinforceieray1and2areinphaseHowever rays357etcwillbeoutofphasewithrays246etcbutsinceray2ismoreintensethanray3andray4moreintensethanray5thesepairswillnot cancel out so there will be a maximum of intensity

Ifthepathdifferenceissuchthatwave1and2areoutofphasesincewave1ismoreintensethanwave2theywillnotcompletelyannulHoweveritcanbeshownthattheintensitiesofwaves2345hellipaddtoequaltheintensityofwave1Sincewaves345hellipareinphasewithwave2there will be complete cancellation

The path difference will be determined by the angle at which ray1 is incident andalsoon the thickness (andtheactualrefractiveindexaswell)ofthefilmSincethesourceisanextendedsourcethelightwillreachtheeyefrom many different angles and so a system of bright and darkfringeswillbeseenYouwillonlyseefringesifthefilmisverythin(exceptifviewedatnormalincidence)sinceincreasingthethicknessofthefilmwillcausethereflected rays to get so far apart that they will not becollected by the pupil of the eye

From the argument above to find the conditions forconstructive and destructive interferenceweneedonlyfindthepathdifferencebetweenray1andray2Figure921showsthe geometry of the situation

A BE

F

C

D

1 2

φrsquo φrsquo

φ

φ

d

Figure 921 The geometry of interference

Thefilmisofthicknessd and refractive index n and the light has wavelength λ If the line BF is perpendicular to ray 1 then the optical path difference (opd) between ray 1 andray2whenbroughttoafocusis

opd = n(AC+CB)ndashAF

We have to multiple by the refractive index for the path travelledbythelightinthefilmsincethelighttravelsmoreslowly in thefilm If the light travels say adistancex in a material of refractive index n then in the time that it takes to travelthisdistancethelightwouldtraveladistancenx in air

IfthelineCEisatrightanglestoray2thenweseethat

AF = nBE

FromthediagramAC=CDsowecanwrite

opd = n(CD+CB)ndashnBE=nDE

Alsofromthediagramweseethatwhereφ is the angle of refraction

DE=2cosφ

From which opd=2ndcosφ

Bearinginmindthechangeinphaseofray1onreflectionwe have therefore that the condition for constructive interference is

2nd φcos m 12---+

λ m 1 2 hellip = =

Andfordestructiveinterference 2ndcosφ = mλ

Eachfringecorrespondstoaparticularopd for a particular value of the integer m and for any fringe the value of the angle φ is fixedThismeans that itwill be in the formof an arc of a circle with the centre of the circle at the point where the perpendicular drawn from the eye meets the surface of the film Such fringes are called fringesofequal inclinationSince theeyehasa smallaperturethesefringesunlessviewedatneartonormalincidence (φ = 0) will only be observed if the film is very thinThisisbecauseasthethicknessofthefilmincreasesthereflected rayswill get further and further apart and sovery few will enter the eye

Ifwhitelightisshoneontothefilmthenwecanseewhywe get multi-coloured fringes since a series of maxima and minima will be formed for each wavelength present in the white light However when viewed at normalincidenceitispossiblethatonlylightofonecolourwillundergoconstructiveinterferenceandthefilmwilltakeon this colour

PHYSICS 2015indb 297 150514 337 PM

Chapter 9

298

AH

L

Thickness of oil films

The exercise to follow will help explain this use

Non-reflecting films

A very important but simple application of thin filminterferenceisintheproductionofnon-reflectingsurfaces

Athinfilmofthicknessd and refractive index n1 is coated onto glass of refractive index n where n1 lt n Light ofwavelength λ that is reflected at normal incidence willundergo destructive interference if 12

2n d λ

= thatis

14d

nλ

=

(remember that there will now no phase change at the glass interface ie we have a rare to dense reflection)

The use of such films can greatly reduce the loss oflightby reflection at the various surfacesof a systemoflenses or prisms Optical parts of high quality systemsare usually all coatedwithnon-reflecting films in orderto reduce stray reflections The films are usually madebyevaporatingcalciumormagnesiumfluorideonto thesurfacesinvacuumorbychemicaltreatmentwithacidsthat leave a thin layer of silica on the surface of the glass ThecoatedsurfaceshaveapurplishhuebyreflectedlightThis is because the condition for destructive interference fromaparticularfilmthicknesscanonlybeobtainedforone wavelength The wavelength chosen is one that has a value corresponding to light near the middle of the visible spectrumThismeansthatreflectionofredandvioletlightis greater combining to give the purple colour Because of the factor cosφatanglesotherthannormalincidencethepath difference will change butnotsignificantlyuntilsayabout30deg(egcos25deg=090)

It should be borne in mind that no light is actually lost by anon-reflectingfilmthedecreaseofreflectedintensityiscompensated by increase of transmitted intensity

Non-reflecting films can be painted onto aircraft tosuppressreflectionofradarThethicknessofthefilmis determined by

4nd λ

= where λ is the wavelength of

the radar waves and ntherefractiveindexofthefilmat this wavelength

Example

A plane-parallel glass plate of thickness 20 mm isilluminated with light from an extended source The refractiveindexoftheglassis15andthewavelengthofthelight is 600 nm Calculate how many fringes are formed

Solution

We assume that the fringes are formed by light incident at all angles from normal to grazing incidence

At normal incidence we have 2nd = mλ

From which

m 2 15 2 10 3ndashtimestimestimes6 10 7ndashtimes

------------------------------------------- 10 000= =

At grazing incidence the angle of refraction φ is in fact the critical angle

Therefore φφ arcsin 115------- 42 deg= =

ie cosφ = 075

At grazing incidence 2ndcosφ = m λ

From which (and using cos(sinndash1(115) = 075)

m 2 15 2 10 3ndash 075timestimestimestimes6 10 7ndashtimes

------------------------------------------------------------ 7500= =

The total number of fringes seen is equal to m ndash m = 2500

Exercise 95When viewed from above the colour of an oil film onwater appears red in colour Use the data below to estimate theminimumthicknessoftheoilfilm

average wavelength of red light = 630 nm

refractiveindexofoilforredlight=15

refractive index of water for red light = 13

PHYSICS 2015indb 298 150514 337 PM

Wave Phenomena

299

AH

L

94 Resolution

Essential idea Resolution places an absolute limit on the extent to which an optical or other system can separate images of objects

Understandings

bull Thesizeofadiffractingaperturebull Theresolutionofsimplemonochromatic

two-source systems

The size of a diffracting apertureOur discussion concerning diffraction so far has been for rectangular slits When light from a point source enters a smallcircularapertureitdoesnotproduceabrightdotasanimagebutacirculardiscknownasAiryrsquosdiscsurroundedbyfainterconcentriccircularringsasshowninFigure922

Laser

Figure 922 Diffraction at a circular aperture

Diffractionatanapertureisofgreatimportancebecausetheeye and many optical instruments have circular apertures

What is the half-angular width of the central maximum of the diffraction formed by a circular aperture This is not easy to calculate since it involves some advanced mathematics The problemwasfirstsolvedbytheEnglishAstronomerRoyalGeorgeAiryin1835whoshowedthatforcircularapertures

θ 122 λb--------------= where b is the diameter of the aperture

Example 2

Inthefollowingdiagramparallellightfromadistantpointsource(suchasastar)isbroughttofocusonthescreenSbyaconverginglens(the lens is shown as a vertical arrow)

Thefocal length(distance fromlens toscreen) is25cmand the diameter of the lens is 30 cm The wavelength of thelightfromthestaris560nmCalculatethediameterofthe diameter of the image on the screen

Solution

The lens actually acts as a circular aperture of diameter 30 cm The half angular width of central maximum of the diffraction pattern that it forms on the screen is

75

2

122 122 56 10 23 1030 10b

λθminus

minusminus

times times= = = times

times rad

The diameter of the central maxima is therefore

25 times 10-2 times 23 times 10-5

= 57 times 10-6 m

Although this is small it is still finite and is effectively the image of the star as the intensity of the secondary maxima are small compared to that of the central maximum

The resolution of simple monochromatic two-source systemsThe astronomers tell us that many of the stars that we observe with the naked eye are in fact binary stars That iswhatwe see as a single star actually consistsof twostarsinorbitaboutacommoncentreFurthermoretheastronomerstellusthatifweusealdquogoodrdquotelescopethenwewillactuallyseethetwostarsThatiswewillresolve the single point source into its two component parts Sowhatisitthatdetermineswhetherornotweseethetwo stars as a single point source ie what determines whether or not two sources can be resolved It canrsquot just bethatthetelescopemagnifiesthestarssinceiftheyareacting as point sources magnifying them is not going to make a great deal of difference

Ineachofoureyesthereisanaperturethepupilthroughwhich the light enters This light is then focused by the eye lens onto the retina But we have seen that when light passes through an aperture it is diffracted and so when we look at

NATURE OF SCIENCEImproved technology The Rayleigh criterion is the limit of resolution Continuing advancement in technology such as large diameter dishes or lenses or the use of smaller wavelength lasers pushes the limits of what we can resolve (18)

copy IBO 2014

PHYSICS 2015indb 299 150514 337 PM

Chapter 9

300

AH

L

apointsourceadiffractionpatternwillbeformedontheretina If we look at two point sources then two diffraction patterns will be formed on the retina and these patterns will overlap The width of our pupil and the wavelength of the light emitted by the sources will determine the amount by which they overlap But the degree of overlap will also depend on the angular separation of the two point sources WecanseethisfromFigures923

pupilretina

S1

S2

P2

P1

θ

Figure 923

LightfromthesourceS1 enters the eye and is diffracted by the pupil such that the central maximum of the diffraction patternisformedontheretinaatP1SimilarlylightfromS2producesamaximumatP2 If the two central maxima are well separated then there is a fair chance that we will see the two sources as separate sources If they overlap then we will not be able to distinguish one source from another From the diagram we see as the sources are movedclosertotheeyethentheangleθ increases and so does the separation of the central maxima

Figures924925926and927showsthedifferentdiffractionpatternsandtheintensitydistributionthatmightresultonthe retina as a result of light from two point sources

S1 S21 Well resolved

Figure 924 Very well resolved

S1 S22 Well resolved

Figure 925 Well resolved

S1 S23 Just resolved Rayleigh criterion Minimum of S2 coincides with maximum peak of S1

Figure 926 Just resolved

4 Not resolved

Figure 927 Not resolved

We have suggested that if the central maxima of the two diffraction patterns are reasonably separated then we should be able to resolve two point sources In the late 19th century Lord Rayleigh suggested by how much they should be separated in order for the two sources to be just resolved Ifthecentralmaximumofonediffractionpattern coincides with the firstminima of the otherdiffraction pattern then the two sources will just beresolved This is known as the Rayleigh Criterion

Figure926showsjustthissituationThetwosourcesarejust resolved according to the Rayleigh criterion since the peak of the central maximum of one diffraction pattern coincideswiththefirstminimumoftheotherdiffractionpattern This means that the angular separation of the peaks of the two central maxima formed by each source is just the half angular width of one central maximum ie

θ λb---

=

where b is the width of the slit through which the light fromthesourcespassesHoweverweseefromFigure1117that θ is the angle that the two sources subtend at the slit Henceweconcludethattwosourceswillberesolvedbyaslit if the angle that they subtend at the slit is greater than or equaltoλfraslb

PHYSICS 2015indb 300 150514 337 PM

Wave Phenomena

301

AH

L

Sofarwehavebeenassumingthattheeyeisarectangularslit whereas clearly it is a circular aperture and so we must use the formula

θ 122 λb--------------=

As mentioned above the angle θ is sometimes called the resolving power but should more accurately be called the minimum angle of resolution (θmin)

Clearly the smaller θ the greater will be the resolving power

The diffraction grating is a useful for differentiating closely spacedlinesinemissionspectraLikeaprismspectrometerit can disperse a spectrum into its components but it is better suited because it has higher resolution than the prism spectrometer

If λ1 and λ2 are two nearly equal wavelengths that canbarelybedistinguishedtheresolvanceorresolvingpowerofthegratingisdefinedas

R = λ _______ (λ2 ndash λ 1) = λ ___ Δλ

Where λ asymp λ1 asymp λ2

Therefore thediffractiongratingwithahighresolvancewill be better suited in determining small differences in wavelength

If the diffraction grating has Nlinesbeingilluminateditcan be shown that the resolving power of the mth order diffraction is given by

R = mN

Obviouslytheresolvancebecomesgreaterwiththeordernumber m and with a greater number of illuminated slits

Example

A benchmark for the resolving power of a grating is the sodium doublet in the sodium emission spectrum Two yellow lines in itsspectrumhavewavelengthsof58900nmand58959nm

(a) Determinetheresolvanceofagratingifthegivenwavelengths are to be distinguished from each other

(b) Howmanylinesinthegratingmustbeilluminated in order to resolve these lines in the second order spectrum

Solution

(a) R = λ ___ Δλ = 58900nm __________________ (58900ndash58959)nm

= 589nm _______ 059nm = 10 times 103

(b) N = R __ m = 10 times 10 3 ________ 2 =50times102 lines

It has been seen that diffraction effectively limits the resolving power of optical systems This includes such systems as the eye telescopes and microscopes Theresolving power of these systems is dealt with in the next section This section looks at links between technology and resolving power when looking at the very distant and when looking at the very small

Radio telescopes

The average diameter of the pupil of the human eye is about 25 mm This means that the eye will justresolve two point sources emitting light of wavelength 500 nm if their angular separation at the eye is 7

43

50 10122 24 1025 10

θminus

minusminus

times= times = times

times rad

If the eye were to be able to detect radio waves of wavelength 015m then tohave the same resolvingpower thepupilwould have to have a diameter of about 600 m Clearly this is nonsense but it does illustrate a problem facingastronomers who wish to view very distant objects such as quasarsandgalaxies thatemitradiowavesConventionalradio telescopes consist of a large dish typically 25 min diameter Even with such a large diameter the radiowavelength resolving power of the telescope is much less thantheopticalresolvingpowerofthehumaneyeLetuslook at an example

Example

The Galaxy Cygnus A can be resolved optically as an elliptically shaped galaxy However it is also a strongemitterof radiowavesofwavelength015mTheGalaxyisestimatedtobe50times1024mfromEarthUseofaradiotelescope shows that the radio emission is from two sources separated by a distance of 30 times 1021m Estimate thediameterofthedishrequiredtojustresolvethesources

Solution

The angle θ that the sources resolve at the telescope is given by

θ 21

424

30 10 60 1050 10

θ minustimes= = times

times rad

and 4

122 122 015 300060 10

d λθ minus

times= = =

timesm = 30 km

A radio telescope dish of this size would be impossible tomake letalonesupportThisshowsthatasingledishtype radio telescope cannot be used to resolve the sources and yet they were resolved To get round the problemastronomers use two radio telescopes separated by a large

PHYSICS 2015indb 301 150514 337 PM

Chapter 9

302

AH

L

distance The telescopes view the same objects at the same time and the signals that each receive from the objects are simultaneously superimposed The result of the superposition of the two signals is a two-slit interference pattern The pattern has much narrower fringe spacing than that of the diffraction pattern produced by either telescope on its own hence producing a much higherresolvingpowerWhentelescopesareusedlikethistheyare called a stellar interferometer

InSocorroinNewMexicothereisastellarinterferometerthat consists of 27 parabolic dishes each of diameter25 m arranged in a Y-shape that covers an area of 570km2Thisisaso-calledVeryLargeArray(VLA)Evenhigher resolution can be obtained by using an array of radio telescopes in observatories thousands of kilometres apart A system that uses this so-called technique oflsquovery-long-baseline interferometryrsquo (VLBI) is known asalsquovery-long-baselinearrayrsquo(VLBA)WithVLBAaradiowavelength resolving power can be achieved that is 100 timesgreaterthanthebestopticaltelescopesEvenhigherresolving power can be achieved by using a telescope that isinasatelliteorbitingEarthJapanrsquosInstituteofSpaceandAstronautical Science (ISAS) launched sucha system inFebruary 1997TheNationalAstronomicalObservatoryof Japan the National Science Foundationrsquos NationalRadio Astronomy Observatory (NRAO) the CanadianSpaceAgency theAustraliaTelescopeNationalFacilitythe European VLBI Network and the Joint Instituteback this project forVeryLongBaseline Interferometryin EuropeThis project is a very good example of howInternationalismcanoperateinPhysics

Electron microscope

Telescopes are used to look at very distant objects that are very large but because of their distance fromus appearverysmallMicroscopesontheotherhandareusedtolookat objects that are close to us but are physically very small Aswehave seen justmagnifyingobjects that ismakingthemappearlargerisnotsufficientonitsowntogaindetailabouttheobjectfordetailhighresolutionisneeded

Figure928isaschematicofhowanoptical microscope is usedtoviewanobjectandFigure929isaschematicofatransmission electron microscope(TEM)

Optical lens system

bright light source

glass slide containing specimen (object)

eye

Figure 928 The principle of a light microscope

magnetic lens system

electron source

thin wafer of material

ScreenCCD

Figure 929 The principle of an electron microscope

In the optical microscope the resolving power isdetermined by the particular lens system employed and the wavelength λ of the light used For example twopoints in the sample separated by a distance d will just be resolved if

2d mλ=

where m is a property of the lens system know as the numerical aperture In practice the largest value of m obtainable is about 16 Hence if the microscope slideis illuminatedwith light ofwavelength 480 nm a goodmicroscope will resolve two points separated by a distance dasymp15times10-7masymp015micromPointscloser together thanthiswillnotberesolvedHoweverthisisgoodenoughtodistinguishsomevirusessuchastheEbolavirus

Clearlythesmallerλ the higher the resolving power and this is where the electron microscope comes to the fore The electron microscope makes use of the wave nature of electrons(see1315)IntheTEMelectronspassthrougha wafer thin sample and are then focused by a magnetic field onto a fluorescent screen orCCD (charge coupleddevicesee142)Electronsusedinanelectronmicroscopehavewavelengthstypicallyofabout5times10-12mHoweverthe numerical aperture of electron microscopes is considerablysmaller thanthatofanopticalmicroscopetypicallyabout002NonethelessthismeansthataTEMcanresolvetwopointsthatareabout025nmapartThisresolving power is certainly high enough to make out the shape of large molecules

Another type of electron microscope uses a techniqueby which electrons are scattered from the surface of the sample The scattered electrons and then focused as in theTEM to form an image of the surfaceThese so-calledscanningelectronmicroscopes(SEM)havealowerresolving power than TEMrsquos but give very good threedimensional images

The eye

We saw in the last section that the resolving power of the human eye is about 2 times 10-4 rad Suppose that youare looking at car headlights on a dark night and the car

PHYSICS 2015indb 302 150514 337 PM

Wave Phenomena

303

AH

L

is a distance D away If the separation of the headlight is say 15 m then the headlights will subtend an angle θ 15

Dθ =

at your eye Assuming an average wavelength of

500 nm your eye will resolve the headlights into twoseparatesourcesifthisangleequals2times10-4 rad This gives D=75kmInotherwordsifthecarisapproachingyouona straight road then you will be able to distinguish the two headlightsasseparatesourceswhenthecaris75kmawayfrom you Actually because of the structure of the retina and optical defects the resolving power of the average eye is about 6 times 10-4 rad This means that the car is more likely tobe25kmawaybeforeyouresolvetheheadlights

Astronomical telescope

Letusreturntotheexampleofthebinarystarsdiscussedatthe beginning of this section on resolution The stars Kruger A and B form a binary system The average separation of the stars is 14 times 1012mandtheiraveragedistancefromEarthis12times1017mWhenviewedthroughatelescopeonEarththe system will therefore subtend an angle

12

17

14 1012 10

θtimes

=times

= 12times10-5 rad at the objective lens of the telescope Assuming that the average wavelength of the light emitted bythestarsis500nmthenifthetelescopeistoresolvethesystem into two separate images it must have a minimum diameter Dwhere12times10-5 =

7122 500 10D

minustimes times

This gives D=0050m which is about 5 cm So thisparticular system is easily resolved with a small astronomical telescope

Exercise 961 ItissuggestedthatusingtheISASVLBAitwould

bepossibletoldquoseerdquoagrainofriceatadistanceof5000kmEstimatetheresolvingpoweroftheVLBA

2 Thedistancefromtheeyelenstotheretinais20mm The light receptors in the central part of the retinaareabout5times10-6apartDeterminewhetherthe spacing of the receptors will allow for the eye to resolve the headlights in the above discussion whentheyare25kmfromtheeye

3 ThediameterofPlutois23times106 m and its averagedistancefromEarthis60times1012 m EstimatetheminimumdiameteroftheobjectiveofatelescopethatwillenablePlutotobeseenasadisc as opposed to a point source

95 Doppler effect

Essential ideaTheDopplereffectdescribesthephenomenonofwavelengthfrequencyshiftwhenrelativemotionoccurs

Understandingsbull TheDopplereffectforsoundwavesandlightwaves

The Doppler effect for sound waves and light waves

Consider two observers A and B at rest with respect to asoundsourcethatemitsasoundofconstantfrequencyf Clearly both observers will hear a sound of the same frequencyHoweversupposethatthesourcenowmovesat constant speed towards A A will now hear a sound of frequencyfA that is greater than f and B will hear a sound of frequency fB that is less than f This phenomenon is known as the Doppler Effect orDopplerPrinciple after C J Doppler(1803-1853)

The same effect arises for an observer who is either moving towards or away from a stationary source

Figure 930 shows the waves spreading out from a stationary source that emits a sound of constant frequency f The observersAandBhearasoundofthesamefrequency

A B

wavefront

S

Figure 930 Sound waves from a stationary source

Suppose now that the source moves towards A withconstant speed vFigure9310(a)showsasnapshotofthenew wave pattern

NATURE OF SCIENCETechnology Although originally based on physical observations of the pitch of fast moving sources of sound the Doppler effect has an important role in many different areas such as evidence for the expansion of the universe and generating images used in weather reports and in medicine (55)

copy IBO 2014

PHYSICS 2015indb 303 150514 337 PM

Chapter 9

304

AH

L

sourceA B

smallerwavelength

largerwavelength

Figure 931 (a) Sound waves from a moving source

The wavefronts are now crowded together in the direction of travel of the source and stretched out in the opposite direction This is why now the two observers will now hear notesofdifferentfrequenciesHowmuchthewavesbunchtogether and how much they stretch out will depend on the speed vEssentially A

A

cfλ

= and BB

cfλ

= where λA lt λB and v is the speed of sound

IfthesourceisstationaryandAismovingtowardsitthenthe waves from the source incident on A will be bunched up If A is moving away from the stationary source then the waves from the source incident on A will be stretched out

Christian Doppler(1803ndash1853)actuallyappliedtheprinciple(incorrectlyasithappens)totryandexplainthecolourofstars However the Doppler effect does apply to light aswellastosoundIfalightsourceemitsalightoffrequencyf then if it is moving away from an observer the observer willmeasurethelightemittedashavingalowerfrequencythan fSincethesensationofcolourvisionisrelatedtothefrequencyoflight(bluelightisofahigherfrequencythanred light) light emitted by objects moving way from anobserver is often referred to as being red-shifted whereas if the object is moving toward the observer it is referred to as blue-shiftedThisideaisusedinOptionE(Chapter16)

We do not need to consider here the situations where either the source or the observer is accelerating In a situation for example where an accelerating source is approaching astationaryobserverthentheobserverwillhearasoundof ever increasing frequency This sometimes leads toconfusion in describing what is heard when a source approaches passes and then recedes from a stationaryobserver Suppose for example that you are standingon a station platform and a train sounding its whistle is approaching at constant speed What will you hear as the train approaches and then passes through the station As the train approaches you will hear a sound of constant pitch but increasing loudness The pitch of the sound will be greater than if the train were stationary As the train passes through the station you will hear the pitch change at the momentthetrainpassesyoutoasoundagainofconstantpitch The pitch of this sound will be lower than the sound of the approaching train and its intensity will decrease as the train recedes from you What you do not hear is a sound of increasing pitch and then decreasing pitch

The Doppler equations for sound

Although you will not be expected in an IB examination to derivetheequationsassociatedwithaspectsoftheDopplereffectyouwillbeexpectedtoapplythemForcompletenesstherefore thederivationoftheequationsassociatedwiththeDopplereffectasoutlinedaboveisgivenhere

S primeS O

vs

Figure 931 (b)

InFigure931(b)theobserverOisatrestwithrespecttoasourceofsoundSismovingwithconstanspeedvs directly towards O The source is emitting a note of constant frequencyf and the speed of the emitted sound is v

S shows the position of the source ∆t later When the sourceisatresttheninatime∆t the observer will receive f∆t waves and these waves will occupy a distance v∆t ie

v f

λ =f∆tv∆t =

(Because of the motion of the source this number of waves will now occupy a distance (v∆t ndash vs∆t)The lsquonewrsquowavelength is therefore

v ndash vs

fλrsquo =

f∆tv∆t ndash vs∆t

=

If fisthefrequencyheardbyOthen

vfλ

= or

vf

λ = = sv vfminus

From which

s

vf fv v

=minus

Dividingthroughbyv gives

Equation1111

1 sf f v

v=

minus

If the source is moving away from the observer then we have

s

vf fv v

=minus

1

1 sf vv

=+

Equation112

PHYSICS 2015indb 304 150514 337 PM

Wave Phenomena

305

AH

L

We now consider the case where the source is stationary and the observerismoving towardsthesource with speed v0 In this situation the speed of the sound waves as measured by the observer will be v0 + v We therefore have that

v0 + v = ffraslλ = vfλ

times

From which

o1 vf fv

= +

Equation113

If the observerismovingawayfromthesource then

o1 vf fv

= minus

Fromequation(113)wehavethat

o o1 v vf f f f f fv v

∆ = minus = + minus =

Equation114

Thevelocitiesthatwerefertointheaboveequationsarethevelocities with respect to the medium in which the waves fromthesourcetravelHoweverwhenwearedealingwithalight source it is the relative velocity between the source and the observer that we must consider The reason for this is that lightisuniqueintherespectthatthespeedofthelightwavesdoes not depend on the speed of the source All observers irrespective of their speed or the speed of the source will measure the same velocity for the speed of light This is oneof thecornerstonesof theSpecialTheoryofRelativitywhichisdiscussedinmoredetailinOptionH(Chapter18)When applying the Doppler effect to light we aremainlyconcerned with the motion of the source We look here only at the situation where the speed of the source v is much smaller than the speed of light cinfreespace(v ltlt c)Underthesecircumstanceswhenthesourceismovingtowardstheobserverequation111becomes

vf f f fc

minus = ∆ = Equation115

andwhenthesourceismovingawayfromtheobserverequation112becomes vf f f f

cminus = ∆ = minus

Provided that v ltlt c these same equations apply for astationary source and moving observer

We look at the following example and exercise

Example

Asourceemitsasoundoffrequency440HzItmovesinastraight line towards a stationary observer with a speed of 30ms-1Theobserverhearsasoundoffrequency484HzCalculate the speed of sound in air

Solution

We use equation 111 and substitute f= 484 Hz f = 440 Hz

and vs = 30 m s-1

therefore 1

484 440 301v

=minus

such that 30 4401

484vminus = to

give v = 330 m s-1

Example

A particular radio signal from a galaxy is measured as havingafrequencyof139times109HzThesamesignalfromasourceinalaboratoryhasafrequencyof142times109Hz

Suggestwhy thegalaxy ismovingaway fromEarthandcalculateitsrecessionspeed(iethespeedwithwhichitismovingawayfromEarth)

Solution

The fact that the frequency from the moving source is less than that when it is stationary indicates that it is moving away from the stationary observer ie Earth

Using vf fc

∆ = we have

8 96

9

3 10 (142 139) 10 634 10142 10

c fvf∆ times times minus times

= = = timestimes

m s-1

It is usual when dealing with the Doppler effect of light to express speeds as a fraction of c So in this instance we have v = 0021 c

Using the Doppler effectWe have seen in the above example and exercise how the Dopplereffectmaybeusedtomeasuretherecessionspeedof distant galaxies The effect is also used to measure speed inothersituationsHerewewilllookatthegeneralprincipleinvolved in using the Doppler effect to measure speedFigure932showsasource(thetransmitter)thatemitseithersound or em waves of constant frequency f The waves

PHYSICS 2015indb 305 150514 337 PM

Chapter 9

306

AH

L

fromthesourceare incidentonareflectorthat ismovingtowards the transmitter with speed vThereflectedwavesare detected by the receiver placed alongside the transmitter

reector

v

transmitter

receiver

f f

f f

Figure 932 Using the Doppler effect to measure speed

We shall consider the situation where v ltlt c where c is the speed of the waves from the transmitter

For the reflector receiving waves from the transmitterit is effectively an observer moving towards a stationary sourceFromequation(114) itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation116

For the receiver receivingwaves from the reflector it iseffectively a stationary observer receiving waves from a movingsourceFromequation(115)itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation117

Ifweaddequations(116)and(117)wegetthatthetotalDopplershiftatthereceiver∆f is

v vf f f f fc c

minus = ∆ = +

But 1 vf fc

= +

hence

1 v v vf f fc c c

∆ = + +

But since v ltlt cwe can ignore the term2

2

vc when we

expandthebracketintheaboveequation

Therefore we have2vf fc

∆ = Equation118

If v asymp c then we must use the full Doppler equationsHowever for em radiationwewill alwaysonly considersituations in which v ltlt c

Example

Thespeedofsoundinbloodis1500times103 m s-1 Ultrasound offrequency100MHzisreflectedfrombloodflowinginan arteryThe frequencyof the reflectedwaves receivedbackatthetransmitteris105MHzEstimatethespeedofthebloodflowintheartery

Solution

Using equation (118) we have

6 63

2005 10 1015 10

vtimes = times

timesto give v asymp 36 m s-1 (We have assumed that the ultrasound is incident at right angles to the blood flow)

Exercise 971 Judy is standing on the platform of a station A

high speed train is approaching the station in a straight line at constant speed and is sounding its whistleAsthetrainpassesbyJudythefrequencyof the sound emitted by the whistle as heard by Judychangesfrom640Hzto430HzDetermine

(a) thespeedofthetrain

(b) thefrequencyofthesoundemittedbythewhistle as heard by a person on the train (Speedofsound=330ms-1)

2 AgalaxyismovingawayfromEarthwithaspeedof00500cThewavelengthofaparticularspectral line in light emitted by atomic hydrogen inalaboratoryis656times10-7 m Calculate the valueofthewavelengthofthislinemeasuredinthelaboratoryinlightemittedfromasourceofatomic hydrogen in the galaxy

PHYSICS 2015indb 306 150514 337 PM

10 Fields

Contents

101 ndash Describing fields

102 ndash Fields at work

Essential Ideas

Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Similar approaches can be taken in analysing electrical and gravitational potential problems copy IBO 2014

PHYSICS 2015indb 307 150514 337 PM

Chapter 10

308

AH

L

101 Describing fields

Essential idea Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Understandingsbull Gravitationalfieldsbull Electrostaticfieldsbull Electricpotentialandgravitationalpotentialbull Fieldlinesbull Equipotentialsurfaces

Gravitational potentialWehaveseenthatifweliftanobjectofmassm to a height h abovethesurfaceoftheEarththenitsgainingravitationalpotential energy is mghHoweverthisisbynomeansthefullstoryForastartwenowknowthatgvarieswithh and also the expression really gives a difference in potentialenergybetweenthevaluethattheobjecthasattheEarthrsquossurfaceandthevaluethatithasatheighthSowhatwereallyneedisazeropointCanwefindapointwherethepotentialenergyiszeroandusethispointfromwhichtomeasure changes in potential energy

The point that is chosen is in fact infinity At infinity the gravitationalfieldstrengthofanyobjectwillbezeroSoletusseeifwecandeduceanexpressionforthegaininpotentialenergyofanobjectwhenitisldquoliftedrdquofromthesurfaceoftheEarthtoinfinityThisineffectmeansfindingtheworknecessarytoperformthistask

δr

r r δr+

r infin=

mg

AB

Me

Figure 1001 Gravitational forces

Inthediagramweconsidertheworknecessarytomovethe particle of mass m a distance δr in thegravitationalfield of the Earth

The force on the particle at A is FG Mem

r2----------------=

IftheparticleismovedtoB then since δr isverysmallwe can assume that the field remains constant over thedistance ABTheworkδW done againstthegravitationalfieldoftheEarthinmovingthedistanceABis

δWGMem

r2----------------δrndash=

(rememberthatworkdoneagainstaforceisnegative)

TofindthetotalworkdoneW in going from the surface oftheEarthtoinfinitywehavetoaddalltheselittlebitsofworkThisisdonemathematicallybyusingintegralcalculus

WG Me m

r2----------------ndash

rdR

infin

int G Me m 1

r2----- rd

R

infin

intndash G Me m 1r---ndash

R

infinndash= = =

GMe m 0 1R---ndash

ndashndash=

G MemR----------------ndash=

HencewehavewhereRistheradiusoftheEarththattheworkdonebythegravitationalfieldinmovinganobjectofmass m from R(surfaceoftheEarth)toinfinityisgivenby

WGMe m

R---------------- ndash=

We can generalise the result by calculating the worknecessary per unit mass to take a small mass fromthe surface of the Earth to infinity This we call thegravitational potential Vie

Wewouldgetexactlythesameresultifwecalculatedtheworkdonebythefieldtobringthepointmassfrominfinityto the surface of Earth In this respect the formal definition ofgravitationalpotentialatapointinagravitationalfieldis therefore defined as the work done per unit mass in bringing a point mass from infinity to that point

Clearly then the gravitational potential at any point inthe Earthrsquos field distance r from the centre of the Earth (providingr gt R)is

VGMe

r------------ ndash=

Thepotentialisthereforeameasureoftheamountofworkthathastobedonetomoveparticlesbetweenpointsinagravitationalfieldanditsunit is theJkgndash1 We also note thatthepotentialisnegativesothatthepotentialenergy

NATURE OF SCIENCEParadigm shift The move from direct observable actions being responsible for influence on an object to acceptance of a fieldrsquos ldquoaction at a distancerdquo required a paradigm shift in the world of science (23)

copy IBO 2014

PHYSICS 2015indb 308 150514 337 PM

Fields

309

AH

L

aswemoveawayfromtheEarthrsquossurfaceincreasesuntilitreachesthevalueofzeroatinfinity

IfthegravitationalfieldisduetoapointmassofmassmthenwehavethesameexpressionasaboveexceptthatMe is replaced by m andmustalsoexclude thevalueof thepotential at the point mass itself ie at r = 0

WecanexpressthegravitationalpotentialduetotheEarth(orduetoanysphericalmass)intermsofthegravitationalfield strength at its surface

AtthesurfaceoftheEarthwehave

g 0ndash Re GMRe

---------ndash= e

Sothat

g 0Re2 GM= e

Hence at a distance r from the centre of the Earth the gravitationalpotentialVcanbewrittenas

VGMe

r------------ndashg 0 Re

2

r------------ndash= =

The potential at the surface of the Earth

(r = Re)istherefore-g0Re

It is interesting to see how the expression for thegravitational potential ties in with the expression mgh ThepotentialatthesurfaceoftheEarthis-g0Re (see the exampleabove)andataheighthwillbe-g0g 0 Re h+( )ndash ifweassume that g0doesnotchangeover thedistanceh The differenceinpotentialbetweenthesurfaceandtheheighth is therefore g0hSotheworkneededtoraiseanobjectof mass m to a height h is mgh ie m timesdifference ingravitationalpotential

This we have referred to as the gain in gravitationalpotentialenergy(see235)

However this expression can be extended to any twopoints inanygravitationalfieldsuch that ifanobjectofmass mmovesbetween twopointswhosepotentialsareV1 and V2 respectively then the change in gravitationalpotentialenergyoftheobjectism(V1 ndash V2)

Gravitational potential gradient

Let us consider now a region in space where thegravitationalfieldisconstantInFigure912thetwopointsAandBareseparatedbythedistance∆x

A B

direction of uniform gravitational eld of strength I

∆x

Figure 1002 The gravitational potential gradient

The gravitational field is of strength I and is in the direction shownThe gravitational potential at A is V andatBisV +∆V

Theworkdone is takingapointmassmfromAtoB isF∆x = mI∆x

Howeverbydefinitionthisworkisalsoequalto-m∆V

Therefore mI∆x=-m∆V

or VIx

∆= minus

∆Effectivelythissaysthatthemagnitudeofthegravitationalfield strength is equal to the negative gradient of thepotential If I is constant then V is a linear function of x and Iisequaltothenegativegradientofthestraightlinegraph formed by plotting V against x If I is not constant (asusuallythecase)thenthemagnitudeofI at any point in the field can be found by find the gradient of the V-x graph at that point An example of such a calculation can befoundinSection929

For those of you who do HL maths the relationship between field and potential is seen to follow from the expression for the potential of a point mass viz

mV Gr

= minus

2

ddV mG Ir r

minus = + =

Potential due to one or more point masses

Gravitationalpotentialisascalarquantitysocalculatingthepotential due to more than one point mass is a matter of simpleadditionSoforexamplethepotentialV due to the Moon and Earth and a distance x fromthecentreofEarthonastraightlinebetweenthemisgivenbytheexpression

E MM MV Gx r x

= minus + minus

whereME=massofEarthMM= mass of Moon and r = distancebetweencentreofEarthandMoon

PHYSICS 2015indb 309 150514 337 PM

Chapter 10

310

AH

L

Equipotentials and field linesIfthegravitationalpotentialhasthesamevalueatallpointson a surface the surface is said to be an equipotential surfaceSoforexampleifweimagineasphericalshellaboutEarthwhosecentrecoincideswiththecentreofEarththisshellwillbeanequipotentialsurfaceClearlyifwerepresentthegravitationalfieldstrengthbyfieldlinessincethelinesldquoradiaterdquooutfromthecentreofEarththentheselineswillbeatrightanglestothesurfaceIfthefieldlineswerenotnormaltotheequipotentialsurfacethentherewouldbeacomponentof the field parallel to the surfaceThis wouldmean thatpointsonthesurfacewouldbeatdifferentpotentialsandsoitwouldnolongerbeanequipotentialsurfaceThisofcourseholdstrueforanyequipotentialsurface

Figure 1003 shows the field lines and equipotentials fortwopointmassesm

m m

Figure 1003 Equipotentials for two point masses

It is worth noting that we would get exactly the samepattern ifwewere to replace thepointmasseswith twoequalpointcharges(See935)

Escape speedThe potential at the surface of Earth is which meansthattheenergyrequiredtotakeaparticleofmass mfromthesurfacetoinfinityisequalto

ButwhatdoesitactuallymeantotakesomethingtoinfinityWhentheparticleisonthesurfaceoftheEarthwecanthinkofitassittingatthebottomofaldquopotentialwellrdquoasinfigure1004

particle

surface of Earth

innity

GMR---------

FIgure 1004 A potential well

Theldquodepthrdquoofthewellis and if the particle gains an amount of kinetic energy equal to wherem is its massthenitwillhavejustenoughenergytoldquoliftrdquoitoutofthewell

In reality it doesnrsquot actually go to infinity it justmeansthat the particle is effectively free of the gravitationalattractionof theEarthWe say that ithas ldquoescapedrdquo theEarthrsquosgravitationalpullWemeetthisideainconnectionwithmolecular forcesTwomolecules in a solidwill sitat their equilibrium position the separation where therepulsiveforceisequaltotheattractiveforceIfwesupplyjust enough energy to increase the separation of themolecules such that they are an infinite distance apart then intermolecularforcesnolongeraffectthemoleculesandthesolidwillhavebecomealiquidThereisnoincreaseinthekineticenergyofthemoleculesandsothesolidmeltsat constant temperature

Wecancalculatetheescapespeedofanobjectveryeasilyby equating the kinetic energy to the potential energysuch that

12---mvesc ape

2 GMe mRe

----------------=

vesc aperArr2GMe

Re--------------- 2g0 Re = =

Substituting for g0 and Regivesavalueforvescape of about 11kmsndash1 from the surface of the Earth

Youwillnotethattheescapespeeddoesnotdependonthemassoftheobjectsincebothkineticenergyandpotentialenergy are proportional to the mass

IntheoryifyouwanttogetarockettothemoonitcanbedonewithoutreachingtheescapespeedHoweverthiswouldnecessitate anenormousamountof fuel and it islikely that therocketplus fuelwouldbesoheavythat itwouldnevergetoffthegroundItismuchmorepracticalto accelerate the rocket to the escape speed fromEarthorbitandthenintheoryjustlaunchittotheMoon

Example

Use the followingdata todetermine thepotential at thesurface of Mars and the magnitude of the acceleration of free fall

massofMars =64times1023kg

radiusofMars =34times106 m

Determinealsothegravitationalfieldstrengthatadistanceof 68 times 106mabovethesurfaceofMars

PHYSICS 2015indb 310 150514 337 PM

Fields

311

AH

L

Solution

= minus = minus times times = times23

11 76

64 1067 10 ndash13 10

34 10MV GR

minus timestimes

Nkg-1

But V = -g0R

Therefore 7

0 6

13 10 3834 10

VgR

times= minus = =

timesm s-2

To determine the field strength gh at 68 times 106 m above the surface we use the fact that 0 2

Mg GR

= such that GM = g0R2

Therefore 2 2

0h 2 2 2

h h

38 (34) 042(102)

g RGMgR R

times= = = = m s-2

(the distance from the centre is 34 times 106 + 68 times 106 = 102 times 106 m)

Exercise 1011 Thegraphbelowshowshowthegravitational

potentialoutsideoftheEarthvarieswithdistancefrom the centre

0 10 20 30 40 50 60 70

ndash1

ndash2

ndash3

ndash4

ndash5

ndash6

ndash7

r m times 106

V

Jkgndash1

times 10

7

(a) Usethegraphtodeterminethegainingravitationalpotentialenergyofasatelliteofmass200kgasitmovesfromthesurfaceof the Earth to a height of 30 times 107mabovetheEarthrsquossurface

(b) Calculatetheenergyrequiredtotakeit to infinity

(c) DeterminetheslopeofthegraphatthesurfaceoftheEarthmCommenton youranswer

Electric potential energy

The concept of electric potential energy was developedwith gravitational potential energy in mind Just as anobjectnearthesurfaceoftheEarthhaspotentialenergybecause of its gravitational interaction with the Earthsotoothereiselectricalpotentialenergyassociatedwithinteracting charges

Letusfirstlookatacaseoftwopositivepointchargeseachof 1μC that are initially bound together by a thread in a vacuuminspacewithadistancebetweenthemof10cmasshowninFigure1005Whenthethreadiscutthepointchargesinitiallyatrestwouldmoveinoppositedirectionsmovingwithvelocitiesv1 and v2 along the direction of the electrostatic force of repulsion

BEFORE AFTER

v v21

Figure 1005 Interaction of two positive particles

The electric potential energybetweentwopointchargescan be found by simply adding up the energy associated witheachpairofpointchargesForapairofinteractingchargestheelectricpotentialenergyisgivenby

∆U = ∆Ep + ∆Ek = ∆W = ∆Fr = kqQ

____ r2 times r = kqQ

____ r

Becausenoexternalforceisactingonthesystemtheenergyand momentum must be conserved Initially Ek = 0 and Ep =kqQr=9times109 times 1 times 10-1201m=009JWhentheyareagreatdistance fromeachotherEpwillbenegligibleThefinalenergywillbeequaltofrac12mv1

2+frac12mv22=009J

Momentum isalsoconservedand thevelocitieswouldbethe same magnitude but in opposite directions

Electric potential energyismoreoftendefinedintermsofapointchargemovinginanelectricfieldas

lsquothe electricpotential energybetweenany twopoints inan electric field is defined as negativeoftheworkdone by anelectricfieldinmovingapoint electric chargebetweentwolocationsintheelectricfieldrsquo

ΔU = ΔEp = -ΔW = -Fd = qEx

ΔU = ΔEp +ΔEk = ΔW = ΔFr = kqQ r2 x r = kqQ

wherexisthedistancemovedalong(oroppositeto)thedirection of the electric field

Electric potential energyismeasuredinjoule(J)Justasworkisascalarquantitysotooelectricalpotentialenergyis a scalar quantity The negative of the work done by an electric field in moving a unit electric charge between two points is independent of the path taken Inphysicswesaytheelectricfieldisalsquoconservativersquo field

PHYSICS 2015indb 311 150514 337 PM

Chapter 10

312

AH

L

Supposeanexternalforcesuchasyourhandmovesasmallpositive point test charge in the direction of a uniformelectric fieldAs it ismoving itmust be gaining kineticenergyIfthisoccursthentheelectricpotentialenergyofthe unit charge is changing

InFigure1006apointcharge+qismovedbetweenpointsAandBthroughadistancex in a uniform electric field

B

A

x

+ q

Figure 1006 Movement of a positive point charge in a uniform field

InordertomoveapositivepointchargefrompointAtopointBanexternal forcemustbeapplied to thechargeequaltoqE (F = qE)

Since the force is applied through a distance xthennegativeworkhas to bedone tomove the charge because energyis gainedmeaning there is an increaseelectric potential energybetweenthetwopointsRememberthattheworkdoneisequivalenttotheenergygainedorlostinmovingthe charge through the electric field The concept of electric potential energy is only meaningful as the electric field whichgeneratestheforceinquestionisconservative

W F xtimes E q x times= =

θ x cos θ x

Figure 1007 Charge moved at an angle to the field

If a chargemoves at an angle θ to an electric field thecomponent of the displacement parallel to the electric fieldisusedasshowninFigure918

W F x Eq x θcostimes= =

Theelectricpotentialenergyisstoredintheelectricfieldand theelectricfieldwill return theenergy to thepointcharge when required so as not to violate the Law ofconservationofenergy

Electric potential

The electric potential at a point in an electric field is defined as being the work done per unit charge in bringing a small positive point charge from infinity to that point

ΔV = Vinfin ndash Vf = ndash W ___ q

Ifwedesignatethepotentialenergytobezeroatinfinitythen it follows thatelectricpotentialmustalsobezeroat infinity and the electric potential at any point in an electricfieldwillbe

ΔV = ndash W ___ q

NowsupposeweapplyanexternalforcetoasmallpositivetestchargeasitismovedtowardsanisolatedpositivechargeTheexternalforcemustdoworkonthepositivetestchargetomoveittowardstheisolatedpositivechargeandtheworkmustbepositivewhiletheworkdonebytheelectricfieldmustthereforebenegativeSotheelectricpotentialatthatpointmustbepositiveaccordingtotheaboveequationIfanegativeisolatedchargeisusedtheelectricpotentialatapointonthepositivetestchargewouldbenegativePositivepoint chargesof their ownaccordmove fromaplaceofhighelectricpotential toaplaceof lowelectricpotentialNegative point charges move the other way from lowpotentialtohighpotentialInmovingfrompointAtopointBinthediagramthepositivecharge+qismovingfromalowelectricpotentialtoahighelectricpotential

In thedefinitiongiven the termldquoworkperunitchargerdquohas significance If the test charge is +16 times 10-19Cwherethe charge has a potential energy of 32 times 10-17Jthenthepotentialwouldbe32times10-17J+16times10-19C=200JC-1 Now if the charge was doubled the potential wouldbecome 64times10-17 J However the potential per unitchargewouldbethesame

Electric potential is a scalar quantityandithasunitsJC-1 orvoltswhere1voltequalsone joulepercoloumbThevoltallowsustoadoptaunitfortheelectricfieldintermsofthevolt

PreviouslytheunitfortheelectricfieldwasNCndash1

W = qV and F = qEsoW ___ V = F __ E

E = FV ___ W = FV ___ Fm V mndash1

That is the units of the electric field E can also beexpressed as V mndash1

The work done per unit charge in moving a point charge between two points in an electric field is again independent of the path taken

PHYSICS 2015indb 312 150514 337 PM

Fields

313

AH

L

Electric potential due to a point charge

Letus takeapoint rmetres froma chargedobjectThepotentialatthispointcanbecalculatedusingthefollowing

W = ndashFr = ndashqV and F = ndash q1 q2 _____ 4πε0r

2

Therefore

W = ndash q1 q2 _____ 4πε0r

2 times r = ndash q1 q2 _____ 4πε0r

= ndashq1 times q2 _____ 4πε0r

= ndashq1V

That is V q

4πε0r--------------- =

Orsimply

V kqr------

=

Example

Determinehowmuchworkisdonebytheelectricfieldofpointcharge150μCwhenachargeof200μCismovedfrom infinity to apoint 0400m from thepoint charge(Assumenoaccelerationofthecharges)

Solution

The work done by the electric field is W = -qV = -14πε0 times q times (Q rinfin - Q r0400)

W = (- 200 times 10-6 C times 900 times 109 NmC-2 times 150 times 10-6 C) divide 0400 m = - 0675 J

An external force would have to do +0675 J of work

Electric field strength and electric potential gradient

Letus lookbackatFigure1006Supposeagain that thecharge+qismovedasmalldistancebyaforceFfromAtoBsothattheforcecanbeconsideredconstantTheworkdoneisgivenby

∆W F x∆times=

TheforceFandtheelectricfieldEareoppositelydirectedandweknowthat

F = -qE and ΔW = q ΔV

Thereforetheworkdonecanbegivenas

q ΔV = -q E Δx

Therefore E V∆x∆-------

ndash=

The rate of change of potential ΔV at a point with respect to distance Δx in the direction in which the change is maximum is called the potential gradient We say that the electric field = - the potential gradient and the units are Vm-

1 From the equation we can see that in a graph of electric potential versus distance the gradient of the straight line equals the electric field strength

In reality if a charged particle enters a uniform electric field it will be accelerated uniformly by the field and its kinetic energy will increase This is why we had to assume no acceleration in the last worked example

Ek∆ 12---

mv2 q E xsdot sdot q Vx---

xsdot sdot q Vsdot= = = =

Example

Determine how far apart two parallel plates must besituated so that a potential difference of 150 times 102 V produces an electric field strength of 100 times 103 NC-1

Solution

Using E V∆x------- xhArrndash V∆

E-------15 102 Vtimes

100 103 N C 1ndashtimes------------------------------------------= = =

= 150 times 10-1

The plates are 150 times 10-1 m apart

The electric field and the electric potential at a point due to an evenly distributed charge +q on a sphere can be represented graphically as in Figure 1008

r

r

V

E

r0

r0

r0

Charge of +Q evenly distributed over surface

E 14πε0------------ Q

r2----- r r0gt=

V 14πε0------------Q

r----r r0gt=

Figure 1008 Electric field and potential due to a charged sphere

PHYSICS 2015indb 313 150514 337 PM

Chapter 10

314

AH

L

When the sphere becomes charged we know that thechargedistributes itselfevenlyoverthesurfaceThereforeeverypartofthematerialoftheconductorisatthesamepotential As the electric potential at a point is defined as beingnumericallyequaltotheworkdoneinbringingaunitpositivechargefrominfinitytothatpointithasaconstantvalueineverypartofthematerialoftheconductor

Since the potential is the same at all points on the conducting surfacethen∆V∆xiszeroButE=ndash∆V∆xThereforethe electric field inside the conductor is zero There is no electric field inside the conductor

SomefurtherobservationsofthegraphsinFigure1009are

bull OutsidethespherethegraphsobeytherelationshipsgivenasE α 1 r2 and V α 1 r

bull Atthesurfacer = r0Thereforetheelectricfieldandpotentialhavetheminimumvalueforr at this point and this infers a maximum field and potential

bull Insidethespheretheelectricfieldiszerobull Insidethespherenoworkisdonetomoveacharge

fromapointinsidetothesurfaceThereforethereisno potential difference and the potential is the same asitiswhenr = r0

Similargraphscanbedrawnfortheelectricfieldintensityandthe electric potential as a function of distance from conducting parallelplatesandsurfacesandthesearegiveninFigure1009

Potential plot E eld plotE eld

++++

ndashndashndashndash

xx x

+

x x

ndashx x

+ ndashx

x

+ +

x

x

Figure 1009 Electric field and electric potential at a distance from a charged surface

Potential due to one or more point charges

The potential due to one point charge can be determined byusingtheequationformula

V = kq r

Example 1

Determine the electric potential at a point 20 times 10-1 m fromthecentreofan isolatedconducting spherewithapointchargeof40pCinair

Solution

Using the formula

V = kq r we have

V90 10 9times( ) 40 10 12ndashtimes( )times

20 10 1ndashtimes( )------------------------------------------------------------------ 018 V= =

the potential at the point is 180 times 10-1 V

The potential due to a number of point charges can be determined by adding up the potentials due to individual point charges because the electric potential at any point outside a conducting sphere will be the same as if all the charge was concentrated at its centre

Example 2

Threepointchargesofareplacedattheverticesofaright-angledtriangleasshowninthediagrambelowDeterminetheabsolutepotentialat the+20μCchargeduetothetwoothercharges

- 6μC

+3μC +2μC

4 m

3 m

Solution

The electric potential of the +2 μC charge due to the ndash 6 μC charge is

V = (9 times 109 Nm2C-2 times -6 times 10-6 C) divide (radic 32 + 42) m = - 108 times 104 V

PHYSICS 2015indb 314 150514 338 PM

Fields

315

AH

L

Insummarywecanconcludethat

bull No work is done to move a charge along an equipotential

bull Equipotentials are always perpendicular to the electric lines of force

Figure1011and1012showsomeequipotential lines fortwo oppositely charged and identically positive spheresseparated by a distance

+ve ndashve

equipotential lines

Figure 1011 Equipotential lines between two opposite charges

equipotential lines

Figure 1012 Equipotential lines between two charges which are the same

+

ndash ndash ndash ndash

+ + +

40 V

30 V

20 V

10 V

50 V

Figure 1013 Equipotential lines between charged parallel plates

Figure1013showstheequipotentiallinesbetweenchargedparallel plates Throughout this chapter the similarities and differences between gravitational fields and electric fieldshavebeendiscussedTherelationshipsthatexistsbetweengravitationalandelectricquantitiesandtheeffectsofpointmassesandchargesissummarisedinTable1014

The electric potential of the +2 μC charge due to the +3 μC charge is

V = (9 times 109 Nm2C-2 times 3 times 10-6 C) divide 3m = 9 times 103 V

The net absolute potential is the sum of the 2 potentials- 108 times 104 V + 9 times 103 V =- 18 times 103 V

The absolute potential at the point is - 18 times 103 V

Equipotential surfacesRegions in space where the electric potential of a chargedistributionhasaconstantvaluearecalledequipotentials The placeswherethepotentialisconstantinthreedimensionsarecalled equipotential surfacesandwheretheyareconstantintwodimensionstheyarecalledequipotential lines

TheyareinsomewaysanalogoustothecontourlinesontopographicmapsInthiscasethegravitationalpotentialenergy is constant as amassmoves around the contourlines because the mass remains at the same elevationabovetheEarthrsquossurfaceThegravitationalfieldstrengthacts in a direction perpendicular to a contour line

Similarlybecausetheelectricpotentialonanequipotentiallinehas thesamevalueanelectric forcecandonoworkwhena testchargemovesonanequipotentialThereforethe electric field cannot have a component along anequipotentialandthusitmustbeeverywhereperpendiculartotheequipotentialsurfaceorequipotentiallineThisfactmakesiteasytoplotequipotentialsifthelinesofforceorlinesofelectricfluxofanelectricfieldareknown

Forexamplethereareaseriesofequipotentiallinesbetweentwoparallelplateconductorsthatareperpendiculartotheelectric fieldTherewill be a series of concentric circles(eachcirclefurtherapartthanthepreviousone)thatmapouttheequipotentialsaroundanisolatedpositivesphereThe lines of force and some equipotential lines for anisolatedpositivesphereareshowninFigure1010

Lines ofequipotential

Figure 1010 Equipotentials around an isolated positive sphere

PHYSICS 2015indb 315 150514 338 PM

Chapter 10

316

AH

L

Gravitational quantity Electrical quantityQ

uant

ities

g V∆x∆-------ndash= E V∆

x∆-------ndash=

Poin

t mas

ses a

nd ch

arge

s

V Gmr----

ndash= V 14πε0------------q

r---=

g G m

r2-----ndash= E 14πε0------------ q

r2-----=

F Gm1 m2

r2--------------= F 1

4πε0------------

q1q2

r2-----------=

Figure 1014 Formulas

Example

Deduce the electric potential on the surface of a gold nucleus that has a radius of 62 fm

Solution

Using the formula

V = kq r and knowing the atomic number of gold is 79 We will assume the nucleus is spherical and it behaves as if it were a point charge at its centre (relative to outside points)

V = 90 times 109 Nm2C-2 times 79 times 16 times 10-19 C divide 62 times 10-15 m = 18 times 107 V

The potential at the point is 18 MV

Example

Deduce the ionisation energy in electron-volts of theelectron in the hydrogen atom if the electron is in its ground state and it is in a circular orbit at a distance of 53 times 10-11 m from the proton

Solution

This problem is an energy coulombic circular motion question based on Bohrrsquos model of the atom (not the accepted quantum mechanics model) The ionisation energy is the energy required to remove the electron from the ground state to infinity The electron travels in a circular orbit and therefore has a centripetal acceleration The ionisation energy will counteract the coulombic force and the movement of the electron will be in the opposite direction to the centripetal force

Total energy = Ek electron + Ep due to the proton-electron interaction

ΣF = kqQ r2 = mv2 r and as such mv2 = = kqQ r

Therefore Ek electron = frac12 kqQ r

Ep due to the proton-electron interaction = - kqQ r

Total energy = frac12 kqQ r + - kqQ r = -frac12 kqQ r

= - 90 times 109 Nm2C-2 times (16 times 10-19 C)2 divide 53 times 10-11 m = -217 times 10-18 J

= -217 times 10-18 J divide 16 times 10-19 = -136 eV

The ionisation energy is 136 eV

Exercise 1021 ApointchargePisplacedmidwaybetweentwo

identicalnegativechargesWhichoneofthefollowingiscorrectwithregardstoelectricfieldandelectricpotentialatpointP

Electric field Electric potentialA non-zero zeroB zero non-zeroC non-zero non-zeroD zero zero

2 Twopositivechargedspheresaretiedtogetherinavacuumsomewhereinspacewheretherearenoexternal forces A has a mass of 25 g and a charge of20μCandBhasamassof15gandachargeof30μCThedistancebetweenthemis40cmTheyarethenreleasedasshowninthediagram

BEFORE AFTER

v1 v2

A B

(a) Determinetheirinitialelectricpotentialenergy in the before situation

(b) DeterminethespeedofsphereBafterrelease

3 Thediagrambelowrepresentstwoequipotentiallines in separated by a distance of 5 cm in a uniform electric field

+ + + + + + + +

ndash ndash ndash ndash ndash ndash ndash ndash

40 V

20 V 5 cm

PHYSICS 2015indb 316 150514 338 PM

Fields

317

AH

L

Determine the strength of the electric field

4 Thisquestionisabouttheelectricfieldduetoacharged sphere and the motion of electrons in that fieldThediagrambelowshowsanisolatedmetalsphereinavacuumthatcarriesanegativeelectriccharge of 60 μC

ndash

(a) Onthediagramdrawtheconventionalwayto represent the electric field pattern due to the charged sphere and lines to represent threeequipotentialsurfacesintheregionoutside the sphere

(b) Explainhowthelinesrepresentingtheequipotentialsurfacesthatyouhavesketchedindicatethatthestrengthoftheelectricfieldisdecreasingwithdistancefrom the centre of the sphere

(c) Theelectricfieldstrengthatthesurfaceofthe sphere and at points outside the sphere can be determined by assuming that the sphere acts as a point charge of magnitude 60 μC at its centre The radius of the sphere is 25 times 10ndash2 m Deduce that the magnitude of the field strength at the surface of the sphere is 86 times 107 Vmndash1

An electron is initially at rest on the surface of the sphere

(d) (i) Describethepathfollowedbytheelectronasitleavesthesurfaceofthesphere

(ii) Calculatetheinitialaccelerationofthe electron

5 Determinetheamountofworkthatisdoneinmovingachargeof100nCthroughapotentialdifferenceof150times102 V

6 Three identical 20 μC conducting spheres are placedatthecornersofanequilateraltriangleofsides 25 cm The triangle has one apex C pointing upthepageand2baseanglesAandBDeterminetheabsolutepotentialatB

7 Determinehowfaraparttwoparallelplatesmustbesituatedsothatapotentialdifferenceof250 times 102 V produces an electric field strength of 200 times 103 NC-1

8 Thegapbetweentwoparallelplatesis10times10-3mandthereisapotentialdifferenceof10times104 V betweentheplatesCalculate

i theworkdonebyanelectroninmovingfrom one plate to the other

ii thespeedwithwhichtheelectronreachesthe second plate if released from rest

iii theelectricfieldintensitybetweentheplates

9 Anelectronguninapicturetubeisacceleratedby a potential 25 times 103VDeterminethekineticenergygainedbytheelectroninelectron-volts

10 Determine the electric potential 20 times 10-2 m from achargeof-10times10-5 C

11 Determinetheelectricpotentialatapointmid-waybetweenachargeofndash20pCandanotherof+5pConthelinejoiningtheircentresifthecharges are 10 cm apart

12 During a thunderstorm the electric potential differencebetweenacloudandthegroundis10 times 109 V Determine the magnitude of the change in electric potential energy of an electron thatmovesbetweenthesepointsinelectron-volts

13 A charge of 15 μC is placed in a uniform electric fieldoftwooppositelychargedparallelplateswithamagnitudeof14times103 NC-1

(a) Determinetheworkthatmustbedoneagainstthefieldtomovethepointchargeadistance of 55 cm

(b) Calculatethepotentialdifferencebetweenthe final and initial positions of the charge

(c) Determinethepotentialdifferencebetweenthe plates if their separation distance is 15 cm

14 Duringaflashoflightningthepotentialdifferencebetweenacloudandthegroundwas12 times 109 V and the amount of transferred charge was32C

(a) Determinethechangeinenergyofthetransferred charge

(b) Iftheenergyreleasedwasallusedtoacceleratea1tonnecardeduceitsfinalspeed

(c) Iftheenergyreleasedcouldbeusedtomelticeat0degCdeducetheamountoficethatcould be melted

PHYSICS 2015indb 317 150514 338 PM

Chapter 10

318

AH

L

15 SupposethatwhenanelectronmovedfromAtoBin the diagram along an electric field line that the electric field does 36 times 10-19Jofworkonit

Determinethedifferencesinelectricpotential

(a) VB ndash VA(b) VC ndash VA(c) VC ndash VB

16 DeterminethepotentialatpointPthatislocatedatthecentreofthesquareasshowninthediagrambelow

- 6μC

+3μC +2μC

1m

1m

5μC

P

102 Fields at work

Essential idea Similar approaches can be taken inanalysingelectricalandgravitationalpotentialproblems

Understandingsbull Potentialandpotentialenergybull Potentialgradientbull Potentialdifferencebull Escapespeedbull Orbitalmotionorbitalspeedandorbitalenergybull Forcesandinverse-squarelawbehaviour

Orbital motion orbital speed and orbital energy

Although orbital motion may be circular elliptical orparabolic this sub-topic only dealswith circular orbitsThis sub-topic is not fundamentally new physics butan application that synthesizes ideas from gravitationcircularmotiondynamicsandenergy

TheMoonorbitstheEarthandinthissenseitisoftenreferredtoasasatelliteoftheEarthBefore1957itwastheonlyEarthsatelliteHoweverin1957theRussianslaunchedthefirstmanmadesatelliteSputnik1SincethisdatemanymoresatelliteshavebeenlaunchedandtherearenowliterallythousandsofthemorbitingtheEarthSomeareusedtomonitortheweathersome used to enable people to find accurately their position on thesurfaceoftheEarthmanyareusedincommunicationsandnodoubtsomeareusedtospyonothercountriesFigure1015showshowinprincipleasatellitecanbeputintoorbit

Theperson(whosesizeisgreatlyexaggeratedwithrespectto Earth) standing on the surface on the Earth throwssome stonesThe greater the speed with which a stoneis thrown the further it will land from her The pathsfollowedbythethrownstonesareparabolasByastretchof the imaginationwecanvisualisea situation inwhichastoneis thrownwithsuchaspeedthatbecauseof thecurvatureoftheEarthitwillnotlandonthesurfaceoftheEarthbutgointoldquoorbitrdquo(Path4onFigure1015)

NATURE OF SCIENCECommunication of scientific explanations The ability to apply field theory to the unobservable (charges) and the massively scaled (motion of satellites) required scientists to develop new ways to investigate analyse and report findings to a general public used to scientific discoveries based on tangible and discernible evidence (51)

copy IBO 2014

PHYSICS 2015indb 318 150514 338 PM

Fields

319

AH

L

1

2

34Earth

Figure 1015 Throwing a stone into orbit

TheforcethatcausesthestonestofollowaparabolicpathandtofalltoEarthisgravityandsimilarlytheforcethatkeeps the stone in orbit is gravity For circular motionto occur we have seen that a force must act at rightanglestothevelocityofanobjectthatistheremustbeacentripetalforceHenceinthesituationwedescribeherethe centripetal force for circular orbital motion about the EarthisprovidedbygravitationalattractionoftheEarth

We can calculate the speedwithwhich a stonemustbethrowninordertoputitintoorbitjustabovethesurfaceof the Earth

If the stone has mass m and speed v thenwehave fromNewtonrsquos2ndlaw

mv2

RE---------- G

MEm

RE2------------=

whereRE is the radius of the Earth and ME is the mass of the Earth

Bearinginmindthatg 0 GME

RE2--------= then

v g RE 10 64 106timestimes 8 103times= = =

Thatisthestonemustbethrownat8times103m sndash1

Clearly we are not going to get a satellite into orbit soclosetothesurfaceoftheEarthMovingatthisspeedthefrictionduetoairresistancewouldmeltthesatellitebeforeithadtravelledacoupleofkilometresInrealitythereforeasatelliteisputintoorbitabouttheEarthbysendingitattachedtoarocketbeyondtheEarthrsquosatmosphereandthengivingitacomponentofvelocityperpendiculartoaradialvectorfromtheEarthSeeFigure1016

Earth Satellite carried by rocket to hereSatellite orbit

Tangential component of velocity

Figure 1016 Getting a satellite into orbit

Keplerrsquos third law(This work of Kepler and Newtonrsquos synthesis of the work is an excellent example of the scientific method and makes for a good TOK discussion)

In 1627 Johannes Kepler (1571-1630)publishedhis lawsof planetary motion The laws are empirical in natureandwere deduced from the observations of theDanishastronomer Tycho de Brahe (1546-1601)The third lawgives a relationship between the radius of orbit R of a planet and its period TofrevolutionabouttheSunThelawisexpressedmathematicallyas

2

3 constantTR

=

We shall nowuseNewtonrsquos LawofGravitation to showhowitisthattheplanetsmoveinaccordancewithKeplerrsquosthirdlaw

InessenceNewtonwasabletousehislawofgravitytopredict the motion of the planets since all he had to do was factor theF givenby this law intohis second law F = ma to find their accelerations and hence their future positions

InFigure1017theEarthisshownorbitingtheSunandthedistancebetweentheircentresisR

RSun

Earth

F es

F se

Figure 1017 Planets move according to Keplerrsquos third law

Fes is the force that the Earth exerts on the Sun and Fse is the force that the Sun exerts on the Earth The forces are equalandoppositeandtheSunandtheEarthwillactuallyorbitaboutacommoncentreHoweversince theSun issoverymuchmoremassivethantheEarththiscommoncentrewillbeclosetothecentreoftheSunandsowecanregard the Earth as orbiting about the centre of the Sun Theotherthingthatweshallassumeisthatwecanignorethe forces that the other planets exert on the Earth (This wouldnotbeawisethingtodoifyouwereplanningtosend a space ship to theMoon for example)We shallalso assume that we have followed Newtonrsquos exampleandindeedprovedthataspherewillactasapointmasssituated at the centre of the sphere

Kepler had postulated that the orbits of the planets areellipticalbut since theeccentricityof theEarthrsquosorbit issmallweshallassumeacircularorbit

PHYSICS 2015indb 319 150514 338 PM

Chapter 10

320

AH

L

TheaccelerationoftheEarthtowardstheSunisa = Rω2

whereω ω 2πT------=

Hencea R 2π

T------ times

2 4π2 RT 2------------- = =

But the acceleration is given by Newtonrsquos Second Law F = mawhereFisnowgivenbytheLawofGravitationSo in this situation

F maGMs Me

R2-------------------= = butwealsohavethat

a a R 2πT

------ times

2 4π2 RT 2------------- = = and m = Me so that

G MsMe

R2------------------- Me4π2R

T2-------------times

GMs

4π2-----------rArr R3

T2------= =

Butthequantity

isaconstantthathasthesamevalueforeachoftheplanetssowehaveforalltheplanetsnotjustEarththat

R3

T2------ k =

wherekisaconstantWhichisofcourseKeplerrsquosthirdlaw

ThisisindeedanamazingbreakthroughItisdifficulttorefutethe idea that all particles attract each other in accordance withtheLawofGravitationwhenthelawisabletoaccountfortheobservedmotionoftheplanetsabouttheSun

The gravitational effects of the planets upon each othershould produce perturbations in their orbits Such is the predictivepoweroftheUniversalGravitationalLawthatit enabled physicists to compute these perturbations The telescopehadbeeninventedin1608andbythemiddleofthe 18th Century had reached a degree of perfection in design that enabled astronomers to actually measure the orbital perturbations of the planets Their measurements werealways inagreementwith thepredictionsmadebyNewtonrsquoslawHoweverin1781anewplanetUranuswasdiscovered and the orbit of this planet did not fit withthe orbit predicted by Universal Gravitation Such wasthe physicistrsquos faith in theNewtonianmethod that theysuspectedthatthediscrepancywasduetothepresenceofayetundetectedplanetUsingtheLawofGravitationtheFrenchastronomerJLeverrier and the English astronomer J C Adamswereable tocalculate justhowmassive thisnewplanetmustbeandalsowhereitshouldbeIn1846the planetNeptunewas discovered justwhere they hadpredicted Ina similarwaydiscrepancies in theorbitofNeptune ledto thepredictionandsubsequentdiscoveryin1930oftheplanetPlutoNewtonrsquosLawofGravitation

had passed the ultimate test of any theory it is not only abletoexplainexistingdatabutalsotomakepredictions

Satellite energyWhenasatelliteisinorbitaboutaplanetitwillhavebothkineticenergyandgravitationalpotentialenergySupposewe consider a satellite ofmass m that is in an orbit of radius r about a planet of mass M

Thegravitationalpotentialduetotheplanetatdistancer from its centre is

ThegravitationalpotentialenergyofthesatelliteVsat

is therefore GMe m

r----------------ndash

ThatisVsat GMem

r----------------ndash=

Thegravitationalfieldstrengthatthesurfaceoftheplanetisgivenby

g 0GMe

Re2------------=

Hencewecanwrite

Vsat g0 Re

2

rndash=

m

ThekineticenergyofthesatelliteKsatisequaltofrac12mv2wherev is its orbital speed

Byequatingthegravitationalforceactingonthesatellitetoitscentripetalaccelerationwehave

G Mem

r2---------------- mv2

r---------- mv2hArr

GMemr----------------= =

Fromwhich12---

mv2 12---

GMemr----------------times=

= 0 e

2g R

r

2m

Which is actually quite interesting since it shows thatirrespective of the orbital radius the KE is numericallyequal to half the PE Also the total energy E tot K sat Vsat+ 1

2---G Me m

r----------------times G Me m

r----------------ndash + 1

2---G Me m

r----------------ndash= = =of the satelliteisalwaysnegativesince

E tot K sat Vsat+ 12---

G Me mr----------------times

G Me mr----------------ndash

+ 12---

G Me mr----------------ndash= = =

PHYSICS 2015indb 320 150514 338 PM

Fields

321

AH

L

The energies of an orbiting satellite as a function of radial distancefromthecentreofaplanetareshownplottedinFigure1018

12

100806

0402

0ndash02

ndash04

ndash06ndash08ndash10

ndash12

2 4 6 8 10 12

ener

gya

rbita

ry u

nits

distance timesR

kinetic energytotal energy

potential energy

Figure 1018 Energy of an orbiting satellite as a function of distance from the centre of a planet

WeightlessnessSuppose that you are inanelevator(lift)thatisdescendingat constant speedandyou let goof abook thatyouareholdinginyourhandThebookwillfalltothefloorwithaccelerationequaltotheaccelerationduetogravityIfthecablethatsupportstheelevatorweretosnap(asituationthatItrustwillneverhappentoanyofyou)andyounowletgothebookthatyouareholdinginyourotherhandthis book will not fall to the floor - it will stay exactlyin linewithyourhandThis isbecause thebook isnowfallingwith the sameaccelerationas theelevatorandassuch the book cannot ldquocatchrdquo up with the floor of theelevatorFurthermoreifyouhappenedtobestandingonasetofbathroomscalesthescaleswouldnowreadzero-youwouldbeapparentlyweightlessItisthisideaoffreefallthatexplainstheapparentweightlessnessofastronautsin an orbiting satellite These astronauts are in free fall inthesensethattheyareacceleratingtowardsthecentre of the Earth

It is actually possible to define theweight of a body inseveral different waysWe can define it for example asthegravitationalforceexertedonthebodybyaspecifiedobject such as the EarthThiswe have seen thatwe doinlotsofsituationswherewedefinetheweightasbeingequaltomgIfweusethisdefinitionthenanobjectinfreefallcannotbydefinitionbeweightlesssinceitisstillinagravitationalfieldHoweverifwedefinetheweightofanobjectintermsofaldquoweighingrdquoprocesssuchasthereadingonasetofbathroomscaleswhichineffectmeasuresthecontact force between the object and the scales thenclearlyobjectsinfreefallareweightlessOnenowhasto

askthequestionwhetherornotitispossibleForexampletomeasurethegravitationalforceactingonanastronautin orbit about the Earth

WecanalsodefineweightintermsofthenetgravitationalforceactingonabodyduetoseveraldifferentobjectsForexample for an object out in space its weight could bedefined in terms of the resultant of the forces exerted on it bytheSuntheMoontheEarthandalltheotherplanetsin the Solar System If this resultant is zero at a particular pointthenthebodyisweightlessatthispoint

Inviewofthevariousdefinitionsofweightthatareavailabletousitisimportantthatwhenweusethewordldquoweightrdquoweareawareofthecontextinwhichitisbeingused

Example

Calculate the height above the surface of the Earth atwhichageo-stationarysatelliteorbits

Solution

A geo-stationary satellite is one that orbits the Earth in such a way that it is stationary with respect to a point on the surface of the Earth This means that its orbital period must be the same as the time for the Earth to spin once on its axis ie 24 hours

From Keplerrsquos third law we have G Ms

4π2----------- R3

T2------=

That is

RMe

mh

using the fact that the force of attraction between the satellite and the Earth is given by

FG Mem

R2----------------=

and that F = ma

where m is the mass of the satellite and a 4π2R

T2-------------=

we have

GMem

R2---------------- m 4π2R

T2-------------times

GMe

4π2------------rArr R3

T2------= =

PHYSICS 2015indb 321 150514 338 PM

Chapter 10

322

AH

L

Now the mass of the Earth is 60 times 1024 kg and the period T measured in seconds is given by T = 86400 s

So substitution gives R = 42 times 106 m

The radius of the Earth is 64 times 106 m so that the orbital height h is about 36 times 107 m

Example

Calculatetheminimumenergyrequiredtoputasatelliteofmass500kgintoanorbitthatisasaheightequaltotheEarthrsquosradiusabovethesurfaceoftheEarth

Solution

We have seen that when dealing with gravitational fields and potential it is useful to remember that

g 0G M

Re2---------= or g0 Re

2 GM=

The gravitational potential at the surface of the Earth is

g 0ndash Re GMRe

---------ndash=

The gravitational potential at a distance R

from the centre of the Earth is

=

The difference in potential between the surface of the Earth and a point distance R from the centre is therefore

V∆ g 0Re 1ReR------ndash

=

If R 2Re= then V∆g0 Re

2------------=

This means that the work required to ldquoliftrdquo the satellite into orbit is g0Rm where m is the mass of the satellite This is equal to

10 32 106 500timestimestimes = 16000 MJ

However the satellite must also have kinetic energy in order to orbit Earth This will be equal to

g 0 mRe2

2R-----------------g 0mRe

2

4----------------- 8000 MJ= =

The minimum energy required is therefore

24000 MJ

Exercise 1031 The speed needed to put a satellite in orbit does

not depend on

A the radius of the orbitB theshapeoftheorbitC thevalueofgattheorbitD the mass of the satellite

2 EstimatethespeedofanEarthsatellitewhoseorbitis400kmabovetheEarthrsquossurfaceAlsodetermine the period of the orbit

3 Calculatethespeedofa200kgsatelliteorbitingthe Earth at a height of 70 times 106 m

Assume that g = 82 m sndash2 for this orbit

4 TheradiioftwosatellitesXandYorbitingthe Earth are 2r and 8rwherer is the radius of the Earth Calculate the ratio of the periods of revolutionofXtoY

5 A satellite of mass mkgissentfromEarthrsquossurfaceinto an orbit of radius 5RwhereR is the radius of theEarthWritedownanexpressionfor

(a) thepotentialenergyofthesatelliteinorbit

(b) thekineticenergyofthesatelliteinorbit

(c) theminimumworkrequiredtosendthesatellitefromrestattheEarthrsquossurfaceintoits orbit

6 A satellite in an orbit of 10rfallsbacktoEarth(radius r)afteramalfunctionDeterminethespeedwithwhichitwillhittheEarthrsquossurface

7 The radius of the moon is frac14 that of the Earth AssumingEarthandtheMoontohavethesamedensitycomparetheaccelerationsoffreefallatthe surface of Earth to that at the surface of the Moon

8 UsethefollowingdatatodeterminethegravitationalfieldstrengthatthesurfaceoftheMoon and hence determine the escape speed from the surface of the Moon

Mass of the Moon = 73 times 1022kg

RadiusoftheMoon=17times106 m

PHYSICS 2015indb 322 150514 338 PM

Chapter 9

286

AH

L

Boundary ConditionsThe two solutions to the general SHM equation are

and Which solution applies to aparticularsystemdependsasmentionedaboveontheboundary conditions for that system For systems such astheharmonicoscillatorandthesimplependulumtheboundary condition that gives the solution is that the displacement x = x0 when t = 0 For some other systems it might turn out that x = 0 when t =0 This will lead to the solution From a practical pointofviewthetwosolutionsareessentiallythesamefor example when timing the oscillations of a simple pendulumyoumightdecidetostartthetimingwhenthependulumbobpasses through theequilibriumpositionIneffectthetwosolutionsdifferinphaseby

The table in Figure 904 summarises the solutions we have forSHM

x = x 0 cosωt x = x 0 sinωt

ν = minus ν 0 sinωt ν = ν 0 cosωt

ν = ndash ωx 0 sinωt ν = ndash ωx 0 cosωt

ν = plusmn ω radic______

x0 2 ndash x2 ν = plusmn ω radic

______ x0

2 ndash x2

Figure 904 Common equations

We should mention that since the general solution to the SHMequation is there are infactthreesolutionstotheequationThisdemonstratesa fundamental property of second order differential equationsthatoneofthesolutionstotheequationisthesum of all the other solutions This is the mathematical basis of the so-called principle of superposition

SHM is a very good example in which to apply the NewtonianmethoddiscussedinChapter2ieiftheforcesthatactonasystemareknownthenthefuturebehaviourofthesystemcanbepredictedHerewehaveasituationinwhich the force is given by ndashkx From Newtonrsquos second law therefore ndashkx = mawherem is the mass of the system and a istheaccelerationofthesystemHowevertheaccelerationis not constant For those of you who have a mathematical benttherelationbetweentheforceandtheaccelerationis written as

2dd

xkx mt

minus =2

This is what is called a ldquosecond orderdifferentialequationrdquoThesolutionoftheequationgives x as a function of tTheactualsolutionisoftheSHMequationis

x = Pcosωt + Qsinωt where P and Q are constants and ω is theangularfrequencyofthesystemandisequalto

Whether a particular solution involves the sine function or thecosinefunctiondependsontheso-calledlsquoboundaryconditionsrsquo If for example x = x0 (the amplitude)whent =0thenthesolutionisx = x0cosωt

The beauty of thismathematical approach is that oncethe general equation has been solved the solution forall systemsexecutingSHM isknownAll thathas tobeshowntoknowifasystemwillexecuteSHMisthattheaccelerationofthesystemisgivenbyEquation45ortheforceisgivenbyequation47Thephysicalquantitiesthatω will depend on is determined by the particular system Forexampleforaweightofmassm oscillating at the end of a vertically supported spring whose spring constant is k then ω = or from equation 44 For a simple pendulum ω = where l is the length of the pendulum and g is the acceleration of free fall such

that T=2πradic_

l _ g

Examples

ThegraphinFigure905showsthevariationwithtime t of the displacement x ofasystemexecutingSHM(SomequestionpartswerealreadydoneintheSLchapter4)

-10

-8

-6

-4

-2

0

2

4

6

8

10

05 1 15 2 25 3 35t s

xcm

Figure 905 Displacement ndash time graph for SHM

Use the graph to determine the

(i) periodofoscillation(ii) amplitudeofoscillation(iii) maximumspeed(iv) speedatt = 13 s(v) maximumacceleration

Solutions

(i) 20 s (time for one cycle)(ii) 80 cm(iii) Using gives

(remember that ) = 25 cm s-1

PHYSICS 2015indb 286 150514 337 PM

Wave Phenomena

287

AH

L

(iv) v = minusv0sinωt = minus25sin (13π) To find the value of the sine function we have to convert the 13π into degrees (remember ω and hence ωt is measured in radians)

1 deg therefore 13π = 13 times 180 = 234deg therefore v1 = minus25sin (234deg) = +20 cm s ndash1

Or we can solve using ωradic_______

( x0 2 ndash x2 )

from the graph at t = 13 s x = ndash 48 cm

therefore v = π times = 20 cm s-1

(v) Using = π2 times 80 = 79 m s-2

Energy changesWe must now look at the energy changes involved in SHMTodosowewillagainconcentrateontheharmonicoscillator The mass is stationary at x = +x0 (maximumextension)andalsoatx = ndashx0 (maximumcompression)At these two positions the energy of the system is all potential energy and is in fact the elastic potential energy stored in the spring This is the total energy of the system ET and clearly

Equation412

ThatisthatforanysystemperformingSHMtheenergyofthesystemisproportionaltothesquareoftheamplitudeThis is an important result and one that we shall return to when we discuss wave motion

At x =0thespringisatitsequilibriumextensionandthemagnitude of the velocity v of the oscillating mass is a maximum v0TheenergyisallkineticandagainisequaltoET We can see that this is indeed the case as the expression for the maximum kinetic energy Emax in terms of v0 is

Clearly ET and Emax areequalsuchthat

ET = 1 __ 2 kx0 2 = Emax = 1 __ 2 mv0

2

From which

v0 2 = k __ m x0

2(asv0ge0)

Therefore

v0 = radic__

k __ m x0 = ωx0

whichtiesinwiththevelocitybeingequaltothegradientofthedisplacement-timegraph(see415)Asthesystemoscillates there is a continual interchange between kinetic energy and potential energy such that the loss in kinetic energyequalsthegaininpotentialenergyandET = EK + EP

Remembering that wehavethat

Clearly thepotentialenergyEP at any displacement x is given by

At any displacement xthekineticenergyEK is EK = 1__2mv2

Hencerememberingthat

v_______( x0

2 ndash x2 ) wehave

EK = 1__2m 2 ( x0

2 ndash x2 )

Although we have derived these equations for a harmonicoscillatortheyarevalidforanysystemoscillatingwithSHMThe sketch graph in Figure 906 shows the variation with displacement x of EK and EP for one period

displacement

ener

gy potentialkinetic

Figure 906 Energy and displacement

Example

The amplitude of oscillation of a mass suspended by a vertical spring is 80 cm The spring constant of the spring is74Nmndash1Determine

(a) thetotalenergyoftheoscillator

(b) thepotentialandthekineticenergyoftheoscillator at a displacement of 48 cm from equilibrium

PHYSICS 2015indb 287 150514 337 PM

Chapter 9

288

AH

L

Solution

(a) Spring constant k = 74 N m-1 and

x0 = 80 times 10-2 m

ET = frac12 kx02

= frac12 times 74 times 64 times 10-4 = 024 J

(b) At x = 48 cm

Therefore EP = frac12 times 74 times (48 times 10-2)2

= 0085 times 10-4 J

EK = ET minus EP = 024 minus 0085 = 016 J

EXERCISE 911 Which graph shows the relationship between the

acceleration a and the displacement x from the equilibriumpositionofanobjectundergoingsimple harmonic motion

Acceleration

Displacement

Acceleration

Displacement

A B

Acceleration

Displacement

Acceleration

Displacement

C D

2 AnobjectisundergoingsimpleharmonicmotionaboutafixedpointPandthemagnitudeofitsdisplacementfromPisx Which one of the following statements is correct

Magnitudeoftheresultant force

Directionoftheresultant force

A Proportionaltox AwayfrompointP

B Inversely proportional to x AwayfrompointP

C Proportionaltox TowardspointP

D Inversely proportional to x TowardspointP

3 Theangularspeedoftheldquominuterdquohandofananalogue watch is

A π1800rads-1 C π30rads-1

B π60rads-1 D 120rads-1

4 Which of the following is true of the magnitude of the acceleration of the object that is undergoing simple harmonic motion

A It is greatest at the midpoint of the motionB It is greatest at the end points of the motionC It is uniform throughout the motionD Itisgreatestatthemidpointsandthe

endpoints

5 Aparticleoscillateswithsimpleharmonicmotionwith a period T

At time t=0theparticlehasitsmaximumdisplacement Which graph The variation with time t of the kinetic energy of the particle is shown in which diagram below

A

00

Ek

T t

D

00

Ek

T t

C

00

Ek

T t

B

0

Ek

t0 T

6 Figure907belowshowshowthedisplacementofthe magnet of mass 03 kg hanging from a spring varies with time for two oscillations

0 02 04 06 08 10

0

2

-2

-4

4

Disp

lace

men

t (cm

)

Time (s)

Figure 907

PHYSICS 2015indb 288 150514 337 PM

Wave Phenomena

289

AH

L

Using information from this graph calculate the

(a) valueofthespringconstant(b) maximumkineticenergyofthemagnet

7 AsystemisoscillatingwithSHMasdescribedbythe graph in the Figure 908 below

Disp

lace

men

t (cm

)

005 1 15 2 25 3 35

2

-2

-6

-4

4

6

Time (s)

Figure 908

(a) Usethegraphtodeterminethe

(i) periodofoscillation(ii) amplitudeofoscillation(iii) maximumspeed(iv) speedatt=13s(v) maximumacceleration

(b) Statetwovaluesoftforwhenthemagnitudeofthevelocity is a maximum and two values of t for when the magnitude of the acceleration is a maximum

9 Figure 909 shows the relationship that exists between the acceleration and the displacement from the equilibriumpositionforaharmonicoscillator

Acce

lera

tion

(m-2

)

Displacement (cm)0

-4000

-2000

2000

-05 050

Figure 909

(a) Stateandexplaintworeasonswhythegraphopposite indicates that the object is executing simple harmonic motion

(b) Determinethefrequencyofoscillation

92 Single-slit diffraction

Essential idea Single-slitdiffractionoccurswhenawaveis incident upon a slit of approximately the same size as the wavelength

Understandingsbull Thenatureofsingle-slitdiffraction

The nature of single-slit diffractionSingle slit diffraction intensity distribution

When plane wavefronts pass through a small aperture they spread out as discussed in chapter 4 This is an example ofthephenomenoncalleddiffractionLightwavesarenoexception to this and ways for observing the diffraction of light have also been discussed previously

Howeverwhenwelookatthediffractionpatternproducedbylightweobserveafringepatternthatisonthescreenthere is a bright central maximum with ldquosecondaryrdquomaxima either side of it There are also regions where there is no illumination and these minima separate the maxima If we were to actually plot how the intensity of illumination varies along the screen then we would obtain a graph similar to that as in Figure 910

intensity

distance along screen

Figure 910 Intensity distribution for single-slit diffraction

We would get the same intensity distribution if we were to plot the intensity against the angle of diffraction θ(Seenextsection)

NATURE OF SCIENCEDevelopment of theories When light passes through an aperture the summation of all parts of the wave leads to an intensity pattern that is far removed from the geometrical shadow that simple theory predicts (19)

copy IBO 2014

PHYSICS 2015indb 289 150514 337 PM

Chapter 9

290

AH

L

This intensity pattern arises from the fact that each point ontheslitactsinaccordancewithHuygenrsquosprincipleasa source of secondary wavefronts It is the interference between these secondary wavefronts that produces the typical diffraction pattern

Obtaining an expression for the intensity distribution is mathematically a little tricky and it is not something that we are going to attempt here However we can deducea useful relationship from a simple argument In this argument we deal with a phenomenon called Fraunhofer diffractionthatisthelightsourceandthescreenareaninfinitedistanceawayformtheslitThiscanbeachievedwith the set up shown in Figure 911

lens 1 lens 2

single slit

source

screen

Figure 911 Apparatus for viewing Fraunhofer diffraction

The source is placed at the principal focus of lens 1 and the screen isplacedat theprincipal focusof lens2Lens1 ensures that parallel wavefronts fall on the single slit and lens2ensuresthattheparallelraysarebroughttoafocuson the screen The same effect can be achieved using a laser and placing the screen some distance from the slit If the light and screenarenot an infinitedistance from the slitthen we are dealing with a phenomenon called Fresnel diffraction and such diffraction is very difficult to analyse mathematically To obtain a good idea of how the single slit patterncomesaboutweconsiderthediagramFigure912

Xθ2

Yλ

d

P

b

f

screen

θ1

Figure 912 Single slit diffraction

In particular we consider the light from one edge of the slittothepointPwherethispointis justonewavelengthfurther from the lower edge of the slit than it is from the

upper edge The secondary wavefront from the upper edge willtraveladistanceλ2furtherthanasecondarywavefrontfrom a point at the centre of the slitHencewhen thesewavefronts arrive at P theywill be out of phase andwillinterfere destructively The wavefronts from the next point below the upper edge will similarly interfere destructively with the wavefront from the next point below the centre of the slit In this way we can pair the sources across the whole width of the slit If the screen is a long way from the slit then the angles θ1 and θ2becomenearlyequal (If the screen is at infinity then they are equal and the two lines PX and XY are at right angles to each other)FromFigure1116weseethereforethattherewillbeaminimumatPif

λ b θ1 sin= where b is the width of the slit

Howeverbothanglesareverysmallequaltoθ saywhereθ is the angle of diffraction

Soitcanbewritten θ λb---

=

This actually gives us the half-angular width of the central maximum We can calculate the actual width of the maximum along the screen if we know the focal length of the lens focussing the light onto the screen If this is f then we have that

θ df---

= Suchthat d fλb-----

=

To obtain the position of the next maximum in the pattern we note that the path difference is 3

2---λ We therefore divide

theslitintothreeequalpartstwoofwhichwillproducewavefronts that will cancel and the other producing wavefronts that reinforce The intensity of the second maximum is therefore much less than the intensity of thecentralmaximumof theorderof005or120of theoriginalintensity(Muchlessthanonethirdinfactsincethewavefrontsthatreinforcewillhavedifferingphases)

We can also see now how diffraction effects become more and more noticeable the narrower the slit becomes If light of wavelength 430 nm was to pass through a slit of width say10cmandfallonascreen30mawaythenthehalfangular width of the central maximum would be 013 microm

d = f λ

___ b = 3 times 430 times 10 ndash9 ____________ 01 = 013 μm

There will be lots of maxima of nearly the same intensity and the maxima will be packed very closely together (Thefirstminimumoccursatadistanceof012micromfromthe centre of the central maximum and the next occurs effectivelyatadistanceof024microm)Weeffectivelyobservethe geometric pattern Refer to Example

We also see now how for diffraction effects to be noticeable the wavelength must be of the order of the slit width The width of the pattern increases in proportion to the wavelength and decreases inversely with the width of the

PHYSICS 2015indb 290 150514 337 PM

Wave Phenomena

291

AH

L

93 Interference

Essential idea Interference patterns from multiple slits andthinfilmsproduceaccuratelyrepeatablepatterns

Understandingsbull Youngrsquosdouble-slitexperimentbull Modulationoftwo-slitinterferencepatternbyone-slit

diffraction effectbull Multipleslitanddiffractiongratinginterferencepatternsbull Thinfilminterference

Youngrsquos double slit experimentPlease refer to interference in chapter 4 However wewill reiterate the condition for the interference of waves from two sources to be observed The two sources must be coherentthatistheymusthavethesamephaseorthephase difference between them must remain constant

Also to reinforce topics concerning the principle ofsuperpositionandpathdifferenceandphasedifferenceincluded in Figure 913 is another example of two source interference

We can obtain evidence for the wave nature of sound by showing that sound produces an interference pattern Figure 913 shows the set up for demonstrating this

CRO

speaker

speaker

signalgenerator

microphoneX

Y

Figure 913 Interference using sound waves

The two speakers are connected to the same output of the signal generator andplaced about 50 cmapartThesignalgeneratorfrequencyissettoabout600Hzandthe

slit If the slit width is much greater than the wavelength then the width of the central maxima is very small

Insummary

for destructive interference and minima

dsinθ=nλ wheren=123hellip

for constructive interference and maxima

dsinθ=(n+frac12)λ wheren=123hellip

Note that waves from point sources along the slit arrive in phase at the centre of the fringe pattern and constructively interferetoproducethecentralmaximumcalledthezero ordermaximum(n=0)

Example

LightfromalaserisusedtoformasingleslitdiffractionpatternThe width of the slit is 010 mm and the screen is placed 30 m from the slit The width of the central maximum is measured as26cmCalculatethewavelengthofthelaserlight

Solution

Since the screen is a long way from the slit we can use the small angle approximation such that the f is equal to 30 m

The half-width of the central maximum is 13 cm so we have

λ13 10 2ndashtimes( ) 10 10 4ndashtimes( )times

30------------------------------------------------------------------=

To give λ = 430 nm

This example demonstrates why the image of a point source formed by a thin converging lens will always have a finite width

Exercise 921 Aparallelbeamoflightofwavelength500nmis

incidentonaslitofwidth025mmThelightisbroughttofocusonascreenplaced150mfromthe slit Calculate the angular width and the linear width of the central diffraction maximum

2 Lightfromalaserisusedtoformasingleslitdiffraction pattern on a screen The width of the slit is 010 mm and the screen is 30 m from the slit The width of the central diffraction maximum is 26cmCalculatethewavelengthofthelaserlight

3 Determinetheanglethatyouwouldexpecttofindconstructiveinterference(forn=0n=1andn=2) fromasingleslitofwidth20μmandlightofwavelength 600 nm

NATURE OF SCIENCECuriosity Observed patterns of iridescence in animals such as the shimmer of peacock feathers led scientists to develop the theory of thin film interference (15)

Serendipity The first laboratory production of thin films was accidental (15)

copy IBO 2014

PHYSICS 2015indb 291 150514 337 PM

Chapter 9

292

AH

L

microphoneismovedalongthelineXYthatisaboutonemetre from the speakers As the microphone is moved along XY the trace on the cathode ray oscilloscope isseen to go through a series of maxima and minima corresponding to points of constructive and destructive interference of the sound waves

Aninterestinginvestigationistofindhowtheseparationof the points of maximum interference depends on the frequency of the source and also the distance apart ofthe speakers

In the demonstration of the interference between two soundsourcesdescribedaboveifweweretomoveoneofthe speakers from side to side or backwards and forwards the sound emitted from the two speakers would no longer be in phase No permanent points of constructive or destructive interference will now be located since the phase difference between the waves from the two sources is nolongerconstantiethesourcesarenolongercoherent

Light from an incandescent source is emitted with acompletely random phase Although the light from two separate sources will interfere because of the randomlychanging phase no permanent points of constructive or destructive interference will be observed This is why a singleslit isneededintheYoungrsquosdoubleslitexperimentByactingasapointsourceitessentiallybecomesacoherentlight source The light emitted from a laser is also very nearly coherent and this is why it is so easy to demonstrate optical interference and diffraction with a laser

Youngrsquos double slit experiment is one of the great classic experiments of physics and did much to reinforce the wave theoryoflightThomasYoungcarriedouttheexperimentin about 1830

It is essentially the demonstration with the ripple tank and the sound experiment previously described butusing light The essential features of the experiment are shown in Figure 914

screendouble slit

single slit

coloured lter

S 1

S2

Figure 914 Youngrsquos double slit experiment

YoungallowedsunlighttofallontoanarrowsingleslitA few centimetres from the single slit he placed a double slit The slits are very narrow and separated by a distance equal to about fourteen slitwidthsA screen is placed

about ametre from thedouble slitsYoungobservedapatternofmulticolouredldquofringesrdquo in the screenWhenhe placed a coloured filter between the single slit anddouble slit he obtained a pattern that consisted of bright coloured fringes separated by darkness

The single slit essentially ensures that the light falling on the double slit is coherent The two slits then act as the two speakers in the sound experiment or the two dippers in the ripple tank The light waves from each slit interfere and produce the interference pattern on the screen Without thefilterapatternisformedforeachwavelengthpresentinthesunlightHencethemulticolouredfringepattern

You can demonstrate optical interference for yourselfSmoking a small piece of glass and then drawing twoparallel lines on it can make a double slit If you then look throughthedoubleslitatasingletungstenfilamentlampyouwillseethefringepatternByplacingfiltersbetweenthe lamps and the slits you will see the monochromatic fringe pattern

YoucanalsoseetheeffectsofopticalinterferencebylookingatnetcurtainsEachlsquoholersquointhenetactsasapointsourceand the light from all these separate sources interferes and producesquiteacomplicatedinterferencepattern

A laser can also be used to demonstrate optical interference Sincethelightfromthelaseriscoherentitisveryeasytodemonstrate interference Just point the laser at a screen and place a double slit in the path of the laser beam

LetusnowlookattheYoungrsquosdoubleslitexperimentinmore detail The geometry of the situation is shown in Figure915

S1

S2 X

P

Q

D

y

d

screen

θ

θrsquo

Figure 915 The geometry of Youngrsquos double slit experiment

S1andS2 are the two narrow slits that we shall regard as twocoherentmonochromaticpointsourcesThedistancefrom the sources to the screen is D and the distance between the slits is d

The waves from the two sources will be in phase at Q and therewill be abright fringehereWewish tofindtheconditionfortheretobeabrightfringeatPdistancey from Q

PHYSICS 2015indb 292 150514 337 PM

Wave Phenomena

293

AH

L

We note that that D (asymp1metre)isverymuchgreaterthaneither y or d(asymp fewmillimetres)Thismeansthatbothθ and θʹ areverysmallanglesandforintentsandpurposesequal

From the diagram we have that

θ = Dy

And

2S Xd

θʹ =

(Remembertheanglesareverysmall)

Sinceθ asymp θʹ then

2S XyD d

=

ButS2X is the path difference between the waves as they

traveltoPWehavethereforethat

path difference = Dyd

TherewillthereforebeabrightfringeatPif

Dyd

= nλ

Supposethatthereisabrightfringeaty = y1 corresponding

to n = n1 then

Ddy1 = n

1λ

If the next bright fringe occurs at y = y2 this will correspond

to n = n1+1Hence

Ddy2 =(n

1+1)λ

This means that the spacing between the fringes y2 ndash y

1 is

given by

y2 ndash y

1 = d

Dλ

Youngactuallyusethisexpressiontomeasurethewavelengthof the light he used and it is a method still used today

We see for instance that if in a given set up we move the slits closer together then the spacing between the fringes will get greaterEffectivelyourinterferencepatternspreadsoutthatis there will be fewer fringes in a given distance We can also increase the fringe spacing by increasing the distance betweentheslitsandthescreenYouwillalsonotethatfora

givensetupusinglightofdifferentwavelengthsthenldquoredrdquofringeswillspacefurtherapartthanldquobluerdquofringes

In this analysis we have assumed that the slits act as point sources and as such the fringes will be uniformly spaced andofequalintensityAmorethoroughanalysisshouldtakeintoaccountthefinitewidthoftheslits

ReturningtoFigure915weseethatwecanwritethepathdifferenceS2X as

S2X = dtan θʹ

But since θʹ is a small angle the sine and tangent will be nearlyequalsothat

S2X = dsinθʹ

And since θʹ asymp θ then

S2X = dsinθ

The condition therefore for a bright fringe to be found at a point of the screen can therefore be written as

dsinθ = nλ

Insummary

for constructive interference and minima

d sinθ = nλ wheren=123hellip

for destructive interference and maxima

d sinθ = ( n + frac12 )λ wheren=123hellip

Figure 916 shows the intensity distribution of the fringes on the screen when the separation of the slits is large comparedtotheirwidthThefringesareofequalintensityandofequalseparation

Intensity

distance along screen

Figure 916 The intensity distribution of the fringes

It isworthnoting that if the slits are close together theintensity of the fringes is modulated by the intensity distribution of the diffraction pattern of one of the slits

PHYSICS 2015indb 293 150514 337 PM

Chapter 9

294

AH

L

Example

Light of wavelength 500nm is incident on two smallparallel slits separated by 10mmDetermine the anglewherethefirstmaximumisformed

If after passing through the slits the light is brought toa focusona screen15m fromtheslitscalculate theobserved fringe spacing on the screen

Solution

Using the small angle approximation we have

3

7

10105minus

minustimes==

dλθ

= 5 times 10-4 rad

The fringe spacing is given by

3

7

1010551

minus

minustimestimes==

dDy λ

= 075 mm

Modulation of two-slit interference pattern by one-slit diffraction effect

In practice the intensity pattern of the maxima as shown in Figure 917 is not constant but fluctuates while decreasingsymmetrically on either side of the central maximum as shown inFigure917Theintensitypatternisacombinationofboththe single-slit diffractionenvelope and the double slit pattern That is the amplitude of the two-slit interference pattern is modulated(ieadjustedto)byasingleslitdiffractionenvelope

2λD

___ λD

D

___λd

d

___

Double slitnite width

Figure 917 Double slit pattern

ForagivenslitseparationdwavelengthoflightandfixedslittoscreendistancethevariationinintensitydependsonthewidthoftheslitDAlthoughincreasingthewidthoftheslitsincreasestheintensityoflightinthefringesitalsomakesthem less sharp and they become blurred As the slit width is widenedthefringesgraduallydisappearbecausenumerouspoint sources along the widened slit give rise to their own dark and bright fringes that then overlap with each other

Exercise 931 InFigure913thedistancebetweenthespeakers

is050mandthedistancebetweenthelineofthespeakersandthescreenis20mAsthemicrophoneismovedalongthelineXYthedistance between successive points of maximum soundintensityis030mThefrequencyofthesoundwavesis44times103HzCalculateavalueforthe speed of sound

2 Laserlightofwavelength610nmfallsontwoslits10x10-5apartDeterminetheseparationofthefirstordermaximumformedonascreen20maway

3 Twoparallelslits012mmapartareilluminatedwith red light with a wavelength of 600 nm An interferencepatternfallsonascreen15maway

Calculate

(a) thedistancefromthecentralmaximumtothefirstbrightfringe

(b) thedistancetotheseconddarkline

Multiple slit and diffraction grating interference patterns

If we examine the interference pattern produced when monochromatic light passes through a different number of slits we notice that as the number of slits increases the number of observed fringes decreases the spacingbetween them increases and the individual fringes become much sharper We can get some idea of how this comes about by looking at the way light behaves when a parallel beam passes through a large number of slits The diagram for this is shown in Figure 918

θθd

1

2

3

telescope

Figure 918 A parallel beam passing through several slits

The slits are very small so that they can be considered to act as point sources They are also very close together such that d issmall(10ndash6 m)EachslitbecomesasourceofcircularwavefrontsandthewavesfromeachslitwillinterfereLetus

PHYSICS 2015indb 294 150514 337 PM

Wave Phenomena

295

AH

L

consider the light that leaves the slit at an angle θ as shown Thepathdifferencebetweenwave1andwave2isdsinθ and ifthisisequaltoanintegralnumberofwavelengthsthenthe two waves will interfere constructively in this direction Similarly wave 2 will interfere constructively with wave3 at this angle andwave 3with 4 etc across thewholegratingHenceifwelookatthelightthroughatelescopethatisbringittoafocusthenwhenthetelescopemakesanangle θ to the grating a bright fringe will be observed The condition for observing a bright fringe is therefore

dsinθ = mλ

Supposeweuselightofwavelength500nmandsupposethat d = 16 times 10ndash6 m

Obviously we will see a bright fringe in the straight on position θ=0(thezero order)

The next position will be when m =1(thefirst order)andsubstitutionintheaboveequationgivesθ = 18deg

The next position will be when m=2(thesecond order)and this give θ = 38deg

For m=3sinθ is greater than 1 so with this set up we only obtain5fringesonezeroorderandtwoeithersideofthezero order

The calculation shows that the separation of the orders is relatively large At any angles other than 18deg or 38deg the light leaving the slits interferes destructively We can see that the fringes will be sharp since if we move just a small angle away from 18deg the light from the slits will interfere destructively

An array of narrow slits is usually made by cutting narrow transparent lines very close together into the emulsion on aphotographicplate(typically200linespermillimetre)Suchanarrangementiscalledadiffraction grating

The diffraction grating is of great use in examining the spectral characteristics of light sources

All elements have their own characteristic spectrum An element can be made to emit light either by heating it until it is incandescent or by causing an electric discharge through itwhen it is in a gaseous state Yourschool probably has some discharge tubes and diffraction gratingsIfithasthenlookattheglowingdischargetubethroughadiffractiongratingIfforexampletheelementthat you are looking has three distinct wavelengths then each wavelength will be diffracted by a different amount inaccordancewiththeequationdsinθ = nλ

Alsoiflaserlightisshonethroughagratingontoascreenyou will see just how sharp and spaced out are the maxima By measuring the line spacing and the distance of the screen fromthelaserthewavelengthofthelasercanbemeasured

If your school has a set of multiple slits say from a single slit toeight slits then it isalsoaworthwhileexercise toexamine how the diffraction pattern changes when laser light is shone through increasing numbers of slits

Consider a grating of NparallelequidistantslitsofwidthaseparatedbyanopaqueregionofwidthD We know that light diffracted from one slit will be superimposed and interfere with light diffracted from all the other slits The intensityprofile of the resultant wave will have a shape determined by

bull theintensityoflightincidentontheslitsbull thewavelengthofthislightbull theslitwidthbull theslitseparation

Typical intensity profiles formed when a plane wavemonochromaticlightatnormalincidenceononetwoandsix slits are shown in Figure 919 In this example the slit widthisonequarterofthegratingspacing

Six slits sharperfringes

d = 4a

I α N 2

(N ndash 2)= 4 subsidiary maxima

order0 4321

I = 36

Two slits

order0

single-slitdiraction

envelope

d = 4a4th principal maximum missing

principa maxima

I = 4 (nominal units)

4321

intensity

order

ISS

One slit

0 1

single-slitdiraction

pattern

N = number of slitsslit width a

Figure 919 Intensity profiles

PHYSICS 2015indb 295 150514 337 PM

Chapter 9

296

AH

L

1 For Nge2thefringepatterncontainsprincipalmaximathat are modulated by the diffraction envelope

2 ForNgt2thefringepatterncontains(Nndash2)subsidiary maxima

3 Where the grating spacing d is four time the slit width a(4xa =d)thenthefourthprincipalmaximumismissing from the fringe pattern Where the grating spacingdisthreetimetheslitwidtha(3xa =d)thenthe third principal maximum is missing from the fringepatternandsoon

4 Increasing N increases the absolute intensities of the diffraction pattern (I = N2)

If white light is shone through a grating then the central image will be white but for the other orders each will be spread out intoacontinuousspectrumcomposedofaninfinitenumberof adjacent images of the slit formed by the wavelength of the different wavelengths present in the white light At any given point in the continuous spectrum the light will be very nearly monochromatic because of the narrowness of the images of the slit formed by the grating This is in contrast to the doubleslitwhereifwhitelightisusedtheimagesarebroadand the spectral colours are not separated

In summary as is the case of interference at a doubleslit there is a path difference between the rays fromadjacent slits of d sinθ Where there is a whole number of wavelengthsconstructiveinterferenceoccurs

For constructive interference and minima

d sinθ = mλ wheren=123hellip

When there is a path difference between adjacent rays of halfawavelengthdestructiveinterferenceoccurs

For and maxima

d sinθ = ( m + frac12 )λ wheren=123hellip

Exercise 941 Lightfromalaserisshonethroughadiffraction

grating on to a screen The screen is a distance of 20mfromthelaserThedistancebetweenthecentraldiffractionmaximumandthefirstprincipalmaximum formed on the screen is 094 m The diffractiongratinghas680linespermmEstimatethe wavelength of the light emitted by the laser

2 Monochromaticlightfromalaserisnormallyincident on a six-slit grating The slit spacing is three times the slit width An interference patternisformedonascreen2montheotherside of the grating

(a) Determineandexplaintheorderoftheprincipal maxima missing from the fringe pattern

(b) Howmanysubsidiarymaximaaretherebetween the principal maxima

(c) Howmanyprincipalmaximaarethere

(i) Betweenthetwofirstorderminimain the diffraction envelope

(ii) Betweenthefirstandsecondorderminima in the diffraction envelope

(d) Drawthepatternformedbythebrightfringesin(c)

(e) Sketchtheintensityprofileofthefringepattern

Thin film interferenceYoumightwellbe familiarwith the colouredpatternoffringes that can be seen when light is reflected off thesurfaceofwateruponwhichathinoilfilmhasbeenspiltorfromlightreflectedfrombubblesWecanseehowthesepatternsarisebylookingatFigure920

Extendedlight source air

A

B

C

12

3

4

Oil lm

Water

Figure 920 Reflection from an oil film

Consider light from an extended source incident on a thinfilmWealsoconsiderawave fromonepointofthe source whose direction is represented by the ray shownSomeofthislightwillbereflectedatAandsometransmitted through the filmwhere somewill again bereflectedatB(somewillalsobetransmittedintotheair)SomeofthelightreflectedatBwillthenbetransmittedatCandsomereflectedandsoonIfweconsiderjustrays1and2thenthesewillbenotbeinphasewhenthearebrought to a focus by the eye since they have travelled

PHYSICS 2015indb 296 150514 337 PM

Wave Phenomena

297

AH

L

different distances If the path difference between them is an integral number of half wavelengths then we might expect the twowaves to be out of phaseHowever ray1 undergoes a phase change of π on reflection but ray2 does not since it is reflected at a boundary betweena more dense and less dense medium (See chapter 4)Henceifthepathdifferenceisanintegralnumberofhalf-wavelengthsrays1and2willreinforceieray1and2areinphaseHowever rays357etcwillbeoutofphasewithrays246etcbutsinceray2ismoreintensethanray3andray4moreintensethanray5thesepairswillnot cancel out so there will be a maximum of intensity

Ifthepathdifferenceissuchthatwave1and2areoutofphasesincewave1ismoreintensethanwave2theywillnotcompletelyannulHoweveritcanbeshownthattheintensitiesofwaves2345hellipaddtoequaltheintensityofwave1Sincewaves345hellipareinphasewithwave2there will be complete cancellation

The path difference will be determined by the angle at which ray1 is incident andalsoon the thickness (andtheactualrefractiveindexaswell)ofthefilmSincethesourceisanextendedsourcethelightwillreachtheeyefrom many different angles and so a system of bright and darkfringeswillbeseenYouwillonlyseefringesifthefilmisverythin(exceptifviewedatnormalincidence)sinceincreasingthethicknessofthefilmwillcausethereflected rays to get so far apart that they will not becollected by the pupil of the eye

From the argument above to find the conditions forconstructive and destructive interferenceweneedonlyfindthepathdifferencebetweenray1andray2Figure921showsthe geometry of the situation

A BE

F

C

D

1 2

φrsquo φrsquo

φ

φ

d

Figure 921 The geometry of interference

Thefilmisofthicknessd and refractive index n and the light has wavelength λ If the line BF is perpendicular to ray 1 then the optical path difference (opd) between ray 1 andray2whenbroughttoafocusis

opd = n(AC+CB)ndashAF

We have to multiple by the refractive index for the path travelledbythelightinthefilmsincethelighttravelsmoreslowly in thefilm If the light travels say adistancex in a material of refractive index n then in the time that it takes to travelthisdistancethelightwouldtraveladistancenx in air

IfthelineCEisatrightanglestoray2thenweseethat

AF = nBE

FromthediagramAC=CDsowecanwrite

opd = n(CD+CB)ndashnBE=nDE

Alsofromthediagramweseethatwhereφ is the angle of refraction

DE=2cosφ

From which opd=2ndcosφ

Bearinginmindthechangeinphaseofray1onreflectionwe have therefore that the condition for constructive interference is

2nd φcos m 12---+

λ m 1 2 hellip = =

Andfordestructiveinterference 2ndcosφ = mλ

Eachfringecorrespondstoaparticularopd for a particular value of the integer m and for any fringe the value of the angle φ is fixedThismeans that itwill be in the formof an arc of a circle with the centre of the circle at the point where the perpendicular drawn from the eye meets the surface of the film Such fringes are called fringesofequal inclinationSince theeyehasa smallaperturethesefringesunlessviewedatneartonormalincidence (φ = 0) will only be observed if the film is very thinThisisbecauseasthethicknessofthefilmincreasesthereflected rayswill get further and further apart and sovery few will enter the eye

Ifwhitelightisshoneontothefilmthenwecanseewhywe get multi-coloured fringes since a series of maxima and minima will be formed for each wavelength present in the white light However when viewed at normalincidenceitispossiblethatonlylightofonecolourwillundergoconstructiveinterferenceandthefilmwilltakeon this colour

PHYSICS 2015indb 297 150514 337 PM

Chapter 9

298

AH

L

Thickness of oil films

The exercise to follow will help explain this use

Non-reflecting films

A very important but simple application of thin filminterferenceisintheproductionofnon-reflectingsurfaces

Athinfilmofthicknessd and refractive index n1 is coated onto glass of refractive index n where n1 lt n Light ofwavelength λ that is reflected at normal incidence willundergo destructive interference if 12

2n d λ

= thatis

14d

nλ

=

(remember that there will now no phase change at the glass interface ie we have a rare to dense reflection)

The use of such films can greatly reduce the loss oflightby reflection at the various surfacesof a systemoflenses or prisms Optical parts of high quality systemsare usually all coatedwithnon-reflecting films in orderto reduce stray reflections The films are usually madebyevaporatingcalciumormagnesiumfluorideonto thesurfacesinvacuumorbychemicaltreatmentwithacidsthat leave a thin layer of silica on the surface of the glass ThecoatedsurfaceshaveapurplishhuebyreflectedlightThis is because the condition for destructive interference fromaparticularfilmthicknesscanonlybeobtainedforone wavelength The wavelength chosen is one that has a value corresponding to light near the middle of the visible spectrumThismeansthatreflectionofredandvioletlightis greater combining to give the purple colour Because of the factor cosφatanglesotherthannormalincidencethepath difference will change butnotsignificantlyuntilsayabout30deg(egcos25deg=090)

It should be borne in mind that no light is actually lost by anon-reflectingfilmthedecreaseofreflectedintensityiscompensated by increase of transmitted intensity

Non-reflecting films can be painted onto aircraft tosuppressreflectionofradarThethicknessofthefilmis determined by

4nd λ

= where λ is the wavelength of

the radar waves and ntherefractiveindexofthefilmat this wavelength

Example

A plane-parallel glass plate of thickness 20 mm isilluminated with light from an extended source The refractiveindexoftheglassis15andthewavelengthofthelight is 600 nm Calculate how many fringes are formed

Solution

We assume that the fringes are formed by light incident at all angles from normal to grazing incidence

At normal incidence we have 2nd = mλ

From which

m 2 15 2 10 3ndashtimestimestimes6 10 7ndashtimes

------------------------------------------- 10 000= =

At grazing incidence the angle of refraction φ is in fact the critical angle

Therefore φφ arcsin 115------- 42 deg= =

ie cosφ = 075

At grazing incidence 2ndcosφ = m λ

From which (and using cos(sinndash1(115) = 075)

m 2 15 2 10 3ndash 075timestimestimestimes6 10 7ndashtimes

------------------------------------------------------------ 7500= =

The total number of fringes seen is equal to m ndash m = 2500

Exercise 95When viewed from above the colour of an oil film onwater appears red in colour Use the data below to estimate theminimumthicknessoftheoilfilm

average wavelength of red light = 630 nm

refractiveindexofoilforredlight=15

refractive index of water for red light = 13

PHYSICS 2015indb 298 150514 337 PM

Wave Phenomena

299

AH

L

94 Resolution

Essential idea Resolution places an absolute limit on the extent to which an optical or other system can separate images of objects

Understandings

bull Thesizeofadiffractingaperturebull Theresolutionofsimplemonochromatic

two-source systems

The size of a diffracting apertureOur discussion concerning diffraction so far has been for rectangular slits When light from a point source enters a smallcircularapertureitdoesnotproduceabrightdotasanimagebutacirculardiscknownasAiryrsquosdiscsurroundedbyfainterconcentriccircularringsasshowninFigure922

Laser

Figure 922 Diffraction at a circular aperture

Diffractionatanapertureisofgreatimportancebecausetheeye and many optical instruments have circular apertures

What is the half-angular width of the central maximum of the diffraction formed by a circular aperture This is not easy to calculate since it involves some advanced mathematics The problemwasfirstsolvedbytheEnglishAstronomerRoyalGeorgeAiryin1835whoshowedthatforcircularapertures

θ 122 λb--------------= where b is the diameter of the aperture

Example 2

Inthefollowingdiagramparallellightfromadistantpointsource(suchasastar)isbroughttofocusonthescreenSbyaconverginglens(the lens is shown as a vertical arrow)

Thefocal length(distance fromlens toscreen) is25cmand the diameter of the lens is 30 cm The wavelength of thelightfromthestaris560nmCalculatethediameterofthe diameter of the image on the screen

Solution

The lens actually acts as a circular aperture of diameter 30 cm The half angular width of central maximum of the diffraction pattern that it forms on the screen is

75

2

122 122 56 10 23 1030 10b

λθminus

minusminus

times times= = = times

times rad

The diameter of the central maxima is therefore

25 times 10-2 times 23 times 10-5

= 57 times 10-6 m

Although this is small it is still finite and is effectively the image of the star as the intensity of the secondary maxima are small compared to that of the central maximum

The resolution of simple monochromatic two-source systemsThe astronomers tell us that many of the stars that we observe with the naked eye are in fact binary stars That iswhatwe see as a single star actually consistsof twostarsinorbitaboutacommoncentreFurthermoretheastronomerstellusthatifweusealdquogoodrdquotelescopethenwewillactuallyseethetwostarsThatiswewillresolve the single point source into its two component parts Sowhatisitthatdetermineswhetherornotweseethetwo stars as a single point source ie what determines whether or not two sources can be resolved It canrsquot just bethatthetelescopemagnifiesthestarssinceiftheyareacting as point sources magnifying them is not going to make a great deal of difference

Ineachofoureyesthereisanaperturethepupilthroughwhich the light enters This light is then focused by the eye lens onto the retina But we have seen that when light passes through an aperture it is diffracted and so when we look at

NATURE OF SCIENCEImproved technology The Rayleigh criterion is the limit of resolution Continuing advancement in technology such as large diameter dishes or lenses or the use of smaller wavelength lasers pushes the limits of what we can resolve (18)

copy IBO 2014

PHYSICS 2015indb 299 150514 337 PM

Chapter 9

300

AH

L

apointsourceadiffractionpatternwillbeformedontheretina If we look at two point sources then two diffraction patterns will be formed on the retina and these patterns will overlap The width of our pupil and the wavelength of the light emitted by the sources will determine the amount by which they overlap But the degree of overlap will also depend on the angular separation of the two point sources WecanseethisfromFigures923

pupilretina

S1

S2

P2

P1

θ

Figure 923

LightfromthesourceS1 enters the eye and is diffracted by the pupil such that the central maximum of the diffraction patternisformedontheretinaatP1SimilarlylightfromS2producesamaximumatP2 If the two central maxima are well separated then there is a fair chance that we will see the two sources as separate sources If they overlap then we will not be able to distinguish one source from another From the diagram we see as the sources are movedclosertotheeyethentheangleθ increases and so does the separation of the central maxima

Figures924925926and927showsthedifferentdiffractionpatternsandtheintensitydistributionthatmightresultonthe retina as a result of light from two point sources

S1 S21 Well resolved

Figure 924 Very well resolved

S1 S22 Well resolved

Figure 925 Well resolved

S1 S23 Just resolved Rayleigh criterion Minimum of S2 coincides with maximum peak of S1

Figure 926 Just resolved

4 Not resolved

Figure 927 Not resolved

We have suggested that if the central maxima of the two diffraction patterns are reasonably separated then we should be able to resolve two point sources In the late 19th century Lord Rayleigh suggested by how much they should be separated in order for the two sources to be just resolved Ifthecentralmaximumofonediffractionpattern coincides with the firstminima of the otherdiffraction pattern then the two sources will just beresolved This is known as the Rayleigh Criterion

Figure926showsjustthissituationThetwosourcesarejust resolved according to the Rayleigh criterion since the peak of the central maximum of one diffraction pattern coincideswiththefirstminimumoftheotherdiffractionpattern This means that the angular separation of the peaks of the two central maxima formed by each source is just the half angular width of one central maximum ie

θ λb---

=

where b is the width of the slit through which the light fromthesourcespassesHoweverweseefromFigure1117that θ is the angle that the two sources subtend at the slit Henceweconcludethattwosourceswillberesolvedbyaslit if the angle that they subtend at the slit is greater than or equaltoλfraslb

PHYSICS 2015indb 300 150514 337 PM

Wave Phenomena

301

AH

L

Sofarwehavebeenassumingthattheeyeisarectangularslit whereas clearly it is a circular aperture and so we must use the formula

θ 122 λb--------------=

As mentioned above the angle θ is sometimes called the resolving power but should more accurately be called the minimum angle of resolution (θmin)

Clearly the smaller θ the greater will be the resolving power

The diffraction grating is a useful for differentiating closely spacedlinesinemissionspectraLikeaprismspectrometerit can disperse a spectrum into its components but it is better suited because it has higher resolution than the prism spectrometer

If λ1 and λ2 are two nearly equal wavelengths that canbarelybedistinguishedtheresolvanceorresolvingpowerofthegratingisdefinedas

R = λ _______ (λ2 ndash λ 1) = λ ___ Δλ

Where λ asymp λ1 asymp λ2

Therefore thediffractiongratingwithahighresolvancewill be better suited in determining small differences in wavelength

If the diffraction grating has Nlinesbeingilluminateditcan be shown that the resolving power of the mth order diffraction is given by

R = mN

Obviouslytheresolvancebecomesgreaterwiththeordernumber m and with a greater number of illuminated slits

Example

A benchmark for the resolving power of a grating is the sodium doublet in the sodium emission spectrum Two yellow lines in itsspectrumhavewavelengthsof58900nmand58959nm

(a) Determinetheresolvanceofagratingifthegivenwavelengths are to be distinguished from each other

(b) Howmanylinesinthegratingmustbeilluminated in order to resolve these lines in the second order spectrum

Solution

(a) R = λ ___ Δλ = 58900nm __________________ (58900ndash58959)nm

= 589nm _______ 059nm = 10 times 103

(b) N = R __ m = 10 times 10 3 ________ 2 =50times102 lines

It has been seen that diffraction effectively limits the resolving power of optical systems This includes such systems as the eye telescopes and microscopes Theresolving power of these systems is dealt with in the next section This section looks at links between technology and resolving power when looking at the very distant and when looking at the very small

Radio telescopes

The average diameter of the pupil of the human eye is about 25 mm This means that the eye will justresolve two point sources emitting light of wavelength 500 nm if their angular separation at the eye is 7

43

50 10122 24 1025 10

θminus

minusminus

times= times = times

times rad

If the eye were to be able to detect radio waves of wavelength 015m then tohave the same resolvingpower thepupilwould have to have a diameter of about 600 m Clearly this is nonsense but it does illustrate a problem facingastronomers who wish to view very distant objects such as quasarsandgalaxies thatemitradiowavesConventionalradio telescopes consist of a large dish typically 25 min diameter Even with such a large diameter the radiowavelength resolving power of the telescope is much less thantheopticalresolvingpowerofthehumaneyeLetuslook at an example

Example

The Galaxy Cygnus A can be resolved optically as an elliptically shaped galaxy However it is also a strongemitterof radiowavesofwavelength015mTheGalaxyisestimatedtobe50times1024mfromEarthUseofaradiotelescope shows that the radio emission is from two sources separated by a distance of 30 times 1021m Estimate thediameterofthedishrequiredtojustresolvethesources

Solution

The angle θ that the sources resolve at the telescope is given by

θ 21

424

30 10 60 1050 10

θ minustimes= = times

times rad

and 4

122 122 015 300060 10

d λθ minus

times= = =

timesm = 30 km

A radio telescope dish of this size would be impossible tomake letalonesupportThisshowsthatasingledishtype radio telescope cannot be used to resolve the sources and yet they were resolved To get round the problemastronomers use two radio telescopes separated by a large

PHYSICS 2015indb 301 150514 337 PM

Chapter 9

302

AH

L

distance The telescopes view the same objects at the same time and the signals that each receive from the objects are simultaneously superimposed The result of the superposition of the two signals is a two-slit interference pattern The pattern has much narrower fringe spacing than that of the diffraction pattern produced by either telescope on its own hence producing a much higherresolvingpowerWhentelescopesareusedlikethistheyare called a stellar interferometer

InSocorroinNewMexicothereisastellarinterferometerthat consists of 27 parabolic dishes each of diameter25 m arranged in a Y-shape that covers an area of 570km2Thisisaso-calledVeryLargeArray(VLA)Evenhigher resolution can be obtained by using an array of radio telescopes in observatories thousands of kilometres apart A system that uses this so-called technique oflsquovery-long-baseline interferometryrsquo (VLBI) is known asalsquovery-long-baselinearrayrsquo(VLBA)WithVLBAaradiowavelength resolving power can be achieved that is 100 timesgreaterthanthebestopticaltelescopesEvenhigherresolving power can be achieved by using a telescope that isinasatelliteorbitingEarthJapanrsquosInstituteofSpaceandAstronautical Science (ISAS) launched sucha system inFebruary 1997TheNationalAstronomicalObservatoryof Japan the National Science Foundationrsquos NationalRadio Astronomy Observatory (NRAO) the CanadianSpaceAgency theAustraliaTelescopeNationalFacilitythe European VLBI Network and the Joint Instituteback this project forVeryLongBaseline Interferometryin EuropeThis project is a very good example of howInternationalismcanoperateinPhysics

Electron microscope

Telescopes are used to look at very distant objects that are very large but because of their distance fromus appearverysmallMicroscopesontheotherhandareusedtolookat objects that are close to us but are physically very small Aswehave seen justmagnifyingobjects that ismakingthemappearlargerisnotsufficientonitsowntogaindetailabouttheobjectfordetailhighresolutionisneeded

Figure928isaschematicofhowanoptical microscope is usedtoviewanobjectandFigure929isaschematicofatransmission electron microscope(TEM)

Optical lens system

bright light source

glass slide containing specimen (object)

eye

Figure 928 The principle of a light microscope

magnetic lens system

electron source

thin wafer of material

ScreenCCD

Figure 929 The principle of an electron microscope

In the optical microscope the resolving power isdetermined by the particular lens system employed and the wavelength λ of the light used For example twopoints in the sample separated by a distance d will just be resolved if

2d mλ=

where m is a property of the lens system know as the numerical aperture In practice the largest value of m obtainable is about 16 Hence if the microscope slideis illuminatedwith light ofwavelength 480 nm a goodmicroscope will resolve two points separated by a distance dasymp15times10-7masymp015micromPointscloser together thanthiswillnotberesolvedHoweverthisisgoodenoughtodistinguishsomevirusessuchastheEbolavirus

Clearlythesmallerλ the higher the resolving power and this is where the electron microscope comes to the fore The electron microscope makes use of the wave nature of electrons(see1315)IntheTEMelectronspassthrougha wafer thin sample and are then focused by a magnetic field onto a fluorescent screen orCCD (charge coupleddevicesee142)Electronsusedinanelectronmicroscopehavewavelengthstypicallyofabout5times10-12mHoweverthe numerical aperture of electron microscopes is considerablysmaller thanthatofanopticalmicroscopetypicallyabout002NonethelessthismeansthataTEMcanresolvetwopointsthatareabout025nmapartThisresolving power is certainly high enough to make out the shape of large molecules

Another type of electron microscope uses a techniqueby which electrons are scattered from the surface of the sample The scattered electrons and then focused as in theTEM to form an image of the surfaceThese so-calledscanningelectronmicroscopes(SEM)havealowerresolving power than TEMrsquos but give very good threedimensional images

The eye

We saw in the last section that the resolving power of the human eye is about 2 times 10-4 rad Suppose that youare looking at car headlights on a dark night and the car

PHYSICS 2015indb 302 150514 337 PM

Wave Phenomena

303

AH

L

is a distance D away If the separation of the headlight is say 15 m then the headlights will subtend an angle θ 15

Dθ =

at your eye Assuming an average wavelength of

500 nm your eye will resolve the headlights into twoseparatesourcesifthisangleequals2times10-4 rad This gives D=75kmInotherwordsifthecarisapproachingyouona straight road then you will be able to distinguish the two headlightsasseparatesourceswhenthecaris75kmawayfrom you Actually because of the structure of the retina and optical defects the resolving power of the average eye is about 6 times 10-4 rad This means that the car is more likely tobe25kmawaybeforeyouresolvetheheadlights

Astronomical telescope

Letusreturntotheexampleofthebinarystarsdiscussedatthe beginning of this section on resolution The stars Kruger A and B form a binary system The average separation of the stars is 14 times 1012mandtheiraveragedistancefromEarthis12times1017mWhenviewedthroughatelescopeonEarththe system will therefore subtend an angle

12

17

14 1012 10

θtimes

=times

= 12times10-5 rad at the objective lens of the telescope Assuming that the average wavelength of the light emitted bythestarsis500nmthenifthetelescopeistoresolvethesystem into two separate images it must have a minimum diameter Dwhere12times10-5 =

7122 500 10D

minustimes times

This gives D=0050m which is about 5 cm So thisparticular system is easily resolved with a small astronomical telescope

Exercise 961 ItissuggestedthatusingtheISASVLBAitwould

bepossibletoldquoseerdquoagrainofriceatadistanceof5000kmEstimatetheresolvingpoweroftheVLBA

2 Thedistancefromtheeyelenstotheretinais20mm The light receptors in the central part of the retinaareabout5times10-6apartDeterminewhetherthe spacing of the receptors will allow for the eye to resolve the headlights in the above discussion whentheyare25kmfromtheeye

3 ThediameterofPlutois23times106 m and its averagedistancefromEarthis60times1012 m EstimatetheminimumdiameteroftheobjectiveofatelescopethatwillenablePlutotobeseenasadisc as opposed to a point source

95 Doppler effect

Essential ideaTheDopplereffectdescribesthephenomenonofwavelengthfrequencyshiftwhenrelativemotionoccurs

Understandingsbull TheDopplereffectforsoundwavesandlightwaves

The Doppler effect for sound waves and light waves

Consider two observers A and B at rest with respect to asoundsourcethatemitsasoundofconstantfrequencyf Clearly both observers will hear a sound of the same frequencyHoweversupposethatthesourcenowmovesat constant speed towards A A will now hear a sound of frequencyfA that is greater than f and B will hear a sound of frequency fB that is less than f This phenomenon is known as the Doppler Effect orDopplerPrinciple after C J Doppler(1803-1853)

The same effect arises for an observer who is either moving towards or away from a stationary source

Figure 930 shows the waves spreading out from a stationary source that emits a sound of constant frequency f The observersAandBhearasoundofthesamefrequency

A B

wavefront

S

Figure 930 Sound waves from a stationary source

Suppose now that the source moves towards A withconstant speed vFigure9310(a)showsasnapshotofthenew wave pattern

NATURE OF SCIENCETechnology Although originally based on physical observations of the pitch of fast moving sources of sound the Doppler effect has an important role in many different areas such as evidence for the expansion of the universe and generating images used in weather reports and in medicine (55)

copy IBO 2014

PHYSICS 2015indb 303 150514 337 PM

Chapter 9

304

AH

L

sourceA B

smallerwavelength

largerwavelength

Figure 931 (a) Sound waves from a moving source

The wavefronts are now crowded together in the direction of travel of the source and stretched out in the opposite direction This is why now the two observers will now hear notesofdifferentfrequenciesHowmuchthewavesbunchtogether and how much they stretch out will depend on the speed vEssentially A

A

cfλ

= and BB

cfλ

= where λA lt λB and v is the speed of sound

IfthesourceisstationaryandAismovingtowardsitthenthe waves from the source incident on A will be bunched up If A is moving away from the stationary source then the waves from the source incident on A will be stretched out

Christian Doppler(1803ndash1853)actuallyappliedtheprinciple(incorrectlyasithappens)totryandexplainthecolourofstars However the Doppler effect does apply to light aswellastosoundIfalightsourceemitsalightoffrequencyf then if it is moving away from an observer the observer willmeasurethelightemittedashavingalowerfrequencythan fSincethesensationofcolourvisionisrelatedtothefrequencyoflight(bluelightisofahigherfrequencythanred light) light emitted by objects moving way from anobserver is often referred to as being red-shifted whereas if the object is moving toward the observer it is referred to as blue-shiftedThisideaisusedinOptionE(Chapter16)

We do not need to consider here the situations where either the source or the observer is accelerating In a situation for example where an accelerating source is approaching astationaryobserverthentheobserverwillhearasoundof ever increasing frequency This sometimes leads toconfusion in describing what is heard when a source approaches passes and then recedes from a stationaryobserver Suppose for example that you are standingon a station platform and a train sounding its whistle is approaching at constant speed What will you hear as the train approaches and then passes through the station As the train approaches you will hear a sound of constant pitch but increasing loudness The pitch of the sound will be greater than if the train were stationary As the train passes through the station you will hear the pitch change at the momentthetrainpassesyoutoasoundagainofconstantpitch The pitch of this sound will be lower than the sound of the approaching train and its intensity will decrease as the train recedes from you What you do not hear is a sound of increasing pitch and then decreasing pitch

The Doppler equations for sound

Although you will not be expected in an IB examination to derivetheequationsassociatedwithaspectsoftheDopplereffectyouwillbeexpectedtoapplythemForcompletenesstherefore thederivationoftheequationsassociatedwiththeDopplereffectasoutlinedaboveisgivenhere

S primeS O

vs

Figure 931 (b)

InFigure931(b)theobserverOisatrestwithrespecttoasourceofsoundSismovingwithconstanspeedvs directly towards O The source is emitting a note of constant frequencyf and the speed of the emitted sound is v

S shows the position of the source ∆t later When the sourceisatresttheninatime∆t the observer will receive f∆t waves and these waves will occupy a distance v∆t ie

v f

λ =f∆tv∆t =

(Because of the motion of the source this number of waves will now occupy a distance (v∆t ndash vs∆t)The lsquonewrsquowavelength is therefore

v ndash vs

fλrsquo =

f∆tv∆t ndash vs∆t

=

If fisthefrequencyheardbyOthen

vfλ

= or

vf

λ = = sv vfminus

From which

s

vf fv v

=minus

Dividingthroughbyv gives

Equation1111

1 sf f v

v=

minus

If the source is moving away from the observer then we have

s

vf fv v

=minus

1

1 sf vv

=+

Equation112

PHYSICS 2015indb 304 150514 337 PM

Wave Phenomena

305

AH

L

We now consider the case where the source is stationary and the observerismoving towardsthesource with speed v0 In this situation the speed of the sound waves as measured by the observer will be v0 + v We therefore have that

v0 + v = ffraslλ = vfλ

times

From which

o1 vf fv

= +

Equation113

If the observerismovingawayfromthesource then

o1 vf fv

= minus

Fromequation(113)wehavethat

o o1 v vf f f f f fv v

∆ = minus = + minus =

Equation114

Thevelocitiesthatwerefertointheaboveequationsarethevelocities with respect to the medium in which the waves fromthesourcetravelHoweverwhenwearedealingwithalight source it is the relative velocity between the source and the observer that we must consider The reason for this is that lightisuniqueintherespectthatthespeedofthelightwavesdoes not depend on the speed of the source All observers irrespective of their speed or the speed of the source will measure the same velocity for the speed of light This is oneof thecornerstonesof theSpecialTheoryofRelativitywhichisdiscussedinmoredetailinOptionH(Chapter18)When applying the Doppler effect to light we aremainlyconcerned with the motion of the source We look here only at the situation where the speed of the source v is much smaller than the speed of light cinfreespace(v ltlt c)Underthesecircumstanceswhenthesourceismovingtowardstheobserverequation111becomes

vf f f fc

minus = ∆ = Equation115

andwhenthesourceismovingawayfromtheobserverequation112becomes vf f f f

cminus = ∆ = minus

Provided that v ltlt c these same equations apply for astationary source and moving observer

We look at the following example and exercise

Example

Asourceemitsasoundoffrequency440HzItmovesinastraight line towards a stationary observer with a speed of 30ms-1Theobserverhearsasoundoffrequency484HzCalculate the speed of sound in air

Solution

We use equation 111 and substitute f= 484 Hz f = 440 Hz

and vs = 30 m s-1

therefore 1

484 440 301v

=minus

such that 30 4401

484vminus = to

give v = 330 m s-1

Example

A particular radio signal from a galaxy is measured as havingafrequencyof139times109HzThesamesignalfromasourceinalaboratoryhasafrequencyof142times109Hz

Suggestwhy thegalaxy ismovingaway fromEarthandcalculateitsrecessionspeed(iethespeedwithwhichitismovingawayfromEarth)

Solution

The fact that the frequency from the moving source is less than that when it is stationary indicates that it is moving away from the stationary observer ie Earth

Using vf fc

∆ = we have

8 96

9

3 10 (142 139) 10 634 10142 10

c fvf∆ times times minus times

= = = timestimes

m s-1

It is usual when dealing with the Doppler effect of light to express speeds as a fraction of c So in this instance we have v = 0021 c

Using the Doppler effectWe have seen in the above example and exercise how the Dopplereffectmaybeusedtomeasuretherecessionspeedof distant galaxies The effect is also used to measure speed inothersituationsHerewewilllookatthegeneralprincipleinvolved in using the Doppler effect to measure speedFigure932showsasource(thetransmitter)thatemitseithersound or em waves of constant frequency f The waves

PHYSICS 2015indb 305 150514 337 PM

Chapter 9

306

AH

L

fromthesourceare incidentonareflectorthat ismovingtowards the transmitter with speed vThereflectedwavesare detected by the receiver placed alongside the transmitter

reector

v

transmitter

receiver

f f

f f

Figure 932 Using the Doppler effect to measure speed

We shall consider the situation where v ltlt c where c is the speed of the waves from the transmitter

For the reflector receiving waves from the transmitterit is effectively an observer moving towards a stationary sourceFromequation(114) itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation116

For the receiver receivingwaves from the reflector it iseffectively a stationary observer receiving waves from a movingsourceFromequation(115)itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation117

Ifweaddequations(116)and(117)wegetthatthetotalDopplershiftatthereceiver∆f is

v vf f f f fc c

minus = ∆ = +

But 1 vf fc

= +

hence

1 v v vf f fc c c

∆ = + +

But since v ltlt cwe can ignore the term2

2

vc when we

expandthebracketintheaboveequation

Therefore we have2vf fc

∆ = Equation118

If v asymp c then we must use the full Doppler equationsHowever for em radiationwewill alwaysonly considersituations in which v ltlt c

Example

Thespeedofsoundinbloodis1500times103 m s-1 Ultrasound offrequency100MHzisreflectedfrombloodflowinginan arteryThe frequencyof the reflectedwaves receivedbackatthetransmitteris105MHzEstimatethespeedofthebloodflowintheartery

Solution

Using equation (118) we have

6 63

2005 10 1015 10

vtimes = times

timesto give v asymp 36 m s-1 (We have assumed that the ultrasound is incident at right angles to the blood flow)

Exercise 971 Judy is standing on the platform of a station A

high speed train is approaching the station in a straight line at constant speed and is sounding its whistleAsthetrainpassesbyJudythefrequencyof the sound emitted by the whistle as heard by Judychangesfrom640Hzto430HzDetermine

(a) thespeedofthetrain

(b) thefrequencyofthesoundemittedbythewhistle as heard by a person on the train (Speedofsound=330ms-1)

2 AgalaxyismovingawayfromEarthwithaspeedof00500cThewavelengthofaparticularspectral line in light emitted by atomic hydrogen inalaboratoryis656times10-7 m Calculate the valueofthewavelengthofthislinemeasuredinthelaboratoryinlightemittedfromasourceofatomic hydrogen in the galaxy

PHYSICS 2015indb 306 150514 337 PM

10 Fields

Contents

101 ndash Describing fields

102 ndash Fields at work

Essential Ideas

Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Similar approaches can be taken in analysing electrical and gravitational potential problems copy IBO 2014

PHYSICS 2015indb 307 150514 337 PM

Chapter 10

308

AH

L

101 Describing fields

Essential idea Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Understandingsbull Gravitationalfieldsbull Electrostaticfieldsbull Electricpotentialandgravitationalpotentialbull Fieldlinesbull Equipotentialsurfaces

Gravitational potentialWehaveseenthatifweliftanobjectofmassm to a height h abovethesurfaceoftheEarththenitsgainingravitationalpotential energy is mghHoweverthisisbynomeansthefullstoryForastartwenowknowthatgvarieswithh and also the expression really gives a difference in potentialenergybetweenthevaluethattheobjecthasattheEarthrsquossurfaceandthevaluethatithasatheighthSowhatwereallyneedisazeropointCanwefindapointwherethepotentialenergyiszeroandusethispointfromwhichtomeasure changes in potential energy

The point that is chosen is in fact infinity At infinity the gravitationalfieldstrengthofanyobjectwillbezeroSoletusseeifwecandeduceanexpressionforthegaininpotentialenergyofanobjectwhenitisldquoliftedrdquofromthesurfaceoftheEarthtoinfinityThisineffectmeansfindingtheworknecessarytoperformthistask

δr

r r δr+

r infin=

mg

AB

Me

Figure 1001 Gravitational forces

Inthediagramweconsidertheworknecessarytomovethe particle of mass m a distance δr in thegravitationalfield of the Earth

The force on the particle at A is FG Mem

r2----------------=

IftheparticleismovedtoB then since δr isverysmallwe can assume that the field remains constant over thedistance ABTheworkδW done againstthegravitationalfieldoftheEarthinmovingthedistanceABis

δWGMem

r2----------------δrndash=

(rememberthatworkdoneagainstaforceisnegative)

TofindthetotalworkdoneW in going from the surface oftheEarthtoinfinitywehavetoaddalltheselittlebitsofworkThisisdonemathematicallybyusingintegralcalculus

WG Me m

r2----------------ndash

rdR

infin

int G Me m 1

r2----- rd

R

infin

intndash G Me m 1r---ndash

R

infinndash= = =

GMe m 0 1R---ndash

ndashndash=

G MemR----------------ndash=

HencewehavewhereRistheradiusoftheEarththattheworkdonebythegravitationalfieldinmovinganobjectofmass m from R(surfaceoftheEarth)toinfinityisgivenby

WGMe m

R---------------- ndash=

We can generalise the result by calculating the worknecessary per unit mass to take a small mass fromthe surface of the Earth to infinity This we call thegravitational potential Vie

Wewouldgetexactlythesameresultifwecalculatedtheworkdonebythefieldtobringthepointmassfrominfinityto the surface of Earth In this respect the formal definition ofgravitationalpotentialatapointinagravitationalfieldis therefore defined as the work done per unit mass in bringing a point mass from infinity to that point

Clearly then the gravitational potential at any point inthe Earthrsquos field distance r from the centre of the Earth (providingr gt R)is

VGMe

r------------ ndash=

Thepotentialisthereforeameasureoftheamountofworkthathastobedonetomoveparticlesbetweenpointsinagravitationalfieldanditsunit is theJkgndash1 We also note thatthepotentialisnegativesothatthepotentialenergy

NATURE OF SCIENCEParadigm shift The move from direct observable actions being responsible for influence on an object to acceptance of a fieldrsquos ldquoaction at a distancerdquo required a paradigm shift in the world of science (23)

copy IBO 2014

PHYSICS 2015indb 308 150514 337 PM

Fields

309

AH

L

aswemoveawayfromtheEarthrsquossurfaceincreasesuntilitreachesthevalueofzeroatinfinity

IfthegravitationalfieldisduetoapointmassofmassmthenwehavethesameexpressionasaboveexceptthatMe is replaced by m andmustalsoexclude thevalueof thepotential at the point mass itself ie at r = 0

WecanexpressthegravitationalpotentialduetotheEarth(orduetoanysphericalmass)intermsofthegravitationalfield strength at its surface

AtthesurfaceoftheEarthwehave

g 0ndash Re GMRe

---------ndash= e

Sothat

g 0Re2 GM= e

Hence at a distance r from the centre of the Earth the gravitationalpotentialVcanbewrittenas

VGMe

r------------ndashg 0 Re

2

r------------ndash= =

The potential at the surface of the Earth

(r = Re)istherefore-g0Re

It is interesting to see how the expression for thegravitational potential ties in with the expression mgh ThepotentialatthesurfaceoftheEarthis-g0Re (see the exampleabove)andataheighthwillbe-g0g 0 Re h+( )ndash ifweassume that g0doesnotchangeover thedistanceh The differenceinpotentialbetweenthesurfaceandtheheighth is therefore g0hSotheworkneededtoraiseanobjectof mass m to a height h is mgh ie m timesdifference ingravitationalpotential

This we have referred to as the gain in gravitationalpotentialenergy(see235)

However this expression can be extended to any twopoints inanygravitationalfieldsuch that ifanobjectofmass mmovesbetween twopointswhosepotentialsareV1 and V2 respectively then the change in gravitationalpotentialenergyoftheobjectism(V1 ndash V2)

Gravitational potential gradient

Let us consider now a region in space where thegravitationalfieldisconstantInFigure912thetwopointsAandBareseparatedbythedistance∆x

A B

direction of uniform gravitational eld of strength I

∆x

Figure 1002 The gravitational potential gradient

The gravitational field is of strength I and is in the direction shownThe gravitational potential at A is V andatBisV +∆V

Theworkdone is takingapointmassmfromAtoB isF∆x = mI∆x

Howeverbydefinitionthisworkisalsoequalto-m∆V

Therefore mI∆x=-m∆V

or VIx

∆= minus

∆Effectivelythissaysthatthemagnitudeofthegravitationalfield strength is equal to the negative gradient of thepotential If I is constant then V is a linear function of x and Iisequaltothenegativegradientofthestraightlinegraph formed by plotting V against x If I is not constant (asusuallythecase)thenthemagnitudeofI at any point in the field can be found by find the gradient of the V-x graph at that point An example of such a calculation can befoundinSection929

For those of you who do HL maths the relationship between field and potential is seen to follow from the expression for the potential of a point mass viz

mV Gr

= minus

2

ddV mG Ir r

minus = + =

Potential due to one or more point masses

Gravitationalpotentialisascalarquantitysocalculatingthepotential due to more than one point mass is a matter of simpleadditionSoforexamplethepotentialV due to the Moon and Earth and a distance x fromthecentreofEarthonastraightlinebetweenthemisgivenbytheexpression

E MM MV Gx r x

= minus + minus

whereME=massofEarthMM= mass of Moon and r = distancebetweencentreofEarthandMoon

PHYSICS 2015indb 309 150514 337 PM

Chapter 10

310

AH

L

Equipotentials and field linesIfthegravitationalpotentialhasthesamevalueatallpointson a surface the surface is said to be an equipotential surfaceSoforexampleifweimagineasphericalshellaboutEarthwhosecentrecoincideswiththecentreofEarththisshellwillbeanequipotentialsurfaceClearlyifwerepresentthegravitationalfieldstrengthbyfieldlinessincethelinesldquoradiaterdquooutfromthecentreofEarththentheselineswillbeatrightanglestothesurfaceIfthefieldlineswerenotnormaltotheequipotentialsurfacethentherewouldbeacomponentof the field parallel to the surfaceThis wouldmean thatpointsonthesurfacewouldbeatdifferentpotentialsandsoitwouldnolongerbeanequipotentialsurfaceThisofcourseholdstrueforanyequipotentialsurface

Figure 1003 shows the field lines and equipotentials fortwopointmassesm

m m

Figure 1003 Equipotentials for two point masses

It is worth noting that we would get exactly the samepattern ifwewere to replace thepointmasseswith twoequalpointcharges(See935)

Escape speedThe potential at the surface of Earth is which meansthattheenergyrequiredtotakeaparticleofmass mfromthesurfacetoinfinityisequalto

ButwhatdoesitactuallymeantotakesomethingtoinfinityWhentheparticleisonthesurfaceoftheEarthwecanthinkofitassittingatthebottomofaldquopotentialwellrdquoasinfigure1004

particle

surface of Earth

innity

GMR---------

FIgure 1004 A potential well

Theldquodepthrdquoofthewellis and if the particle gains an amount of kinetic energy equal to wherem is its massthenitwillhavejustenoughenergytoldquoliftrdquoitoutofthewell

In reality it doesnrsquot actually go to infinity it justmeansthat the particle is effectively free of the gravitationalattractionof theEarthWe say that ithas ldquoescapedrdquo theEarthrsquosgravitationalpullWemeetthisideainconnectionwithmolecular forcesTwomolecules in a solidwill sitat their equilibrium position the separation where therepulsiveforceisequaltotheattractiveforceIfwesupplyjust enough energy to increase the separation of themolecules such that they are an infinite distance apart then intermolecularforcesnolongeraffectthemoleculesandthesolidwillhavebecomealiquidThereisnoincreaseinthekineticenergyofthemoleculesandsothesolidmeltsat constant temperature

Wecancalculatetheescapespeedofanobjectveryeasilyby equating the kinetic energy to the potential energysuch that

12---mvesc ape

2 GMe mRe

----------------=

vesc aperArr2GMe

Re--------------- 2g0 Re = =

Substituting for g0 and Regivesavalueforvescape of about 11kmsndash1 from the surface of the Earth

Youwillnotethattheescapespeeddoesnotdependonthemassoftheobjectsincebothkineticenergyandpotentialenergy are proportional to the mass

IntheoryifyouwanttogetarockettothemoonitcanbedonewithoutreachingtheescapespeedHoweverthiswouldnecessitate anenormousamountof fuel and it islikely that therocketplus fuelwouldbesoheavythat itwouldnevergetoffthegroundItismuchmorepracticalto accelerate the rocket to the escape speed fromEarthorbitandthenintheoryjustlaunchittotheMoon

Example

Use the followingdata todetermine thepotential at thesurface of Mars and the magnitude of the acceleration of free fall

massofMars =64times1023kg

radiusofMars =34times106 m

Determinealsothegravitationalfieldstrengthatadistanceof 68 times 106mabovethesurfaceofMars

PHYSICS 2015indb 310 150514 337 PM

Fields

311

AH

L

Solution

= minus = minus times times = times23

11 76

64 1067 10 ndash13 10

34 10MV GR

minus timestimes

Nkg-1

But V = -g0R

Therefore 7

0 6

13 10 3834 10

VgR

times= minus = =

timesm s-2

To determine the field strength gh at 68 times 106 m above the surface we use the fact that 0 2

Mg GR

= such that GM = g0R2

Therefore 2 2

0h 2 2 2

h h

38 (34) 042(102)

g RGMgR R

times= = = = m s-2

(the distance from the centre is 34 times 106 + 68 times 106 = 102 times 106 m)

Exercise 1011 Thegraphbelowshowshowthegravitational

potentialoutsideoftheEarthvarieswithdistancefrom the centre

0 10 20 30 40 50 60 70

ndash1

ndash2

ndash3

ndash4

ndash5

ndash6

ndash7

r m times 106

V

Jkgndash1

times 10

7

(a) Usethegraphtodeterminethegainingravitationalpotentialenergyofasatelliteofmass200kgasitmovesfromthesurfaceof the Earth to a height of 30 times 107mabovetheEarthrsquossurface

(b) Calculatetheenergyrequiredtotakeit to infinity

(c) DeterminetheslopeofthegraphatthesurfaceoftheEarthmCommenton youranswer

Electric potential energy

The concept of electric potential energy was developedwith gravitational potential energy in mind Just as anobjectnearthesurfaceoftheEarthhaspotentialenergybecause of its gravitational interaction with the Earthsotoothereiselectricalpotentialenergyassociatedwithinteracting charges

Letusfirstlookatacaseoftwopositivepointchargeseachof 1μC that are initially bound together by a thread in a vacuuminspacewithadistancebetweenthemof10cmasshowninFigure1005Whenthethreadiscutthepointchargesinitiallyatrestwouldmoveinoppositedirectionsmovingwithvelocitiesv1 and v2 along the direction of the electrostatic force of repulsion

BEFORE AFTER

v v21

Figure 1005 Interaction of two positive particles

The electric potential energybetweentwopointchargescan be found by simply adding up the energy associated witheachpairofpointchargesForapairofinteractingchargestheelectricpotentialenergyisgivenby

∆U = ∆Ep + ∆Ek = ∆W = ∆Fr = kqQ

____ r2 times r = kqQ

____ r

Becausenoexternalforceisactingonthesystemtheenergyand momentum must be conserved Initially Ek = 0 and Ep =kqQr=9times109 times 1 times 10-1201m=009JWhentheyareagreatdistance fromeachotherEpwillbenegligibleThefinalenergywillbeequaltofrac12mv1

2+frac12mv22=009J

Momentum isalsoconservedand thevelocitieswouldbethe same magnitude but in opposite directions

Electric potential energyismoreoftendefinedintermsofapointchargemovinginanelectricfieldas

lsquothe electricpotential energybetweenany twopoints inan electric field is defined as negativeoftheworkdone by anelectricfieldinmovingapoint electric chargebetweentwolocationsintheelectricfieldrsquo

ΔU = ΔEp = -ΔW = -Fd = qEx

ΔU = ΔEp +ΔEk = ΔW = ΔFr = kqQ r2 x r = kqQ

wherexisthedistancemovedalong(oroppositeto)thedirection of the electric field

Electric potential energyismeasuredinjoule(J)Justasworkisascalarquantitysotooelectricalpotentialenergyis a scalar quantity The negative of the work done by an electric field in moving a unit electric charge between two points is independent of the path taken Inphysicswesaytheelectricfieldisalsquoconservativersquo field

PHYSICS 2015indb 311 150514 337 PM

Chapter 10

312

AH

L

Supposeanexternalforcesuchasyourhandmovesasmallpositive point test charge in the direction of a uniformelectric fieldAs it ismoving itmust be gaining kineticenergyIfthisoccursthentheelectricpotentialenergyofthe unit charge is changing

InFigure1006apointcharge+qismovedbetweenpointsAandBthroughadistancex in a uniform electric field

B

A

x

+ q

Figure 1006 Movement of a positive point charge in a uniform field

InordertomoveapositivepointchargefrompointAtopointBanexternal forcemustbeapplied to thechargeequaltoqE (F = qE)

Since the force is applied through a distance xthennegativeworkhas to bedone tomove the charge because energyis gainedmeaning there is an increaseelectric potential energybetweenthetwopointsRememberthattheworkdoneisequivalenttotheenergygainedorlostinmovingthe charge through the electric field The concept of electric potential energy is only meaningful as the electric field whichgeneratestheforceinquestionisconservative

W F xtimes E q x times= =

θ x cos θ x

Figure 1007 Charge moved at an angle to the field

If a chargemoves at an angle θ to an electric field thecomponent of the displacement parallel to the electric fieldisusedasshowninFigure918

W F x Eq x θcostimes= =

Theelectricpotentialenergyisstoredintheelectricfieldand theelectricfieldwill return theenergy to thepointcharge when required so as not to violate the Law ofconservationofenergy

Electric potential

The electric potential at a point in an electric field is defined as being the work done per unit charge in bringing a small positive point charge from infinity to that point

ΔV = Vinfin ndash Vf = ndash W ___ q

Ifwedesignatethepotentialenergytobezeroatinfinitythen it follows thatelectricpotentialmustalsobezeroat infinity and the electric potential at any point in an electricfieldwillbe

ΔV = ndash W ___ q

NowsupposeweapplyanexternalforcetoasmallpositivetestchargeasitismovedtowardsanisolatedpositivechargeTheexternalforcemustdoworkonthepositivetestchargetomoveittowardstheisolatedpositivechargeandtheworkmustbepositivewhiletheworkdonebytheelectricfieldmustthereforebenegativeSotheelectricpotentialatthatpointmustbepositiveaccordingtotheaboveequationIfanegativeisolatedchargeisusedtheelectricpotentialatapointonthepositivetestchargewouldbenegativePositivepoint chargesof their ownaccordmove fromaplaceofhighelectricpotential toaplaceof lowelectricpotentialNegative point charges move the other way from lowpotentialtohighpotentialInmovingfrompointAtopointBinthediagramthepositivecharge+qismovingfromalowelectricpotentialtoahighelectricpotential

In thedefinitiongiven the termldquoworkperunitchargerdquohas significance If the test charge is +16 times 10-19Cwherethe charge has a potential energy of 32 times 10-17Jthenthepotentialwouldbe32times10-17J+16times10-19C=200JC-1 Now if the charge was doubled the potential wouldbecome 64times10-17 J However the potential per unitchargewouldbethesame

Electric potential is a scalar quantityandithasunitsJC-1 orvoltswhere1voltequalsone joulepercoloumbThevoltallowsustoadoptaunitfortheelectricfieldintermsofthevolt

PreviouslytheunitfortheelectricfieldwasNCndash1

W = qV and F = qEsoW ___ V = F __ E

E = FV ___ W = FV ___ Fm V mndash1

That is the units of the electric field E can also beexpressed as V mndash1

The work done per unit charge in moving a point charge between two points in an electric field is again independent of the path taken

PHYSICS 2015indb 312 150514 337 PM

Fields

313

AH

L

Electric potential due to a point charge

Letus takeapoint rmetres froma chargedobjectThepotentialatthispointcanbecalculatedusingthefollowing

W = ndashFr = ndashqV and F = ndash q1 q2 _____ 4πε0r

2

Therefore

W = ndash q1 q2 _____ 4πε0r

2 times r = ndash q1 q2 _____ 4πε0r

= ndashq1 times q2 _____ 4πε0r

= ndashq1V

That is V q

4πε0r--------------- =

Orsimply

V kqr------

=

Example

Determinehowmuchworkisdonebytheelectricfieldofpointcharge150μCwhenachargeof200μCismovedfrom infinity to apoint 0400m from thepoint charge(Assumenoaccelerationofthecharges)

Solution

The work done by the electric field is W = -qV = -14πε0 times q times (Q rinfin - Q r0400)

W = (- 200 times 10-6 C times 900 times 109 NmC-2 times 150 times 10-6 C) divide 0400 m = - 0675 J

An external force would have to do +0675 J of work

Electric field strength and electric potential gradient

Letus lookbackatFigure1006Supposeagain that thecharge+qismovedasmalldistancebyaforceFfromAtoBsothattheforcecanbeconsideredconstantTheworkdoneisgivenby

∆W F x∆times=

TheforceFandtheelectricfieldEareoppositelydirectedandweknowthat

F = -qE and ΔW = q ΔV

Thereforetheworkdonecanbegivenas

q ΔV = -q E Δx

Therefore E V∆x∆-------

ndash=

The rate of change of potential ΔV at a point with respect to distance Δx in the direction in which the change is maximum is called the potential gradient We say that the electric field = - the potential gradient and the units are Vm-

1 From the equation we can see that in a graph of electric potential versus distance the gradient of the straight line equals the electric field strength

In reality if a charged particle enters a uniform electric field it will be accelerated uniformly by the field and its kinetic energy will increase This is why we had to assume no acceleration in the last worked example

Ek∆ 12---

mv2 q E xsdot sdot q Vx---

xsdot sdot q Vsdot= = = =

Example

Determine how far apart two parallel plates must besituated so that a potential difference of 150 times 102 V produces an electric field strength of 100 times 103 NC-1

Solution

Using E V∆x------- xhArrndash V∆

E-------15 102 Vtimes

100 103 N C 1ndashtimes------------------------------------------= = =

= 150 times 10-1

The plates are 150 times 10-1 m apart

The electric field and the electric potential at a point due to an evenly distributed charge +q on a sphere can be represented graphically as in Figure 1008

r

r

V

E

r0

r0

r0

Charge of +Q evenly distributed over surface

E 14πε0------------ Q

r2----- r r0gt=

V 14πε0------------Q

r----r r0gt=

Figure 1008 Electric field and potential due to a charged sphere

PHYSICS 2015indb 313 150514 337 PM

Chapter 10

314

AH

L

When the sphere becomes charged we know that thechargedistributes itselfevenlyoverthesurfaceThereforeeverypartofthematerialoftheconductorisatthesamepotential As the electric potential at a point is defined as beingnumericallyequaltotheworkdoneinbringingaunitpositivechargefrominfinitytothatpointithasaconstantvalueineverypartofthematerialoftheconductor

Since the potential is the same at all points on the conducting surfacethen∆V∆xiszeroButE=ndash∆V∆xThereforethe electric field inside the conductor is zero There is no electric field inside the conductor

SomefurtherobservationsofthegraphsinFigure1009are

bull OutsidethespherethegraphsobeytherelationshipsgivenasE α 1 r2 and V α 1 r

bull Atthesurfacer = r0Thereforetheelectricfieldandpotentialhavetheminimumvalueforr at this point and this infers a maximum field and potential

bull Insidethespheretheelectricfieldiszerobull Insidethespherenoworkisdonetomoveacharge

fromapointinsidetothesurfaceThereforethereisno potential difference and the potential is the same asitiswhenr = r0

Similargraphscanbedrawnfortheelectricfieldintensityandthe electric potential as a function of distance from conducting parallelplatesandsurfacesandthesearegiveninFigure1009

Potential plot E eld plotE eld

++++

ndashndashndashndash

xx x

+

x x

ndashx x

+ ndashx

x

+ +

x

x

Figure 1009 Electric field and electric potential at a distance from a charged surface

Potential due to one or more point charges

The potential due to one point charge can be determined byusingtheequationformula

V = kq r

Example 1

Determine the electric potential at a point 20 times 10-1 m fromthecentreofan isolatedconducting spherewithapointchargeof40pCinair

Solution

Using the formula

V = kq r we have

V90 10 9times( ) 40 10 12ndashtimes( )times

20 10 1ndashtimes( )------------------------------------------------------------------ 018 V= =

the potential at the point is 180 times 10-1 V

The potential due to a number of point charges can be determined by adding up the potentials due to individual point charges because the electric potential at any point outside a conducting sphere will be the same as if all the charge was concentrated at its centre

Example 2

Threepointchargesofareplacedattheverticesofaright-angledtriangleasshowninthediagrambelowDeterminetheabsolutepotentialat the+20μCchargeduetothetwoothercharges

- 6μC

+3μC +2μC

4 m

3 m

Solution

The electric potential of the +2 μC charge due to the ndash 6 μC charge is

V = (9 times 109 Nm2C-2 times -6 times 10-6 C) divide (radic 32 + 42) m = - 108 times 104 V

PHYSICS 2015indb 314 150514 338 PM

Fields

315

AH

L

Insummarywecanconcludethat

bull No work is done to move a charge along an equipotential

bull Equipotentials are always perpendicular to the electric lines of force

Figure1011and1012showsomeequipotential lines fortwo oppositely charged and identically positive spheresseparated by a distance

+ve ndashve

equipotential lines

Figure 1011 Equipotential lines between two opposite charges

equipotential lines

Figure 1012 Equipotential lines between two charges which are the same

+

ndash ndash ndash ndash

+ + +

40 V

30 V

20 V

10 V

50 V

Figure 1013 Equipotential lines between charged parallel plates

Figure1013showstheequipotentiallinesbetweenchargedparallel plates Throughout this chapter the similarities and differences between gravitational fields and electric fieldshavebeendiscussedTherelationshipsthatexistsbetweengravitationalandelectricquantitiesandtheeffectsofpointmassesandchargesissummarisedinTable1014

The electric potential of the +2 μC charge due to the +3 μC charge is

V = (9 times 109 Nm2C-2 times 3 times 10-6 C) divide 3m = 9 times 103 V

The net absolute potential is the sum of the 2 potentials- 108 times 104 V + 9 times 103 V =- 18 times 103 V

The absolute potential at the point is - 18 times 103 V

Equipotential surfacesRegions in space where the electric potential of a chargedistributionhasaconstantvaluearecalledequipotentials The placeswherethepotentialisconstantinthreedimensionsarecalled equipotential surfacesandwheretheyareconstantintwodimensionstheyarecalledequipotential lines

TheyareinsomewaysanalogoustothecontourlinesontopographicmapsInthiscasethegravitationalpotentialenergy is constant as amassmoves around the contourlines because the mass remains at the same elevationabovetheEarthrsquossurfaceThegravitationalfieldstrengthacts in a direction perpendicular to a contour line

Similarlybecausetheelectricpotentialonanequipotentiallinehas thesamevalueanelectric forcecandonoworkwhena testchargemovesonanequipotentialThereforethe electric field cannot have a component along anequipotentialandthusitmustbeeverywhereperpendiculartotheequipotentialsurfaceorequipotentiallineThisfactmakesiteasytoplotequipotentialsifthelinesofforceorlinesofelectricfluxofanelectricfieldareknown

Forexamplethereareaseriesofequipotentiallinesbetweentwoparallelplateconductorsthatareperpendiculartotheelectric fieldTherewill be a series of concentric circles(eachcirclefurtherapartthanthepreviousone)thatmapouttheequipotentialsaroundanisolatedpositivesphereThe lines of force and some equipotential lines for anisolatedpositivesphereareshowninFigure1010

Lines ofequipotential

Figure 1010 Equipotentials around an isolated positive sphere

PHYSICS 2015indb 315 150514 338 PM

Chapter 10

316

AH

L

Gravitational quantity Electrical quantityQ

uant

ities

g V∆x∆-------ndash= E V∆

x∆-------ndash=

Poin

t mas

ses a

nd ch

arge

s

V Gmr----

ndash= V 14πε0------------q

r---=

g G m

r2-----ndash= E 14πε0------------ q

r2-----=

F Gm1 m2

r2--------------= F 1

4πε0------------

q1q2

r2-----------=

Figure 1014 Formulas

Example

Deduce the electric potential on the surface of a gold nucleus that has a radius of 62 fm

Solution

Using the formula

V = kq r and knowing the atomic number of gold is 79 We will assume the nucleus is spherical and it behaves as if it were a point charge at its centre (relative to outside points)

V = 90 times 109 Nm2C-2 times 79 times 16 times 10-19 C divide 62 times 10-15 m = 18 times 107 V

The potential at the point is 18 MV

Example

Deduce the ionisation energy in electron-volts of theelectron in the hydrogen atom if the electron is in its ground state and it is in a circular orbit at a distance of 53 times 10-11 m from the proton

Solution

This problem is an energy coulombic circular motion question based on Bohrrsquos model of the atom (not the accepted quantum mechanics model) The ionisation energy is the energy required to remove the electron from the ground state to infinity The electron travels in a circular orbit and therefore has a centripetal acceleration The ionisation energy will counteract the coulombic force and the movement of the electron will be in the opposite direction to the centripetal force

Total energy = Ek electron + Ep due to the proton-electron interaction

ΣF = kqQ r2 = mv2 r and as such mv2 = = kqQ r

Therefore Ek electron = frac12 kqQ r

Ep due to the proton-electron interaction = - kqQ r

Total energy = frac12 kqQ r + - kqQ r = -frac12 kqQ r

= - 90 times 109 Nm2C-2 times (16 times 10-19 C)2 divide 53 times 10-11 m = -217 times 10-18 J

= -217 times 10-18 J divide 16 times 10-19 = -136 eV

The ionisation energy is 136 eV

Exercise 1021 ApointchargePisplacedmidwaybetweentwo

identicalnegativechargesWhichoneofthefollowingiscorrectwithregardstoelectricfieldandelectricpotentialatpointP

Electric field Electric potentialA non-zero zeroB zero non-zeroC non-zero non-zeroD zero zero

2 Twopositivechargedspheresaretiedtogetherinavacuumsomewhereinspacewheretherearenoexternal forces A has a mass of 25 g and a charge of20μCandBhasamassof15gandachargeof30μCThedistancebetweenthemis40cmTheyarethenreleasedasshowninthediagram

BEFORE AFTER

v1 v2

A B

(a) Determinetheirinitialelectricpotentialenergy in the before situation

(b) DeterminethespeedofsphereBafterrelease

3 Thediagrambelowrepresentstwoequipotentiallines in separated by a distance of 5 cm in a uniform electric field

+ + + + + + + +

ndash ndash ndash ndash ndash ndash ndash ndash

40 V

20 V 5 cm

PHYSICS 2015indb 316 150514 338 PM

Fields

317

AH

L

Determine the strength of the electric field

4 Thisquestionisabouttheelectricfieldduetoacharged sphere and the motion of electrons in that fieldThediagrambelowshowsanisolatedmetalsphereinavacuumthatcarriesanegativeelectriccharge of 60 μC

ndash

(a) Onthediagramdrawtheconventionalwayto represent the electric field pattern due to the charged sphere and lines to represent threeequipotentialsurfacesintheregionoutside the sphere

(b) Explainhowthelinesrepresentingtheequipotentialsurfacesthatyouhavesketchedindicatethatthestrengthoftheelectricfieldisdecreasingwithdistancefrom the centre of the sphere

(c) Theelectricfieldstrengthatthesurfaceofthe sphere and at points outside the sphere can be determined by assuming that the sphere acts as a point charge of magnitude 60 μC at its centre The radius of the sphere is 25 times 10ndash2 m Deduce that the magnitude of the field strength at the surface of the sphere is 86 times 107 Vmndash1

An electron is initially at rest on the surface of the sphere

(d) (i) Describethepathfollowedbytheelectronasitleavesthesurfaceofthesphere

(ii) Calculatetheinitialaccelerationofthe electron

5 Determinetheamountofworkthatisdoneinmovingachargeof100nCthroughapotentialdifferenceof150times102 V

6 Three identical 20 μC conducting spheres are placedatthecornersofanequilateraltriangleofsides 25 cm The triangle has one apex C pointing upthepageand2baseanglesAandBDeterminetheabsolutepotentialatB

7 Determinehowfaraparttwoparallelplatesmustbesituatedsothatapotentialdifferenceof250 times 102 V produces an electric field strength of 200 times 103 NC-1

8 Thegapbetweentwoparallelplatesis10times10-3mandthereisapotentialdifferenceof10times104 V betweentheplatesCalculate

i theworkdonebyanelectroninmovingfrom one plate to the other

ii thespeedwithwhichtheelectronreachesthe second plate if released from rest

iii theelectricfieldintensitybetweentheplates

9 Anelectronguninapicturetubeisacceleratedby a potential 25 times 103VDeterminethekineticenergygainedbytheelectroninelectron-volts

10 Determine the electric potential 20 times 10-2 m from achargeof-10times10-5 C

11 Determinetheelectricpotentialatapointmid-waybetweenachargeofndash20pCandanotherof+5pConthelinejoiningtheircentresifthecharges are 10 cm apart

12 During a thunderstorm the electric potential differencebetweenacloudandthegroundis10 times 109 V Determine the magnitude of the change in electric potential energy of an electron thatmovesbetweenthesepointsinelectron-volts

13 A charge of 15 μC is placed in a uniform electric fieldoftwooppositelychargedparallelplateswithamagnitudeof14times103 NC-1

(a) Determinetheworkthatmustbedoneagainstthefieldtomovethepointchargeadistance of 55 cm

(b) Calculatethepotentialdifferencebetweenthe final and initial positions of the charge

(c) Determinethepotentialdifferencebetweenthe plates if their separation distance is 15 cm

14 Duringaflashoflightningthepotentialdifferencebetweenacloudandthegroundwas12 times 109 V and the amount of transferred charge was32C

(a) Determinethechangeinenergyofthetransferred charge

(b) Iftheenergyreleasedwasallusedtoacceleratea1tonnecardeduceitsfinalspeed

(c) Iftheenergyreleasedcouldbeusedtomelticeat0degCdeducetheamountoficethatcould be melted

PHYSICS 2015indb 317 150514 338 PM

Chapter 10

318

AH

L

15 SupposethatwhenanelectronmovedfromAtoBin the diagram along an electric field line that the electric field does 36 times 10-19Jofworkonit

Determinethedifferencesinelectricpotential

(a) VB ndash VA(b) VC ndash VA(c) VC ndash VB

16 DeterminethepotentialatpointPthatislocatedatthecentreofthesquareasshowninthediagrambelow

- 6μC

+3μC +2μC

1m

1m

5μC

P

102 Fields at work

Essential idea Similar approaches can be taken inanalysingelectricalandgravitationalpotentialproblems

Understandingsbull Potentialandpotentialenergybull Potentialgradientbull Potentialdifferencebull Escapespeedbull Orbitalmotionorbitalspeedandorbitalenergybull Forcesandinverse-squarelawbehaviour

Orbital motion orbital speed and orbital energy

Although orbital motion may be circular elliptical orparabolic this sub-topic only dealswith circular orbitsThis sub-topic is not fundamentally new physics butan application that synthesizes ideas from gravitationcircularmotiondynamicsandenergy

TheMoonorbitstheEarthandinthissenseitisoftenreferredtoasasatelliteoftheEarthBefore1957itwastheonlyEarthsatelliteHoweverin1957theRussianslaunchedthefirstmanmadesatelliteSputnik1SincethisdatemanymoresatelliteshavebeenlaunchedandtherearenowliterallythousandsofthemorbitingtheEarthSomeareusedtomonitortheweathersome used to enable people to find accurately their position on thesurfaceoftheEarthmanyareusedincommunicationsandnodoubtsomeareusedtospyonothercountriesFigure1015showshowinprincipleasatellitecanbeputintoorbit

Theperson(whosesizeisgreatlyexaggeratedwithrespectto Earth) standing on the surface on the Earth throwssome stonesThe greater the speed with which a stoneis thrown the further it will land from her The pathsfollowedbythethrownstonesareparabolasByastretchof the imaginationwecanvisualisea situation inwhichastoneis thrownwithsuchaspeedthatbecauseof thecurvatureoftheEarthitwillnotlandonthesurfaceoftheEarthbutgointoldquoorbitrdquo(Path4onFigure1015)

NATURE OF SCIENCECommunication of scientific explanations The ability to apply field theory to the unobservable (charges) and the massively scaled (motion of satellites) required scientists to develop new ways to investigate analyse and report findings to a general public used to scientific discoveries based on tangible and discernible evidence (51)

copy IBO 2014

PHYSICS 2015indb 318 150514 338 PM

Fields

319

AH

L

1

2

34Earth

Figure 1015 Throwing a stone into orbit

TheforcethatcausesthestonestofollowaparabolicpathandtofalltoEarthisgravityandsimilarlytheforcethatkeeps the stone in orbit is gravity For circular motionto occur we have seen that a force must act at rightanglestothevelocityofanobjectthatistheremustbeacentripetalforceHenceinthesituationwedescribeherethe centripetal force for circular orbital motion about the EarthisprovidedbygravitationalattractionoftheEarth

We can calculate the speedwithwhich a stonemustbethrowninordertoputitintoorbitjustabovethesurfaceof the Earth

If the stone has mass m and speed v thenwehave fromNewtonrsquos2ndlaw

mv2

RE---------- G

MEm

RE2------------=

whereRE is the radius of the Earth and ME is the mass of the Earth

Bearinginmindthatg 0 GME

RE2--------= then

v g RE 10 64 106timestimes 8 103times= = =

Thatisthestonemustbethrownat8times103m sndash1

Clearly we are not going to get a satellite into orbit soclosetothesurfaceoftheEarthMovingatthisspeedthefrictionduetoairresistancewouldmeltthesatellitebeforeithadtravelledacoupleofkilometresInrealitythereforeasatelliteisputintoorbitabouttheEarthbysendingitattachedtoarocketbeyondtheEarthrsquosatmosphereandthengivingitacomponentofvelocityperpendiculartoaradialvectorfromtheEarthSeeFigure1016

Earth Satellite carried by rocket to hereSatellite orbit

Tangential component of velocity

Figure 1016 Getting a satellite into orbit

Keplerrsquos third law(This work of Kepler and Newtonrsquos synthesis of the work is an excellent example of the scientific method and makes for a good TOK discussion)

In 1627 Johannes Kepler (1571-1630)publishedhis lawsof planetary motion The laws are empirical in natureandwere deduced from the observations of theDanishastronomer Tycho de Brahe (1546-1601)The third lawgives a relationship between the radius of orbit R of a planet and its period TofrevolutionabouttheSunThelawisexpressedmathematicallyas

2

3 constantTR

=

We shall nowuseNewtonrsquos LawofGravitation to showhowitisthattheplanetsmoveinaccordancewithKeplerrsquosthirdlaw

InessenceNewtonwasabletousehislawofgravitytopredict the motion of the planets since all he had to do was factor theF givenby this law intohis second law F = ma to find their accelerations and hence their future positions

InFigure1017theEarthisshownorbitingtheSunandthedistancebetweentheircentresisR

RSun

Earth

F es

F se

Figure 1017 Planets move according to Keplerrsquos third law

Fes is the force that the Earth exerts on the Sun and Fse is the force that the Sun exerts on the Earth The forces are equalandoppositeandtheSunandtheEarthwillactuallyorbitaboutacommoncentreHoweversince theSun issoverymuchmoremassivethantheEarththiscommoncentrewillbeclosetothecentreoftheSunandsowecanregard the Earth as orbiting about the centre of the Sun Theotherthingthatweshallassumeisthatwecanignorethe forces that the other planets exert on the Earth (This wouldnotbeawisethingtodoifyouwereplanningtosend a space ship to theMoon for example)We shallalso assume that we have followed Newtonrsquos exampleandindeedprovedthataspherewillactasapointmasssituated at the centre of the sphere

Kepler had postulated that the orbits of the planets areellipticalbut since theeccentricityof theEarthrsquosorbit issmallweshallassumeacircularorbit

PHYSICS 2015indb 319 150514 338 PM

Chapter 10

320

AH

L

TheaccelerationoftheEarthtowardstheSunisa = Rω2

whereω ω 2πT------=

Hencea R 2π

T------ times

2 4π2 RT 2------------- = =

But the acceleration is given by Newtonrsquos Second Law F = mawhereFisnowgivenbytheLawofGravitationSo in this situation

F maGMs Me

R2-------------------= = butwealsohavethat

a a R 2πT

------ times

2 4π2 RT 2------------- = = and m = Me so that

G MsMe

R2------------------- Me4π2R

T2-------------times

GMs

4π2-----------rArr R3

T2------= =

Butthequantity

isaconstantthathasthesamevalueforeachoftheplanetssowehaveforalltheplanetsnotjustEarththat

R3

T2------ k =

wherekisaconstantWhichisofcourseKeplerrsquosthirdlaw

ThisisindeedanamazingbreakthroughItisdifficulttorefutethe idea that all particles attract each other in accordance withtheLawofGravitationwhenthelawisabletoaccountfortheobservedmotionoftheplanetsabouttheSun

The gravitational effects of the planets upon each othershould produce perturbations in their orbits Such is the predictivepoweroftheUniversalGravitationalLawthatit enabled physicists to compute these perturbations The telescopehadbeeninventedin1608andbythemiddleofthe 18th Century had reached a degree of perfection in design that enabled astronomers to actually measure the orbital perturbations of the planets Their measurements werealways inagreementwith thepredictionsmadebyNewtonrsquoslawHoweverin1781anewplanetUranuswasdiscovered and the orbit of this planet did not fit withthe orbit predicted by Universal Gravitation Such wasthe physicistrsquos faith in theNewtonianmethod that theysuspectedthatthediscrepancywasduetothepresenceofayetundetectedplanetUsingtheLawofGravitationtheFrenchastronomerJLeverrier and the English astronomer J C Adamswereable tocalculate justhowmassive thisnewplanetmustbeandalsowhereitshouldbeIn1846the planetNeptunewas discovered justwhere they hadpredicted Ina similarwaydiscrepancies in theorbitofNeptune ledto thepredictionandsubsequentdiscoveryin1930oftheplanetPlutoNewtonrsquosLawofGravitation

had passed the ultimate test of any theory it is not only abletoexplainexistingdatabutalsotomakepredictions

Satellite energyWhenasatelliteisinorbitaboutaplanetitwillhavebothkineticenergyandgravitationalpotentialenergySupposewe consider a satellite ofmass m that is in an orbit of radius r about a planet of mass M

Thegravitationalpotentialduetotheplanetatdistancer from its centre is

ThegravitationalpotentialenergyofthesatelliteVsat

is therefore GMe m

r----------------ndash

ThatisVsat GMem

r----------------ndash=

Thegravitationalfieldstrengthatthesurfaceoftheplanetisgivenby

g 0GMe

Re2------------=

Hencewecanwrite

Vsat g0 Re

2

rndash=

m

ThekineticenergyofthesatelliteKsatisequaltofrac12mv2wherev is its orbital speed

Byequatingthegravitationalforceactingonthesatellitetoitscentripetalaccelerationwehave

G Mem

r2---------------- mv2

r---------- mv2hArr

GMemr----------------= =

Fromwhich12---

mv2 12---

GMemr----------------times=

= 0 e

2g R

r

2m

Which is actually quite interesting since it shows thatirrespective of the orbital radius the KE is numericallyequal to half the PE Also the total energy E tot K sat Vsat+ 1

2---G Me m

r----------------times G Me m

r----------------ndash + 1

2---G Me m

r----------------ndash= = =of the satelliteisalwaysnegativesince

E tot K sat Vsat+ 12---

G Me mr----------------times

G Me mr----------------ndash

+ 12---

G Me mr----------------ndash= = =

PHYSICS 2015indb 320 150514 338 PM

Fields

321

AH

L

The energies of an orbiting satellite as a function of radial distancefromthecentreofaplanetareshownplottedinFigure1018

12

100806

0402

0ndash02

ndash04

ndash06ndash08ndash10

ndash12

2 4 6 8 10 12

ener

gya

rbita

ry u

nits

distance timesR

kinetic energytotal energy

potential energy

Figure 1018 Energy of an orbiting satellite as a function of distance from the centre of a planet

WeightlessnessSuppose that you are inanelevator(lift)thatisdescendingat constant speedandyou let goof abook thatyouareholdinginyourhandThebookwillfalltothefloorwithaccelerationequaltotheaccelerationduetogravityIfthecablethatsupportstheelevatorweretosnap(asituationthatItrustwillneverhappentoanyofyou)andyounowletgothebookthatyouareholdinginyourotherhandthis book will not fall to the floor - it will stay exactlyin linewithyourhandThis isbecause thebook isnowfallingwith the sameaccelerationas theelevatorandassuch the book cannot ldquocatchrdquo up with the floor of theelevatorFurthermoreifyouhappenedtobestandingonasetofbathroomscalesthescaleswouldnowreadzero-youwouldbeapparentlyweightlessItisthisideaoffreefallthatexplainstheapparentweightlessnessofastronautsin an orbiting satellite These astronauts are in free fall inthesensethattheyareacceleratingtowardsthecentre of the Earth

It is actually possible to define theweight of a body inseveral different waysWe can define it for example asthegravitationalforceexertedonthebodybyaspecifiedobject such as the EarthThiswe have seen thatwe doinlotsofsituationswherewedefinetheweightasbeingequaltomgIfweusethisdefinitionthenanobjectinfreefallcannotbydefinitionbeweightlesssinceitisstillinagravitationalfieldHoweverifwedefinetheweightofanobjectintermsofaldquoweighingrdquoprocesssuchasthereadingonasetofbathroomscaleswhichineffectmeasuresthecontact force between the object and the scales thenclearlyobjectsinfreefallareweightlessOnenowhasto

askthequestionwhetherornotitispossibleForexampletomeasurethegravitationalforceactingonanastronautin orbit about the Earth

WecanalsodefineweightintermsofthenetgravitationalforceactingonabodyduetoseveraldifferentobjectsForexample for an object out in space its weight could bedefined in terms of the resultant of the forces exerted on it bytheSuntheMoontheEarthandalltheotherplanetsin the Solar System If this resultant is zero at a particular pointthenthebodyisweightlessatthispoint

Inviewofthevariousdefinitionsofweightthatareavailabletousitisimportantthatwhenweusethewordldquoweightrdquoweareawareofthecontextinwhichitisbeingused

Example

Calculate the height above the surface of the Earth atwhichageo-stationarysatelliteorbits

Solution

A geo-stationary satellite is one that orbits the Earth in such a way that it is stationary with respect to a point on the surface of the Earth This means that its orbital period must be the same as the time for the Earth to spin once on its axis ie 24 hours

From Keplerrsquos third law we have G Ms

4π2----------- R3

T2------=

That is

RMe

mh

using the fact that the force of attraction between the satellite and the Earth is given by

FG Mem

R2----------------=

and that F = ma

where m is the mass of the satellite and a 4π2R

T2-------------=

we have

GMem

R2---------------- m 4π2R

T2-------------times

GMe

4π2------------rArr R3

T2------= =

PHYSICS 2015indb 321 150514 338 PM

Chapter 10

322

AH

L

Now the mass of the Earth is 60 times 1024 kg and the period T measured in seconds is given by T = 86400 s

So substitution gives R = 42 times 106 m

The radius of the Earth is 64 times 106 m so that the orbital height h is about 36 times 107 m

Example

Calculatetheminimumenergyrequiredtoputasatelliteofmass500kgintoanorbitthatisasaheightequaltotheEarthrsquosradiusabovethesurfaceoftheEarth

Solution

We have seen that when dealing with gravitational fields and potential it is useful to remember that

g 0G M

Re2---------= or g0 Re

2 GM=

The gravitational potential at the surface of the Earth is

g 0ndash Re GMRe

---------ndash=

The gravitational potential at a distance R

from the centre of the Earth is

=

The difference in potential between the surface of the Earth and a point distance R from the centre is therefore

V∆ g 0Re 1ReR------ndash

=

If R 2Re= then V∆g0 Re

2------------=

This means that the work required to ldquoliftrdquo the satellite into orbit is g0Rm where m is the mass of the satellite This is equal to

10 32 106 500timestimestimes = 16000 MJ

However the satellite must also have kinetic energy in order to orbit Earth This will be equal to

g 0 mRe2

2R-----------------g 0mRe

2

4----------------- 8000 MJ= =

The minimum energy required is therefore

24000 MJ

Exercise 1031 The speed needed to put a satellite in orbit does

not depend on

A the radius of the orbitB theshapeoftheorbitC thevalueofgattheorbitD the mass of the satellite

2 EstimatethespeedofanEarthsatellitewhoseorbitis400kmabovetheEarthrsquossurfaceAlsodetermine the period of the orbit

3 Calculatethespeedofa200kgsatelliteorbitingthe Earth at a height of 70 times 106 m

Assume that g = 82 m sndash2 for this orbit

4 TheradiioftwosatellitesXandYorbitingthe Earth are 2r and 8rwherer is the radius of the Earth Calculate the ratio of the periods of revolutionofXtoY

5 A satellite of mass mkgissentfromEarthrsquossurfaceinto an orbit of radius 5RwhereR is the radius of theEarthWritedownanexpressionfor

(a) thepotentialenergyofthesatelliteinorbit

(b) thekineticenergyofthesatelliteinorbit

(c) theminimumworkrequiredtosendthesatellitefromrestattheEarthrsquossurfaceintoits orbit

6 A satellite in an orbit of 10rfallsbacktoEarth(radius r)afteramalfunctionDeterminethespeedwithwhichitwillhittheEarthrsquossurface

7 The radius of the moon is frac14 that of the Earth AssumingEarthandtheMoontohavethesamedensitycomparetheaccelerationsoffreefallatthe surface of Earth to that at the surface of the Moon

8 UsethefollowingdatatodeterminethegravitationalfieldstrengthatthesurfaceoftheMoon and hence determine the escape speed from the surface of the Moon

Mass of the Moon = 73 times 1022kg

RadiusoftheMoon=17times106 m

PHYSICS 2015indb 322 150514 338 PM

Wave Phenomena

287

AH

L

(iv) v = minusv0sinωt = minus25sin (13π) To find the value of the sine function we have to convert the 13π into degrees (remember ω and hence ωt is measured in radians)

1 deg therefore 13π = 13 times 180 = 234deg therefore v1 = minus25sin (234deg) = +20 cm s ndash1

Or we can solve using ωradic_______

( x0 2 ndash x2 )

from the graph at t = 13 s x = ndash 48 cm

therefore v = π times = 20 cm s-1

(v) Using = π2 times 80 = 79 m s-2

Energy changesWe must now look at the energy changes involved in SHMTodosowewillagainconcentrateontheharmonicoscillator The mass is stationary at x = +x0 (maximumextension)andalsoatx = ndashx0 (maximumcompression)At these two positions the energy of the system is all potential energy and is in fact the elastic potential energy stored in the spring This is the total energy of the system ET and clearly

Equation412

ThatisthatforanysystemperformingSHMtheenergyofthesystemisproportionaltothesquareoftheamplitudeThis is an important result and one that we shall return to when we discuss wave motion

At x =0thespringisatitsequilibriumextensionandthemagnitude of the velocity v of the oscillating mass is a maximum v0TheenergyisallkineticandagainisequaltoET We can see that this is indeed the case as the expression for the maximum kinetic energy Emax in terms of v0 is

Clearly ET and Emax areequalsuchthat

ET = 1 __ 2 kx0 2 = Emax = 1 __ 2 mv0

2

From which

v0 2 = k __ m x0

2(asv0ge0)

Therefore

v0 = radic__

k __ m x0 = ωx0

whichtiesinwiththevelocitybeingequaltothegradientofthedisplacement-timegraph(see415)Asthesystemoscillates there is a continual interchange between kinetic energy and potential energy such that the loss in kinetic energyequalsthegaininpotentialenergyandET = EK + EP

Remembering that wehavethat

Clearly thepotentialenergyEP at any displacement x is given by

At any displacement xthekineticenergyEK is EK = 1__2mv2

Hencerememberingthat

v_______( x0

2 ndash x2 ) wehave

EK = 1__2m 2 ( x0

2 ndash x2 )

Although we have derived these equations for a harmonicoscillatortheyarevalidforanysystemoscillatingwithSHMThe sketch graph in Figure 906 shows the variation with displacement x of EK and EP for one period

displacement

ener

gy potentialkinetic

Figure 906 Energy and displacement

Example

The amplitude of oscillation of a mass suspended by a vertical spring is 80 cm The spring constant of the spring is74Nmndash1Determine

(a) thetotalenergyoftheoscillator

(b) thepotentialandthekineticenergyoftheoscillator at a displacement of 48 cm from equilibrium

PHYSICS 2015indb 287 150514 337 PM

Chapter 9

288

AH

L

Solution

(a) Spring constant k = 74 N m-1 and

x0 = 80 times 10-2 m

ET = frac12 kx02

= frac12 times 74 times 64 times 10-4 = 024 J

(b) At x = 48 cm

Therefore EP = frac12 times 74 times (48 times 10-2)2

= 0085 times 10-4 J

EK = ET minus EP = 024 minus 0085 = 016 J

EXERCISE 911 Which graph shows the relationship between the

acceleration a and the displacement x from the equilibriumpositionofanobjectundergoingsimple harmonic motion

Acceleration

Displacement

Acceleration

Displacement

A B

Acceleration

Displacement

Acceleration

Displacement

C D

2 AnobjectisundergoingsimpleharmonicmotionaboutafixedpointPandthemagnitudeofitsdisplacementfromPisx Which one of the following statements is correct

Magnitudeoftheresultant force

Directionoftheresultant force

A Proportionaltox AwayfrompointP

B Inversely proportional to x AwayfrompointP

C Proportionaltox TowardspointP

D Inversely proportional to x TowardspointP

3 Theangularspeedoftheldquominuterdquohandofananalogue watch is

A π1800rads-1 C π30rads-1

B π60rads-1 D 120rads-1

4 Which of the following is true of the magnitude of the acceleration of the object that is undergoing simple harmonic motion

A It is greatest at the midpoint of the motionB It is greatest at the end points of the motionC It is uniform throughout the motionD Itisgreatestatthemidpointsandthe

endpoints

5 Aparticleoscillateswithsimpleharmonicmotionwith a period T

At time t=0theparticlehasitsmaximumdisplacement Which graph The variation with time t of the kinetic energy of the particle is shown in which diagram below

A

00

Ek

T t

D

00

Ek

T t

C

00

Ek

T t

B

0

Ek

t0 T

6 Figure907belowshowshowthedisplacementofthe magnet of mass 03 kg hanging from a spring varies with time for two oscillations

0 02 04 06 08 10

0

2

-2

-4

4

Disp

lace

men

t (cm

)

Time (s)

Figure 907

PHYSICS 2015indb 288 150514 337 PM

Wave Phenomena

289

AH

L

Using information from this graph calculate the

(a) valueofthespringconstant(b) maximumkineticenergyofthemagnet

7 AsystemisoscillatingwithSHMasdescribedbythe graph in the Figure 908 below

Disp

lace

men

t (cm

)

005 1 15 2 25 3 35

2

-2

-6

-4

4

6

Time (s)

Figure 908

(a) Usethegraphtodeterminethe

(i) periodofoscillation(ii) amplitudeofoscillation(iii) maximumspeed(iv) speedatt=13s(v) maximumacceleration

(b) Statetwovaluesoftforwhenthemagnitudeofthevelocity is a maximum and two values of t for when the magnitude of the acceleration is a maximum

9 Figure 909 shows the relationship that exists between the acceleration and the displacement from the equilibriumpositionforaharmonicoscillator

Acce

lera

tion

(m-2

)

Displacement (cm)0

-4000

-2000

2000

-05 050

Figure 909

(a) Stateandexplaintworeasonswhythegraphopposite indicates that the object is executing simple harmonic motion

(b) Determinethefrequencyofoscillation

92 Single-slit diffraction

Essential idea Single-slitdiffractionoccurswhenawaveis incident upon a slit of approximately the same size as the wavelength

Understandingsbull Thenatureofsingle-slitdiffraction

The nature of single-slit diffractionSingle slit diffraction intensity distribution

When plane wavefronts pass through a small aperture they spread out as discussed in chapter 4 This is an example ofthephenomenoncalleddiffractionLightwavesarenoexception to this and ways for observing the diffraction of light have also been discussed previously

Howeverwhenwelookatthediffractionpatternproducedbylightweobserveafringepatternthatisonthescreenthere is a bright central maximum with ldquosecondaryrdquomaxima either side of it There are also regions where there is no illumination and these minima separate the maxima If we were to actually plot how the intensity of illumination varies along the screen then we would obtain a graph similar to that as in Figure 910

intensity

distance along screen

Figure 910 Intensity distribution for single-slit diffraction

We would get the same intensity distribution if we were to plot the intensity against the angle of diffraction θ(Seenextsection)

NATURE OF SCIENCEDevelopment of theories When light passes through an aperture the summation of all parts of the wave leads to an intensity pattern that is far removed from the geometrical shadow that simple theory predicts (19)

copy IBO 2014

PHYSICS 2015indb 289 150514 337 PM

Chapter 9

290

AH

L

This intensity pattern arises from the fact that each point ontheslitactsinaccordancewithHuygenrsquosprincipleasa source of secondary wavefronts It is the interference between these secondary wavefronts that produces the typical diffraction pattern

Obtaining an expression for the intensity distribution is mathematically a little tricky and it is not something that we are going to attempt here However we can deducea useful relationship from a simple argument In this argument we deal with a phenomenon called Fraunhofer diffractionthatisthelightsourceandthescreenareaninfinitedistanceawayformtheslitThiscanbeachievedwith the set up shown in Figure 911

lens 1 lens 2

single slit

source

screen

Figure 911 Apparatus for viewing Fraunhofer diffraction

The source is placed at the principal focus of lens 1 and the screen isplacedat theprincipal focusof lens2Lens1 ensures that parallel wavefronts fall on the single slit and lens2ensuresthattheparallelraysarebroughttoafocuson the screen The same effect can be achieved using a laser and placing the screen some distance from the slit If the light and screenarenot an infinitedistance from the slitthen we are dealing with a phenomenon called Fresnel diffraction and such diffraction is very difficult to analyse mathematically To obtain a good idea of how the single slit patterncomesaboutweconsiderthediagramFigure912

Xθ2

Yλ

d

P

b

f

screen

θ1

Figure 912 Single slit diffraction

In particular we consider the light from one edge of the slittothepointPwherethispointis justonewavelengthfurther from the lower edge of the slit than it is from the

upper edge The secondary wavefront from the upper edge willtraveladistanceλ2furtherthanasecondarywavefrontfrom a point at the centre of the slitHencewhen thesewavefronts arrive at P theywill be out of phase andwillinterfere destructively The wavefronts from the next point below the upper edge will similarly interfere destructively with the wavefront from the next point below the centre of the slit In this way we can pair the sources across the whole width of the slit If the screen is a long way from the slit then the angles θ1 and θ2becomenearlyequal (If the screen is at infinity then they are equal and the two lines PX and XY are at right angles to each other)FromFigure1116weseethereforethattherewillbeaminimumatPif

λ b θ1 sin= where b is the width of the slit

Howeverbothanglesareverysmallequaltoθ saywhereθ is the angle of diffraction

Soitcanbewritten θ λb---

=

This actually gives us the half-angular width of the central maximum We can calculate the actual width of the maximum along the screen if we know the focal length of the lens focussing the light onto the screen If this is f then we have that

θ df---

= Suchthat d fλb-----

=

To obtain the position of the next maximum in the pattern we note that the path difference is 3

2---λ We therefore divide

theslitintothreeequalpartstwoofwhichwillproducewavefronts that will cancel and the other producing wavefronts that reinforce The intensity of the second maximum is therefore much less than the intensity of thecentralmaximumof theorderof005or120of theoriginalintensity(Muchlessthanonethirdinfactsincethewavefrontsthatreinforcewillhavedifferingphases)

We can also see now how diffraction effects become more and more noticeable the narrower the slit becomes If light of wavelength 430 nm was to pass through a slit of width say10cmandfallonascreen30mawaythenthehalfangular width of the central maximum would be 013 microm

d = f λ

___ b = 3 times 430 times 10 ndash9 ____________ 01 = 013 μm

There will be lots of maxima of nearly the same intensity and the maxima will be packed very closely together (Thefirstminimumoccursatadistanceof012micromfromthe centre of the central maximum and the next occurs effectivelyatadistanceof024microm)Weeffectivelyobservethe geometric pattern Refer to Example

We also see now how for diffraction effects to be noticeable the wavelength must be of the order of the slit width The width of the pattern increases in proportion to the wavelength and decreases inversely with the width of the

PHYSICS 2015indb 290 150514 337 PM

Wave Phenomena

291

AH

L

93 Interference

Essential idea Interference patterns from multiple slits andthinfilmsproduceaccuratelyrepeatablepatterns

Understandingsbull Youngrsquosdouble-slitexperimentbull Modulationoftwo-slitinterferencepatternbyone-slit

diffraction effectbull Multipleslitanddiffractiongratinginterferencepatternsbull Thinfilminterference

Youngrsquos double slit experimentPlease refer to interference in chapter 4 However wewill reiterate the condition for the interference of waves from two sources to be observed The two sources must be coherentthatistheymusthavethesamephaseorthephase difference between them must remain constant

Also to reinforce topics concerning the principle ofsuperpositionandpathdifferenceandphasedifferenceincluded in Figure 913 is another example of two source interference

We can obtain evidence for the wave nature of sound by showing that sound produces an interference pattern Figure 913 shows the set up for demonstrating this

CRO

speaker

speaker

signalgenerator

microphoneX

Y

Figure 913 Interference using sound waves

The two speakers are connected to the same output of the signal generator andplaced about 50 cmapartThesignalgeneratorfrequencyissettoabout600Hzandthe

slit If the slit width is much greater than the wavelength then the width of the central maxima is very small

Insummary

for destructive interference and minima

dsinθ=nλ wheren=123hellip

for constructive interference and maxima

dsinθ=(n+frac12)λ wheren=123hellip

Note that waves from point sources along the slit arrive in phase at the centre of the fringe pattern and constructively interferetoproducethecentralmaximumcalledthezero ordermaximum(n=0)

Example

LightfromalaserisusedtoformasingleslitdiffractionpatternThe width of the slit is 010 mm and the screen is placed 30 m from the slit The width of the central maximum is measured as26cmCalculatethewavelengthofthelaserlight

Solution

Since the screen is a long way from the slit we can use the small angle approximation such that the f is equal to 30 m

The half-width of the central maximum is 13 cm so we have

λ13 10 2ndashtimes( ) 10 10 4ndashtimes( )times

30------------------------------------------------------------------=

To give λ = 430 nm

This example demonstrates why the image of a point source formed by a thin converging lens will always have a finite width

Exercise 921 Aparallelbeamoflightofwavelength500nmis

incidentonaslitofwidth025mmThelightisbroughttofocusonascreenplaced150mfromthe slit Calculate the angular width and the linear width of the central diffraction maximum

2 Lightfromalaserisusedtoformasingleslitdiffraction pattern on a screen The width of the slit is 010 mm and the screen is 30 m from the slit The width of the central diffraction maximum is 26cmCalculatethewavelengthofthelaserlight

3 Determinetheanglethatyouwouldexpecttofindconstructiveinterference(forn=0n=1andn=2) fromasingleslitofwidth20μmandlightofwavelength 600 nm

NATURE OF SCIENCECuriosity Observed patterns of iridescence in animals such as the shimmer of peacock feathers led scientists to develop the theory of thin film interference (15)

Serendipity The first laboratory production of thin films was accidental (15)

copy IBO 2014

PHYSICS 2015indb 291 150514 337 PM

Chapter 9

292

AH

L

microphoneismovedalongthelineXYthatisaboutonemetre from the speakers As the microphone is moved along XY the trace on the cathode ray oscilloscope isseen to go through a series of maxima and minima corresponding to points of constructive and destructive interference of the sound waves

Aninterestinginvestigationistofindhowtheseparationof the points of maximum interference depends on the frequency of the source and also the distance apart ofthe speakers

In the demonstration of the interference between two soundsourcesdescribedaboveifweweretomoveoneofthe speakers from side to side or backwards and forwards the sound emitted from the two speakers would no longer be in phase No permanent points of constructive or destructive interference will now be located since the phase difference between the waves from the two sources is nolongerconstantiethesourcesarenolongercoherent

Light from an incandescent source is emitted with acompletely random phase Although the light from two separate sources will interfere because of the randomlychanging phase no permanent points of constructive or destructive interference will be observed This is why a singleslit isneededintheYoungrsquosdoubleslitexperimentByactingasapointsourceitessentiallybecomesacoherentlight source The light emitted from a laser is also very nearly coherent and this is why it is so easy to demonstrate optical interference and diffraction with a laser

Youngrsquos double slit experiment is one of the great classic experiments of physics and did much to reinforce the wave theoryoflightThomasYoungcarriedouttheexperimentin about 1830

It is essentially the demonstration with the ripple tank and the sound experiment previously described butusing light The essential features of the experiment are shown in Figure 914

screendouble slit

single slit

coloured lter

S 1

S2

Figure 914 Youngrsquos double slit experiment

YoungallowedsunlighttofallontoanarrowsingleslitA few centimetres from the single slit he placed a double slit The slits are very narrow and separated by a distance equal to about fourteen slitwidthsA screen is placed

about ametre from thedouble slitsYoungobservedapatternofmulticolouredldquofringesrdquo in the screenWhenhe placed a coloured filter between the single slit anddouble slit he obtained a pattern that consisted of bright coloured fringes separated by darkness

The single slit essentially ensures that the light falling on the double slit is coherent The two slits then act as the two speakers in the sound experiment or the two dippers in the ripple tank The light waves from each slit interfere and produce the interference pattern on the screen Without thefilterapatternisformedforeachwavelengthpresentinthesunlightHencethemulticolouredfringepattern

You can demonstrate optical interference for yourselfSmoking a small piece of glass and then drawing twoparallel lines on it can make a double slit If you then look throughthedoubleslitatasingletungstenfilamentlampyouwillseethefringepatternByplacingfiltersbetweenthe lamps and the slits you will see the monochromatic fringe pattern

YoucanalsoseetheeffectsofopticalinterferencebylookingatnetcurtainsEachlsquoholersquointhenetactsasapointsourceand the light from all these separate sources interferes and producesquiteacomplicatedinterferencepattern

A laser can also be used to demonstrate optical interference Sincethelightfromthelaseriscoherentitisveryeasytodemonstrate interference Just point the laser at a screen and place a double slit in the path of the laser beam

LetusnowlookattheYoungrsquosdoubleslitexperimentinmore detail The geometry of the situation is shown in Figure915

S1

S2 X

P

Q

D

y

d

screen

θ

θrsquo

Figure 915 The geometry of Youngrsquos double slit experiment

S1andS2 are the two narrow slits that we shall regard as twocoherentmonochromaticpointsourcesThedistancefrom the sources to the screen is D and the distance between the slits is d

The waves from the two sources will be in phase at Q and therewill be abright fringehereWewish tofindtheconditionfortheretobeabrightfringeatPdistancey from Q

PHYSICS 2015indb 292 150514 337 PM

Wave Phenomena

293

AH

L

We note that that D (asymp1metre)isverymuchgreaterthaneither y or d(asymp fewmillimetres)Thismeansthatbothθ and θʹ areverysmallanglesandforintentsandpurposesequal

From the diagram we have that

θ = Dy

And

2S Xd

θʹ =

(Remembertheanglesareverysmall)

Sinceθ asymp θʹ then

2S XyD d

=

ButS2X is the path difference between the waves as they

traveltoPWehavethereforethat

path difference = Dyd

TherewillthereforebeabrightfringeatPif

Dyd

= nλ

Supposethatthereisabrightfringeaty = y1 corresponding

to n = n1 then

Ddy1 = n

1λ

If the next bright fringe occurs at y = y2 this will correspond

to n = n1+1Hence

Ddy2 =(n

1+1)λ

This means that the spacing between the fringes y2 ndash y

1 is

given by

y2 ndash y

1 = d

Dλ

Youngactuallyusethisexpressiontomeasurethewavelengthof the light he used and it is a method still used today

We see for instance that if in a given set up we move the slits closer together then the spacing between the fringes will get greaterEffectivelyourinterferencepatternspreadsoutthatis there will be fewer fringes in a given distance We can also increase the fringe spacing by increasing the distance betweentheslitsandthescreenYouwillalsonotethatfora

givensetupusinglightofdifferentwavelengthsthenldquoredrdquofringeswillspacefurtherapartthanldquobluerdquofringes

In this analysis we have assumed that the slits act as point sources and as such the fringes will be uniformly spaced andofequalintensityAmorethoroughanalysisshouldtakeintoaccountthefinitewidthoftheslits

ReturningtoFigure915weseethatwecanwritethepathdifferenceS2X as

S2X = dtan θʹ

But since θʹ is a small angle the sine and tangent will be nearlyequalsothat

S2X = dsinθʹ

And since θʹ asymp θ then

S2X = dsinθ

The condition therefore for a bright fringe to be found at a point of the screen can therefore be written as

dsinθ = nλ

Insummary

for constructive interference and minima

d sinθ = nλ wheren=123hellip

for destructive interference and maxima

d sinθ = ( n + frac12 )λ wheren=123hellip

Figure 916 shows the intensity distribution of the fringes on the screen when the separation of the slits is large comparedtotheirwidthThefringesareofequalintensityandofequalseparation

Intensity

distance along screen

Figure 916 The intensity distribution of the fringes

It isworthnoting that if the slits are close together theintensity of the fringes is modulated by the intensity distribution of the diffraction pattern of one of the slits

PHYSICS 2015indb 293 150514 337 PM

Chapter 9

294

AH

L

Example

Light of wavelength 500nm is incident on two smallparallel slits separated by 10mmDetermine the anglewherethefirstmaximumisformed

If after passing through the slits the light is brought toa focusona screen15m fromtheslitscalculate theobserved fringe spacing on the screen

Solution

Using the small angle approximation we have

3

7

10105minus

minustimes==

dλθ

= 5 times 10-4 rad

The fringe spacing is given by

3

7

1010551

minus

minustimestimes==

dDy λ

= 075 mm

Modulation of two-slit interference pattern by one-slit diffraction effect

In practice the intensity pattern of the maxima as shown in Figure 917 is not constant but fluctuates while decreasingsymmetrically on either side of the central maximum as shown inFigure917Theintensitypatternisacombinationofboththe single-slit diffractionenvelope and the double slit pattern That is the amplitude of the two-slit interference pattern is modulated(ieadjustedto)byasingleslitdiffractionenvelope

2λD

___ λD

D

___λd

d

___

Double slitnite width

Figure 917 Double slit pattern

ForagivenslitseparationdwavelengthoflightandfixedslittoscreendistancethevariationinintensitydependsonthewidthoftheslitDAlthoughincreasingthewidthoftheslitsincreasestheintensityoflightinthefringesitalsomakesthem less sharp and they become blurred As the slit width is widenedthefringesgraduallydisappearbecausenumerouspoint sources along the widened slit give rise to their own dark and bright fringes that then overlap with each other

Exercise 931 InFigure913thedistancebetweenthespeakers

is050mandthedistancebetweenthelineofthespeakersandthescreenis20mAsthemicrophoneismovedalongthelineXYthedistance between successive points of maximum soundintensityis030mThefrequencyofthesoundwavesis44times103HzCalculateavalueforthe speed of sound

2 Laserlightofwavelength610nmfallsontwoslits10x10-5apartDeterminetheseparationofthefirstordermaximumformedonascreen20maway

3 Twoparallelslits012mmapartareilluminatedwith red light with a wavelength of 600 nm An interferencepatternfallsonascreen15maway

Calculate

(a) thedistancefromthecentralmaximumtothefirstbrightfringe

(b) thedistancetotheseconddarkline

Multiple slit and diffraction grating interference patterns

If we examine the interference pattern produced when monochromatic light passes through a different number of slits we notice that as the number of slits increases the number of observed fringes decreases the spacingbetween them increases and the individual fringes become much sharper We can get some idea of how this comes about by looking at the way light behaves when a parallel beam passes through a large number of slits The diagram for this is shown in Figure 918

θθd

1

2

3

telescope

Figure 918 A parallel beam passing through several slits

The slits are very small so that they can be considered to act as point sources They are also very close together such that d issmall(10ndash6 m)EachslitbecomesasourceofcircularwavefrontsandthewavesfromeachslitwillinterfereLetus

PHYSICS 2015indb 294 150514 337 PM

Wave Phenomena

295

AH

L

consider the light that leaves the slit at an angle θ as shown Thepathdifferencebetweenwave1andwave2isdsinθ and ifthisisequaltoanintegralnumberofwavelengthsthenthe two waves will interfere constructively in this direction Similarly wave 2 will interfere constructively with wave3 at this angle andwave 3with 4 etc across thewholegratingHenceifwelookatthelightthroughatelescopethatisbringittoafocusthenwhenthetelescopemakesanangle θ to the grating a bright fringe will be observed The condition for observing a bright fringe is therefore

dsinθ = mλ

Supposeweuselightofwavelength500nmandsupposethat d = 16 times 10ndash6 m

Obviously we will see a bright fringe in the straight on position θ=0(thezero order)

The next position will be when m =1(thefirst order)andsubstitutionintheaboveequationgivesθ = 18deg

The next position will be when m=2(thesecond order)and this give θ = 38deg

For m=3sinθ is greater than 1 so with this set up we only obtain5fringesonezeroorderandtwoeithersideofthezero order

The calculation shows that the separation of the orders is relatively large At any angles other than 18deg or 38deg the light leaving the slits interferes destructively We can see that the fringes will be sharp since if we move just a small angle away from 18deg the light from the slits will interfere destructively

An array of narrow slits is usually made by cutting narrow transparent lines very close together into the emulsion on aphotographicplate(typically200linespermillimetre)Suchanarrangementiscalledadiffraction grating

The diffraction grating is of great use in examining the spectral characteristics of light sources

All elements have their own characteristic spectrum An element can be made to emit light either by heating it until it is incandescent or by causing an electric discharge through itwhen it is in a gaseous state Yourschool probably has some discharge tubes and diffraction gratingsIfithasthenlookattheglowingdischargetubethroughadiffractiongratingIfforexampletheelementthat you are looking has three distinct wavelengths then each wavelength will be diffracted by a different amount inaccordancewiththeequationdsinθ = nλ

Alsoiflaserlightisshonethroughagratingontoascreenyou will see just how sharp and spaced out are the maxima By measuring the line spacing and the distance of the screen fromthelaserthewavelengthofthelasercanbemeasured

If your school has a set of multiple slits say from a single slit toeight slits then it isalsoaworthwhileexercise toexamine how the diffraction pattern changes when laser light is shone through increasing numbers of slits

Consider a grating of NparallelequidistantslitsofwidthaseparatedbyanopaqueregionofwidthD We know that light diffracted from one slit will be superimposed and interfere with light diffracted from all the other slits The intensityprofile of the resultant wave will have a shape determined by

bull theintensityoflightincidentontheslitsbull thewavelengthofthislightbull theslitwidthbull theslitseparation

Typical intensity profiles formed when a plane wavemonochromaticlightatnormalincidenceononetwoandsix slits are shown in Figure 919 In this example the slit widthisonequarterofthegratingspacing

Six slits sharperfringes

d = 4a

I α N 2

(N ndash 2)= 4 subsidiary maxima

order0 4321

I = 36

Two slits

order0

single-slitdiraction

envelope

d = 4a4th principal maximum missing

principa maxima

I = 4 (nominal units)

4321

intensity

order

ISS

One slit

0 1

single-slitdiraction

pattern

N = number of slitsslit width a

Figure 919 Intensity profiles

PHYSICS 2015indb 295 150514 337 PM

Chapter 9

296

AH

L

1 For Nge2thefringepatterncontainsprincipalmaximathat are modulated by the diffraction envelope

2 ForNgt2thefringepatterncontains(Nndash2)subsidiary maxima

3 Where the grating spacing d is four time the slit width a(4xa =d)thenthefourthprincipalmaximumismissing from the fringe pattern Where the grating spacingdisthreetimetheslitwidtha(3xa =d)thenthe third principal maximum is missing from the fringepatternandsoon

4 Increasing N increases the absolute intensities of the diffraction pattern (I = N2)

If white light is shone through a grating then the central image will be white but for the other orders each will be spread out intoacontinuousspectrumcomposedofaninfinitenumberof adjacent images of the slit formed by the wavelength of the different wavelengths present in the white light At any given point in the continuous spectrum the light will be very nearly monochromatic because of the narrowness of the images of the slit formed by the grating This is in contrast to the doubleslitwhereifwhitelightisusedtheimagesarebroadand the spectral colours are not separated

In summary as is the case of interference at a doubleslit there is a path difference between the rays fromadjacent slits of d sinθ Where there is a whole number of wavelengthsconstructiveinterferenceoccurs

For constructive interference and minima

d sinθ = mλ wheren=123hellip

When there is a path difference between adjacent rays of halfawavelengthdestructiveinterferenceoccurs

For and maxima

d sinθ = ( m + frac12 )λ wheren=123hellip

Exercise 941 Lightfromalaserisshonethroughadiffraction

grating on to a screen The screen is a distance of 20mfromthelaserThedistancebetweenthecentraldiffractionmaximumandthefirstprincipalmaximum formed on the screen is 094 m The diffractiongratinghas680linespermmEstimatethe wavelength of the light emitted by the laser

2 Monochromaticlightfromalaserisnormallyincident on a six-slit grating The slit spacing is three times the slit width An interference patternisformedonascreen2montheotherside of the grating

(a) Determineandexplaintheorderoftheprincipal maxima missing from the fringe pattern

(b) Howmanysubsidiarymaximaaretherebetween the principal maxima

(c) Howmanyprincipalmaximaarethere

(i) Betweenthetwofirstorderminimain the diffraction envelope

(ii) Betweenthefirstandsecondorderminima in the diffraction envelope

(d) Drawthepatternformedbythebrightfringesin(c)

(e) Sketchtheintensityprofileofthefringepattern

Thin film interferenceYoumightwellbe familiarwith the colouredpatternoffringes that can be seen when light is reflected off thesurfaceofwateruponwhichathinoilfilmhasbeenspiltorfromlightreflectedfrombubblesWecanseehowthesepatternsarisebylookingatFigure920

Extendedlight source air

A

B

C

12

3

4

Oil lm

Water

Figure 920 Reflection from an oil film

Consider light from an extended source incident on a thinfilmWealsoconsiderawave fromonepointofthe source whose direction is represented by the ray shownSomeofthislightwillbereflectedatAandsometransmitted through the filmwhere somewill again bereflectedatB(somewillalsobetransmittedintotheair)SomeofthelightreflectedatBwillthenbetransmittedatCandsomereflectedandsoonIfweconsiderjustrays1and2thenthesewillbenotbeinphasewhenthearebrought to a focus by the eye since they have travelled

PHYSICS 2015indb 296 150514 337 PM

Wave Phenomena

297

AH

L

different distances If the path difference between them is an integral number of half wavelengths then we might expect the twowaves to be out of phaseHowever ray1 undergoes a phase change of π on reflection but ray2 does not since it is reflected at a boundary betweena more dense and less dense medium (See chapter 4)Henceifthepathdifferenceisanintegralnumberofhalf-wavelengthsrays1and2willreinforceieray1and2areinphaseHowever rays357etcwillbeoutofphasewithrays246etcbutsinceray2ismoreintensethanray3andray4moreintensethanray5thesepairswillnot cancel out so there will be a maximum of intensity

Ifthepathdifferenceissuchthatwave1and2areoutofphasesincewave1ismoreintensethanwave2theywillnotcompletelyannulHoweveritcanbeshownthattheintensitiesofwaves2345hellipaddtoequaltheintensityofwave1Sincewaves345hellipareinphasewithwave2there will be complete cancellation

The path difference will be determined by the angle at which ray1 is incident andalsoon the thickness (andtheactualrefractiveindexaswell)ofthefilmSincethesourceisanextendedsourcethelightwillreachtheeyefrom many different angles and so a system of bright and darkfringeswillbeseenYouwillonlyseefringesifthefilmisverythin(exceptifviewedatnormalincidence)sinceincreasingthethicknessofthefilmwillcausethereflected rays to get so far apart that they will not becollected by the pupil of the eye

From the argument above to find the conditions forconstructive and destructive interferenceweneedonlyfindthepathdifferencebetweenray1andray2Figure921showsthe geometry of the situation

A BE

F

C

D

1 2

φrsquo φrsquo

φ

φ

d

Figure 921 The geometry of interference

Thefilmisofthicknessd and refractive index n and the light has wavelength λ If the line BF is perpendicular to ray 1 then the optical path difference (opd) between ray 1 andray2whenbroughttoafocusis

opd = n(AC+CB)ndashAF

We have to multiple by the refractive index for the path travelledbythelightinthefilmsincethelighttravelsmoreslowly in thefilm If the light travels say adistancex in a material of refractive index n then in the time that it takes to travelthisdistancethelightwouldtraveladistancenx in air

IfthelineCEisatrightanglestoray2thenweseethat

AF = nBE

FromthediagramAC=CDsowecanwrite

opd = n(CD+CB)ndashnBE=nDE

Alsofromthediagramweseethatwhereφ is the angle of refraction

DE=2cosφ

From which opd=2ndcosφ

Bearinginmindthechangeinphaseofray1onreflectionwe have therefore that the condition for constructive interference is

2nd φcos m 12---+

λ m 1 2 hellip = =

Andfordestructiveinterference 2ndcosφ = mλ

Eachfringecorrespondstoaparticularopd for a particular value of the integer m and for any fringe the value of the angle φ is fixedThismeans that itwill be in the formof an arc of a circle with the centre of the circle at the point where the perpendicular drawn from the eye meets the surface of the film Such fringes are called fringesofequal inclinationSince theeyehasa smallaperturethesefringesunlessviewedatneartonormalincidence (φ = 0) will only be observed if the film is very thinThisisbecauseasthethicknessofthefilmincreasesthereflected rayswill get further and further apart and sovery few will enter the eye

Ifwhitelightisshoneontothefilmthenwecanseewhywe get multi-coloured fringes since a series of maxima and minima will be formed for each wavelength present in the white light However when viewed at normalincidenceitispossiblethatonlylightofonecolourwillundergoconstructiveinterferenceandthefilmwilltakeon this colour

PHYSICS 2015indb 297 150514 337 PM

Chapter 9

298

AH

L

Thickness of oil films

The exercise to follow will help explain this use

Non-reflecting films

A very important but simple application of thin filminterferenceisintheproductionofnon-reflectingsurfaces

Athinfilmofthicknessd and refractive index n1 is coated onto glass of refractive index n where n1 lt n Light ofwavelength λ that is reflected at normal incidence willundergo destructive interference if 12

2n d λ

= thatis

14d

nλ

=

(remember that there will now no phase change at the glass interface ie we have a rare to dense reflection)

The use of such films can greatly reduce the loss oflightby reflection at the various surfacesof a systemoflenses or prisms Optical parts of high quality systemsare usually all coatedwithnon-reflecting films in orderto reduce stray reflections The films are usually madebyevaporatingcalciumormagnesiumfluorideonto thesurfacesinvacuumorbychemicaltreatmentwithacidsthat leave a thin layer of silica on the surface of the glass ThecoatedsurfaceshaveapurplishhuebyreflectedlightThis is because the condition for destructive interference fromaparticularfilmthicknesscanonlybeobtainedforone wavelength The wavelength chosen is one that has a value corresponding to light near the middle of the visible spectrumThismeansthatreflectionofredandvioletlightis greater combining to give the purple colour Because of the factor cosφatanglesotherthannormalincidencethepath difference will change butnotsignificantlyuntilsayabout30deg(egcos25deg=090)

It should be borne in mind that no light is actually lost by anon-reflectingfilmthedecreaseofreflectedintensityiscompensated by increase of transmitted intensity

Non-reflecting films can be painted onto aircraft tosuppressreflectionofradarThethicknessofthefilmis determined by

4nd λ

= where λ is the wavelength of

the radar waves and ntherefractiveindexofthefilmat this wavelength

Example

A plane-parallel glass plate of thickness 20 mm isilluminated with light from an extended source The refractiveindexoftheglassis15andthewavelengthofthelight is 600 nm Calculate how many fringes are formed

Solution

We assume that the fringes are formed by light incident at all angles from normal to grazing incidence

At normal incidence we have 2nd = mλ

From which

m 2 15 2 10 3ndashtimestimestimes6 10 7ndashtimes

------------------------------------------- 10 000= =

At grazing incidence the angle of refraction φ is in fact the critical angle

Therefore φφ arcsin 115------- 42 deg= =

ie cosφ = 075

At grazing incidence 2ndcosφ = m λ

From which (and using cos(sinndash1(115) = 075)

m 2 15 2 10 3ndash 075timestimestimestimes6 10 7ndashtimes

------------------------------------------------------------ 7500= =

The total number of fringes seen is equal to m ndash m = 2500

Exercise 95When viewed from above the colour of an oil film onwater appears red in colour Use the data below to estimate theminimumthicknessoftheoilfilm

average wavelength of red light = 630 nm

refractiveindexofoilforredlight=15

refractive index of water for red light = 13

PHYSICS 2015indb 298 150514 337 PM

Wave Phenomena

299

AH

L

94 Resolution

Essential idea Resolution places an absolute limit on the extent to which an optical or other system can separate images of objects

Understandings

bull Thesizeofadiffractingaperturebull Theresolutionofsimplemonochromatic

two-source systems

The size of a diffracting apertureOur discussion concerning diffraction so far has been for rectangular slits When light from a point source enters a smallcircularapertureitdoesnotproduceabrightdotasanimagebutacirculardiscknownasAiryrsquosdiscsurroundedbyfainterconcentriccircularringsasshowninFigure922

Laser

Figure 922 Diffraction at a circular aperture

Diffractionatanapertureisofgreatimportancebecausetheeye and many optical instruments have circular apertures

What is the half-angular width of the central maximum of the diffraction formed by a circular aperture This is not easy to calculate since it involves some advanced mathematics The problemwasfirstsolvedbytheEnglishAstronomerRoyalGeorgeAiryin1835whoshowedthatforcircularapertures

θ 122 λb--------------= where b is the diameter of the aperture

Example 2

Inthefollowingdiagramparallellightfromadistantpointsource(suchasastar)isbroughttofocusonthescreenSbyaconverginglens(the lens is shown as a vertical arrow)

Thefocal length(distance fromlens toscreen) is25cmand the diameter of the lens is 30 cm The wavelength of thelightfromthestaris560nmCalculatethediameterofthe diameter of the image on the screen

Solution

The lens actually acts as a circular aperture of diameter 30 cm The half angular width of central maximum of the diffraction pattern that it forms on the screen is

75

2

122 122 56 10 23 1030 10b

λθminus

minusminus

times times= = = times

times rad

The diameter of the central maxima is therefore

25 times 10-2 times 23 times 10-5

= 57 times 10-6 m

Although this is small it is still finite and is effectively the image of the star as the intensity of the secondary maxima are small compared to that of the central maximum

The resolution of simple monochromatic two-source systemsThe astronomers tell us that many of the stars that we observe with the naked eye are in fact binary stars That iswhatwe see as a single star actually consistsof twostarsinorbitaboutacommoncentreFurthermoretheastronomerstellusthatifweusealdquogoodrdquotelescopethenwewillactuallyseethetwostarsThatiswewillresolve the single point source into its two component parts Sowhatisitthatdetermineswhetherornotweseethetwo stars as a single point source ie what determines whether or not two sources can be resolved It canrsquot just bethatthetelescopemagnifiesthestarssinceiftheyareacting as point sources magnifying them is not going to make a great deal of difference

Ineachofoureyesthereisanaperturethepupilthroughwhich the light enters This light is then focused by the eye lens onto the retina But we have seen that when light passes through an aperture it is diffracted and so when we look at

NATURE OF SCIENCEImproved technology The Rayleigh criterion is the limit of resolution Continuing advancement in technology such as large diameter dishes or lenses or the use of smaller wavelength lasers pushes the limits of what we can resolve (18)

copy IBO 2014

PHYSICS 2015indb 299 150514 337 PM

Chapter 9

300

AH

L

apointsourceadiffractionpatternwillbeformedontheretina If we look at two point sources then two diffraction patterns will be formed on the retina and these patterns will overlap The width of our pupil and the wavelength of the light emitted by the sources will determine the amount by which they overlap But the degree of overlap will also depend on the angular separation of the two point sources WecanseethisfromFigures923

pupilretina

S1

S2

P2

P1

θ

Figure 923

LightfromthesourceS1 enters the eye and is diffracted by the pupil such that the central maximum of the diffraction patternisformedontheretinaatP1SimilarlylightfromS2producesamaximumatP2 If the two central maxima are well separated then there is a fair chance that we will see the two sources as separate sources If they overlap then we will not be able to distinguish one source from another From the diagram we see as the sources are movedclosertotheeyethentheangleθ increases and so does the separation of the central maxima

Figures924925926and927showsthedifferentdiffractionpatternsandtheintensitydistributionthatmightresultonthe retina as a result of light from two point sources

S1 S21 Well resolved

Figure 924 Very well resolved

S1 S22 Well resolved

Figure 925 Well resolved

S1 S23 Just resolved Rayleigh criterion Minimum of S2 coincides with maximum peak of S1

Figure 926 Just resolved

4 Not resolved

Figure 927 Not resolved

We have suggested that if the central maxima of the two diffraction patterns are reasonably separated then we should be able to resolve two point sources In the late 19th century Lord Rayleigh suggested by how much they should be separated in order for the two sources to be just resolved Ifthecentralmaximumofonediffractionpattern coincides with the firstminima of the otherdiffraction pattern then the two sources will just beresolved This is known as the Rayleigh Criterion

Figure926showsjustthissituationThetwosourcesarejust resolved according to the Rayleigh criterion since the peak of the central maximum of one diffraction pattern coincideswiththefirstminimumoftheotherdiffractionpattern This means that the angular separation of the peaks of the two central maxima formed by each source is just the half angular width of one central maximum ie

θ λb---

=

where b is the width of the slit through which the light fromthesourcespassesHoweverweseefromFigure1117that θ is the angle that the two sources subtend at the slit Henceweconcludethattwosourceswillberesolvedbyaslit if the angle that they subtend at the slit is greater than or equaltoλfraslb

PHYSICS 2015indb 300 150514 337 PM

Wave Phenomena

301

AH

L

Sofarwehavebeenassumingthattheeyeisarectangularslit whereas clearly it is a circular aperture and so we must use the formula

θ 122 λb--------------=

As mentioned above the angle θ is sometimes called the resolving power but should more accurately be called the minimum angle of resolution (θmin)

Clearly the smaller θ the greater will be the resolving power

The diffraction grating is a useful for differentiating closely spacedlinesinemissionspectraLikeaprismspectrometerit can disperse a spectrum into its components but it is better suited because it has higher resolution than the prism spectrometer

If λ1 and λ2 are two nearly equal wavelengths that canbarelybedistinguishedtheresolvanceorresolvingpowerofthegratingisdefinedas

R = λ _______ (λ2 ndash λ 1) = λ ___ Δλ

Where λ asymp λ1 asymp λ2

Therefore thediffractiongratingwithahighresolvancewill be better suited in determining small differences in wavelength

If the diffraction grating has Nlinesbeingilluminateditcan be shown that the resolving power of the mth order diffraction is given by

R = mN

Obviouslytheresolvancebecomesgreaterwiththeordernumber m and with a greater number of illuminated slits

Example

A benchmark for the resolving power of a grating is the sodium doublet in the sodium emission spectrum Two yellow lines in itsspectrumhavewavelengthsof58900nmand58959nm

(a) Determinetheresolvanceofagratingifthegivenwavelengths are to be distinguished from each other

(b) Howmanylinesinthegratingmustbeilluminated in order to resolve these lines in the second order spectrum

Solution

(a) R = λ ___ Δλ = 58900nm __________________ (58900ndash58959)nm

= 589nm _______ 059nm = 10 times 103

(b) N = R __ m = 10 times 10 3 ________ 2 =50times102 lines

It has been seen that diffraction effectively limits the resolving power of optical systems This includes such systems as the eye telescopes and microscopes Theresolving power of these systems is dealt with in the next section This section looks at links between technology and resolving power when looking at the very distant and when looking at the very small

Radio telescopes

The average diameter of the pupil of the human eye is about 25 mm This means that the eye will justresolve two point sources emitting light of wavelength 500 nm if their angular separation at the eye is 7

43

50 10122 24 1025 10

θminus

minusminus

times= times = times

times rad

If the eye were to be able to detect radio waves of wavelength 015m then tohave the same resolvingpower thepupilwould have to have a diameter of about 600 m Clearly this is nonsense but it does illustrate a problem facingastronomers who wish to view very distant objects such as quasarsandgalaxies thatemitradiowavesConventionalradio telescopes consist of a large dish typically 25 min diameter Even with such a large diameter the radiowavelength resolving power of the telescope is much less thantheopticalresolvingpowerofthehumaneyeLetuslook at an example

Example

The Galaxy Cygnus A can be resolved optically as an elliptically shaped galaxy However it is also a strongemitterof radiowavesofwavelength015mTheGalaxyisestimatedtobe50times1024mfromEarthUseofaradiotelescope shows that the radio emission is from two sources separated by a distance of 30 times 1021m Estimate thediameterofthedishrequiredtojustresolvethesources

Solution

The angle θ that the sources resolve at the telescope is given by

θ 21

424

30 10 60 1050 10

θ minustimes= = times

times rad

and 4

122 122 015 300060 10

d λθ minus

times= = =

timesm = 30 km

A radio telescope dish of this size would be impossible tomake letalonesupportThisshowsthatasingledishtype radio telescope cannot be used to resolve the sources and yet they were resolved To get round the problemastronomers use two radio telescopes separated by a large

PHYSICS 2015indb 301 150514 337 PM

Chapter 9

302

AH

L

distance The telescopes view the same objects at the same time and the signals that each receive from the objects are simultaneously superimposed The result of the superposition of the two signals is a two-slit interference pattern The pattern has much narrower fringe spacing than that of the diffraction pattern produced by either telescope on its own hence producing a much higherresolvingpowerWhentelescopesareusedlikethistheyare called a stellar interferometer

InSocorroinNewMexicothereisastellarinterferometerthat consists of 27 parabolic dishes each of diameter25 m arranged in a Y-shape that covers an area of 570km2Thisisaso-calledVeryLargeArray(VLA)Evenhigher resolution can be obtained by using an array of radio telescopes in observatories thousands of kilometres apart A system that uses this so-called technique oflsquovery-long-baseline interferometryrsquo (VLBI) is known asalsquovery-long-baselinearrayrsquo(VLBA)WithVLBAaradiowavelength resolving power can be achieved that is 100 timesgreaterthanthebestopticaltelescopesEvenhigherresolving power can be achieved by using a telescope that isinasatelliteorbitingEarthJapanrsquosInstituteofSpaceandAstronautical Science (ISAS) launched sucha system inFebruary 1997TheNationalAstronomicalObservatoryof Japan the National Science Foundationrsquos NationalRadio Astronomy Observatory (NRAO) the CanadianSpaceAgency theAustraliaTelescopeNationalFacilitythe European VLBI Network and the Joint Instituteback this project forVeryLongBaseline Interferometryin EuropeThis project is a very good example of howInternationalismcanoperateinPhysics

Electron microscope

Telescopes are used to look at very distant objects that are very large but because of their distance fromus appearverysmallMicroscopesontheotherhandareusedtolookat objects that are close to us but are physically very small Aswehave seen justmagnifyingobjects that ismakingthemappearlargerisnotsufficientonitsowntogaindetailabouttheobjectfordetailhighresolutionisneeded

Figure928isaschematicofhowanoptical microscope is usedtoviewanobjectandFigure929isaschematicofatransmission electron microscope(TEM)

Optical lens system

bright light source

glass slide containing specimen (object)

eye

Figure 928 The principle of a light microscope

magnetic lens system

electron source

thin wafer of material

ScreenCCD

Figure 929 The principle of an electron microscope

In the optical microscope the resolving power isdetermined by the particular lens system employed and the wavelength λ of the light used For example twopoints in the sample separated by a distance d will just be resolved if

2d mλ=

where m is a property of the lens system know as the numerical aperture In practice the largest value of m obtainable is about 16 Hence if the microscope slideis illuminatedwith light ofwavelength 480 nm a goodmicroscope will resolve two points separated by a distance dasymp15times10-7masymp015micromPointscloser together thanthiswillnotberesolvedHoweverthisisgoodenoughtodistinguishsomevirusessuchastheEbolavirus

Clearlythesmallerλ the higher the resolving power and this is where the electron microscope comes to the fore The electron microscope makes use of the wave nature of electrons(see1315)IntheTEMelectronspassthrougha wafer thin sample and are then focused by a magnetic field onto a fluorescent screen orCCD (charge coupleddevicesee142)Electronsusedinanelectronmicroscopehavewavelengthstypicallyofabout5times10-12mHoweverthe numerical aperture of electron microscopes is considerablysmaller thanthatofanopticalmicroscopetypicallyabout002NonethelessthismeansthataTEMcanresolvetwopointsthatareabout025nmapartThisresolving power is certainly high enough to make out the shape of large molecules

Another type of electron microscope uses a techniqueby which electrons are scattered from the surface of the sample The scattered electrons and then focused as in theTEM to form an image of the surfaceThese so-calledscanningelectronmicroscopes(SEM)havealowerresolving power than TEMrsquos but give very good threedimensional images

The eye

We saw in the last section that the resolving power of the human eye is about 2 times 10-4 rad Suppose that youare looking at car headlights on a dark night and the car

PHYSICS 2015indb 302 150514 337 PM

Wave Phenomena

303

AH

L

is a distance D away If the separation of the headlight is say 15 m then the headlights will subtend an angle θ 15

Dθ =

at your eye Assuming an average wavelength of

500 nm your eye will resolve the headlights into twoseparatesourcesifthisangleequals2times10-4 rad This gives D=75kmInotherwordsifthecarisapproachingyouona straight road then you will be able to distinguish the two headlightsasseparatesourceswhenthecaris75kmawayfrom you Actually because of the structure of the retina and optical defects the resolving power of the average eye is about 6 times 10-4 rad This means that the car is more likely tobe25kmawaybeforeyouresolvetheheadlights

Astronomical telescope

Letusreturntotheexampleofthebinarystarsdiscussedatthe beginning of this section on resolution The stars Kruger A and B form a binary system The average separation of the stars is 14 times 1012mandtheiraveragedistancefromEarthis12times1017mWhenviewedthroughatelescopeonEarththe system will therefore subtend an angle

12

17

14 1012 10

θtimes

=times

= 12times10-5 rad at the objective lens of the telescope Assuming that the average wavelength of the light emitted bythestarsis500nmthenifthetelescopeistoresolvethesystem into two separate images it must have a minimum diameter Dwhere12times10-5 =

7122 500 10D

minustimes times

This gives D=0050m which is about 5 cm So thisparticular system is easily resolved with a small astronomical telescope

Exercise 961 ItissuggestedthatusingtheISASVLBAitwould

bepossibletoldquoseerdquoagrainofriceatadistanceof5000kmEstimatetheresolvingpoweroftheVLBA

2 Thedistancefromtheeyelenstotheretinais20mm The light receptors in the central part of the retinaareabout5times10-6apartDeterminewhetherthe spacing of the receptors will allow for the eye to resolve the headlights in the above discussion whentheyare25kmfromtheeye

3 ThediameterofPlutois23times106 m and its averagedistancefromEarthis60times1012 m EstimatetheminimumdiameteroftheobjectiveofatelescopethatwillenablePlutotobeseenasadisc as opposed to a point source

95 Doppler effect

Essential ideaTheDopplereffectdescribesthephenomenonofwavelengthfrequencyshiftwhenrelativemotionoccurs

Understandingsbull TheDopplereffectforsoundwavesandlightwaves

The Doppler effect for sound waves and light waves

Consider two observers A and B at rest with respect to asoundsourcethatemitsasoundofconstantfrequencyf Clearly both observers will hear a sound of the same frequencyHoweversupposethatthesourcenowmovesat constant speed towards A A will now hear a sound of frequencyfA that is greater than f and B will hear a sound of frequency fB that is less than f This phenomenon is known as the Doppler Effect orDopplerPrinciple after C J Doppler(1803-1853)

The same effect arises for an observer who is either moving towards or away from a stationary source

Figure 930 shows the waves spreading out from a stationary source that emits a sound of constant frequency f The observersAandBhearasoundofthesamefrequency

A B

wavefront

S

Figure 930 Sound waves from a stationary source

Suppose now that the source moves towards A withconstant speed vFigure9310(a)showsasnapshotofthenew wave pattern

NATURE OF SCIENCETechnology Although originally based on physical observations of the pitch of fast moving sources of sound the Doppler effect has an important role in many different areas such as evidence for the expansion of the universe and generating images used in weather reports and in medicine (55)

copy IBO 2014

PHYSICS 2015indb 303 150514 337 PM

Chapter 9

304

AH

L

sourceA B

smallerwavelength

largerwavelength

Figure 931 (a) Sound waves from a moving source

The wavefronts are now crowded together in the direction of travel of the source and stretched out in the opposite direction This is why now the two observers will now hear notesofdifferentfrequenciesHowmuchthewavesbunchtogether and how much they stretch out will depend on the speed vEssentially A

A

cfλ

= and BB

cfλ

= where λA lt λB and v is the speed of sound

IfthesourceisstationaryandAismovingtowardsitthenthe waves from the source incident on A will be bunched up If A is moving away from the stationary source then the waves from the source incident on A will be stretched out

Christian Doppler(1803ndash1853)actuallyappliedtheprinciple(incorrectlyasithappens)totryandexplainthecolourofstars However the Doppler effect does apply to light aswellastosoundIfalightsourceemitsalightoffrequencyf then if it is moving away from an observer the observer willmeasurethelightemittedashavingalowerfrequencythan fSincethesensationofcolourvisionisrelatedtothefrequencyoflight(bluelightisofahigherfrequencythanred light) light emitted by objects moving way from anobserver is often referred to as being red-shifted whereas if the object is moving toward the observer it is referred to as blue-shiftedThisideaisusedinOptionE(Chapter16)

We do not need to consider here the situations where either the source or the observer is accelerating In a situation for example where an accelerating source is approaching astationaryobserverthentheobserverwillhearasoundof ever increasing frequency This sometimes leads toconfusion in describing what is heard when a source approaches passes and then recedes from a stationaryobserver Suppose for example that you are standingon a station platform and a train sounding its whistle is approaching at constant speed What will you hear as the train approaches and then passes through the station As the train approaches you will hear a sound of constant pitch but increasing loudness The pitch of the sound will be greater than if the train were stationary As the train passes through the station you will hear the pitch change at the momentthetrainpassesyoutoasoundagainofconstantpitch The pitch of this sound will be lower than the sound of the approaching train and its intensity will decrease as the train recedes from you What you do not hear is a sound of increasing pitch and then decreasing pitch

The Doppler equations for sound

Although you will not be expected in an IB examination to derivetheequationsassociatedwithaspectsoftheDopplereffectyouwillbeexpectedtoapplythemForcompletenesstherefore thederivationoftheequationsassociatedwiththeDopplereffectasoutlinedaboveisgivenhere

S primeS O

vs

Figure 931 (b)

InFigure931(b)theobserverOisatrestwithrespecttoasourceofsoundSismovingwithconstanspeedvs directly towards O The source is emitting a note of constant frequencyf and the speed of the emitted sound is v

S shows the position of the source ∆t later When the sourceisatresttheninatime∆t the observer will receive f∆t waves and these waves will occupy a distance v∆t ie

v f

λ =f∆tv∆t =

(Because of the motion of the source this number of waves will now occupy a distance (v∆t ndash vs∆t)The lsquonewrsquowavelength is therefore

v ndash vs

fλrsquo =

f∆tv∆t ndash vs∆t

=

If fisthefrequencyheardbyOthen

vfλ

= or

vf

λ = = sv vfminus

From which

s

vf fv v

=minus

Dividingthroughbyv gives

Equation1111

1 sf f v

v=

minus

If the source is moving away from the observer then we have

s

vf fv v

=minus

1

1 sf vv

=+

Equation112

PHYSICS 2015indb 304 150514 337 PM

Wave Phenomena

305

AH

L

We now consider the case where the source is stationary and the observerismoving towardsthesource with speed v0 In this situation the speed of the sound waves as measured by the observer will be v0 + v We therefore have that

v0 + v = ffraslλ = vfλ

times

From which

o1 vf fv

= +

Equation113

If the observerismovingawayfromthesource then

o1 vf fv

= minus

Fromequation(113)wehavethat

o o1 v vf f f f f fv v

∆ = minus = + minus =

Equation114

Thevelocitiesthatwerefertointheaboveequationsarethevelocities with respect to the medium in which the waves fromthesourcetravelHoweverwhenwearedealingwithalight source it is the relative velocity between the source and the observer that we must consider The reason for this is that lightisuniqueintherespectthatthespeedofthelightwavesdoes not depend on the speed of the source All observers irrespective of their speed or the speed of the source will measure the same velocity for the speed of light This is oneof thecornerstonesof theSpecialTheoryofRelativitywhichisdiscussedinmoredetailinOptionH(Chapter18)When applying the Doppler effect to light we aremainlyconcerned with the motion of the source We look here only at the situation where the speed of the source v is much smaller than the speed of light cinfreespace(v ltlt c)Underthesecircumstanceswhenthesourceismovingtowardstheobserverequation111becomes

vf f f fc

minus = ∆ = Equation115

andwhenthesourceismovingawayfromtheobserverequation112becomes vf f f f

cminus = ∆ = minus

Provided that v ltlt c these same equations apply for astationary source and moving observer

We look at the following example and exercise

Example

Asourceemitsasoundoffrequency440HzItmovesinastraight line towards a stationary observer with a speed of 30ms-1Theobserverhearsasoundoffrequency484HzCalculate the speed of sound in air

Solution

We use equation 111 and substitute f= 484 Hz f = 440 Hz

and vs = 30 m s-1

therefore 1

484 440 301v

=minus

such that 30 4401

484vminus = to

give v = 330 m s-1

Example

A particular radio signal from a galaxy is measured as havingafrequencyof139times109HzThesamesignalfromasourceinalaboratoryhasafrequencyof142times109Hz

Suggestwhy thegalaxy ismovingaway fromEarthandcalculateitsrecessionspeed(iethespeedwithwhichitismovingawayfromEarth)

Solution

The fact that the frequency from the moving source is less than that when it is stationary indicates that it is moving away from the stationary observer ie Earth

Using vf fc

∆ = we have

8 96

9

3 10 (142 139) 10 634 10142 10

c fvf∆ times times minus times

= = = timestimes

m s-1

It is usual when dealing with the Doppler effect of light to express speeds as a fraction of c So in this instance we have v = 0021 c

Using the Doppler effectWe have seen in the above example and exercise how the Dopplereffectmaybeusedtomeasuretherecessionspeedof distant galaxies The effect is also used to measure speed inothersituationsHerewewilllookatthegeneralprincipleinvolved in using the Doppler effect to measure speedFigure932showsasource(thetransmitter)thatemitseithersound or em waves of constant frequency f The waves

PHYSICS 2015indb 305 150514 337 PM

Chapter 9

306

AH

L

fromthesourceare incidentonareflectorthat ismovingtowards the transmitter with speed vThereflectedwavesare detected by the receiver placed alongside the transmitter

reector

v

transmitter

receiver

f f

f f

Figure 932 Using the Doppler effect to measure speed

We shall consider the situation where v ltlt c where c is the speed of the waves from the transmitter

For the reflector receiving waves from the transmitterit is effectively an observer moving towards a stationary sourceFromequation(114) itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation116

For the receiver receivingwaves from the reflector it iseffectively a stationary observer receiving waves from a movingsourceFromequation(115)itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation117

Ifweaddequations(116)and(117)wegetthatthetotalDopplershiftatthereceiver∆f is

v vf f f f fc c

minus = ∆ = +

But 1 vf fc

= +

hence

1 v v vf f fc c c

∆ = + +

But since v ltlt cwe can ignore the term2

2

vc when we

expandthebracketintheaboveequation

Therefore we have2vf fc

∆ = Equation118

If v asymp c then we must use the full Doppler equationsHowever for em radiationwewill alwaysonly considersituations in which v ltlt c

Example

Thespeedofsoundinbloodis1500times103 m s-1 Ultrasound offrequency100MHzisreflectedfrombloodflowinginan arteryThe frequencyof the reflectedwaves receivedbackatthetransmitteris105MHzEstimatethespeedofthebloodflowintheartery

Solution

Using equation (118) we have

6 63

2005 10 1015 10

vtimes = times

timesto give v asymp 36 m s-1 (We have assumed that the ultrasound is incident at right angles to the blood flow)

Exercise 971 Judy is standing on the platform of a station A

high speed train is approaching the station in a straight line at constant speed and is sounding its whistleAsthetrainpassesbyJudythefrequencyof the sound emitted by the whistle as heard by Judychangesfrom640Hzto430HzDetermine

(a) thespeedofthetrain

(b) thefrequencyofthesoundemittedbythewhistle as heard by a person on the train (Speedofsound=330ms-1)

2 AgalaxyismovingawayfromEarthwithaspeedof00500cThewavelengthofaparticularspectral line in light emitted by atomic hydrogen inalaboratoryis656times10-7 m Calculate the valueofthewavelengthofthislinemeasuredinthelaboratoryinlightemittedfromasourceofatomic hydrogen in the galaxy

PHYSICS 2015indb 306 150514 337 PM

10 Fields

Contents

101 ndash Describing fields

102 ndash Fields at work

Essential Ideas

Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Similar approaches can be taken in analysing electrical and gravitational potential problems copy IBO 2014

PHYSICS 2015indb 307 150514 337 PM

Chapter 10

308

AH

L

101 Describing fields

Essential idea Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Understandingsbull Gravitationalfieldsbull Electrostaticfieldsbull Electricpotentialandgravitationalpotentialbull Fieldlinesbull Equipotentialsurfaces

Gravitational potentialWehaveseenthatifweliftanobjectofmassm to a height h abovethesurfaceoftheEarththenitsgainingravitationalpotential energy is mghHoweverthisisbynomeansthefullstoryForastartwenowknowthatgvarieswithh and also the expression really gives a difference in potentialenergybetweenthevaluethattheobjecthasattheEarthrsquossurfaceandthevaluethatithasatheighthSowhatwereallyneedisazeropointCanwefindapointwherethepotentialenergyiszeroandusethispointfromwhichtomeasure changes in potential energy

The point that is chosen is in fact infinity At infinity the gravitationalfieldstrengthofanyobjectwillbezeroSoletusseeifwecandeduceanexpressionforthegaininpotentialenergyofanobjectwhenitisldquoliftedrdquofromthesurfaceoftheEarthtoinfinityThisineffectmeansfindingtheworknecessarytoperformthistask

δr

r r δr+

r infin=

mg

AB

Me

Figure 1001 Gravitational forces

Inthediagramweconsidertheworknecessarytomovethe particle of mass m a distance δr in thegravitationalfield of the Earth

The force on the particle at A is FG Mem

r2----------------=

IftheparticleismovedtoB then since δr isverysmallwe can assume that the field remains constant over thedistance ABTheworkδW done againstthegravitationalfieldoftheEarthinmovingthedistanceABis

δWGMem

r2----------------δrndash=

(rememberthatworkdoneagainstaforceisnegative)

TofindthetotalworkdoneW in going from the surface oftheEarthtoinfinitywehavetoaddalltheselittlebitsofworkThisisdonemathematicallybyusingintegralcalculus

WG Me m

r2----------------ndash

rdR

infin

int G Me m 1

r2----- rd

R

infin

intndash G Me m 1r---ndash

R

infinndash= = =

GMe m 0 1R---ndash

ndashndash=

G MemR----------------ndash=

HencewehavewhereRistheradiusoftheEarththattheworkdonebythegravitationalfieldinmovinganobjectofmass m from R(surfaceoftheEarth)toinfinityisgivenby

WGMe m

R---------------- ndash=

We can generalise the result by calculating the worknecessary per unit mass to take a small mass fromthe surface of the Earth to infinity This we call thegravitational potential Vie

Wewouldgetexactlythesameresultifwecalculatedtheworkdonebythefieldtobringthepointmassfrominfinityto the surface of Earth In this respect the formal definition ofgravitationalpotentialatapointinagravitationalfieldis therefore defined as the work done per unit mass in bringing a point mass from infinity to that point

Clearly then the gravitational potential at any point inthe Earthrsquos field distance r from the centre of the Earth (providingr gt R)is

VGMe

r------------ ndash=

Thepotentialisthereforeameasureoftheamountofworkthathastobedonetomoveparticlesbetweenpointsinagravitationalfieldanditsunit is theJkgndash1 We also note thatthepotentialisnegativesothatthepotentialenergy

NATURE OF SCIENCEParadigm shift The move from direct observable actions being responsible for influence on an object to acceptance of a fieldrsquos ldquoaction at a distancerdquo required a paradigm shift in the world of science (23)

copy IBO 2014

PHYSICS 2015indb 308 150514 337 PM

Fields

309

AH

L

aswemoveawayfromtheEarthrsquossurfaceincreasesuntilitreachesthevalueofzeroatinfinity

IfthegravitationalfieldisduetoapointmassofmassmthenwehavethesameexpressionasaboveexceptthatMe is replaced by m andmustalsoexclude thevalueof thepotential at the point mass itself ie at r = 0

WecanexpressthegravitationalpotentialduetotheEarth(orduetoanysphericalmass)intermsofthegravitationalfield strength at its surface

AtthesurfaceoftheEarthwehave

g 0ndash Re GMRe

---------ndash= e

Sothat

g 0Re2 GM= e

Hence at a distance r from the centre of the Earth the gravitationalpotentialVcanbewrittenas

VGMe

r------------ndashg 0 Re

2

r------------ndash= =

The potential at the surface of the Earth

(r = Re)istherefore-g0Re

It is interesting to see how the expression for thegravitational potential ties in with the expression mgh ThepotentialatthesurfaceoftheEarthis-g0Re (see the exampleabove)andataheighthwillbe-g0g 0 Re h+( )ndash ifweassume that g0doesnotchangeover thedistanceh The differenceinpotentialbetweenthesurfaceandtheheighth is therefore g0hSotheworkneededtoraiseanobjectof mass m to a height h is mgh ie m timesdifference ingravitationalpotential

This we have referred to as the gain in gravitationalpotentialenergy(see235)

However this expression can be extended to any twopoints inanygravitationalfieldsuch that ifanobjectofmass mmovesbetween twopointswhosepotentialsareV1 and V2 respectively then the change in gravitationalpotentialenergyoftheobjectism(V1 ndash V2)

Gravitational potential gradient

Let us consider now a region in space where thegravitationalfieldisconstantInFigure912thetwopointsAandBareseparatedbythedistance∆x

A B

direction of uniform gravitational eld of strength I

∆x

Figure 1002 The gravitational potential gradient

The gravitational field is of strength I and is in the direction shownThe gravitational potential at A is V andatBisV +∆V

Theworkdone is takingapointmassmfromAtoB isF∆x = mI∆x

Howeverbydefinitionthisworkisalsoequalto-m∆V

Therefore mI∆x=-m∆V

or VIx

∆= minus

∆Effectivelythissaysthatthemagnitudeofthegravitationalfield strength is equal to the negative gradient of thepotential If I is constant then V is a linear function of x and Iisequaltothenegativegradientofthestraightlinegraph formed by plotting V against x If I is not constant (asusuallythecase)thenthemagnitudeofI at any point in the field can be found by find the gradient of the V-x graph at that point An example of such a calculation can befoundinSection929

For those of you who do HL maths the relationship between field and potential is seen to follow from the expression for the potential of a point mass viz

mV Gr

= minus

2

ddV mG Ir r

minus = + =

Potential due to one or more point masses

Gravitationalpotentialisascalarquantitysocalculatingthepotential due to more than one point mass is a matter of simpleadditionSoforexamplethepotentialV due to the Moon and Earth and a distance x fromthecentreofEarthonastraightlinebetweenthemisgivenbytheexpression

E MM MV Gx r x

= minus + minus

whereME=massofEarthMM= mass of Moon and r = distancebetweencentreofEarthandMoon

PHYSICS 2015indb 309 150514 337 PM

Chapter 10

310

AH

L

Equipotentials and field linesIfthegravitationalpotentialhasthesamevalueatallpointson a surface the surface is said to be an equipotential surfaceSoforexampleifweimagineasphericalshellaboutEarthwhosecentrecoincideswiththecentreofEarththisshellwillbeanequipotentialsurfaceClearlyifwerepresentthegravitationalfieldstrengthbyfieldlinessincethelinesldquoradiaterdquooutfromthecentreofEarththentheselineswillbeatrightanglestothesurfaceIfthefieldlineswerenotnormaltotheequipotentialsurfacethentherewouldbeacomponentof the field parallel to the surfaceThis wouldmean thatpointsonthesurfacewouldbeatdifferentpotentialsandsoitwouldnolongerbeanequipotentialsurfaceThisofcourseholdstrueforanyequipotentialsurface

Figure 1003 shows the field lines and equipotentials fortwopointmassesm

m m

Figure 1003 Equipotentials for two point masses

It is worth noting that we would get exactly the samepattern ifwewere to replace thepointmasseswith twoequalpointcharges(See935)

Escape speedThe potential at the surface of Earth is which meansthattheenergyrequiredtotakeaparticleofmass mfromthesurfacetoinfinityisequalto

ButwhatdoesitactuallymeantotakesomethingtoinfinityWhentheparticleisonthesurfaceoftheEarthwecanthinkofitassittingatthebottomofaldquopotentialwellrdquoasinfigure1004

particle

surface of Earth

innity

GMR---------

FIgure 1004 A potential well

Theldquodepthrdquoofthewellis and if the particle gains an amount of kinetic energy equal to wherem is its massthenitwillhavejustenoughenergytoldquoliftrdquoitoutofthewell

In reality it doesnrsquot actually go to infinity it justmeansthat the particle is effectively free of the gravitationalattractionof theEarthWe say that ithas ldquoescapedrdquo theEarthrsquosgravitationalpullWemeetthisideainconnectionwithmolecular forcesTwomolecules in a solidwill sitat their equilibrium position the separation where therepulsiveforceisequaltotheattractiveforceIfwesupplyjust enough energy to increase the separation of themolecules such that they are an infinite distance apart then intermolecularforcesnolongeraffectthemoleculesandthesolidwillhavebecomealiquidThereisnoincreaseinthekineticenergyofthemoleculesandsothesolidmeltsat constant temperature

Wecancalculatetheescapespeedofanobjectveryeasilyby equating the kinetic energy to the potential energysuch that

12---mvesc ape

2 GMe mRe

----------------=

vesc aperArr2GMe

Re--------------- 2g0 Re = =

Substituting for g0 and Regivesavalueforvescape of about 11kmsndash1 from the surface of the Earth

Youwillnotethattheescapespeeddoesnotdependonthemassoftheobjectsincebothkineticenergyandpotentialenergy are proportional to the mass

IntheoryifyouwanttogetarockettothemoonitcanbedonewithoutreachingtheescapespeedHoweverthiswouldnecessitate anenormousamountof fuel and it islikely that therocketplus fuelwouldbesoheavythat itwouldnevergetoffthegroundItismuchmorepracticalto accelerate the rocket to the escape speed fromEarthorbitandthenintheoryjustlaunchittotheMoon

Example

Use the followingdata todetermine thepotential at thesurface of Mars and the magnitude of the acceleration of free fall

massofMars =64times1023kg

radiusofMars =34times106 m

Determinealsothegravitationalfieldstrengthatadistanceof 68 times 106mabovethesurfaceofMars

PHYSICS 2015indb 310 150514 337 PM

Fields

311

AH

L

Solution

= minus = minus times times = times23

11 76

64 1067 10 ndash13 10

34 10MV GR

minus timestimes

Nkg-1

But V = -g0R

Therefore 7

0 6

13 10 3834 10

VgR

times= minus = =

timesm s-2

To determine the field strength gh at 68 times 106 m above the surface we use the fact that 0 2

Mg GR

= such that GM = g0R2

Therefore 2 2

0h 2 2 2

h h

38 (34) 042(102)

g RGMgR R

times= = = = m s-2

(the distance from the centre is 34 times 106 + 68 times 106 = 102 times 106 m)

Exercise 1011 Thegraphbelowshowshowthegravitational

potentialoutsideoftheEarthvarieswithdistancefrom the centre

0 10 20 30 40 50 60 70

ndash1

ndash2

ndash3

ndash4

ndash5

ndash6

ndash7

r m times 106

V

Jkgndash1

times 10

7

(a) Usethegraphtodeterminethegainingravitationalpotentialenergyofasatelliteofmass200kgasitmovesfromthesurfaceof the Earth to a height of 30 times 107mabovetheEarthrsquossurface

(b) Calculatetheenergyrequiredtotakeit to infinity

(c) DeterminetheslopeofthegraphatthesurfaceoftheEarthmCommenton youranswer

Electric potential energy

The concept of electric potential energy was developedwith gravitational potential energy in mind Just as anobjectnearthesurfaceoftheEarthhaspotentialenergybecause of its gravitational interaction with the Earthsotoothereiselectricalpotentialenergyassociatedwithinteracting charges

Letusfirstlookatacaseoftwopositivepointchargeseachof 1μC that are initially bound together by a thread in a vacuuminspacewithadistancebetweenthemof10cmasshowninFigure1005Whenthethreadiscutthepointchargesinitiallyatrestwouldmoveinoppositedirectionsmovingwithvelocitiesv1 and v2 along the direction of the electrostatic force of repulsion

BEFORE AFTER

v v21

Figure 1005 Interaction of two positive particles

The electric potential energybetweentwopointchargescan be found by simply adding up the energy associated witheachpairofpointchargesForapairofinteractingchargestheelectricpotentialenergyisgivenby

∆U = ∆Ep + ∆Ek = ∆W = ∆Fr = kqQ

____ r2 times r = kqQ

____ r

Becausenoexternalforceisactingonthesystemtheenergyand momentum must be conserved Initially Ek = 0 and Ep =kqQr=9times109 times 1 times 10-1201m=009JWhentheyareagreatdistance fromeachotherEpwillbenegligibleThefinalenergywillbeequaltofrac12mv1

2+frac12mv22=009J

Momentum isalsoconservedand thevelocitieswouldbethe same magnitude but in opposite directions

Electric potential energyismoreoftendefinedintermsofapointchargemovinginanelectricfieldas

lsquothe electricpotential energybetweenany twopoints inan electric field is defined as negativeoftheworkdone by anelectricfieldinmovingapoint electric chargebetweentwolocationsintheelectricfieldrsquo

ΔU = ΔEp = -ΔW = -Fd = qEx

ΔU = ΔEp +ΔEk = ΔW = ΔFr = kqQ r2 x r = kqQ

wherexisthedistancemovedalong(oroppositeto)thedirection of the electric field

Electric potential energyismeasuredinjoule(J)Justasworkisascalarquantitysotooelectricalpotentialenergyis a scalar quantity The negative of the work done by an electric field in moving a unit electric charge between two points is independent of the path taken Inphysicswesaytheelectricfieldisalsquoconservativersquo field

PHYSICS 2015indb 311 150514 337 PM

Chapter 10

312

AH

L

Supposeanexternalforcesuchasyourhandmovesasmallpositive point test charge in the direction of a uniformelectric fieldAs it ismoving itmust be gaining kineticenergyIfthisoccursthentheelectricpotentialenergyofthe unit charge is changing

InFigure1006apointcharge+qismovedbetweenpointsAandBthroughadistancex in a uniform electric field

B

A

x

+ q

Figure 1006 Movement of a positive point charge in a uniform field

InordertomoveapositivepointchargefrompointAtopointBanexternal forcemustbeapplied to thechargeequaltoqE (F = qE)

Since the force is applied through a distance xthennegativeworkhas to bedone tomove the charge because energyis gainedmeaning there is an increaseelectric potential energybetweenthetwopointsRememberthattheworkdoneisequivalenttotheenergygainedorlostinmovingthe charge through the electric field The concept of electric potential energy is only meaningful as the electric field whichgeneratestheforceinquestionisconservative

W F xtimes E q x times= =

θ x cos θ x

Figure 1007 Charge moved at an angle to the field

If a chargemoves at an angle θ to an electric field thecomponent of the displacement parallel to the electric fieldisusedasshowninFigure918

W F x Eq x θcostimes= =

Theelectricpotentialenergyisstoredintheelectricfieldand theelectricfieldwill return theenergy to thepointcharge when required so as not to violate the Law ofconservationofenergy

Electric potential

The electric potential at a point in an electric field is defined as being the work done per unit charge in bringing a small positive point charge from infinity to that point

ΔV = Vinfin ndash Vf = ndash W ___ q

Ifwedesignatethepotentialenergytobezeroatinfinitythen it follows thatelectricpotentialmustalsobezeroat infinity and the electric potential at any point in an electricfieldwillbe

ΔV = ndash W ___ q

NowsupposeweapplyanexternalforcetoasmallpositivetestchargeasitismovedtowardsanisolatedpositivechargeTheexternalforcemustdoworkonthepositivetestchargetomoveittowardstheisolatedpositivechargeandtheworkmustbepositivewhiletheworkdonebytheelectricfieldmustthereforebenegativeSotheelectricpotentialatthatpointmustbepositiveaccordingtotheaboveequationIfanegativeisolatedchargeisusedtheelectricpotentialatapointonthepositivetestchargewouldbenegativePositivepoint chargesof their ownaccordmove fromaplaceofhighelectricpotential toaplaceof lowelectricpotentialNegative point charges move the other way from lowpotentialtohighpotentialInmovingfrompointAtopointBinthediagramthepositivecharge+qismovingfromalowelectricpotentialtoahighelectricpotential

In thedefinitiongiven the termldquoworkperunitchargerdquohas significance If the test charge is +16 times 10-19Cwherethe charge has a potential energy of 32 times 10-17Jthenthepotentialwouldbe32times10-17J+16times10-19C=200JC-1 Now if the charge was doubled the potential wouldbecome 64times10-17 J However the potential per unitchargewouldbethesame

Electric potential is a scalar quantityandithasunitsJC-1 orvoltswhere1voltequalsone joulepercoloumbThevoltallowsustoadoptaunitfortheelectricfieldintermsofthevolt

PreviouslytheunitfortheelectricfieldwasNCndash1

W = qV and F = qEsoW ___ V = F __ E

E = FV ___ W = FV ___ Fm V mndash1

That is the units of the electric field E can also beexpressed as V mndash1

The work done per unit charge in moving a point charge between two points in an electric field is again independent of the path taken

PHYSICS 2015indb 312 150514 337 PM

Fields

313

AH

L

Electric potential due to a point charge

Letus takeapoint rmetres froma chargedobjectThepotentialatthispointcanbecalculatedusingthefollowing

W = ndashFr = ndashqV and F = ndash q1 q2 _____ 4πε0r

2

Therefore

W = ndash q1 q2 _____ 4πε0r

2 times r = ndash q1 q2 _____ 4πε0r

= ndashq1 times q2 _____ 4πε0r

= ndashq1V

That is V q

4πε0r--------------- =

Orsimply

V kqr------

=

Example

Determinehowmuchworkisdonebytheelectricfieldofpointcharge150μCwhenachargeof200μCismovedfrom infinity to apoint 0400m from thepoint charge(Assumenoaccelerationofthecharges)

Solution

The work done by the electric field is W = -qV = -14πε0 times q times (Q rinfin - Q r0400)

W = (- 200 times 10-6 C times 900 times 109 NmC-2 times 150 times 10-6 C) divide 0400 m = - 0675 J

An external force would have to do +0675 J of work

Electric field strength and electric potential gradient

Letus lookbackatFigure1006Supposeagain that thecharge+qismovedasmalldistancebyaforceFfromAtoBsothattheforcecanbeconsideredconstantTheworkdoneisgivenby

∆W F x∆times=

TheforceFandtheelectricfieldEareoppositelydirectedandweknowthat

F = -qE and ΔW = q ΔV

Thereforetheworkdonecanbegivenas

q ΔV = -q E Δx

Therefore E V∆x∆-------

ndash=

The rate of change of potential ΔV at a point with respect to distance Δx in the direction in which the change is maximum is called the potential gradient We say that the electric field = - the potential gradient and the units are Vm-

1 From the equation we can see that in a graph of electric potential versus distance the gradient of the straight line equals the electric field strength

In reality if a charged particle enters a uniform electric field it will be accelerated uniformly by the field and its kinetic energy will increase This is why we had to assume no acceleration in the last worked example

Ek∆ 12---

mv2 q E xsdot sdot q Vx---

xsdot sdot q Vsdot= = = =

Example

Determine how far apart two parallel plates must besituated so that a potential difference of 150 times 102 V produces an electric field strength of 100 times 103 NC-1

Solution

Using E V∆x------- xhArrndash V∆

E-------15 102 Vtimes

100 103 N C 1ndashtimes------------------------------------------= = =

= 150 times 10-1

The plates are 150 times 10-1 m apart

The electric field and the electric potential at a point due to an evenly distributed charge +q on a sphere can be represented graphically as in Figure 1008

r

r

V

E

r0

r0

r0

Charge of +Q evenly distributed over surface

E 14πε0------------ Q

r2----- r r0gt=

V 14πε0------------Q

r----r r0gt=

Figure 1008 Electric field and potential due to a charged sphere

PHYSICS 2015indb 313 150514 337 PM

Chapter 10

314

AH

L

When the sphere becomes charged we know that thechargedistributes itselfevenlyoverthesurfaceThereforeeverypartofthematerialoftheconductorisatthesamepotential As the electric potential at a point is defined as beingnumericallyequaltotheworkdoneinbringingaunitpositivechargefrominfinitytothatpointithasaconstantvalueineverypartofthematerialoftheconductor

Since the potential is the same at all points on the conducting surfacethen∆V∆xiszeroButE=ndash∆V∆xThereforethe electric field inside the conductor is zero There is no electric field inside the conductor

SomefurtherobservationsofthegraphsinFigure1009are

bull OutsidethespherethegraphsobeytherelationshipsgivenasE α 1 r2 and V α 1 r

bull Atthesurfacer = r0Thereforetheelectricfieldandpotentialhavetheminimumvalueforr at this point and this infers a maximum field and potential

bull Insidethespheretheelectricfieldiszerobull Insidethespherenoworkisdonetomoveacharge

fromapointinsidetothesurfaceThereforethereisno potential difference and the potential is the same asitiswhenr = r0

Similargraphscanbedrawnfortheelectricfieldintensityandthe electric potential as a function of distance from conducting parallelplatesandsurfacesandthesearegiveninFigure1009

Potential plot E eld plotE eld

++++

ndashndashndashndash

xx x

+

x x

ndashx x

+ ndashx

x

+ +

x

x

Figure 1009 Electric field and electric potential at a distance from a charged surface

Potential due to one or more point charges

The potential due to one point charge can be determined byusingtheequationformula

V = kq r

Example 1

Determine the electric potential at a point 20 times 10-1 m fromthecentreofan isolatedconducting spherewithapointchargeof40pCinair

Solution

Using the formula

V = kq r we have

V90 10 9times( ) 40 10 12ndashtimes( )times

20 10 1ndashtimes( )------------------------------------------------------------------ 018 V= =

the potential at the point is 180 times 10-1 V

The potential due to a number of point charges can be determined by adding up the potentials due to individual point charges because the electric potential at any point outside a conducting sphere will be the same as if all the charge was concentrated at its centre

Example 2

Threepointchargesofareplacedattheverticesofaright-angledtriangleasshowninthediagrambelowDeterminetheabsolutepotentialat the+20μCchargeduetothetwoothercharges

- 6μC

+3μC +2μC

4 m

3 m

Solution

The electric potential of the +2 μC charge due to the ndash 6 μC charge is

V = (9 times 109 Nm2C-2 times -6 times 10-6 C) divide (radic 32 + 42) m = - 108 times 104 V

PHYSICS 2015indb 314 150514 338 PM

Fields

315

AH

L

Insummarywecanconcludethat

bull No work is done to move a charge along an equipotential

bull Equipotentials are always perpendicular to the electric lines of force

Figure1011and1012showsomeequipotential lines fortwo oppositely charged and identically positive spheresseparated by a distance

+ve ndashve

equipotential lines

Figure 1011 Equipotential lines between two opposite charges

equipotential lines

Figure 1012 Equipotential lines between two charges which are the same

+

ndash ndash ndash ndash

+ + +

40 V

30 V

20 V

10 V

50 V

Figure 1013 Equipotential lines between charged parallel plates

Figure1013showstheequipotentiallinesbetweenchargedparallel plates Throughout this chapter the similarities and differences between gravitational fields and electric fieldshavebeendiscussedTherelationshipsthatexistsbetweengravitationalandelectricquantitiesandtheeffectsofpointmassesandchargesissummarisedinTable1014

The electric potential of the +2 μC charge due to the +3 μC charge is

V = (9 times 109 Nm2C-2 times 3 times 10-6 C) divide 3m = 9 times 103 V

The net absolute potential is the sum of the 2 potentials- 108 times 104 V + 9 times 103 V =- 18 times 103 V

The absolute potential at the point is - 18 times 103 V

Equipotential surfacesRegions in space where the electric potential of a chargedistributionhasaconstantvaluearecalledequipotentials The placeswherethepotentialisconstantinthreedimensionsarecalled equipotential surfacesandwheretheyareconstantintwodimensionstheyarecalledequipotential lines

TheyareinsomewaysanalogoustothecontourlinesontopographicmapsInthiscasethegravitationalpotentialenergy is constant as amassmoves around the contourlines because the mass remains at the same elevationabovetheEarthrsquossurfaceThegravitationalfieldstrengthacts in a direction perpendicular to a contour line

Similarlybecausetheelectricpotentialonanequipotentiallinehas thesamevalueanelectric forcecandonoworkwhena testchargemovesonanequipotentialThereforethe electric field cannot have a component along anequipotentialandthusitmustbeeverywhereperpendiculartotheequipotentialsurfaceorequipotentiallineThisfactmakesiteasytoplotequipotentialsifthelinesofforceorlinesofelectricfluxofanelectricfieldareknown

Forexamplethereareaseriesofequipotentiallinesbetweentwoparallelplateconductorsthatareperpendiculartotheelectric fieldTherewill be a series of concentric circles(eachcirclefurtherapartthanthepreviousone)thatmapouttheequipotentialsaroundanisolatedpositivesphereThe lines of force and some equipotential lines for anisolatedpositivesphereareshowninFigure1010

Lines ofequipotential

Figure 1010 Equipotentials around an isolated positive sphere

PHYSICS 2015indb 315 150514 338 PM

Chapter 10

316

AH

L

Gravitational quantity Electrical quantityQ

uant

ities

g V∆x∆-------ndash= E V∆

x∆-------ndash=

Poin

t mas

ses a

nd ch

arge

s

V Gmr----

ndash= V 14πε0------------q

r---=

g G m

r2-----ndash= E 14πε0------------ q

r2-----=

F Gm1 m2

r2--------------= F 1

4πε0------------

q1q2

r2-----------=

Figure 1014 Formulas

Example

Deduce the electric potential on the surface of a gold nucleus that has a radius of 62 fm

Solution

Using the formula

V = kq r and knowing the atomic number of gold is 79 We will assume the nucleus is spherical and it behaves as if it were a point charge at its centre (relative to outside points)

V = 90 times 109 Nm2C-2 times 79 times 16 times 10-19 C divide 62 times 10-15 m = 18 times 107 V

The potential at the point is 18 MV

Example

Deduce the ionisation energy in electron-volts of theelectron in the hydrogen atom if the electron is in its ground state and it is in a circular orbit at a distance of 53 times 10-11 m from the proton

Solution

This problem is an energy coulombic circular motion question based on Bohrrsquos model of the atom (not the accepted quantum mechanics model) The ionisation energy is the energy required to remove the electron from the ground state to infinity The electron travels in a circular orbit and therefore has a centripetal acceleration The ionisation energy will counteract the coulombic force and the movement of the electron will be in the opposite direction to the centripetal force

Total energy = Ek electron + Ep due to the proton-electron interaction

ΣF = kqQ r2 = mv2 r and as such mv2 = = kqQ r

Therefore Ek electron = frac12 kqQ r

Ep due to the proton-electron interaction = - kqQ r

Total energy = frac12 kqQ r + - kqQ r = -frac12 kqQ r

= - 90 times 109 Nm2C-2 times (16 times 10-19 C)2 divide 53 times 10-11 m = -217 times 10-18 J

= -217 times 10-18 J divide 16 times 10-19 = -136 eV

The ionisation energy is 136 eV

Exercise 1021 ApointchargePisplacedmidwaybetweentwo

identicalnegativechargesWhichoneofthefollowingiscorrectwithregardstoelectricfieldandelectricpotentialatpointP

Electric field Electric potentialA non-zero zeroB zero non-zeroC non-zero non-zeroD zero zero

2 Twopositivechargedspheresaretiedtogetherinavacuumsomewhereinspacewheretherearenoexternal forces A has a mass of 25 g and a charge of20μCandBhasamassof15gandachargeof30μCThedistancebetweenthemis40cmTheyarethenreleasedasshowninthediagram

BEFORE AFTER

v1 v2

A B

(a) Determinetheirinitialelectricpotentialenergy in the before situation

(b) DeterminethespeedofsphereBafterrelease

3 Thediagrambelowrepresentstwoequipotentiallines in separated by a distance of 5 cm in a uniform electric field

+ + + + + + + +

ndash ndash ndash ndash ndash ndash ndash ndash

40 V

20 V 5 cm

PHYSICS 2015indb 316 150514 338 PM

Fields

317

AH

L

Determine the strength of the electric field

4 Thisquestionisabouttheelectricfieldduetoacharged sphere and the motion of electrons in that fieldThediagrambelowshowsanisolatedmetalsphereinavacuumthatcarriesanegativeelectriccharge of 60 μC

ndash

(a) Onthediagramdrawtheconventionalwayto represent the electric field pattern due to the charged sphere and lines to represent threeequipotentialsurfacesintheregionoutside the sphere

(b) Explainhowthelinesrepresentingtheequipotentialsurfacesthatyouhavesketchedindicatethatthestrengthoftheelectricfieldisdecreasingwithdistancefrom the centre of the sphere

(c) Theelectricfieldstrengthatthesurfaceofthe sphere and at points outside the sphere can be determined by assuming that the sphere acts as a point charge of magnitude 60 μC at its centre The radius of the sphere is 25 times 10ndash2 m Deduce that the magnitude of the field strength at the surface of the sphere is 86 times 107 Vmndash1

An electron is initially at rest on the surface of the sphere

(d) (i) Describethepathfollowedbytheelectronasitleavesthesurfaceofthesphere

(ii) Calculatetheinitialaccelerationofthe electron

5 Determinetheamountofworkthatisdoneinmovingachargeof100nCthroughapotentialdifferenceof150times102 V

6 Three identical 20 μC conducting spheres are placedatthecornersofanequilateraltriangleofsides 25 cm The triangle has one apex C pointing upthepageand2baseanglesAandBDeterminetheabsolutepotentialatB

7 Determinehowfaraparttwoparallelplatesmustbesituatedsothatapotentialdifferenceof250 times 102 V produces an electric field strength of 200 times 103 NC-1

8 Thegapbetweentwoparallelplatesis10times10-3mandthereisapotentialdifferenceof10times104 V betweentheplatesCalculate

i theworkdonebyanelectroninmovingfrom one plate to the other

ii thespeedwithwhichtheelectronreachesthe second plate if released from rest

iii theelectricfieldintensitybetweentheplates

9 Anelectronguninapicturetubeisacceleratedby a potential 25 times 103VDeterminethekineticenergygainedbytheelectroninelectron-volts

10 Determine the electric potential 20 times 10-2 m from achargeof-10times10-5 C

11 Determinetheelectricpotentialatapointmid-waybetweenachargeofndash20pCandanotherof+5pConthelinejoiningtheircentresifthecharges are 10 cm apart

12 During a thunderstorm the electric potential differencebetweenacloudandthegroundis10 times 109 V Determine the magnitude of the change in electric potential energy of an electron thatmovesbetweenthesepointsinelectron-volts

13 A charge of 15 μC is placed in a uniform electric fieldoftwooppositelychargedparallelplateswithamagnitudeof14times103 NC-1

(a) Determinetheworkthatmustbedoneagainstthefieldtomovethepointchargeadistance of 55 cm

(b) Calculatethepotentialdifferencebetweenthe final and initial positions of the charge

(c) Determinethepotentialdifferencebetweenthe plates if their separation distance is 15 cm

14 Duringaflashoflightningthepotentialdifferencebetweenacloudandthegroundwas12 times 109 V and the amount of transferred charge was32C

(a) Determinethechangeinenergyofthetransferred charge

(b) Iftheenergyreleasedwasallusedtoacceleratea1tonnecardeduceitsfinalspeed

(c) Iftheenergyreleasedcouldbeusedtomelticeat0degCdeducetheamountoficethatcould be melted

PHYSICS 2015indb 317 150514 338 PM

Chapter 10

318

AH

L

15 SupposethatwhenanelectronmovedfromAtoBin the diagram along an electric field line that the electric field does 36 times 10-19Jofworkonit

Determinethedifferencesinelectricpotential

(a) VB ndash VA(b) VC ndash VA(c) VC ndash VB

16 DeterminethepotentialatpointPthatislocatedatthecentreofthesquareasshowninthediagrambelow

- 6μC

+3μC +2μC

1m

1m

5μC

P

102 Fields at work

Essential idea Similar approaches can be taken inanalysingelectricalandgravitationalpotentialproblems

Understandingsbull Potentialandpotentialenergybull Potentialgradientbull Potentialdifferencebull Escapespeedbull Orbitalmotionorbitalspeedandorbitalenergybull Forcesandinverse-squarelawbehaviour

Orbital motion orbital speed and orbital energy

Although orbital motion may be circular elliptical orparabolic this sub-topic only dealswith circular orbitsThis sub-topic is not fundamentally new physics butan application that synthesizes ideas from gravitationcircularmotiondynamicsandenergy

TheMoonorbitstheEarthandinthissenseitisoftenreferredtoasasatelliteoftheEarthBefore1957itwastheonlyEarthsatelliteHoweverin1957theRussianslaunchedthefirstmanmadesatelliteSputnik1SincethisdatemanymoresatelliteshavebeenlaunchedandtherearenowliterallythousandsofthemorbitingtheEarthSomeareusedtomonitortheweathersome used to enable people to find accurately their position on thesurfaceoftheEarthmanyareusedincommunicationsandnodoubtsomeareusedtospyonothercountriesFigure1015showshowinprincipleasatellitecanbeputintoorbit

Theperson(whosesizeisgreatlyexaggeratedwithrespectto Earth) standing on the surface on the Earth throwssome stonesThe greater the speed with which a stoneis thrown the further it will land from her The pathsfollowedbythethrownstonesareparabolasByastretchof the imaginationwecanvisualisea situation inwhichastoneis thrownwithsuchaspeedthatbecauseof thecurvatureoftheEarthitwillnotlandonthesurfaceoftheEarthbutgointoldquoorbitrdquo(Path4onFigure1015)

NATURE OF SCIENCECommunication of scientific explanations The ability to apply field theory to the unobservable (charges) and the massively scaled (motion of satellites) required scientists to develop new ways to investigate analyse and report findings to a general public used to scientific discoveries based on tangible and discernible evidence (51)

copy IBO 2014

PHYSICS 2015indb 318 150514 338 PM

Fields

319

AH

L

1

2

34Earth

Figure 1015 Throwing a stone into orbit

TheforcethatcausesthestonestofollowaparabolicpathandtofalltoEarthisgravityandsimilarlytheforcethatkeeps the stone in orbit is gravity For circular motionto occur we have seen that a force must act at rightanglestothevelocityofanobjectthatistheremustbeacentripetalforceHenceinthesituationwedescribeherethe centripetal force for circular orbital motion about the EarthisprovidedbygravitationalattractionoftheEarth

We can calculate the speedwithwhich a stonemustbethrowninordertoputitintoorbitjustabovethesurfaceof the Earth

If the stone has mass m and speed v thenwehave fromNewtonrsquos2ndlaw

mv2

RE---------- G

MEm

RE2------------=

whereRE is the radius of the Earth and ME is the mass of the Earth

Bearinginmindthatg 0 GME

RE2--------= then

v g RE 10 64 106timestimes 8 103times= = =

Thatisthestonemustbethrownat8times103m sndash1

Clearly we are not going to get a satellite into orbit soclosetothesurfaceoftheEarthMovingatthisspeedthefrictionduetoairresistancewouldmeltthesatellitebeforeithadtravelledacoupleofkilometresInrealitythereforeasatelliteisputintoorbitabouttheEarthbysendingitattachedtoarocketbeyondtheEarthrsquosatmosphereandthengivingitacomponentofvelocityperpendiculartoaradialvectorfromtheEarthSeeFigure1016

Earth Satellite carried by rocket to hereSatellite orbit

Tangential component of velocity

Figure 1016 Getting a satellite into orbit

Keplerrsquos third law(This work of Kepler and Newtonrsquos synthesis of the work is an excellent example of the scientific method and makes for a good TOK discussion)

In 1627 Johannes Kepler (1571-1630)publishedhis lawsof planetary motion The laws are empirical in natureandwere deduced from the observations of theDanishastronomer Tycho de Brahe (1546-1601)The third lawgives a relationship between the radius of orbit R of a planet and its period TofrevolutionabouttheSunThelawisexpressedmathematicallyas

2

3 constantTR

=

We shall nowuseNewtonrsquos LawofGravitation to showhowitisthattheplanetsmoveinaccordancewithKeplerrsquosthirdlaw

InessenceNewtonwasabletousehislawofgravitytopredict the motion of the planets since all he had to do was factor theF givenby this law intohis second law F = ma to find their accelerations and hence their future positions

InFigure1017theEarthisshownorbitingtheSunandthedistancebetweentheircentresisR

RSun

Earth

F es

F se

Figure 1017 Planets move according to Keplerrsquos third law

Fes is the force that the Earth exerts on the Sun and Fse is the force that the Sun exerts on the Earth The forces are equalandoppositeandtheSunandtheEarthwillactuallyorbitaboutacommoncentreHoweversince theSun issoverymuchmoremassivethantheEarththiscommoncentrewillbeclosetothecentreoftheSunandsowecanregard the Earth as orbiting about the centre of the Sun Theotherthingthatweshallassumeisthatwecanignorethe forces that the other planets exert on the Earth (This wouldnotbeawisethingtodoifyouwereplanningtosend a space ship to theMoon for example)We shallalso assume that we have followed Newtonrsquos exampleandindeedprovedthataspherewillactasapointmasssituated at the centre of the sphere

Kepler had postulated that the orbits of the planets areellipticalbut since theeccentricityof theEarthrsquosorbit issmallweshallassumeacircularorbit

PHYSICS 2015indb 319 150514 338 PM

Chapter 10

320

AH

L

TheaccelerationoftheEarthtowardstheSunisa = Rω2

whereω ω 2πT------=

Hencea R 2π

T------ times

2 4π2 RT 2------------- = =

But the acceleration is given by Newtonrsquos Second Law F = mawhereFisnowgivenbytheLawofGravitationSo in this situation

F maGMs Me

R2-------------------= = butwealsohavethat

a a R 2πT

------ times

2 4π2 RT 2------------- = = and m = Me so that

G MsMe

R2------------------- Me4π2R

T2-------------times

GMs

4π2-----------rArr R3

T2------= =

Butthequantity

isaconstantthathasthesamevalueforeachoftheplanetssowehaveforalltheplanetsnotjustEarththat

R3

T2------ k =

wherekisaconstantWhichisofcourseKeplerrsquosthirdlaw

ThisisindeedanamazingbreakthroughItisdifficulttorefutethe idea that all particles attract each other in accordance withtheLawofGravitationwhenthelawisabletoaccountfortheobservedmotionoftheplanetsabouttheSun

The gravitational effects of the planets upon each othershould produce perturbations in their orbits Such is the predictivepoweroftheUniversalGravitationalLawthatit enabled physicists to compute these perturbations The telescopehadbeeninventedin1608andbythemiddleofthe 18th Century had reached a degree of perfection in design that enabled astronomers to actually measure the orbital perturbations of the planets Their measurements werealways inagreementwith thepredictionsmadebyNewtonrsquoslawHoweverin1781anewplanetUranuswasdiscovered and the orbit of this planet did not fit withthe orbit predicted by Universal Gravitation Such wasthe physicistrsquos faith in theNewtonianmethod that theysuspectedthatthediscrepancywasduetothepresenceofayetundetectedplanetUsingtheLawofGravitationtheFrenchastronomerJLeverrier and the English astronomer J C Adamswereable tocalculate justhowmassive thisnewplanetmustbeandalsowhereitshouldbeIn1846the planetNeptunewas discovered justwhere they hadpredicted Ina similarwaydiscrepancies in theorbitofNeptune ledto thepredictionandsubsequentdiscoveryin1930oftheplanetPlutoNewtonrsquosLawofGravitation

had passed the ultimate test of any theory it is not only abletoexplainexistingdatabutalsotomakepredictions

Satellite energyWhenasatelliteisinorbitaboutaplanetitwillhavebothkineticenergyandgravitationalpotentialenergySupposewe consider a satellite ofmass m that is in an orbit of radius r about a planet of mass M

Thegravitationalpotentialduetotheplanetatdistancer from its centre is

ThegravitationalpotentialenergyofthesatelliteVsat

is therefore GMe m

r----------------ndash

ThatisVsat GMem

r----------------ndash=

Thegravitationalfieldstrengthatthesurfaceoftheplanetisgivenby

g 0GMe

Re2------------=

Hencewecanwrite

Vsat g0 Re

2

rndash=

m

ThekineticenergyofthesatelliteKsatisequaltofrac12mv2wherev is its orbital speed

Byequatingthegravitationalforceactingonthesatellitetoitscentripetalaccelerationwehave

G Mem

r2---------------- mv2

r---------- mv2hArr

GMemr----------------= =

Fromwhich12---

mv2 12---

GMemr----------------times=

= 0 e

2g R

r

2m

Which is actually quite interesting since it shows thatirrespective of the orbital radius the KE is numericallyequal to half the PE Also the total energy E tot K sat Vsat+ 1

2---G Me m

r----------------times G Me m

r----------------ndash + 1

2---G Me m

r----------------ndash= = =of the satelliteisalwaysnegativesince

E tot K sat Vsat+ 12---

G Me mr----------------times

G Me mr----------------ndash

+ 12---

G Me mr----------------ndash= = =

PHYSICS 2015indb 320 150514 338 PM

Fields

321

AH

L

The energies of an orbiting satellite as a function of radial distancefromthecentreofaplanetareshownplottedinFigure1018

12

100806

0402

0ndash02

ndash04

ndash06ndash08ndash10

ndash12

2 4 6 8 10 12

ener

gya

rbita

ry u

nits

distance timesR

kinetic energytotal energy

potential energy

Figure 1018 Energy of an orbiting satellite as a function of distance from the centre of a planet

WeightlessnessSuppose that you are inanelevator(lift)thatisdescendingat constant speedandyou let goof abook thatyouareholdinginyourhandThebookwillfalltothefloorwithaccelerationequaltotheaccelerationduetogravityIfthecablethatsupportstheelevatorweretosnap(asituationthatItrustwillneverhappentoanyofyou)andyounowletgothebookthatyouareholdinginyourotherhandthis book will not fall to the floor - it will stay exactlyin linewithyourhandThis isbecause thebook isnowfallingwith the sameaccelerationas theelevatorandassuch the book cannot ldquocatchrdquo up with the floor of theelevatorFurthermoreifyouhappenedtobestandingonasetofbathroomscalesthescaleswouldnowreadzero-youwouldbeapparentlyweightlessItisthisideaoffreefallthatexplainstheapparentweightlessnessofastronautsin an orbiting satellite These astronauts are in free fall inthesensethattheyareacceleratingtowardsthecentre of the Earth

It is actually possible to define theweight of a body inseveral different waysWe can define it for example asthegravitationalforceexertedonthebodybyaspecifiedobject such as the EarthThiswe have seen thatwe doinlotsofsituationswherewedefinetheweightasbeingequaltomgIfweusethisdefinitionthenanobjectinfreefallcannotbydefinitionbeweightlesssinceitisstillinagravitationalfieldHoweverifwedefinetheweightofanobjectintermsofaldquoweighingrdquoprocesssuchasthereadingonasetofbathroomscaleswhichineffectmeasuresthecontact force between the object and the scales thenclearlyobjectsinfreefallareweightlessOnenowhasto

askthequestionwhetherornotitispossibleForexampletomeasurethegravitationalforceactingonanastronautin orbit about the Earth

WecanalsodefineweightintermsofthenetgravitationalforceactingonabodyduetoseveraldifferentobjectsForexample for an object out in space its weight could bedefined in terms of the resultant of the forces exerted on it bytheSuntheMoontheEarthandalltheotherplanetsin the Solar System If this resultant is zero at a particular pointthenthebodyisweightlessatthispoint

Inviewofthevariousdefinitionsofweightthatareavailabletousitisimportantthatwhenweusethewordldquoweightrdquoweareawareofthecontextinwhichitisbeingused

Example

Calculate the height above the surface of the Earth atwhichageo-stationarysatelliteorbits

Solution

A geo-stationary satellite is one that orbits the Earth in such a way that it is stationary with respect to a point on the surface of the Earth This means that its orbital period must be the same as the time for the Earth to spin once on its axis ie 24 hours

From Keplerrsquos third law we have G Ms

4π2----------- R3

T2------=

That is

RMe

mh

using the fact that the force of attraction between the satellite and the Earth is given by

FG Mem

R2----------------=

and that F = ma

where m is the mass of the satellite and a 4π2R

T2-------------=

we have

GMem

R2---------------- m 4π2R

T2-------------times

GMe

4π2------------rArr R3

T2------= =

PHYSICS 2015indb 321 150514 338 PM

Chapter 10

322

AH

L

Now the mass of the Earth is 60 times 1024 kg and the period T measured in seconds is given by T = 86400 s

So substitution gives R = 42 times 106 m

The radius of the Earth is 64 times 106 m so that the orbital height h is about 36 times 107 m

Example

Calculatetheminimumenergyrequiredtoputasatelliteofmass500kgintoanorbitthatisasaheightequaltotheEarthrsquosradiusabovethesurfaceoftheEarth

Solution

We have seen that when dealing with gravitational fields and potential it is useful to remember that

g 0G M

Re2---------= or g0 Re

2 GM=

The gravitational potential at the surface of the Earth is

g 0ndash Re GMRe

---------ndash=

The gravitational potential at a distance R

from the centre of the Earth is

=

The difference in potential between the surface of the Earth and a point distance R from the centre is therefore

V∆ g 0Re 1ReR------ndash

=

If R 2Re= then V∆g0 Re

2------------=

This means that the work required to ldquoliftrdquo the satellite into orbit is g0Rm where m is the mass of the satellite This is equal to

10 32 106 500timestimestimes = 16000 MJ

However the satellite must also have kinetic energy in order to orbit Earth This will be equal to

g 0 mRe2

2R-----------------g 0mRe

2

4----------------- 8000 MJ= =

The minimum energy required is therefore

24000 MJ

Exercise 1031 The speed needed to put a satellite in orbit does

not depend on

A the radius of the orbitB theshapeoftheorbitC thevalueofgattheorbitD the mass of the satellite

2 EstimatethespeedofanEarthsatellitewhoseorbitis400kmabovetheEarthrsquossurfaceAlsodetermine the period of the orbit

3 Calculatethespeedofa200kgsatelliteorbitingthe Earth at a height of 70 times 106 m

Assume that g = 82 m sndash2 for this orbit

4 TheradiioftwosatellitesXandYorbitingthe Earth are 2r and 8rwherer is the radius of the Earth Calculate the ratio of the periods of revolutionofXtoY

5 A satellite of mass mkgissentfromEarthrsquossurfaceinto an orbit of radius 5RwhereR is the radius of theEarthWritedownanexpressionfor

(a) thepotentialenergyofthesatelliteinorbit

(b) thekineticenergyofthesatelliteinorbit

(c) theminimumworkrequiredtosendthesatellitefromrestattheEarthrsquossurfaceintoits orbit

6 A satellite in an orbit of 10rfallsbacktoEarth(radius r)afteramalfunctionDeterminethespeedwithwhichitwillhittheEarthrsquossurface

7 The radius of the moon is frac14 that of the Earth AssumingEarthandtheMoontohavethesamedensitycomparetheaccelerationsoffreefallatthe surface of Earth to that at the surface of the Moon

8 UsethefollowingdatatodeterminethegravitationalfieldstrengthatthesurfaceoftheMoon and hence determine the escape speed from the surface of the Moon

Mass of the Moon = 73 times 1022kg

RadiusoftheMoon=17times106 m

PHYSICS 2015indb 322 150514 338 PM

Chapter 9

288

AH

L

Solution

(a) Spring constant k = 74 N m-1 and

x0 = 80 times 10-2 m

ET = frac12 kx02

= frac12 times 74 times 64 times 10-4 = 024 J

(b) At x = 48 cm

Therefore EP = frac12 times 74 times (48 times 10-2)2

= 0085 times 10-4 J

EK = ET minus EP = 024 minus 0085 = 016 J

EXERCISE 911 Which graph shows the relationship between the

acceleration a and the displacement x from the equilibriumpositionofanobjectundergoingsimple harmonic motion

Acceleration

Displacement

Acceleration

Displacement

A B

Acceleration

Displacement

Acceleration

Displacement

C D

2 AnobjectisundergoingsimpleharmonicmotionaboutafixedpointPandthemagnitudeofitsdisplacementfromPisx Which one of the following statements is correct

Magnitudeoftheresultant force

Directionoftheresultant force

A Proportionaltox AwayfrompointP

B Inversely proportional to x AwayfrompointP

C Proportionaltox TowardspointP

D Inversely proportional to x TowardspointP

3 Theangularspeedoftheldquominuterdquohandofananalogue watch is

A π1800rads-1 C π30rads-1

B π60rads-1 D 120rads-1

4 Which of the following is true of the magnitude of the acceleration of the object that is undergoing simple harmonic motion

A It is greatest at the midpoint of the motionB It is greatest at the end points of the motionC It is uniform throughout the motionD Itisgreatestatthemidpointsandthe

endpoints

5 Aparticleoscillateswithsimpleharmonicmotionwith a period T

At time t=0theparticlehasitsmaximumdisplacement Which graph The variation with time t of the kinetic energy of the particle is shown in which diagram below

A

00

Ek

T t

D

00

Ek

T t

C

00

Ek

T t

B

0

Ek

t0 T

6 Figure907belowshowshowthedisplacementofthe magnet of mass 03 kg hanging from a spring varies with time for two oscillations

0 02 04 06 08 10

0

2

-2

-4

4

Disp

lace

men

t (cm

)

Time (s)

Figure 907

PHYSICS 2015indb 288 150514 337 PM

Wave Phenomena

289

AH

L

Using information from this graph calculate the

(a) valueofthespringconstant(b) maximumkineticenergyofthemagnet

7 AsystemisoscillatingwithSHMasdescribedbythe graph in the Figure 908 below

Disp

lace

men

t (cm

)

005 1 15 2 25 3 35

2

-2

-6

-4

4

6

Time (s)

Figure 908

(a) Usethegraphtodeterminethe

(i) periodofoscillation(ii) amplitudeofoscillation(iii) maximumspeed(iv) speedatt=13s(v) maximumacceleration

(b) Statetwovaluesoftforwhenthemagnitudeofthevelocity is a maximum and two values of t for when the magnitude of the acceleration is a maximum

9 Figure 909 shows the relationship that exists between the acceleration and the displacement from the equilibriumpositionforaharmonicoscillator

Acce

lera

tion

(m-2

)

Displacement (cm)0

-4000

-2000

2000

-05 050

Figure 909

(a) Stateandexplaintworeasonswhythegraphopposite indicates that the object is executing simple harmonic motion

(b) Determinethefrequencyofoscillation

92 Single-slit diffraction

Essential idea Single-slitdiffractionoccurswhenawaveis incident upon a slit of approximately the same size as the wavelength

Understandingsbull Thenatureofsingle-slitdiffraction

The nature of single-slit diffractionSingle slit diffraction intensity distribution

When plane wavefronts pass through a small aperture they spread out as discussed in chapter 4 This is an example ofthephenomenoncalleddiffractionLightwavesarenoexception to this and ways for observing the diffraction of light have also been discussed previously

Howeverwhenwelookatthediffractionpatternproducedbylightweobserveafringepatternthatisonthescreenthere is a bright central maximum with ldquosecondaryrdquomaxima either side of it There are also regions where there is no illumination and these minima separate the maxima If we were to actually plot how the intensity of illumination varies along the screen then we would obtain a graph similar to that as in Figure 910

intensity

distance along screen

Figure 910 Intensity distribution for single-slit diffraction

We would get the same intensity distribution if we were to plot the intensity against the angle of diffraction θ(Seenextsection)

NATURE OF SCIENCEDevelopment of theories When light passes through an aperture the summation of all parts of the wave leads to an intensity pattern that is far removed from the geometrical shadow that simple theory predicts (19)

copy IBO 2014

PHYSICS 2015indb 289 150514 337 PM

Chapter 9

290

AH

L

This intensity pattern arises from the fact that each point ontheslitactsinaccordancewithHuygenrsquosprincipleasa source of secondary wavefronts It is the interference between these secondary wavefronts that produces the typical diffraction pattern

Obtaining an expression for the intensity distribution is mathematically a little tricky and it is not something that we are going to attempt here However we can deducea useful relationship from a simple argument In this argument we deal with a phenomenon called Fraunhofer diffractionthatisthelightsourceandthescreenareaninfinitedistanceawayformtheslitThiscanbeachievedwith the set up shown in Figure 911

lens 1 lens 2

single slit

source

screen

Figure 911 Apparatus for viewing Fraunhofer diffraction

The source is placed at the principal focus of lens 1 and the screen isplacedat theprincipal focusof lens2Lens1 ensures that parallel wavefronts fall on the single slit and lens2ensuresthattheparallelraysarebroughttoafocuson the screen The same effect can be achieved using a laser and placing the screen some distance from the slit If the light and screenarenot an infinitedistance from the slitthen we are dealing with a phenomenon called Fresnel diffraction and such diffraction is very difficult to analyse mathematically To obtain a good idea of how the single slit patterncomesaboutweconsiderthediagramFigure912

Xθ2

Yλ

d

P

b

f

screen

θ1

Figure 912 Single slit diffraction

In particular we consider the light from one edge of the slittothepointPwherethispointis justonewavelengthfurther from the lower edge of the slit than it is from the

upper edge The secondary wavefront from the upper edge willtraveladistanceλ2furtherthanasecondarywavefrontfrom a point at the centre of the slitHencewhen thesewavefronts arrive at P theywill be out of phase andwillinterfere destructively The wavefronts from the next point below the upper edge will similarly interfere destructively with the wavefront from the next point below the centre of the slit In this way we can pair the sources across the whole width of the slit If the screen is a long way from the slit then the angles θ1 and θ2becomenearlyequal (If the screen is at infinity then they are equal and the two lines PX and XY are at right angles to each other)FromFigure1116weseethereforethattherewillbeaminimumatPif

λ b θ1 sin= where b is the width of the slit

Howeverbothanglesareverysmallequaltoθ saywhereθ is the angle of diffraction

Soitcanbewritten θ λb---

=

This actually gives us the half-angular width of the central maximum We can calculate the actual width of the maximum along the screen if we know the focal length of the lens focussing the light onto the screen If this is f then we have that

θ df---

= Suchthat d fλb-----

=

To obtain the position of the next maximum in the pattern we note that the path difference is 3

2---λ We therefore divide

theslitintothreeequalpartstwoofwhichwillproducewavefronts that will cancel and the other producing wavefronts that reinforce The intensity of the second maximum is therefore much less than the intensity of thecentralmaximumof theorderof005or120of theoriginalintensity(Muchlessthanonethirdinfactsincethewavefrontsthatreinforcewillhavedifferingphases)

We can also see now how diffraction effects become more and more noticeable the narrower the slit becomes If light of wavelength 430 nm was to pass through a slit of width say10cmandfallonascreen30mawaythenthehalfangular width of the central maximum would be 013 microm

d = f λ

___ b = 3 times 430 times 10 ndash9 ____________ 01 = 013 μm

There will be lots of maxima of nearly the same intensity and the maxima will be packed very closely together (Thefirstminimumoccursatadistanceof012micromfromthe centre of the central maximum and the next occurs effectivelyatadistanceof024microm)Weeffectivelyobservethe geometric pattern Refer to Example

We also see now how for diffraction effects to be noticeable the wavelength must be of the order of the slit width The width of the pattern increases in proportion to the wavelength and decreases inversely with the width of the

PHYSICS 2015indb 290 150514 337 PM

Wave Phenomena

291

AH

L

93 Interference

Essential idea Interference patterns from multiple slits andthinfilmsproduceaccuratelyrepeatablepatterns

Understandingsbull Youngrsquosdouble-slitexperimentbull Modulationoftwo-slitinterferencepatternbyone-slit

diffraction effectbull Multipleslitanddiffractiongratinginterferencepatternsbull Thinfilminterference

Youngrsquos double slit experimentPlease refer to interference in chapter 4 However wewill reiterate the condition for the interference of waves from two sources to be observed The two sources must be coherentthatistheymusthavethesamephaseorthephase difference between them must remain constant

Also to reinforce topics concerning the principle ofsuperpositionandpathdifferenceandphasedifferenceincluded in Figure 913 is another example of two source interference

We can obtain evidence for the wave nature of sound by showing that sound produces an interference pattern Figure 913 shows the set up for demonstrating this

CRO

speaker

speaker

signalgenerator

microphoneX

Y

Figure 913 Interference using sound waves

The two speakers are connected to the same output of the signal generator andplaced about 50 cmapartThesignalgeneratorfrequencyissettoabout600Hzandthe

slit If the slit width is much greater than the wavelength then the width of the central maxima is very small

Insummary

for destructive interference and minima

dsinθ=nλ wheren=123hellip

for constructive interference and maxima

dsinθ=(n+frac12)λ wheren=123hellip

Note that waves from point sources along the slit arrive in phase at the centre of the fringe pattern and constructively interferetoproducethecentralmaximumcalledthezero ordermaximum(n=0)

Example

LightfromalaserisusedtoformasingleslitdiffractionpatternThe width of the slit is 010 mm and the screen is placed 30 m from the slit The width of the central maximum is measured as26cmCalculatethewavelengthofthelaserlight

Solution

Since the screen is a long way from the slit we can use the small angle approximation such that the f is equal to 30 m

The half-width of the central maximum is 13 cm so we have

λ13 10 2ndashtimes( ) 10 10 4ndashtimes( )times

30------------------------------------------------------------------=

To give λ = 430 nm

This example demonstrates why the image of a point source formed by a thin converging lens will always have a finite width

Exercise 921 Aparallelbeamoflightofwavelength500nmis

incidentonaslitofwidth025mmThelightisbroughttofocusonascreenplaced150mfromthe slit Calculate the angular width and the linear width of the central diffraction maximum

2 Lightfromalaserisusedtoformasingleslitdiffraction pattern on a screen The width of the slit is 010 mm and the screen is 30 m from the slit The width of the central diffraction maximum is 26cmCalculatethewavelengthofthelaserlight

3 Determinetheanglethatyouwouldexpecttofindconstructiveinterference(forn=0n=1andn=2) fromasingleslitofwidth20μmandlightofwavelength 600 nm

NATURE OF SCIENCECuriosity Observed patterns of iridescence in animals such as the shimmer of peacock feathers led scientists to develop the theory of thin film interference (15)

Serendipity The first laboratory production of thin films was accidental (15)

copy IBO 2014

PHYSICS 2015indb 291 150514 337 PM

Chapter 9

292

AH

L

microphoneismovedalongthelineXYthatisaboutonemetre from the speakers As the microphone is moved along XY the trace on the cathode ray oscilloscope isseen to go through a series of maxima and minima corresponding to points of constructive and destructive interference of the sound waves

Aninterestinginvestigationistofindhowtheseparationof the points of maximum interference depends on the frequency of the source and also the distance apart ofthe speakers

In the demonstration of the interference between two soundsourcesdescribedaboveifweweretomoveoneofthe speakers from side to side or backwards and forwards the sound emitted from the two speakers would no longer be in phase No permanent points of constructive or destructive interference will now be located since the phase difference between the waves from the two sources is nolongerconstantiethesourcesarenolongercoherent

Light from an incandescent source is emitted with acompletely random phase Although the light from two separate sources will interfere because of the randomlychanging phase no permanent points of constructive or destructive interference will be observed This is why a singleslit isneededintheYoungrsquosdoubleslitexperimentByactingasapointsourceitessentiallybecomesacoherentlight source The light emitted from a laser is also very nearly coherent and this is why it is so easy to demonstrate optical interference and diffraction with a laser

Youngrsquos double slit experiment is one of the great classic experiments of physics and did much to reinforce the wave theoryoflightThomasYoungcarriedouttheexperimentin about 1830

It is essentially the demonstration with the ripple tank and the sound experiment previously described butusing light The essential features of the experiment are shown in Figure 914

screendouble slit

single slit

coloured lter

S 1

S2

Figure 914 Youngrsquos double slit experiment

YoungallowedsunlighttofallontoanarrowsingleslitA few centimetres from the single slit he placed a double slit The slits are very narrow and separated by a distance equal to about fourteen slitwidthsA screen is placed

about ametre from thedouble slitsYoungobservedapatternofmulticolouredldquofringesrdquo in the screenWhenhe placed a coloured filter between the single slit anddouble slit he obtained a pattern that consisted of bright coloured fringes separated by darkness

The single slit essentially ensures that the light falling on the double slit is coherent The two slits then act as the two speakers in the sound experiment or the two dippers in the ripple tank The light waves from each slit interfere and produce the interference pattern on the screen Without thefilterapatternisformedforeachwavelengthpresentinthesunlightHencethemulticolouredfringepattern

You can demonstrate optical interference for yourselfSmoking a small piece of glass and then drawing twoparallel lines on it can make a double slit If you then look throughthedoubleslitatasingletungstenfilamentlampyouwillseethefringepatternByplacingfiltersbetweenthe lamps and the slits you will see the monochromatic fringe pattern

YoucanalsoseetheeffectsofopticalinterferencebylookingatnetcurtainsEachlsquoholersquointhenetactsasapointsourceand the light from all these separate sources interferes and producesquiteacomplicatedinterferencepattern

A laser can also be used to demonstrate optical interference Sincethelightfromthelaseriscoherentitisveryeasytodemonstrate interference Just point the laser at a screen and place a double slit in the path of the laser beam

LetusnowlookattheYoungrsquosdoubleslitexperimentinmore detail The geometry of the situation is shown in Figure915

S1

S2 X

P

Q

D

y

d

screen

θ

θrsquo

Figure 915 The geometry of Youngrsquos double slit experiment

S1andS2 are the two narrow slits that we shall regard as twocoherentmonochromaticpointsourcesThedistancefrom the sources to the screen is D and the distance between the slits is d

The waves from the two sources will be in phase at Q and therewill be abright fringehereWewish tofindtheconditionfortheretobeabrightfringeatPdistancey from Q

PHYSICS 2015indb 292 150514 337 PM

Wave Phenomena

293

AH

L

We note that that D (asymp1metre)isverymuchgreaterthaneither y or d(asymp fewmillimetres)Thismeansthatbothθ and θʹ areverysmallanglesandforintentsandpurposesequal

From the diagram we have that

θ = Dy

And

2S Xd

θʹ =

(Remembertheanglesareverysmall)

Sinceθ asymp θʹ then

2S XyD d

=

ButS2X is the path difference between the waves as they

traveltoPWehavethereforethat

path difference = Dyd

TherewillthereforebeabrightfringeatPif

Dyd

= nλ

Supposethatthereisabrightfringeaty = y1 corresponding

to n = n1 then

Ddy1 = n

1λ

If the next bright fringe occurs at y = y2 this will correspond

to n = n1+1Hence

Ddy2 =(n

1+1)λ

This means that the spacing between the fringes y2 ndash y

1 is

given by

y2 ndash y

1 = d

Dλ

Youngactuallyusethisexpressiontomeasurethewavelengthof the light he used and it is a method still used today

We see for instance that if in a given set up we move the slits closer together then the spacing between the fringes will get greaterEffectivelyourinterferencepatternspreadsoutthatis there will be fewer fringes in a given distance We can also increase the fringe spacing by increasing the distance betweentheslitsandthescreenYouwillalsonotethatfora

givensetupusinglightofdifferentwavelengthsthenldquoredrdquofringeswillspacefurtherapartthanldquobluerdquofringes

In this analysis we have assumed that the slits act as point sources and as such the fringes will be uniformly spaced andofequalintensityAmorethoroughanalysisshouldtakeintoaccountthefinitewidthoftheslits

ReturningtoFigure915weseethatwecanwritethepathdifferenceS2X as

S2X = dtan θʹ

But since θʹ is a small angle the sine and tangent will be nearlyequalsothat

S2X = dsinθʹ

And since θʹ asymp θ then

S2X = dsinθ

The condition therefore for a bright fringe to be found at a point of the screen can therefore be written as

dsinθ = nλ

Insummary

for constructive interference and minima

d sinθ = nλ wheren=123hellip

for destructive interference and maxima

d sinθ = ( n + frac12 )λ wheren=123hellip

Figure 916 shows the intensity distribution of the fringes on the screen when the separation of the slits is large comparedtotheirwidthThefringesareofequalintensityandofequalseparation

Intensity

distance along screen

Figure 916 The intensity distribution of the fringes

It isworthnoting that if the slits are close together theintensity of the fringes is modulated by the intensity distribution of the diffraction pattern of one of the slits

PHYSICS 2015indb 293 150514 337 PM

Chapter 9

294

AH

L

Example

Light of wavelength 500nm is incident on two smallparallel slits separated by 10mmDetermine the anglewherethefirstmaximumisformed

If after passing through the slits the light is brought toa focusona screen15m fromtheslitscalculate theobserved fringe spacing on the screen

Solution

Using the small angle approximation we have

3

7

10105minus

minustimes==

dλθ

= 5 times 10-4 rad

The fringe spacing is given by

3

7

1010551

minus

minustimestimes==

dDy λ

= 075 mm

Modulation of two-slit interference pattern by one-slit diffraction effect

In practice the intensity pattern of the maxima as shown in Figure 917 is not constant but fluctuates while decreasingsymmetrically on either side of the central maximum as shown inFigure917Theintensitypatternisacombinationofboththe single-slit diffractionenvelope and the double slit pattern That is the amplitude of the two-slit interference pattern is modulated(ieadjustedto)byasingleslitdiffractionenvelope

2λD

___ λD

D

___λd

d

___

Double slitnite width

Figure 917 Double slit pattern

ForagivenslitseparationdwavelengthoflightandfixedslittoscreendistancethevariationinintensitydependsonthewidthoftheslitDAlthoughincreasingthewidthoftheslitsincreasestheintensityoflightinthefringesitalsomakesthem less sharp and they become blurred As the slit width is widenedthefringesgraduallydisappearbecausenumerouspoint sources along the widened slit give rise to their own dark and bright fringes that then overlap with each other

Exercise 931 InFigure913thedistancebetweenthespeakers

is050mandthedistancebetweenthelineofthespeakersandthescreenis20mAsthemicrophoneismovedalongthelineXYthedistance between successive points of maximum soundintensityis030mThefrequencyofthesoundwavesis44times103HzCalculateavalueforthe speed of sound

2 Laserlightofwavelength610nmfallsontwoslits10x10-5apartDeterminetheseparationofthefirstordermaximumformedonascreen20maway

3 Twoparallelslits012mmapartareilluminatedwith red light with a wavelength of 600 nm An interferencepatternfallsonascreen15maway

Calculate

(a) thedistancefromthecentralmaximumtothefirstbrightfringe

(b) thedistancetotheseconddarkline

Multiple slit and diffraction grating interference patterns

If we examine the interference pattern produced when monochromatic light passes through a different number of slits we notice that as the number of slits increases the number of observed fringes decreases the spacingbetween them increases and the individual fringes become much sharper We can get some idea of how this comes about by looking at the way light behaves when a parallel beam passes through a large number of slits The diagram for this is shown in Figure 918

θθd

1

2

3

telescope

Figure 918 A parallel beam passing through several slits

The slits are very small so that they can be considered to act as point sources They are also very close together such that d issmall(10ndash6 m)EachslitbecomesasourceofcircularwavefrontsandthewavesfromeachslitwillinterfereLetus

PHYSICS 2015indb 294 150514 337 PM

Wave Phenomena

295

AH

L

consider the light that leaves the slit at an angle θ as shown Thepathdifferencebetweenwave1andwave2isdsinθ and ifthisisequaltoanintegralnumberofwavelengthsthenthe two waves will interfere constructively in this direction Similarly wave 2 will interfere constructively with wave3 at this angle andwave 3with 4 etc across thewholegratingHenceifwelookatthelightthroughatelescopethatisbringittoafocusthenwhenthetelescopemakesanangle θ to the grating a bright fringe will be observed The condition for observing a bright fringe is therefore

dsinθ = mλ

Supposeweuselightofwavelength500nmandsupposethat d = 16 times 10ndash6 m

Obviously we will see a bright fringe in the straight on position θ=0(thezero order)

The next position will be when m =1(thefirst order)andsubstitutionintheaboveequationgivesθ = 18deg

The next position will be when m=2(thesecond order)and this give θ = 38deg

For m=3sinθ is greater than 1 so with this set up we only obtain5fringesonezeroorderandtwoeithersideofthezero order

The calculation shows that the separation of the orders is relatively large At any angles other than 18deg or 38deg the light leaving the slits interferes destructively We can see that the fringes will be sharp since if we move just a small angle away from 18deg the light from the slits will interfere destructively

An array of narrow slits is usually made by cutting narrow transparent lines very close together into the emulsion on aphotographicplate(typically200linespermillimetre)Suchanarrangementiscalledadiffraction grating

The diffraction grating is of great use in examining the spectral characteristics of light sources

All elements have their own characteristic spectrum An element can be made to emit light either by heating it until it is incandescent or by causing an electric discharge through itwhen it is in a gaseous state Yourschool probably has some discharge tubes and diffraction gratingsIfithasthenlookattheglowingdischargetubethroughadiffractiongratingIfforexampletheelementthat you are looking has three distinct wavelengths then each wavelength will be diffracted by a different amount inaccordancewiththeequationdsinθ = nλ

Alsoiflaserlightisshonethroughagratingontoascreenyou will see just how sharp and spaced out are the maxima By measuring the line spacing and the distance of the screen fromthelaserthewavelengthofthelasercanbemeasured

If your school has a set of multiple slits say from a single slit toeight slits then it isalsoaworthwhileexercise toexamine how the diffraction pattern changes when laser light is shone through increasing numbers of slits

Consider a grating of NparallelequidistantslitsofwidthaseparatedbyanopaqueregionofwidthD We know that light diffracted from one slit will be superimposed and interfere with light diffracted from all the other slits The intensityprofile of the resultant wave will have a shape determined by

bull theintensityoflightincidentontheslitsbull thewavelengthofthislightbull theslitwidthbull theslitseparation

Typical intensity profiles formed when a plane wavemonochromaticlightatnormalincidenceononetwoandsix slits are shown in Figure 919 In this example the slit widthisonequarterofthegratingspacing

Six slits sharperfringes

d = 4a

I α N 2

(N ndash 2)= 4 subsidiary maxima

order0 4321

I = 36

Two slits

order0

single-slitdiraction

envelope

d = 4a4th principal maximum missing

principa maxima

I = 4 (nominal units)

4321

intensity

order

ISS

One slit

0 1

single-slitdiraction

pattern

N = number of slitsslit width a

Figure 919 Intensity profiles

PHYSICS 2015indb 295 150514 337 PM

Chapter 9

296

AH

L

1 For Nge2thefringepatterncontainsprincipalmaximathat are modulated by the diffraction envelope

2 ForNgt2thefringepatterncontains(Nndash2)subsidiary maxima

3 Where the grating spacing d is four time the slit width a(4xa =d)thenthefourthprincipalmaximumismissing from the fringe pattern Where the grating spacingdisthreetimetheslitwidtha(3xa =d)thenthe third principal maximum is missing from the fringepatternandsoon

4 Increasing N increases the absolute intensities of the diffraction pattern (I = N2)

If white light is shone through a grating then the central image will be white but for the other orders each will be spread out intoacontinuousspectrumcomposedofaninfinitenumberof adjacent images of the slit formed by the wavelength of the different wavelengths present in the white light At any given point in the continuous spectrum the light will be very nearly monochromatic because of the narrowness of the images of the slit formed by the grating This is in contrast to the doubleslitwhereifwhitelightisusedtheimagesarebroadand the spectral colours are not separated

In summary as is the case of interference at a doubleslit there is a path difference between the rays fromadjacent slits of d sinθ Where there is a whole number of wavelengthsconstructiveinterferenceoccurs

For constructive interference and minima

d sinθ = mλ wheren=123hellip

When there is a path difference between adjacent rays of halfawavelengthdestructiveinterferenceoccurs

For and maxima

d sinθ = ( m + frac12 )λ wheren=123hellip

Exercise 941 Lightfromalaserisshonethroughadiffraction

grating on to a screen The screen is a distance of 20mfromthelaserThedistancebetweenthecentraldiffractionmaximumandthefirstprincipalmaximum formed on the screen is 094 m The diffractiongratinghas680linespermmEstimatethe wavelength of the light emitted by the laser

2 Monochromaticlightfromalaserisnormallyincident on a six-slit grating The slit spacing is three times the slit width An interference patternisformedonascreen2montheotherside of the grating

(a) Determineandexplaintheorderoftheprincipal maxima missing from the fringe pattern

(b) Howmanysubsidiarymaximaaretherebetween the principal maxima

(c) Howmanyprincipalmaximaarethere

(i) Betweenthetwofirstorderminimain the diffraction envelope

(ii) Betweenthefirstandsecondorderminima in the diffraction envelope

(d) Drawthepatternformedbythebrightfringesin(c)

(e) Sketchtheintensityprofileofthefringepattern

Thin film interferenceYoumightwellbe familiarwith the colouredpatternoffringes that can be seen when light is reflected off thesurfaceofwateruponwhichathinoilfilmhasbeenspiltorfromlightreflectedfrombubblesWecanseehowthesepatternsarisebylookingatFigure920

Extendedlight source air

A

B

C

12

3

4

Oil lm

Water

Figure 920 Reflection from an oil film

Consider light from an extended source incident on a thinfilmWealsoconsiderawave fromonepointofthe source whose direction is represented by the ray shownSomeofthislightwillbereflectedatAandsometransmitted through the filmwhere somewill again bereflectedatB(somewillalsobetransmittedintotheair)SomeofthelightreflectedatBwillthenbetransmittedatCandsomereflectedandsoonIfweconsiderjustrays1and2thenthesewillbenotbeinphasewhenthearebrought to a focus by the eye since they have travelled

PHYSICS 2015indb 296 150514 337 PM

Wave Phenomena

297

AH

L

different distances If the path difference between them is an integral number of half wavelengths then we might expect the twowaves to be out of phaseHowever ray1 undergoes a phase change of π on reflection but ray2 does not since it is reflected at a boundary betweena more dense and less dense medium (See chapter 4)Henceifthepathdifferenceisanintegralnumberofhalf-wavelengthsrays1and2willreinforceieray1and2areinphaseHowever rays357etcwillbeoutofphasewithrays246etcbutsinceray2ismoreintensethanray3andray4moreintensethanray5thesepairswillnot cancel out so there will be a maximum of intensity

Ifthepathdifferenceissuchthatwave1and2areoutofphasesincewave1ismoreintensethanwave2theywillnotcompletelyannulHoweveritcanbeshownthattheintensitiesofwaves2345hellipaddtoequaltheintensityofwave1Sincewaves345hellipareinphasewithwave2there will be complete cancellation

The path difference will be determined by the angle at which ray1 is incident andalsoon the thickness (andtheactualrefractiveindexaswell)ofthefilmSincethesourceisanextendedsourcethelightwillreachtheeyefrom many different angles and so a system of bright and darkfringeswillbeseenYouwillonlyseefringesifthefilmisverythin(exceptifviewedatnormalincidence)sinceincreasingthethicknessofthefilmwillcausethereflected rays to get so far apart that they will not becollected by the pupil of the eye

From the argument above to find the conditions forconstructive and destructive interferenceweneedonlyfindthepathdifferencebetweenray1andray2Figure921showsthe geometry of the situation

A BE

F

C

D

1 2

φrsquo φrsquo

φ

φ

d

Figure 921 The geometry of interference

Thefilmisofthicknessd and refractive index n and the light has wavelength λ If the line BF is perpendicular to ray 1 then the optical path difference (opd) between ray 1 andray2whenbroughttoafocusis

opd = n(AC+CB)ndashAF

We have to multiple by the refractive index for the path travelledbythelightinthefilmsincethelighttravelsmoreslowly in thefilm If the light travels say adistancex in a material of refractive index n then in the time that it takes to travelthisdistancethelightwouldtraveladistancenx in air

IfthelineCEisatrightanglestoray2thenweseethat

AF = nBE

FromthediagramAC=CDsowecanwrite

opd = n(CD+CB)ndashnBE=nDE

Alsofromthediagramweseethatwhereφ is the angle of refraction

DE=2cosφ

From which opd=2ndcosφ

Bearinginmindthechangeinphaseofray1onreflectionwe have therefore that the condition for constructive interference is

2nd φcos m 12---+

λ m 1 2 hellip = =

Andfordestructiveinterference 2ndcosφ = mλ

Eachfringecorrespondstoaparticularopd for a particular value of the integer m and for any fringe the value of the angle φ is fixedThismeans that itwill be in the formof an arc of a circle with the centre of the circle at the point where the perpendicular drawn from the eye meets the surface of the film Such fringes are called fringesofequal inclinationSince theeyehasa smallaperturethesefringesunlessviewedatneartonormalincidence (φ = 0) will only be observed if the film is very thinThisisbecauseasthethicknessofthefilmincreasesthereflected rayswill get further and further apart and sovery few will enter the eye

Ifwhitelightisshoneontothefilmthenwecanseewhywe get multi-coloured fringes since a series of maxima and minima will be formed for each wavelength present in the white light However when viewed at normalincidenceitispossiblethatonlylightofonecolourwillundergoconstructiveinterferenceandthefilmwilltakeon this colour

PHYSICS 2015indb 297 150514 337 PM

Chapter 9

298

AH

L

Thickness of oil films

The exercise to follow will help explain this use

Non-reflecting films

A very important but simple application of thin filminterferenceisintheproductionofnon-reflectingsurfaces

Athinfilmofthicknessd and refractive index n1 is coated onto glass of refractive index n where n1 lt n Light ofwavelength λ that is reflected at normal incidence willundergo destructive interference if 12

2n d λ

= thatis

14d

nλ

=

(remember that there will now no phase change at the glass interface ie we have a rare to dense reflection)

The use of such films can greatly reduce the loss oflightby reflection at the various surfacesof a systemoflenses or prisms Optical parts of high quality systemsare usually all coatedwithnon-reflecting films in orderto reduce stray reflections The films are usually madebyevaporatingcalciumormagnesiumfluorideonto thesurfacesinvacuumorbychemicaltreatmentwithacidsthat leave a thin layer of silica on the surface of the glass ThecoatedsurfaceshaveapurplishhuebyreflectedlightThis is because the condition for destructive interference fromaparticularfilmthicknesscanonlybeobtainedforone wavelength The wavelength chosen is one that has a value corresponding to light near the middle of the visible spectrumThismeansthatreflectionofredandvioletlightis greater combining to give the purple colour Because of the factor cosφatanglesotherthannormalincidencethepath difference will change butnotsignificantlyuntilsayabout30deg(egcos25deg=090)

It should be borne in mind that no light is actually lost by anon-reflectingfilmthedecreaseofreflectedintensityiscompensated by increase of transmitted intensity

Non-reflecting films can be painted onto aircraft tosuppressreflectionofradarThethicknessofthefilmis determined by

4nd λ

= where λ is the wavelength of

the radar waves and ntherefractiveindexofthefilmat this wavelength

Example

A plane-parallel glass plate of thickness 20 mm isilluminated with light from an extended source The refractiveindexoftheglassis15andthewavelengthofthelight is 600 nm Calculate how many fringes are formed

Solution

We assume that the fringes are formed by light incident at all angles from normal to grazing incidence

At normal incidence we have 2nd = mλ

From which

m 2 15 2 10 3ndashtimestimestimes6 10 7ndashtimes

------------------------------------------- 10 000= =

At grazing incidence the angle of refraction φ is in fact the critical angle

Therefore φφ arcsin 115------- 42 deg= =

ie cosφ = 075

At grazing incidence 2ndcosφ = m λ

From which (and using cos(sinndash1(115) = 075)

m 2 15 2 10 3ndash 075timestimestimestimes6 10 7ndashtimes

------------------------------------------------------------ 7500= =

The total number of fringes seen is equal to m ndash m = 2500

Exercise 95When viewed from above the colour of an oil film onwater appears red in colour Use the data below to estimate theminimumthicknessoftheoilfilm

average wavelength of red light = 630 nm

refractiveindexofoilforredlight=15

refractive index of water for red light = 13

PHYSICS 2015indb 298 150514 337 PM

Wave Phenomena

299

AH

L

94 Resolution

Essential idea Resolution places an absolute limit on the extent to which an optical or other system can separate images of objects

Understandings

bull Thesizeofadiffractingaperturebull Theresolutionofsimplemonochromatic

two-source systems

The size of a diffracting apertureOur discussion concerning diffraction so far has been for rectangular slits When light from a point source enters a smallcircularapertureitdoesnotproduceabrightdotasanimagebutacirculardiscknownasAiryrsquosdiscsurroundedbyfainterconcentriccircularringsasshowninFigure922

Laser

Figure 922 Diffraction at a circular aperture

Diffractionatanapertureisofgreatimportancebecausetheeye and many optical instruments have circular apertures

What is the half-angular width of the central maximum of the diffraction formed by a circular aperture This is not easy to calculate since it involves some advanced mathematics The problemwasfirstsolvedbytheEnglishAstronomerRoyalGeorgeAiryin1835whoshowedthatforcircularapertures

θ 122 λb--------------= where b is the diameter of the aperture

Example 2

Inthefollowingdiagramparallellightfromadistantpointsource(suchasastar)isbroughttofocusonthescreenSbyaconverginglens(the lens is shown as a vertical arrow)

Thefocal length(distance fromlens toscreen) is25cmand the diameter of the lens is 30 cm The wavelength of thelightfromthestaris560nmCalculatethediameterofthe diameter of the image on the screen

Solution

The lens actually acts as a circular aperture of diameter 30 cm The half angular width of central maximum of the diffraction pattern that it forms on the screen is

75

2

122 122 56 10 23 1030 10b

λθminus

minusminus

times times= = = times

times rad

The diameter of the central maxima is therefore

25 times 10-2 times 23 times 10-5

= 57 times 10-6 m

Although this is small it is still finite and is effectively the image of the star as the intensity of the secondary maxima are small compared to that of the central maximum

The resolution of simple monochromatic two-source systemsThe astronomers tell us that many of the stars that we observe with the naked eye are in fact binary stars That iswhatwe see as a single star actually consistsof twostarsinorbitaboutacommoncentreFurthermoretheastronomerstellusthatifweusealdquogoodrdquotelescopethenwewillactuallyseethetwostarsThatiswewillresolve the single point source into its two component parts Sowhatisitthatdetermineswhetherornotweseethetwo stars as a single point source ie what determines whether or not two sources can be resolved It canrsquot just bethatthetelescopemagnifiesthestarssinceiftheyareacting as point sources magnifying them is not going to make a great deal of difference

Ineachofoureyesthereisanaperturethepupilthroughwhich the light enters This light is then focused by the eye lens onto the retina But we have seen that when light passes through an aperture it is diffracted and so when we look at

NATURE OF SCIENCEImproved technology The Rayleigh criterion is the limit of resolution Continuing advancement in technology such as large diameter dishes or lenses or the use of smaller wavelength lasers pushes the limits of what we can resolve (18)

copy IBO 2014

PHYSICS 2015indb 299 150514 337 PM

Chapter 9

300

AH

L

apointsourceadiffractionpatternwillbeformedontheretina If we look at two point sources then two diffraction patterns will be formed on the retina and these patterns will overlap The width of our pupil and the wavelength of the light emitted by the sources will determine the amount by which they overlap But the degree of overlap will also depend on the angular separation of the two point sources WecanseethisfromFigures923

pupilretina

S1

S2

P2

P1

θ

Figure 923

LightfromthesourceS1 enters the eye and is diffracted by the pupil such that the central maximum of the diffraction patternisformedontheretinaatP1SimilarlylightfromS2producesamaximumatP2 If the two central maxima are well separated then there is a fair chance that we will see the two sources as separate sources If they overlap then we will not be able to distinguish one source from another From the diagram we see as the sources are movedclosertotheeyethentheangleθ increases and so does the separation of the central maxima

Figures924925926and927showsthedifferentdiffractionpatternsandtheintensitydistributionthatmightresultonthe retina as a result of light from two point sources

S1 S21 Well resolved

Figure 924 Very well resolved

S1 S22 Well resolved

Figure 925 Well resolved

S1 S23 Just resolved Rayleigh criterion Minimum of S2 coincides with maximum peak of S1

Figure 926 Just resolved

4 Not resolved

Figure 927 Not resolved

We have suggested that if the central maxima of the two diffraction patterns are reasonably separated then we should be able to resolve two point sources In the late 19th century Lord Rayleigh suggested by how much they should be separated in order for the two sources to be just resolved Ifthecentralmaximumofonediffractionpattern coincides with the firstminima of the otherdiffraction pattern then the two sources will just beresolved This is known as the Rayleigh Criterion

Figure926showsjustthissituationThetwosourcesarejust resolved according to the Rayleigh criterion since the peak of the central maximum of one diffraction pattern coincideswiththefirstminimumoftheotherdiffractionpattern This means that the angular separation of the peaks of the two central maxima formed by each source is just the half angular width of one central maximum ie

θ λb---

=

where b is the width of the slit through which the light fromthesourcespassesHoweverweseefromFigure1117that θ is the angle that the two sources subtend at the slit Henceweconcludethattwosourceswillberesolvedbyaslit if the angle that they subtend at the slit is greater than or equaltoλfraslb

PHYSICS 2015indb 300 150514 337 PM

Wave Phenomena

301

AH

L

Sofarwehavebeenassumingthattheeyeisarectangularslit whereas clearly it is a circular aperture and so we must use the formula

θ 122 λb--------------=

As mentioned above the angle θ is sometimes called the resolving power but should more accurately be called the minimum angle of resolution (θmin)

Clearly the smaller θ the greater will be the resolving power

The diffraction grating is a useful for differentiating closely spacedlinesinemissionspectraLikeaprismspectrometerit can disperse a spectrum into its components but it is better suited because it has higher resolution than the prism spectrometer

If λ1 and λ2 are two nearly equal wavelengths that canbarelybedistinguishedtheresolvanceorresolvingpowerofthegratingisdefinedas

R = λ _______ (λ2 ndash λ 1) = λ ___ Δλ

Where λ asymp λ1 asymp λ2

Therefore thediffractiongratingwithahighresolvancewill be better suited in determining small differences in wavelength

If the diffraction grating has Nlinesbeingilluminateditcan be shown that the resolving power of the mth order diffraction is given by

R = mN

Obviouslytheresolvancebecomesgreaterwiththeordernumber m and with a greater number of illuminated slits

Example

A benchmark for the resolving power of a grating is the sodium doublet in the sodium emission spectrum Two yellow lines in itsspectrumhavewavelengthsof58900nmand58959nm

(a) Determinetheresolvanceofagratingifthegivenwavelengths are to be distinguished from each other

(b) Howmanylinesinthegratingmustbeilluminated in order to resolve these lines in the second order spectrum

Solution

(a) R = λ ___ Δλ = 58900nm __________________ (58900ndash58959)nm

= 589nm _______ 059nm = 10 times 103

(b) N = R __ m = 10 times 10 3 ________ 2 =50times102 lines

It has been seen that diffraction effectively limits the resolving power of optical systems This includes such systems as the eye telescopes and microscopes Theresolving power of these systems is dealt with in the next section This section looks at links between technology and resolving power when looking at the very distant and when looking at the very small

Radio telescopes

The average diameter of the pupil of the human eye is about 25 mm This means that the eye will justresolve two point sources emitting light of wavelength 500 nm if their angular separation at the eye is 7

43

50 10122 24 1025 10

θminus

minusminus

times= times = times

times rad

If the eye were to be able to detect radio waves of wavelength 015m then tohave the same resolvingpower thepupilwould have to have a diameter of about 600 m Clearly this is nonsense but it does illustrate a problem facingastronomers who wish to view very distant objects such as quasarsandgalaxies thatemitradiowavesConventionalradio telescopes consist of a large dish typically 25 min diameter Even with such a large diameter the radiowavelength resolving power of the telescope is much less thantheopticalresolvingpowerofthehumaneyeLetuslook at an example

Example

The Galaxy Cygnus A can be resolved optically as an elliptically shaped galaxy However it is also a strongemitterof radiowavesofwavelength015mTheGalaxyisestimatedtobe50times1024mfromEarthUseofaradiotelescope shows that the radio emission is from two sources separated by a distance of 30 times 1021m Estimate thediameterofthedishrequiredtojustresolvethesources

Solution

The angle θ that the sources resolve at the telescope is given by

θ 21

424

30 10 60 1050 10

θ minustimes= = times

times rad

and 4

122 122 015 300060 10

d λθ minus

times= = =

timesm = 30 km

A radio telescope dish of this size would be impossible tomake letalonesupportThisshowsthatasingledishtype radio telescope cannot be used to resolve the sources and yet they were resolved To get round the problemastronomers use two radio telescopes separated by a large

PHYSICS 2015indb 301 150514 337 PM

Chapter 9

302

AH

L

distance The telescopes view the same objects at the same time and the signals that each receive from the objects are simultaneously superimposed The result of the superposition of the two signals is a two-slit interference pattern The pattern has much narrower fringe spacing than that of the diffraction pattern produced by either telescope on its own hence producing a much higherresolvingpowerWhentelescopesareusedlikethistheyare called a stellar interferometer

InSocorroinNewMexicothereisastellarinterferometerthat consists of 27 parabolic dishes each of diameter25 m arranged in a Y-shape that covers an area of 570km2Thisisaso-calledVeryLargeArray(VLA)Evenhigher resolution can be obtained by using an array of radio telescopes in observatories thousands of kilometres apart A system that uses this so-called technique oflsquovery-long-baseline interferometryrsquo (VLBI) is known asalsquovery-long-baselinearrayrsquo(VLBA)WithVLBAaradiowavelength resolving power can be achieved that is 100 timesgreaterthanthebestopticaltelescopesEvenhigherresolving power can be achieved by using a telescope that isinasatelliteorbitingEarthJapanrsquosInstituteofSpaceandAstronautical Science (ISAS) launched sucha system inFebruary 1997TheNationalAstronomicalObservatoryof Japan the National Science Foundationrsquos NationalRadio Astronomy Observatory (NRAO) the CanadianSpaceAgency theAustraliaTelescopeNationalFacilitythe European VLBI Network and the Joint Instituteback this project forVeryLongBaseline Interferometryin EuropeThis project is a very good example of howInternationalismcanoperateinPhysics

Electron microscope

Telescopes are used to look at very distant objects that are very large but because of their distance fromus appearverysmallMicroscopesontheotherhandareusedtolookat objects that are close to us but are physically very small Aswehave seen justmagnifyingobjects that ismakingthemappearlargerisnotsufficientonitsowntogaindetailabouttheobjectfordetailhighresolutionisneeded

Figure928isaschematicofhowanoptical microscope is usedtoviewanobjectandFigure929isaschematicofatransmission electron microscope(TEM)

Optical lens system

bright light source

glass slide containing specimen (object)

eye

Figure 928 The principle of a light microscope

magnetic lens system

electron source

thin wafer of material

ScreenCCD

Figure 929 The principle of an electron microscope

In the optical microscope the resolving power isdetermined by the particular lens system employed and the wavelength λ of the light used For example twopoints in the sample separated by a distance d will just be resolved if

2d mλ=

where m is a property of the lens system know as the numerical aperture In practice the largest value of m obtainable is about 16 Hence if the microscope slideis illuminatedwith light ofwavelength 480 nm a goodmicroscope will resolve two points separated by a distance dasymp15times10-7masymp015micromPointscloser together thanthiswillnotberesolvedHoweverthisisgoodenoughtodistinguishsomevirusessuchastheEbolavirus

Clearlythesmallerλ the higher the resolving power and this is where the electron microscope comes to the fore The electron microscope makes use of the wave nature of electrons(see1315)IntheTEMelectronspassthrougha wafer thin sample and are then focused by a magnetic field onto a fluorescent screen orCCD (charge coupleddevicesee142)Electronsusedinanelectronmicroscopehavewavelengthstypicallyofabout5times10-12mHoweverthe numerical aperture of electron microscopes is considerablysmaller thanthatofanopticalmicroscopetypicallyabout002NonethelessthismeansthataTEMcanresolvetwopointsthatareabout025nmapartThisresolving power is certainly high enough to make out the shape of large molecules

Another type of electron microscope uses a techniqueby which electrons are scattered from the surface of the sample The scattered electrons and then focused as in theTEM to form an image of the surfaceThese so-calledscanningelectronmicroscopes(SEM)havealowerresolving power than TEMrsquos but give very good threedimensional images

The eye

We saw in the last section that the resolving power of the human eye is about 2 times 10-4 rad Suppose that youare looking at car headlights on a dark night and the car

PHYSICS 2015indb 302 150514 337 PM

Wave Phenomena

303

AH

L

is a distance D away If the separation of the headlight is say 15 m then the headlights will subtend an angle θ 15

Dθ =

at your eye Assuming an average wavelength of

500 nm your eye will resolve the headlights into twoseparatesourcesifthisangleequals2times10-4 rad This gives D=75kmInotherwordsifthecarisapproachingyouona straight road then you will be able to distinguish the two headlightsasseparatesourceswhenthecaris75kmawayfrom you Actually because of the structure of the retina and optical defects the resolving power of the average eye is about 6 times 10-4 rad This means that the car is more likely tobe25kmawaybeforeyouresolvetheheadlights

Astronomical telescope

Letusreturntotheexampleofthebinarystarsdiscussedatthe beginning of this section on resolution The stars Kruger A and B form a binary system The average separation of the stars is 14 times 1012mandtheiraveragedistancefromEarthis12times1017mWhenviewedthroughatelescopeonEarththe system will therefore subtend an angle

12

17

14 1012 10

θtimes

=times

= 12times10-5 rad at the objective lens of the telescope Assuming that the average wavelength of the light emitted bythestarsis500nmthenifthetelescopeistoresolvethesystem into two separate images it must have a minimum diameter Dwhere12times10-5 =

7122 500 10D

minustimes times

This gives D=0050m which is about 5 cm So thisparticular system is easily resolved with a small astronomical telescope

Exercise 961 ItissuggestedthatusingtheISASVLBAitwould

bepossibletoldquoseerdquoagrainofriceatadistanceof5000kmEstimatetheresolvingpoweroftheVLBA

2 Thedistancefromtheeyelenstotheretinais20mm The light receptors in the central part of the retinaareabout5times10-6apartDeterminewhetherthe spacing of the receptors will allow for the eye to resolve the headlights in the above discussion whentheyare25kmfromtheeye

3 ThediameterofPlutois23times106 m and its averagedistancefromEarthis60times1012 m EstimatetheminimumdiameteroftheobjectiveofatelescopethatwillenablePlutotobeseenasadisc as opposed to a point source

95 Doppler effect

Essential ideaTheDopplereffectdescribesthephenomenonofwavelengthfrequencyshiftwhenrelativemotionoccurs

Understandingsbull TheDopplereffectforsoundwavesandlightwaves

The Doppler effect for sound waves and light waves

Consider two observers A and B at rest with respect to asoundsourcethatemitsasoundofconstantfrequencyf Clearly both observers will hear a sound of the same frequencyHoweversupposethatthesourcenowmovesat constant speed towards A A will now hear a sound of frequencyfA that is greater than f and B will hear a sound of frequency fB that is less than f This phenomenon is known as the Doppler Effect orDopplerPrinciple after C J Doppler(1803-1853)

The same effect arises for an observer who is either moving towards or away from a stationary source

Figure 930 shows the waves spreading out from a stationary source that emits a sound of constant frequency f The observersAandBhearasoundofthesamefrequency

A B

wavefront

S

Figure 930 Sound waves from a stationary source

Suppose now that the source moves towards A withconstant speed vFigure9310(a)showsasnapshotofthenew wave pattern

NATURE OF SCIENCETechnology Although originally based on physical observations of the pitch of fast moving sources of sound the Doppler effect has an important role in many different areas such as evidence for the expansion of the universe and generating images used in weather reports and in medicine (55)

copy IBO 2014

PHYSICS 2015indb 303 150514 337 PM

Chapter 9

304

AH

L

sourceA B

smallerwavelength

largerwavelength

Figure 931 (a) Sound waves from a moving source

The wavefronts are now crowded together in the direction of travel of the source and stretched out in the opposite direction This is why now the two observers will now hear notesofdifferentfrequenciesHowmuchthewavesbunchtogether and how much they stretch out will depend on the speed vEssentially A

A

cfλ

= and BB

cfλ

= where λA lt λB and v is the speed of sound

IfthesourceisstationaryandAismovingtowardsitthenthe waves from the source incident on A will be bunched up If A is moving away from the stationary source then the waves from the source incident on A will be stretched out

Christian Doppler(1803ndash1853)actuallyappliedtheprinciple(incorrectlyasithappens)totryandexplainthecolourofstars However the Doppler effect does apply to light aswellastosoundIfalightsourceemitsalightoffrequencyf then if it is moving away from an observer the observer willmeasurethelightemittedashavingalowerfrequencythan fSincethesensationofcolourvisionisrelatedtothefrequencyoflight(bluelightisofahigherfrequencythanred light) light emitted by objects moving way from anobserver is often referred to as being red-shifted whereas if the object is moving toward the observer it is referred to as blue-shiftedThisideaisusedinOptionE(Chapter16)

We do not need to consider here the situations where either the source or the observer is accelerating In a situation for example where an accelerating source is approaching astationaryobserverthentheobserverwillhearasoundof ever increasing frequency This sometimes leads toconfusion in describing what is heard when a source approaches passes and then recedes from a stationaryobserver Suppose for example that you are standingon a station platform and a train sounding its whistle is approaching at constant speed What will you hear as the train approaches and then passes through the station As the train approaches you will hear a sound of constant pitch but increasing loudness The pitch of the sound will be greater than if the train were stationary As the train passes through the station you will hear the pitch change at the momentthetrainpassesyoutoasoundagainofconstantpitch The pitch of this sound will be lower than the sound of the approaching train and its intensity will decrease as the train recedes from you What you do not hear is a sound of increasing pitch and then decreasing pitch

The Doppler equations for sound

Although you will not be expected in an IB examination to derivetheequationsassociatedwithaspectsoftheDopplereffectyouwillbeexpectedtoapplythemForcompletenesstherefore thederivationoftheequationsassociatedwiththeDopplereffectasoutlinedaboveisgivenhere

S primeS O

vs

Figure 931 (b)

InFigure931(b)theobserverOisatrestwithrespecttoasourceofsoundSismovingwithconstanspeedvs directly towards O The source is emitting a note of constant frequencyf and the speed of the emitted sound is v

S shows the position of the source ∆t later When the sourceisatresttheninatime∆t the observer will receive f∆t waves and these waves will occupy a distance v∆t ie

v f

λ =f∆tv∆t =

(Because of the motion of the source this number of waves will now occupy a distance (v∆t ndash vs∆t)The lsquonewrsquowavelength is therefore

v ndash vs

fλrsquo =

f∆tv∆t ndash vs∆t

=

If fisthefrequencyheardbyOthen

vfλ

= or

vf

λ = = sv vfminus

From which

s

vf fv v

=minus

Dividingthroughbyv gives

Equation1111

1 sf f v

v=

minus

If the source is moving away from the observer then we have

s

vf fv v

=minus

1

1 sf vv

=+

Equation112

PHYSICS 2015indb 304 150514 337 PM

Wave Phenomena

305

AH

L

We now consider the case where the source is stationary and the observerismoving towardsthesource with speed v0 In this situation the speed of the sound waves as measured by the observer will be v0 + v We therefore have that

v0 + v = ffraslλ = vfλ

times

From which

o1 vf fv

= +

Equation113

If the observerismovingawayfromthesource then

o1 vf fv

= minus

Fromequation(113)wehavethat

o o1 v vf f f f f fv v

∆ = minus = + minus =

Equation114

Thevelocitiesthatwerefertointheaboveequationsarethevelocities with respect to the medium in which the waves fromthesourcetravelHoweverwhenwearedealingwithalight source it is the relative velocity between the source and the observer that we must consider The reason for this is that lightisuniqueintherespectthatthespeedofthelightwavesdoes not depend on the speed of the source All observers irrespective of their speed or the speed of the source will measure the same velocity for the speed of light This is oneof thecornerstonesof theSpecialTheoryofRelativitywhichisdiscussedinmoredetailinOptionH(Chapter18)When applying the Doppler effect to light we aremainlyconcerned with the motion of the source We look here only at the situation where the speed of the source v is much smaller than the speed of light cinfreespace(v ltlt c)Underthesecircumstanceswhenthesourceismovingtowardstheobserverequation111becomes

vf f f fc

minus = ∆ = Equation115

andwhenthesourceismovingawayfromtheobserverequation112becomes vf f f f

cminus = ∆ = minus

Provided that v ltlt c these same equations apply for astationary source and moving observer

We look at the following example and exercise

Example

Asourceemitsasoundoffrequency440HzItmovesinastraight line towards a stationary observer with a speed of 30ms-1Theobserverhearsasoundoffrequency484HzCalculate the speed of sound in air

Solution

We use equation 111 and substitute f= 484 Hz f = 440 Hz

and vs = 30 m s-1

therefore 1

484 440 301v

=minus

such that 30 4401

484vminus = to

give v = 330 m s-1

Example

A particular radio signal from a galaxy is measured as havingafrequencyof139times109HzThesamesignalfromasourceinalaboratoryhasafrequencyof142times109Hz

Suggestwhy thegalaxy ismovingaway fromEarthandcalculateitsrecessionspeed(iethespeedwithwhichitismovingawayfromEarth)

Solution

The fact that the frequency from the moving source is less than that when it is stationary indicates that it is moving away from the stationary observer ie Earth

Using vf fc

∆ = we have

8 96

9

3 10 (142 139) 10 634 10142 10

c fvf∆ times times minus times

= = = timestimes

m s-1

It is usual when dealing with the Doppler effect of light to express speeds as a fraction of c So in this instance we have v = 0021 c

Using the Doppler effectWe have seen in the above example and exercise how the Dopplereffectmaybeusedtomeasuretherecessionspeedof distant galaxies The effect is also used to measure speed inothersituationsHerewewilllookatthegeneralprincipleinvolved in using the Doppler effect to measure speedFigure932showsasource(thetransmitter)thatemitseithersound or em waves of constant frequency f The waves

PHYSICS 2015indb 305 150514 337 PM

Chapter 9

306

AH

L

fromthesourceare incidentonareflectorthat ismovingtowards the transmitter with speed vThereflectedwavesare detected by the receiver placed alongside the transmitter

reector

v

transmitter

receiver

f f

f f

Figure 932 Using the Doppler effect to measure speed

We shall consider the situation where v ltlt c where c is the speed of the waves from the transmitter

For the reflector receiving waves from the transmitterit is effectively an observer moving towards a stationary sourceFromequation(114) itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation116

For the receiver receivingwaves from the reflector it iseffectively a stationary observer receiving waves from a movingsourceFromequation(115)itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation117

Ifweaddequations(116)and(117)wegetthatthetotalDopplershiftatthereceiver∆f is

v vf f f f fc c

minus = ∆ = +

But 1 vf fc

= +

hence

1 v v vf f fc c c

∆ = + +

But since v ltlt cwe can ignore the term2

2

vc when we

expandthebracketintheaboveequation

Therefore we have2vf fc

∆ = Equation118

If v asymp c then we must use the full Doppler equationsHowever for em radiationwewill alwaysonly considersituations in which v ltlt c

Example

Thespeedofsoundinbloodis1500times103 m s-1 Ultrasound offrequency100MHzisreflectedfrombloodflowinginan arteryThe frequencyof the reflectedwaves receivedbackatthetransmitteris105MHzEstimatethespeedofthebloodflowintheartery

Solution

Using equation (118) we have

6 63

2005 10 1015 10

vtimes = times

timesto give v asymp 36 m s-1 (We have assumed that the ultrasound is incident at right angles to the blood flow)

Exercise 971 Judy is standing on the platform of a station A

high speed train is approaching the station in a straight line at constant speed and is sounding its whistleAsthetrainpassesbyJudythefrequencyof the sound emitted by the whistle as heard by Judychangesfrom640Hzto430HzDetermine

(a) thespeedofthetrain

(b) thefrequencyofthesoundemittedbythewhistle as heard by a person on the train (Speedofsound=330ms-1)

2 AgalaxyismovingawayfromEarthwithaspeedof00500cThewavelengthofaparticularspectral line in light emitted by atomic hydrogen inalaboratoryis656times10-7 m Calculate the valueofthewavelengthofthislinemeasuredinthelaboratoryinlightemittedfromasourceofatomic hydrogen in the galaxy

PHYSICS 2015indb 306 150514 337 PM

10 Fields

Contents

101 ndash Describing fields

102 ndash Fields at work

Essential Ideas

Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Similar approaches can be taken in analysing electrical and gravitational potential problems copy IBO 2014

PHYSICS 2015indb 307 150514 337 PM

Chapter 10

308

AH

L

101 Describing fields

Essential idea Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Understandingsbull Gravitationalfieldsbull Electrostaticfieldsbull Electricpotentialandgravitationalpotentialbull Fieldlinesbull Equipotentialsurfaces

Gravitational potentialWehaveseenthatifweliftanobjectofmassm to a height h abovethesurfaceoftheEarththenitsgainingravitationalpotential energy is mghHoweverthisisbynomeansthefullstoryForastartwenowknowthatgvarieswithh and also the expression really gives a difference in potentialenergybetweenthevaluethattheobjecthasattheEarthrsquossurfaceandthevaluethatithasatheighthSowhatwereallyneedisazeropointCanwefindapointwherethepotentialenergyiszeroandusethispointfromwhichtomeasure changes in potential energy

The point that is chosen is in fact infinity At infinity the gravitationalfieldstrengthofanyobjectwillbezeroSoletusseeifwecandeduceanexpressionforthegaininpotentialenergyofanobjectwhenitisldquoliftedrdquofromthesurfaceoftheEarthtoinfinityThisineffectmeansfindingtheworknecessarytoperformthistask

δr

r r δr+

r infin=

mg

AB

Me

Figure 1001 Gravitational forces

Inthediagramweconsidertheworknecessarytomovethe particle of mass m a distance δr in thegravitationalfield of the Earth

The force on the particle at A is FG Mem

r2----------------=

IftheparticleismovedtoB then since δr isverysmallwe can assume that the field remains constant over thedistance ABTheworkδW done againstthegravitationalfieldoftheEarthinmovingthedistanceABis

δWGMem

r2----------------δrndash=

(rememberthatworkdoneagainstaforceisnegative)

TofindthetotalworkdoneW in going from the surface oftheEarthtoinfinitywehavetoaddalltheselittlebitsofworkThisisdonemathematicallybyusingintegralcalculus

WG Me m

r2----------------ndash

rdR

infin

int G Me m 1

r2----- rd

R

infin

intndash G Me m 1r---ndash

R

infinndash= = =

GMe m 0 1R---ndash

ndashndash=

G MemR----------------ndash=

HencewehavewhereRistheradiusoftheEarththattheworkdonebythegravitationalfieldinmovinganobjectofmass m from R(surfaceoftheEarth)toinfinityisgivenby

WGMe m

R---------------- ndash=

We can generalise the result by calculating the worknecessary per unit mass to take a small mass fromthe surface of the Earth to infinity This we call thegravitational potential Vie

Wewouldgetexactlythesameresultifwecalculatedtheworkdonebythefieldtobringthepointmassfrominfinityto the surface of Earth In this respect the formal definition ofgravitationalpotentialatapointinagravitationalfieldis therefore defined as the work done per unit mass in bringing a point mass from infinity to that point

Clearly then the gravitational potential at any point inthe Earthrsquos field distance r from the centre of the Earth (providingr gt R)is

VGMe

r------------ ndash=

Thepotentialisthereforeameasureoftheamountofworkthathastobedonetomoveparticlesbetweenpointsinagravitationalfieldanditsunit is theJkgndash1 We also note thatthepotentialisnegativesothatthepotentialenergy

NATURE OF SCIENCEParadigm shift The move from direct observable actions being responsible for influence on an object to acceptance of a fieldrsquos ldquoaction at a distancerdquo required a paradigm shift in the world of science (23)

copy IBO 2014

PHYSICS 2015indb 308 150514 337 PM

Fields

309

AH

L

aswemoveawayfromtheEarthrsquossurfaceincreasesuntilitreachesthevalueofzeroatinfinity

IfthegravitationalfieldisduetoapointmassofmassmthenwehavethesameexpressionasaboveexceptthatMe is replaced by m andmustalsoexclude thevalueof thepotential at the point mass itself ie at r = 0

WecanexpressthegravitationalpotentialduetotheEarth(orduetoanysphericalmass)intermsofthegravitationalfield strength at its surface

AtthesurfaceoftheEarthwehave

g 0ndash Re GMRe

---------ndash= e

Sothat

g 0Re2 GM= e

Hence at a distance r from the centre of the Earth the gravitationalpotentialVcanbewrittenas

VGMe

r------------ndashg 0 Re

2

r------------ndash= =

The potential at the surface of the Earth

(r = Re)istherefore-g0Re

It is interesting to see how the expression for thegravitational potential ties in with the expression mgh ThepotentialatthesurfaceoftheEarthis-g0Re (see the exampleabove)andataheighthwillbe-g0g 0 Re h+( )ndash ifweassume that g0doesnotchangeover thedistanceh The differenceinpotentialbetweenthesurfaceandtheheighth is therefore g0hSotheworkneededtoraiseanobjectof mass m to a height h is mgh ie m timesdifference ingravitationalpotential

This we have referred to as the gain in gravitationalpotentialenergy(see235)

However this expression can be extended to any twopoints inanygravitationalfieldsuch that ifanobjectofmass mmovesbetween twopointswhosepotentialsareV1 and V2 respectively then the change in gravitationalpotentialenergyoftheobjectism(V1 ndash V2)

Gravitational potential gradient

Let us consider now a region in space where thegravitationalfieldisconstantInFigure912thetwopointsAandBareseparatedbythedistance∆x

A B

direction of uniform gravitational eld of strength I

∆x

Figure 1002 The gravitational potential gradient

The gravitational field is of strength I and is in the direction shownThe gravitational potential at A is V andatBisV +∆V

Theworkdone is takingapointmassmfromAtoB isF∆x = mI∆x

Howeverbydefinitionthisworkisalsoequalto-m∆V

Therefore mI∆x=-m∆V

or VIx

∆= minus

∆Effectivelythissaysthatthemagnitudeofthegravitationalfield strength is equal to the negative gradient of thepotential If I is constant then V is a linear function of x and Iisequaltothenegativegradientofthestraightlinegraph formed by plotting V against x If I is not constant (asusuallythecase)thenthemagnitudeofI at any point in the field can be found by find the gradient of the V-x graph at that point An example of such a calculation can befoundinSection929

For those of you who do HL maths the relationship between field and potential is seen to follow from the expression for the potential of a point mass viz

mV Gr

= minus

2

ddV mG Ir r

minus = + =

Potential due to one or more point masses

Gravitationalpotentialisascalarquantitysocalculatingthepotential due to more than one point mass is a matter of simpleadditionSoforexamplethepotentialV due to the Moon and Earth and a distance x fromthecentreofEarthonastraightlinebetweenthemisgivenbytheexpression

E MM MV Gx r x

= minus + minus

whereME=massofEarthMM= mass of Moon and r = distancebetweencentreofEarthandMoon

PHYSICS 2015indb 309 150514 337 PM

Chapter 10

310

AH

L

Equipotentials and field linesIfthegravitationalpotentialhasthesamevalueatallpointson a surface the surface is said to be an equipotential surfaceSoforexampleifweimagineasphericalshellaboutEarthwhosecentrecoincideswiththecentreofEarththisshellwillbeanequipotentialsurfaceClearlyifwerepresentthegravitationalfieldstrengthbyfieldlinessincethelinesldquoradiaterdquooutfromthecentreofEarththentheselineswillbeatrightanglestothesurfaceIfthefieldlineswerenotnormaltotheequipotentialsurfacethentherewouldbeacomponentof the field parallel to the surfaceThis wouldmean thatpointsonthesurfacewouldbeatdifferentpotentialsandsoitwouldnolongerbeanequipotentialsurfaceThisofcourseholdstrueforanyequipotentialsurface

Figure 1003 shows the field lines and equipotentials fortwopointmassesm

m m

Figure 1003 Equipotentials for two point masses

It is worth noting that we would get exactly the samepattern ifwewere to replace thepointmasseswith twoequalpointcharges(See935)

Escape speedThe potential at the surface of Earth is which meansthattheenergyrequiredtotakeaparticleofmass mfromthesurfacetoinfinityisequalto

ButwhatdoesitactuallymeantotakesomethingtoinfinityWhentheparticleisonthesurfaceoftheEarthwecanthinkofitassittingatthebottomofaldquopotentialwellrdquoasinfigure1004

particle

surface of Earth

innity

GMR---------

FIgure 1004 A potential well

Theldquodepthrdquoofthewellis and if the particle gains an amount of kinetic energy equal to wherem is its massthenitwillhavejustenoughenergytoldquoliftrdquoitoutofthewell

In reality it doesnrsquot actually go to infinity it justmeansthat the particle is effectively free of the gravitationalattractionof theEarthWe say that ithas ldquoescapedrdquo theEarthrsquosgravitationalpullWemeetthisideainconnectionwithmolecular forcesTwomolecules in a solidwill sitat their equilibrium position the separation where therepulsiveforceisequaltotheattractiveforceIfwesupplyjust enough energy to increase the separation of themolecules such that they are an infinite distance apart then intermolecularforcesnolongeraffectthemoleculesandthesolidwillhavebecomealiquidThereisnoincreaseinthekineticenergyofthemoleculesandsothesolidmeltsat constant temperature

Wecancalculatetheescapespeedofanobjectveryeasilyby equating the kinetic energy to the potential energysuch that

12---mvesc ape

2 GMe mRe

----------------=

vesc aperArr2GMe

Re--------------- 2g0 Re = =

Substituting for g0 and Regivesavalueforvescape of about 11kmsndash1 from the surface of the Earth

Youwillnotethattheescapespeeddoesnotdependonthemassoftheobjectsincebothkineticenergyandpotentialenergy are proportional to the mass

IntheoryifyouwanttogetarockettothemoonitcanbedonewithoutreachingtheescapespeedHoweverthiswouldnecessitate anenormousamountof fuel and it islikely that therocketplus fuelwouldbesoheavythat itwouldnevergetoffthegroundItismuchmorepracticalto accelerate the rocket to the escape speed fromEarthorbitandthenintheoryjustlaunchittotheMoon

Example

Use the followingdata todetermine thepotential at thesurface of Mars and the magnitude of the acceleration of free fall

massofMars =64times1023kg

radiusofMars =34times106 m

Determinealsothegravitationalfieldstrengthatadistanceof 68 times 106mabovethesurfaceofMars

PHYSICS 2015indb 310 150514 337 PM

Fields

311

AH

L

Solution

= minus = minus times times = times23

11 76

64 1067 10 ndash13 10

34 10MV GR

minus timestimes

Nkg-1

But V = -g0R

Therefore 7

0 6

13 10 3834 10

VgR

times= minus = =

timesm s-2

To determine the field strength gh at 68 times 106 m above the surface we use the fact that 0 2

Mg GR

= such that GM = g0R2

Therefore 2 2

0h 2 2 2

h h

38 (34) 042(102)

g RGMgR R

times= = = = m s-2

(the distance from the centre is 34 times 106 + 68 times 106 = 102 times 106 m)

Exercise 1011 Thegraphbelowshowshowthegravitational

potentialoutsideoftheEarthvarieswithdistancefrom the centre

0 10 20 30 40 50 60 70

ndash1

ndash2

ndash3

ndash4

ndash5

ndash6

ndash7

r m times 106

V

Jkgndash1

times 10

7

(a) Usethegraphtodeterminethegainingravitationalpotentialenergyofasatelliteofmass200kgasitmovesfromthesurfaceof the Earth to a height of 30 times 107mabovetheEarthrsquossurface

(b) Calculatetheenergyrequiredtotakeit to infinity

(c) DeterminetheslopeofthegraphatthesurfaceoftheEarthmCommenton youranswer

Electric potential energy

The concept of electric potential energy was developedwith gravitational potential energy in mind Just as anobjectnearthesurfaceoftheEarthhaspotentialenergybecause of its gravitational interaction with the Earthsotoothereiselectricalpotentialenergyassociatedwithinteracting charges

Letusfirstlookatacaseoftwopositivepointchargeseachof 1μC that are initially bound together by a thread in a vacuuminspacewithadistancebetweenthemof10cmasshowninFigure1005Whenthethreadiscutthepointchargesinitiallyatrestwouldmoveinoppositedirectionsmovingwithvelocitiesv1 and v2 along the direction of the electrostatic force of repulsion

BEFORE AFTER

v v21

Figure 1005 Interaction of two positive particles

The electric potential energybetweentwopointchargescan be found by simply adding up the energy associated witheachpairofpointchargesForapairofinteractingchargestheelectricpotentialenergyisgivenby

∆U = ∆Ep + ∆Ek = ∆W = ∆Fr = kqQ

____ r2 times r = kqQ

____ r

Becausenoexternalforceisactingonthesystemtheenergyand momentum must be conserved Initially Ek = 0 and Ep =kqQr=9times109 times 1 times 10-1201m=009JWhentheyareagreatdistance fromeachotherEpwillbenegligibleThefinalenergywillbeequaltofrac12mv1

2+frac12mv22=009J

Momentum isalsoconservedand thevelocitieswouldbethe same magnitude but in opposite directions

Electric potential energyismoreoftendefinedintermsofapointchargemovinginanelectricfieldas

lsquothe electricpotential energybetweenany twopoints inan electric field is defined as negativeoftheworkdone by anelectricfieldinmovingapoint electric chargebetweentwolocationsintheelectricfieldrsquo

ΔU = ΔEp = -ΔW = -Fd = qEx

ΔU = ΔEp +ΔEk = ΔW = ΔFr = kqQ r2 x r = kqQ

wherexisthedistancemovedalong(oroppositeto)thedirection of the electric field

Electric potential energyismeasuredinjoule(J)Justasworkisascalarquantitysotooelectricalpotentialenergyis a scalar quantity The negative of the work done by an electric field in moving a unit electric charge between two points is independent of the path taken Inphysicswesaytheelectricfieldisalsquoconservativersquo field

PHYSICS 2015indb 311 150514 337 PM

Chapter 10

312

AH

L

Supposeanexternalforcesuchasyourhandmovesasmallpositive point test charge in the direction of a uniformelectric fieldAs it ismoving itmust be gaining kineticenergyIfthisoccursthentheelectricpotentialenergyofthe unit charge is changing

InFigure1006apointcharge+qismovedbetweenpointsAandBthroughadistancex in a uniform electric field

B

A

x

+ q

Figure 1006 Movement of a positive point charge in a uniform field

InordertomoveapositivepointchargefrompointAtopointBanexternal forcemustbeapplied to thechargeequaltoqE (F = qE)

Since the force is applied through a distance xthennegativeworkhas to bedone tomove the charge because energyis gainedmeaning there is an increaseelectric potential energybetweenthetwopointsRememberthattheworkdoneisequivalenttotheenergygainedorlostinmovingthe charge through the electric field The concept of electric potential energy is only meaningful as the electric field whichgeneratestheforceinquestionisconservative

W F xtimes E q x times= =

θ x cos θ x

Figure 1007 Charge moved at an angle to the field

If a chargemoves at an angle θ to an electric field thecomponent of the displacement parallel to the electric fieldisusedasshowninFigure918

W F x Eq x θcostimes= =

Theelectricpotentialenergyisstoredintheelectricfieldand theelectricfieldwill return theenergy to thepointcharge when required so as not to violate the Law ofconservationofenergy

Electric potential

The electric potential at a point in an electric field is defined as being the work done per unit charge in bringing a small positive point charge from infinity to that point

ΔV = Vinfin ndash Vf = ndash W ___ q

Ifwedesignatethepotentialenergytobezeroatinfinitythen it follows thatelectricpotentialmustalsobezeroat infinity and the electric potential at any point in an electricfieldwillbe

ΔV = ndash W ___ q

NowsupposeweapplyanexternalforcetoasmallpositivetestchargeasitismovedtowardsanisolatedpositivechargeTheexternalforcemustdoworkonthepositivetestchargetomoveittowardstheisolatedpositivechargeandtheworkmustbepositivewhiletheworkdonebytheelectricfieldmustthereforebenegativeSotheelectricpotentialatthatpointmustbepositiveaccordingtotheaboveequationIfanegativeisolatedchargeisusedtheelectricpotentialatapointonthepositivetestchargewouldbenegativePositivepoint chargesof their ownaccordmove fromaplaceofhighelectricpotential toaplaceof lowelectricpotentialNegative point charges move the other way from lowpotentialtohighpotentialInmovingfrompointAtopointBinthediagramthepositivecharge+qismovingfromalowelectricpotentialtoahighelectricpotential

In thedefinitiongiven the termldquoworkperunitchargerdquohas significance If the test charge is +16 times 10-19Cwherethe charge has a potential energy of 32 times 10-17Jthenthepotentialwouldbe32times10-17J+16times10-19C=200JC-1 Now if the charge was doubled the potential wouldbecome 64times10-17 J However the potential per unitchargewouldbethesame

Electric potential is a scalar quantityandithasunitsJC-1 orvoltswhere1voltequalsone joulepercoloumbThevoltallowsustoadoptaunitfortheelectricfieldintermsofthevolt

PreviouslytheunitfortheelectricfieldwasNCndash1

W = qV and F = qEsoW ___ V = F __ E

E = FV ___ W = FV ___ Fm V mndash1

That is the units of the electric field E can also beexpressed as V mndash1

The work done per unit charge in moving a point charge between two points in an electric field is again independent of the path taken

PHYSICS 2015indb 312 150514 337 PM

Fields

313

AH

L

Electric potential due to a point charge

Letus takeapoint rmetres froma chargedobjectThepotentialatthispointcanbecalculatedusingthefollowing

W = ndashFr = ndashqV and F = ndash q1 q2 _____ 4πε0r

2

Therefore

W = ndash q1 q2 _____ 4πε0r

2 times r = ndash q1 q2 _____ 4πε0r

= ndashq1 times q2 _____ 4πε0r

= ndashq1V

That is V q

4πε0r--------------- =

Orsimply

V kqr------

=

Example

Determinehowmuchworkisdonebytheelectricfieldofpointcharge150μCwhenachargeof200μCismovedfrom infinity to apoint 0400m from thepoint charge(Assumenoaccelerationofthecharges)

Solution

The work done by the electric field is W = -qV = -14πε0 times q times (Q rinfin - Q r0400)

W = (- 200 times 10-6 C times 900 times 109 NmC-2 times 150 times 10-6 C) divide 0400 m = - 0675 J

An external force would have to do +0675 J of work

Electric field strength and electric potential gradient

Letus lookbackatFigure1006Supposeagain that thecharge+qismovedasmalldistancebyaforceFfromAtoBsothattheforcecanbeconsideredconstantTheworkdoneisgivenby

∆W F x∆times=

TheforceFandtheelectricfieldEareoppositelydirectedandweknowthat

F = -qE and ΔW = q ΔV

Thereforetheworkdonecanbegivenas

q ΔV = -q E Δx

Therefore E V∆x∆-------

ndash=

The rate of change of potential ΔV at a point with respect to distance Δx in the direction in which the change is maximum is called the potential gradient We say that the electric field = - the potential gradient and the units are Vm-

1 From the equation we can see that in a graph of electric potential versus distance the gradient of the straight line equals the electric field strength

In reality if a charged particle enters a uniform electric field it will be accelerated uniformly by the field and its kinetic energy will increase This is why we had to assume no acceleration in the last worked example

Ek∆ 12---

mv2 q E xsdot sdot q Vx---

xsdot sdot q Vsdot= = = =

Example

Determine how far apart two parallel plates must besituated so that a potential difference of 150 times 102 V produces an electric field strength of 100 times 103 NC-1

Solution

Using E V∆x------- xhArrndash V∆

E-------15 102 Vtimes

100 103 N C 1ndashtimes------------------------------------------= = =

= 150 times 10-1

The plates are 150 times 10-1 m apart

The electric field and the electric potential at a point due to an evenly distributed charge +q on a sphere can be represented graphically as in Figure 1008

r

r

V

E

r0

r0

r0

Charge of +Q evenly distributed over surface

E 14πε0------------ Q

r2----- r r0gt=

V 14πε0------------Q

r----r r0gt=

Figure 1008 Electric field and potential due to a charged sphere

PHYSICS 2015indb 313 150514 337 PM

Chapter 10

314

AH

L

When the sphere becomes charged we know that thechargedistributes itselfevenlyoverthesurfaceThereforeeverypartofthematerialoftheconductorisatthesamepotential As the electric potential at a point is defined as beingnumericallyequaltotheworkdoneinbringingaunitpositivechargefrominfinitytothatpointithasaconstantvalueineverypartofthematerialoftheconductor

Since the potential is the same at all points on the conducting surfacethen∆V∆xiszeroButE=ndash∆V∆xThereforethe electric field inside the conductor is zero There is no electric field inside the conductor

SomefurtherobservationsofthegraphsinFigure1009are

bull OutsidethespherethegraphsobeytherelationshipsgivenasE α 1 r2 and V α 1 r

bull Atthesurfacer = r0Thereforetheelectricfieldandpotentialhavetheminimumvalueforr at this point and this infers a maximum field and potential

bull Insidethespheretheelectricfieldiszerobull Insidethespherenoworkisdonetomoveacharge

fromapointinsidetothesurfaceThereforethereisno potential difference and the potential is the same asitiswhenr = r0

Similargraphscanbedrawnfortheelectricfieldintensityandthe electric potential as a function of distance from conducting parallelplatesandsurfacesandthesearegiveninFigure1009

Potential plot E eld plotE eld

++++

ndashndashndashndash

xx x

+

x x

ndashx x

+ ndashx

x

+ +

x

x

Figure 1009 Electric field and electric potential at a distance from a charged surface

Potential due to one or more point charges

The potential due to one point charge can be determined byusingtheequationformula

V = kq r

Example 1

Determine the electric potential at a point 20 times 10-1 m fromthecentreofan isolatedconducting spherewithapointchargeof40pCinair

Solution

Using the formula

V = kq r we have

V90 10 9times( ) 40 10 12ndashtimes( )times

20 10 1ndashtimes( )------------------------------------------------------------------ 018 V= =

the potential at the point is 180 times 10-1 V

The potential due to a number of point charges can be determined by adding up the potentials due to individual point charges because the electric potential at any point outside a conducting sphere will be the same as if all the charge was concentrated at its centre

Example 2

Threepointchargesofareplacedattheverticesofaright-angledtriangleasshowninthediagrambelowDeterminetheabsolutepotentialat the+20μCchargeduetothetwoothercharges

- 6μC

+3μC +2μC

4 m

3 m

Solution

The electric potential of the +2 μC charge due to the ndash 6 μC charge is

V = (9 times 109 Nm2C-2 times -6 times 10-6 C) divide (radic 32 + 42) m = - 108 times 104 V

PHYSICS 2015indb 314 150514 338 PM

Fields

315

AH

L

Insummarywecanconcludethat

bull No work is done to move a charge along an equipotential

bull Equipotentials are always perpendicular to the electric lines of force

Figure1011and1012showsomeequipotential lines fortwo oppositely charged and identically positive spheresseparated by a distance

+ve ndashve

equipotential lines

Figure 1011 Equipotential lines between two opposite charges

equipotential lines

Figure 1012 Equipotential lines between two charges which are the same

+

ndash ndash ndash ndash

+ + +

40 V

30 V

20 V

10 V

50 V

Figure 1013 Equipotential lines between charged parallel plates

Figure1013showstheequipotentiallinesbetweenchargedparallel plates Throughout this chapter the similarities and differences between gravitational fields and electric fieldshavebeendiscussedTherelationshipsthatexistsbetweengravitationalandelectricquantitiesandtheeffectsofpointmassesandchargesissummarisedinTable1014

The electric potential of the +2 μC charge due to the +3 μC charge is

V = (9 times 109 Nm2C-2 times 3 times 10-6 C) divide 3m = 9 times 103 V

The net absolute potential is the sum of the 2 potentials- 108 times 104 V + 9 times 103 V =- 18 times 103 V

The absolute potential at the point is - 18 times 103 V

Equipotential surfacesRegions in space where the electric potential of a chargedistributionhasaconstantvaluearecalledequipotentials The placeswherethepotentialisconstantinthreedimensionsarecalled equipotential surfacesandwheretheyareconstantintwodimensionstheyarecalledequipotential lines

TheyareinsomewaysanalogoustothecontourlinesontopographicmapsInthiscasethegravitationalpotentialenergy is constant as amassmoves around the contourlines because the mass remains at the same elevationabovetheEarthrsquossurfaceThegravitationalfieldstrengthacts in a direction perpendicular to a contour line

Similarlybecausetheelectricpotentialonanequipotentiallinehas thesamevalueanelectric forcecandonoworkwhena testchargemovesonanequipotentialThereforethe electric field cannot have a component along anequipotentialandthusitmustbeeverywhereperpendiculartotheequipotentialsurfaceorequipotentiallineThisfactmakesiteasytoplotequipotentialsifthelinesofforceorlinesofelectricfluxofanelectricfieldareknown

Forexamplethereareaseriesofequipotentiallinesbetweentwoparallelplateconductorsthatareperpendiculartotheelectric fieldTherewill be a series of concentric circles(eachcirclefurtherapartthanthepreviousone)thatmapouttheequipotentialsaroundanisolatedpositivesphereThe lines of force and some equipotential lines for anisolatedpositivesphereareshowninFigure1010

Lines ofequipotential

Figure 1010 Equipotentials around an isolated positive sphere

PHYSICS 2015indb 315 150514 338 PM

Chapter 10

316

AH

L

Gravitational quantity Electrical quantityQ

uant

ities

g V∆x∆-------ndash= E V∆

x∆-------ndash=

Poin

t mas

ses a

nd ch

arge

s

V Gmr----

ndash= V 14πε0------------q

r---=

g G m

r2-----ndash= E 14πε0------------ q

r2-----=

F Gm1 m2

r2--------------= F 1

4πε0------------

q1q2

r2-----------=

Figure 1014 Formulas

Example

Deduce the electric potential on the surface of a gold nucleus that has a radius of 62 fm

Solution

Using the formula

V = kq r and knowing the atomic number of gold is 79 We will assume the nucleus is spherical and it behaves as if it were a point charge at its centre (relative to outside points)

V = 90 times 109 Nm2C-2 times 79 times 16 times 10-19 C divide 62 times 10-15 m = 18 times 107 V

The potential at the point is 18 MV

Example

Deduce the ionisation energy in electron-volts of theelectron in the hydrogen atom if the electron is in its ground state and it is in a circular orbit at a distance of 53 times 10-11 m from the proton

Solution

This problem is an energy coulombic circular motion question based on Bohrrsquos model of the atom (not the accepted quantum mechanics model) The ionisation energy is the energy required to remove the electron from the ground state to infinity The electron travels in a circular orbit and therefore has a centripetal acceleration The ionisation energy will counteract the coulombic force and the movement of the electron will be in the opposite direction to the centripetal force

Total energy = Ek electron + Ep due to the proton-electron interaction

ΣF = kqQ r2 = mv2 r and as such mv2 = = kqQ r

Therefore Ek electron = frac12 kqQ r

Ep due to the proton-electron interaction = - kqQ r

Total energy = frac12 kqQ r + - kqQ r = -frac12 kqQ r

= - 90 times 109 Nm2C-2 times (16 times 10-19 C)2 divide 53 times 10-11 m = -217 times 10-18 J

= -217 times 10-18 J divide 16 times 10-19 = -136 eV

The ionisation energy is 136 eV

Exercise 1021 ApointchargePisplacedmidwaybetweentwo

identicalnegativechargesWhichoneofthefollowingiscorrectwithregardstoelectricfieldandelectricpotentialatpointP

Electric field Electric potentialA non-zero zeroB zero non-zeroC non-zero non-zeroD zero zero

2 Twopositivechargedspheresaretiedtogetherinavacuumsomewhereinspacewheretherearenoexternal forces A has a mass of 25 g and a charge of20μCandBhasamassof15gandachargeof30μCThedistancebetweenthemis40cmTheyarethenreleasedasshowninthediagram

BEFORE AFTER

v1 v2

A B

(a) Determinetheirinitialelectricpotentialenergy in the before situation

(b) DeterminethespeedofsphereBafterrelease

3 Thediagrambelowrepresentstwoequipotentiallines in separated by a distance of 5 cm in a uniform electric field

+ + + + + + + +

ndash ndash ndash ndash ndash ndash ndash ndash

40 V

20 V 5 cm

PHYSICS 2015indb 316 150514 338 PM

Fields

317

AH

L

Determine the strength of the electric field

4 Thisquestionisabouttheelectricfieldduetoacharged sphere and the motion of electrons in that fieldThediagrambelowshowsanisolatedmetalsphereinavacuumthatcarriesanegativeelectriccharge of 60 μC

ndash

(a) Onthediagramdrawtheconventionalwayto represent the electric field pattern due to the charged sphere and lines to represent threeequipotentialsurfacesintheregionoutside the sphere

(b) Explainhowthelinesrepresentingtheequipotentialsurfacesthatyouhavesketchedindicatethatthestrengthoftheelectricfieldisdecreasingwithdistancefrom the centre of the sphere

(c) Theelectricfieldstrengthatthesurfaceofthe sphere and at points outside the sphere can be determined by assuming that the sphere acts as a point charge of magnitude 60 μC at its centre The radius of the sphere is 25 times 10ndash2 m Deduce that the magnitude of the field strength at the surface of the sphere is 86 times 107 Vmndash1

An electron is initially at rest on the surface of the sphere

(d) (i) Describethepathfollowedbytheelectronasitleavesthesurfaceofthesphere

(ii) Calculatetheinitialaccelerationofthe electron

5 Determinetheamountofworkthatisdoneinmovingachargeof100nCthroughapotentialdifferenceof150times102 V

6 Three identical 20 μC conducting spheres are placedatthecornersofanequilateraltriangleofsides 25 cm The triangle has one apex C pointing upthepageand2baseanglesAandBDeterminetheabsolutepotentialatB

7 Determinehowfaraparttwoparallelplatesmustbesituatedsothatapotentialdifferenceof250 times 102 V produces an electric field strength of 200 times 103 NC-1

8 Thegapbetweentwoparallelplatesis10times10-3mandthereisapotentialdifferenceof10times104 V betweentheplatesCalculate

i theworkdonebyanelectroninmovingfrom one plate to the other

ii thespeedwithwhichtheelectronreachesthe second plate if released from rest

iii theelectricfieldintensitybetweentheplates

9 Anelectronguninapicturetubeisacceleratedby a potential 25 times 103VDeterminethekineticenergygainedbytheelectroninelectron-volts

10 Determine the electric potential 20 times 10-2 m from achargeof-10times10-5 C

11 Determinetheelectricpotentialatapointmid-waybetweenachargeofndash20pCandanotherof+5pConthelinejoiningtheircentresifthecharges are 10 cm apart

12 During a thunderstorm the electric potential differencebetweenacloudandthegroundis10 times 109 V Determine the magnitude of the change in electric potential energy of an electron thatmovesbetweenthesepointsinelectron-volts

13 A charge of 15 μC is placed in a uniform electric fieldoftwooppositelychargedparallelplateswithamagnitudeof14times103 NC-1

(a) Determinetheworkthatmustbedoneagainstthefieldtomovethepointchargeadistance of 55 cm

(b) Calculatethepotentialdifferencebetweenthe final and initial positions of the charge

(c) Determinethepotentialdifferencebetweenthe plates if their separation distance is 15 cm

14 Duringaflashoflightningthepotentialdifferencebetweenacloudandthegroundwas12 times 109 V and the amount of transferred charge was32C

(a) Determinethechangeinenergyofthetransferred charge

(b) Iftheenergyreleasedwasallusedtoacceleratea1tonnecardeduceitsfinalspeed

(c) Iftheenergyreleasedcouldbeusedtomelticeat0degCdeducetheamountoficethatcould be melted

PHYSICS 2015indb 317 150514 338 PM

Chapter 10

318

AH

L

15 SupposethatwhenanelectronmovedfromAtoBin the diagram along an electric field line that the electric field does 36 times 10-19Jofworkonit

Determinethedifferencesinelectricpotential

(a) VB ndash VA(b) VC ndash VA(c) VC ndash VB

16 DeterminethepotentialatpointPthatislocatedatthecentreofthesquareasshowninthediagrambelow

- 6μC

+3μC +2μC

1m

1m

5μC

P

102 Fields at work

Essential idea Similar approaches can be taken inanalysingelectricalandgravitationalpotentialproblems

Understandingsbull Potentialandpotentialenergybull Potentialgradientbull Potentialdifferencebull Escapespeedbull Orbitalmotionorbitalspeedandorbitalenergybull Forcesandinverse-squarelawbehaviour

Orbital motion orbital speed and orbital energy

Although orbital motion may be circular elliptical orparabolic this sub-topic only dealswith circular orbitsThis sub-topic is not fundamentally new physics butan application that synthesizes ideas from gravitationcircularmotiondynamicsandenergy

TheMoonorbitstheEarthandinthissenseitisoftenreferredtoasasatelliteoftheEarthBefore1957itwastheonlyEarthsatelliteHoweverin1957theRussianslaunchedthefirstmanmadesatelliteSputnik1SincethisdatemanymoresatelliteshavebeenlaunchedandtherearenowliterallythousandsofthemorbitingtheEarthSomeareusedtomonitortheweathersome used to enable people to find accurately their position on thesurfaceoftheEarthmanyareusedincommunicationsandnodoubtsomeareusedtospyonothercountriesFigure1015showshowinprincipleasatellitecanbeputintoorbit

Theperson(whosesizeisgreatlyexaggeratedwithrespectto Earth) standing on the surface on the Earth throwssome stonesThe greater the speed with which a stoneis thrown the further it will land from her The pathsfollowedbythethrownstonesareparabolasByastretchof the imaginationwecanvisualisea situation inwhichastoneis thrownwithsuchaspeedthatbecauseof thecurvatureoftheEarthitwillnotlandonthesurfaceoftheEarthbutgointoldquoorbitrdquo(Path4onFigure1015)

NATURE OF SCIENCECommunication of scientific explanations The ability to apply field theory to the unobservable (charges) and the massively scaled (motion of satellites) required scientists to develop new ways to investigate analyse and report findings to a general public used to scientific discoveries based on tangible and discernible evidence (51)

copy IBO 2014

PHYSICS 2015indb 318 150514 338 PM

Fields

319

AH

L

1

2

34Earth

Figure 1015 Throwing a stone into orbit

TheforcethatcausesthestonestofollowaparabolicpathandtofalltoEarthisgravityandsimilarlytheforcethatkeeps the stone in orbit is gravity For circular motionto occur we have seen that a force must act at rightanglestothevelocityofanobjectthatistheremustbeacentripetalforceHenceinthesituationwedescribeherethe centripetal force for circular orbital motion about the EarthisprovidedbygravitationalattractionoftheEarth

We can calculate the speedwithwhich a stonemustbethrowninordertoputitintoorbitjustabovethesurfaceof the Earth

If the stone has mass m and speed v thenwehave fromNewtonrsquos2ndlaw

mv2

RE---------- G

MEm

RE2------------=

whereRE is the radius of the Earth and ME is the mass of the Earth

Bearinginmindthatg 0 GME

RE2--------= then

v g RE 10 64 106timestimes 8 103times= = =

Thatisthestonemustbethrownat8times103m sndash1

Clearly we are not going to get a satellite into orbit soclosetothesurfaceoftheEarthMovingatthisspeedthefrictionduetoairresistancewouldmeltthesatellitebeforeithadtravelledacoupleofkilometresInrealitythereforeasatelliteisputintoorbitabouttheEarthbysendingitattachedtoarocketbeyondtheEarthrsquosatmosphereandthengivingitacomponentofvelocityperpendiculartoaradialvectorfromtheEarthSeeFigure1016

Earth Satellite carried by rocket to hereSatellite orbit

Tangential component of velocity

Figure 1016 Getting a satellite into orbit

Keplerrsquos third law(This work of Kepler and Newtonrsquos synthesis of the work is an excellent example of the scientific method and makes for a good TOK discussion)

In 1627 Johannes Kepler (1571-1630)publishedhis lawsof planetary motion The laws are empirical in natureandwere deduced from the observations of theDanishastronomer Tycho de Brahe (1546-1601)The third lawgives a relationship between the radius of orbit R of a planet and its period TofrevolutionabouttheSunThelawisexpressedmathematicallyas

2

3 constantTR

=

We shall nowuseNewtonrsquos LawofGravitation to showhowitisthattheplanetsmoveinaccordancewithKeplerrsquosthirdlaw

InessenceNewtonwasabletousehislawofgravitytopredict the motion of the planets since all he had to do was factor theF givenby this law intohis second law F = ma to find their accelerations and hence their future positions

InFigure1017theEarthisshownorbitingtheSunandthedistancebetweentheircentresisR

RSun

Earth

F es

F se

Figure 1017 Planets move according to Keplerrsquos third law

Fes is the force that the Earth exerts on the Sun and Fse is the force that the Sun exerts on the Earth The forces are equalandoppositeandtheSunandtheEarthwillactuallyorbitaboutacommoncentreHoweversince theSun issoverymuchmoremassivethantheEarththiscommoncentrewillbeclosetothecentreoftheSunandsowecanregard the Earth as orbiting about the centre of the Sun Theotherthingthatweshallassumeisthatwecanignorethe forces that the other planets exert on the Earth (This wouldnotbeawisethingtodoifyouwereplanningtosend a space ship to theMoon for example)We shallalso assume that we have followed Newtonrsquos exampleandindeedprovedthataspherewillactasapointmasssituated at the centre of the sphere

Kepler had postulated that the orbits of the planets areellipticalbut since theeccentricityof theEarthrsquosorbit issmallweshallassumeacircularorbit

PHYSICS 2015indb 319 150514 338 PM

Chapter 10

320

AH

L

TheaccelerationoftheEarthtowardstheSunisa = Rω2

whereω ω 2πT------=

Hencea R 2π

T------ times

2 4π2 RT 2------------- = =

But the acceleration is given by Newtonrsquos Second Law F = mawhereFisnowgivenbytheLawofGravitationSo in this situation

F maGMs Me

R2-------------------= = butwealsohavethat

a a R 2πT

------ times

2 4π2 RT 2------------- = = and m = Me so that

G MsMe

R2------------------- Me4π2R

T2-------------times

GMs

4π2-----------rArr R3

T2------= =

Butthequantity

isaconstantthathasthesamevalueforeachoftheplanetssowehaveforalltheplanetsnotjustEarththat

R3

T2------ k =

wherekisaconstantWhichisofcourseKeplerrsquosthirdlaw

ThisisindeedanamazingbreakthroughItisdifficulttorefutethe idea that all particles attract each other in accordance withtheLawofGravitationwhenthelawisabletoaccountfortheobservedmotionoftheplanetsabouttheSun

The gravitational effects of the planets upon each othershould produce perturbations in their orbits Such is the predictivepoweroftheUniversalGravitationalLawthatit enabled physicists to compute these perturbations The telescopehadbeeninventedin1608andbythemiddleofthe 18th Century had reached a degree of perfection in design that enabled astronomers to actually measure the orbital perturbations of the planets Their measurements werealways inagreementwith thepredictionsmadebyNewtonrsquoslawHoweverin1781anewplanetUranuswasdiscovered and the orbit of this planet did not fit withthe orbit predicted by Universal Gravitation Such wasthe physicistrsquos faith in theNewtonianmethod that theysuspectedthatthediscrepancywasduetothepresenceofayetundetectedplanetUsingtheLawofGravitationtheFrenchastronomerJLeverrier and the English astronomer J C Adamswereable tocalculate justhowmassive thisnewplanetmustbeandalsowhereitshouldbeIn1846the planetNeptunewas discovered justwhere they hadpredicted Ina similarwaydiscrepancies in theorbitofNeptune ledto thepredictionandsubsequentdiscoveryin1930oftheplanetPlutoNewtonrsquosLawofGravitation

had passed the ultimate test of any theory it is not only abletoexplainexistingdatabutalsotomakepredictions

Satellite energyWhenasatelliteisinorbitaboutaplanetitwillhavebothkineticenergyandgravitationalpotentialenergySupposewe consider a satellite ofmass m that is in an orbit of radius r about a planet of mass M

Thegravitationalpotentialduetotheplanetatdistancer from its centre is

ThegravitationalpotentialenergyofthesatelliteVsat

is therefore GMe m

r----------------ndash

ThatisVsat GMem

r----------------ndash=

Thegravitationalfieldstrengthatthesurfaceoftheplanetisgivenby

g 0GMe

Re2------------=

Hencewecanwrite

Vsat g0 Re

2

rndash=

m

ThekineticenergyofthesatelliteKsatisequaltofrac12mv2wherev is its orbital speed

Byequatingthegravitationalforceactingonthesatellitetoitscentripetalaccelerationwehave

G Mem

r2---------------- mv2

r---------- mv2hArr

GMemr----------------= =

Fromwhich12---

mv2 12---

GMemr----------------times=

= 0 e

2g R

r

2m

Which is actually quite interesting since it shows thatirrespective of the orbital radius the KE is numericallyequal to half the PE Also the total energy E tot K sat Vsat+ 1

2---G Me m

r----------------times G Me m

r----------------ndash + 1

2---G Me m

r----------------ndash= = =of the satelliteisalwaysnegativesince

E tot K sat Vsat+ 12---

G Me mr----------------times

G Me mr----------------ndash

+ 12---

G Me mr----------------ndash= = =

PHYSICS 2015indb 320 150514 338 PM

Fields

321

AH

L

The energies of an orbiting satellite as a function of radial distancefromthecentreofaplanetareshownplottedinFigure1018

12

100806

0402

0ndash02

ndash04

ndash06ndash08ndash10

ndash12

2 4 6 8 10 12

ener

gya

rbita

ry u

nits

distance timesR

kinetic energytotal energy

potential energy

Figure 1018 Energy of an orbiting satellite as a function of distance from the centre of a planet

WeightlessnessSuppose that you are inanelevator(lift)thatisdescendingat constant speedandyou let goof abook thatyouareholdinginyourhandThebookwillfalltothefloorwithaccelerationequaltotheaccelerationduetogravityIfthecablethatsupportstheelevatorweretosnap(asituationthatItrustwillneverhappentoanyofyou)andyounowletgothebookthatyouareholdinginyourotherhandthis book will not fall to the floor - it will stay exactlyin linewithyourhandThis isbecause thebook isnowfallingwith the sameaccelerationas theelevatorandassuch the book cannot ldquocatchrdquo up with the floor of theelevatorFurthermoreifyouhappenedtobestandingonasetofbathroomscalesthescaleswouldnowreadzero-youwouldbeapparentlyweightlessItisthisideaoffreefallthatexplainstheapparentweightlessnessofastronautsin an orbiting satellite These astronauts are in free fall inthesensethattheyareacceleratingtowardsthecentre of the Earth

It is actually possible to define theweight of a body inseveral different waysWe can define it for example asthegravitationalforceexertedonthebodybyaspecifiedobject such as the EarthThiswe have seen thatwe doinlotsofsituationswherewedefinetheweightasbeingequaltomgIfweusethisdefinitionthenanobjectinfreefallcannotbydefinitionbeweightlesssinceitisstillinagravitationalfieldHoweverifwedefinetheweightofanobjectintermsofaldquoweighingrdquoprocesssuchasthereadingonasetofbathroomscaleswhichineffectmeasuresthecontact force between the object and the scales thenclearlyobjectsinfreefallareweightlessOnenowhasto

askthequestionwhetherornotitispossibleForexampletomeasurethegravitationalforceactingonanastronautin orbit about the Earth

WecanalsodefineweightintermsofthenetgravitationalforceactingonabodyduetoseveraldifferentobjectsForexample for an object out in space its weight could bedefined in terms of the resultant of the forces exerted on it bytheSuntheMoontheEarthandalltheotherplanetsin the Solar System If this resultant is zero at a particular pointthenthebodyisweightlessatthispoint

Inviewofthevariousdefinitionsofweightthatareavailabletousitisimportantthatwhenweusethewordldquoweightrdquoweareawareofthecontextinwhichitisbeingused

Example

Calculate the height above the surface of the Earth atwhichageo-stationarysatelliteorbits

Solution

A geo-stationary satellite is one that orbits the Earth in such a way that it is stationary with respect to a point on the surface of the Earth This means that its orbital period must be the same as the time for the Earth to spin once on its axis ie 24 hours

From Keplerrsquos third law we have G Ms

4π2----------- R3

T2------=

That is

RMe

mh

using the fact that the force of attraction between the satellite and the Earth is given by

FG Mem

R2----------------=

and that F = ma

where m is the mass of the satellite and a 4π2R

T2-------------=

we have

GMem

R2---------------- m 4π2R

T2-------------times

GMe

4π2------------rArr R3

T2------= =

PHYSICS 2015indb 321 150514 338 PM

Chapter 10

322

AH

L

Now the mass of the Earth is 60 times 1024 kg and the period T measured in seconds is given by T = 86400 s

So substitution gives R = 42 times 106 m

The radius of the Earth is 64 times 106 m so that the orbital height h is about 36 times 107 m

Example

Calculatetheminimumenergyrequiredtoputasatelliteofmass500kgintoanorbitthatisasaheightequaltotheEarthrsquosradiusabovethesurfaceoftheEarth

Solution

We have seen that when dealing with gravitational fields and potential it is useful to remember that

g 0G M

Re2---------= or g0 Re

2 GM=

The gravitational potential at the surface of the Earth is

g 0ndash Re GMRe

---------ndash=

The gravitational potential at a distance R

from the centre of the Earth is

=

The difference in potential between the surface of the Earth and a point distance R from the centre is therefore

V∆ g 0Re 1ReR------ndash

=

If R 2Re= then V∆g0 Re

2------------=

This means that the work required to ldquoliftrdquo the satellite into orbit is g0Rm where m is the mass of the satellite This is equal to

10 32 106 500timestimestimes = 16000 MJ

However the satellite must also have kinetic energy in order to orbit Earth This will be equal to

g 0 mRe2

2R-----------------g 0mRe

2

4----------------- 8000 MJ= =

The minimum energy required is therefore

24000 MJ

Exercise 1031 The speed needed to put a satellite in orbit does

not depend on

A the radius of the orbitB theshapeoftheorbitC thevalueofgattheorbitD the mass of the satellite

2 EstimatethespeedofanEarthsatellitewhoseorbitis400kmabovetheEarthrsquossurfaceAlsodetermine the period of the orbit

3 Calculatethespeedofa200kgsatelliteorbitingthe Earth at a height of 70 times 106 m

Assume that g = 82 m sndash2 for this orbit

4 TheradiioftwosatellitesXandYorbitingthe Earth are 2r and 8rwherer is the radius of the Earth Calculate the ratio of the periods of revolutionofXtoY

5 A satellite of mass mkgissentfromEarthrsquossurfaceinto an orbit of radius 5RwhereR is the radius of theEarthWritedownanexpressionfor

(a) thepotentialenergyofthesatelliteinorbit

(b) thekineticenergyofthesatelliteinorbit

(c) theminimumworkrequiredtosendthesatellitefromrestattheEarthrsquossurfaceintoits orbit

6 A satellite in an orbit of 10rfallsbacktoEarth(radius r)afteramalfunctionDeterminethespeedwithwhichitwillhittheEarthrsquossurface

7 The radius of the moon is frac14 that of the Earth AssumingEarthandtheMoontohavethesamedensitycomparetheaccelerationsoffreefallatthe surface of Earth to that at the surface of the Moon

8 UsethefollowingdatatodeterminethegravitationalfieldstrengthatthesurfaceoftheMoon and hence determine the escape speed from the surface of the Moon

Mass of the Moon = 73 times 1022kg

RadiusoftheMoon=17times106 m

PHYSICS 2015indb 322 150514 338 PM

Wave Phenomena

289

AH

L

Using information from this graph calculate the

(a) valueofthespringconstant(b) maximumkineticenergyofthemagnet

7 AsystemisoscillatingwithSHMasdescribedbythe graph in the Figure 908 below

Disp

lace

men

t (cm

)

005 1 15 2 25 3 35

2

-2

-6

-4

4

6

Time (s)

Figure 908

(a) Usethegraphtodeterminethe

(i) periodofoscillation(ii) amplitudeofoscillation(iii) maximumspeed(iv) speedatt=13s(v) maximumacceleration

(b) Statetwovaluesoftforwhenthemagnitudeofthevelocity is a maximum and two values of t for when the magnitude of the acceleration is a maximum

9 Figure 909 shows the relationship that exists between the acceleration and the displacement from the equilibriumpositionforaharmonicoscillator

Acce

lera

tion

(m-2

)

Displacement (cm)0

-4000

-2000

2000

-05 050

Figure 909

(a) Stateandexplaintworeasonswhythegraphopposite indicates that the object is executing simple harmonic motion

(b) Determinethefrequencyofoscillation

92 Single-slit diffraction

Essential idea Single-slitdiffractionoccurswhenawaveis incident upon a slit of approximately the same size as the wavelength

Understandingsbull Thenatureofsingle-slitdiffraction

The nature of single-slit diffractionSingle slit diffraction intensity distribution

When plane wavefronts pass through a small aperture they spread out as discussed in chapter 4 This is an example ofthephenomenoncalleddiffractionLightwavesarenoexception to this and ways for observing the diffraction of light have also been discussed previously

Howeverwhenwelookatthediffractionpatternproducedbylightweobserveafringepatternthatisonthescreenthere is a bright central maximum with ldquosecondaryrdquomaxima either side of it There are also regions where there is no illumination and these minima separate the maxima If we were to actually plot how the intensity of illumination varies along the screen then we would obtain a graph similar to that as in Figure 910

intensity

distance along screen

Figure 910 Intensity distribution for single-slit diffraction

We would get the same intensity distribution if we were to plot the intensity against the angle of diffraction θ(Seenextsection)

NATURE OF SCIENCEDevelopment of theories When light passes through an aperture the summation of all parts of the wave leads to an intensity pattern that is far removed from the geometrical shadow that simple theory predicts (19)

copy IBO 2014

PHYSICS 2015indb 289 150514 337 PM

Chapter 9

290

AH

L

This intensity pattern arises from the fact that each point ontheslitactsinaccordancewithHuygenrsquosprincipleasa source of secondary wavefronts It is the interference between these secondary wavefronts that produces the typical diffraction pattern

Obtaining an expression for the intensity distribution is mathematically a little tricky and it is not something that we are going to attempt here However we can deducea useful relationship from a simple argument In this argument we deal with a phenomenon called Fraunhofer diffractionthatisthelightsourceandthescreenareaninfinitedistanceawayformtheslitThiscanbeachievedwith the set up shown in Figure 911

lens 1 lens 2

single slit

source

screen

Figure 911 Apparatus for viewing Fraunhofer diffraction

The source is placed at the principal focus of lens 1 and the screen isplacedat theprincipal focusof lens2Lens1 ensures that parallel wavefronts fall on the single slit and lens2ensuresthattheparallelraysarebroughttoafocuson the screen The same effect can be achieved using a laser and placing the screen some distance from the slit If the light and screenarenot an infinitedistance from the slitthen we are dealing with a phenomenon called Fresnel diffraction and such diffraction is very difficult to analyse mathematically To obtain a good idea of how the single slit patterncomesaboutweconsiderthediagramFigure912

Xθ2

Yλ

d

P

b

f

screen

θ1

Figure 912 Single slit diffraction

In particular we consider the light from one edge of the slittothepointPwherethispointis justonewavelengthfurther from the lower edge of the slit than it is from the

upper edge The secondary wavefront from the upper edge willtraveladistanceλ2furtherthanasecondarywavefrontfrom a point at the centre of the slitHencewhen thesewavefronts arrive at P theywill be out of phase andwillinterfere destructively The wavefronts from the next point below the upper edge will similarly interfere destructively with the wavefront from the next point below the centre of the slit In this way we can pair the sources across the whole width of the slit If the screen is a long way from the slit then the angles θ1 and θ2becomenearlyequal (If the screen is at infinity then they are equal and the two lines PX and XY are at right angles to each other)FromFigure1116weseethereforethattherewillbeaminimumatPif

λ b θ1 sin= where b is the width of the slit

Howeverbothanglesareverysmallequaltoθ saywhereθ is the angle of diffraction

Soitcanbewritten θ λb---

=

This actually gives us the half-angular width of the central maximum We can calculate the actual width of the maximum along the screen if we know the focal length of the lens focussing the light onto the screen If this is f then we have that

θ df---

= Suchthat d fλb-----

=

To obtain the position of the next maximum in the pattern we note that the path difference is 3

2---λ We therefore divide

theslitintothreeequalpartstwoofwhichwillproducewavefronts that will cancel and the other producing wavefronts that reinforce The intensity of the second maximum is therefore much less than the intensity of thecentralmaximumof theorderof005or120of theoriginalintensity(Muchlessthanonethirdinfactsincethewavefrontsthatreinforcewillhavedifferingphases)

We can also see now how diffraction effects become more and more noticeable the narrower the slit becomes If light of wavelength 430 nm was to pass through a slit of width say10cmandfallonascreen30mawaythenthehalfangular width of the central maximum would be 013 microm

d = f λ

___ b = 3 times 430 times 10 ndash9 ____________ 01 = 013 μm

There will be lots of maxima of nearly the same intensity and the maxima will be packed very closely together (Thefirstminimumoccursatadistanceof012micromfromthe centre of the central maximum and the next occurs effectivelyatadistanceof024microm)Weeffectivelyobservethe geometric pattern Refer to Example

We also see now how for diffraction effects to be noticeable the wavelength must be of the order of the slit width The width of the pattern increases in proportion to the wavelength and decreases inversely with the width of the

PHYSICS 2015indb 290 150514 337 PM

Wave Phenomena

291

AH

L

93 Interference

Essential idea Interference patterns from multiple slits andthinfilmsproduceaccuratelyrepeatablepatterns

Understandingsbull Youngrsquosdouble-slitexperimentbull Modulationoftwo-slitinterferencepatternbyone-slit

diffraction effectbull Multipleslitanddiffractiongratinginterferencepatternsbull Thinfilminterference

Youngrsquos double slit experimentPlease refer to interference in chapter 4 However wewill reiterate the condition for the interference of waves from two sources to be observed The two sources must be coherentthatistheymusthavethesamephaseorthephase difference between them must remain constant

Also to reinforce topics concerning the principle ofsuperpositionandpathdifferenceandphasedifferenceincluded in Figure 913 is another example of two source interference

We can obtain evidence for the wave nature of sound by showing that sound produces an interference pattern Figure 913 shows the set up for demonstrating this

CRO

speaker

speaker

signalgenerator

microphoneX

Y

Figure 913 Interference using sound waves

The two speakers are connected to the same output of the signal generator andplaced about 50 cmapartThesignalgeneratorfrequencyissettoabout600Hzandthe

slit If the slit width is much greater than the wavelength then the width of the central maxima is very small

Insummary

for destructive interference and minima

dsinθ=nλ wheren=123hellip

for constructive interference and maxima

dsinθ=(n+frac12)λ wheren=123hellip

Note that waves from point sources along the slit arrive in phase at the centre of the fringe pattern and constructively interferetoproducethecentralmaximumcalledthezero ordermaximum(n=0)

Example

LightfromalaserisusedtoformasingleslitdiffractionpatternThe width of the slit is 010 mm and the screen is placed 30 m from the slit The width of the central maximum is measured as26cmCalculatethewavelengthofthelaserlight

Solution

Since the screen is a long way from the slit we can use the small angle approximation such that the f is equal to 30 m

The half-width of the central maximum is 13 cm so we have

λ13 10 2ndashtimes( ) 10 10 4ndashtimes( )times

30------------------------------------------------------------------=

To give λ = 430 nm

This example demonstrates why the image of a point source formed by a thin converging lens will always have a finite width

Exercise 921 Aparallelbeamoflightofwavelength500nmis

incidentonaslitofwidth025mmThelightisbroughttofocusonascreenplaced150mfromthe slit Calculate the angular width and the linear width of the central diffraction maximum

2 Lightfromalaserisusedtoformasingleslitdiffraction pattern on a screen The width of the slit is 010 mm and the screen is 30 m from the slit The width of the central diffraction maximum is 26cmCalculatethewavelengthofthelaserlight

3 Determinetheanglethatyouwouldexpecttofindconstructiveinterference(forn=0n=1andn=2) fromasingleslitofwidth20μmandlightofwavelength 600 nm

NATURE OF SCIENCECuriosity Observed patterns of iridescence in animals such as the shimmer of peacock feathers led scientists to develop the theory of thin film interference (15)

Serendipity The first laboratory production of thin films was accidental (15)

copy IBO 2014

PHYSICS 2015indb 291 150514 337 PM

Chapter 9

292

AH

L

microphoneismovedalongthelineXYthatisaboutonemetre from the speakers As the microphone is moved along XY the trace on the cathode ray oscilloscope isseen to go through a series of maxima and minima corresponding to points of constructive and destructive interference of the sound waves

Aninterestinginvestigationistofindhowtheseparationof the points of maximum interference depends on the frequency of the source and also the distance apart ofthe speakers

In the demonstration of the interference between two soundsourcesdescribedaboveifweweretomoveoneofthe speakers from side to side or backwards and forwards the sound emitted from the two speakers would no longer be in phase No permanent points of constructive or destructive interference will now be located since the phase difference between the waves from the two sources is nolongerconstantiethesourcesarenolongercoherent

Light from an incandescent source is emitted with acompletely random phase Although the light from two separate sources will interfere because of the randomlychanging phase no permanent points of constructive or destructive interference will be observed This is why a singleslit isneededintheYoungrsquosdoubleslitexperimentByactingasapointsourceitessentiallybecomesacoherentlight source The light emitted from a laser is also very nearly coherent and this is why it is so easy to demonstrate optical interference and diffraction with a laser

Youngrsquos double slit experiment is one of the great classic experiments of physics and did much to reinforce the wave theoryoflightThomasYoungcarriedouttheexperimentin about 1830

It is essentially the demonstration with the ripple tank and the sound experiment previously described butusing light The essential features of the experiment are shown in Figure 914

screendouble slit

single slit

coloured lter

S 1

S2

Figure 914 Youngrsquos double slit experiment

YoungallowedsunlighttofallontoanarrowsingleslitA few centimetres from the single slit he placed a double slit The slits are very narrow and separated by a distance equal to about fourteen slitwidthsA screen is placed

about ametre from thedouble slitsYoungobservedapatternofmulticolouredldquofringesrdquo in the screenWhenhe placed a coloured filter between the single slit anddouble slit he obtained a pattern that consisted of bright coloured fringes separated by darkness

The single slit essentially ensures that the light falling on the double slit is coherent The two slits then act as the two speakers in the sound experiment or the two dippers in the ripple tank The light waves from each slit interfere and produce the interference pattern on the screen Without thefilterapatternisformedforeachwavelengthpresentinthesunlightHencethemulticolouredfringepattern

You can demonstrate optical interference for yourselfSmoking a small piece of glass and then drawing twoparallel lines on it can make a double slit If you then look throughthedoubleslitatasingletungstenfilamentlampyouwillseethefringepatternByplacingfiltersbetweenthe lamps and the slits you will see the monochromatic fringe pattern

YoucanalsoseetheeffectsofopticalinterferencebylookingatnetcurtainsEachlsquoholersquointhenetactsasapointsourceand the light from all these separate sources interferes and producesquiteacomplicatedinterferencepattern

A laser can also be used to demonstrate optical interference Sincethelightfromthelaseriscoherentitisveryeasytodemonstrate interference Just point the laser at a screen and place a double slit in the path of the laser beam

LetusnowlookattheYoungrsquosdoubleslitexperimentinmore detail The geometry of the situation is shown in Figure915

S1

S2 X

P

Q

D

y

d

screen

θ

θrsquo

Figure 915 The geometry of Youngrsquos double slit experiment

S1andS2 are the two narrow slits that we shall regard as twocoherentmonochromaticpointsourcesThedistancefrom the sources to the screen is D and the distance between the slits is d

The waves from the two sources will be in phase at Q and therewill be abright fringehereWewish tofindtheconditionfortheretobeabrightfringeatPdistancey from Q

PHYSICS 2015indb 292 150514 337 PM

Wave Phenomena

293

AH

L

We note that that D (asymp1metre)isverymuchgreaterthaneither y or d(asymp fewmillimetres)Thismeansthatbothθ and θʹ areverysmallanglesandforintentsandpurposesequal

From the diagram we have that

θ = Dy

And

2S Xd

θʹ =

(Remembertheanglesareverysmall)

Sinceθ asymp θʹ then

2S XyD d

=

ButS2X is the path difference between the waves as they

traveltoPWehavethereforethat

path difference = Dyd

TherewillthereforebeabrightfringeatPif

Dyd

= nλ

Supposethatthereisabrightfringeaty = y1 corresponding

to n = n1 then

Ddy1 = n

1λ

If the next bright fringe occurs at y = y2 this will correspond

to n = n1+1Hence

Ddy2 =(n

1+1)λ

This means that the spacing between the fringes y2 ndash y

1 is

given by

y2 ndash y

1 = d

Dλ

Youngactuallyusethisexpressiontomeasurethewavelengthof the light he used and it is a method still used today

We see for instance that if in a given set up we move the slits closer together then the spacing between the fringes will get greaterEffectivelyourinterferencepatternspreadsoutthatis there will be fewer fringes in a given distance We can also increase the fringe spacing by increasing the distance betweentheslitsandthescreenYouwillalsonotethatfora

givensetupusinglightofdifferentwavelengthsthenldquoredrdquofringeswillspacefurtherapartthanldquobluerdquofringes

In this analysis we have assumed that the slits act as point sources and as such the fringes will be uniformly spaced andofequalintensityAmorethoroughanalysisshouldtakeintoaccountthefinitewidthoftheslits

ReturningtoFigure915weseethatwecanwritethepathdifferenceS2X as

S2X = dtan θʹ

But since θʹ is a small angle the sine and tangent will be nearlyequalsothat

S2X = dsinθʹ

And since θʹ asymp θ then

S2X = dsinθ

The condition therefore for a bright fringe to be found at a point of the screen can therefore be written as

dsinθ = nλ

Insummary

for constructive interference and minima

d sinθ = nλ wheren=123hellip

for destructive interference and maxima

d sinθ = ( n + frac12 )λ wheren=123hellip

Figure 916 shows the intensity distribution of the fringes on the screen when the separation of the slits is large comparedtotheirwidthThefringesareofequalintensityandofequalseparation

Intensity

distance along screen

Figure 916 The intensity distribution of the fringes

It isworthnoting that if the slits are close together theintensity of the fringes is modulated by the intensity distribution of the diffraction pattern of one of the slits

PHYSICS 2015indb 293 150514 337 PM

Chapter 9

294

AH

L

Example

Light of wavelength 500nm is incident on two smallparallel slits separated by 10mmDetermine the anglewherethefirstmaximumisformed

If after passing through the slits the light is brought toa focusona screen15m fromtheslitscalculate theobserved fringe spacing on the screen

Solution

Using the small angle approximation we have

3

7

10105minus

minustimes==

dλθ

= 5 times 10-4 rad

The fringe spacing is given by

3

7

1010551

minus

minustimestimes==

dDy λ

= 075 mm

Modulation of two-slit interference pattern by one-slit diffraction effect

In practice the intensity pattern of the maxima as shown in Figure 917 is not constant but fluctuates while decreasingsymmetrically on either side of the central maximum as shown inFigure917Theintensitypatternisacombinationofboththe single-slit diffractionenvelope and the double slit pattern That is the amplitude of the two-slit interference pattern is modulated(ieadjustedto)byasingleslitdiffractionenvelope

2λD

___ λD

D

___λd

d

___

Double slitnite width

Figure 917 Double slit pattern

ForagivenslitseparationdwavelengthoflightandfixedslittoscreendistancethevariationinintensitydependsonthewidthoftheslitDAlthoughincreasingthewidthoftheslitsincreasestheintensityoflightinthefringesitalsomakesthem less sharp and they become blurred As the slit width is widenedthefringesgraduallydisappearbecausenumerouspoint sources along the widened slit give rise to their own dark and bright fringes that then overlap with each other

Exercise 931 InFigure913thedistancebetweenthespeakers

is050mandthedistancebetweenthelineofthespeakersandthescreenis20mAsthemicrophoneismovedalongthelineXYthedistance between successive points of maximum soundintensityis030mThefrequencyofthesoundwavesis44times103HzCalculateavalueforthe speed of sound

2 Laserlightofwavelength610nmfallsontwoslits10x10-5apartDeterminetheseparationofthefirstordermaximumformedonascreen20maway

3 Twoparallelslits012mmapartareilluminatedwith red light with a wavelength of 600 nm An interferencepatternfallsonascreen15maway

Calculate

(a) thedistancefromthecentralmaximumtothefirstbrightfringe

(b) thedistancetotheseconddarkline

Multiple slit and diffraction grating interference patterns

If we examine the interference pattern produced when monochromatic light passes through a different number of slits we notice that as the number of slits increases the number of observed fringes decreases the spacingbetween them increases and the individual fringes become much sharper We can get some idea of how this comes about by looking at the way light behaves when a parallel beam passes through a large number of slits The diagram for this is shown in Figure 918

θθd

1

2

3

telescope

Figure 918 A parallel beam passing through several slits

The slits are very small so that they can be considered to act as point sources They are also very close together such that d issmall(10ndash6 m)EachslitbecomesasourceofcircularwavefrontsandthewavesfromeachslitwillinterfereLetus

PHYSICS 2015indb 294 150514 337 PM

Wave Phenomena

295

AH

L

consider the light that leaves the slit at an angle θ as shown Thepathdifferencebetweenwave1andwave2isdsinθ and ifthisisequaltoanintegralnumberofwavelengthsthenthe two waves will interfere constructively in this direction Similarly wave 2 will interfere constructively with wave3 at this angle andwave 3with 4 etc across thewholegratingHenceifwelookatthelightthroughatelescopethatisbringittoafocusthenwhenthetelescopemakesanangle θ to the grating a bright fringe will be observed The condition for observing a bright fringe is therefore

dsinθ = mλ

Supposeweuselightofwavelength500nmandsupposethat d = 16 times 10ndash6 m

Obviously we will see a bright fringe in the straight on position θ=0(thezero order)

The next position will be when m =1(thefirst order)andsubstitutionintheaboveequationgivesθ = 18deg

The next position will be when m=2(thesecond order)and this give θ = 38deg

For m=3sinθ is greater than 1 so with this set up we only obtain5fringesonezeroorderandtwoeithersideofthezero order

The calculation shows that the separation of the orders is relatively large At any angles other than 18deg or 38deg the light leaving the slits interferes destructively We can see that the fringes will be sharp since if we move just a small angle away from 18deg the light from the slits will interfere destructively

An array of narrow slits is usually made by cutting narrow transparent lines very close together into the emulsion on aphotographicplate(typically200linespermillimetre)Suchanarrangementiscalledadiffraction grating

The diffraction grating is of great use in examining the spectral characteristics of light sources

All elements have their own characteristic spectrum An element can be made to emit light either by heating it until it is incandescent or by causing an electric discharge through itwhen it is in a gaseous state Yourschool probably has some discharge tubes and diffraction gratingsIfithasthenlookattheglowingdischargetubethroughadiffractiongratingIfforexampletheelementthat you are looking has three distinct wavelengths then each wavelength will be diffracted by a different amount inaccordancewiththeequationdsinθ = nλ

Alsoiflaserlightisshonethroughagratingontoascreenyou will see just how sharp and spaced out are the maxima By measuring the line spacing and the distance of the screen fromthelaserthewavelengthofthelasercanbemeasured

If your school has a set of multiple slits say from a single slit toeight slits then it isalsoaworthwhileexercise toexamine how the diffraction pattern changes when laser light is shone through increasing numbers of slits

Consider a grating of NparallelequidistantslitsofwidthaseparatedbyanopaqueregionofwidthD We know that light diffracted from one slit will be superimposed and interfere with light diffracted from all the other slits The intensityprofile of the resultant wave will have a shape determined by

bull theintensityoflightincidentontheslitsbull thewavelengthofthislightbull theslitwidthbull theslitseparation

Typical intensity profiles formed when a plane wavemonochromaticlightatnormalincidenceononetwoandsix slits are shown in Figure 919 In this example the slit widthisonequarterofthegratingspacing

Six slits sharperfringes

d = 4a

I α N 2

(N ndash 2)= 4 subsidiary maxima

order0 4321

I = 36

Two slits

order0

single-slitdiraction

envelope

d = 4a4th principal maximum missing

principa maxima

I = 4 (nominal units)

4321

intensity

order

ISS

One slit

0 1

single-slitdiraction

pattern

N = number of slitsslit width a

Figure 919 Intensity profiles

PHYSICS 2015indb 295 150514 337 PM

Chapter 9

296

AH

L

1 For Nge2thefringepatterncontainsprincipalmaximathat are modulated by the diffraction envelope

2 ForNgt2thefringepatterncontains(Nndash2)subsidiary maxima

3 Where the grating spacing d is four time the slit width a(4xa =d)thenthefourthprincipalmaximumismissing from the fringe pattern Where the grating spacingdisthreetimetheslitwidtha(3xa =d)thenthe third principal maximum is missing from the fringepatternandsoon

4 Increasing N increases the absolute intensities of the diffraction pattern (I = N2)

If white light is shone through a grating then the central image will be white but for the other orders each will be spread out intoacontinuousspectrumcomposedofaninfinitenumberof adjacent images of the slit formed by the wavelength of the different wavelengths present in the white light At any given point in the continuous spectrum the light will be very nearly monochromatic because of the narrowness of the images of the slit formed by the grating This is in contrast to the doubleslitwhereifwhitelightisusedtheimagesarebroadand the spectral colours are not separated

In summary as is the case of interference at a doubleslit there is a path difference between the rays fromadjacent slits of d sinθ Where there is a whole number of wavelengthsconstructiveinterferenceoccurs

For constructive interference and minima

d sinθ = mλ wheren=123hellip

When there is a path difference between adjacent rays of halfawavelengthdestructiveinterferenceoccurs

For and maxima

d sinθ = ( m + frac12 )λ wheren=123hellip

Exercise 941 Lightfromalaserisshonethroughadiffraction

grating on to a screen The screen is a distance of 20mfromthelaserThedistancebetweenthecentraldiffractionmaximumandthefirstprincipalmaximum formed on the screen is 094 m The diffractiongratinghas680linespermmEstimatethe wavelength of the light emitted by the laser

2 Monochromaticlightfromalaserisnormallyincident on a six-slit grating The slit spacing is three times the slit width An interference patternisformedonascreen2montheotherside of the grating

(a) Determineandexplaintheorderoftheprincipal maxima missing from the fringe pattern

(b) Howmanysubsidiarymaximaaretherebetween the principal maxima

(c) Howmanyprincipalmaximaarethere

(i) Betweenthetwofirstorderminimain the diffraction envelope

(ii) Betweenthefirstandsecondorderminima in the diffraction envelope

(d) Drawthepatternformedbythebrightfringesin(c)

(e) Sketchtheintensityprofileofthefringepattern

Thin film interferenceYoumightwellbe familiarwith the colouredpatternoffringes that can be seen when light is reflected off thesurfaceofwateruponwhichathinoilfilmhasbeenspiltorfromlightreflectedfrombubblesWecanseehowthesepatternsarisebylookingatFigure920

Extendedlight source air

A

B

C

12

3

4

Oil lm

Water

Figure 920 Reflection from an oil film

Consider light from an extended source incident on a thinfilmWealsoconsiderawave fromonepointofthe source whose direction is represented by the ray shownSomeofthislightwillbereflectedatAandsometransmitted through the filmwhere somewill again bereflectedatB(somewillalsobetransmittedintotheair)SomeofthelightreflectedatBwillthenbetransmittedatCandsomereflectedandsoonIfweconsiderjustrays1and2thenthesewillbenotbeinphasewhenthearebrought to a focus by the eye since they have travelled

PHYSICS 2015indb 296 150514 337 PM

Wave Phenomena

297

AH

L

different distances If the path difference between them is an integral number of half wavelengths then we might expect the twowaves to be out of phaseHowever ray1 undergoes a phase change of π on reflection but ray2 does not since it is reflected at a boundary betweena more dense and less dense medium (See chapter 4)Henceifthepathdifferenceisanintegralnumberofhalf-wavelengthsrays1and2willreinforceieray1and2areinphaseHowever rays357etcwillbeoutofphasewithrays246etcbutsinceray2ismoreintensethanray3andray4moreintensethanray5thesepairswillnot cancel out so there will be a maximum of intensity

Ifthepathdifferenceissuchthatwave1and2areoutofphasesincewave1ismoreintensethanwave2theywillnotcompletelyannulHoweveritcanbeshownthattheintensitiesofwaves2345hellipaddtoequaltheintensityofwave1Sincewaves345hellipareinphasewithwave2there will be complete cancellation

The path difference will be determined by the angle at which ray1 is incident andalsoon the thickness (andtheactualrefractiveindexaswell)ofthefilmSincethesourceisanextendedsourcethelightwillreachtheeyefrom many different angles and so a system of bright and darkfringeswillbeseenYouwillonlyseefringesifthefilmisverythin(exceptifviewedatnormalincidence)sinceincreasingthethicknessofthefilmwillcausethereflected rays to get so far apart that they will not becollected by the pupil of the eye

From the argument above to find the conditions forconstructive and destructive interferenceweneedonlyfindthepathdifferencebetweenray1andray2Figure921showsthe geometry of the situation

A BE

F

C

D

1 2

φrsquo φrsquo

φ

φ

d

Figure 921 The geometry of interference

Thefilmisofthicknessd and refractive index n and the light has wavelength λ If the line BF is perpendicular to ray 1 then the optical path difference (opd) between ray 1 andray2whenbroughttoafocusis

opd = n(AC+CB)ndashAF

We have to multiple by the refractive index for the path travelledbythelightinthefilmsincethelighttravelsmoreslowly in thefilm If the light travels say adistancex in a material of refractive index n then in the time that it takes to travelthisdistancethelightwouldtraveladistancenx in air

IfthelineCEisatrightanglestoray2thenweseethat

AF = nBE

FromthediagramAC=CDsowecanwrite

opd = n(CD+CB)ndashnBE=nDE

Alsofromthediagramweseethatwhereφ is the angle of refraction

DE=2cosφ

From which opd=2ndcosφ

Bearinginmindthechangeinphaseofray1onreflectionwe have therefore that the condition for constructive interference is

2nd φcos m 12---+

λ m 1 2 hellip = =

Andfordestructiveinterference 2ndcosφ = mλ

Eachfringecorrespondstoaparticularopd for a particular value of the integer m and for any fringe the value of the angle φ is fixedThismeans that itwill be in the formof an arc of a circle with the centre of the circle at the point where the perpendicular drawn from the eye meets the surface of the film Such fringes are called fringesofequal inclinationSince theeyehasa smallaperturethesefringesunlessviewedatneartonormalincidence (φ = 0) will only be observed if the film is very thinThisisbecauseasthethicknessofthefilmincreasesthereflected rayswill get further and further apart and sovery few will enter the eye

Ifwhitelightisshoneontothefilmthenwecanseewhywe get multi-coloured fringes since a series of maxima and minima will be formed for each wavelength present in the white light However when viewed at normalincidenceitispossiblethatonlylightofonecolourwillundergoconstructiveinterferenceandthefilmwilltakeon this colour

PHYSICS 2015indb 297 150514 337 PM

Chapter 9

298

AH

L

Thickness of oil films

The exercise to follow will help explain this use

Non-reflecting films

A very important but simple application of thin filminterferenceisintheproductionofnon-reflectingsurfaces

Athinfilmofthicknessd and refractive index n1 is coated onto glass of refractive index n where n1 lt n Light ofwavelength λ that is reflected at normal incidence willundergo destructive interference if 12

2n d λ

= thatis

14d

nλ

=

(remember that there will now no phase change at the glass interface ie we have a rare to dense reflection)

The use of such films can greatly reduce the loss oflightby reflection at the various surfacesof a systemoflenses or prisms Optical parts of high quality systemsare usually all coatedwithnon-reflecting films in orderto reduce stray reflections The films are usually madebyevaporatingcalciumormagnesiumfluorideonto thesurfacesinvacuumorbychemicaltreatmentwithacidsthat leave a thin layer of silica on the surface of the glass ThecoatedsurfaceshaveapurplishhuebyreflectedlightThis is because the condition for destructive interference fromaparticularfilmthicknesscanonlybeobtainedforone wavelength The wavelength chosen is one that has a value corresponding to light near the middle of the visible spectrumThismeansthatreflectionofredandvioletlightis greater combining to give the purple colour Because of the factor cosφatanglesotherthannormalincidencethepath difference will change butnotsignificantlyuntilsayabout30deg(egcos25deg=090)

It should be borne in mind that no light is actually lost by anon-reflectingfilmthedecreaseofreflectedintensityiscompensated by increase of transmitted intensity

Non-reflecting films can be painted onto aircraft tosuppressreflectionofradarThethicknessofthefilmis determined by

4nd λ

= where λ is the wavelength of

the radar waves and ntherefractiveindexofthefilmat this wavelength

Example

A plane-parallel glass plate of thickness 20 mm isilluminated with light from an extended source The refractiveindexoftheglassis15andthewavelengthofthelight is 600 nm Calculate how many fringes are formed

Solution

We assume that the fringes are formed by light incident at all angles from normal to grazing incidence

At normal incidence we have 2nd = mλ

From which

m 2 15 2 10 3ndashtimestimestimes6 10 7ndashtimes

------------------------------------------- 10 000= =

At grazing incidence the angle of refraction φ is in fact the critical angle

Therefore φφ arcsin 115------- 42 deg= =

ie cosφ = 075

At grazing incidence 2ndcosφ = m λ

From which (and using cos(sinndash1(115) = 075)

m 2 15 2 10 3ndash 075timestimestimestimes6 10 7ndashtimes

------------------------------------------------------------ 7500= =

The total number of fringes seen is equal to m ndash m = 2500

Exercise 95When viewed from above the colour of an oil film onwater appears red in colour Use the data below to estimate theminimumthicknessoftheoilfilm

average wavelength of red light = 630 nm

refractiveindexofoilforredlight=15

refractive index of water for red light = 13

PHYSICS 2015indb 298 150514 337 PM

Wave Phenomena

299

AH

L

94 Resolution

Essential idea Resolution places an absolute limit on the extent to which an optical or other system can separate images of objects

Understandings

bull Thesizeofadiffractingaperturebull Theresolutionofsimplemonochromatic

two-source systems

The size of a diffracting apertureOur discussion concerning diffraction so far has been for rectangular slits When light from a point source enters a smallcircularapertureitdoesnotproduceabrightdotasanimagebutacirculardiscknownasAiryrsquosdiscsurroundedbyfainterconcentriccircularringsasshowninFigure922

Laser

Figure 922 Diffraction at a circular aperture

Diffractionatanapertureisofgreatimportancebecausetheeye and many optical instruments have circular apertures

What is the half-angular width of the central maximum of the diffraction formed by a circular aperture This is not easy to calculate since it involves some advanced mathematics The problemwasfirstsolvedbytheEnglishAstronomerRoyalGeorgeAiryin1835whoshowedthatforcircularapertures

θ 122 λb--------------= where b is the diameter of the aperture

Example 2

Inthefollowingdiagramparallellightfromadistantpointsource(suchasastar)isbroughttofocusonthescreenSbyaconverginglens(the lens is shown as a vertical arrow)

Thefocal length(distance fromlens toscreen) is25cmand the diameter of the lens is 30 cm The wavelength of thelightfromthestaris560nmCalculatethediameterofthe diameter of the image on the screen

Solution

The lens actually acts as a circular aperture of diameter 30 cm The half angular width of central maximum of the diffraction pattern that it forms on the screen is

75

2

122 122 56 10 23 1030 10b

λθminus

minusminus

times times= = = times

times rad

The diameter of the central maxima is therefore

25 times 10-2 times 23 times 10-5

= 57 times 10-6 m

Although this is small it is still finite and is effectively the image of the star as the intensity of the secondary maxima are small compared to that of the central maximum

The resolution of simple monochromatic two-source systemsThe astronomers tell us that many of the stars that we observe with the naked eye are in fact binary stars That iswhatwe see as a single star actually consistsof twostarsinorbitaboutacommoncentreFurthermoretheastronomerstellusthatifweusealdquogoodrdquotelescopethenwewillactuallyseethetwostarsThatiswewillresolve the single point source into its two component parts Sowhatisitthatdetermineswhetherornotweseethetwo stars as a single point source ie what determines whether or not two sources can be resolved It canrsquot just bethatthetelescopemagnifiesthestarssinceiftheyareacting as point sources magnifying them is not going to make a great deal of difference

Ineachofoureyesthereisanaperturethepupilthroughwhich the light enters This light is then focused by the eye lens onto the retina But we have seen that when light passes through an aperture it is diffracted and so when we look at

NATURE OF SCIENCEImproved technology The Rayleigh criterion is the limit of resolution Continuing advancement in technology such as large diameter dishes or lenses or the use of smaller wavelength lasers pushes the limits of what we can resolve (18)

copy IBO 2014

PHYSICS 2015indb 299 150514 337 PM

Chapter 9

300

AH

L

apointsourceadiffractionpatternwillbeformedontheretina If we look at two point sources then two diffraction patterns will be formed on the retina and these patterns will overlap The width of our pupil and the wavelength of the light emitted by the sources will determine the amount by which they overlap But the degree of overlap will also depend on the angular separation of the two point sources WecanseethisfromFigures923

pupilretina

S1

S2

P2

P1

θ

Figure 923

LightfromthesourceS1 enters the eye and is diffracted by the pupil such that the central maximum of the diffraction patternisformedontheretinaatP1SimilarlylightfromS2producesamaximumatP2 If the two central maxima are well separated then there is a fair chance that we will see the two sources as separate sources If they overlap then we will not be able to distinguish one source from another From the diagram we see as the sources are movedclosertotheeyethentheangleθ increases and so does the separation of the central maxima

Figures924925926and927showsthedifferentdiffractionpatternsandtheintensitydistributionthatmightresultonthe retina as a result of light from two point sources

S1 S21 Well resolved

Figure 924 Very well resolved

S1 S22 Well resolved

Figure 925 Well resolved

S1 S23 Just resolved Rayleigh criterion Minimum of S2 coincides with maximum peak of S1

Figure 926 Just resolved

4 Not resolved

Figure 927 Not resolved

We have suggested that if the central maxima of the two diffraction patterns are reasonably separated then we should be able to resolve two point sources In the late 19th century Lord Rayleigh suggested by how much they should be separated in order for the two sources to be just resolved Ifthecentralmaximumofonediffractionpattern coincides with the firstminima of the otherdiffraction pattern then the two sources will just beresolved This is known as the Rayleigh Criterion

Figure926showsjustthissituationThetwosourcesarejust resolved according to the Rayleigh criterion since the peak of the central maximum of one diffraction pattern coincideswiththefirstminimumoftheotherdiffractionpattern This means that the angular separation of the peaks of the two central maxima formed by each source is just the half angular width of one central maximum ie

θ λb---

=

where b is the width of the slit through which the light fromthesourcespassesHoweverweseefromFigure1117that θ is the angle that the two sources subtend at the slit Henceweconcludethattwosourceswillberesolvedbyaslit if the angle that they subtend at the slit is greater than or equaltoλfraslb

PHYSICS 2015indb 300 150514 337 PM

Wave Phenomena

301

AH

L

Sofarwehavebeenassumingthattheeyeisarectangularslit whereas clearly it is a circular aperture and so we must use the formula

θ 122 λb--------------=

As mentioned above the angle θ is sometimes called the resolving power but should more accurately be called the minimum angle of resolution (θmin)

Clearly the smaller θ the greater will be the resolving power

The diffraction grating is a useful for differentiating closely spacedlinesinemissionspectraLikeaprismspectrometerit can disperse a spectrum into its components but it is better suited because it has higher resolution than the prism spectrometer

If λ1 and λ2 are two nearly equal wavelengths that canbarelybedistinguishedtheresolvanceorresolvingpowerofthegratingisdefinedas

R = λ _______ (λ2 ndash λ 1) = λ ___ Δλ

Where λ asymp λ1 asymp λ2

Therefore thediffractiongratingwithahighresolvancewill be better suited in determining small differences in wavelength

If the diffraction grating has Nlinesbeingilluminateditcan be shown that the resolving power of the mth order diffraction is given by

R = mN

Obviouslytheresolvancebecomesgreaterwiththeordernumber m and with a greater number of illuminated slits

Example

A benchmark for the resolving power of a grating is the sodium doublet in the sodium emission spectrum Two yellow lines in itsspectrumhavewavelengthsof58900nmand58959nm

(a) Determinetheresolvanceofagratingifthegivenwavelengths are to be distinguished from each other

(b) Howmanylinesinthegratingmustbeilluminated in order to resolve these lines in the second order spectrum

Solution

(a) R = λ ___ Δλ = 58900nm __________________ (58900ndash58959)nm

= 589nm _______ 059nm = 10 times 103

(b) N = R __ m = 10 times 10 3 ________ 2 =50times102 lines

It has been seen that diffraction effectively limits the resolving power of optical systems This includes such systems as the eye telescopes and microscopes Theresolving power of these systems is dealt with in the next section This section looks at links between technology and resolving power when looking at the very distant and when looking at the very small

Radio telescopes

The average diameter of the pupil of the human eye is about 25 mm This means that the eye will justresolve two point sources emitting light of wavelength 500 nm if their angular separation at the eye is 7

43

50 10122 24 1025 10

θminus

minusminus

times= times = times

times rad

If the eye were to be able to detect radio waves of wavelength 015m then tohave the same resolvingpower thepupilwould have to have a diameter of about 600 m Clearly this is nonsense but it does illustrate a problem facingastronomers who wish to view very distant objects such as quasarsandgalaxies thatemitradiowavesConventionalradio telescopes consist of a large dish typically 25 min diameter Even with such a large diameter the radiowavelength resolving power of the telescope is much less thantheopticalresolvingpowerofthehumaneyeLetuslook at an example

Example

The Galaxy Cygnus A can be resolved optically as an elliptically shaped galaxy However it is also a strongemitterof radiowavesofwavelength015mTheGalaxyisestimatedtobe50times1024mfromEarthUseofaradiotelescope shows that the radio emission is from two sources separated by a distance of 30 times 1021m Estimate thediameterofthedishrequiredtojustresolvethesources

Solution

The angle θ that the sources resolve at the telescope is given by

θ 21

424

30 10 60 1050 10

θ minustimes= = times

times rad

and 4

122 122 015 300060 10

d λθ minus

times= = =

timesm = 30 km

A radio telescope dish of this size would be impossible tomake letalonesupportThisshowsthatasingledishtype radio telescope cannot be used to resolve the sources and yet they were resolved To get round the problemastronomers use two radio telescopes separated by a large

PHYSICS 2015indb 301 150514 337 PM

Chapter 9

302

AH

L

distance The telescopes view the same objects at the same time and the signals that each receive from the objects are simultaneously superimposed The result of the superposition of the two signals is a two-slit interference pattern The pattern has much narrower fringe spacing than that of the diffraction pattern produced by either telescope on its own hence producing a much higherresolvingpowerWhentelescopesareusedlikethistheyare called a stellar interferometer

InSocorroinNewMexicothereisastellarinterferometerthat consists of 27 parabolic dishes each of diameter25 m arranged in a Y-shape that covers an area of 570km2Thisisaso-calledVeryLargeArray(VLA)Evenhigher resolution can be obtained by using an array of radio telescopes in observatories thousands of kilometres apart A system that uses this so-called technique oflsquovery-long-baseline interferometryrsquo (VLBI) is known asalsquovery-long-baselinearrayrsquo(VLBA)WithVLBAaradiowavelength resolving power can be achieved that is 100 timesgreaterthanthebestopticaltelescopesEvenhigherresolving power can be achieved by using a telescope that isinasatelliteorbitingEarthJapanrsquosInstituteofSpaceandAstronautical Science (ISAS) launched sucha system inFebruary 1997TheNationalAstronomicalObservatoryof Japan the National Science Foundationrsquos NationalRadio Astronomy Observatory (NRAO) the CanadianSpaceAgency theAustraliaTelescopeNationalFacilitythe European VLBI Network and the Joint Instituteback this project forVeryLongBaseline Interferometryin EuropeThis project is a very good example of howInternationalismcanoperateinPhysics

Electron microscope

Telescopes are used to look at very distant objects that are very large but because of their distance fromus appearverysmallMicroscopesontheotherhandareusedtolookat objects that are close to us but are physically very small Aswehave seen justmagnifyingobjects that ismakingthemappearlargerisnotsufficientonitsowntogaindetailabouttheobjectfordetailhighresolutionisneeded

Figure928isaschematicofhowanoptical microscope is usedtoviewanobjectandFigure929isaschematicofatransmission electron microscope(TEM)

Optical lens system

bright light source

glass slide containing specimen (object)

eye

Figure 928 The principle of a light microscope

magnetic lens system

electron source

thin wafer of material

ScreenCCD

Figure 929 The principle of an electron microscope

In the optical microscope the resolving power isdetermined by the particular lens system employed and the wavelength λ of the light used For example twopoints in the sample separated by a distance d will just be resolved if

2d mλ=

where m is a property of the lens system know as the numerical aperture In practice the largest value of m obtainable is about 16 Hence if the microscope slideis illuminatedwith light ofwavelength 480 nm a goodmicroscope will resolve two points separated by a distance dasymp15times10-7masymp015micromPointscloser together thanthiswillnotberesolvedHoweverthisisgoodenoughtodistinguishsomevirusessuchastheEbolavirus

Clearlythesmallerλ the higher the resolving power and this is where the electron microscope comes to the fore The electron microscope makes use of the wave nature of electrons(see1315)IntheTEMelectronspassthrougha wafer thin sample and are then focused by a magnetic field onto a fluorescent screen orCCD (charge coupleddevicesee142)Electronsusedinanelectronmicroscopehavewavelengthstypicallyofabout5times10-12mHoweverthe numerical aperture of electron microscopes is considerablysmaller thanthatofanopticalmicroscopetypicallyabout002NonethelessthismeansthataTEMcanresolvetwopointsthatareabout025nmapartThisresolving power is certainly high enough to make out the shape of large molecules

Another type of electron microscope uses a techniqueby which electrons are scattered from the surface of the sample The scattered electrons and then focused as in theTEM to form an image of the surfaceThese so-calledscanningelectronmicroscopes(SEM)havealowerresolving power than TEMrsquos but give very good threedimensional images

The eye

We saw in the last section that the resolving power of the human eye is about 2 times 10-4 rad Suppose that youare looking at car headlights on a dark night and the car

PHYSICS 2015indb 302 150514 337 PM

Wave Phenomena

303

AH

L

is a distance D away If the separation of the headlight is say 15 m then the headlights will subtend an angle θ 15

Dθ =

at your eye Assuming an average wavelength of

500 nm your eye will resolve the headlights into twoseparatesourcesifthisangleequals2times10-4 rad This gives D=75kmInotherwordsifthecarisapproachingyouona straight road then you will be able to distinguish the two headlightsasseparatesourceswhenthecaris75kmawayfrom you Actually because of the structure of the retina and optical defects the resolving power of the average eye is about 6 times 10-4 rad This means that the car is more likely tobe25kmawaybeforeyouresolvetheheadlights

Astronomical telescope

Letusreturntotheexampleofthebinarystarsdiscussedatthe beginning of this section on resolution The stars Kruger A and B form a binary system The average separation of the stars is 14 times 1012mandtheiraveragedistancefromEarthis12times1017mWhenviewedthroughatelescopeonEarththe system will therefore subtend an angle

12

17

14 1012 10

θtimes

=times

= 12times10-5 rad at the objective lens of the telescope Assuming that the average wavelength of the light emitted bythestarsis500nmthenifthetelescopeistoresolvethesystem into two separate images it must have a minimum diameter Dwhere12times10-5 =

7122 500 10D

minustimes times

This gives D=0050m which is about 5 cm So thisparticular system is easily resolved with a small astronomical telescope

Exercise 961 ItissuggestedthatusingtheISASVLBAitwould

bepossibletoldquoseerdquoagrainofriceatadistanceof5000kmEstimatetheresolvingpoweroftheVLBA

2 Thedistancefromtheeyelenstotheretinais20mm The light receptors in the central part of the retinaareabout5times10-6apartDeterminewhetherthe spacing of the receptors will allow for the eye to resolve the headlights in the above discussion whentheyare25kmfromtheeye

3 ThediameterofPlutois23times106 m and its averagedistancefromEarthis60times1012 m EstimatetheminimumdiameteroftheobjectiveofatelescopethatwillenablePlutotobeseenasadisc as opposed to a point source

95 Doppler effect

Essential ideaTheDopplereffectdescribesthephenomenonofwavelengthfrequencyshiftwhenrelativemotionoccurs

Understandingsbull TheDopplereffectforsoundwavesandlightwaves

The Doppler effect for sound waves and light waves

Consider two observers A and B at rest with respect to asoundsourcethatemitsasoundofconstantfrequencyf Clearly both observers will hear a sound of the same frequencyHoweversupposethatthesourcenowmovesat constant speed towards A A will now hear a sound of frequencyfA that is greater than f and B will hear a sound of frequency fB that is less than f This phenomenon is known as the Doppler Effect orDopplerPrinciple after C J Doppler(1803-1853)

The same effect arises for an observer who is either moving towards or away from a stationary source

Figure 930 shows the waves spreading out from a stationary source that emits a sound of constant frequency f The observersAandBhearasoundofthesamefrequency

A B

wavefront

S

Figure 930 Sound waves from a stationary source

Suppose now that the source moves towards A withconstant speed vFigure9310(a)showsasnapshotofthenew wave pattern

NATURE OF SCIENCETechnology Although originally based on physical observations of the pitch of fast moving sources of sound the Doppler effect has an important role in many different areas such as evidence for the expansion of the universe and generating images used in weather reports and in medicine (55)

copy IBO 2014

PHYSICS 2015indb 303 150514 337 PM

Chapter 9

304

AH

L

sourceA B

smallerwavelength

largerwavelength

Figure 931 (a) Sound waves from a moving source

The wavefronts are now crowded together in the direction of travel of the source and stretched out in the opposite direction This is why now the two observers will now hear notesofdifferentfrequenciesHowmuchthewavesbunchtogether and how much they stretch out will depend on the speed vEssentially A

A

cfλ

= and BB

cfλ

= where λA lt λB and v is the speed of sound

IfthesourceisstationaryandAismovingtowardsitthenthe waves from the source incident on A will be bunched up If A is moving away from the stationary source then the waves from the source incident on A will be stretched out

Christian Doppler(1803ndash1853)actuallyappliedtheprinciple(incorrectlyasithappens)totryandexplainthecolourofstars However the Doppler effect does apply to light aswellastosoundIfalightsourceemitsalightoffrequencyf then if it is moving away from an observer the observer willmeasurethelightemittedashavingalowerfrequencythan fSincethesensationofcolourvisionisrelatedtothefrequencyoflight(bluelightisofahigherfrequencythanred light) light emitted by objects moving way from anobserver is often referred to as being red-shifted whereas if the object is moving toward the observer it is referred to as blue-shiftedThisideaisusedinOptionE(Chapter16)

We do not need to consider here the situations where either the source or the observer is accelerating In a situation for example where an accelerating source is approaching astationaryobserverthentheobserverwillhearasoundof ever increasing frequency This sometimes leads toconfusion in describing what is heard when a source approaches passes and then recedes from a stationaryobserver Suppose for example that you are standingon a station platform and a train sounding its whistle is approaching at constant speed What will you hear as the train approaches and then passes through the station As the train approaches you will hear a sound of constant pitch but increasing loudness The pitch of the sound will be greater than if the train were stationary As the train passes through the station you will hear the pitch change at the momentthetrainpassesyoutoasoundagainofconstantpitch The pitch of this sound will be lower than the sound of the approaching train and its intensity will decrease as the train recedes from you What you do not hear is a sound of increasing pitch and then decreasing pitch

The Doppler equations for sound

Although you will not be expected in an IB examination to derivetheequationsassociatedwithaspectsoftheDopplereffectyouwillbeexpectedtoapplythemForcompletenesstherefore thederivationoftheequationsassociatedwiththeDopplereffectasoutlinedaboveisgivenhere

S primeS O

vs

Figure 931 (b)

InFigure931(b)theobserverOisatrestwithrespecttoasourceofsoundSismovingwithconstanspeedvs directly towards O The source is emitting a note of constant frequencyf and the speed of the emitted sound is v

S shows the position of the source ∆t later When the sourceisatresttheninatime∆t the observer will receive f∆t waves and these waves will occupy a distance v∆t ie

v f

λ =f∆tv∆t =

(Because of the motion of the source this number of waves will now occupy a distance (v∆t ndash vs∆t)The lsquonewrsquowavelength is therefore

v ndash vs

fλrsquo =

f∆tv∆t ndash vs∆t

=

If fisthefrequencyheardbyOthen

vfλ

= or

vf

λ = = sv vfminus

From which

s

vf fv v

=minus

Dividingthroughbyv gives

Equation1111

1 sf f v

v=

minus

If the source is moving away from the observer then we have

s

vf fv v

=minus

1

1 sf vv

=+

Equation112

PHYSICS 2015indb 304 150514 337 PM

Wave Phenomena

305

AH

L

We now consider the case where the source is stationary and the observerismoving towardsthesource with speed v0 In this situation the speed of the sound waves as measured by the observer will be v0 + v We therefore have that

v0 + v = ffraslλ = vfλ

times

From which

o1 vf fv

= +

Equation113

If the observerismovingawayfromthesource then

o1 vf fv

= minus

Fromequation(113)wehavethat

o o1 v vf f f f f fv v

∆ = minus = + minus =

Equation114

Thevelocitiesthatwerefertointheaboveequationsarethevelocities with respect to the medium in which the waves fromthesourcetravelHoweverwhenwearedealingwithalight source it is the relative velocity between the source and the observer that we must consider The reason for this is that lightisuniqueintherespectthatthespeedofthelightwavesdoes not depend on the speed of the source All observers irrespective of their speed or the speed of the source will measure the same velocity for the speed of light This is oneof thecornerstonesof theSpecialTheoryofRelativitywhichisdiscussedinmoredetailinOptionH(Chapter18)When applying the Doppler effect to light we aremainlyconcerned with the motion of the source We look here only at the situation where the speed of the source v is much smaller than the speed of light cinfreespace(v ltlt c)Underthesecircumstanceswhenthesourceismovingtowardstheobserverequation111becomes

vf f f fc

minus = ∆ = Equation115

andwhenthesourceismovingawayfromtheobserverequation112becomes vf f f f

cminus = ∆ = minus

Provided that v ltlt c these same equations apply for astationary source and moving observer

We look at the following example and exercise

Example

Asourceemitsasoundoffrequency440HzItmovesinastraight line towards a stationary observer with a speed of 30ms-1Theobserverhearsasoundoffrequency484HzCalculate the speed of sound in air

Solution

We use equation 111 and substitute f= 484 Hz f = 440 Hz

and vs = 30 m s-1

therefore 1

484 440 301v

=minus

such that 30 4401

484vminus = to

give v = 330 m s-1

Example

A particular radio signal from a galaxy is measured as havingafrequencyof139times109HzThesamesignalfromasourceinalaboratoryhasafrequencyof142times109Hz

Suggestwhy thegalaxy ismovingaway fromEarthandcalculateitsrecessionspeed(iethespeedwithwhichitismovingawayfromEarth)

Solution

The fact that the frequency from the moving source is less than that when it is stationary indicates that it is moving away from the stationary observer ie Earth

Using vf fc

∆ = we have

8 96

9

3 10 (142 139) 10 634 10142 10

c fvf∆ times times minus times

= = = timestimes

m s-1

It is usual when dealing with the Doppler effect of light to express speeds as a fraction of c So in this instance we have v = 0021 c

Using the Doppler effectWe have seen in the above example and exercise how the Dopplereffectmaybeusedtomeasuretherecessionspeedof distant galaxies The effect is also used to measure speed inothersituationsHerewewilllookatthegeneralprincipleinvolved in using the Doppler effect to measure speedFigure932showsasource(thetransmitter)thatemitseithersound or em waves of constant frequency f The waves

PHYSICS 2015indb 305 150514 337 PM

Chapter 9

306

AH

L

fromthesourceare incidentonareflectorthat ismovingtowards the transmitter with speed vThereflectedwavesare detected by the receiver placed alongside the transmitter

reector

v

transmitter

receiver

f f

f f

Figure 932 Using the Doppler effect to measure speed

We shall consider the situation where v ltlt c where c is the speed of the waves from the transmitter

For the reflector receiving waves from the transmitterit is effectively an observer moving towards a stationary sourceFromequation(114) itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation116

For the receiver receivingwaves from the reflector it iseffectively a stationary observer receiving waves from a movingsourceFromequation(115)itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation117

Ifweaddequations(116)and(117)wegetthatthetotalDopplershiftatthereceiver∆f is

v vf f f f fc c

minus = ∆ = +

But 1 vf fc

= +

hence

1 v v vf f fc c c

∆ = + +

But since v ltlt cwe can ignore the term2

2

vc when we

expandthebracketintheaboveequation

Therefore we have2vf fc

∆ = Equation118

If v asymp c then we must use the full Doppler equationsHowever for em radiationwewill alwaysonly considersituations in which v ltlt c

Example

Thespeedofsoundinbloodis1500times103 m s-1 Ultrasound offrequency100MHzisreflectedfrombloodflowinginan arteryThe frequencyof the reflectedwaves receivedbackatthetransmitteris105MHzEstimatethespeedofthebloodflowintheartery

Solution

Using equation (118) we have

6 63

2005 10 1015 10

vtimes = times

timesto give v asymp 36 m s-1 (We have assumed that the ultrasound is incident at right angles to the blood flow)

Exercise 971 Judy is standing on the platform of a station A

high speed train is approaching the station in a straight line at constant speed and is sounding its whistleAsthetrainpassesbyJudythefrequencyof the sound emitted by the whistle as heard by Judychangesfrom640Hzto430HzDetermine

(a) thespeedofthetrain

(b) thefrequencyofthesoundemittedbythewhistle as heard by a person on the train (Speedofsound=330ms-1)

2 AgalaxyismovingawayfromEarthwithaspeedof00500cThewavelengthofaparticularspectral line in light emitted by atomic hydrogen inalaboratoryis656times10-7 m Calculate the valueofthewavelengthofthislinemeasuredinthelaboratoryinlightemittedfromasourceofatomic hydrogen in the galaxy

PHYSICS 2015indb 306 150514 337 PM

10 Fields

Contents

101 ndash Describing fields

102 ndash Fields at work

Essential Ideas

Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Similar approaches can be taken in analysing electrical and gravitational potential problems copy IBO 2014

PHYSICS 2015indb 307 150514 337 PM

Chapter 10

308

AH

L

101 Describing fields

Essential idea Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Understandingsbull Gravitationalfieldsbull Electrostaticfieldsbull Electricpotentialandgravitationalpotentialbull Fieldlinesbull Equipotentialsurfaces

Gravitational potentialWehaveseenthatifweliftanobjectofmassm to a height h abovethesurfaceoftheEarththenitsgainingravitationalpotential energy is mghHoweverthisisbynomeansthefullstoryForastartwenowknowthatgvarieswithh and also the expression really gives a difference in potentialenergybetweenthevaluethattheobjecthasattheEarthrsquossurfaceandthevaluethatithasatheighthSowhatwereallyneedisazeropointCanwefindapointwherethepotentialenergyiszeroandusethispointfromwhichtomeasure changes in potential energy

The point that is chosen is in fact infinity At infinity the gravitationalfieldstrengthofanyobjectwillbezeroSoletusseeifwecandeduceanexpressionforthegaininpotentialenergyofanobjectwhenitisldquoliftedrdquofromthesurfaceoftheEarthtoinfinityThisineffectmeansfindingtheworknecessarytoperformthistask

δr

r r δr+

r infin=

mg

AB

Me

Figure 1001 Gravitational forces

Inthediagramweconsidertheworknecessarytomovethe particle of mass m a distance δr in thegravitationalfield of the Earth

The force on the particle at A is FG Mem

r2----------------=

IftheparticleismovedtoB then since δr isverysmallwe can assume that the field remains constant over thedistance ABTheworkδW done againstthegravitationalfieldoftheEarthinmovingthedistanceABis

δWGMem

r2----------------δrndash=

(rememberthatworkdoneagainstaforceisnegative)

TofindthetotalworkdoneW in going from the surface oftheEarthtoinfinitywehavetoaddalltheselittlebitsofworkThisisdonemathematicallybyusingintegralcalculus

WG Me m

r2----------------ndash

rdR

infin

int G Me m 1

r2----- rd

R

infin

intndash G Me m 1r---ndash

R

infinndash= = =

GMe m 0 1R---ndash

ndashndash=

G MemR----------------ndash=

HencewehavewhereRistheradiusoftheEarththattheworkdonebythegravitationalfieldinmovinganobjectofmass m from R(surfaceoftheEarth)toinfinityisgivenby

WGMe m

R---------------- ndash=

We can generalise the result by calculating the worknecessary per unit mass to take a small mass fromthe surface of the Earth to infinity This we call thegravitational potential Vie

Wewouldgetexactlythesameresultifwecalculatedtheworkdonebythefieldtobringthepointmassfrominfinityto the surface of Earth In this respect the formal definition ofgravitationalpotentialatapointinagravitationalfieldis therefore defined as the work done per unit mass in bringing a point mass from infinity to that point

Clearly then the gravitational potential at any point inthe Earthrsquos field distance r from the centre of the Earth (providingr gt R)is

VGMe

r------------ ndash=

Thepotentialisthereforeameasureoftheamountofworkthathastobedonetomoveparticlesbetweenpointsinagravitationalfieldanditsunit is theJkgndash1 We also note thatthepotentialisnegativesothatthepotentialenergy

NATURE OF SCIENCEParadigm shift The move from direct observable actions being responsible for influence on an object to acceptance of a fieldrsquos ldquoaction at a distancerdquo required a paradigm shift in the world of science (23)

copy IBO 2014

PHYSICS 2015indb 308 150514 337 PM

Fields

309

AH

L

aswemoveawayfromtheEarthrsquossurfaceincreasesuntilitreachesthevalueofzeroatinfinity

IfthegravitationalfieldisduetoapointmassofmassmthenwehavethesameexpressionasaboveexceptthatMe is replaced by m andmustalsoexclude thevalueof thepotential at the point mass itself ie at r = 0

WecanexpressthegravitationalpotentialduetotheEarth(orduetoanysphericalmass)intermsofthegravitationalfield strength at its surface

AtthesurfaceoftheEarthwehave

g 0ndash Re GMRe

---------ndash= e

Sothat

g 0Re2 GM= e

Hence at a distance r from the centre of the Earth the gravitationalpotentialVcanbewrittenas

VGMe

r------------ndashg 0 Re

2

r------------ndash= =

The potential at the surface of the Earth

(r = Re)istherefore-g0Re

It is interesting to see how the expression for thegravitational potential ties in with the expression mgh ThepotentialatthesurfaceoftheEarthis-g0Re (see the exampleabove)andataheighthwillbe-g0g 0 Re h+( )ndash ifweassume that g0doesnotchangeover thedistanceh The differenceinpotentialbetweenthesurfaceandtheheighth is therefore g0hSotheworkneededtoraiseanobjectof mass m to a height h is mgh ie m timesdifference ingravitationalpotential

This we have referred to as the gain in gravitationalpotentialenergy(see235)

However this expression can be extended to any twopoints inanygravitationalfieldsuch that ifanobjectofmass mmovesbetween twopointswhosepotentialsareV1 and V2 respectively then the change in gravitationalpotentialenergyoftheobjectism(V1 ndash V2)

Gravitational potential gradient

Let us consider now a region in space where thegravitationalfieldisconstantInFigure912thetwopointsAandBareseparatedbythedistance∆x

A B

direction of uniform gravitational eld of strength I

∆x

Figure 1002 The gravitational potential gradient

The gravitational field is of strength I and is in the direction shownThe gravitational potential at A is V andatBisV +∆V

Theworkdone is takingapointmassmfromAtoB isF∆x = mI∆x

Howeverbydefinitionthisworkisalsoequalto-m∆V

Therefore mI∆x=-m∆V

or VIx

∆= minus

∆Effectivelythissaysthatthemagnitudeofthegravitationalfield strength is equal to the negative gradient of thepotential If I is constant then V is a linear function of x and Iisequaltothenegativegradientofthestraightlinegraph formed by plotting V against x If I is not constant (asusuallythecase)thenthemagnitudeofI at any point in the field can be found by find the gradient of the V-x graph at that point An example of such a calculation can befoundinSection929

For those of you who do HL maths the relationship between field and potential is seen to follow from the expression for the potential of a point mass viz

mV Gr

= minus

2

ddV mG Ir r

minus = + =

Potential due to one or more point masses

Gravitationalpotentialisascalarquantitysocalculatingthepotential due to more than one point mass is a matter of simpleadditionSoforexamplethepotentialV due to the Moon and Earth and a distance x fromthecentreofEarthonastraightlinebetweenthemisgivenbytheexpression

E MM MV Gx r x

= minus + minus

whereME=massofEarthMM= mass of Moon and r = distancebetweencentreofEarthandMoon

PHYSICS 2015indb 309 150514 337 PM

Chapter 10

310

AH

L

Equipotentials and field linesIfthegravitationalpotentialhasthesamevalueatallpointson a surface the surface is said to be an equipotential surfaceSoforexampleifweimagineasphericalshellaboutEarthwhosecentrecoincideswiththecentreofEarththisshellwillbeanequipotentialsurfaceClearlyifwerepresentthegravitationalfieldstrengthbyfieldlinessincethelinesldquoradiaterdquooutfromthecentreofEarththentheselineswillbeatrightanglestothesurfaceIfthefieldlineswerenotnormaltotheequipotentialsurfacethentherewouldbeacomponentof the field parallel to the surfaceThis wouldmean thatpointsonthesurfacewouldbeatdifferentpotentialsandsoitwouldnolongerbeanequipotentialsurfaceThisofcourseholdstrueforanyequipotentialsurface

Figure 1003 shows the field lines and equipotentials fortwopointmassesm

m m

Figure 1003 Equipotentials for two point masses

It is worth noting that we would get exactly the samepattern ifwewere to replace thepointmasseswith twoequalpointcharges(See935)

Escape speedThe potential at the surface of Earth is which meansthattheenergyrequiredtotakeaparticleofmass mfromthesurfacetoinfinityisequalto

ButwhatdoesitactuallymeantotakesomethingtoinfinityWhentheparticleisonthesurfaceoftheEarthwecanthinkofitassittingatthebottomofaldquopotentialwellrdquoasinfigure1004

particle

surface of Earth

innity

GMR---------

FIgure 1004 A potential well

Theldquodepthrdquoofthewellis and if the particle gains an amount of kinetic energy equal to wherem is its massthenitwillhavejustenoughenergytoldquoliftrdquoitoutofthewell

In reality it doesnrsquot actually go to infinity it justmeansthat the particle is effectively free of the gravitationalattractionof theEarthWe say that ithas ldquoescapedrdquo theEarthrsquosgravitationalpullWemeetthisideainconnectionwithmolecular forcesTwomolecules in a solidwill sitat their equilibrium position the separation where therepulsiveforceisequaltotheattractiveforceIfwesupplyjust enough energy to increase the separation of themolecules such that they are an infinite distance apart then intermolecularforcesnolongeraffectthemoleculesandthesolidwillhavebecomealiquidThereisnoincreaseinthekineticenergyofthemoleculesandsothesolidmeltsat constant temperature

Wecancalculatetheescapespeedofanobjectveryeasilyby equating the kinetic energy to the potential energysuch that

12---mvesc ape

2 GMe mRe

----------------=

vesc aperArr2GMe

Re--------------- 2g0 Re = =

Substituting for g0 and Regivesavalueforvescape of about 11kmsndash1 from the surface of the Earth

Youwillnotethattheescapespeeddoesnotdependonthemassoftheobjectsincebothkineticenergyandpotentialenergy are proportional to the mass

IntheoryifyouwanttogetarockettothemoonitcanbedonewithoutreachingtheescapespeedHoweverthiswouldnecessitate anenormousamountof fuel and it islikely that therocketplus fuelwouldbesoheavythat itwouldnevergetoffthegroundItismuchmorepracticalto accelerate the rocket to the escape speed fromEarthorbitandthenintheoryjustlaunchittotheMoon

Example

Use the followingdata todetermine thepotential at thesurface of Mars and the magnitude of the acceleration of free fall

massofMars =64times1023kg

radiusofMars =34times106 m

Determinealsothegravitationalfieldstrengthatadistanceof 68 times 106mabovethesurfaceofMars

PHYSICS 2015indb 310 150514 337 PM

Fields

311

AH

L

Solution

= minus = minus times times = times23

11 76

64 1067 10 ndash13 10

34 10MV GR

minus timestimes

Nkg-1

But V = -g0R

Therefore 7

0 6

13 10 3834 10

VgR

times= minus = =

timesm s-2

To determine the field strength gh at 68 times 106 m above the surface we use the fact that 0 2

Mg GR

= such that GM = g0R2

Therefore 2 2

0h 2 2 2

h h

38 (34) 042(102)

g RGMgR R

times= = = = m s-2

(the distance from the centre is 34 times 106 + 68 times 106 = 102 times 106 m)

Exercise 1011 Thegraphbelowshowshowthegravitational

potentialoutsideoftheEarthvarieswithdistancefrom the centre

0 10 20 30 40 50 60 70

ndash1

ndash2

ndash3

ndash4

ndash5

ndash6

ndash7

r m times 106

V

Jkgndash1

times 10

7

(a) Usethegraphtodeterminethegainingravitationalpotentialenergyofasatelliteofmass200kgasitmovesfromthesurfaceof the Earth to a height of 30 times 107mabovetheEarthrsquossurface

(b) Calculatetheenergyrequiredtotakeit to infinity

(c) DeterminetheslopeofthegraphatthesurfaceoftheEarthmCommenton youranswer

Electric potential energy

The concept of electric potential energy was developedwith gravitational potential energy in mind Just as anobjectnearthesurfaceoftheEarthhaspotentialenergybecause of its gravitational interaction with the Earthsotoothereiselectricalpotentialenergyassociatedwithinteracting charges

Letusfirstlookatacaseoftwopositivepointchargeseachof 1μC that are initially bound together by a thread in a vacuuminspacewithadistancebetweenthemof10cmasshowninFigure1005Whenthethreadiscutthepointchargesinitiallyatrestwouldmoveinoppositedirectionsmovingwithvelocitiesv1 and v2 along the direction of the electrostatic force of repulsion

BEFORE AFTER

v v21

Figure 1005 Interaction of two positive particles

The electric potential energybetweentwopointchargescan be found by simply adding up the energy associated witheachpairofpointchargesForapairofinteractingchargestheelectricpotentialenergyisgivenby

∆U = ∆Ep + ∆Ek = ∆W = ∆Fr = kqQ

____ r2 times r = kqQ

____ r

Becausenoexternalforceisactingonthesystemtheenergyand momentum must be conserved Initially Ek = 0 and Ep =kqQr=9times109 times 1 times 10-1201m=009JWhentheyareagreatdistance fromeachotherEpwillbenegligibleThefinalenergywillbeequaltofrac12mv1

2+frac12mv22=009J

Momentum isalsoconservedand thevelocitieswouldbethe same magnitude but in opposite directions

Electric potential energyismoreoftendefinedintermsofapointchargemovinginanelectricfieldas

lsquothe electricpotential energybetweenany twopoints inan electric field is defined as negativeoftheworkdone by anelectricfieldinmovingapoint electric chargebetweentwolocationsintheelectricfieldrsquo

ΔU = ΔEp = -ΔW = -Fd = qEx

ΔU = ΔEp +ΔEk = ΔW = ΔFr = kqQ r2 x r = kqQ

wherexisthedistancemovedalong(oroppositeto)thedirection of the electric field

Electric potential energyismeasuredinjoule(J)Justasworkisascalarquantitysotooelectricalpotentialenergyis a scalar quantity The negative of the work done by an electric field in moving a unit electric charge between two points is independent of the path taken Inphysicswesaytheelectricfieldisalsquoconservativersquo field

PHYSICS 2015indb 311 150514 337 PM

Chapter 10

312

AH

L

Supposeanexternalforcesuchasyourhandmovesasmallpositive point test charge in the direction of a uniformelectric fieldAs it ismoving itmust be gaining kineticenergyIfthisoccursthentheelectricpotentialenergyofthe unit charge is changing

InFigure1006apointcharge+qismovedbetweenpointsAandBthroughadistancex in a uniform electric field

B

A

x

+ q

Figure 1006 Movement of a positive point charge in a uniform field

InordertomoveapositivepointchargefrompointAtopointBanexternal forcemustbeapplied to thechargeequaltoqE (F = qE)

Since the force is applied through a distance xthennegativeworkhas to bedone tomove the charge because energyis gainedmeaning there is an increaseelectric potential energybetweenthetwopointsRememberthattheworkdoneisequivalenttotheenergygainedorlostinmovingthe charge through the electric field The concept of electric potential energy is only meaningful as the electric field whichgeneratestheforceinquestionisconservative

W F xtimes E q x times= =

θ x cos θ x

Figure 1007 Charge moved at an angle to the field

If a chargemoves at an angle θ to an electric field thecomponent of the displacement parallel to the electric fieldisusedasshowninFigure918

W F x Eq x θcostimes= =

Theelectricpotentialenergyisstoredintheelectricfieldand theelectricfieldwill return theenergy to thepointcharge when required so as not to violate the Law ofconservationofenergy

Electric potential

The electric potential at a point in an electric field is defined as being the work done per unit charge in bringing a small positive point charge from infinity to that point

ΔV = Vinfin ndash Vf = ndash W ___ q

Ifwedesignatethepotentialenergytobezeroatinfinitythen it follows thatelectricpotentialmustalsobezeroat infinity and the electric potential at any point in an electricfieldwillbe

ΔV = ndash W ___ q

NowsupposeweapplyanexternalforcetoasmallpositivetestchargeasitismovedtowardsanisolatedpositivechargeTheexternalforcemustdoworkonthepositivetestchargetomoveittowardstheisolatedpositivechargeandtheworkmustbepositivewhiletheworkdonebytheelectricfieldmustthereforebenegativeSotheelectricpotentialatthatpointmustbepositiveaccordingtotheaboveequationIfanegativeisolatedchargeisusedtheelectricpotentialatapointonthepositivetestchargewouldbenegativePositivepoint chargesof their ownaccordmove fromaplaceofhighelectricpotential toaplaceof lowelectricpotentialNegative point charges move the other way from lowpotentialtohighpotentialInmovingfrompointAtopointBinthediagramthepositivecharge+qismovingfromalowelectricpotentialtoahighelectricpotential

In thedefinitiongiven the termldquoworkperunitchargerdquohas significance If the test charge is +16 times 10-19Cwherethe charge has a potential energy of 32 times 10-17Jthenthepotentialwouldbe32times10-17J+16times10-19C=200JC-1 Now if the charge was doubled the potential wouldbecome 64times10-17 J However the potential per unitchargewouldbethesame

Electric potential is a scalar quantityandithasunitsJC-1 orvoltswhere1voltequalsone joulepercoloumbThevoltallowsustoadoptaunitfortheelectricfieldintermsofthevolt

PreviouslytheunitfortheelectricfieldwasNCndash1

W = qV and F = qEsoW ___ V = F __ E

E = FV ___ W = FV ___ Fm V mndash1

That is the units of the electric field E can also beexpressed as V mndash1

The work done per unit charge in moving a point charge between two points in an electric field is again independent of the path taken

PHYSICS 2015indb 312 150514 337 PM

Fields

313

AH

L

Electric potential due to a point charge

Letus takeapoint rmetres froma chargedobjectThepotentialatthispointcanbecalculatedusingthefollowing

W = ndashFr = ndashqV and F = ndash q1 q2 _____ 4πε0r

2

Therefore

W = ndash q1 q2 _____ 4πε0r

2 times r = ndash q1 q2 _____ 4πε0r

= ndashq1 times q2 _____ 4πε0r

= ndashq1V

That is V q

4πε0r--------------- =

Orsimply

V kqr------

=

Example

Determinehowmuchworkisdonebytheelectricfieldofpointcharge150μCwhenachargeof200μCismovedfrom infinity to apoint 0400m from thepoint charge(Assumenoaccelerationofthecharges)

Solution

The work done by the electric field is W = -qV = -14πε0 times q times (Q rinfin - Q r0400)

W = (- 200 times 10-6 C times 900 times 109 NmC-2 times 150 times 10-6 C) divide 0400 m = - 0675 J

An external force would have to do +0675 J of work

Electric field strength and electric potential gradient

Letus lookbackatFigure1006Supposeagain that thecharge+qismovedasmalldistancebyaforceFfromAtoBsothattheforcecanbeconsideredconstantTheworkdoneisgivenby

∆W F x∆times=

TheforceFandtheelectricfieldEareoppositelydirectedandweknowthat

F = -qE and ΔW = q ΔV

Thereforetheworkdonecanbegivenas

q ΔV = -q E Δx

Therefore E V∆x∆-------

ndash=

The rate of change of potential ΔV at a point with respect to distance Δx in the direction in which the change is maximum is called the potential gradient We say that the electric field = - the potential gradient and the units are Vm-

1 From the equation we can see that in a graph of electric potential versus distance the gradient of the straight line equals the electric field strength

In reality if a charged particle enters a uniform electric field it will be accelerated uniformly by the field and its kinetic energy will increase This is why we had to assume no acceleration in the last worked example

Ek∆ 12---

mv2 q E xsdot sdot q Vx---

xsdot sdot q Vsdot= = = =

Example

Determine how far apart two parallel plates must besituated so that a potential difference of 150 times 102 V produces an electric field strength of 100 times 103 NC-1

Solution

Using E V∆x------- xhArrndash V∆

E-------15 102 Vtimes

100 103 N C 1ndashtimes------------------------------------------= = =

= 150 times 10-1

The plates are 150 times 10-1 m apart

The electric field and the electric potential at a point due to an evenly distributed charge +q on a sphere can be represented graphically as in Figure 1008

r

r

V

E

r0

r0

r0

Charge of +Q evenly distributed over surface

E 14πε0------------ Q

r2----- r r0gt=

V 14πε0------------Q

r----r r0gt=

Figure 1008 Electric field and potential due to a charged sphere

PHYSICS 2015indb 313 150514 337 PM

Chapter 10

314

AH

L

When the sphere becomes charged we know that thechargedistributes itselfevenlyoverthesurfaceThereforeeverypartofthematerialoftheconductorisatthesamepotential As the electric potential at a point is defined as beingnumericallyequaltotheworkdoneinbringingaunitpositivechargefrominfinitytothatpointithasaconstantvalueineverypartofthematerialoftheconductor

Since the potential is the same at all points on the conducting surfacethen∆V∆xiszeroButE=ndash∆V∆xThereforethe electric field inside the conductor is zero There is no electric field inside the conductor

SomefurtherobservationsofthegraphsinFigure1009are

bull OutsidethespherethegraphsobeytherelationshipsgivenasE α 1 r2 and V α 1 r

bull Atthesurfacer = r0Thereforetheelectricfieldandpotentialhavetheminimumvalueforr at this point and this infers a maximum field and potential

bull Insidethespheretheelectricfieldiszerobull Insidethespherenoworkisdonetomoveacharge

fromapointinsidetothesurfaceThereforethereisno potential difference and the potential is the same asitiswhenr = r0

Similargraphscanbedrawnfortheelectricfieldintensityandthe electric potential as a function of distance from conducting parallelplatesandsurfacesandthesearegiveninFigure1009

Potential plot E eld plotE eld

++++

ndashndashndashndash

xx x

+

x x

ndashx x

+ ndashx

x

+ +

x

x

Figure 1009 Electric field and electric potential at a distance from a charged surface

Potential due to one or more point charges

The potential due to one point charge can be determined byusingtheequationformula

V = kq r

Example 1

Determine the electric potential at a point 20 times 10-1 m fromthecentreofan isolatedconducting spherewithapointchargeof40pCinair

Solution

Using the formula

V = kq r we have

V90 10 9times( ) 40 10 12ndashtimes( )times

20 10 1ndashtimes( )------------------------------------------------------------------ 018 V= =

the potential at the point is 180 times 10-1 V

The potential due to a number of point charges can be determined by adding up the potentials due to individual point charges because the electric potential at any point outside a conducting sphere will be the same as if all the charge was concentrated at its centre

Example 2

Threepointchargesofareplacedattheverticesofaright-angledtriangleasshowninthediagrambelowDeterminetheabsolutepotentialat the+20μCchargeduetothetwoothercharges

- 6μC

+3μC +2μC

4 m

3 m

Solution

The electric potential of the +2 μC charge due to the ndash 6 μC charge is

V = (9 times 109 Nm2C-2 times -6 times 10-6 C) divide (radic 32 + 42) m = - 108 times 104 V

PHYSICS 2015indb 314 150514 338 PM

Fields

315

AH

L

Insummarywecanconcludethat

bull No work is done to move a charge along an equipotential

bull Equipotentials are always perpendicular to the electric lines of force

Figure1011and1012showsomeequipotential lines fortwo oppositely charged and identically positive spheresseparated by a distance

+ve ndashve

equipotential lines

Figure 1011 Equipotential lines between two opposite charges

equipotential lines

Figure 1012 Equipotential lines between two charges which are the same

+

ndash ndash ndash ndash

+ + +

40 V

30 V

20 V

10 V

50 V

Figure 1013 Equipotential lines between charged parallel plates

Figure1013showstheequipotentiallinesbetweenchargedparallel plates Throughout this chapter the similarities and differences between gravitational fields and electric fieldshavebeendiscussedTherelationshipsthatexistsbetweengravitationalandelectricquantitiesandtheeffectsofpointmassesandchargesissummarisedinTable1014

The electric potential of the +2 μC charge due to the +3 μC charge is

V = (9 times 109 Nm2C-2 times 3 times 10-6 C) divide 3m = 9 times 103 V

The net absolute potential is the sum of the 2 potentials- 108 times 104 V + 9 times 103 V =- 18 times 103 V

The absolute potential at the point is - 18 times 103 V

Equipotential surfacesRegions in space where the electric potential of a chargedistributionhasaconstantvaluearecalledequipotentials The placeswherethepotentialisconstantinthreedimensionsarecalled equipotential surfacesandwheretheyareconstantintwodimensionstheyarecalledequipotential lines

TheyareinsomewaysanalogoustothecontourlinesontopographicmapsInthiscasethegravitationalpotentialenergy is constant as amassmoves around the contourlines because the mass remains at the same elevationabovetheEarthrsquossurfaceThegravitationalfieldstrengthacts in a direction perpendicular to a contour line

Similarlybecausetheelectricpotentialonanequipotentiallinehas thesamevalueanelectric forcecandonoworkwhena testchargemovesonanequipotentialThereforethe electric field cannot have a component along anequipotentialandthusitmustbeeverywhereperpendiculartotheequipotentialsurfaceorequipotentiallineThisfactmakesiteasytoplotequipotentialsifthelinesofforceorlinesofelectricfluxofanelectricfieldareknown

Forexamplethereareaseriesofequipotentiallinesbetweentwoparallelplateconductorsthatareperpendiculartotheelectric fieldTherewill be a series of concentric circles(eachcirclefurtherapartthanthepreviousone)thatmapouttheequipotentialsaroundanisolatedpositivesphereThe lines of force and some equipotential lines for anisolatedpositivesphereareshowninFigure1010

Lines ofequipotential

Figure 1010 Equipotentials around an isolated positive sphere

PHYSICS 2015indb 315 150514 338 PM

Chapter 10

316

AH

L

Gravitational quantity Electrical quantityQ

uant

ities

g V∆x∆-------ndash= E V∆

x∆-------ndash=

Poin

t mas

ses a

nd ch

arge

s

V Gmr----

ndash= V 14πε0------------q

r---=

g G m

r2-----ndash= E 14πε0------------ q

r2-----=

F Gm1 m2

r2--------------= F 1

4πε0------------

q1q2

r2-----------=

Figure 1014 Formulas

Example

Deduce the electric potential on the surface of a gold nucleus that has a radius of 62 fm

Solution

Using the formula

V = kq r and knowing the atomic number of gold is 79 We will assume the nucleus is spherical and it behaves as if it were a point charge at its centre (relative to outside points)

V = 90 times 109 Nm2C-2 times 79 times 16 times 10-19 C divide 62 times 10-15 m = 18 times 107 V

The potential at the point is 18 MV

Example

Deduce the ionisation energy in electron-volts of theelectron in the hydrogen atom if the electron is in its ground state and it is in a circular orbit at a distance of 53 times 10-11 m from the proton

Solution

This problem is an energy coulombic circular motion question based on Bohrrsquos model of the atom (not the accepted quantum mechanics model) The ionisation energy is the energy required to remove the electron from the ground state to infinity The electron travels in a circular orbit and therefore has a centripetal acceleration The ionisation energy will counteract the coulombic force and the movement of the electron will be in the opposite direction to the centripetal force

Total energy = Ek electron + Ep due to the proton-electron interaction

ΣF = kqQ r2 = mv2 r and as such mv2 = = kqQ r

Therefore Ek electron = frac12 kqQ r

Ep due to the proton-electron interaction = - kqQ r

Total energy = frac12 kqQ r + - kqQ r = -frac12 kqQ r

= - 90 times 109 Nm2C-2 times (16 times 10-19 C)2 divide 53 times 10-11 m = -217 times 10-18 J

= -217 times 10-18 J divide 16 times 10-19 = -136 eV

The ionisation energy is 136 eV

Exercise 1021 ApointchargePisplacedmidwaybetweentwo

identicalnegativechargesWhichoneofthefollowingiscorrectwithregardstoelectricfieldandelectricpotentialatpointP

Electric field Electric potentialA non-zero zeroB zero non-zeroC non-zero non-zeroD zero zero

2 Twopositivechargedspheresaretiedtogetherinavacuumsomewhereinspacewheretherearenoexternal forces A has a mass of 25 g and a charge of20μCandBhasamassof15gandachargeof30μCThedistancebetweenthemis40cmTheyarethenreleasedasshowninthediagram

BEFORE AFTER

v1 v2

A B

(a) Determinetheirinitialelectricpotentialenergy in the before situation

(b) DeterminethespeedofsphereBafterrelease

3 Thediagrambelowrepresentstwoequipotentiallines in separated by a distance of 5 cm in a uniform electric field

+ + + + + + + +

ndash ndash ndash ndash ndash ndash ndash ndash

40 V

20 V 5 cm

PHYSICS 2015indb 316 150514 338 PM

Fields

317

AH

L

Determine the strength of the electric field

4 Thisquestionisabouttheelectricfieldduetoacharged sphere and the motion of electrons in that fieldThediagrambelowshowsanisolatedmetalsphereinavacuumthatcarriesanegativeelectriccharge of 60 μC

ndash

(a) Onthediagramdrawtheconventionalwayto represent the electric field pattern due to the charged sphere and lines to represent threeequipotentialsurfacesintheregionoutside the sphere

(b) Explainhowthelinesrepresentingtheequipotentialsurfacesthatyouhavesketchedindicatethatthestrengthoftheelectricfieldisdecreasingwithdistancefrom the centre of the sphere

(c) Theelectricfieldstrengthatthesurfaceofthe sphere and at points outside the sphere can be determined by assuming that the sphere acts as a point charge of magnitude 60 μC at its centre The radius of the sphere is 25 times 10ndash2 m Deduce that the magnitude of the field strength at the surface of the sphere is 86 times 107 Vmndash1

An electron is initially at rest on the surface of the sphere

(d) (i) Describethepathfollowedbytheelectronasitleavesthesurfaceofthesphere

(ii) Calculatetheinitialaccelerationofthe electron

5 Determinetheamountofworkthatisdoneinmovingachargeof100nCthroughapotentialdifferenceof150times102 V

6 Three identical 20 μC conducting spheres are placedatthecornersofanequilateraltriangleofsides 25 cm The triangle has one apex C pointing upthepageand2baseanglesAandBDeterminetheabsolutepotentialatB

7 Determinehowfaraparttwoparallelplatesmustbesituatedsothatapotentialdifferenceof250 times 102 V produces an electric field strength of 200 times 103 NC-1

8 Thegapbetweentwoparallelplatesis10times10-3mandthereisapotentialdifferenceof10times104 V betweentheplatesCalculate

i theworkdonebyanelectroninmovingfrom one plate to the other

ii thespeedwithwhichtheelectronreachesthe second plate if released from rest

iii theelectricfieldintensitybetweentheplates

9 Anelectronguninapicturetubeisacceleratedby a potential 25 times 103VDeterminethekineticenergygainedbytheelectroninelectron-volts

10 Determine the electric potential 20 times 10-2 m from achargeof-10times10-5 C

11 Determinetheelectricpotentialatapointmid-waybetweenachargeofndash20pCandanotherof+5pConthelinejoiningtheircentresifthecharges are 10 cm apart

12 During a thunderstorm the electric potential differencebetweenacloudandthegroundis10 times 109 V Determine the magnitude of the change in electric potential energy of an electron thatmovesbetweenthesepointsinelectron-volts

13 A charge of 15 μC is placed in a uniform electric fieldoftwooppositelychargedparallelplateswithamagnitudeof14times103 NC-1

(a) Determinetheworkthatmustbedoneagainstthefieldtomovethepointchargeadistance of 55 cm

(b) Calculatethepotentialdifferencebetweenthe final and initial positions of the charge

(c) Determinethepotentialdifferencebetweenthe plates if their separation distance is 15 cm

14 Duringaflashoflightningthepotentialdifferencebetweenacloudandthegroundwas12 times 109 V and the amount of transferred charge was32C

(a) Determinethechangeinenergyofthetransferred charge

(b) Iftheenergyreleasedwasallusedtoacceleratea1tonnecardeduceitsfinalspeed

(c) Iftheenergyreleasedcouldbeusedtomelticeat0degCdeducetheamountoficethatcould be melted

PHYSICS 2015indb 317 150514 338 PM

Chapter 10

318

AH

L

15 SupposethatwhenanelectronmovedfromAtoBin the diagram along an electric field line that the electric field does 36 times 10-19Jofworkonit

Determinethedifferencesinelectricpotential

(a) VB ndash VA(b) VC ndash VA(c) VC ndash VB

16 DeterminethepotentialatpointPthatislocatedatthecentreofthesquareasshowninthediagrambelow

- 6μC

+3μC +2μC

1m

1m

5μC

P

102 Fields at work

Essential idea Similar approaches can be taken inanalysingelectricalandgravitationalpotentialproblems

Understandingsbull Potentialandpotentialenergybull Potentialgradientbull Potentialdifferencebull Escapespeedbull Orbitalmotionorbitalspeedandorbitalenergybull Forcesandinverse-squarelawbehaviour

Orbital motion orbital speed and orbital energy

Although orbital motion may be circular elliptical orparabolic this sub-topic only dealswith circular orbitsThis sub-topic is not fundamentally new physics butan application that synthesizes ideas from gravitationcircularmotiondynamicsandenergy

TheMoonorbitstheEarthandinthissenseitisoftenreferredtoasasatelliteoftheEarthBefore1957itwastheonlyEarthsatelliteHoweverin1957theRussianslaunchedthefirstmanmadesatelliteSputnik1SincethisdatemanymoresatelliteshavebeenlaunchedandtherearenowliterallythousandsofthemorbitingtheEarthSomeareusedtomonitortheweathersome used to enable people to find accurately their position on thesurfaceoftheEarthmanyareusedincommunicationsandnodoubtsomeareusedtospyonothercountriesFigure1015showshowinprincipleasatellitecanbeputintoorbit

Theperson(whosesizeisgreatlyexaggeratedwithrespectto Earth) standing on the surface on the Earth throwssome stonesThe greater the speed with which a stoneis thrown the further it will land from her The pathsfollowedbythethrownstonesareparabolasByastretchof the imaginationwecanvisualisea situation inwhichastoneis thrownwithsuchaspeedthatbecauseof thecurvatureoftheEarthitwillnotlandonthesurfaceoftheEarthbutgointoldquoorbitrdquo(Path4onFigure1015)

NATURE OF SCIENCECommunication of scientific explanations The ability to apply field theory to the unobservable (charges) and the massively scaled (motion of satellites) required scientists to develop new ways to investigate analyse and report findings to a general public used to scientific discoveries based on tangible and discernible evidence (51)

copy IBO 2014

PHYSICS 2015indb 318 150514 338 PM

Fields

319

AH

L

1

2

34Earth

Figure 1015 Throwing a stone into orbit

TheforcethatcausesthestonestofollowaparabolicpathandtofalltoEarthisgravityandsimilarlytheforcethatkeeps the stone in orbit is gravity For circular motionto occur we have seen that a force must act at rightanglestothevelocityofanobjectthatistheremustbeacentripetalforceHenceinthesituationwedescribeherethe centripetal force for circular orbital motion about the EarthisprovidedbygravitationalattractionoftheEarth

We can calculate the speedwithwhich a stonemustbethrowninordertoputitintoorbitjustabovethesurfaceof the Earth

If the stone has mass m and speed v thenwehave fromNewtonrsquos2ndlaw

mv2

RE---------- G

MEm

RE2------------=

whereRE is the radius of the Earth and ME is the mass of the Earth

Bearinginmindthatg 0 GME

RE2--------= then

v g RE 10 64 106timestimes 8 103times= = =

Thatisthestonemustbethrownat8times103m sndash1

Clearly we are not going to get a satellite into orbit soclosetothesurfaceoftheEarthMovingatthisspeedthefrictionduetoairresistancewouldmeltthesatellitebeforeithadtravelledacoupleofkilometresInrealitythereforeasatelliteisputintoorbitabouttheEarthbysendingitattachedtoarocketbeyondtheEarthrsquosatmosphereandthengivingitacomponentofvelocityperpendiculartoaradialvectorfromtheEarthSeeFigure1016

Earth Satellite carried by rocket to hereSatellite orbit

Tangential component of velocity

Figure 1016 Getting a satellite into orbit

Keplerrsquos third law(This work of Kepler and Newtonrsquos synthesis of the work is an excellent example of the scientific method and makes for a good TOK discussion)

In 1627 Johannes Kepler (1571-1630)publishedhis lawsof planetary motion The laws are empirical in natureandwere deduced from the observations of theDanishastronomer Tycho de Brahe (1546-1601)The third lawgives a relationship between the radius of orbit R of a planet and its period TofrevolutionabouttheSunThelawisexpressedmathematicallyas

2

3 constantTR

=

We shall nowuseNewtonrsquos LawofGravitation to showhowitisthattheplanetsmoveinaccordancewithKeplerrsquosthirdlaw

InessenceNewtonwasabletousehislawofgravitytopredict the motion of the planets since all he had to do was factor theF givenby this law intohis second law F = ma to find their accelerations and hence their future positions

InFigure1017theEarthisshownorbitingtheSunandthedistancebetweentheircentresisR

RSun

Earth

F es

F se

Figure 1017 Planets move according to Keplerrsquos third law

Fes is the force that the Earth exerts on the Sun and Fse is the force that the Sun exerts on the Earth The forces are equalandoppositeandtheSunandtheEarthwillactuallyorbitaboutacommoncentreHoweversince theSun issoverymuchmoremassivethantheEarththiscommoncentrewillbeclosetothecentreoftheSunandsowecanregard the Earth as orbiting about the centre of the Sun Theotherthingthatweshallassumeisthatwecanignorethe forces that the other planets exert on the Earth (This wouldnotbeawisethingtodoifyouwereplanningtosend a space ship to theMoon for example)We shallalso assume that we have followed Newtonrsquos exampleandindeedprovedthataspherewillactasapointmasssituated at the centre of the sphere

Kepler had postulated that the orbits of the planets areellipticalbut since theeccentricityof theEarthrsquosorbit issmallweshallassumeacircularorbit

PHYSICS 2015indb 319 150514 338 PM

Chapter 10

320

AH

L

TheaccelerationoftheEarthtowardstheSunisa = Rω2

whereω ω 2πT------=

Hencea R 2π

T------ times

2 4π2 RT 2------------- = =

But the acceleration is given by Newtonrsquos Second Law F = mawhereFisnowgivenbytheLawofGravitationSo in this situation

F maGMs Me

R2-------------------= = butwealsohavethat

a a R 2πT

------ times

2 4π2 RT 2------------- = = and m = Me so that

G MsMe

R2------------------- Me4π2R

T2-------------times

GMs

4π2-----------rArr R3

T2------= =

Butthequantity

isaconstantthathasthesamevalueforeachoftheplanetssowehaveforalltheplanetsnotjustEarththat

R3

T2------ k =

wherekisaconstantWhichisofcourseKeplerrsquosthirdlaw

ThisisindeedanamazingbreakthroughItisdifficulttorefutethe idea that all particles attract each other in accordance withtheLawofGravitationwhenthelawisabletoaccountfortheobservedmotionoftheplanetsabouttheSun

The gravitational effects of the planets upon each othershould produce perturbations in their orbits Such is the predictivepoweroftheUniversalGravitationalLawthatit enabled physicists to compute these perturbations The telescopehadbeeninventedin1608andbythemiddleofthe 18th Century had reached a degree of perfection in design that enabled astronomers to actually measure the orbital perturbations of the planets Their measurements werealways inagreementwith thepredictionsmadebyNewtonrsquoslawHoweverin1781anewplanetUranuswasdiscovered and the orbit of this planet did not fit withthe orbit predicted by Universal Gravitation Such wasthe physicistrsquos faith in theNewtonianmethod that theysuspectedthatthediscrepancywasduetothepresenceofayetundetectedplanetUsingtheLawofGravitationtheFrenchastronomerJLeverrier and the English astronomer J C Adamswereable tocalculate justhowmassive thisnewplanetmustbeandalsowhereitshouldbeIn1846the planetNeptunewas discovered justwhere they hadpredicted Ina similarwaydiscrepancies in theorbitofNeptune ledto thepredictionandsubsequentdiscoveryin1930oftheplanetPlutoNewtonrsquosLawofGravitation

had passed the ultimate test of any theory it is not only abletoexplainexistingdatabutalsotomakepredictions

Satellite energyWhenasatelliteisinorbitaboutaplanetitwillhavebothkineticenergyandgravitationalpotentialenergySupposewe consider a satellite ofmass m that is in an orbit of radius r about a planet of mass M

Thegravitationalpotentialduetotheplanetatdistancer from its centre is

ThegravitationalpotentialenergyofthesatelliteVsat

is therefore GMe m

r----------------ndash

ThatisVsat GMem

r----------------ndash=

Thegravitationalfieldstrengthatthesurfaceoftheplanetisgivenby

g 0GMe

Re2------------=

Hencewecanwrite

Vsat g0 Re

2

rndash=

m

ThekineticenergyofthesatelliteKsatisequaltofrac12mv2wherev is its orbital speed

Byequatingthegravitationalforceactingonthesatellitetoitscentripetalaccelerationwehave

G Mem

r2---------------- mv2

r---------- mv2hArr

GMemr----------------= =

Fromwhich12---

mv2 12---

GMemr----------------times=

= 0 e

2g R

r

2m

Which is actually quite interesting since it shows thatirrespective of the orbital radius the KE is numericallyequal to half the PE Also the total energy E tot K sat Vsat+ 1

2---G Me m

r----------------times G Me m

r----------------ndash + 1

2---G Me m

r----------------ndash= = =of the satelliteisalwaysnegativesince

E tot K sat Vsat+ 12---

G Me mr----------------times

G Me mr----------------ndash

+ 12---

G Me mr----------------ndash= = =

PHYSICS 2015indb 320 150514 338 PM

Fields

321

AH

L

The energies of an orbiting satellite as a function of radial distancefromthecentreofaplanetareshownplottedinFigure1018

12

100806

0402

0ndash02

ndash04

ndash06ndash08ndash10

ndash12

2 4 6 8 10 12

ener

gya

rbita

ry u

nits

distance timesR

kinetic energytotal energy

potential energy

Figure 1018 Energy of an orbiting satellite as a function of distance from the centre of a planet

WeightlessnessSuppose that you are inanelevator(lift)thatisdescendingat constant speedandyou let goof abook thatyouareholdinginyourhandThebookwillfalltothefloorwithaccelerationequaltotheaccelerationduetogravityIfthecablethatsupportstheelevatorweretosnap(asituationthatItrustwillneverhappentoanyofyou)andyounowletgothebookthatyouareholdinginyourotherhandthis book will not fall to the floor - it will stay exactlyin linewithyourhandThis isbecause thebook isnowfallingwith the sameaccelerationas theelevatorandassuch the book cannot ldquocatchrdquo up with the floor of theelevatorFurthermoreifyouhappenedtobestandingonasetofbathroomscalesthescaleswouldnowreadzero-youwouldbeapparentlyweightlessItisthisideaoffreefallthatexplainstheapparentweightlessnessofastronautsin an orbiting satellite These astronauts are in free fall inthesensethattheyareacceleratingtowardsthecentre of the Earth

It is actually possible to define theweight of a body inseveral different waysWe can define it for example asthegravitationalforceexertedonthebodybyaspecifiedobject such as the EarthThiswe have seen thatwe doinlotsofsituationswherewedefinetheweightasbeingequaltomgIfweusethisdefinitionthenanobjectinfreefallcannotbydefinitionbeweightlesssinceitisstillinagravitationalfieldHoweverifwedefinetheweightofanobjectintermsofaldquoweighingrdquoprocesssuchasthereadingonasetofbathroomscaleswhichineffectmeasuresthecontact force between the object and the scales thenclearlyobjectsinfreefallareweightlessOnenowhasto

askthequestionwhetherornotitispossibleForexampletomeasurethegravitationalforceactingonanastronautin orbit about the Earth

WecanalsodefineweightintermsofthenetgravitationalforceactingonabodyduetoseveraldifferentobjectsForexample for an object out in space its weight could bedefined in terms of the resultant of the forces exerted on it bytheSuntheMoontheEarthandalltheotherplanetsin the Solar System If this resultant is zero at a particular pointthenthebodyisweightlessatthispoint

Inviewofthevariousdefinitionsofweightthatareavailabletousitisimportantthatwhenweusethewordldquoweightrdquoweareawareofthecontextinwhichitisbeingused

Example

Calculate the height above the surface of the Earth atwhichageo-stationarysatelliteorbits

Solution

A geo-stationary satellite is one that orbits the Earth in such a way that it is stationary with respect to a point on the surface of the Earth This means that its orbital period must be the same as the time for the Earth to spin once on its axis ie 24 hours

From Keplerrsquos third law we have G Ms

4π2----------- R3

T2------=

That is

RMe

mh

using the fact that the force of attraction between the satellite and the Earth is given by

FG Mem

R2----------------=

and that F = ma

where m is the mass of the satellite and a 4π2R

T2-------------=

we have

GMem

R2---------------- m 4π2R

T2-------------times

GMe

4π2------------rArr R3

T2------= =

PHYSICS 2015indb 321 150514 338 PM

Chapter 10

322

AH

L

Now the mass of the Earth is 60 times 1024 kg and the period T measured in seconds is given by T = 86400 s

So substitution gives R = 42 times 106 m

The radius of the Earth is 64 times 106 m so that the orbital height h is about 36 times 107 m

Example

Calculatetheminimumenergyrequiredtoputasatelliteofmass500kgintoanorbitthatisasaheightequaltotheEarthrsquosradiusabovethesurfaceoftheEarth

Solution

We have seen that when dealing with gravitational fields and potential it is useful to remember that

g 0G M

Re2---------= or g0 Re

2 GM=

The gravitational potential at the surface of the Earth is

g 0ndash Re GMRe

---------ndash=

The gravitational potential at a distance R

from the centre of the Earth is

=

The difference in potential between the surface of the Earth and a point distance R from the centre is therefore

V∆ g 0Re 1ReR------ndash

=

If R 2Re= then V∆g0 Re

2------------=

This means that the work required to ldquoliftrdquo the satellite into orbit is g0Rm where m is the mass of the satellite This is equal to

10 32 106 500timestimestimes = 16000 MJ

However the satellite must also have kinetic energy in order to orbit Earth This will be equal to

g 0 mRe2

2R-----------------g 0mRe

2

4----------------- 8000 MJ= =

The minimum energy required is therefore

24000 MJ

Exercise 1031 The speed needed to put a satellite in orbit does

not depend on

A the radius of the orbitB theshapeoftheorbitC thevalueofgattheorbitD the mass of the satellite

2 EstimatethespeedofanEarthsatellitewhoseorbitis400kmabovetheEarthrsquossurfaceAlsodetermine the period of the orbit

3 Calculatethespeedofa200kgsatelliteorbitingthe Earth at a height of 70 times 106 m

Assume that g = 82 m sndash2 for this orbit

4 TheradiioftwosatellitesXandYorbitingthe Earth are 2r and 8rwherer is the radius of the Earth Calculate the ratio of the periods of revolutionofXtoY

5 A satellite of mass mkgissentfromEarthrsquossurfaceinto an orbit of radius 5RwhereR is the radius of theEarthWritedownanexpressionfor

(a) thepotentialenergyofthesatelliteinorbit

(b) thekineticenergyofthesatelliteinorbit

(c) theminimumworkrequiredtosendthesatellitefromrestattheEarthrsquossurfaceintoits orbit

6 A satellite in an orbit of 10rfallsbacktoEarth(radius r)afteramalfunctionDeterminethespeedwithwhichitwillhittheEarthrsquossurface

7 The radius of the moon is frac14 that of the Earth AssumingEarthandtheMoontohavethesamedensitycomparetheaccelerationsoffreefallatthe surface of Earth to that at the surface of the Moon

8 UsethefollowingdatatodeterminethegravitationalfieldstrengthatthesurfaceoftheMoon and hence determine the escape speed from the surface of the Moon

Mass of the Moon = 73 times 1022kg

RadiusoftheMoon=17times106 m

PHYSICS 2015indb 322 150514 338 PM

Chapter 9

290

AH

L

This intensity pattern arises from the fact that each point ontheslitactsinaccordancewithHuygenrsquosprincipleasa source of secondary wavefronts It is the interference between these secondary wavefronts that produces the typical diffraction pattern

Obtaining an expression for the intensity distribution is mathematically a little tricky and it is not something that we are going to attempt here However we can deducea useful relationship from a simple argument In this argument we deal with a phenomenon called Fraunhofer diffractionthatisthelightsourceandthescreenareaninfinitedistanceawayformtheslitThiscanbeachievedwith the set up shown in Figure 911

lens 1 lens 2

single slit

source

screen

Figure 911 Apparatus for viewing Fraunhofer diffraction

The source is placed at the principal focus of lens 1 and the screen isplacedat theprincipal focusof lens2Lens1 ensures that parallel wavefronts fall on the single slit and lens2ensuresthattheparallelraysarebroughttoafocuson the screen The same effect can be achieved using a laser and placing the screen some distance from the slit If the light and screenarenot an infinitedistance from the slitthen we are dealing with a phenomenon called Fresnel diffraction and such diffraction is very difficult to analyse mathematically To obtain a good idea of how the single slit patterncomesaboutweconsiderthediagramFigure912

Xθ2

Yλ

d

P

b

f

screen

θ1

Figure 912 Single slit diffraction

In particular we consider the light from one edge of the slittothepointPwherethispointis justonewavelengthfurther from the lower edge of the slit than it is from the

upper edge The secondary wavefront from the upper edge willtraveladistanceλ2furtherthanasecondarywavefrontfrom a point at the centre of the slitHencewhen thesewavefronts arrive at P theywill be out of phase andwillinterfere destructively The wavefronts from the next point below the upper edge will similarly interfere destructively with the wavefront from the next point below the centre of the slit In this way we can pair the sources across the whole width of the slit If the screen is a long way from the slit then the angles θ1 and θ2becomenearlyequal (If the screen is at infinity then they are equal and the two lines PX and XY are at right angles to each other)FromFigure1116weseethereforethattherewillbeaminimumatPif

λ b θ1 sin= where b is the width of the slit

Howeverbothanglesareverysmallequaltoθ saywhereθ is the angle of diffraction

Soitcanbewritten θ λb---

=

This actually gives us the half-angular width of the central maximum We can calculate the actual width of the maximum along the screen if we know the focal length of the lens focussing the light onto the screen If this is f then we have that

θ df---

= Suchthat d fλb-----

=

To obtain the position of the next maximum in the pattern we note that the path difference is 3

2---λ We therefore divide

theslitintothreeequalpartstwoofwhichwillproducewavefronts that will cancel and the other producing wavefronts that reinforce The intensity of the second maximum is therefore much less than the intensity of thecentralmaximumof theorderof005or120of theoriginalintensity(Muchlessthanonethirdinfactsincethewavefrontsthatreinforcewillhavedifferingphases)

We can also see now how diffraction effects become more and more noticeable the narrower the slit becomes If light of wavelength 430 nm was to pass through a slit of width say10cmandfallonascreen30mawaythenthehalfangular width of the central maximum would be 013 microm

d = f λ

___ b = 3 times 430 times 10 ndash9 ____________ 01 = 013 μm

There will be lots of maxima of nearly the same intensity and the maxima will be packed very closely together (Thefirstminimumoccursatadistanceof012micromfromthe centre of the central maximum and the next occurs effectivelyatadistanceof024microm)Weeffectivelyobservethe geometric pattern Refer to Example

We also see now how for diffraction effects to be noticeable the wavelength must be of the order of the slit width The width of the pattern increases in proportion to the wavelength and decreases inversely with the width of the

PHYSICS 2015indb 290 150514 337 PM

Wave Phenomena

291

AH

L

93 Interference

Essential idea Interference patterns from multiple slits andthinfilmsproduceaccuratelyrepeatablepatterns

Understandingsbull Youngrsquosdouble-slitexperimentbull Modulationoftwo-slitinterferencepatternbyone-slit

diffraction effectbull Multipleslitanddiffractiongratinginterferencepatternsbull Thinfilminterference

Youngrsquos double slit experimentPlease refer to interference in chapter 4 However wewill reiterate the condition for the interference of waves from two sources to be observed The two sources must be coherentthatistheymusthavethesamephaseorthephase difference between them must remain constant

Also to reinforce topics concerning the principle ofsuperpositionandpathdifferenceandphasedifferenceincluded in Figure 913 is another example of two source interference

We can obtain evidence for the wave nature of sound by showing that sound produces an interference pattern Figure 913 shows the set up for demonstrating this

CRO

speaker

speaker

signalgenerator

microphoneX

Y

Figure 913 Interference using sound waves

The two speakers are connected to the same output of the signal generator andplaced about 50 cmapartThesignalgeneratorfrequencyissettoabout600Hzandthe

slit If the slit width is much greater than the wavelength then the width of the central maxima is very small

Insummary

for destructive interference and minima

dsinθ=nλ wheren=123hellip

for constructive interference and maxima

dsinθ=(n+frac12)λ wheren=123hellip

Note that waves from point sources along the slit arrive in phase at the centre of the fringe pattern and constructively interferetoproducethecentralmaximumcalledthezero ordermaximum(n=0)

Example

LightfromalaserisusedtoformasingleslitdiffractionpatternThe width of the slit is 010 mm and the screen is placed 30 m from the slit The width of the central maximum is measured as26cmCalculatethewavelengthofthelaserlight

Solution

Since the screen is a long way from the slit we can use the small angle approximation such that the f is equal to 30 m

The half-width of the central maximum is 13 cm so we have

λ13 10 2ndashtimes( ) 10 10 4ndashtimes( )times

30------------------------------------------------------------------=

To give λ = 430 nm

This example demonstrates why the image of a point source formed by a thin converging lens will always have a finite width

Exercise 921 Aparallelbeamoflightofwavelength500nmis

incidentonaslitofwidth025mmThelightisbroughttofocusonascreenplaced150mfromthe slit Calculate the angular width and the linear width of the central diffraction maximum

2 Lightfromalaserisusedtoformasingleslitdiffraction pattern on a screen The width of the slit is 010 mm and the screen is 30 m from the slit The width of the central diffraction maximum is 26cmCalculatethewavelengthofthelaserlight

3 Determinetheanglethatyouwouldexpecttofindconstructiveinterference(forn=0n=1andn=2) fromasingleslitofwidth20μmandlightofwavelength 600 nm

NATURE OF SCIENCECuriosity Observed patterns of iridescence in animals such as the shimmer of peacock feathers led scientists to develop the theory of thin film interference (15)

Serendipity The first laboratory production of thin films was accidental (15)

copy IBO 2014

PHYSICS 2015indb 291 150514 337 PM

Chapter 9

292

AH

L

microphoneismovedalongthelineXYthatisaboutonemetre from the speakers As the microphone is moved along XY the trace on the cathode ray oscilloscope isseen to go through a series of maxima and minima corresponding to points of constructive and destructive interference of the sound waves

Aninterestinginvestigationistofindhowtheseparationof the points of maximum interference depends on the frequency of the source and also the distance apart ofthe speakers

In the demonstration of the interference between two soundsourcesdescribedaboveifweweretomoveoneofthe speakers from side to side or backwards and forwards the sound emitted from the two speakers would no longer be in phase No permanent points of constructive or destructive interference will now be located since the phase difference between the waves from the two sources is nolongerconstantiethesourcesarenolongercoherent

Light from an incandescent source is emitted with acompletely random phase Although the light from two separate sources will interfere because of the randomlychanging phase no permanent points of constructive or destructive interference will be observed This is why a singleslit isneededintheYoungrsquosdoubleslitexperimentByactingasapointsourceitessentiallybecomesacoherentlight source The light emitted from a laser is also very nearly coherent and this is why it is so easy to demonstrate optical interference and diffraction with a laser

Youngrsquos double slit experiment is one of the great classic experiments of physics and did much to reinforce the wave theoryoflightThomasYoungcarriedouttheexperimentin about 1830

It is essentially the demonstration with the ripple tank and the sound experiment previously described butusing light The essential features of the experiment are shown in Figure 914

screendouble slit

single slit

coloured lter

S 1

S2

Figure 914 Youngrsquos double slit experiment

YoungallowedsunlighttofallontoanarrowsingleslitA few centimetres from the single slit he placed a double slit The slits are very narrow and separated by a distance equal to about fourteen slitwidthsA screen is placed

about ametre from thedouble slitsYoungobservedapatternofmulticolouredldquofringesrdquo in the screenWhenhe placed a coloured filter between the single slit anddouble slit he obtained a pattern that consisted of bright coloured fringes separated by darkness

The single slit essentially ensures that the light falling on the double slit is coherent The two slits then act as the two speakers in the sound experiment or the two dippers in the ripple tank The light waves from each slit interfere and produce the interference pattern on the screen Without thefilterapatternisformedforeachwavelengthpresentinthesunlightHencethemulticolouredfringepattern

You can demonstrate optical interference for yourselfSmoking a small piece of glass and then drawing twoparallel lines on it can make a double slit If you then look throughthedoubleslitatasingletungstenfilamentlampyouwillseethefringepatternByplacingfiltersbetweenthe lamps and the slits you will see the monochromatic fringe pattern

YoucanalsoseetheeffectsofopticalinterferencebylookingatnetcurtainsEachlsquoholersquointhenetactsasapointsourceand the light from all these separate sources interferes and producesquiteacomplicatedinterferencepattern

A laser can also be used to demonstrate optical interference Sincethelightfromthelaseriscoherentitisveryeasytodemonstrate interference Just point the laser at a screen and place a double slit in the path of the laser beam

LetusnowlookattheYoungrsquosdoubleslitexperimentinmore detail The geometry of the situation is shown in Figure915

S1

S2 X

P

Q

D

y

d

screen

θ

θrsquo

Figure 915 The geometry of Youngrsquos double slit experiment

S1andS2 are the two narrow slits that we shall regard as twocoherentmonochromaticpointsourcesThedistancefrom the sources to the screen is D and the distance between the slits is d

The waves from the two sources will be in phase at Q and therewill be abright fringehereWewish tofindtheconditionfortheretobeabrightfringeatPdistancey from Q

PHYSICS 2015indb 292 150514 337 PM

Wave Phenomena

293

AH

L

We note that that D (asymp1metre)isverymuchgreaterthaneither y or d(asymp fewmillimetres)Thismeansthatbothθ and θʹ areverysmallanglesandforintentsandpurposesequal

From the diagram we have that

θ = Dy

And

2S Xd

θʹ =

(Remembertheanglesareverysmall)

Sinceθ asymp θʹ then

2S XyD d

=

ButS2X is the path difference between the waves as they

traveltoPWehavethereforethat

path difference = Dyd

TherewillthereforebeabrightfringeatPif

Dyd

= nλ

Supposethatthereisabrightfringeaty = y1 corresponding

to n = n1 then

Ddy1 = n

1λ

If the next bright fringe occurs at y = y2 this will correspond

to n = n1+1Hence

Ddy2 =(n

1+1)λ

This means that the spacing between the fringes y2 ndash y

1 is

given by

y2 ndash y

1 = d

Dλ

Youngactuallyusethisexpressiontomeasurethewavelengthof the light he used and it is a method still used today

We see for instance that if in a given set up we move the slits closer together then the spacing between the fringes will get greaterEffectivelyourinterferencepatternspreadsoutthatis there will be fewer fringes in a given distance We can also increase the fringe spacing by increasing the distance betweentheslitsandthescreenYouwillalsonotethatfora

givensetupusinglightofdifferentwavelengthsthenldquoredrdquofringeswillspacefurtherapartthanldquobluerdquofringes

In this analysis we have assumed that the slits act as point sources and as such the fringes will be uniformly spaced andofequalintensityAmorethoroughanalysisshouldtakeintoaccountthefinitewidthoftheslits

ReturningtoFigure915weseethatwecanwritethepathdifferenceS2X as

S2X = dtan θʹ

But since θʹ is a small angle the sine and tangent will be nearlyequalsothat

S2X = dsinθʹ

And since θʹ asymp θ then

S2X = dsinθ

The condition therefore for a bright fringe to be found at a point of the screen can therefore be written as

dsinθ = nλ

Insummary

for constructive interference and minima

d sinθ = nλ wheren=123hellip

for destructive interference and maxima

d sinθ = ( n + frac12 )λ wheren=123hellip

Figure 916 shows the intensity distribution of the fringes on the screen when the separation of the slits is large comparedtotheirwidthThefringesareofequalintensityandofequalseparation

Intensity

distance along screen

Figure 916 The intensity distribution of the fringes

It isworthnoting that if the slits are close together theintensity of the fringes is modulated by the intensity distribution of the diffraction pattern of one of the slits

PHYSICS 2015indb 293 150514 337 PM

Chapter 9

294

AH

L

Example

Light of wavelength 500nm is incident on two smallparallel slits separated by 10mmDetermine the anglewherethefirstmaximumisformed

If after passing through the slits the light is brought toa focusona screen15m fromtheslitscalculate theobserved fringe spacing on the screen

Solution

Using the small angle approximation we have

3

7

10105minus

minustimes==

dλθ

= 5 times 10-4 rad

The fringe spacing is given by

3

7

1010551

minus

minustimestimes==

dDy λ

= 075 mm

Modulation of two-slit interference pattern by one-slit diffraction effect

In practice the intensity pattern of the maxima as shown in Figure 917 is not constant but fluctuates while decreasingsymmetrically on either side of the central maximum as shown inFigure917Theintensitypatternisacombinationofboththe single-slit diffractionenvelope and the double slit pattern That is the amplitude of the two-slit interference pattern is modulated(ieadjustedto)byasingleslitdiffractionenvelope

2λD

___ λD

D

___λd

d

___

Double slitnite width

Figure 917 Double slit pattern

ForagivenslitseparationdwavelengthoflightandfixedslittoscreendistancethevariationinintensitydependsonthewidthoftheslitDAlthoughincreasingthewidthoftheslitsincreasestheintensityoflightinthefringesitalsomakesthem less sharp and they become blurred As the slit width is widenedthefringesgraduallydisappearbecausenumerouspoint sources along the widened slit give rise to their own dark and bright fringes that then overlap with each other

Exercise 931 InFigure913thedistancebetweenthespeakers

is050mandthedistancebetweenthelineofthespeakersandthescreenis20mAsthemicrophoneismovedalongthelineXYthedistance between successive points of maximum soundintensityis030mThefrequencyofthesoundwavesis44times103HzCalculateavalueforthe speed of sound

2 Laserlightofwavelength610nmfallsontwoslits10x10-5apartDeterminetheseparationofthefirstordermaximumformedonascreen20maway

3 Twoparallelslits012mmapartareilluminatedwith red light with a wavelength of 600 nm An interferencepatternfallsonascreen15maway

Calculate

(a) thedistancefromthecentralmaximumtothefirstbrightfringe

(b) thedistancetotheseconddarkline

Multiple slit and diffraction grating interference patterns

If we examine the interference pattern produced when monochromatic light passes through a different number of slits we notice that as the number of slits increases the number of observed fringes decreases the spacingbetween them increases and the individual fringes become much sharper We can get some idea of how this comes about by looking at the way light behaves when a parallel beam passes through a large number of slits The diagram for this is shown in Figure 918

θθd

1

2

3

telescope

Figure 918 A parallel beam passing through several slits

The slits are very small so that they can be considered to act as point sources They are also very close together such that d issmall(10ndash6 m)EachslitbecomesasourceofcircularwavefrontsandthewavesfromeachslitwillinterfereLetus

PHYSICS 2015indb 294 150514 337 PM

Wave Phenomena

295

AH

L

consider the light that leaves the slit at an angle θ as shown Thepathdifferencebetweenwave1andwave2isdsinθ and ifthisisequaltoanintegralnumberofwavelengthsthenthe two waves will interfere constructively in this direction Similarly wave 2 will interfere constructively with wave3 at this angle andwave 3with 4 etc across thewholegratingHenceifwelookatthelightthroughatelescopethatisbringittoafocusthenwhenthetelescopemakesanangle θ to the grating a bright fringe will be observed The condition for observing a bright fringe is therefore

dsinθ = mλ

Supposeweuselightofwavelength500nmandsupposethat d = 16 times 10ndash6 m

Obviously we will see a bright fringe in the straight on position θ=0(thezero order)

The next position will be when m =1(thefirst order)andsubstitutionintheaboveequationgivesθ = 18deg

The next position will be when m=2(thesecond order)and this give θ = 38deg

For m=3sinθ is greater than 1 so with this set up we only obtain5fringesonezeroorderandtwoeithersideofthezero order

The calculation shows that the separation of the orders is relatively large At any angles other than 18deg or 38deg the light leaving the slits interferes destructively We can see that the fringes will be sharp since if we move just a small angle away from 18deg the light from the slits will interfere destructively

An array of narrow slits is usually made by cutting narrow transparent lines very close together into the emulsion on aphotographicplate(typically200linespermillimetre)Suchanarrangementiscalledadiffraction grating

The diffraction grating is of great use in examining the spectral characteristics of light sources

All elements have their own characteristic spectrum An element can be made to emit light either by heating it until it is incandescent or by causing an electric discharge through itwhen it is in a gaseous state Yourschool probably has some discharge tubes and diffraction gratingsIfithasthenlookattheglowingdischargetubethroughadiffractiongratingIfforexampletheelementthat you are looking has three distinct wavelengths then each wavelength will be diffracted by a different amount inaccordancewiththeequationdsinθ = nλ

Alsoiflaserlightisshonethroughagratingontoascreenyou will see just how sharp and spaced out are the maxima By measuring the line spacing and the distance of the screen fromthelaserthewavelengthofthelasercanbemeasured

If your school has a set of multiple slits say from a single slit toeight slits then it isalsoaworthwhileexercise toexamine how the diffraction pattern changes when laser light is shone through increasing numbers of slits

Consider a grating of NparallelequidistantslitsofwidthaseparatedbyanopaqueregionofwidthD We know that light diffracted from one slit will be superimposed and interfere with light diffracted from all the other slits The intensityprofile of the resultant wave will have a shape determined by

bull theintensityoflightincidentontheslitsbull thewavelengthofthislightbull theslitwidthbull theslitseparation

Typical intensity profiles formed when a plane wavemonochromaticlightatnormalincidenceononetwoandsix slits are shown in Figure 919 In this example the slit widthisonequarterofthegratingspacing

Six slits sharperfringes

d = 4a

I α N 2

(N ndash 2)= 4 subsidiary maxima

order0 4321

I = 36

Two slits

order0

single-slitdiraction

envelope

d = 4a4th principal maximum missing

principa maxima

I = 4 (nominal units)

4321

intensity

order

ISS

One slit

0 1

single-slitdiraction

pattern

N = number of slitsslit width a

Figure 919 Intensity profiles

PHYSICS 2015indb 295 150514 337 PM

Chapter 9

296

AH

L

1 For Nge2thefringepatterncontainsprincipalmaximathat are modulated by the diffraction envelope

2 ForNgt2thefringepatterncontains(Nndash2)subsidiary maxima

3 Where the grating spacing d is four time the slit width a(4xa =d)thenthefourthprincipalmaximumismissing from the fringe pattern Where the grating spacingdisthreetimetheslitwidtha(3xa =d)thenthe third principal maximum is missing from the fringepatternandsoon

4 Increasing N increases the absolute intensities of the diffraction pattern (I = N2)

If white light is shone through a grating then the central image will be white but for the other orders each will be spread out intoacontinuousspectrumcomposedofaninfinitenumberof adjacent images of the slit formed by the wavelength of the different wavelengths present in the white light At any given point in the continuous spectrum the light will be very nearly monochromatic because of the narrowness of the images of the slit formed by the grating This is in contrast to the doubleslitwhereifwhitelightisusedtheimagesarebroadand the spectral colours are not separated

In summary as is the case of interference at a doubleslit there is a path difference between the rays fromadjacent slits of d sinθ Where there is a whole number of wavelengthsconstructiveinterferenceoccurs

For constructive interference and minima

d sinθ = mλ wheren=123hellip

When there is a path difference between adjacent rays of halfawavelengthdestructiveinterferenceoccurs

For and maxima

d sinθ = ( m + frac12 )λ wheren=123hellip

Exercise 941 Lightfromalaserisshonethroughadiffraction

grating on to a screen The screen is a distance of 20mfromthelaserThedistancebetweenthecentraldiffractionmaximumandthefirstprincipalmaximum formed on the screen is 094 m The diffractiongratinghas680linespermmEstimatethe wavelength of the light emitted by the laser

2 Monochromaticlightfromalaserisnormallyincident on a six-slit grating The slit spacing is three times the slit width An interference patternisformedonascreen2montheotherside of the grating

(a) Determineandexplaintheorderoftheprincipal maxima missing from the fringe pattern

(b) Howmanysubsidiarymaximaaretherebetween the principal maxima

(c) Howmanyprincipalmaximaarethere

(i) Betweenthetwofirstorderminimain the diffraction envelope

(ii) Betweenthefirstandsecondorderminima in the diffraction envelope

(d) Drawthepatternformedbythebrightfringesin(c)

(e) Sketchtheintensityprofileofthefringepattern

Thin film interferenceYoumightwellbe familiarwith the colouredpatternoffringes that can be seen when light is reflected off thesurfaceofwateruponwhichathinoilfilmhasbeenspiltorfromlightreflectedfrombubblesWecanseehowthesepatternsarisebylookingatFigure920

Extendedlight source air

A

B

C

12

3

4

Oil lm

Water

Figure 920 Reflection from an oil film

Consider light from an extended source incident on a thinfilmWealsoconsiderawave fromonepointofthe source whose direction is represented by the ray shownSomeofthislightwillbereflectedatAandsometransmitted through the filmwhere somewill again bereflectedatB(somewillalsobetransmittedintotheair)SomeofthelightreflectedatBwillthenbetransmittedatCandsomereflectedandsoonIfweconsiderjustrays1and2thenthesewillbenotbeinphasewhenthearebrought to a focus by the eye since they have travelled

PHYSICS 2015indb 296 150514 337 PM

Wave Phenomena

297

AH

L

different distances If the path difference between them is an integral number of half wavelengths then we might expect the twowaves to be out of phaseHowever ray1 undergoes a phase change of π on reflection but ray2 does not since it is reflected at a boundary betweena more dense and less dense medium (See chapter 4)Henceifthepathdifferenceisanintegralnumberofhalf-wavelengthsrays1and2willreinforceieray1and2areinphaseHowever rays357etcwillbeoutofphasewithrays246etcbutsinceray2ismoreintensethanray3andray4moreintensethanray5thesepairswillnot cancel out so there will be a maximum of intensity

Ifthepathdifferenceissuchthatwave1and2areoutofphasesincewave1ismoreintensethanwave2theywillnotcompletelyannulHoweveritcanbeshownthattheintensitiesofwaves2345hellipaddtoequaltheintensityofwave1Sincewaves345hellipareinphasewithwave2there will be complete cancellation

The path difference will be determined by the angle at which ray1 is incident andalsoon the thickness (andtheactualrefractiveindexaswell)ofthefilmSincethesourceisanextendedsourcethelightwillreachtheeyefrom many different angles and so a system of bright and darkfringeswillbeseenYouwillonlyseefringesifthefilmisverythin(exceptifviewedatnormalincidence)sinceincreasingthethicknessofthefilmwillcausethereflected rays to get so far apart that they will not becollected by the pupil of the eye

From the argument above to find the conditions forconstructive and destructive interferenceweneedonlyfindthepathdifferencebetweenray1andray2Figure921showsthe geometry of the situation

A BE

F

C

D

1 2

φrsquo φrsquo

φ

φ

d

Figure 921 The geometry of interference

Thefilmisofthicknessd and refractive index n and the light has wavelength λ If the line BF is perpendicular to ray 1 then the optical path difference (opd) between ray 1 andray2whenbroughttoafocusis

opd = n(AC+CB)ndashAF

We have to multiple by the refractive index for the path travelledbythelightinthefilmsincethelighttravelsmoreslowly in thefilm If the light travels say adistancex in a material of refractive index n then in the time that it takes to travelthisdistancethelightwouldtraveladistancenx in air

IfthelineCEisatrightanglestoray2thenweseethat

AF = nBE

FromthediagramAC=CDsowecanwrite

opd = n(CD+CB)ndashnBE=nDE

Alsofromthediagramweseethatwhereφ is the angle of refraction

DE=2cosφ

From which opd=2ndcosφ

Bearinginmindthechangeinphaseofray1onreflectionwe have therefore that the condition for constructive interference is

2nd φcos m 12---+

λ m 1 2 hellip = =

Andfordestructiveinterference 2ndcosφ = mλ

Eachfringecorrespondstoaparticularopd for a particular value of the integer m and for any fringe the value of the angle φ is fixedThismeans that itwill be in the formof an arc of a circle with the centre of the circle at the point where the perpendicular drawn from the eye meets the surface of the film Such fringes are called fringesofequal inclinationSince theeyehasa smallaperturethesefringesunlessviewedatneartonormalincidence (φ = 0) will only be observed if the film is very thinThisisbecauseasthethicknessofthefilmincreasesthereflected rayswill get further and further apart and sovery few will enter the eye

Ifwhitelightisshoneontothefilmthenwecanseewhywe get multi-coloured fringes since a series of maxima and minima will be formed for each wavelength present in the white light However when viewed at normalincidenceitispossiblethatonlylightofonecolourwillundergoconstructiveinterferenceandthefilmwilltakeon this colour

PHYSICS 2015indb 297 150514 337 PM

Chapter 9

298

AH

L

Thickness of oil films

The exercise to follow will help explain this use

Non-reflecting films

A very important but simple application of thin filminterferenceisintheproductionofnon-reflectingsurfaces

Athinfilmofthicknessd and refractive index n1 is coated onto glass of refractive index n where n1 lt n Light ofwavelength λ that is reflected at normal incidence willundergo destructive interference if 12

2n d λ

= thatis

14d

nλ

=

(remember that there will now no phase change at the glass interface ie we have a rare to dense reflection)

The use of such films can greatly reduce the loss oflightby reflection at the various surfacesof a systemoflenses or prisms Optical parts of high quality systemsare usually all coatedwithnon-reflecting films in orderto reduce stray reflections The films are usually madebyevaporatingcalciumormagnesiumfluorideonto thesurfacesinvacuumorbychemicaltreatmentwithacidsthat leave a thin layer of silica on the surface of the glass ThecoatedsurfaceshaveapurplishhuebyreflectedlightThis is because the condition for destructive interference fromaparticularfilmthicknesscanonlybeobtainedforone wavelength The wavelength chosen is one that has a value corresponding to light near the middle of the visible spectrumThismeansthatreflectionofredandvioletlightis greater combining to give the purple colour Because of the factor cosφatanglesotherthannormalincidencethepath difference will change butnotsignificantlyuntilsayabout30deg(egcos25deg=090)

It should be borne in mind that no light is actually lost by anon-reflectingfilmthedecreaseofreflectedintensityiscompensated by increase of transmitted intensity

Non-reflecting films can be painted onto aircraft tosuppressreflectionofradarThethicknessofthefilmis determined by

4nd λ

= where λ is the wavelength of

the radar waves and ntherefractiveindexofthefilmat this wavelength

Example

A plane-parallel glass plate of thickness 20 mm isilluminated with light from an extended source The refractiveindexoftheglassis15andthewavelengthofthelight is 600 nm Calculate how many fringes are formed

Solution

We assume that the fringes are formed by light incident at all angles from normal to grazing incidence

At normal incidence we have 2nd = mλ

From which

m 2 15 2 10 3ndashtimestimestimes6 10 7ndashtimes

------------------------------------------- 10 000= =

At grazing incidence the angle of refraction φ is in fact the critical angle

Therefore φφ arcsin 115------- 42 deg= =

ie cosφ = 075

At grazing incidence 2ndcosφ = m λ

From which (and using cos(sinndash1(115) = 075)

m 2 15 2 10 3ndash 075timestimestimestimes6 10 7ndashtimes

------------------------------------------------------------ 7500= =

The total number of fringes seen is equal to m ndash m = 2500

Exercise 95When viewed from above the colour of an oil film onwater appears red in colour Use the data below to estimate theminimumthicknessoftheoilfilm

average wavelength of red light = 630 nm

refractiveindexofoilforredlight=15

refractive index of water for red light = 13

PHYSICS 2015indb 298 150514 337 PM

Wave Phenomena

299

AH

L

94 Resolution

Essential idea Resolution places an absolute limit on the extent to which an optical or other system can separate images of objects

Understandings

bull Thesizeofadiffractingaperturebull Theresolutionofsimplemonochromatic

two-source systems

The size of a diffracting apertureOur discussion concerning diffraction so far has been for rectangular slits When light from a point source enters a smallcircularapertureitdoesnotproduceabrightdotasanimagebutacirculardiscknownasAiryrsquosdiscsurroundedbyfainterconcentriccircularringsasshowninFigure922

Laser

Figure 922 Diffraction at a circular aperture

Diffractionatanapertureisofgreatimportancebecausetheeye and many optical instruments have circular apertures

What is the half-angular width of the central maximum of the diffraction formed by a circular aperture This is not easy to calculate since it involves some advanced mathematics The problemwasfirstsolvedbytheEnglishAstronomerRoyalGeorgeAiryin1835whoshowedthatforcircularapertures

θ 122 λb--------------= where b is the diameter of the aperture

Example 2

Inthefollowingdiagramparallellightfromadistantpointsource(suchasastar)isbroughttofocusonthescreenSbyaconverginglens(the lens is shown as a vertical arrow)

Thefocal length(distance fromlens toscreen) is25cmand the diameter of the lens is 30 cm The wavelength of thelightfromthestaris560nmCalculatethediameterofthe diameter of the image on the screen

Solution

The lens actually acts as a circular aperture of diameter 30 cm The half angular width of central maximum of the diffraction pattern that it forms on the screen is

75

2

122 122 56 10 23 1030 10b

λθminus

minusminus

times times= = = times

times rad

The diameter of the central maxima is therefore

25 times 10-2 times 23 times 10-5

= 57 times 10-6 m

Although this is small it is still finite and is effectively the image of the star as the intensity of the secondary maxima are small compared to that of the central maximum

The resolution of simple monochromatic two-source systemsThe astronomers tell us that many of the stars that we observe with the naked eye are in fact binary stars That iswhatwe see as a single star actually consistsof twostarsinorbitaboutacommoncentreFurthermoretheastronomerstellusthatifweusealdquogoodrdquotelescopethenwewillactuallyseethetwostarsThatiswewillresolve the single point source into its two component parts Sowhatisitthatdetermineswhetherornotweseethetwo stars as a single point source ie what determines whether or not two sources can be resolved It canrsquot just bethatthetelescopemagnifiesthestarssinceiftheyareacting as point sources magnifying them is not going to make a great deal of difference

Ineachofoureyesthereisanaperturethepupilthroughwhich the light enters This light is then focused by the eye lens onto the retina But we have seen that when light passes through an aperture it is diffracted and so when we look at

NATURE OF SCIENCEImproved technology The Rayleigh criterion is the limit of resolution Continuing advancement in technology such as large diameter dishes or lenses or the use of smaller wavelength lasers pushes the limits of what we can resolve (18)

copy IBO 2014

PHYSICS 2015indb 299 150514 337 PM

Chapter 9

300

AH

L

apointsourceadiffractionpatternwillbeformedontheretina If we look at two point sources then two diffraction patterns will be formed on the retina and these patterns will overlap The width of our pupil and the wavelength of the light emitted by the sources will determine the amount by which they overlap But the degree of overlap will also depend on the angular separation of the two point sources WecanseethisfromFigures923

pupilretina

S1

S2

P2

P1

θ

Figure 923

LightfromthesourceS1 enters the eye and is diffracted by the pupil such that the central maximum of the diffraction patternisformedontheretinaatP1SimilarlylightfromS2producesamaximumatP2 If the two central maxima are well separated then there is a fair chance that we will see the two sources as separate sources If they overlap then we will not be able to distinguish one source from another From the diagram we see as the sources are movedclosertotheeyethentheangleθ increases and so does the separation of the central maxima

Figures924925926and927showsthedifferentdiffractionpatternsandtheintensitydistributionthatmightresultonthe retina as a result of light from two point sources

S1 S21 Well resolved

Figure 924 Very well resolved

S1 S22 Well resolved

Figure 925 Well resolved

S1 S23 Just resolved Rayleigh criterion Minimum of S2 coincides with maximum peak of S1

Figure 926 Just resolved

4 Not resolved

Figure 927 Not resolved

We have suggested that if the central maxima of the two diffraction patterns are reasonably separated then we should be able to resolve two point sources In the late 19th century Lord Rayleigh suggested by how much they should be separated in order for the two sources to be just resolved Ifthecentralmaximumofonediffractionpattern coincides with the firstminima of the otherdiffraction pattern then the two sources will just beresolved This is known as the Rayleigh Criterion

Figure926showsjustthissituationThetwosourcesarejust resolved according to the Rayleigh criterion since the peak of the central maximum of one diffraction pattern coincideswiththefirstminimumoftheotherdiffractionpattern This means that the angular separation of the peaks of the two central maxima formed by each source is just the half angular width of one central maximum ie

θ λb---

=

where b is the width of the slit through which the light fromthesourcespassesHoweverweseefromFigure1117that θ is the angle that the two sources subtend at the slit Henceweconcludethattwosourceswillberesolvedbyaslit if the angle that they subtend at the slit is greater than or equaltoλfraslb

PHYSICS 2015indb 300 150514 337 PM

Wave Phenomena

301

AH

L

Sofarwehavebeenassumingthattheeyeisarectangularslit whereas clearly it is a circular aperture and so we must use the formula

θ 122 λb--------------=

As mentioned above the angle θ is sometimes called the resolving power but should more accurately be called the minimum angle of resolution (θmin)

Clearly the smaller θ the greater will be the resolving power

The diffraction grating is a useful for differentiating closely spacedlinesinemissionspectraLikeaprismspectrometerit can disperse a spectrum into its components but it is better suited because it has higher resolution than the prism spectrometer

If λ1 and λ2 are two nearly equal wavelengths that canbarelybedistinguishedtheresolvanceorresolvingpowerofthegratingisdefinedas

R = λ _______ (λ2 ndash λ 1) = λ ___ Δλ

Where λ asymp λ1 asymp λ2

Therefore thediffractiongratingwithahighresolvancewill be better suited in determining small differences in wavelength

If the diffraction grating has Nlinesbeingilluminateditcan be shown that the resolving power of the mth order diffraction is given by

R = mN

Obviouslytheresolvancebecomesgreaterwiththeordernumber m and with a greater number of illuminated slits

Example

A benchmark for the resolving power of a grating is the sodium doublet in the sodium emission spectrum Two yellow lines in itsspectrumhavewavelengthsof58900nmand58959nm

(a) Determinetheresolvanceofagratingifthegivenwavelengths are to be distinguished from each other

(b) Howmanylinesinthegratingmustbeilluminated in order to resolve these lines in the second order spectrum

Solution

(a) R = λ ___ Δλ = 58900nm __________________ (58900ndash58959)nm

= 589nm _______ 059nm = 10 times 103

(b) N = R __ m = 10 times 10 3 ________ 2 =50times102 lines

It has been seen that diffraction effectively limits the resolving power of optical systems This includes such systems as the eye telescopes and microscopes Theresolving power of these systems is dealt with in the next section This section looks at links between technology and resolving power when looking at the very distant and when looking at the very small

Radio telescopes

The average diameter of the pupil of the human eye is about 25 mm This means that the eye will justresolve two point sources emitting light of wavelength 500 nm if their angular separation at the eye is 7

43

50 10122 24 1025 10

θminus

minusminus

times= times = times

times rad

If the eye were to be able to detect radio waves of wavelength 015m then tohave the same resolvingpower thepupilwould have to have a diameter of about 600 m Clearly this is nonsense but it does illustrate a problem facingastronomers who wish to view very distant objects such as quasarsandgalaxies thatemitradiowavesConventionalradio telescopes consist of a large dish typically 25 min diameter Even with such a large diameter the radiowavelength resolving power of the telescope is much less thantheopticalresolvingpowerofthehumaneyeLetuslook at an example

Example

The Galaxy Cygnus A can be resolved optically as an elliptically shaped galaxy However it is also a strongemitterof radiowavesofwavelength015mTheGalaxyisestimatedtobe50times1024mfromEarthUseofaradiotelescope shows that the radio emission is from two sources separated by a distance of 30 times 1021m Estimate thediameterofthedishrequiredtojustresolvethesources

Solution

The angle θ that the sources resolve at the telescope is given by

θ 21

424

30 10 60 1050 10

θ minustimes= = times

times rad

and 4

122 122 015 300060 10

d λθ minus

times= = =

timesm = 30 km

A radio telescope dish of this size would be impossible tomake letalonesupportThisshowsthatasingledishtype radio telescope cannot be used to resolve the sources and yet they were resolved To get round the problemastronomers use two radio telescopes separated by a large

PHYSICS 2015indb 301 150514 337 PM

Chapter 9

302

AH

L

distance The telescopes view the same objects at the same time and the signals that each receive from the objects are simultaneously superimposed The result of the superposition of the two signals is a two-slit interference pattern The pattern has much narrower fringe spacing than that of the diffraction pattern produced by either telescope on its own hence producing a much higherresolvingpowerWhentelescopesareusedlikethistheyare called a stellar interferometer

InSocorroinNewMexicothereisastellarinterferometerthat consists of 27 parabolic dishes each of diameter25 m arranged in a Y-shape that covers an area of 570km2Thisisaso-calledVeryLargeArray(VLA)Evenhigher resolution can be obtained by using an array of radio telescopes in observatories thousands of kilometres apart A system that uses this so-called technique oflsquovery-long-baseline interferometryrsquo (VLBI) is known asalsquovery-long-baselinearrayrsquo(VLBA)WithVLBAaradiowavelength resolving power can be achieved that is 100 timesgreaterthanthebestopticaltelescopesEvenhigherresolving power can be achieved by using a telescope that isinasatelliteorbitingEarthJapanrsquosInstituteofSpaceandAstronautical Science (ISAS) launched sucha system inFebruary 1997TheNationalAstronomicalObservatoryof Japan the National Science Foundationrsquos NationalRadio Astronomy Observatory (NRAO) the CanadianSpaceAgency theAustraliaTelescopeNationalFacilitythe European VLBI Network and the Joint Instituteback this project forVeryLongBaseline Interferometryin EuropeThis project is a very good example of howInternationalismcanoperateinPhysics

Electron microscope

Telescopes are used to look at very distant objects that are very large but because of their distance fromus appearverysmallMicroscopesontheotherhandareusedtolookat objects that are close to us but are physically very small Aswehave seen justmagnifyingobjects that ismakingthemappearlargerisnotsufficientonitsowntogaindetailabouttheobjectfordetailhighresolutionisneeded

Figure928isaschematicofhowanoptical microscope is usedtoviewanobjectandFigure929isaschematicofatransmission electron microscope(TEM)

Optical lens system

bright light source

glass slide containing specimen (object)

eye

Figure 928 The principle of a light microscope

magnetic lens system

electron source

thin wafer of material

ScreenCCD

Figure 929 The principle of an electron microscope

In the optical microscope the resolving power isdetermined by the particular lens system employed and the wavelength λ of the light used For example twopoints in the sample separated by a distance d will just be resolved if

2d mλ=

where m is a property of the lens system know as the numerical aperture In practice the largest value of m obtainable is about 16 Hence if the microscope slideis illuminatedwith light ofwavelength 480 nm a goodmicroscope will resolve two points separated by a distance dasymp15times10-7masymp015micromPointscloser together thanthiswillnotberesolvedHoweverthisisgoodenoughtodistinguishsomevirusessuchastheEbolavirus

Clearlythesmallerλ the higher the resolving power and this is where the electron microscope comes to the fore The electron microscope makes use of the wave nature of electrons(see1315)IntheTEMelectronspassthrougha wafer thin sample and are then focused by a magnetic field onto a fluorescent screen orCCD (charge coupleddevicesee142)Electronsusedinanelectronmicroscopehavewavelengthstypicallyofabout5times10-12mHoweverthe numerical aperture of electron microscopes is considerablysmaller thanthatofanopticalmicroscopetypicallyabout002NonethelessthismeansthataTEMcanresolvetwopointsthatareabout025nmapartThisresolving power is certainly high enough to make out the shape of large molecules

Another type of electron microscope uses a techniqueby which electrons are scattered from the surface of the sample The scattered electrons and then focused as in theTEM to form an image of the surfaceThese so-calledscanningelectronmicroscopes(SEM)havealowerresolving power than TEMrsquos but give very good threedimensional images

The eye

We saw in the last section that the resolving power of the human eye is about 2 times 10-4 rad Suppose that youare looking at car headlights on a dark night and the car

PHYSICS 2015indb 302 150514 337 PM

Wave Phenomena

303

AH

L

is a distance D away If the separation of the headlight is say 15 m then the headlights will subtend an angle θ 15

Dθ =

at your eye Assuming an average wavelength of

500 nm your eye will resolve the headlights into twoseparatesourcesifthisangleequals2times10-4 rad This gives D=75kmInotherwordsifthecarisapproachingyouona straight road then you will be able to distinguish the two headlightsasseparatesourceswhenthecaris75kmawayfrom you Actually because of the structure of the retina and optical defects the resolving power of the average eye is about 6 times 10-4 rad This means that the car is more likely tobe25kmawaybeforeyouresolvetheheadlights

Astronomical telescope

Letusreturntotheexampleofthebinarystarsdiscussedatthe beginning of this section on resolution The stars Kruger A and B form a binary system The average separation of the stars is 14 times 1012mandtheiraveragedistancefromEarthis12times1017mWhenviewedthroughatelescopeonEarththe system will therefore subtend an angle

12

17

14 1012 10

θtimes

=times

= 12times10-5 rad at the objective lens of the telescope Assuming that the average wavelength of the light emitted bythestarsis500nmthenifthetelescopeistoresolvethesystem into two separate images it must have a minimum diameter Dwhere12times10-5 =

7122 500 10D

minustimes times

This gives D=0050m which is about 5 cm So thisparticular system is easily resolved with a small astronomical telescope

Exercise 961 ItissuggestedthatusingtheISASVLBAitwould

bepossibletoldquoseerdquoagrainofriceatadistanceof5000kmEstimatetheresolvingpoweroftheVLBA

2 Thedistancefromtheeyelenstotheretinais20mm The light receptors in the central part of the retinaareabout5times10-6apartDeterminewhetherthe spacing of the receptors will allow for the eye to resolve the headlights in the above discussion whentheyare25kmfromtheeye

3 ThediameterofPlutois23times106 m and its averagedistancefromEarthis60times1012 m EstimatetheminimumdiameteroftheobjectiveofatelescopethatwillenablePlutotobeseenasadisc as opposed to a point source

95 Doppler effect

Essential ideaTheDopplereffectdescribesthephenomenonofwavelengthfrequencyshiftwhenrelativemotionoccurs

Understandingsbull TheDopplereffectforsoundwavesandlightwaves

The Doppler effect for sound waves and light waves

Consider two observers A and B at rest with respect to asoundsourcethatemitsasoundofconstantfrequencyf Clearly both observers will hear a sound of the same frequencyHoweversupposethatthesourcenowmovesat constant speed towards A A will now hear a sound of frequencyfA that is greater than f and B will hear a sound of frequency fB that is less than f This phenomenon is known as the Doppler Effect orDopplerPrinciple after C J Doppler(1803-1853)

The same effect arises for an observer who is either moving towards or away from a stationary source

Figure 930 shows the waves spreading out from a stationary source that emits a sound of constant frequency f The observersAandBhearasoundofthesamefrequency

A B

wavefront

S

Figure 930 Sound waves from a stationary source

Suppose now that the source moves towards A withconstant speed vFigure9310(a)showsasnapshotofthenew wave pattern

NATURE OF SCIENCETechnology Although originally based on physical observations of the pitch of fast moving sources of sound the Doppler effect has an important role in many different areas such as evidence for the expansion of the universe and generating images used in weather reports and in medicine (55)

copy IBO 2014

PHYSICS 2015indb 303 150514 337 PM

Chapter 9

304

AH

L

sourceA B

smallerwavelength

largerwavelength

Figure 931 (a) Sound waves from a moving source

The wavefronts are now crowded together in the direction of travel of the source and stretched out in the opposite direction This is why now the two observers will now hear notesofdifferentfrequenciesHowmuchthewavesbunchtogether and how much they stretch out will depend on the speed vEssentially A

A

cfλ

= and BB

cfλ

= where λA lt λB and v is the speed of sound

IfthesourceisstationaryandAismovingtowardsitthenthe waves from the source incident on A will be bunched up If A is moving away from the stationary source then the waves from the source incident on A will be stretched out

Christian Doppler(1803ndash1853)actuallyappliedtheprinciple(incorrectlyasithappens)totryandexplainthecolourofstars However the Doppler effect does apply to light aswellastosoundIfalightsourceemitsalightoffrequencyf then if it is moving away from an observer the observer willmeasurethelightemittedashavingalowerfrequencythan fSincethesensationofcolourvisionisrelatedtothefrequencyoflight(bluelightisofahigherfrequencythanred light) light emitted by objects moving way from anobserver is often referred to as being red-shifted whereas if the object is moving toward the observer it is referred to as blue-shiftedThisideaisusedinOptionE(Chapter16)

We do not need to consider here the situations where either the source or the observer is accelerating In a situation for example where an accelerating source is approaching astationaryobserverthentheobserverwillhearasoundof ever increasing frequency This sometimes leads toconfusion in describing what is heard when a source approaches passes and then recedes from a stationaryobserver Suppose for example that you are standingon a station platform and a train sounding its whistle is approaching at constant speed What will you hear as the train approaches and then passes through the station As the train approaches you will hear a sound of constant pitch but increasing loudness The pitch of the sound will be greater than if the train were stationary As the train passes through the station you will hear the pitch change at the momentthetrainpassesyoutoasoundagainofconstantpitch The pitch of this sound will be lower than the sound of the approaching train and its intensity will decrease as the train recedes from you What you do not hear is a sound of increasing pitch and then decreasing pitch

The Doppler equations for sound

Although you will not be expected in an IB examination to derivetheequationsassociatedwithaspectsoftheDopplereffectyouwillbeexpectedtoapplythemForcompletenesstherefore thederivationoftheequationsassociatedwiththeDopplereffectasoutlinedaboveisgivenhere

S primeS O

vs

Figure 931 (b)

InFigure931(b)theobserverOisatrestwithrespecttoasourceofsoundSismovingwithconstanspeedvs directly towards O The source is emitting a note of constant frequencyf and the speed of the emitted sound is v

S shows the position of the source ∆t later When the sourceisatresttheninatime∆t the observer will receive f∆t waves and these waves will occupy a distance v∆t ie

v f

λ =f∆tv∆t =

(Because of the motion of the source this number of waves will now occupy a distance (v∆t ndash vs∆t)The lsquonewrsquowavelength is therefore

v ndash vs

fλrsquo =

f∆tv∆t ndash vs∆t

=

If fisthefrequencyheardbyOthen

vfλ

= or

vf

λ = = sv vfminus

From which

s

vf fv v

=minus

Dividingthroughbyv gives

Equation1111

1 sf f v

v=

minus

If the source is moving away from the observer then we have

s

vf fv v

=minus

1

1 sf vv

=+

Equation112

PHYSICS 2015indb 304 150514 337 PM

Wave Phenomena

305

AH

L

We now consider the case where the source is stationary and the observerismoving towardsthesource with speed v0 In this situation the speed of the sound waves as measured by the observer will be v0 + v We therefore have that

v0 + v = ffraslλ = vfλ

times

From which

o1 vf fv

= +

Equation113

If the observerismovingawayfromthesource then

o1 vf fv

= minus

Fromequation(113)wehavethat

o o1 v vf f f f f fv v

∆ = minus = + minus =

Equation114

Thevelocitiesthatwerefertointheaboveequationsarethevelocities with respect to the medium in which the waves fromthesourcetravelHoweverwhenwearedealingwithalight source it is the relative velocity between the source and the observer that we must consider The reason for this is that lightisuniqueintherespectthatthespeedofthelightwavesdoes not depend on the speed of the source All observers irrespective of their speed or the speed of the source will measure the same velocity for the speed of light This is oneof thecornerstonesof theSpecialTheoryofRelativitywhichisdiscussedinmoredetailinOptionH(Chapter18)When applying the Doppler effect to light we aremainlyconcerned with the motion of the source We look here only at the situation where the speed of the source v is much smaller than the speed of light cinfreespace(v ltlt c)Underthesecircumstanceswhenthesourceismovingtowardstheobserverequation111becomes

vf f f fc

minus = ∆ = Equation115

andwhenthesourceismovingawayfromtheobserverequation112becomes vf f f f

cminus = ∆ = minus

Provided that v ltlt c these same equations apply for astationary source and moving observer

We look at the following example and exercise

Example

Asourceemitsasoundoffrequency440HzItmovesinastraight line towards a stationary observer with a speed of 30ms-1Theobserverhearsasoundoffrequency484HzCalculate the speed of sound in air

Solution

We use equation 111 and substitute f= 484 Hz f = 440 Hz

and vs = 30 m s-1

therefore 1

484 440 301v

=minus

such that 30 4401

484vminus = to

give v = 330 m s-1

Example

A particular radio signal from a galaxy is measured as havingafrequencyof139times109HzThesamesignalfromasourceinalaboratoryhasafrequencyof142times109Hz

Suggestwhy thegalaxy ismovingaway fromEarthandcalculateitsrecessionspeed(iethespeedwithwhichitismovingawayfromEarth)

Solution

The fact that the frequency from the moving source is less than that when it is stationary indicates that it is moving away from the stationary observer ie Earth

Using vf fc

∆ = we have

8 96

9

3 10 (142 139) 10 634 10142 10

c fvf∆ times times minus times

= = = timestimes

m s-1

It is usual when dealing with the Doppler effect of light to express speeds as a fraction of c So in this instance we have v = 0021 c

Using the Doppler effectWe have seen in the above example and exercise how the Dopplereffectmaybeusedtomeasuretherecessionspeedof distant galaxies The effect is also used to measure speed inothersituationsHerewewilllookatthegeneralprincipleinvolved in using the Doppler effect to measure speedFigure932showsasource(thetransmitter)thatemitseithersound or em waves of constant frequency f The waves

PHYSICS 2015indb 305 150514 337 PM

Chapter 9

306

AH

L

fromthesourceare incidentonareflectorthat ismovingtowards the transmitter with speed vThereflectedwavesare detected by the receiver placed alongside the transmitter

reector

v

transmitter

receiver

f f

f f

Figure 932 Using the Doppler effect to measure speed

We shall consider the situation where v ltlt c where c is the speed of the waves from the transmitter

For the reflector receiving waves from the transmitterit is effectively an observer moving towards a stationary sourceFromequation(114) itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation116

For the receiver receivingwaves from the reflector it iseffectively a stationary observer receiving waves from a movingsourceFromequation(115)itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation117

Ifweaddequations(116)and(117)wegetthatthetotalDopplershiftatthereceiver∆f is

v vf f f f fc c

minus = ∆ = +

But 1 vf fc

= +

hence

1 v v vf f fc c c

∆ = + +

But since v ltlt cwe can ignore the term2

2

vc when we

expandthebracketintheaboveequation

Therefore we have2vf fc

∆ = Equation118

If v asymp c then we must use the full Doppler equationsHowever for em radiationwewill alwaysonly considersituations in which v ltlt c

Example

Thespeedofsoundinbloodis1500times103 m s-1 Ultrasound offrequency100MHzisreflectedfrombloodflowinginan arteryThe frequencyof the reflectedwaves receivedbackatthetransmitteris105MHzEstimatethespeedofthebloodflowintheartery

Solution

Using equation (118) we have

6 63

2005 10 1015 10

vtimes = times

timesto give v asymp 36 m s-1 (We have assumed that the ultrasound is incident at right angles to the blood flow)

Exercise 971 Judy is standing on the platform of a station A

high speed train is approaching the station in a straight line at constant speed and is sounding its whistleAsthetrainpassesbyJudythefrequencyof the sound emitted by the whistle as heard by Judychangesfrom640Hzto430HzDetermine

(a) thespeedofthetrain

(b) thefrequencyofthesoundemittedbythewhistle as heard by a person on the train (Speedofsound=330ms-1)

2 AgalaxyismovingawayfromEarthwithaspeedof00500cThewavelengthofaparticularspectral line in light emitted by atomic hydrogen inalaboratoryis656times10-7 m Calculate the valueofthewavelengthofthislinemeasuredinthelaboratoryinlightemittedfromasourceofatomic hydrogen in the galaxy

PHYSICS 2015indb 306 150514 337 PM

10 Fields

Contents

101 ndash Describing fields

102 ndash Fields at work

Essential Ideas

Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Similar approaches can be taken in analysing electrical and gravitational potential problems copy IBO 2014

PHYSICS 2015indb 307 150514 337 PM

Chapter 10

308

AH

L

101 Describing fields

Essential idea Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Understandingsbull Gravitationalfieldsbull Electrostaticfieldsbull Electricpotentialandgravitationalpotentialbull Fieldlinesbull Equipotentialsurfaces

Gravitational potentialWehaveseenthatifweliftanobjectofmassm to a height h abovethesurfaceoftheEarththenitsgainingravitationalpotential energy is mghHoweverthisisbynomeansthefullstoryForastartwenowknowthatgvarieswithh and also the expression really gives a difference in potentialenergybetweenthevaluethattheobjecthasattheEarthrsquossurfaceandthevaluethatithasatheighthSowhatwereallyneedisazeropointCanwefindapointwherethepotentialenergyiszeroandusethispointfromwhichtomeasure changes in potential energy

The point that is chosen is in fact infinity At infinity the gravitationalfieldstrengthofanyobjectwillbezeroSoletusseeifwecandeduceanexpressionforthegaininpotentialenergyofanobjectwhenitisldquoliftedrdquofromthesurfaceoftheEarthtoinfinityThisineffectmeansfindingtheworknecessarytoperformthistask

δr

r r δr+

r infin=

mg

AB

Me

Figure 1001 Gravitational forces

Inthediagramweconsidertheworknecessarytomovethe particle of mass m a distance δr in thegravitationalfield of the Earth

The force on the particle at A is FG Mem

r2----------------=

IftheparticleismovedtoB then since δr isverysmallwe can assume that the field remains constant over thedistance ABTheworkδW done againstthegravitationalfieldoftheEarthinmovingthedistanceABis

δWGMem

r2----------------δrndash=

(rememberthatworkdoneagainstaforceisnegative)

TofindthetotalworkdoneW in going from the surface oftheEarthtoinfinitywehavetoaddalltheselittlebitsofworkThisisdonemathematicallybyusingintegralcalculus

WG Me m

r2----------------ndash

rdR

infin

int G Me m 1

r2----- rd

R

infin

intndash G Me m 1r---ndash

R

infinndash= = =

GMe m 0 1R---ndash

ndashndash=

G MemR----------------ndash=

HencewehavewhereRistheradiusoftheEarththattheworkdonebythegravitationalfieldinmovinganobjectofmass m from R(surfaceoftheEarth)toinfinityisgivenby

WGMe m

R---------------- ndash=

We can generalise the result by calculating the worknecessary per unit mass to take a small mass fromthe surface of the Earth to infinity This we call thegravitational potential Vie

Wewouldgetexactlythesameresultifwecalculatedtheworkdonebythefieldtobringthepointmassfrominfinityto the surface of Earth In this respect the formal definition ofgravitationalpotentialatapointinagravitationalfieldis therefore defined as the work done per unit mass in bringing a point mass from infinity to that point

Clearly then the gravitational potential at any point inthe Earthrsquos field distance r from the centre of the Earth (providingr gt R)is

VGMe

r------------ ndash=

Thepotentialisthereforeameasureoftheamountofworkthathastobedonetomoveparticlesbetweenpointsinagravitationalfieldanditsunit is theJkgndash1 We also note thatthepotentialisnegativesothatthepotentialenergy

NATURE OF SCIENCEParadigm shift The move from direct observable actions being responsible for influence on an object to acceptance of a fieldrsquos ldquoaction at a distancerdquo required a paradigm shift in the world of science (23)

copy IBO 2014

PHYSICS 2015indb 308 150514 337 PM

Fields

309

AH

L

aswemoveawayfromtheEarthrsquossurfaceincreasesuntilitreachesthevalueofzeroatinfinity

IfthegravitationalfieldisduetoapointmassofmassmthenwehavethesameexpressionasaboveexceptthatMe is replaced by m andmustalsoexclude thevalueof thepotential at the point mass itself ie at r = 0

WecanexpressthegravitationalpotentialduetotheEarth(orduetoanysphericalmass)intermsofthegravitationalfield strength at its surface

AtthesurfaceoftheEarthwehave

g 0ndash Re GMRe

---------ndash= e

Sothat

g 0Re2 GM= e

Hence at a distance r from the centre of the Earth the gravitationalpotentialVcanbewrittenas

VGMe

r------------ndashg 0 Re

2

r------------ndash= =

The potential at the surface of the Earth

(r = Re)istherefore-g0Re

It is interesting to see how the expression for thegravitational potential ties in with the expression mgh ThepotentialatthesurfaceoftheEarthis-g0Re (see the exampleabove)andataheighthwillbe-g0g 0 Re h+( )ndash ifweassume that g0doesnotchangeover thedistanceh The differenceinpotentialbetweenthesurfaceandtheheighth is therefore g0hSotheworkneededtoraiseanobjectof mass m to a height h is mgh ie m timesdifference ingravitationalpotential

This we have referred to as the gain in gravitationalpotentialenergy(see235)

However this expression can be extended to any twopoints inanygravitationalfieldsuch that ifanobjectofmass mmovesbetween twopointswhosepotentialsareV1 and V2 respectively then the change in gravitationalpotentialenergyoftheobjectism(V1 ndash V2)

Gravitational potential gradient

Let us consider now a region in space where thegravitationalfieldisconstantInFigure912thetwopointsAandBareseparatedbythedistance∆x

A B

direction of uniform gravitational eld of strength I

∆x

Figure 1002 The gravitational potential gradient

The gravitational field is of strength I and is in the direction shownThe gravitational potential at A is V andatBisV +∆V

Theworkdone is takingapointmassmfromAtoB isF∆x = mI∆x

Howeverbydefinitionthisworkisalsoequalto-m∆V

Therefore mI∆x=-m∆V

or VIx

∆= minus

∆Effectivelythissaysthatthemagnitudeofthegravitationalfield strength is equal to the negative gradient of thepotential If I is constant then V is a linear function of x and Iisequaltothenegativegradientofthestraightlinegraph formed by plotting V against x If I is not constant (asusuallythecase)thenthemagnitudeofI at any point in the field can be found by find the gradient of the V-x graph at that point An example of such a calculation can befoundinSection929

For those of you who do HL maths the relationship between field and potential is seen to follow from the expression for the potential of a point mass viz

mV Gr

= minus

2

ddV mG Ir r

minus = + =

Potential due to one or more point masses

Gravitationalpotentialisascalarquantitysocalculatingthepotential due to more than one point mass is a matter of simpleadditionSoforexamplethepotentialV due to the Moon and Earth and a distance x fromthecentreofEarthonastraightlinebetweenthemisgivenbytheexpression

E MM MV Gx r x

= minus + minus

whereME=massofEarthMM= mass of Moon and r = distancebetweencentreofEarthandMoon

PHYSICS 2015indb 309 150514 337 PM

Chapter 10

310

AH

L

Equipotentials and field linesIfthegravitationalpotentialhasthesamevalueatallpointson a surface the surface is said to be an equipotential surfaceSoforexampleifweimagineasphericalshellaboutEarthwhosecentrecoincideswiththecentreofEarththisshellwillbeanequipotentialsurfaceClearlyifwerepresentthegravitationalfieldstrengthbyfieldlinessincethelinesldquoradiaterdquooutfromthecentreofEarththentheselineswillbeatrightanglestothesurfaceIfthefieldlineswerenotnormaltotheequipotentialsurfacethentherewouldbeacomponentof the field parallel to the surfaceThis wouldmean thatpointsonthesurfacewouldbeatdifferentpotentialsandsoitwouldnolongerbeanequipotentialsurfaceThisofcourseholdstrueforanyequipotentialsurface

Figure 1003 shows the field lines and equipotentials fortwopointmassesm

m m

Figure 1003 Equipotentials for two point masses

It is worth noting that we would get exactly the samepattern ifwewere to replace thepointmasseswith twoequalpointcharges(See935)

Escape speedThe potential at the surface of Earth is which meansthattheenergyrequiredtotakeaparticleofmass mfromthesurfacetoinfinityisequalto

ButwhatdoesitactuallymeantotakesomethingtoinfinityWhentheparticleisonthesurfaceoftheEarthwecanthinkofitassittingatthebottomofaldquopotentialwellrdquoasinfigure1004

particle

surface of Earth

innity

GMR---------

FIgure 1004 A potential well

Theldquodepthrdquoofthewellis and if the particle gains an amount of kinetic energy equal to wherem is its massthenitwillhavejustenoughenergytoldquoliftrdquoitoutofthewell

In reality it doesnrsquot actually go to infinity it justmeansthat the particle is effectively free of the gravitationalattractionof theEarthWe say that ithas ldquoescapedrdquo theEarthrsquosgravitationalpullWemeetthisideainconnectionwithmolecular forcesTwomolecules in a solidwill sitat their equilibrium position the separation where therepulsiveforceisequaltotheattractiveforceIfwesupplyjust enough energy to increase the separation of themolecules such that they are an infinite distance apart then intermolecularforcesnolongeraffectthemoleculesandthesolidwillhavebecomealiquidThereisnoincreaseinthekineticenergyofthemoleculesandsothesolidmeltsat constant temperature

Wecancalculatetheescapespeedofanobjectveryeasilyby equating the kinetic energy to the potential energysuch that

12---mvesc ape

2 GMe mRe

----------------=

vesc aperArr2GMe

Re--------------- 2g0 Re = =

Substituting for g0 and Regivesavalueforvescape of about 11kmsndash1 from the surface of the Earth

Youwillnotethattheescapespeeddoesnotdependonthemassoftheobjectsincebothkineticenergyandpotentialenergy are proportional to the mass

IntheoryifyouwanttogetarockettothemoonitcanbedonewithoutreachingtheescapespeedHoweverthiswouldnecessitate anenormousamountof fuel and it islikely that therocketplus fuelwouldbesoheavythat itwouldnevergetoffthegroundItismuchmorepracticalto accelerate the rocket to the escape speed fromEarthorbitandthenintheoryjustlaunchittotheMoon

Example

Use the followingdata todetermine thepotential at thesurface of Mars and the magnitude of the acceleration of free fall

massofMars =64times1023kg

radiusofMars =34times106 m

Determinealsothegravitationalfieldstrengthatadistanceof 68 times 106mabovethesurfaceofMars

PHYSICS 2015indb 310 150514 337 PM

Fields

311

AH

L

Solution

= minus = minus times times = times23

11 76

64 1067 10 ndash13 10

34 10MV GR

minus timestimes

Nkg-1

But V = -g0R

Therefore 7

0 6

13 10 3834 10

VgR

times= minus = =

timesm s-2

To determine the field strength gh at 68 times 106 m above the surface we use the fact that 0 2

Mg GR

= such that GM = g0R2

Therefore 2 2

0h 2 2 2

h h

38 (34) 042(102)

g RGMgR R

times= = = = m s-2

(the distance from the centre is 34 times 106 + 68 times 106 = 102 times 106 m)

Exercise 1011 Thegraphbelowshowshowthegravitational

potentialoutsideoftheEarthvarieswithdistancefrom the centre

0 10 20 30 40 50 60 70

ndash1

ndash2

ndash3

ndash4

ndash5

ndash6

ndash7

r m times 106

V

Jkgndash1

times 10

7

(a) Usethegraphtodeterminethegainingravitationalpotentialenergyofasatelliteofmass200kgasitmovesfromthesurfaceof the Earth to a height of 30 times 107mabovetheEarthrsquossurface

(b) Calculatetheenergyrequiredtotakeit to infinity

(c) DeterminetheslopeofthegraphatthesurfaceoftheEarthmCommenton youranswer

Electric potential energy

The concept of electric potential energy was developedwith gravitational potential energy in mind Just as anobjectnearthesurfaceoftheEarthhaspotentialenergybecause of its gravitational interaction with the Earthsotoothereiselectricalpotentialenergyassociatedwithinteracting charges

Letusfirstlookatacaseoftwopositivepointchargeseachof 1μC that are initially bound together by a thread in a vacuuminspacewithadistancebetweenthemof10cmasshowninFigure1005Whenthethreadiscutthepointchargesinitiallyatrestwouldmoveinoppositedirectionsmovingwithvelocitiesv1 and v2 along the direction of the electrostatic force of repulsion

BEFORE AFTER

v v21

Figure 1005 Interaction of two positive particles

The electric potential energybetweentwopointchargescan be found by simply adding up the energy associated witheachpairofpointchargesForapairofinteractingchargestheelectricpotentialenergyisgivenby

∆U = ∆Ep + ∆Ek = ∆W = ∆Fr = kqQ

____ r2 times r = kqQ

____ r

Becausenoexternalforceisactingonthesystemtheenergyand momentum must be conserved Initially Ek = 0 and Ep =kqQr=9times109 times 1 times 10-1201m=009JWhentheyareagreatdistance fromeachotherEpwillbenegligibleThefinalenergywillbeequaltofrac12mv1

2+frac12mv22=009J

Momentum isalsoconservedand thevelocitieswouldbethe same magnitude but in opposite directions

Electric potential energyismoreoftendefinedintermsofapointchargemovinginanelectricfieldas

lsquothe electricpotential energybetweenany twopoints inan electric field is defined as negativeoftheworkdone by anelectricfieldinmovingapoint electric chargebetweentwolocationsintheelectricfieldrsquo

ΔU = ΔEp = -ΔW = -Fd = qEx

ΔU = ΔEp +ΔEk = ΔW = ΔFr = kqQ r2 x r = kqQ

wherexisthedistancemovedalong(oroppositeto)thedirection of the electric field

Electric potential energyismeasuredinjoule(J)Justasworkisascalarquantitysotooelectricalpotentialenergyis a scalar quantity The negative of the work done by an electric field in moving a unit electric charge between two points is independent of the path taken Inphysicswesaytheelectricfieldisalsquoconservativersquo field

PHYSICS 2015indb 311 150514 337 PM

Chapter 10

312

AH

L

Supposeanexternalforcesuchasyourhandmovesasmallpositive point test charge in the direction of a uniformelectric fieldAs it ismoving itmust be gaining kineticenergyIfthisoccursthentheelectricpotentialenergyofthe unit charge is changing

InFigure1006apointcharge+qismovedbetweenpointsAandBthroughadistancex in a uniform electric field

B

A

x

+ q

Figure 1006 Movement of a positive point charge in a uniform field

InordertomoveapositivepointchargefrompointAtopointBanexternal forcemustbeapplied to thechargeequaltoqE (F = qE)

Since the force is applied through a distance xthennegativeworkhas to bedone tomove the charge because energyis gainedmeaning there is an increaseelectric potential energybetweenthetwopointsRememberthattheworkdoneisequivalenttotheenergygainedorlostinmovingthe charge through the electric field The concept of electric potential energy is only meaningful as the electric field whichgeneratestheforceinquestionisconservative

W F xtimes E q x times= =

θ x cos θ x

Figure 1007 Charge moved at an angle to the field

If a chargemoves at an angle θ to an electric field thecomponent of the displacement parallel to the electric fieldisusedasshowninFigure918

W F x Eq x θcostimes= =

Theelectricpotentialenergyisstoredintheelectricfieldand theelectricfieldwill return theenergy to thepointcharge when required so as not to violate the Law ofconservationofenergy

Electric potential

The electric potential at a point in an electric field is defined as being the work done per unit charge in bringing a small positive point charge from infinity to that point

ΔV = Vinfin ndash Vf = ndash W ___ q

Ifwedesignatethepotentialenergytobezeroatinfinitythen it follows thatelectricpotentialmustalsobezeroat infinity and the electric potential at any point in an electricfieldwillbe

ΔV = ndash W ___ q

NowsupposeweapplyanexternalforcetoasmallpositivetestchargeasitismovedtowardsanisolatedpositivechargeTheexternalforcemustdoworkonthepositivetestchargetomoveittowardstheisolatedpositivechargeandtheworkmustbepositivewhiletheworkdonebytheelectricfieldmustthereforebenegativeSotheelectricpotentialatthatpointmustbepositiveaccordingtotheaboveequationIfanegativeisolatedchargeisusedtheelectricpotentialatapointonthepositivetestchargewouldbenegativePositivepoint chargesof their ownaccordmove fromaplaceofhighelectricpotential toaplaceof lowelectricpotentialNegative point charges move the other way from lowpotentialtohighpotentialInmovingfrompointAtopointBinthediagramthepositivecharge+qismovingfromalowelectricpotentialtoahighelectricpotential

In thedefinitiongiven the termldquoworkperunitchargerdquohas significance If the test charge is +16 times 10-19Cwherethe charge has a potential energy of 32 times 10-17Jthenthepotentialwouldbe32times10-17J+16times10-19C=200JC-1 Now if the charge was doubled the potential wouldbecome 64times10-17 J However the potential per unitchargewouldbethesame

Electric potential is a scalar quantityandithasunitsJC-1 orvoltswhere1voltequalsone joulepercoloumbThevoltallowsustoadoptaunitfortheelectricfieldintermsofthevolt

PreviouslytheunitfortheelectricfieldwasNCndash1

W = qV and F = qEsoW ___ V = F __ E

E = FV ___ W = FV ___ Fm V mndash1

That is the units of the electric field E can also beexpressed as V mndash1

The work done per unit charge in moving a point charge between two points in an electric field is again independent of the path taken

PHYSICS 2015indb 312 150514 337 PM

Fields

313

AH

L

Electric potential due to a point charge

Letus takeapoint rmetres froma chargedobjectThepotentialatthispointcanbecalculatedusingthefollowing

W = ndashFr = ndashqV and F = ndash q1 q2 _____ 4πε0r

2

Therefore

W = ndash q1 q2 _____ 4πε0r

2 times r = ndash q1 q2 _____ 4πε0r

= ndashq1 times q2 _____ 4πε0r

= ndashq1V

That is V q

4πε0r--------------- =

Orsimply

V kqr------

=

Example

Determinehowmuchworkisdonebytheelectricfieldofpointcharge150μCwhenachargeof200μCismovedfrom infinity to apoint 0400m from thepoint charge(Assumenoaccelerationofthecharges)

Solution

The work done by the electric field is W = -qV = -14πε0 times q times (Q rinfin - Q r0400)

W = (- 200 times 10-6 C times 900 times 109 NmC-2 times 150 times 10-6 C) divide 0400 m = - 0675 J

An external force would have to do +0675 J of work

Electric field strength and electric potential gradient

Letus lookbackatFigure1006Supposeagain that thecharge+qismovedasmalldistancebyaforceFfromAtoBsothattheforcecanbeconsideredconstantTheworkdoneisgivenby

∆W F x∆times=

TheforceFandtheelectricfieldEareoppositelydirectedandweknowthat

F = -qE and ΔW = q ΔV

Thereforetheworkdonecanbegivenas

q ΔV = -q E Δx

Therefore E V∆x∆-------

ndash=

The rate of change of potential ΔV at a point with respect to distance Δx in the direction in which the change is maximum is called the potential gradient We say that the electric field = - the potential gradient and the units are Vm-

1 From the equation we can see that in a graph of electric potential versus distance the gradient of the straight line equals the electric field strength

In reality if a charged particle enters a uniform electric field it will be accelerated uniformly by the field and its kinetic energy will increase This is why we had to assume no acceleration in the last worked example

Ek∆ 12---

mv2 q E xsdot sdot q Vx---

xsdot sdot q Vsdot= = = =

Example

Determine how far apart two parallel plates must besituated so that a potential difference of 150 times 102 V produces an electric field strength of 100 times 103 NC-1

Solution

Using E V∆x------- xhArrndash V∆

E-------15 102 Vtimes

100 103 N C 1ndashtimes------------------------------------------= = =

= 150 times 10-1

The plates are 150 times 10-1 m apart

The electric field and the electric potential at a point due to an evenly distributed charge +q on a sphere can be represented graphically as in Figure 1008

r

r

V

E

r0

r0

r0

Charge of +Q evenly distributed over surface

E 14πε0------------ Q

r2----- r r0gt=

V 14πε0------------Q

r----r r0gt=

Figure 1008 Electric field and potential due to a charged sphere

PHYSICS 2015indb 313 150514 337 PM

Chapter 10

314

AH

L

When the sphere becomes charged we know that thechargedistributes itselfevenlyoverthesurfaceThereforeeverypartofthematerialoftheconductorisatthesamepotential As the electric potential at a point is defined as beingnumericallyequaltotheworkdoneinbringingaunitpositivechargefrominfinitytothatpointithasaconstantvalueineverypartofthematerialoftheconductor

Since the potential is the same at all points on the conducting surfacethen∆V∆xiszeroButE=ndash∆V∆xThereforethe electric field inside the conductor is zero There is no electric field inside the conductor

SomefurtherobservationsofthegraphsinFigure1009are

bull OutsidethespherethegraphsobeytherelationshipsgivenasE α 1 r2 and V α 1 r

bull Atthesurfacer = r0Thereforetheelectricfieldandpotentialhavetheminimumvalueforr at this point and this infers a maximum field and potential

bull Insidethespheretheelectricfieldiszerobull Insidethespherenoworkisdonetomoveacharge

fromapointinsidetothesurfaceThereforethereisno potential difference and the potential is the same asitiswhenr = r0

Similargraphscanbedrawnfortheelectricfieldintensityandthe electric potential as a function of distance from conducting parallelplatesandsurfacesandthesearegiveninFigure1009

Potential plot E eld plotE eld

++++

ndashndashndashndash

xx x

+

x x

ndashx x

+ ndashx

x

+ +

x

x

Figure 1009 Electric field and electric potential at a distance from a charged surface

Potential due to one or more point charges

The potential due to one point charge can be determined byusingtheequationformula

V = kq r

Example 1

Determine the electric potential at a point 20 times 10-1 m fromthecentreofan isolatedconducting spherewithapointchargeof40pCinair

Solution

Using the formula

V = kq r we have

V90 10 9times( ) 40 10 12ndashtimes( )times

20 10 1ndashtimes( )------------------------------------------------------------------ 018 V= =

the potential at the point is 180 times 10-1 V

The potential due to a number of point charges can be determined by adding up the potentials due to individual point charges because the electric potential at any point outside a conducting sphere will be the same as if all the charge was concentrated at its centre

Example 2

Threepointchargesofareplacedattheverticesofaright-angledtriangleasshowninthediagrambelowDeterminetheabsolutepotentialat the+20μCchargeduetothetwoothercharges

- 6μC

+3μC +2μC

4 m

3 m

Solution

The electric potential of the +2 μC charge due to the ndash 6 μC charge is

V = (9 times 109 Nm2C-2 times -6 times 10-6 C) divide (radic 32 + 42) m = - 108 times 104 V

PHYSICS 2015indb 314 150514 338 PM

Fields

315

AH

L

Insummarywecanconcludethat

bull No work is done to move a charge along an equipotential

bull Equipotentials are always perpendicular to the electric lines of force

Figure1011and1012showsomeequipotential lines fortwo oppositely charged and identically positive spheresseparated by a distance

+ve ndashve

equipotential lines

Figure 1011 Equipotential lines between two opposite charges

equipotential lines

Figure 1012 Equipotential lines between two charges which are the same

+

ndash ndash ndash ndash

+ + +

40 V

30 V

20 V

10 V

50 V

Figure 1013 Equipotential lines between charged parallel plates

Figure1013showstheequipotentiallinesbetweenchargedparallel plates Throughout this chapter the similarities and differences between gravitational fields and electric fieldshavebeendiscussedTherelationshipsthatexistsbetweengravitationalandelectricquantitiesandtheeffectsofpointmassesandchargesissummarisedinTable1014

The electric potential of the +2 μC charge due to the +3 μC charge is

V = (9 times 109 Nm2C-2 times 3 times 10-6 C) divide 3m = 9 times 103 V

The net absolute potential is the sum of the 2 potentials- 108 times 104 V + 9 times 103 V =- 18 times 103 V

The absolute potential at the point is - 18 times 103 V

Equipotential surfacesRegions in space where the electric potential of a chargedistributionhasaconstantvaluearecalledequipotentials The placeswherethepotentialisconstantinthreedimensionsarecalled equipotential surfacesandwheretheyareconstantintwodimensionstheyarecalledequipotential lines

TheyareinsomewaysanalogoustothecontourlinesontopographicmapsInthiscasethegravitationalpotentialenergy is constant as amassmoves around the contourlines because the mass remains at the same elevationabovetheEarthrsquossurfaceThegravitationalfieldstrengthacts in a direction perpendicular to a contour line

Similarlybecausetheelectricpotentialonanequipotentiallinehas thesamevalueanelectric forcecandonoworkwhena testchargemovesonanequipotentialThereforethe electric field cannot have a component along anequipotentialandthusitmustbeeverywhereperpendiculartotheequipotentialsurfaceorequipotentiallineThisfactmakesiteasytoplotequipotentialsifthelinesofforceorlinesofelectricfluxofanelectricfieldareknown

Forexamplethereareaseriesofequipotentiallinesbetweentwoparallelplateconductorsthatareperpendiculartotheelectric fieldTherewill be a series of concentric circles(eachcirclefurtherapartthanthepreviousone)thatmapouttheequipotentialsaroundanisolatedpositivesphereThe lines of force and some equipotential lines for anisolatedpositivesphereareshowninFigure1010

Lines ofequipotential

Figure 1010 Equipotentials around an isolated positive sphere

PHYSICS 2015indb 315 150514 338 PM

Chapter 10

316

AH

L

Gravitational quantity Electrical quantityQ

uant

ities

g V∆x∆-------ndash= E V∆

x∆-------ndash=

Poin

t mas

ses a

nd ch

arge

s

V Gmr----

ndash= V 14πε0------------q

r---=

g G m

r2-----ndash= E 14πε0------------ q

r2-----=

F Gm1 m2

r2--------------= F 1

4πε0------------

q1q2

r2-----------=

Figure 1014 Formulas

Example

Deduce the electric potential on the surface of a gold nucleus that has a radius of 62 fm

Solution

Using the formula

V = kq r and knowing the atomic number of gold is 79 We will assume the nucleus is spherical and it behaves as if it were a point charge at its centre (relative to outside points)

V = 90 times 109 Nm2C-2 times 79 times 16 times 10-19 C divide 62 times 10-15 m = 18 times 107 V

The potential at the point is 18 MV

Example

Deduce the ionisation energy in electron-volts of theelectron in the hydrogen atom if the electron is in its ground state and it is in a circular orbit at a distance of 53 times 10-11 m from the proton

Solution

This problem is an energy coulombic circular motion question based on Bohrrsquos model of the atom (not the accepted quantum mechanics model) The ionisation energy is the energy required to remove the electron from the ground state to infinity The electron travels in a circular orbit and therefore has a centripetal acceleration The ionisation energy will counteract the coulombic force and the movement of the electron will be in the opposite direction to the centripetal force

Total energy = Ek electron + Ep due to the proton-electron interaction

ΣF = kqQ r2 = mv2 r and as such mv2 = = kqQ r

Therefore Ek electron = frac12 kqQ r

Ep due to the proton-electron interaction = - kqQ r

Total energy = frac12 kqQ r + - kqQ r = -frac12 kqQ r

= - 90 times 109 Nm2C-2 times (16 times 10-19 C)2 divide 53 times 10-11 m = -217 times 10-18 J

= -217 times 10-18 J divide 16 times 10-19 = -136 eV

The ionisation energy is 136 eV

Exercise 1021 ApointchargePisplacedmidwaybetweentwo

identicalnegativechargesWhichoneofthefollowingiscorrectwithregardstoelectricfieldandelectricpotentialatpointP

Electric field Electric potentialA non-zero zeroB zero non-zeroC non-zero non-zeroD zero zero

2 Twopositivechargedspheresaretiedtogetherinavacuumsomewhereinspacewheretherearenoexternal forces A has a mass of 25 g and a charge of20μCandBhasamassof15gandachargeof30μCThedistancebetweenthemis40cmTheyarethenreleasedasshowninthediagram

BEFORE AFTER

v1 v2

A B

(a) Determinetheirinitialelectricpotentialenergy in the before situation

(b) DeterminethespeedofsphereBafterrelease

3 Thediagrambelowrepresentstwoequipotentiallines in separated by a distance of 5 cm in a uniform electric field

+ + + + + + + +

ndash ndash ndash ndash ndash ndash ndash ndash

40 V

20 V 5 cm

PHYSICS 2015indb 316 150514 338 PM

Fields

317

AH

L

Determine the strength of the electric field

4 Thisquestionisabouttheelectricfieldduetoacharged sphere and the motion of electrons in that fieldThediagrambelowshowsanisolatedmetalsphereinavacuumthatcarriesanegativeelectriccharge of 60 μC

ndash

(a) Onthediagramdrawtheconventionalwayto represent the electric field pattern due to the charged sphere and lines to represent threeequipotentialsurfacesintheregionoutside the sphere

(b) Explainhowthelinesrepresentingtheequipotentialsurfacesthatyouhavesketchedindicatethatthestrengthoftheelectricfieldisdecreasingwithdistancefrom the centre of the sphere

(c) Theelectricfieldstrengthatthesurfaceofthe sphere and at points outside the sphere can be determined by assuming that the sphere acts as a point charge of magnitude 60 μC at its centre The radius of the sphere is 25 times 10ndash2 m Deduce that the magnitude of the field strength at the surface of the sphere is 86 times 107 Vmndash1

An electron is initially at rest on the surface of the sphere

(d) (i) Describethepathfollowedbytheelectronasitleavesthesurfaceofthesphere

(ii) Calculatetheinitialaccelerationofthe electron

5 Determinetheamountofworkthatisdoneinmovingachargeof100nCthroughapotentialdifferenceof150times102 V

6 Three identical 20 μC conducting spheres are placedatthecornersofanequilateraltriangleofsides 25 cm The triangle has one apex C pointing upthepageand2baseanglesAandBDeterminetheabsolutepotentialatB

7 Determinehowfaraparttwoparallelplatesmustbesituatedsothatapotentialdifferenceof250 times 102 V produces an electric field strength of 200 times 103 NC-1

8 Thegapbetweentwoparallelplatesis10times10-3mandthereisapotentialdifferenceof10times104 V betweentheplatesCalculate

i theworkdonebyanelectroninmovingfrom one plate to the other

ii thespeedwithwhichtheelectronreachesthe second plate if released from rest

iii theelectricfieldintensitybetweentheplates

9 Anelectronguninapicturetubeisacceleratedby a potential 25 times 103VDeterminethekineticenergygainedbytheelectroninelectron-volts

10 Determine the electric potential 20 times 10-2 m from achargeof-10times10-5 C

11 Determinetheelectricpotentialatapointmid-waybetweenachargeofndash20pCandanotherof+5pConthelinejoiningtheircentresifthecharges are 10 cm apart

12 During a thunderstorm the electric potential differencebetweenacloudandthegroundis10 times 109 V Determine the magnitude of the change in electric potential energy of an electron thatmovesbetweenthesepointsinelectron-volts

13 A charge of 15 μC is placed in a uniform electric fieldoftwooppositelychargedparallelplateswithamagnitudeof14times103 NC-1

(a) Determinetheworkthatmustbedoneagainstthefieldtomovethepointchargeadistance of 55 cm

(b) Calculatethepotentialdifferencebetweenthe final and initial positions of the charge

(c) Determinethepotentialdifferencebetweenthe plates if their separation distance is 15 cm

14 Duringaflashoflightningthepotentialdifferencebetweenacloudandthegroundwas12 times 109 V and the amount of transferred charge was32C

(a) Determinethechangeinenergyofthetransferred charge

(b) Iftheenergyreleasedwasallusedtoacceleratea1tonnecardeduceitsfinalspeed

(c) Iftheenergyreleasedcouldbeusedtomelticeat0degCdeducetheamountoficethatcould be melted

PHYSICS 2015indb 317 150514 338 PM

Chapter 10

318

AH

L

15 SupposethatwhenanelectronmovedfromAtoBin the diagram along an electric field line that the electric field does 36 times 10-19Jofworkonit

Determinethedifferencesinelectricpotential

(a) VB ndash VA(b) VC ndash VA(c) VC ndash VB

16 DeterminethepotentialatpointPthatislocatedatthecentreofthesquareasshowninthediagrambelow

- 6μC

+3μC +2μC

1m

1m

5μC

P

102 Fields at work

Essential idea Similar approaches can be taken inanalysingelectricalandgravitationalpotentialproblems

Understandingsbull Potentialandpotentialenergybull Potentialgradientbull Potentialdifferencebull Escapespeedbull Orbitalmotionorbitalspeedandorbitalenergybull Forcesandinverse-squarelawbehaviour

Orbital motion orbital speed and orbital energy

Although orbital motion may be circular elliptical orparabolic this sub-topic only dealswith circular orbitsThis sub-topic is not fundamentally new physics butan application that synthesizes ideas from gravitationcircularmotiondynamicsandenergy

TheMoonorbitstheEarthandinthissenseitisoftenreferredtoasasatelliteoftheEarthBefore1957itwastheonlyEarthsatelliteHoweverin1957theRussianslaunchedthefirstmanmadesatelliteSputnik1SincethisdatemanymoresatelliteshavebeenlaunchedandtherearenowliterallythousandsofthemorbitingtheEarthSomeareusedtomonitortheweathersome used to enable people to find accurately their position on thesurfaceoftheEarthmanyareusedincommunicationsandnodoubtsomeareusedtospyonothercountriesFigure1015showshowinprincipleasatellitecanbeputintoorbit

Theperson(whosesizeisgreatlyexaggeratedwithrespectto Earth) standing on the surface on the Earth throwssome stonesThe greater the speed with which a stoneis thrown the further it will land from her The pathsfollowedbythethrownstonesareparabolasByastretchof the imaginationwecanvisualisea situation inwhichastoneis thrownwithsuchaspeedthatbecauseof thecurvatureoftheEarthitwillnotlandonthesurfaceoftheEarthbutgointoldquoorbitrdquo(Path4onFigure1015)

NATURE OF SCIENCECommunication of scientific explanations The ability to apply field theory to the unobservable (charges) and the massively scaled (motion of satellites) required scientists to develop new ways to investigate analyse and report findings to a general public used to scientific discoveries based on tangible and discernible evidence (51)

copy IBO 2014

PHYSICS 2015indb 318 150514 338 PM

Fields

319

AH

L

1

2

34Earth

Figure 1015 Throwing a stone into orbit

TheforcethatcausesthestonestofollowaparabolicpathandtofalltoEarthisgravityandsimilarlytheforcethatkeeps the stone in orbit is gravity For circular motionto occur we have seen that a force must act at rightanglestothevelocityofanobjectthatistheremustbeacentripetalforceHenceinthesituationwedescribeherethe centripetal force for circular orbital motion about the EarthisprovidedbygravitationalattractionoftheEarth

We can calculate the speedwithwhich a stonemustbethrowninordertoputitintoorbitjustabovethesurfaceof the Earth

If the stone has mass m and speed v thenwehave fromNewtonrsquos2ndlaw

mv2

RE---------- G

MEm

RE2------------=

whereRE is the radius of the Earth and ME is the mass of the Earth

Bearinginmindthatg 0 GME

RE2--------= then

v g RE 10 64 106timestimes 8 103times= = =

Thatisthestonemustbethrownat8times103m sndash1

Clearly we are not going to get a satellite into orbit soclosetothesurfaceoftheEarthMovingatthisspeedthefrictionduetoairresistancewouldmeltthesatellitebeforeithadtravelledacoupleofkilometresInrealitythereforeasatelliteisputintoorbitabouttheEarthbysendingitattachedtoarocketbeyondtheEarthrsquosatmosphereandthengivingitacomponentofvelocityperpendiculartoaradialvectorfromtheEarthSeeFigure1016

Earth Satellite carried by rocket to hereSatellite orbit

Tangential component of velocity

Figure 1016 Getting a satellite into orbit

Keplerrsquos third law(This work of Kepler and Newtonrsquos synthesis of the work is an excellent example of the scientific method and makes for a good TOK discussion)

In 1627 Johannes Kepler (1571-1630)publishedhis lawsof planetary motion The laws are empirical in natureandwere deduced from the observations of theDanishastronomer Tycho de Brahe (1546-1601)The third lawgives a relationship between the radius of orbit R of a planet and its period TofrevolutionabouttheSunThelawisexpressedmathematicallyas

2

3 constantTR

=

We shall nowuseNewtonrsquos LawofGravitation to showhowitisthattheplanetsmoveinaccordancewithKeplerrsquosthirdlaw

InessenceNewtonwasabletousehislawofgravitytopredict the motion of the planets since all he had to do was factor theF givenby this law intohis second law F = ma to find their accelerations and hence their future positions

InFigure1017theEarthisshownorbitingtheSunandthedistancebetweentheircentresisR

RSun

Earth

F es

F se

Figure 1017 Planets move according to Keplerrsquos third law

Fes is the force that the Earth exerts on the Sun and Fse is the force that the Sun exerts on the Earth The forces are equalandoppositeandtheSunandtheEarthwillactuallyorbitaboutacommoncentreHoweversince theSun issoverymuchmoremassivethantheEarththiscommoncentrewillbeclosetothecentreoftheSunandsowecanregard the Earth as orbiting about the centre of the Sun Theotherthingthatweshallassumeisthatwecanignorethe forces that the other planets exert on the Earth (This wouldnotbeawisethingtodoifyouwereplanningtosend a space ship to theMoon for example)We shallalso assume that we have followed Newtonrsquos exampleandindeedprovedthataspherewillactasapointmasssituated at the centre of the sphere

Kepler had postulated that the orbits of the planets areellipticalbut since theeccentricityof theEarthrsquosorbit issmallweshallassumeacircularorbit

PHYSICS 2015indb 319 150514 338 PM

Chapter 10

320

AH

L

TheaccelerationoftheEarthtowardstheSunisa = Rω2

whereω ω 2πT------=

Hencea R 2π

T------ times

2 4π2 RT 2------------- = =

But the acceleration is given by Newtonrsquos Second Law F = mawhereFisnowgivenbytheLawofGravitationSo in this situation

F maGMs Me

R2-------------------= = butwealsohavethat

a a R 2πT

------ times

2 4π2 RT 2------------- = = and m = Me so that

G MsMe

R2------------------- Me4π2R

T2-------------times

GMs

4π2-----------rArr R3

T2------= =

Butthequantity

isaconstantthathasthesamevalueforeachoftheplanetssowehaveforalltheplanetsnotjustEarththat

R3

T2------ k =

wherekisaconstantWhichisofcourseKeplerrsquosthirdlaw

ThisisindeedanamazingbreakthroughItisdifficulttorefutethe idea that all particles attract each other in accordance withtheLawofGravitationwhenthelawisabletoaccountfortheobservedmotionoftheplanetsabouttheSun

The gravitational effects of the planets upon each othershould produce perturbations in their orbits Such is the predictivepoweroftheUniversalGravitationalLawthatit enabled physicists to compute these perturbations The telescopehadbeeninventedin1608andbythemiddleofthe 18th Century had reached a degree of perfection in design that enabled astronomers to actually measure the orbital perturbations of the planets Their measurements werealways inagreementwith thepredictionsmadebyNewtonrsquoslawHoweverin1781anewplanetUranuswasdiscovered and the orbit of this planet did not fit withthe orbit predicted by Universal Gravitation Such wasthe physicistrsquos faith in theNewtonianmethod that theysuspectedthatthediscrepancywasduetothepresenceofayetundetectedplanetUsingtheLawofGravitationtheFrenchastronomerJLeverrier and the English astronomer J C Adamswereable tocalculate justhowmassive thisnewplanetmustbeandalsowhereitshouldbeIn1846the planetNeptunewas discovered justwhere they hadpredicted Ina similarwaydiscrepancies in theorbitofNeptune ledto thepredictionandsubsequentdiscoveryin1930oftheplanetPlutoNewtonrsquosLawofGravitation

had passed the ultimate test of any theory it is not only abletoexplainexistingdatabutalsotomakepredictions

Satellite energyWhenasatelliteisinorbitaboutaplanetitwillhavebothkineticenergyandgravitationalpotentialenergySupposewe consider a satellite ofmass m that is in an orbit of radius r about a planet of mass M

Thegravitationalpotentialduetotheplanetatdistancer from its centre is

ThegravitationalpotentialenergyofthesatelliteVsat

is therefore GMe m

r----------------ndash

ThatisVsat GMem

r----------------ndash=

Thegravitationalfieldstrengthatthesurfaceoftheplanetisgivenby

g 0GMe

Re2------------=

Hencewecanwrite

Vsat g0 Re

2

rndash=

m

ThekineticenergyofthesatelliteKsatisequaltofrac12mv2wherev is its orbital speed

Byequatingthegravitationalforceactingonthesatellitetoitscentripetalaccelerationwehave

G Mem

r2---------------- mv2

r---------- mv2hArr

GMemr----------------= =

Fromwhich12---

mv2 12---

GMemr----------------times=

= 0 e

2g R

r

2m

Which is actually quite interesting since it shows thatirrespective of the orbital radius the KE is numericallyequal to half the PE Also the total energy E tot K sat Vsat+ 1

2---G Me m

r----------------times G Me m

r----------------ndash + 1

2---G Me m

r----------------ndash= = =of the satelliteisalwaysnegativesince

E tot K sat Vsat+ 12---

G Me mr----------------times

G Me mr----------------ndash

+ 12---

G Me mr----------------ndash= = =

PHYSICS 2015indb 320 150514 338 PM

Fields

321

AH

L

The energies of an orbiting satellite as a function of radial distancefromthecentreofaplanetareshownplottedinFigure1018

12

100806

0402

0ndash02

ndash04

ndash06ndash08ndash10

ndash12

2 4 6 8 10 12

ener

gya

rbita

ry u

nits

distance timesR

kinetic energytotal energy

potential energy

Figure 1018 Energy of an orbiting satellite as a function of distance from the centre of a planet

WeightlessnessSuppose that you are inanelevator(lift)thatisdescendingat constant speedandyou let goof abook thatyouareholdinginyourhandThebookwillfalltothefloorwithaccelerationequaltotheaccelerationduetogravityIfthecablethatsupportstheelevatorweretosnap(asituationthatItrustwillneverhappentoanyofyou)andyounowletgothebookthatyouareholdinginyourotherhandthis book will not fall to the floor - it will stay exactlyin linewithyourhandThis isbecause thebook isnowfallingwith the sameaccelerationas theelevatorandassuch the book cannot ldquocatchrdquo up with the floor of theelevatorFurthermoreifyouhappenedtobestandingonasetofbathroomscalesthescaleswouldnowreadzero-youwouldbeapparentlyweightlessItisthisideaoffreefallthatexplainstheapparentweightlessnessofastronautsin an orbiting satellite These astronauts are in free fall inthesensethattheyareacceleratingtowardsthecentre of the Earth

It is actually possible to define theweight of a body inseveral different waysWe can define it for example asthegravitationalforceexertedonthebodybyaspecifiedobject such as the EarthThiswe have seen thatwe doinlotsofsituationswherewedefinetheweightasbeingequaltomgIfweusethisdefinitionthenanobjectinfreefallcannotbydefinitionbeweightlesssinceitisstillinagravitationalfieldHoweverifwedefinetheweightofanobjectintermsofaldquoweighingrdquoprocesssuchasthereadingonasetofbathroomscaleswhichineffectmeasuresthecontact force between the object and the scales thenclearlyobjectsinfreefallareweightlessOnenowhasto

askthequestionwhetherornotitispossibleForexampletomeasurethegravitationalforceactingonanastronautin orbit about the Earth

WecanalsodefineweightintermsofthenetgravitationalforceactingonabodyduetoseveraldifferentobjectsForexample for an object out in space its weight could bedefined in terms of the resultant of the forces exerted on it bytheSuntheMoontheEarthandalltheotherplanetsin the Solar System If this resultant is zero at a particular pointthenthebodyisweightlessatthispoint

Inviewofthevariousdefinitionsofweightthatareavailabletousitisimportantthatwhenweusethewordldquoweightrdquoweareawareofthecontextinwhichitisbeingused

Example

Calculate the height above the surface of the Earth atwhichageo-stationarysatelliteorbits

Solution

A geo-stationary satellite is one that orbits the Earth in such a way that it is stationary with respect to a point on the surface of the Earth This means that its orbital period must be the same as the time for the Earth to spin once on its axis ie 24 hours

From Keplerrsquos third law we have G Ms

4π2----------- R3

T2------=

That is

RMe

mh

using the fact that the force of attraction between the satellite and the Earth is given by

FG Mem

R2----------------=

and that F = ma

where m is the mass of the satellite and a 4π2R

T2-------------=

we have

GMem

R2---------------- m 4π2R

T2-------------times

GMe

4π2------------rArr R3

T2------= =

PHYSICS 2015indb 321 150514 338 PM

Chapter 10

322

AH

L

Now the mass of the Earth is 60 times 1024 kg and the period T measured in seconds is given by T = 86400 s

So substitution gives R = 42 times 106 m

The radius of the Earth is 64 times 106 m so that the orbital height h is about 36 times 107 m

Example

Calculatetheminimumenergyrequiredtoputasatelliteofmass500kgintoanorbitthatisasaheightequaltotheEarthrsquosradiusabovethesurfaceoftheEarth

Solution

We have seen that when dealing with gravitational fields and potential it is useful to remember that

g 0G M

Re2---------= or g0 Re

2 GM=

The gravitational potential at the surface of the Earth is

g 0ndash Re GMRe

---------ndash=

The gravitational potential at a distance R

from the centre of the Earth is

=

The difference in potential between the surface of the Earth and a point distance R from the centre is therefore

V∆ g 0Re 1ReR------ndash

=

If R 2Re= then V∆g0 Re

2------------=

This means that the work required to ldquoliftrdquo the satellite into orbit is g0Rm where m is the mass of the satellite This is equal to

10 32 106 500timestimestimes = 16000 MJ

However the satellite must also have kinetic energy in order to orbit Earth This will be equal to

g 0 mRe2

2R-----------------g 0mRe

2

4----------------- 8000 MJ= =

The minimum energy required is therefore

24000 MJ

Exercise 1031 The speed needed to put a satellite in orbit does

not depend on

A the radius of the orbitB theshapeoftheorbitC thevalueofgattheorbitD the mass of the satellite

2 EstimatethespeedofanEarthsatellitewhoseorbitis400kmabovetheEarthrsquossurfaceAlsodetermine the period of the orbit

3 Calculatethespeedofa200kgsatelliteorbitingthe Earth at a height of 70 times 106 m

Assume that g = 82 m sndash2 for this orbit

4 TheradiioftwosatellitesXandYorbitingthe Earth are 2r and 8rwherer is the radius of the Earth Calculate the ratio of the periods of revolutionofXtoY

5 A satellite of mass mkgissentfromEarthrsquossurfaceinto an orbit of radius 5RwhereR is the radius of theEarthWritedownanexpressionfor

(a) thepotentialenergyofthesatelliteinorbit

(b) thekineticenergyofthesatelliteinorbit

(c) theminimumworkrequiredtosendthesatellitefromrestattheEarthrsquossurfaceintoits orbit

6 A satellite in an orbit of 10rfallsbacktoEarth(radius r)afteramalfunctionDeterminethespeedwithwhichitwillhittheEarthrsquossurface

7 The radius of the moon is frac14 that of the Earth AssumingEarthandtheMoontohavethesamedensitycomparetheaccelerationsoffreefallatthe surface of Earth to that at the surface of the Moon

8 UsethefollowingdatatodeterminethegravitationalfieldstrengthatthesurfaceoftheMoon and hence determine the escape speed from the surface of the Moon

Mass of the Moon = 73 times 1022kg

RadiusoftheMoon=17times106 m

PHYSICS 2015indb 322 150514 338 PM

Wave Phenomena

291

AH

L

93 Interference

Essential idea Interference patterns from multiple slits andthinfilmsproduceaccuratelyrepeatablepatterns

Understandingsbull Youngrsquosdouble-slitexperimentbull Modulationoftwo-slitinterferencepatternbyone-slit

diffraction effectbull Multipleslitanddiffractiongratinginterferencepatternsbull Thinfilminterference

Youngrsquos double slit experimentPlease refer to interference in chapter 4 However wewill reiterate the condition for the interference of waves from two sources to be observed The two sources must be coherentthatistheymusthavethesamephaseorthephase difference between them must remain constant

Also to reinforce topics concerning the principle ofsuperpositionandpathdifferenceandphasedifferenceincluded in Figure 913 is another example of two source interference

We can obtain evidence for the wave nature of sound by showing that sound produces an interference pattern Figure 913 shows the set up for demonstrating this

CRO

speaker

speaker

signalgenerator

microphoneX

Y

Figure 913 Interference using sound waves

The two speakers are connected to the same output of the signal generator andplaced about 50 cmapartThesignalgeneratorfrequencyissettoabout600Hzandthe

slit If the slit width is much greater than the wavelength then the width of the central maxima is very small

Insummary

for destructive interference and minima

dsinθ=nλ wheren=123hellip

for constructive interference and maxima

dsinθ=(n+frac12)λ wheren=123hellip

Note that waves from point sources along the slit arrive in phase at the centre of the fringe pattern and constructively interferetoproducethecentralmaximumcalledthezero ordermaximum(n=0)

Example

LightfromalaserisusedtoformasingleslitdiffractionpatternThe width of the slit is 010 mm and the screen is placed 30 m from the slit The width of the central maximum is measured as26cmCalculatethewavelengthofthelaserlight

Solution

Since the screen is a long way from the slit we can use the small angle approximation such that the f is equal to 30 m

The half-width of the central maximum is 13 cm so we have

λ13 10 2ndashtimes( ) 10 10 4ndashtimes( )times

30------------------------------------------------------------------=

To give λ = 430 nm

This example demonstrates why the image of a point source formed by a thin converging lens will always have a finite width

Exercise 921 Aparallelbeamoflightofwavelength500nmis

incidentonaslitofwidth025mmThelightisbroughttofocusonascreenplaced150mfromthe slit Calculate the angular width and the linear width of the central diffraction maximum

2 Lightfromalaserisusedtoformasingleslitdiffraction pattern on a screen The width of the slit is 010 mm and the screen is 30 m from the slit The width of the central diffraction maximum is 26cmCalculatethewavelengthofthelaserlight

3 Determinetheanglethatyouwouldexpecttofindconstructiveinterference(forn=0n=1andn=2) fromasingleslitofwidth20μmandlightofwavelength 600 nm

NATURE OF SCIENCECuriosity Observed patterns of iridescence in animals such as the shimmer of peacock feathers led scientists to develop the theory of thin film interference (15)

Serendipity The first laboratory production of thin films was accidental (15)

copy IBO 2014

PHYSICS 2015indb 291 150514 337 PM

Chapter 9

292

AH

L

microphoneismovedalongthelineXYthatisaboutonemetre from the speakers As the microphone is moved along XY the trace on the cathode ray oscilloscope isseen to go through a series of maxima and minima corresponding to points of constructive and destructive interference of the sound waves

Aninterestinginvestigationistofindhowtheseparationof the points of maximum interference depends on the frequency of the source and also the distance apart ofthe speakers

In the demonstration of the interference between two soundsourcesdescribedaboveifweweretomoveoneofthe speakers from side to side or backwards and forwards the sound emitted from the two speakers would no longer be in phase No permanent points of constructive or destructive interference will now be located since the phase difference between the waves from the two sources is nolongerconstantiethesourcesarenolongercoherent

Light from an incandescent source is emitted with acompletely random phase Although the light from two separate sources will interfere because of the randomlychanging phase no permanent points of constructive or destructive interference will be observed This is why a singleslit isneededintheYoungrsquosdoubleslitexperimentByactingasapointsourceitessentiallybecomesacoherentlight source The light emitted from a laser is also very nearly coherent and this is why it is so easy to demonstrate optical interference and diffraction with a laser

Youngrsquos double slit experiment is one of the great classic experiments of physics and did much to reinforce the wave theoryoflightThomasYoungcarriedouttheexperimentin about 1830

It is essentially the demonstration with the ripple tank and the sound experiment previously described butusing light The essential features of the experiment are shown in Figure 914

screendouble slit

single slit

coloured lter

S 1

S2

Figure 914 Youngrsquos double slit experiment

YoungallowedsunlighttofallontoanarrowsingleslitA few centimetres from the single slit he placed a double slit The slits are very narrow and separated by a distance equal to about fourteen slitwidthsA screen is placed

about ametre from thedouble slitsYoungobservedapatternofmulticolouredldquofringesrdquo in the screenWhenhe placed a coloured filter between the single slit anddouble slit he obtained a pattern that consisted of bright coloured fringes separated by darkness

The single slit essentially ensures that the light falling on the double slit is coherent The two slits then act as the two speakers in the sound experiment or the two dippers in the ripple tank The light waves from each slit interfere and produce the interference pattern on the screen Without thefilterapatternisformedforeachwavelengthpresentinthesunlightHencethemulticolouredfringepattern

You can demonstrate optical interference for yourselfSmoking a small piece of glass and then drawing twoparallel lines on it can make a double slit If you then look throughthedoubleslitatasingletungstenfilamentlampyouwillseethefringepatternByplacingfiltersbetweenthe lamps and the slits you will see the monochromatic fringe pattern

YoucanalsoseetheeffectsofopticalinterferencebylookingatnetcurtainsEachlsquoholersquointhenetactsasapointsourceand the light from all these separate sources interferes and producesquiteacomplicatedinterferencepattern

A laser can also be used to demonstrate optical interference Sincethelightfromthelaseriscoherentitisveryeasytodemonstrate interference Just point the laser at a screen and place a double slit in the path of the laser beam

LetusnowlookattheYoungrsquosdoubleslitexperimentinmore detail The geometry of the situation is shown in Figure915

S1

S2 X

P

Q

D

y

d

screen

θ

θrsquo

Figure 915 The geometry of Youngrsquos double slit experiment

S1andS2 are the two narrow slits that we shall regard as twocoherentmonochromaticpointsourcesThedistancefrom the sources to the screen is D and the distance between the slits is d

The waves from the two sources will be in phase at Q and therewill be abright fringehereWewish tofindtheconditionfortheretobeabrightfringeatPdistancey from Q

PHYSICS 2015indb 292 150514 337 PM

Wave Phenomena

293

AH

L

We note that that D (asymp1metre)isverymuchgreaterthaneither y or d(asymp fewmillimetres)Thismeansthatbothθ and θʹ areverysmallanglesandforintentsandpurposesequal

From the diagram we have that

θ = Dy

And

2S Xd

θʹ =

(Remembertheanglesareverysmall)

Sinceθ asymp θʹ then

2S XyD d

=

ButS2X is the path difference between the waves as they

traveltoPWehavethereforethat

path difference = Dyd

TherewillthereforebeabrightfringeatPif

Dyd

= nλ

Supposethatthereisabrightfringeaty = y1 corresponding

to n = n1 then

Ddy1 = n

1λ

If the next bright fringe occurs at y = y2 this will correspond

to n = n1+1Hence

Ddy2 =(n

1+1)λ

This means that the spacing between the fringes y2 ndash y

1 is

given by

y2 ndash y

1 = d

Dλ

Youngactuallyusethisexpressiontomeasurethewavelengthof the light he used and it is a method still used today

We see for instance that if in a given set up we move the slits closer together then the spacing between the fringes will get greaterEffectivelyourinterferencepatternspreadsoutthatis there will be fewer fringes in a given distance We can also increase the fringe spacing by increasing the distance betweentheslitsandthescreenYouwillalsonotethatfora

givensetupusinglightofdifferentwavelengthsthenldquoredrdquofringeswillspacefurtherapartthanldquobluerdquofringes

In this analysis we have assumed that the slits act as point sources and as such the fringes will be uniformly spaced andofequalintensityAmorethoroughanalysisshouldtakeintoaccountthefinitewidthoftheslits

ReturningtoFigure915weseethatwecanwritethepathdifferenceS2X as

S2X = dtan θʹ

But since θʹ is a small angle the sine and tangent will be nearlyequalsothat

S2X = dsinθʹ

And since θʹ asymp θ then

S2X = dsinθ

The condition therefore for a bright fringe to be found at a point of the screen can therefore be written as

dsinθ = nλ

Insummary

for constructive interference and minima

d sinθ = nλ wheren=123hellip

for destructive interference and maxima

d sinθ = ( n + frac12 )λ wheren=123hellip

Figure 916 shows the intensity distribution of the fringes on the screen when the separation of the slits is large comparedtotheirwidthThefringesareofequalintensityandofequalseparation

Intensity

distance along screen

Figure 916 The intensity distribution of the fringes

It isworthnoting that if the slits are close together theintensity of the fringes is modulated by the intensity distribution of the diffraction pattern of one of the slits

PHYSICS 2015indb 293 150514 337 PM

Chapter 9

294

AH

L

Example

Light of wavelength 500nm is incident on two smallparallel slits separated by 10mmDetermine the anglewherethefirstmaximumisformed

If after passing through the slits the light is brought toa focusona screen15m fromtheslitscalculate theobserved fringe spacing on the screen

Solution

Using the small angle approximation we have

3

7

10105minus

minustimes==

dλθ

= 5 times 10-4 rad

The fringe spacing is given by

3

7

1010551

minus

minustimestimes==

dDy λ

= 075 mm

Modulation of two-slit interference pattern by one-slit diffraction effect

In practice the intensity pattern of the maxima as shown in Figure 917 is not constant but fluctuates while decreasingsymmetrically on either side of the central maximum as shown inFigure917Theintensitypatternisacombinationofboththe single-slit diffractionenvelope and the double slit pattern That is the amplitude of the two-slit interference pattern is modulated(ieadjustedto)byasingleslitdiffractionenvelope

2λD

___ λD

D

___λd

d

___

Double slitnite width

Figure 917 Double slit pattern

ForagivenslitseparationdwavelengthoflightandfixedslittoscreendistancethevariationinintensitydependsonthewidthoftheslitDAlthoughincreasingthewidthoftheslitsincreasestheintensityoflightinthefringesitalsomakesthem less sharp and they become blurred As the slit width is widenedthefringesgraduallydisappearbecausenumerouspoint sources along the widened slit give rise to their own dark and bright fringes that then overlap with each other

Exercise 931 InFigure913thedistancebetweenthespeakers

is050mandthedistancebetweenthelineofthespeakersandthescreenis20mAsthemicrophoneismovedalongthelineXYthedistance between successive points of maximum soundintensityis030mThefrequencyofthesoundwavesis44times103HzCalculateavalueforthe speed of sound

2 Laserlightofwavelength610nmfallsontwoslits10x10-5apartDeterminetheseparationofthefirstordermaximumformedonascreen20maway

3 Twoparallelslits012mmapartareilluminatedwith red light with a wavelength of 600 nm An interferencepatternfallsonascreen15maway

Calculate

(a) thedistancefromthecentralmaximumtothefirstbrightfringe

(b) thedistancetotheseconddarkline

Multiple slit and diffraction grating interference patterns

If we examine the interference pattern produced when monochromatic light passes through a different number of slits we notice that as the number of slits increases the number of observed fringes decreases the spacingbetween them increases and the individual fringes become much sharper We can get some idea of how this comes about by looking at the way light behaves when a parallel beam passes through a large number of slits The diagram for this is shown in Figure 918

θθd

1

2

3

telescope

Figure 918 A parallel beam passing through several slits

The slits are very small so that they can be considered to act as point sources They are also very close together such that d issmall(10ndash6 m)EachslitbecomesasourceofcircularwavefrontsandthewavesfromeachslitwillinterfereLetus

PHYSICS 2015indb 294 150514 337 PM

Wave Phenomena

295

AH

L

consider the light that leaves the slit at an angle θ as shown Thepathdifferencebetweenwave1andwave2isdsinθ and ifthisisequaltoanintegralnumberofwavelengthsthenthe two waves will interfere constructively in this direction Similarly wave 2 will interfere constructively with wave3 at this angle andwave 3with 4 etc across thewholegratingHenceifwelookatthelightthroughatelescopethatisbringittoafocusthenwhenthetelescopemakesanangle θ to the grating a bright fringe will be observed The condition for observing a bright fringe is therefore

dsinθ = mλ

Supposeweuselightofwavelength500nmandsupposethat d = 16 times 10ndash6 m

Obviously we will see a bright fringe in the straight on position θ=0(thezero order)

The next position will be when m =1(thefirst order)andsubstitutionintheaboveequationgivesθ = 18deg

The next position will be when m=2(thesecond order)and this give θ = 38deg

For m=3sinθ is greater than 1 so with this set up we only obtain5fringesonezeroorderandtwoeithersideofthezero order

The calculation shows that the separation of the orders is relatively large At any angles other than 18deg or 38deg the light leaving the slits interferes destructively We can see that the fringes will be sharp since if we move just a small angle away from 18deg the light from the slits will interfere destructively

An array of narrow slits is usually made by cutting narrow transparent lines very close together into the emulsion on aphotographicplate(typically200linespermillimetre)Suchanarrangementiscalledadiffraction grating

The diffraction grating is of great use in examining the spectral characteristics of light sources

All elements have their own characteristic spectrum An element can be made to emit light either by heating it until it is incandescent or by causing an electric discharge through itwhen it is in a gaseous state Yourschool probably has some discharge tubes and diffraction gratingsIfithasthenlookattheglowingdischargetubethroughadiffractiongratingIfforexampletheelementthat you are looking has three distinct wavelengths then each wavelength will be diffracted by a different amount inaccordancewiththeequationdsinθ = nλ

Alsoiflaserlightisshonethroughagratingontoascreenyou will see just how sharp and spaced out are the maxima By measuring the line spacing and the distance of the screen fromthelaserthewavelengthofthelasercanbemeasured

If your school has a set of multiple slits say from a single slit toeight slits then it isalsoaworthwhileexercise toexamine how the diffraction pattern changes when laser light is shone through increasing numbers of slits

Consider a grating of NparallelequidistantslitsofwidthaseparatedbyanopaqueregionofwidthD We know that light diffracted from one slit will be superimposed and interfere with light diffracted from all the other slits The intensityprofile of the resultant wave will have a shape determined by

bull theintensityoflightincidentontheslitsbull thewavelengthofthislightbull theslitwidthbull theslitseparation

Typical intensity profiles formed when a plane wavemonochromaticlightatnormalincidenceononetwoandsix slits are shown in Figure 919 In this example the slit widthisonequarterofthegratingspacing

Six slits sharperfringes

d = 4a

I α N 2

(N ndash 2)= 4 subsidiary maxima

order0 4321

I = 36

Two slits

order0

single-slitdiraction

envelope

d = 4a4th principal maximum missing

principa maxima

I = 4 (nominal units)

4321

intensity

order

ISS

One slit

0 1

single-slitdiraction

pattern

N = number of slitsslit width a

Figure 919 Intensity profiles

PHYSICS 2015indb 295 150514 337 PM

Chapter 9

296

AH

L

1 For Nge2thefringepatterncontainsprincipalmaximathat are modulated by the diffraction envelope

2 ForNgt2thefringepatterncontains(Nndash2)subsidiary maxima

3 Where the grating spacing d is four time the slit width a(4xa =d)thenthefourthprincipalmaximumismissing from the fringe pattern Where the grating spacingdisthreetimetheslitwidtha(3xa =d)thenthe third principal maximum is missing from the fringepatternandsoon

4 Increasing N increases the absolute intensities of the diffraction pattern (I = N2)

If white light is shone through a grating then the central image will be white but for the other orders each will be spread out intoacontinuousspectrumcomposedofaninfinitenumberof adjacent images of the slit formed by the wavelength of the different wavelengths present in the white light At any given point in the continuous spectrum the light will be very nearly monochromatic because of the narrowness of the images of the slit formed by the grating This is in contrast to the doubleslitwhereifwhitelightisusedtheimagesarebroadand the spectral colours are not separated

In summary as is the case of interference at a doubleslit there is a path difference between the rays fromadjacent slits of d sinθ Where there is a whole number of wavelengthsconstructiveinterferenceoccurs

For constructive interference and minima

d sinθ = mλ wheren=123hellip

When there is a path difference between adjacent rays of halfawavelengthdestructiveinterferenceoccurs

For and maxima

d sinθ = ( m + frac12 )λ wheren=123hellip

Exercise 941 Lightfromalaserisshonethroughadiffraction

grating on to a screen The screen is a distance of 20mfromthelaserThedistancebetweenthecentraldiffractionmaximumandthefirstprincipalmaximum formed on the screen is 094 m The diffractiongratinghas680linespermmEstimatethe wavelength of the light emitted by the laser

2 Monochromaticlightfromalaserisnormallyincident on a six-slit grating The slit spacing is three times the slit width An interference patternisformedonascreen2montheotherside of the grating

(a) Determineandexplaintheorderoftheprincipal maxima missing from the fringe pattern

(b) Howmanysubsidiarymaximaaretherebetween the principal maxima

(c) Howmanyprincipalmaximaarethere

(i) Betweenthetwofirstorderminimain the diffraction envelope

(ii) Betweenthefirstandsecondorderminima in the diffraction envelope

(d) Drawthepatternformedbythebrightfringesin(c)

(e) Sketchtheintensityprofileofthefringepattern

Thin film interferenceYoumightwellbe familiarwith the colouredpatternoffringes that can be seen when light is reflected off thesurfaceofwateruponwhichathinoilfilmhasbeenspiltorfromlightreflectedfrombubblesWecanseehowthesepatternsarisebylookingatFigure920

Extendedlight source air

A

B

C

12

3

4

Oil lm

Water

Figure 920 Reflection from an oil film

Consider light from an extended source incident on a thinfilmWealsoconsiderawave fromonepointofthe source whose direction is represented by the ray shownSomeofthislightwillbereflectedatAandsometransmitted through the filmwhere somewill again bereflectedatB(somewillalsobetransmittedintotheair)SomeofthelightreflectedatBwillthenbetransmittedatCandsomereflectedandsoonIfweconsiderjustrays1and2thenthesewillbenotbeinphasewhenthearebrought to a focus by the eye since they have travelled

PHYSICS 2015indb 296 150514 337 PM

Wave Phenomena

297

AH

L

different distances If the path difference between them is an integral number of half wavelengths then we might expect the twowaves to be out of phaseHowever ray1 undergoes a phase change of π on reflection but ray2 does not since it is reflected at a boundary betweena more dense and less dense medium (See chapter 4)Henceifthepathdifferenceisanintegralnumberofhalf-wavelengthsrays1and2willreinforceieray1and2areinphaseHowever rays357etcwillbeoutofphasewithrays246etcbutsinceray2ismoreintensethanray3andray4moreintensethanray5thesepairswillnot cancel out so there will be a maximum of intensity

Ifthepathdifferenceissuchthatwave1and2areoutofphasesincewave1ismoreintensethanwave2theywillnotcompletelyannulHoweveritcanbeshownthattheintensitiesofwaves2345hellipaddtoequaltheintensityofwave1Sincewaves345hellipareinphasewithwave2there will be complete cancellation

The path difference will be determined by the angle at which ray1 is incident andalsoon the thickness (andtheactualrefractiveindexaswell)ofthefilmSincethesourceisanextendedsourcethelightwillreachtheeyefrom many different angles and so a system of bright and darkfringeswillbeseenYouwillonlyseefringesifthefilmisverythin(exceptifviewedatnormalincidence)sinceincreasingthethicknessofthefilmwillcausethereflected rays to get so far apart that they will not becollected by the pupil of the eye

From the argument above to find the conditions forconstructive and destructive interferenceweneedonlyfindthepathdifferencebetweenray1andray2Figure921showsthe geometry of the situation

A BE

F

C

D

1 2

φrsquo φrsquo

φ

φ

d

Figure 921 The geometry of interference

Thefilmisofthicknessd and refractive index n and the light has wavelength λ If the line BF is perpendicular to ray 1 then the optical path difference (opd) between ray 1 andray2whenbroughttoafocusis

opd = n(AC+CB)ndashAF

We have to multiple by the refractive index for the path travelledbythelightinthefilmsincethelighttravelsmoreslowly in thefilm If the light travels say adistancex in a material of refractive index n then in the time that it takes to travelthisdistancethelightwouldtraveladistancenx in air

IfthelineCEisatrightanglestoray2thenweseethat

AF = nBE

FromthediagramAC=CDsowecanwrite

opd = n(CD+CB)ndashnBE=nDE

Alsofromthediagramweseethatwhereφ is the angle of refraction

DE=2cosφ

From which opd=2ndcosφ

Bearinginmindthechangeinphaseofray1onreflectionwe have therefore that the condition for constructive interference is

2nd φcos m 12---+

λ m 1 2 hellip = =

Andfordestructiveinterference 2ndcosφ = mλ

Eachfringecorrespondstoaparticularopd for a particular value of the integer m and for any fringe the value of the angle φ is fixedThismeans that itwill be in the formof an arc of a circle with the centre of the circle at the point where the perpendicular drawn from the eye meets the surface of the film Such fringes are called fringesofequal inclinationSince theeyehasa smallaperturethesefringesunlessviewedatneartonormalincidence (φ = 0) will only be observed if the film is very thinThisisbecauseasthethicknessofthefilmincreasesthereflected rayswill get further and further apart and sovery few will enter the eye

Ifwhitelightisshoneontothefilmthenwecanseewhywe get multi-coloured fringes since a series of maxima and minima will be formed for each wavelength present in the white light However when viewed at normalincidenceitispossiblethatonlylightofonecolourwillundergoconstructiveinterferenceandthefilmwilltakeon this colour

PHYSICS 2015indb 297 150514 337 PM

Chapter 9

298

AH

L

Thickness of oil films

The exercise to follow will help explain this use

Non-reflecting films

A very important but simple application of thin filminterferenceisintheproductionofnon-reflectingsurfaces

Athinfilmofthicknessd and refractive index n1 is coated onto glass of refractive index n where n1 lt n Light ofwavelength λ that is reflected at normal incidence willundergo destructive interference if 12

2n d λ

= thatis

14d

nλ

=

(remember that there will now no phase change at the glass interface ie we have a rare to dense reflection)

The use of such films can greatly reduce the loss oflightby reflection at the various surfacesof a systemoflenses or prisms Optical parts of high quality systemsare usually all coatedwithnon-reflecting films in orderto reduce stray reflections The films are usually madebyevaporatingcalciumormagnesiumfluorideonto thesurfacesinvacuumorbychemicaltreatmentwithacidsthat leave a thin layer of silica on the surface of the glass ThecoatedsurfaceshaveapurplishhuebyreflectedlightThis is because the condition for destructive interference fromaparticularfilmthicknesscanonlybeobtainedforone wavelength The wavelength chosen is one that has a value corresponding to light near the middle of the visible spectrumThismeansthatreflectionofredandvioletlightis greater combining to give the purple colour Because of the factor cosφatanglesotherthannormalincidencethepath difference will change butnotsignificantlyuntilsayabout30deg(egcos25deg=090)

It should be borne in mind that no light is actually lost by anon-reflectingfilmthedecreaseofreflectedintensityiscompensated by increase of transmitted intensity

Non-reflecting films can be painted onto aircraft tosuppressreflectionofradarThethicknessofthefilmis determined by

4nd λ

= where λ is the wavelength of

the radar waves and ntherefractiveindexofthefilmat this wavelength

Example

A plane-parallel glass plate of thickness 20 mm isilluminated with light from an extended source The refractiveindexoftheglassis15andthewavelengthofthelight is 600 nm Calculate how many fringes are formed

Solution

We assume that the fringes are formed by light incident at all angles from normal to grazing incidence

At normal incidence we have 2nd = mλ

From which

m 2 15 2 10 3ndashtimestimestimes6 10 7ndashtimes

------------------------------------------- 10 000= =

At grazing incidence the angle of refraction φ is in fact the critical angle

Therefore φφ arcsin 115------- 42 deg= =

ie cosφ = 075

At grazing incidence 2ndcosφ = m λ

From which (and using cos(sinndash1(115) = 075)

m 2 15 2 10 3ndash 075timestimestimestimes6 10 7ndashtimes

------------------------------------------------------------ 7500= =

The total number of fringes seen is equal to m ndash m = 2500

Exercise 95When viewed from above the colour of an oil film onwater appears red in colour Use the data below to estimate theminimumthicknessoftheoilfilm

average wavelength of red light = 630 nm

refractiveindexofoilforredlight=15

refractive index of water for red light = 13

PHYSICS 2015indb 298 150514 337 PM

Wave Phenomena

299

AH

L

94 Resolution

Essential idea Resolution places an absolute limit on the extent to which an optical or other system can separate images of objects

Understandings

bull Thesizeofadiffractingaperturebull Theresolutionofsimplemonochromatic

two-source systems

The size of a diffracting apertureOur discussion concerning diffraction so far has been for rectangular slits When light from a point source enters a smallcircularapertureitdoesnotproduceabrightdotasanimagebutacirculardiscknownasAiryrsquosdiscsurroundedbyfainterconcentriccircularringsasshowninFigure922

Laser

Figure 922 Diffraction at a circular aperture

Diffractionatanapertureisofgreatimportancebecausetheeye and many optical instruments have circular apertures

What is the half-angular width of the central maximum of the diffraction formed by a circular aperture This is not easy to calculate since it involves some advanced mathematics The problemwasfirstsolvedbytheEnglishAstronomerRoyalGeorgeAiryin1835whoshowedthatforcircularapertures

θ 122 λb--------------= where b is the diameter of the aperture

Example 2

Inthefollowingdiagramparallellightfromadistantpointsource(suchasastar)isbroughttofocusonthescreenSbyaconverginglens(the lens is shown as a vertical arrow)

Thefocal length(distance fromlens toscreen) is25cmand the diameter of the lens is 30 cm The wavelength of thelightfromthestaris560nmCalculatethediameterofthe diameter of the image on the screen

Solution

The lens actually acts as a circular aperture of diameter 30 cm The half angular width of central maximum of the diffraction pattern that it forms on the screen is

75

2

122 122 56 10 23 1030 10b

λθminus

minusminus

times times= = = times

times rad

The diameter of the central maxima is therefore

25 times 10-2 times 23 times 10-5

= 57 times 10-6 m

Although this is small it is still finite and is effectively the image of the star as the intensity of the secondary maxima are small compared to that of the central maximum

The resolution of simple monochromatic two-source systemsThe astronomers tell us that many of the stars that we observe with the naked eye are in fact binary stars That iswhatwe see as a single star actually consistsof twostarsinorbitaboutacommoncentreFurthermoretheastronomerstellusthatifweusealdquogoodrdquotelescopethenwewillactuallyseethetwostarsThatiswewillresolve the single point source into its two component parts Sowhatisitthatdetermineswhetherornotweseethetwo stars as a single point source ie what determines whether or not two sources can be resolved It canrsquot just bethatthetelescopemagnifiesthestarssinceiftheyareacting as point sources magnifying them is not going to make a great deal of difference

Ineachofoureyesthereisanaperturethepupilthroughwhich the light enters This light is then focused by the eye lens onto the retina But we have seen that when light passes through an aperture it is diffracted and so when we look at

NATURE OF SCIENCEImproved technology The Rayleigh criterion is the limit of resolution Continuing advancement in technology such as large diameter dishes or lenses or the use of smaller wavelength lasers pushes the limits of what we can resolve (18)

copy IBO 2014

PHYSICS 2015indb 299 150514 337 PM

Chapter 9

300

AH

L

apointsourceadiffractionpatternwillbeformedontheretina If we look at two point sources then two diffraction patterns will be formed on the retina and these patterns will overlap The width of our pupil and the wavelength of the light emitted by the sources will determine the amount by which they overlap But the degree of overlap will also depend on the angular separation of the two point sources WecanseethisfromFigures923

pupilretina

S1

S2

P2

P1

θ

Figure 923

LightfromthesourceS1 enters the eye and is diffracted by the pupil such that the central maximum of the diffraction patternisformedontheretinaatP1SimilarlylightfromS2producesamaximumatP2 If the two central maxima are well separated then there is a fair chance that we will see the two sources as separate sources If they overlap then we will not be able to distinguish one source from another From the diagram we see as the sources are movedclosertotheeyethentheangleθ increases and so does the separation of the central maxima

Figures924925926and927showsthedifferentdiffractionpatternsandtheintensitydistributionthatmightresultonthe retina as a result of light from two point sources

S1 S21 Well resolved

Figure 924 Very well resolved

S1 S22 Well resolved

Figure 925 Well resolved

S1 S23 Just resolved Rayleigh criterion Minimum of S2 coincides with maximum peak of S1

Figure 926 Just resolved

4 Not resolved

Figure 927 Not resolved

We have suggested that if the central maxima of the two diffraction patterns are reasonably separated then we should be able to resolve two point sources In the late 19th century Lord Rayleigh suggested by how much they should be separated in order for the two sources to be just resolved Ifthecentralmaximumofonediffractionpattern coincides with the firstminima of the otherdiffraction pattern then the two sources will just beresolved This is known as the Rayleigh Criterion

Figure926showsjustthissituationThetwosourcesarejust resolved according to the Rayleigh criterion since the peak of the central maximum of one diffraction pattern coincideswiththefirstminimumoftheotherdiffractionpattern This means that the angular separation of the peaks of the two central maxima formed by each source is just the half angular width of one central maximum ie

θ λb---

=

where b is the width of the slit through which the light fromthesourcespassesHoweverweseefromFigure1117that θ is the angle that the two sources subtend at the slit Henceweconcludethattwosourceswillberesolvedbyaslit if the angle that they subtend at the slit is greater than or equaltoλfraslb

PHYSICS 2015indb 300 150514 337 PM

Wave Phenomena

301

AH

L

Sofarwehavebeenassumingthattheeyeisarectangularslit whereas clearly it is a circular aperture and so we must use the formula

θ 122 λb--------------=

As mentioned above the angle θ is sometimes called the resolving power but should more accurately be called the minimum angle of resolution (θmin)

Clearly the smaller θ the greater will be the resolving power

The diffraction grating is a useful for differentiating closely spacedlinesinemissionspectraLikeaprismspectrometerit can disperse a spectrum into its components but it is better suited because it has higher resolution than the prism spectrometer

If λ1 and λ2 are two nearly equal wavelengths that canbarelybedistinguishedtheresolvanceorresolvingpowerofthegratingisdefinedas

R = λ _______ (λ2 ndash λ 1) = λ ___ Δλ

Where λ asymp λ1 asymp λ2

Therefore thediffractiongratingwithahighresolvancewill be better suited in determining small differences in wavelength

If the diffraction grating has Nlinesbeingilluminateditcan be shown that the resolving power of the mth order diffraction is given by

R = mN

Obviouslytheresolvancebecomesgreaterwiththeordernumber m and with a greater number of illuminated slits

Example

A benchmark for the resolving power of a grating is the sodium doublet in the sodium emission spectrum Two yellow lines in itsspectrumhavewavelengthsof58900nmand58959nm

(a) Determinetheresolvanceofagratingifthegivenwavelengths are to be distinguished from each other

(b) Howmanylinesinthegratingmustbeilluminated in order to resolve these lines in the second order spectrum

Solution

(a) R = λ ___ Δλ = 58900nm __________________ (58900ndash58959)nm

= 589nm _______ 059nm = 10 times 103

(b) N = R __ m = 10 times 10 3 ________ 2 =50times102 lines

It has been seen that diffraction effectively limits the resolving power of optical systems This includes such systems as the eye telescopes and microscopes Theresolving power of these systems is dealt with in the next section This section looks at links between technology and resolving power when looking at the very distant and when looking at the very small

Radio telescopes

The average diameter of the pupil of the human eye is about 25 mm This means that the eye will justresolve two point sources emitting light of wavelength 500 nm if their angular separation at the eye is 7

43

50 10122 24 1025 10

θminus

minusminus

times= times = times

times rad

If the eye were to be able to detect radio waves of wavelength 015m then tohave the same resolvingpower thepupilwould have to have a diameter of about 600 m Clearly this is nonsense but it does illustrate a problem facingastronomers who wish to view very distant objects such as quasarsandgalaxies thatemitradiowavesConventionalradio telescopes consist of a large dish typically 25 min diameter Even with such a large diameter the radiowavelength resolving power of the telescope is much less thantheopticalresolvingpowerofthehumaneyeLetuslook at an example

Example

The Galaxy Cygnus A can be resolved optically as an elliptically shaped galaxy However it is also a strongemitterof radiowavesofwavelength015mTheGalaxyisestimatedtobe50times1024mfromEarthUseofaradiotelescope shows that the radio emission is from two sources separated by a distance of 30 times 1021m Estimate thediameterofthedishrequiredtojustresolvethesources

Solution

The angle θ that the sources resolve at the telescope is given by

θ 21

424

30 10 60 1050 10

θ minustimes= = times

times rad

and 4

122 122 015 300060 10

d λθ minus

times= = =

timesm = 30 km

A radio telescope dish of this size would be impossible tomake letalonesupportThisshowsthatasingledishtype radio telescope cannot be used to resolve the sources and yet they were resolved To get round the problemastronomers use two radio telescopes separated by a large

PHYSICS 2015indb 301 150514 337 PM

Chapter 9

302

AH

L

distance The telescopes view the same objects at the same time and the signals that each receive from the objects are simultaneously superimposed The result of the superposition of the two signals is a two-slit interference pattern The pattern has much narrower fringe spacing than that of the diffraction pattern produced by either telescope on its own hence producing a much higherresolvingpowerWhentelescopesareusedlikethistheyare called a stellar interferometer

InSocorroinNewMexicothereisastellarinterferometerthat consists of 27 parabolic dishes each of diameter25 m arranged in a Y-shape that covers an area of 570km2Thisisaso-calledVeryLargeArray(VLA)Evenhigher resolution can be obtained by using an array of radio telescopes in observatories thousands of kilometres apart A system that uses this so-called technique oflsquovery-long-baseline interferometryrsquo (VLBI) is known asalsquovery-long-baselinearrayrsquo(VLBA)WithVLBAaradiowavelength resolving power can be achieved that is 100 timesgreaterthanthebestopticaltelescopesEvenhigherresolving power can be achieved by using a telescope that isinasatelliteorbitingEarthJapanrsquosInstituteofSpaceandAstronautical Science (ISAS) launched sucha system inFebruary 1997TheNationalAstronomicalObservatoryof Japan the National Science Foundationrsquos NationalRadio Astronomy Observatory (NRAO) the CanadianSpaceAgency theAustraliaTelescopeNationalFacilitythe European VLBI Network and the Joint Instituteback this project forVeryLongBaseline Interferometryin EuropeThis project is a very good example of howInternationalismcanoperateinPhysics

Electron microscope

Telescopes are used to look at very distant objects that are very large but because of their distance fromus appearverysmallMicroscopesontheotherhandareusedtolookat objects that are close to us but are physically very small Aswehave seen justmagnifyingobjects that ismakingthemappearlargerisnotsufficientonitsowntogaindetailabouttheobjectfordetailhighresolutionisneeded

Figure928isaschematicofhowanoptical microscope is usedtoviewanobjectandFigure929isaschematicofatransmission electron microscope(TEM)

Optical lens system

bright light source

glass slide containing specimen (object)

eye

Figure 928 The principle of a light microscope

magnetic lens system

electron source

thin wafer of material

ScreenCCD

Figure 929 The principle of an electron microscope

In the optical microscope the resolving power isdetermined by the particular lens system employed and the wavelength λ of the light used For example twopoints in the sample separated by a distance d will just be resolved if

2d mλ=

where m is a property of the lens system know as the numerical aperture In practice the largest value of m obtainable is about 16 Hence if the microscope slideis illuminatedwith light ofwavelength 480 nm a goodmicroscope will resolve two points separated by a distance dasymp15times10-7masymp015micromPointscloser together thanthiswillnotberesolvedHoweverthisisgoodenoughtodistinguishsomevirusessuchastheEbolavirus

Clearlythesmallerλ the higher the resolving power and this is where the electron microscope comes to the fore The electron microscope makes use of the wave nature of electrons(see1315)IntheTEMelectronspassthrougha wafer thin sample and are then focused by a magnetic field onto a fluorescent screen orCCD (charge coupleddevicesee142)Electronsusedinanelectronmicroscopehavewavelengthstypicallyofabout5times10-12mHoweverthe numerical aperture of electron microscopes is considerablysmaller thanthatofanopticalmicroscopetypicallyabout002NonethelessthismeansthataTEMcanresolvetwopointsthatareabout025nmapartThisresolving power is certainly high enough to make out the shape of large molecules

Another type of electron microscope uses a techniqueby which electrons are scattered from the surface of the sample The scattered electrons and then focused as in theTEM to form an image of the surfaceThese so-calledscanningelectronmicroscopes(SEM)havealowerresolving power than TEMrsquos but give very good threedimensional images

The eye

We saw in the last section that the resolving power of the human eye is about 2 times 10-4 rad Suppose that youare looking at car headlights on a dark night and the car

PHYSICS 2015indb 302 150514 337 PM

Wave Phenomena

303

AH

L

is a distance D away If the separation of the headlight is say 15 m then the headlights will subtend an angle θ 15

Dθ =

at your eye Assuming an average wavelength of

500 nm your eye will resolve the headlights into twoseparatesourcesifthisangleequals2times10-4 rad This gives D=75kmInotherwordsifthecarisapproachingyouona straight road then you will be able to distinguish the two headlightsasseparatesourceswhenthecaris75kmawayfrom you Actually because of the structure of the retina and optical defects the resolving power of the average eye is about 6 times 10-4 rad This means that the car is more likely tobe25kmawaybeforeyouresolvetheheadlights

Astronomical telescope

Letusreturntotheexampleofthebinarystarsdiscussedatthe beginning of this section on resolution The stars Kruger A and B form a binary system The average separation of the stars is 14 times 1012mandtheiraveragedistancefromEarthis12times1017mWhenviewedthroughatelescopeonEarththe system will therefore subtend an angle

12

17

14 1012 10

θtimes

=times

= 12times10-5 rad at the objective lens of the telescope Assuming that the average wavelength of the light emitted bythestarsis500nmthenifthetelescopeistoresolvethesystem into two separate images it must have a minimum diameter Dwhere12times10-5 =

7122 500 10D

minustimes times

This gives D=0050m which is about 5 cm So thisparticular system is easily resolved with a small astronomical telescope

Exercise 961 ItissuggestedthatusingtheISASVLBAitwould

bepossibletoldquoseerdquoagrainofriceatadistanceof5000kmEstimatetheresolvingpoweroftheVLBA

2 Thedistancefromtheeyelenstotheretinais20mm The light receptors in the central part of the retinaareabout5times10-6apartDeterminewhetherthe spacing of the receptors will allow for the eye to resolve the headlights in the above discussion whentheyare25kmfromtheeye

3 ThediameterofPlutois23times106 m and its averagedistancefromEarthis60times1012 m EstimatetheminimumdiameteroftheobjectiveofatelescopethatwillenablePlutotobeseenasadisc as opposed to a point source

95 Doppler effect

Essential ideaTheDopplereffectdescribesthephenomenonofwavelengthfrequencyshiftwhenrelativemotionoccurs

Understandingsbull TheDopplereffectforsoundwavesandlightwaves

The Doppler effect for sound waves and light waves

Consider two observers A and B at rest with respect to asoundsourcethatemitsasoundofconstantfrequencyf Clearly both observers will hear a sound of the same frequencyHoweversupposethatthesourcenowmovesat constant speed towards A A will now hear a sound of frequencyfA that is greater than f and B will hear a sound of frequency fB that is less than f This phenomenon is known as the Doppler Effect orDopplerPrinciple after C J Doppler(1803-1853)

The same effect arises for an observer who is either moving towards or away from a stationary source

Figure 930 shows the waves spreading out from a stationary source that emits a sound of constant frequency f The observersAandBhearasoundofthesamefrequency

A B

wavefront

S

Figure 930 Sound waves from a stationary source

Suppose now that the source moves towards A withconstant speed vFigure9310(a)showsasnapshotofthenew wave pattern

NATURE OF SCIENCETechnology Although originally based on physical observations of the pitch of fast moving sources of sound the Doppler effect has an important role in many different areas such as evidence for the expansion of the universe and generating images used in weather reports and in medicine (55)

copy IBO 2014

PHYSICS 2015indb 303 150514 337 PM

Chapter 9

304

AH

L

sourceA B

smallerwavelength

largerwavelength

Figure 931 (a) Sound waves from a moving source

The wavefronts are now crowded together in the direction of travel of the source and stretched out in the opposite direction This is why now the two observers will now hear notesofdifferentfrequenciesHowmuchthewavesbunchtogether and how much they stretch out will depend on the speed vEssentially A

A

cfλ

= and BB

cfλ

= where λA lt λB and v is the speed of sound

IfthesourceisstationaryandAismovingtowardsitthenthe waves from the source incident on A will be bunched up If A is moving away from the stationary source then the waves from the source incident on A will be stretched out

Christian Doppler(1803ndash1853)actuallyappliedtheprinciple(incorrectlyasithappens)totryandexplainthecolourofstars However the Doppler effect does apply to light aswellastosoundIfalightsourceemitsalightoffrequencyf then if it is moving away from an observer the observer willmeasurethelightemittedashavingalowerfrequencythan fSincethesensationofcolourvisionisrelatedtothefrequencyoflight(bluelightisofahigherfrequencythanred light) light emitted by objects moving way from anobserver is often referred to as being red-shifted whereas if the object is moving toward the observer it is referred to as blue-shiftedThisideaisusedinOptionE(Chapter16)

We do not need to consider here the situations where either the source or the observer is accelerating In a situation for example where an accelerating source is approaching astationaryobserverthentheobserverwillhearasoundof ever increasing frequency This sometimes leads toconfusion in describing what is heard when a source approaches passes and then recedes from a stationaryobserver Suppose for example that you are standingon a station platform and a train sounding its whistle is approaching at constant speed What will you hear as the train approaches and then passes through the station As the train approaches you will hear a sound of constant pitch but increasing loudness The pitch of the sound will be greater than if the train were stationary As the train passes through the station you will hear the pitch change at the momentthetrainpassesyoutoasoundagainofconstantpitch The pitch of this sound will be lower than the sound of the approaching train and its intensity will decrease as the train recedes from you What you do not hear is a sound of increasing pitch and then decreasing pitch

The Doppler equations for sound

Although you will not be expected in an IB examination to derivetheequationsassociatedwithaspectsoftheDopplereffectyouwillbeexpectedtoapplythemForcompletenesstherefore thederivationoftheequationsassociatedwiththeDopplereffectasoutlinedaboveisgivenhere

S primeS O

vs

Figure 931 (b)

InFigure931(b)theobserverOisatrestwithrespecttoasourceofsoundSismovingwithconstanspeedvs directly towards O The source is emitting a note of constant frequencyf and the speed of the emitted sound is v

S shows the position of the source ∆t later When the sourceisatresttheninatime∆t the observer will receive f∆t waves and these waves will occupy a distance v∆t ie

v f

λ =f∆tv∆t =

(Because of the motion of the source this number of waves will now occupy a distance (v∆t ndash vs∆t)The lsquonewrsquowavelength is therefore

v ndash vs

fλrsquo =

f∆tv∆t ndash vs∆t

=

If fisthefrequencyheardbyOthen

vfλ

= or

vf

λ = = sv vfminus

From which

s

vf fv v

=minus

Dividingthroughbyv gives

Equation1111

1 sf f v

v=

minus

If the source is moving away from the observer then we have

s

vf fv v

=minus

1

1 sf vv

=+

Equation112

PHYSICS 2015indb 304 150514 337 PM

Wave Phenomena

305

AH

L

We now consider the case where the source is stationary and the observerismoving towardsthesource with speed v0 In this situation the speed of the sound waves as measured by the observer will be v0 + v We therefore have that

v0 + v = ffraslλ = vfλ

times

From which

o1 vf fv

= +

Equation113

If the observerismovingawayfromthesource then

o1 vf fv

= minus

Fromequation(113)wehavethat

o o1 v vf f f f f fv v

∆ = minus = + minus =

Equation114

Thevelocitiesthatwerefertointheaboveequationsarethevelocities with respect to the medium in which the waves fromthesourcetravelHoweverwhenwearedealingwithalight source it is the relative velocity between the source and the observer that we must consider The reason for this is that lightisuniqueintherespectthatthespeedofthelightwavesdoes not depend on the speed of the source All observers irrespective of their speed or the speed of the source will measure the same velocity for the speed of light This is oneof thecornerstonesof theSpecialTheoryofRelativitywhichisdiscussedinmoredetailinOptionH(Chapter18)When applying the Doppler effect to light we aremainlyconcerned with the motion of the source We look here only at the situation where the speed of the source v is much smaller than the speed of light cinfreespace(v ltlt c)Underthesecircumstanceswhenthesourceismovingtowardstheobserverequation111becomes

vf f f fc

minus = ∆ = Equation115

andwhenthesourceismovingawayfromtheobserverequation112becomes vf f f f

cminus = ∆ = minus

Provided that v ltlt c these same equations apply for astationary source and moving observer

We look at the following example and exercise

Example

Asourceemitsasoundoffrequency440HzItmovesinastraight line towards a stationary observer with a speed of 30ms-1Theobserverhearsasoundoffrequency484HzCalculate the speed of sound in air

Solution

We use equation 111 and substitute f= 484 Hz f = 440 Hz

and vs = 30 m s-1

therefore 1

484 440 301v

=minus

such that 30 4401

484vminus = to

give v = 330 m s-1

Example

A particular radio signal from a galaxy is measured as havingafrequencyof139times109HzThesamesignalfromasourceinalaboratoryhasafrequencyof142times109Hz

Suggestwhy thegalaxy ismovingaway fromEarthandcalculateitsrecessionspeed(iethespeedwithwhichitismovingawayfromEarth)

Solution

The fact that the frequency from the moving source is less than that when it is stationary indicates that it is moving away from the stationary observer ie Earth

Using vf fc

∆ = we have

8 96

9

3 10 (142 139) 10 634 10142 10

c fvf∆ times times minus times

= = = timestimes

m s-1

It is usual when dealing with the Doppler effect of light to express speeds as a fraction of c So in this instance we have v = 0021 c

Using the Doppler effectWe have seen in the above example and exercise how the Dopplereffectmaybeusedtomeasuretherecessionspeedof distant galaxies The effect is also used to measure speed inothersituationsHerewewilllookatthegeneralprincipleinvolved in using the Doppler effect to measure speedFigure932showsasource(thetransmitter)thatemitseithersound or em waves of constant frequency f The waves

PHYSICS 2015indb 305 150514 337 PM

Chapter 9

306

AH

L

fromthesourceare incidentonareflectorthat ismovingtowards the transmitter with speed vThereflectedwavesare detected by the receiver placed alongside the transmitter

reector

v

transmitter

receiver

f f

f f

Figure 932 Using the Doppler effect to measure speed

We shall consider the situation where v ltlt c where c is the speed of the waves from the transmitter

For the reflector receiving waves from the transmitterit is effectively an observer moving towards a stationary sourceFromequation(114) itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation116

For the receiver receivingwaves from the reflector it iseffectively a stationary observer receiving waves from a movingsourceFromequation(115)itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation117

Ifweaddequations(116)and(117)wegetthatthetotalDopplershiftatthereceiver∆f is

v vf f f f fc c

minus = ∆ = +

But 1 vf fc

= +

hence

1 v v vf f fc c c

∆ = + +

But since v ltlt cwe can ignore the term2

2

vc when we

expandthebracketintheaboveequation

Therefore we have2vf fc

∆ = Equation118

If v asymp c then we must use the full Doppler equationsHowever for em radiationwewill alwaysonly considersituations in which v ltlt c

Example

Thespeedofsoundinbloodis1500times103 m s-1 Ultrasound offrequency100MHzisreflectedfrombloodflowinginan arteryThe frequencyof the reflectedwaves receivedbackatthetransmitteris105MHzEstimatethespeedofthebloodflowintheartery

Solution

Using equation (118) we have

6 63

2005 10 1015 10

vtimes = times

timesto give v asymp 36 m s-1 (We have assumed that the ultrasound is incident at right angles to the blood flow)

Exercise 971 Judy is standing on the platform of a station A

high speed train is approaching the station in a straight line at constant speed and is sounding its whistleAsthetrainpassesbyJudythefrequencyof the sound emitted by the whistle as heard by Judychangesfrom640Hzto430HzDetermine

(a) thespeedofthetrain

(b) thefrequencyofthesoundemittedbythewhistle as heard by a person on the train (Speedofsound=330ms-1)

2 AgalaxyismovingawayfromEarthwithaspeedof00500cThewavelengthofaparticularspectral line in light emitted by atomic hydrogen inalaboratoryis656times10-7 m Calculate the valueofthewavelengthofthislinemeasuredinthelaboratoryinlightemittedfromasourceofatomic hydrogen in the galaxy

PHYSICS 2015indb 306 150514 337 PM

10 Fields

Contents

101 ndash Describing fields

102 ndash Fields at work

Essential Ideas

Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Similar approaches can be taken in analysing electrical and gravitational potential problems copy IBO 2014

PHYSICS 2015indb 307 150514 337 PM

Chapter 10

308

AH

L

101 Describing fields

Essential idea Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Understandingsbull Gravitationalfieldsbull Electrostaticfieldsbull Electricpotentialandgravitationalpotentialbull Fieldlinesbull Equipotentialsurfaces

Gravitational potentialWehaveseenthatifweliftanobjectofmassm to a height h abovethesurfaceoftheEarththenitsgainingravitationalpotential energy is mghHoweverthisisbynomeansthefullstoryForastartwenowknowthatgvarieswithh and also the expression really gives a difference in potentialenergybetweenthevaluethattheobjecthasattheEarthrsquossurfaceandthevaluethatithasatheighthSowhatwereallyneedisazeropointCanwefindapointwherethepotentialenergyiszeroandusethispointfromwhichtomeasure changes in potential energy

The point that is chosen is in fact infinity At infinity the gravitationalfieldstrengthofanyobjectwillbezeroSoletusseeifwecandeduceanexpressionforthegaininpotentialenergyofanobjectwhenitisldquoliftedrdquofromthesurfaceoftheEarthtoinfinityThisineffectmeansfindingtheworknecessarytoperformthistask

δr

r r δr+

r infin=

mg

AB

Me

Figure 1001 Gravitational forces

Inthediagramweconsidertheworknecessarytomovethe particle of mass m a distance δr in thegravitationalfield of the Earth

The force on the particle at A is FG Mem

r2----------------=

IftheparticleismovedtoB then since δr isverysmallwe can assume that the field remains constant over thedistance ABTheworkδW done againstthegravitationalfieldoftheEarthinmovingthedistanceABis

δWGMem

r2----------------δrndash=

(rememberthatworkdoneagainstaforceisnegative)

TofindthetotalworkdoneW in going from the surface oftheEarthtoinfinitywehavetoaddalltheselittlebitsofworkThisisdonemathematicallybyusingintegralcalculus

WG Me m

r2----------------ndash

rdR

infin

int G Me m 1

r2----- rd

R

infin

intndash G Me m 1r---ndash

R

infinndash= = =

GMe m 0 1R---ndash

ndashndash=

G MemR----------------ndash=

HencewehavewhereRistheradiusoftheEarththattheworkdonebythegravitationalfieldinmovinganobjectofmass m from R(surfaceoftheEarth)toinfinityisgivenby

WGMe m

R---------------- ndash=

We can generalise the result by calculating the worknecessary per unit mass to take a small mass fromthe surface of the Earth to infinity This we call thegravitational potential Vie

Wewouldgetexactlythesameresultifwecalculatedtheworkdonebythefieldtobringthepointmassfrominfinityto the surface of Earth In this respect the formal definition ofgravitationalpotentialatapointinagravitationalfieldis therefore defined as the work done per unit mass in bringing a point mass from infinity to that point

Clearly then the gravitational potential at any point inthe Earthrsquos field distance r from the centre of the Earth (providingr gt R)is

VGMe

r------------ ndash=

Thepotentialisthereforeameasureoftheamountofworkthathastobedonetomoveparticlesbetweenpointsinagravitationalfieldanditsunit is theJkgndash1 We also note thatthepotentialisnegativesothatthepotentialenergy

NATURE OF SCIENCEParadigm shift The move from direct observable actions being responsible for influence on an object to acceptance of a fieldrsquos ldquoaction at a distancerdquo required a paradigm shift in the world of science (23)

copy IBO 2014

PHYSICS 2015indb 308 150514 337 PM

Fields

309

AH

L

aswemoveawayfromtheEarthrsquossurfaceincreasesuntilitreachesthevalueofzeroatinfinity

IfthegravitationalfieldisduetoapointmassofmassmthenwehavethesameexpressionasaboveexceptthatMe is replaced by m andmustalsoexclude thevalueof thepotential at the point mass itself ie at r = 0

WecanexpressthegravitationalpotentialduetotheEarth(orduetoanysphericalmass)intermsofthegravitationalfield strength at its surface

AtthesurfaceoftheEarthwehave

g 0ndash Re GMRe

---------ndash= e

Sothat

g 0Re2 GM= e

Hence at a distance r from the centre of the Earth the gravitationalpotentialVcanbewrittenas

VGMe

r------------ndashg 0 Re

2

r------------ndash= =

The potential at the surface of the Earth

(r = Re)istherefore-g0Re

It is interesting to see how the expression for thegravitational potential ties in with the expression mgh ThepotentialatthesurfaceoftheEarthis-g0Re (see the exampleabove)andataheighthwillbe-g0g 0 Re h+( )ndash ifweassume that g0doesnotchangeover thedistanceh The differenceinpotentialbetweenthesurfaceandtheheighth is therefore g0hSotheworkneededtoraiseanobjectof mass m to a height h is mgh ie m timesdifference ingravitationalpotential

This we have referred to as the gain in gravitationalpotentialenergy(see235)

However this expression can be extended to any twopoints inanygravitationalfieldsuch that ifanobjectofmass mmovesbetween twopointswhosepotentialsareV1 and V2 respectively then the change in gravitationalpotentialenergyoftheobjectism(V1 ndash V2)

Gravitational potential gradient

Let us consider now a region in space where thegravitationalfieldisconstantInFigure912thetwopointsAandBareseparatedbythedistance∆x

A B

direction of uniform gravitational eld of strength I

∆x

Figure 1002 The gravitational potential gradient

The gravitational field is of strength I and is in the direction shownThe gravitational potential at A is V andatBisV +∆V

Theworkdone is takingapointmassmfromAtoB isF∆x = mI∆x

Howeverbydefinitionthisworkisalsoequalto-m∆V

Therefore mI∆x=-m∆V

or VIx

∆= minus

∆Effectivelythissaysthatthemagnitudeofthegravitationalfield strength is equal to the negative gradient of thepotential If I is constant then V is a linear function of x and Iisequaltothenegativegradientofthestraightlinegraph formed by plotting V against x If I is not constant (asusuallythecase)thenthemagnitudeofI at any point in the field can be found by find the gradient of the V-x graph at that point An example of such a calculation can befoundinSection929

For those of you who do HL maths the relationship between field and potential is seen to follow from the expression for the potential of a point mass viz

mV Gr

= minus

2

ddV mG Ir r

minus = + =

Potential due to one or more point masses

Gravitationalpotentialisascalarquantitysocalculatingthepotential due to more than one point mass is a matter of simpleadditionSoforexamplethepotentialV due to the Moon and Earth and a distance x fromthecentreofEarthonastraightlinebetweenthemisgivenbytheexpression

E MM MV Gx r x

= minus + minus

whereME=massofEarthMM= mass of Moon and r = distancebetweencentreofEarthandMoon

PHYSICS 2015indb 309 150514 337 PM

Chapter 10

310

AH

L

Equipotentials and field linesIfthegravitationalpotentialhasthesamevalueatallpointson a surface the surface is said to be an equipotential surfaceSoforexampleifweimagineasphericalshellaboutEarthwhosecentrecoincideswiththecentreofEarththisshellwillbeanequipotentialsurfaceClearlyifwerepresentthegravitationalfieldstrengthbyfieldlinessincethelinesldquoradiaterdquooutfromthecentreofEarththentheselineswillbeatrightanglestothesurfaceIfthefieldlineswerenotnormaltotheequipotentialsurfacethentherewouldbeacomponentof the field parallel to the surfaceThis wouldmean thatpointsonthesurfacewouldbeatdifferentpotentialsandsoitwouldnolongerbeanequipotentialsurfaceThisofcourseholdstrueforanyequipotentialsurface

Figure 1003 shows the field lines and equipotentials fortwopointmassesm

m m

Figure 1003 Equipotentials for two point masses

It is worth noting that we would get exactly the samepattern ifwewere to replace thepointmasseswith twoequalpointcharges(See935)

Escape speedThe potential at the surface of Earth is which meansthattheenergyrequiredtotakeaparticleofmass mfromthesurfacetoinfinityisequalto

ButwhatdoesitactuallymeantotakesomethingtoinfinityWhentheparticleisonthesurfaceoftheEarthwecanthinkofitassittingatthebottomofaldquopotentialwellrdquoasinfigure1004

particle

surface of Earth

innity

GMR---------

FIgure 1004 A potential well

Theldquodepthrdquoofthewellis and if the particle gains an amount of kinetic energy equal to wherem is its massthenitwillhavejustenoughenergytoldquoliftrdquoitoutofthewell

In reality it doesnrsquot actually go to infinity it justmeansthat the particle is effectively free of the gravitationalattractionof theEarthWe say that ithas ldquoescapedrdquo theEarthrsquosgravitationalpullWemeetthisideainconnectionwithmolecular forcesTwomolecules in a solidwill sitat their equilibrium position the separation where therepulsiveforceisequaltotheattractiveforceIfwesupplyjust enough energy to increase the separation of themolecules such that they are an infinite distance apart then intermolecularforcesnolongeraffectthemoleculesandthesolidwillhavebecomealiquidThereisnoincreaseinthekineticenergyofthemoleculesandsothesolidmeltsat constant temperature

Wecancalculatetheescapespeedofanobjectveryeasilyby equating the kinetic energy to the potential energysuch that

12---mvesc ape

2 GMe mRe

----------------=

vesc aperArr2GMe

Re--------------- 2g0 Re = =

Substituting for g0 and Regivesavalueforvescape of about 11kmsndash1 from the surface of the Earth

Youwillnotethattheescapespeeddoesnotdependonthemassoftheobjectsincebothkineticenergyandpotentialenergy are proportional to the mass

IntheoryifyouwanttogetarockettothemoonitcanbedonewithoutreachingtheescapespeedHoweverthiswouldnecessitate anenormousamountof fuel and it islikely that therocketplus fuelwouldbesoheavythat itwouldnevergetoffthegroundItismuchmorepracticalto accelerate the rocket to the escape speed fromEarthorbitandthenintheoryjustlaunchittotheMoon

Example

Use the followingdata todetermine thepotential at thesurface of Mars and the magnitude of the acceleration of free fall

massofMars =64times1023kg

radiusofMars =34times106 m

Determinealsothegravitationalfieldstrengthatadistanceof 68 times 106mabovethesurfaceofMars

PHYSICS 2015indb 310 150514 337 PM

Fields

311

AH

L

Solution

= minus = minus times times = times23

11 76

64 1067 10 ndash13 10

34 10MV GR

minus timestimes

Nkg-1

But V = -g0R

Therefore 7

0 6

13 10 3834 10

VgR

times= minus = =

timesm s-2

To determine the field strength gh at 68 times 106 m above the surface we use the fact that 0 2

Mg GR

= such that GM = g0R2

Therefore 2 2

0h 2 2 2

h h

38 (34) 042(102)

g RGMgR R

times= = = = m s-2

(the distance from the centre is 34 times 106 + 68 times 106 = 102 times 106 m)

Exercise 1011 Thegraphbelowshowshowthegravitational

potentialoutsideoftheEarthvarieswithdistancefrom the centre

0 10 20 30 40 50 60 70

ndash1

ndash2

ndash3

ndash4

ndash5

ndash6

ndash7

r m times 106

V

Jkgndash1

times 10

7

(a) Usethegraphtodeterminethegainingravitationalpotentialenergyofasatelliteofmass200kgasitmovesfromthesurfaceof the Earth to a height of 30 times 107mabovetheEarthrsquossurface

(b) Calculatetheenergyrequiredtotakeit to infinity

(c) DeterminetheslopeofthegraphatthesurfaceoftheEarthmCommenton youranswer

Electric potential energy

The concept of electric potential energy was developedwith gravitational potential energy in mind Just as anobjectnearthesurfaceoftheEarthhaspotentialenergybecause of its gravitational interaction with the Earthsotoothereiselectricalpotentialenergyassociatedwithinteracting charges

Letusfirstlookatacaseoftwopositivepointchargeseachof 1μC that are initially bound together by a thread in a vacuuminspacewithadistancebetweenthemof10cmasshowninFigure1005Whenthethreadiscutthepointchargesinitiallyatrestwouldmoveinoppositedirectionsmovingwithvelocitiesv1 and v2 along the direction of the electrostatic force of repulsion

BEFORE AFTER

v v21

Figure 1005 Interaction of two positive particles

The electric potential energybetweentwopointchargescan be found by simply adding up the energy associated witheachpairofpointchargesForapairofinteractingchargestheelectricpotentialenergyisgivenby

∆U = ∆Ep + ∆Ek = ∆W = ∆Fr = kqQ

____ r2 times r = kqQ

____ r

Becausenoexternalforceisactingonthesystemtheenergyand momentum must be conserved Initially Ek = 0 and Ep =kqQr=9times109 times 1 times 10-1201m=009JWhentheyareagreatdistance fromeachotherEpwillbenegligibleThefinalenergywillbeequaltofrac12mv1

2+frac12mv22=009J

Momentum isalsoconservedand thevelocitieswouldbethe same magnitude but in opposite directions

Electric potential energyismoreoftendefinedintermsofapointchargemovinginanelectricfieldas

lsquothe electricpotential energybetweenany twopoints inan electric field is defined as negativeoftheworkdone by anelectricfieldinmovingapoint electric chargebetweentwolocationsintheelectricfieldrsquo

ΔU = ΔEp = -ΔW = -Fd = qEx

ΔU = ΔEp +ΔEk = ΔW = ΔFr = kqQ r2 x r = kqQ

wherexisthedistancemovedalong(oroppositeto)thedirection of the electric field

Electric potential energyismeasuredinjoule(J)Justasworkisascalarquantitysotooelectricalpotentialenergyis a scalar quantity The negative of the work done by an electric field in moving a unit electric charge between two points is independent of the path taken Inphysicswesaytheelectricfieldisalsquoconservativersquo field

PHYSICS 2015indb 311 150514 337 PM

Chapter 10

312

AH

L

Supposeanexternalforcesuchasyourhandmovesasmallpositive point test charge in the direction of a uniformelectric fieldAs it ismoving itmust be gaining kineticenergyIfthisoccursthentheelectricpotentialenergyofthe unit charge is changing

InFigure1006apointcharge+qismovedbetweenpointsAandBthroughadistancex in a uniform electric field

B

A

x

+ q

Figure 1006 Movement of a positive point charge in a uniform field

InordertomoveapositivepointchargefrompointAtopointBanexternal forcemustbeapplied to thechargeequaltoqE (F = qE)

Since the force is applied through a distance xthennegativeworkhas to bedone tomove the charge because energyis gainedmeaning there is an increaseelectric potential energybetweenthetwopointsRememberthattheworkdoneisequivalenttotheenergygainedorlostinmovingthe charge through the electric field The concept of electric potential energy is only meaningful as the electric field whichgeneratestheforceinquestionisconservative

W F xtimes E q x times= =

θ x cos θ x

Figure 1007 Charge moved at an angle to the field

If a chargemoves at an angle θ to an electric field thecomponent of the displacement parallel to the electric fieldisusedasshowninFigure918

W F x Eq x θcostimes= =

Theelectricpotentialenergyisstoredintheelectricfieldand theelectricfieldwill return theenergy to thepointcharge when required so as not to violate the Law ofconservationofenergy

Electric potential

The electric potential at a point in an electric field is defined as being the work done per unit charge in bringing a small positive point charge from infinity to that point

ΔV = Vinfin ndash Vf = ndash W ___ q

Ifwedesignatethepotentialenergytobezeroatinfinitythen it follows thatelectricpotentialmustalsobezeroat infinity and the electric potential at any point in an electricfieldwillbe

ΔV = ndash W ___ q

NowsupposeweapplyanexternalforcetoasmallpositivetestchargeasitismovedtowardsanisolatedpositivechargeTheexternalforcemustdoworkonthepositivetestchargetomoveittowardstheisolatedpositivechargeandtheworkmustbepositivewhiletheworkdonebytheelectricfieldmustthereforebenegativeSotheelectricpotentialatthatpointmustbepositiveaccordingtotheaboveequationIfanegativeisolatedchargeisusedtheelectricpotentialatapointonthepositivetestchargewouldbenegativePositivepoint chargesof their ownaccordmove fromaplaceofhighelectricpotential toaplaceof lowelectricpotentialNegative point charges move the other way from lowpotentialtohighpotentialInmovingfrompointAtopointBinthediagramthepositivecharge+qismovingfromalowelectricpotentialtoahighelectricpotential

In thedefinitiongiven the termldquoworkperunitchargerdquohas significance If the test charge is +16 times 10-19Cwherethe charge has a potential energy of 32 times 10-17Jthenthepotentialwouldbe32times10-17J+16times10-19C=200JC-1 Now if the charge was doubled the potential wouldbecome 64times10-17 J However the potential per unitchargewouldbethesame

Electric potential is a scalar quantityandithasunitsJC-1 orvoltswhere1voltequalsone joulepercoloumbThevoltallowsustoadoptaunitfortheelectricfieldintermsofthevolt

PreviouslytheunitfortheelectricfieldwasNCndash1

W = qV and F = qEsoW ___ V = F __ E

E = FV ___ W = FV ___ Fm V mndash1

That is the units of the electric field E can also beexpressed as V mndash1

The work done per unit charge in moving a point charge between two points in an electric field is again independent of the path taken

PHYSICS 2015indb 312 150514 337 PM

Fields

313

AH

L

Electric potential due to a point charge

Letus takeapoint rmetres froma chargedobjectThepotentialatthispointcanbecalculatedusingthefollowing

W = ndashFr = ndashqV and F = ndash q1 q2 _____ 4πε0r

2

Therefore

W = ndash q1 q2 _____ 4πε0r

2 times r = ndash q1 q2 _____ 4πε0r

= ndashq1 times q2 _____ 4πε0r

= ndashq1V

That is V q

4πε0r--------------- =

Orsimply

V kqr------

=

Example

Determinehowmuchworkisdonebytheelectricfieldofpointcharge150μCwhenachargeof200μCismovedfrom infinity to apoint 0400m from thepoint charge(Assumenoaccelerationofthecharges)

Solution

The work done by the electric field is W = -qV = -14πε0 times q times (Q rinfin - Q r0400)

W = (- 200 times 10-6 C times 900 times 109 NmC-2 times 150 times 10-6 C) divide 0400 m = - 0675 J

An external force would have to do +0675 J of work

Electric field strength and electric potential gradient

Letus lookbackatFigure1006Supposeagain that thecharge+qismovedasmalldistancebyaforceFfromAtoBsothattheforcecanbeconsideredconstantTheworkdoneisgivenby

∆W F x∆times=

TheforceFandtheelectricfieldEareoppositelydirectedandweknowthat

F = -qE and ΔW = q ΔV

Thereforetheworkdonecanbegivenas

q ΔV = -q E Δx

Therefore E V∆x∆-------

ndash=

The rate of change of potential ΔV at a point with respect to distance Δx in the direction in which the change is maximum is called the potential gradient We say that the electric field = - the potential gradient and the units are Vm-

1 From the equation we can see that in a graph of electric potential versus distance the gradient of the straight line equals the electric field strength

In reality if a charged particle enters a uniform electric field it will be accelerated uniformly by the field and its kinetic energy will increase This is why we had to assume no acceleration in the last worked example

Ek∆ 12---

mv2 q E xsdot sdot q Vx---

xsdot sdot q Vsdot= = = =

Example

Determine how far apart two parallel plates must besituated so that a potential difference of 150 times 102 V produces an electric field strength of 100 times 103 NC-1

Solution

Using E V∆x------- xhArrndash V∆

E-------15 102 Vtimes

100 103 N C 1ndashtimes------------------------------------------= = =

= 150 times 10-1

The plates are 150 times 10-1 m apart

The electric field and the electric potential at a point due to an evenly distributed charge +q on a sphere can be represented graphically as in Figure 1008

r

r

V

E

r0

r0

r0

Charge of +Q evenly distributed over surface

E 14πε0------------ Q

r2----- r r0gt=

V 14πε0------------Q

r----r r0gt=

Figure 1008 Electric field and potential due to a charged sphere

PHYSICS 2015indb 313 150514 337 PM

Chapter 10

314

AH

L

When the sphere becomes charged we know that thechargedistributes itselfevenlyoverthesurfaceThereforeeverypartofthematerialoftheconductorisatthesamepotential As the electric potential at a point is defined as beingnumericallyequaltotheworkdoneinbringingaunitpositivechargefrominfinitytothatpointithasaconstantvalueineverypartofthematerialoftheconductor

Since the potential is the same at all points on the conducting surfacethen∆V∆xiszeroButE=ndash∆V∆xThereforethe electric field inside the conductor is zero There is no electric field inside the conductor

SomefurtherobservationsofthegraphsinFigure1009are

bull OutsidethespherethegraphsobeytherelationshipsgivenasE α 1 r2 and V α 1 r

bull Atthesurfacer = r0Thereforetheelectricfieldandpotentialhavetheminimumvalueforr at this point and this infers a maximum field and potential

bull Insidethespheretheelectricfieldiszerobull Insidethespherenoworkisdonetomoveacharge

fromapointinsidetothesurfaceThereforethereisno potential difference and the potential is the same asitiswhenr = r0

Similargraphscanbedrawnfortheelectricfieldintensityandthe electric potential as a function of distance from conducting parallelplatesandsurfacesandthesearegiveninFigure1009

Potential plot E eld plotE eld

++++

ndashndashndashndash

xx x

+

x x

ndashx x

+ ndashx

x

+ +

x

x

Figure 1009 Electric field and electric potential at a distance from a charged surface

Potential due to one or more point charges

The potential due to one point charge can be determined byusingtheequationformula

V = kq r

Example 1

Determine the electric potential at a point 20 times 10-1 m fromthecentreofan isolatedconducting spherewithapointchargeof40pCinair

Solution

Using the formula

V = kq r we have

V90 10 9times( ) 40 10 12ndashtimes( )times

20 10 1ndashtimes( )------------------------------------------------------------------ 018 V= =

the potential at the point is 180 times 10-1 V

The potential due to a number of point charges can be determined by adding up the potentials due to individual point charges because the electric potential at any point outside a conducting sphere will be the same as if all the charge was concentrated at its centre

Example 2

Threepointchargesofareplacedattheverticesofaright-angledtriangleasshowninthediagrambelowDeterminetheabsolutepotentialat the+20μCchargeduetothetwoothercharges

- 6μC

+3μC +2μC

4 m

3 m

Solution

The electric potential of the +2 μC charge due to the ndash 6 μC charge is

V = (9 times 109 Nm2C-2 times -6 times 10-6 C) divide (radic 32 + 42) m = - 108 times 104 V

PHYSICS 2015indb 314 150514 338 PM

Fields

315

AH

L

Insummarywecanconcludethat

bull No work is done to move a charge along an equipotential

bull Equipotentials are always perpendicular to the electric lines of force

Figure1011and1012showsomeequipotential lines fortwo oppositely charged and identically positive spheresseparated by a distance

+ve ndashve

equipotential lines

Figure 1011 Equipotential lines between two opposite charges

equipotential lines

Figure 1012 Equipotential lines between two charges which are the same

+

ndash ndash ndash ndash

+ + +

40 V

30 V

20 V

10 V

50 V

Figure 1013 Equipotential lines between charged parallel plates

Figure1013showstheequipotentiallinesbetweenchargedparallel plates Throughout this chapter the similarities and differences between gravitational fields and electric fieldshavebeendiscussedTherelationshipsthatexistsbetweengravitationalandelectricquantitiesandtheeffectsofpointmassesandchargesissummarisedinTable1014

The electric potential of the +2 μC charge due to the +3 μC charge is

V = (9 times 109 Nm2C-2 times 3 times 10-6 C) divide 3m = 9 times 103 V

The net absolute potential is the sum of the 2 potentials- 108 times 104 V + 9 times 103 V =- 18 times 103 V

The absolute potential at the point is - 18 times 103 V

Equipotential surfacesRegions in space where the electric potential of a chargedistributionhasaconstantvaluearecalledequipotentials The placeswherethepotentialisconstantinthreedimensionsarecalled equipotential surfacesandwheretheyareconstantintwodimensionstheyarecalledequipotential lines

TheyareinsomewaysanalogoustothecontourlinesontopographicmapsInthiscasethegravitationalpotentialenergy is constant as amassmoves around the contourlines because the mass remains at the same elevationabovetheEarthrsquossurfaceThegravitationalfieldstrengthacts in a direction perpendicular to a contour line

Similarlybecausetheelectricpotentialonanequipotentiallinehas thesamevalueanelectric forcecandonoworkwhena testchargemovesonanequipotentialThereforethe electric field cannot have a component along anequipotentialandthusitmustbeeverywhereperpendiculartotheequipotentialsurfaceorequipotentiallineThisfactmakesiteasytoplotequipotentialsifthelinesofforceorlinesofelectricfluxofanelectricfieldareknown

Forexamplethereareaseriesofequipotentiallinesbetweentwoparallelplateconductorsthatareperpendiculartotheelectric fieldTherewill be a series of concentric circles(eachcirclefurtherapartthanthepreviousone)thatmapouttheequipotentialsaroundanisolatedpositivesphereThe lines of force and some equipotential lines for anisolatedpositivesphereareshowninFigure1010

Lines ofequipotential

Figure 1010 Equipotentials around an isolated positive sphere

PHYSICS 2015indb 315 150514 338 PM

Chapter 10

316

AH

L

Gravitational quantity Electrical quantityQ

uant

ities

g V∆x∆-------ndash= E V∆

x∆-------ndash=

Poin

t mas

ses a

nd ch

arge

s

V Gmr----

ndash= V 14πε0------------q

r---=

g G m

r2-----ndash= E 14πε0------------ q

r2-----=

F Gm1 m2

r2--------------= F 1

4πε0------------

q1q2

r2-----------=

Figure 1014 Formulas

Example

Deduce the electric potential on the surface of a gold nucleus that has a radius of 62 fm

Solution

Using the formula

V = kq r and knowing the atomic number of gold is 79 We will assume the nucleus is spherical and it behaves as if it were a point charge at its centre (relative to outside points)

V = 90 times 109 Nm2C-2 times 79 times 16 times 10-19 C divide 62 times 10-15 m = 18 times 107 V

The potential at the point is 18 MV

Example

Deduce the ionisation energy in electron-volts of theelectron in the hydrogen atom if the electron is in its ground state and it is in a circular orbit at a distance of 53 times 10-11 m from the proton

Solution

This problem is an energy coulombic circular motion question based on Bohrrsquos model of the atom (not the accepted quantum mechanics model) The ionisation energy is the energy required to remove the electron from the ground state to infinity The electron travels in a circular orbit and therefore has a centripetal acceleration The ionisation energy will counteract the coulombic force and the movement of the electron will be in the opposite direction to the centripetal force

Total energy = Ek electron + Ep due to the proton-electron interaction

ΣF = kqQ r2 = mv2 r and as such mv2 = = kqQ r

Therefore Ek electron = frac12 kqQ r

Ep due to the proton-electron interaction = - kqQ r

Total energy = frac12 kqQ r + - kqQ r = -frac12 kqQ r

= - 90 times 109 Nm2C-2 times (16 times 10-19 C)2 divide 53 times 10-11 m = -217 times 10-18 J

= -217 times 10-18 J divide 16 times 10-19 = -136 eV

The ionisation energy is 136 eV

Exercise 1021 ApointchargePisplacedmidwaybetweentwo

identicalnegativechargesWhichoneofthefollowingiscorrectwithregardstoelectricfieldandelectricpotentialatpointP

Electric field Electric potentialA non-zero zeroB zero non-zeroC non-zero non-zeroD zero zero

2 Twopositivechargedspheresaretiedtogetherinavacuumsomewhereinspacewheretherearenoexternal forces A has a mass of 25 g and a charge of20μCandBhasamassof15gandachargeof30μCThedistancebetweenthemis40cmTheyarethenreleasedasshowninthediagram

BEFORE AFTER

v1 v2

A B

(a) Determinetheirinitialelectricpotentialenergy in the before situation

(b) DeterminethespeedofsphereBafterrelease

3 Thediagrambelowrepresentstwoequipotentiallines in separated by a distance of 5 cm in a uniform electric field

+ + + + + + + +

ndash ndash ndash ndash ndash ndash ndash ndash

40 V

20 V 5 cm

PHYSICS 2015indb 316 150514 338 PM

Fields

317

AH

L

Determine the strength of the electric field

4 Thisquestionisabouttheelectricfieldduetoacharged sphere and the motion of electrons in that fieldThediagrambelowshowsanisolatedmetalsphereinavacuumthatcarriesanegativeelectriccharge of 60 μC

ndash

(a) Onthediagramdrawtheconventionalwayto represent the electric field pattern due to the charged sphere and lines to represent threeequipotentialsurfacesintheregionoutside the sphere

(b) Explainhowthelinesrepresentingtheequipotentialsurfacesthatyouhavesketchedindicatethatthestrengthoftheelectricfieldisdecreasingwithdistancefrom the centre of the sphere

(c) Theelectricfieldstrengthatthesurfaceofthe sphere and at points outside the sphere can be determined by assuming that the sphere acts as a point charge of magnitude 60 μC at its centre The radius of the sphere is 25 times 10ndash2 m Deduce that the magnitude of the field strength at the surface of the sphere is 86 times 107 Vmndash1

An electron is initially at rest on the surface of the sphere

(d) (i) Describethepathfollowedbytheelectronasitleavesthesurfaceofthesphere

(ii) Calculatetheinitialaccelerationofthe electron

5 Determinetheamountofworkthatisdoneinmovingachargeof100nCthroughapotentialdifferenceof150times102 V

6 Three identical 20 μC conducting spheres are placedatthecornersofanequilateraltriangleofsides 25 cm The triangle has one apex C pointing upthepageand2baseanglesAandBDeterminetheabsolutepotentialatB

7 Determinehowfaraparttwoparallelplatesmustbesituatedsothatapotentialdifferenceof250 times 102 V produces an electric field strength of 200 times 103 NC-1

8 Thegapbetweentwoparallelplatesis10times10-3mandthereisapotentialdifferenceof10times104 V betweentheplatesCalculate

i theworkdonebyanelectroninmovingfrom one plate to the other

ii thespeedwithwhichtheelectronreachesthe second plate if released from rest

iii theelectricfieldintensitybetweentheplates

9 Anelectronguninapicturetubeisacceleratedby a potential 25 times 103VDeterminethekineticenergygainedbytheelectroninelectron-volts

10 Determine the electric potential 20 times 10-2 m from achargeof-10times10-5 C

11 Determinetheelectricpotentialatapointmid-waybetweenachargeofndash20pCandanotherof+5pConthelinejoiningtheircentresifthecharges are 10 cm apart

12 During a thunderstorm the electric potential differencebetweenacloudandthegroundis10 times 109 V Determine the magnitude of the change in electric potential energy of an electron thatmovesbetweenthesepointsinelectron-volts

13 A charge of 15 μC is placed in a uniform electric fieldoftwooppositelychargedparallelplateswithamagnitudeof14times103 NC-1

(a) Determinetheworkthatmustbedoneagainstthefieldtomovethepointchargeadistance of 55 cm

(b) Calculatethepotentialdifferencebetweenthe final and initial positions of the charge

(c) Determinethepotentialdifferencebetweenthe plates if their separation distance is 15 cm

14 Duringaflashoflightningthepotentialdifferencebetweenacloudandthegroundwas12 times 109 V and the amount of transferred charge was32C

(a) Determinethechangeinenergyofthetransferred charge

(b) Iftheenergyreleasedwasallusedtoacceleratea1tonnecardeduceitsfinalspeed

(c) Iftheenergyreleasedcouldbeusedtomelticeat0degCdeducetheamountoficethatcould be melted

PHYSICS 2015indb 317 150514 338 PM

Chapter 10

318

AH

L

15 SupposethatwhenanelectronmovedfromAtoBin the diagram along an electric field line that the electric field does 36 times 10-19Jofworkonit

Determinethedifferencesinelectricpotential

(a) VB ndash VA(b) VC ndash VA(c) VC ndash VB

16 DeterminethepotentialatpointPthatislocatedatthecentreofthesquareasshowninthediagrambelow

- 6μC

+3μC +2μC

1m

1m

5μC

P

102 Fields at work

Essential idea Similar approaches can be taken inanalysingelectricalandgravitationalpotentialproblems

Understandingsbull Potentialandpotentialenergybull Potentialgradientbull Potentialdifferencebull Escapespeedbull Orbitalmotionorbitalspeedandorbitalenergybull Forcesandinverse-squarelawbehaviour

Orbital motion orbital speed and orbital energy

Although orbital motion may be circular elliptical orparabolic this sub-topic only dealswith circular orbitsThis sub-topic is not fundamentally new physics butan application that synthesizes ideas from gravitationcircularmotiondynamicsandenergy

TheMoonorbitstheEarthandinthissenseitisoftenreferredtoasasatelliteoftheEarthBefore1957itwastheonlyEarthsatelliteHoweverin1957theRussianslaunchedthefirstmanmadesatelliteSputnik1SincethisdatemanymoresatelliteshavebeenlaunchedandtherearenowliterallythousandsofthemorbitingtheEarthSomeareusedtomonitortheweathersome used to enable people to find accurately their position on thesurfaceoftheEarthmanyareusedincommunicationsandnodoubtsomeareusedtospyonothercountriesFigure1015showshowinprincipleasatellitecanbeputintoorbit

Theperson(whosesizeisgreatlyexaggeratedwithrespectto Earth) standing on the surface on the Earth throwssome stonesThe greater the speed with which a stoneis thrown the further it will land from her The pathsfollowedbythethrownstonesareparabolasByastretchof the imaginationwecanvisualisea situation inwhichastoneis thrownwithsuchaspeedthatbecauseof thecurvatureoftheEarthitwillnotlandonthesurfaceoftheEarthbutgointoldquoorbitrdquo(Path4onFigure1015)

NATURE OF SCIENCECommunication of scientific explanations The ability to apply field theory to the unobservable (charges) and the massively scaled (motion of satellites) required scientists to develop new ways to investigate analyse and report findings to a general public used to scientific discoveries based on tangible and discernible evidence (51)

copy IBO 2014

PHYSICS 2015indb 318 150514 338 PM

Fields

319

AH

L

1

2

34Earth

Figure 1015 Throwing a stone into orbit

TheforcethatcausesthestonestofollowaparabolicpathandtofalltoEarthisgravityandsimilarlytheforcethatkeeps the stone in orbit is gravity For circular motionto occur we have seen that a force must act at rightanglestothevelocityofanobjectthatistheremustbeacentripetalforceHenceinthesituationwedescribeherethe centripetal force for circular orbital motion about the EarthisprovidedbygravitationalattractionoftheEarth

We can calculate the speedwithwhich a stonemustbethrowninordertoputitintoorbitjustabovethesurfaceof the Earth

If the stone has mass m and speed v thenwehave fromNewtonrsquos2ndlaw

mv2

RE---------- G

MEm

RE2------------=

whereRE is the radius of the Earth and ME is the mass of the Earth

Bearinginmindthatg 0 GME

RE2--------= then

v g RE 10 64 106timestimes 8 103times= = =

Thatisthestonemustbethrownat8times103m sndash1

Clearly we are not going to get a satellite into orbit soclosetothesurfaceoftheEarthMovingatthisspeedthefrictionduetoairresistancewouldmeltthesatellitebeforeithadtravelledacoupleofkilometresInrealitythereforeasatelliteisputintoorbitabouttheEarthbysendingitattachedtoarocketbeyondtheEarthrsquosatmosphereandthengivingitacomponentofvelocityperpendiculartoaradialvectorfromtheEarthSeeFigure1016

Earth Satellite carried by rocket to hereSatellite orbit

Tangential component of velocity

Figure 1016 Getting a satellite into orbit

Keplerrsquos third law(This work of Kepler and Newtonrsquos synthesis of the work is an excellent example of the scientific method and makes for a good TOK discussion)

In 1627 Johannes Kepler (1571-1630)publishedhis lawsof planetary motion The laws are empirical in natureandwere deduced from the observations of theDanishastronomer Tycho de Brahe (1546-1601)The third lawgives a relationship between the radius of orbit R of a planet and its period TofrevolutionabouttheSunThelawisexpressedmathematicallyas

2

3 constantTR

=

We shall nowuseNewtonrsquos LawofGravitation to showhowitisthattheplanetsmoveinaccordancewithKeplerrsquosthirdlaw

InessenceNewtonwasabletousehislawofgravitytopredict the motion of the planets since all he had to do was factor theF givenby this law intohis second law F = ma to find their accelerations and hence their future positions

InFigure1017theEarthisshownorbitingtheSunandthedistancebetweentheircentresisR

RSun

Earth

F es

F se

Figure 1017 Planets move according to Keplerrsquos third law

Fes is the force that the Earth exerts on the Sun and Fse is the force that the Sun exerts on the Earth The forces are equalandoppositeandtheSunandtheEarthwillactuallyorbitaboutacommoncentreHoweversince theSun issoverymuchmoremassivethantheEarththiscommoncentrewillbeclosetothecentreoftheSunandsowecanregard the Earth as orbiting about the centre of the Sun Theotherthingthatweshallassumeisthatwecanignorethe forces that the other planets exert on the Earth (This wouldnotbeawisethingtodoifyouwereplanningtosend a space ship to theMoon for example)We shallalso assume that we have followed Newtonrsquos exampleandindeedprovedthataspherewillactasapointmasssituated at the centre of the sphere

Kepler had postulated that the orbits of the planets areellipticalbut since theeccentricityof theEarthrsquosorbit issmallweshallassumeacircularorbit

PHYSICS 2015indb 319 150514 338 PM

Chapter 10

320

AH

L

TheaccelerationoftheEarthtowardstheSunisa = Rω2

whereω ω 2πT------=

Hencea R 2π

T------ times

2 4π2 RT 2------------- = =

But the acceleration is given by Newtonrsquos Second Law F = mawhereFisnowgivenbytheLawofGravitationSo in this situation

F maGMs Me

R2-------------------= = butwealsohavethat

a a R 2πT

------ times

2 4π2 RT 2------------- = = and m = Me so that

G MsMe

R2------------------- Me4π2R

T2-------------times

GMs

4π2-----------rArr R3

T2------= =

Butthequantity

isaconstantthathasthesamevalueforeachoftheplanetssowehaveforalltheplanetsnotjustEarththat

R3

T2------ k =

wherekisaconstantWhichisofcourseKeplerrsquosthirdlaw

ThisisindeedanamazingbreakthroughItisdifficulttorefutethe idea that all particles attract each other in accordance withtheLawofGravitationwhenthelawisabletoaccountfortheobservedmotionoftheplanetsabouttheSun

The gravitational effects of the planets upon each othershould produce perturbations in their orbits Such is the predictivepoweroftheUniversalGravitationalLawthatit enabled physicists to compute these perturbations The telescopehadbeeninventedin1608andbythemiddleofthe 18th Century had reached a degree of perfection in design that enabled astronomers to actually measure the orbital perturbations of the planets Their measurements werealways inagreementwith thepredictionsmadebyNewtonrsquoslawHoweverin1781anewplanetUranuswasdiscovered and the orbit of this planet did not fit withthe orbit predicted by Universal Gravitation Such wasthe physicistrsquos faith in theNewtonianmethod that theysuspectedthatthediscrepancywasduetothepresenceofayetundetectedplanetUsingtheLawofGravitationtheFrenchastronomerJLeverrier and the English astronomer J C Adamswereable tocalculate justhowmassive thisnewplanetmustbeandalsowhereitshouldbeIn1846the planetNeptunewas discovered justwhere they hadpredicted Ina similarwaydiscrepancies in theorbitofNeptune ledto thepredictionandsubsequentdiscoveryin1930oftheplanetPlutoNewtonrsquosLawofGravitation

had passed the ultimate test of any theory it is not only abletoexplainexistingdatabutalsotomakepredictions

Satellite energyWhenasatelliteisinorbitaboutaplanetitwillhavebothkineticenergyandgravitationalpotentialenergySupposewe consider a satellite ofmass m that is in an orbit of radius r about a planet of mass M

Thegravitationalpotentialduetotheplanetatdistancer from its centre is

ThegravitationalpotentialenergyofthesatelliteVsat

is therefore GMe m

r----------------ndash

ThatisVsat GMem

r----------------ndash=

Thegravitationalfieldstrengthatthesurfaceoftheplanetisgivenby

g 0GMe

Re2------------=

Hencewecanwrite

Vsat g0 Re

2

rndash=

m

ThekineticenergyofthesatelliteKsatisequaltofrac12mv2wherev is its orbital speed

Byequatingthegravitationalforceactingonthesatellitetoitscentripetalaccelerationwehave

G Mem

r2---------------- mv2

r---------- mv2hArr

GMemr----------------= =

Fromwhich12---

mv2 12---

GMemr----------------times=

= 0 e

2g R

r

2m

Which is actually quite interesting since it shows thatirrespective of the orbital radius the KE is numericallyequal to half the PE Also the total energy E tot K sat Vsat+ 1

2---G Me m

r----------------times G Me m

r----------------ndash + 1

2---G Me m

r----------------ndash= = =of the satelliteisalwaysnegativesince

E tot K sat Vsat+ 12---

G Me mr----------------times

G Me mr----------------ndash

+ 12---

G Me mr----------------ndash= = =

PHYSICS 2015indb 320 150514 338 PM

Fields

321

AH

L

The energies of an orbiting satellite as a function of radial distancefromthecentreofaplanetareshownplottedinFigure1018

12

100806

0402

0ndash02

ndash04

ndash06ndash08ndash10

ndash12

2 4 6 8 10 12

ener

gya

rbita

ry u

nits

distance timesR

kinetic energytotal energy

potential energy

Figure 1018 Energy of an orbiting satellite as a function of distance from the centre of a planet

WeightlessnessSuppose that you are inanelevator(lift)thatisdescendingat constant speedandyou let goof abook thatyouareholdinginyourhandThebookwillfalltothefloorwithaccelerationequaltotheaccelerationduetogravityIfthecablethatsupportstheelevatorweretosnap(asituationthatItrustwillneverhappentoanyofyou)andyounowletgothebookthatyouareholdinginyourotherhandthis book will not fall to the floor - it will stay exactlyin linewithyourhandThis isbecause thebook isnowfallingwith the sameaccelerationas theelevatorandassuch the book cannot ldquocatchrdquo up with the floor of theelevatorFurthermoreifyouhappenedtobestandingonasetofbathroomscalesthescaleswouldnowreadzero-youwouldbeapparentlyweightlessItisthisideaoffreefallthatexplainstheapparentweightlessnessofastronautsin an orbiting satellite These astronauts are in free fall inthesensethattheyareacceleratingtowardsthecentre of the Earth

It is actually possible to define theweight of a body inseveral different waysWe can define it for example asthegravitationalforceexertedonthebodybyaspecifiedobject such as the EarthThiswe have seen thatwe doinlotsofsituationswherewedefinetheweightasbeingequaltomgIfweusethisdefinitionthenanobjectinfreefallcannotbydefinitionbeweightlesssinceitisstillinagravitationalfieldHoweverifwedefinetheweightofanobjectintermsofaldquoweighingrdquoprocesssuchasthereadingonasetofbathroomscaleswhichineffectmeasuresthecontact force between the object and the scales thenclearlyobjectsinfreefallareweightlessOnenowhasto

askthequestionwhetherornotitispossibleForexampletomeasurethegravitationalforceactingonanastronautin orbit about the Earth

WecanalsodefineweightintermsofthenetgravitationalforceactingonabodyduetoseveraldifferentobjectsForexample for an object out in space its weight could bedefined in terms of the resultant of the forces exerted on it bytheSuntheMoontheEarthandalltheotherplanetsin the Solar System If this resultant is zero at a particular pointthenthebodyisweightlessatthispoint

Inviewofthevariousdefinitionsofweightthatareavailabletousitisimportantthatwhenweusethewordldquoweightrdquoweareawareofthecontextinwhichitisbeingused

Example

Calculate the height above the surface of the Earth atwhichageo-stationarysatelliteorbits

Solution

A geo-stationary satellite is one that orbits the Earth in such a way that it is stationary with respect to a point on the surface of the Earth This means that its orbital period must be the same as the time for the Earth to spin once on its axis ie 24 hours

From Keplerrsquos third law we have G Ms

4π2----------- R3

T2------=

That is

RMe

mh

using the fact that the force of attraction between the satellite and the Earth is given by

FG Mem

R2----------------=

and that F = ma

where m is the mass of the satellite and a 4π2R

T2-------------=

we have

GMem

R2---------------- m 4π2R

T2-------------times

GMe

4π2------------rArr R3

T2------= =

PHYSICS 2015indb 321 150514 338 PM

Chapter 10

322

AH

L

Now the mass of the Earth is 60 times 1024 kg and the period T measured in seconds is given by T = 86400 s

So substitution gives R = 42 times 106 m

The radius of the Earth is 64 times 106 m so that the orbital height h is about 36 times 107 m

Example

Calculatetheminimumenergyrequiredtoputasatelliteofmass500kgintoanorbitthatisasaheightequaltotheEarthrsquosradiusabovethesurfaceoftheEarth

Solution

We have seen that when dealing with gravitational fields and potential it is useful to remember that

g 0G M

Re2---------= or g0 Re

2 GM=

The gravitational potential at the surface of the Earth is

g 0ndash Re GMRe

---------ndash=

The gravitational potential at a distance R

from the centre of the Earth is

=

The difference in potential between the surface of the Earth and a point distance R from the centre is therefore

V∆ g 0Re 1ReR------ndash

=

If R 2Re= then V∆g0 Re

2------------=

This means that the work required to ldquoliftrdquo the satellite into orbit is g0Rm where m is the mass of the satellite This is equal to

10 32 106 500timestimestimes = 16000 MJ

However the satellite must also have kinetic energy in order to orbit Earth This will be equal to

g 0 mRe2

2R-----------------g 0mRe

2

4----------------- 8000 MJ= =

The minimum energy required is therefore

24000 MJ

Exercise 1031 The speed needed to put a satellite in orbit does

not depend on

A the radius of the orbitB theshapeoftheorbitC thevalueofgattheorbitD the mass of the satellite

2 EstimatethespeedofanEarthsatellitewhoseorbitis400kmabovetheEarthrsquossurfaceAlsodetermine the period of the orbit

3 Calculatethespeedofa200kgsatelliteorbitingthe Earth at a height of 70 times 106 m

Assume that g = 82 m sndash2 for this orbit

4 TheradiioftwosatellitesXandYorbitingthe Earth are 2r and 8rwherer is the radius of the Earth Calculate the ratio of the periods of revolutionofXtoY

5 A satellite of mass mkgissentfromEarthrsquossurfaceinto an orbit of radius 5RwhereR is the radius of theEarthWritedownanexpressionfor

(a) thepotentialenergyofthesatelliteinorbit

(b) thekineticenergyofthesatelliteinorbit

(c) theminimumworkrequiredtosendthesatellitefromrestattheEarthrsquossurfaceintoits orbit

6 A satellite in an orbit of 10rfallsbacktoEarth(radius r)afteramalfunctionDeterminethespeedwithwhichitwillhittheEarthrsquossurface

7 The radius of the moon is frac14 that of the Earth AssumingEarthandtheMoontohavethesamedensitycomparetheaccelerationsoffreefallatthe surface of Earth to that at the surface of the Moon

8 UsethefollowingdatatodeterminethegravitationalfieldstrengthatthesurfaceoftheMoon and hence determine the escape speed from the surface of the Moon

Mass of the Moon = 73 times 1022kg

RadiusoftheMoon=17times106 m

PHYSICS 2015indb 322 150514 338 PM

Chapter 9

292

AH

L

microphoneismovedalongthelineXYthatisaboutonemetre from the speakers As the microphone is moved along XY the trace on the cathode ray oscilloscope isseen to go through a series of maxima and minima corresponding to points of constructive and destructive interference of the sound waves

Aninterestinginvestigationistofindhowtheseparationof the points of maximum interference depends on the frequency of the source and also the distance apart ofthe speakers

In the demonstration of the interference between two soundsourcesdescribedaboveifweweretomoveoneofthe speakers from side to side or backwards and forwards the sound emitted from the two speakers would no longer be in phase No permanent points of constructive or destructive interference will now be located since the phase difference between the waves from the two sources is nolongerconstantiethesourcesarenolongercoherent

Light from an incandescent source is emitted with acompletely random phase Although the light from two separate sources will interfere because of the randomlychanging phase no permanent points of constructive or destructive interference will be observed This is why a singleslit isneededintheYoungrsquosdoubleslitexperimentByactingasapointsourceitessentiallybecomesacoherentlight source The light emitted from a laser is also very nearly coherent and this is why it is so easy to demonstrate optical interference and diffraction with a laser

Youngrsquos double slit experiment is one of the great classic experiments of physics and did much to reinforce the wave theoryoflightThomasYoungcarriedouttheexperimentin about 1830

It is essentially the demonstration with the ripple tank and the sound experiment previously described butusing light The essential features of the experiment are shown in Figure 914

screendouble slit

single slit

coloured lter

S 1

S2

Figure 914 Youngrsquos double slit experiment

YoungallowedsunlighttofallontoanarrowsingleslitA few centimetres from the single slit he placed a double slit The slits are very narrow and separated by a distance equal to about fourteen slitwidthsA screen is placed

about ametre from thedouble slitsYoungobservedapatternofmulticolouredldquofringesrdquo in the screenWhenhe placed a coloured filter between the single slit anddouble slit he obtained a pattern that consisted of bright coloured fringes separated by darkness

The single slit essentially ensures that the light falling on the double slit is coherent The two slits then act as the two speakers in the sound experiment or the two dippers in the ripple tank The light waves from each slit interfere and produce the interference pattern on the screen Without thefilterapatternisformedforeachwavelengthpresentinthesunlightHencethemulticolouredfringepattern

You can demonstrate optical interference for yourselfSmoking a small piece of glass and then drawing twoparallel lines on it can make a double slit If you then look throughthedoubleslitatasingletungstenfilamentlampyouwillseethefringepatternByplacingfiltersbetweenthe lamps and the slits you will see the monochromatic fringe pattern

YoucanalsoseetheeffectsofopticalinterferencebylookingatnetcurtainsEachlsquoholersquointhenetactsasapointsourceand the light from all these separate sources interferes and producesquiteacomplicatedinterferencepattern

A laser can also be used to demonstrate optical interference Sincethelightfromthelaseriscoherentitisveryeasytodemonstrate interference Just point the laser at a screen and place a double slit in the path of the laser beam

LetusnowlookattheYoungrsquosdoubleslitexperimentinmore detail The geometry of the situation is shown in Figure915

S1

S2 X

P

Q

D

y

d

screen

θ

θrsquo

Figure 915 The geometry of Youngrsquos double slit experiment

S1andS2 are the two narrow slits that we shall regard as twocoherentmonochromaticpointsourcesThedistancefrom the sources to the screen is D and the distance between the slits is d

The waves from the two sources will be in phase at Q and therewill be abright fringehereWewish tofindtheconditionfortheretobeabrightfringeatPdistancey from Q

PHYSICS 2015indb 292 150514 337 PM

Wave Phenomena

293

AH

L

We note that that D (asymp1metre)isverymuchgreaterthaneither y or d(asymp fewmillimetres)Thismeansthatbothθ and θʹ areverysmallanglesandforintentsandpurposesequal

From the diagram we have that

θ = Dy

And

2S Xd

θʹ =

(Remembertheanglesareverysmall)

Sinceθ asymp θʹ then

2S XyD d

=

ButS2X is the path difference between the waves as they

traveltoPWehavethereforethat

path difference = Dyd

TherewillthereforebeabrightfringeatPif

Dyd

= nλ

Supposethatthereisabrightfringeaty = y1 corresponding

to n = n1 then

Ddy1 = n

1λ

If the next bright fringe occurs at y = y2 this will correspond

to n = n1+1Hence

Ddy2 =(n

1+1)λ

This means that the spacing between the fringes y2 ndash y

1 is

given by

y2 ndash y

1 = d

Dλ

Youngactuallyusethisexpressiontomeasurethewavelengthof the light he used and it is a method still used today

We see for instance that if in a given set up we move the slits closer together then the spacing between the fringes will get greaterEffectivelyourinterferencepatternspreadsoutthatis there will be fewer fringes in a given distance We can also increase the fringe spacing by increasing the distance betweentheslitsandthescreenYouwillalsonotethatfora

givensetupusinglightofdifferentwavelengthsthenldquoredrdquofringeswillspacefurtherapartthanldquobluerdquofringes

In this analysis we have assumed that the slits act as point sources and as such the fringes will be uniformly spaced andofequalintensityAmorethoroughanalysisshouldtakeintoaccountthefinitewidthoftheslits

ReturningtoFigure915weseethatwecanwritethepathdifferenceS2X as

S2X = dtan θʹ

But since θʹ is a small angle the sine and tangent will be nearlyequalsothat

S2X = dsinθʹ

And since θʹ asymp θ then

S2X = dsinθ

The condition therefore for a bright fringe to be found at a point of the screen can therefore be written as

dsinθ = nλ

Insummary

for constructive interference and minima

d sinθ = nλ wheren=123hellip

for destructive interference and maxima

d sinθ = ( n + frac12 )λ wheren=123hellip

Figure 916 shows the intensity distribution of the fringes on the screen when the separation of the slits is large comparedtotheirwidthThefringesareofequalintensityandofequalseparation

Intensity

distance along screen

Figure 916 The intensity distribution of the fringes

It isworthnoting that if the slits are close together theintensity of the fringes is modulated by the intensity distribution of the diffraction pattern of one of the slits

PHYSICS 2015indb 293 150514 337 PM

Chapter 9

294

AH

L

Example

Light of wavelength 500nm is incident on two smallparallel slits separated by 10mmDetermine the anglewherethefirstmaximumisformed

If after passing through the slits the light is brought toa focusona screen15m fromtheslitscalculate theobserved fringe spacing on the screen

Solution

Using the small angle approximation we have

3

7

10105minus

minustimes==

dλθ

= 5 times 10-4 rad

The fringe spacing is given by

3

7

1010551

minus

minustimestimes==

dDy λ

= 075 mm

Modulation of two-slit interference pattern by one-slit diffraction effect

In practice the intensity pattern of the maxima as shown in Figure 917 is not constant but fluctuates while decreasingsymmetrically on either side of the central maximum as shown inFigure917Theintensitypatternisacombinationofboththe single-slit diffractionenvelope and the double slit pattern That is the amplitude of the two-slit interference pattern is modulated(ieadjustedto)byasingleslitdiffractionenvelope

2λD

___ λD

D

___λd

d

___

Double slitnite width

Figure 917 Double slit pattern

ForagivenslitseparationdwavelengthoflightandfixedslittoscreendistancethevariationinintensitydependsonthewidthoftheslitDAlthoughincreasingthewidthoftheslitsincreasestheintensityoflightinthefringesitalsomakesthem less sharp and they become blurred As the slit width is widenedthefringesgraduallydisappearbecausenumerouspoint sources along the widened slit give rise to their own dark and bright fringes that then overlap with each other

Exercise 931 InFigure913thedistancebetweenthespeakers

is050mandthedistancebetweenthelineofthespeakersandthescreenis20mAsthemicrophoneismovedalongthelineXYthedistance between successive points of maximum soundintensityis030mThefrequencyofthesoundwavesis44times103HzCalculateavalueforthe speed of sound

2 Laserlightofwavelength610nmfallsontwoslits10x10-5apartDeterminetheseparationofthefirstordermaximumformedonascreen20maway

3 Twoparallelslits012mmapartareilluminatedwith red light with a wavelength of 600 nm An interferencepatternfallsonascreen15maway

Calculate

(a) thedistancefromthecentralmaximumtothefirstbrightfringe

(b) thedistancetotheseconddarkline

Multiple slit and diffraction grating interference patterns

If we examine the interference pattern produced when monochromatic light passes through a different number of slits we notice that as the number of slits increases the number of observed fringes decreases the spacingbetween them increases and the individual fringes become much sharper We can get some idea of how this comes about by looking at the way light behaves when a parallel beam passes through a large number of slits The diagram for this is shown in Figure 918

θθd

1

2

3

telescope

Figure 918 A parallel beam passing through several slits

The slits are very small so that they can be considered to act as point sources They are also very close together such that d issmall(10ndash6 m)EachslitbecomesasourceofcircularwavefrontsandthewavesfromeachslitwillinterfereLetus

PHYSICS 2015indb 294 150514 337 PM

Wave Phenomena

295

AH

L

consider the light that leaves the slit at an angle θ as shown Thepathdifferencebetweenwave1andwave2isdsinθ and ifthisisequaltoanintegralnumberofwavelengthsthenthe two waves will interfere constructively in this direction Similarly wave 2 will interfere constructively with wave3 at this angle andwave 3with 4 etc across thewholegratingHenceifwelookatthelightthroughatelescopethatisbringittoafocusthenwhenthetelescopemakesanangle θ to the grating a bright fringe will be observed The condition for observing a bright fringe is therefore

dsinθ = mλ

Supposeweuselightofwavelength500nmandsupposethat d = 16 times 10ndash6 m

Obviously we will see a bright fringe in the straight on position θ=0(thezero order)

The next position will be when m =1(thefirst order)andsubstitutionintheaboveequationgivesθ = 18deg

The next position will be when m=2(thesecond order)and this give θ = 38deg

For m=3sinθ is greater than 1 so with this set up we only obtain5fringesonezeroorderandtwoeithersideofthezero order

The calculation shows that the separation of the orders is relatively large At any angles other than 18deg or 38deg the light leaving the slits interferes destructively We can see that the fringes will be sharp since if we move just a small angle away from 18deg the light from the slits will interfere destructively

An array of narrow slits is usually made by cutting narrow transparent lines very close together into the emulsion on aphotographicplate(typically200linespermillimetre)Suchanarrangementiscalledadiffraction grating

The diffraction grating is of great use in examining the spectral characteristics of light sources

All elements have their own characteristic spectrum An element can be made to emit light either by heating it until it is incandescent or by causing an electric discharge through itwhen it is in a gaseous state Yourschool probably has some discharge tubes and diffraction gratingsIfithasthenlookattheglowingdischargetubethroughadiffractiongratingIfforexampletheelementthat you are looking has three distinct wavelengths then each wavelength will be diffracted by a different amount inaccordancewiththeequationdsinθ = nλ

Alsoiflaserlightisshonethroughagratingontoascreenyou will see just how sharp and spaced out are the maxima By measuring the line spacing and the distance of the screen fromthelaserthewavelengthofthelasercanbemeasured

If your school has a set of multiple slits say from a single slit toeight slits then it isalsoaworthwhileexercise toexamine how the diffraction pattern changes when laser light is shone through increasing numbers of slits

Consider a grating of NparallelequidistantslitsofwidthaseparatedbyanopaqueregionofwidthD We know that light diffracted from one slit will be superimposed and interfere with light diffracted from all the other slits The intensityprofile of the resultant wave will have a shape determined by

bull theintensityoflightincidentontheslitsbull thewavelengthofthislightbull theslitwidthbull theslitseparation

Typical intensity profiles formed when a plane wavemonochromaticlightatnormalincidenceononetwoandsix slits are shown in Figure 919 In this example the slit widthisonequarterofthegratingspacing

Six slits sharperfringes

d = 4a

I α N 2

(N ndash 2)= 4 subsidiary maxima

order0 4321

I = 36

Two slits

order0

single-slitdiraction

envelope

d = 4a4th principal maximum missing

principa maxima

I = 4 (nominal units)

4321

intensity

order

ISS

One slit

0 1

single-slitdiraction

pattern

N = number of slitsslit width a

Figure 919 Intensity profiles

PHYSICS 2015indb 295 150514 337 PM

Chapter 9

296

AH

L

1 For Nge2thefringepatterncontainsprincipalmaximathat are modulated by the diffraction envelope

2 ForNgt2thefringepatterncontains(Nndash2)subsidiary maxima

3 Where the grating spacing d is four time the slit width a(4xa =d)thenthefourthprincipalmaximumismissing from the fringe pattern Where the grating spacingdisthreetimetheslitwidtha(3xa =d)thenthe third principal maximum is missing from the fringepatternandsoon

4 Increasing N increases the absolute intensities of the diffraction pattern (I = N2)

If white light is shone through a grating then the central image will be white but for the other orders each will be spread out intoacontinuousspectrumcomposedofaninfinitenumberof adjacent images of the slit formed by the wavelength of the different wavelengths present in the white light At any given point in the continuous spectrum the light will be very nearly monochromatic because of the narrowness of the images of the slit formed by the grating This is in contrast to the doubleslitwhereifwhitelightisusedtheimagesarebroadand the spectral colours are not separated

In summary as is the case of interference at a doubleslit there is a path difference between the rays fromadjacent slits of d sinθ Where there is a whole number of wavelengthsconstructiveinterferenceoccurs

For constructive interference and minima

d sinθ = mλ wheren=123hellip

When there is a path difference between adjacent rays of halfawavelengthdestructiveinterferenceoccurs

For and maxima

d sinθ = ( m + frac12 )λ wheren=123hellip

Exercise 941 Lightfromalaserisshonethroughadiffraction

grating on to a screen The screen is a distance of 20mfromthelaserThedistancebetweenthecentraldiffractionmaximumandthefirstprincipalmaximum formed on the screen is 094 m The diffractiongratinghas680linespermmEstimatethe wavelength of the light emitted by the laser

2 Monochromaticlightfromalaserisnormallyincident on a six-slit grating The slit spacing is three times the slit width An interference patternisformedonascreen2montheotherside of the grating

(a) Determineandexplaintheorderoftheprincipal maxima missing from the fringe pattern

(b) Howmanysubsidiarymaximaaretherebetween the principal maxima

(c) Howmanyprincipalmaximaarethere

(i) Betweenthetwofirstorderminimain the diffraction envelope

(ii) Betweenthefirstandsecondorderminima in the diffraction envelope

(d) Drawthepatternformedbythebrightfringesin(c)

(e) Sketchtheintensityprofileofthefringepattern

Thin film interferenceYoumightwellbe familiarwith the colouredpatternoffringes that can be seen when light is reflected off thesurfaceofwateruponwhichathinoilfilmhasbeenspiltorfromlightreflectedfrombubblesWecanseehowthesepatternsarisebylookingatFigure920

Extendedlight source air

A

B

C

12

3

4

Oil lm

Water

Figure 920 Reflection from an oil film

Consider light from an extended source incident on a thinfilmWealsoconsiderawave fromonepointofthe source whose direction is represented by the ray shownSomeofthislightwillbereflectedatAandsometransmitted through the filmwhere somewill again bereflectedatB(somewillalsobetransmittedintotheair)SomeofthelightreflectedatBwillthenbetransmittedatCandsomereflectedandsoonIfweconsiderjustrays1and2thenthesewillbenotbeinphasewhenthearebrought to a focus by the eye since they have travelled

PHYSICS 2015indb 296 150514 337 PM

Wave Phenomena

297

AH

L

different distances If the path difference between them is an integral number of half wavelengths then we might expect the twowaves to be out of phaseHowever ray1 undergoes a phase change of π on reflection but ray2 does not since it is reflected at a boundary betweena more dense and less dense medium (See chapter 4)Henceifthepathdifferenceisanintegralnumberofhalf-wavelengthsrays1and2willreinforceieray1and2areinphaseHowever rays357etcwillbeoutofphasewithrays246etcbutsinceray2ismoreintensethanray3andray4moreintensethanray5thesepairswillnot cancel out so there will be a maximum of intensity

Ifthepathdifferenceissuchthatwave1and2areoutofphasesincewave1ismoreintensethanwave2theywillnotcompletelyannulHoweveritcanbeshownthattheintensitiesofwaves2345hellipaddtoequaltheintensityofwave1Sincewaves345hellipareinphasewithwave2there will be complete cancellation

The path difference will be determined by the angle at which ray1 is incident andalsoon the thickness (andtheactualrefractiveindexaswell)ofthefilmSincethesourceisanextendedsourcethelightwillreachtheeyefrom many different angles and so a system of bright and darkfringeswillbeseenYouwillonlyseefringesifthefilmisverythin(exceptifviewedatnormalincidence)sinceincreasingthethicknessofthefilmwillcausethereflected rays to get so far apart that they will not becollected by the pupil of the eye

From the argument above to find the conditions forconstructive and destructive interferenceweneedonlyfindthepathdifferencebetweenray1andray2Figure921showsthe geometry of the situation

A BE

F

C

D

1 2

φrsquo φrsquo

φ

φ

d

Figure 921 The geometry of interference

Thefilmisofthicknessd and refractive index n and the light has wavelength λ If the line BF is perpendicular to ray 1 then the optical path difference (opd) between ray 1 andray2whenbroughttoafocusis

opd = n(AC+CB)ndashAF

We have to multiple by the refractive index for the path travelledbythelightinthefilmsincethelighttravelsmoreslowly in thefilm If the light travels say adistancex in a material of refractive index n then in the time that it takes to travelthisdistancethelightwouldtraveladistancenx in air

IfthelineCEisatrightanglestoray2thenweseethat

AF = nBE

FromthediagramAC=CDsowecanwrite

opd = n(CD+CB)ndashnBE=nDE

Alsofromthediagramweseethatwhereφ is the angle of refraction

DE=2cosφ

From which opd=2ndcosφ

Bearinginmindthechangeinphaseofray1onreflectionwe have therefore that the condition for constructive interference is

2nd φcos m 12---+

λ m 1 2 hellip = =

Andfordestructiveinterference 2ndcosφ = mλ

Eachfringecorrespondstoaparticularopd for a particular value of the integer m and for any fringe the value of the angle φ is fixedThismeans that itwill be in the formof an arc of a circle with the centre of the circle at the point where the perpendicular drawn from the eye meets the surface of the film Such fringes are called fringesofequal inclinationSince theeyehasa smallaperturethesefringesunlessviewedatneartonormalincidence (φ = 0) will only be observed if the film is very thinThisisbecauseasthethicknessofthefilmincreasesthereflected rayswill get further and further apart and sovery few will enter the eye

Ifwhitelightisshoneontothefilmthenwecanseewhywe get multi-coloured fringes since a series of maxima and minima will be formed for each wavelength present in the white light However when viewed at normalincidenceitispossiblethatonlylightofonecolourwillundergoconstructiveinterferenceandthefilmwilltakeon this colour

PHYSICS 2015indb 297 150514 337 PM

Chapter 9

298

AH

L

Thickness of oil films

The exercise to follow will help explain this use

Non-reflecting films

A very important but simple application of thin filminterferenceisintheproductionofnon-reflectingsurfaces

Athinfilmofthicknessd and refractive index n1 is coated onto glass of refractive index n where n1 lt n Light ofwavelength λ that is reflected at normal incidence willundergo destructive interference if 12

2n d λ

= thatis

14d

nλ

=

(remember that there will now no phase change at the glass interface ie we have a rare to dense reflection)

The use of such films can greatly reduce the loss oflightby reflection at the various surfacesof a systemoflenses or prisms Optical parts of high quality systemsare usually all coatedwithnon-reflecting films in orderto reduce stray reflections The films are usually madebyevaporatingcalciumormagnesiumfluorideonto thesurfacesinvacuumorbychemicaltreatmentwithacidsthat leave a thin layer of silica on the surface of the glass ThecoatedsurfaceshaveapurplishhuebyreflectedlightThis is because the condition for destructive interference fromaparticularfilmthicknesscanonlybeobtainedforone wavelength The wavelength chosen is one that has a value corresponding to light near the middle of the visible spectrumThismeansthatreflectionofredandvioletlightis greater combining to give the purple colour Because of the factor cosφatanglesotherthannormalincidencethepath difference will change butnotsignificantlyuntilsayabout30deg(egcos25deg=090)

It should be borne in mind that no light is actually lost by anon-reflectingfilmthedecreaseofreflectedintensityiscompensated by increase of transmitted intensity

Non-reflecting films can be painted onto aircraft tosuppressreflectionofradarThethicknessofthefilmis determined by

4nd λ

= where λ is the wavelength of

the radar waves and ntherefractiveindexofthefilmat this wavelength

Example

A plane-parallel glass plate of thickness 20 mm isilluminated with light from an extended source The refractiveindexoftheglassis15andthewavelengthofthelight is 600 nm Calculate how many fringes are formed

Solution

We assume that the fringes are formed by light incident at all angles from normal to grazing incidence

At normal incidence we have 2nd = mλ

From which

m 2 15 2 10 3ndashtimestimestimes6 10 7ndashtimes

------------------------------------------- 10 000= =

At grazing incidence the angle of refraction φ is in fact the critical angle

Therefore φφ arcsin 115------- 42 deg= =

ie cosφ = 075

At grazing incidence 2ndcosφ = m λ

From which (and using cos(sinndash1(115) = 075)

m 2 15 2 10 3ndash 075timestimestimestimes6 10 7ndashtimes

------------------------------------------------------------ 7500= =

The total number of fringes seen is equal to m ndash m = 2500

Exercise 95When viewed from above the colour of an oil film onwater appears red in colour Use the data below to estimate theminimumthicknessoftheoilfilm

average wavelength of red light = 630 nm

refractiveindexofoilforredlight=15

refractive index of water for red light = 13

PHYSICS 2015indb 298 150514 337 PM

Wave Phenomena

299

AH

L

94 Resolution

Essential idea Resolution places an absolute limit on the extent to which an optical or other system can separate images of objects

Understandings

bull Thesizeofadiffractingaperturebull Theresolutionofsimplemonochromatic

two-source systems

The size of a diffracting apertureOur discussion concerning diffraction so far has been for rectangular slits When light from a point source enters a smallcircularapertureitdoesnotproduceabrightdotasanimagebutacirculardiscknownasAiryrsquosdiscsurroundedbyfainterconcentriccircularringsasshowninFigure922

Laser

Figure 922 Diffraction at a circular aperture

Diffractionatanapertureisofgreatimportancebecausetheeye and many optical instruments have circular apertures

What is the half-angular width of the central maximum of the diffraction formed by a circular aperture This is not easy to calculate since it involves some advanced mathematics The problemwasfirstsolvedbytheEnglishAstronomerRoyalGeorgeAiryin1835whoshowedthatforcircularapertures

θ 122 λb--------------= where b is the diameter of the aperture

Example 2

Inthefollowingdiagramparallellightfromadistantpointsource(suchasastar)isbroughttofocusonthescreenSbyaconverginglens(the lens is shown as a vertical arrow)

Thefocal length(distance fromlens toscreen) is25cmand the diameter of the lens is 30 cm The wavelength of thelightfromthestaris560nmCalculatethediameterofthe diameter of the image on the screen

Solution

The lens actually acts as a circular aperture of diameter 30 cm The half angular width of central maximum of the diffraction pattern that it forms on the screen is

75

2

122 122 56 10 23 1030 10b

λθminus

minusminus

times times= = = times

times rad

The diameter of the central maxima is therefore

25 times 10-2 times 23 times 10-5

= 57 times 10-6 m

Although this is small it is still finite and is effectively the image of the star as the intensity of the secondary maxima are small compared to that of the central maximum

The resolution of simple monochromatic two-source systemsThe astronomers tell us that many of the stars that we observe with the naked eye are in fact binary stars That iswhatwe see as a single star actually consistsof twostarsinorbitaboutacommoncentreFurthermoretheastronomerstellusthatifweusealdquogoodrdquotelescopethenwewillactuallyseethetwostarsThatiswewillresolve the single point source into its two component parts Sowhatisitthatdetermineswhetherornotweseethetwo stars as a single point source ie what determines whether or not two sources can be resolved It canrsquot just bethatthetelescopemagnifiesthestarssinceiftheyareacting as point sources magnifying them is not going to make a great deal of difference

Ineachofoureyesthereisanaperturethepupilthroughwhich the light enters This light is then focused by the eye lens onto the retina But we have seen that when light passes through an aperture it is diffracted and so when we look at

NATURE OF SCIENCEImproved technology The Rayleigh criterion is the limit of resolution Continuing advancement in technology such as large diameter dishes or lenses or the use of smaller wavelength lasers pushes the limits of what we can resolve (18)

copy IBO 2014

PHYSICS 2015indb 299 150514 337 PM

Chapter 9

300

AH

L

apointsourceadiffractionpatternwillbeformedontheretina If we look at two point sources then two diffraction patterns will be formed on the retina and these patterns will overlap The width of our pupil and the wavelength of the light emitted by the sources will determine the amount by which they overlap But the degree of overlap will also depend on the angular separation of the two point sources WecanseethisfromFigures923

pupilretina

S1

S2

P2

P1

θ

Figure 923

LightfromthesourceS1 enters the eye and is diffracted by the pupil such that the central maximum of the diffraction patternisformedontheretinaatP1SimilarlylightfromS2producesamaximumatP2 If the two central maxima are well separated then there is a fair chance that we will see the two sources as separate sources If they overlap then we will not be able to distinguish one source from another From the diagram we see as the sources are movedclosertotheeyethentheangleθ increases and so does the separation of the central maxima

Figures924925926and927showsthedifferentdiffractionpatternsandtheintensitydistributionthatmightresultonthe retina as a result of light from two point sources

S1 S21 Well resolved

Figure 924 Very well resolved

S1 S22 Well resolved

Figure 925 Well resolved

S1 S23 Just resolved Rayleigh criterion Minimum of S2 coincides with maximum peak of S1

Figure 926 Just resolved

4 Not resolved

Figure 927 Not resolved

We have suggested that if the central maxima of the two diffraction patterns are reasonably separated then we should be able to resolve two point sources In the late 19th century Lord Rayleigh suggested by how much they should be separated in order for the two sources to be just resolved Ifthecentralmaximumofonediffractionpattern coincides with the firstminima of the otherdiffraction pattern then the two sources will just beresolved This is known as the Rayleigh Criterion

Figure926showsjustthissituationThetwosourcesarejust resolved according to the Rayleigh criterion since the peak of the central maximum of one diffraction pattern coincideswiththefirstminimumoftheotherdiffractionpattern This means that the angular separation of the peaks of the two central maxima formed by each source is just the half angular width of one central maximum ie

θ λb---

=

where b is the width of the slit through which the light fromthesourcespassesHoweverweseefromFigure1117that θ is the angle that the two sources subtend at the slit Henceweconcludethattwosourceswillberesolvedbyaslit if the angle that they subtend at the slit is greater than or equaltoλfraslb

PHYSICS 2015indb 300 150514 337 PM

Wave Phenomena

301

AH

L

Sofarwehavebeenassumingthattheeyeisarectangularslit whereas clearly it is a circular aperture and so we must use the formula

θ 122 λb--------------=

As mentioned above the angle θ is sometimes called the resolving power but should more accurately be called the minimum angle of resolution (θmin)

Clearly the smaller θ the greater will be the resolving power

The diffraction grating is a useful for differentiating closely spacedlinesinemissionspectraLikeaprismspectrometerit can disperse a spectrum into its components but it is better suited because it has higher resolution than the prism spectrometer

If λ1 and λ2 are two nearly equal wavelengths that canbarelybedistinguishedtheresolvanceorresolvingpowerofthegratingisdefinedas

R = λ _______ (λ2 ndash λ 1) = λ ___ Δλ

Where λ asymp λ1 asymp λ2

Therefore thediffractiongratingwithahighresolvancewill be better suited in determining small differences in wavelength

If the diffraction grating has Nlinesbeingilluminateditcan be shown that the resolving power of the mth order diffraction is given by

R = mN

Obviouslytheresolvancebecomesgreaterwiththeordernumber m and with a greater number of illuminated slits

Example

A benchmark for the resolving power of a grating is the sodium doublet in the sodium emission spectrum Two yellow lines in itsspectrumhavewavelengthsof58900nmand58959nm

(a) Determinetheresolvanceofagratingifthegivenwavelengths are to be distinguished from each other

(b) Howmanylinesinthegratingmustbeilluminated in order to resolve these lines in the second order spectrum

Solution

(a) R = λ ___ Δλ = 58900nm __________________ (58900ndash58959)nm

= 589nm _______ 059nm = 10 times 103

(b) N = R __ m = 10 times 10 3 ________ 2 =50times102 lines

It has been seen that diffraction effectively limits the resolving power of optical systems This includes such systems as the eye telescopes and microscopes Theresolving power of these systems is dealt with in the next section This section looks at links between technology and resolving power when looking at the very distant and when looking at the very small

Radio telescopes

The average diameter of the pupil of the human eye is about 25 mm This means that the eye will justresolve two point sources emitting light of wavelength 500 nm if their angular separation at the eye is 7

43

50 10122 24 1025 10

θminus

minusminus

times= times = times

times rad

If the eye were to be able to detect radio waves of wavelength 015m then tohave the same resolvingpower thepupilwould have to have a diameter of about 600 m Clearly this is nonsense but it does illustrate a problem facingastronomers who wish to view very distant objects such as quasarsandgalaxies thatemitradiowavesConventionalradio telescopes consist of a large dish typically 25 min diameter Even with such a large diameter the radiowavelength resolving power of the telescope is much less thantheopticalresolvingpowerofthehumaneyeLetuslook at an example

Example

The Galaxy Cygnus A can be resolved optically as an elliptically shaped galaxy However it is also a strongemitterof radiowavesofwavelength015mTheGalaxyisestimatedtobe50times1024mfromEarthUseofaradiotelescope shows that the radio emission is from two sources separated by a distance of 30 times 1021m Estimate thediameterofthedishrequiredtojustresolvethesources

Solution

The angle θ that the sources resolve at the telescope is given by

θ 21

424

30 10 60 1050 10

θ minustimes= = times

times rad

and 4

122 122 015 300060 10

d λθ minus

times= = =

timesm = 30 km

A radio telescope dish of this size would be impossible tomake letalonesupportThisshowsthatasingledishtype radio telescope cannot be used to resolve the sources and yet they were resolved To get round the problemastronomers use two radio telescopes separated by a large

PHYSICS 2015indb 301 150514 337 PM

Chapter 9

302

AH

L

distance The telescopes view the same objects at the same time and the signals that each receive from the objects are simultaneously superimposed The result of the superposition of the two signals is a two-slit interference pattern The pattern has much narrower fringe spacing than that of the diffraction pattern produced by either telescope on its own hence producing a much higherresolvingpowerWhentelescopesareusedlikethistheyare called a stellar interferometer

InSocorroinNewMexicothereisastellarinterferometerthat consists of 27 parabolic dishes each of diameter25 m arranged in a Y-shape that covers an area of 570km2Thisisaso-calledVeryLargeArray(VLA)Evenhigher resolution can be obtained by using an array of radio telescopes in observatories thousands of kilometres apart A system that uses this so-called technique oflsquovery-long-baseline interferometryrsquo (VLBI) is known asalsquovery-long-baselinearrayrsquo(VLBA)WithVLBAaradiowavelength resolving power can be achieved that is 100 timesgreaterthanthebestopticaltelescopesEvenhigherresolving power can be achieved by using a telescope that isinasatelliteorbitingEarthJapanrsquosInstituteofSpaceandAstronautical Science (ISAS) launched sucha system inFebruary 1997TheNationalAstronomicalObservatoryof Japan the National Science Foundationrsquos NationalRadio Astronomy Observatory (NRAO) the CanadianSpaceAgency theAustraliaTelescopeNationalFacilitythe European VLBI Network and the Joint Instituteback this project forVeryLongBaseline Interferometryin EuropeThis project is a very good example of howInternationalismcanoperateinPhysics

Electron microscope

Telescopes are used to look at very distant objects that are very large but because of their distance fromus appearverysmallMicroscopesontheotherhandareusedtolookat objects that are close to us but are physically very small Aswehave seen justmagnifyingobjects that ismakingthemappearlargerisnotsufficientonitsowntogaindetailabouttheobjectfordetailhighresolutionisneeded

Figure928isaschematicofhowanoptical microscope is usedtoviewanobjectandFigure929isaschematicofatransmission electron microscope(TEM)

Optical lens system

bright light source

glass slide containing specimen (object)

eye

Figure 928 The principle of a light microscope

magnetic lens system

electron source

thin wafer of material

ScreenCCD

Figure 929 The principle of an electron microscope

In the optical microscope the resolving power isdetermined by the particular lens system employed and the wavelength λ of the light used For example twopoints in the sample separated by a distance d will just be resolved if

2d mλ=

where m is a property of the lens system know as the numerical aperture In practice the largest value of m obtainable is about 16 Hence if the microscope slideis illuminatedwith light ofwavelength 480 nm a goodmicroscope will resolve two points separated by a distance dasymp15times10-7masymp015micromPointscloser together thanthiswillnotberesolvedHoweverthisisgoodenoughtodistinguishsomevirusessuchastheEbolavirus

Clearlythesmallerλ the higher the resolving power and this is where the electron microscope comes to the fore The electron microscope makes use of the wave nature of electrons(see1315)IntheTEMelectronspassthrougha wafer thin sample and are then focused by a magnetic field onto a fluorescent screen orCCD (charge coupleddevicesee142)Electronsusedinanelectronmicroscopehavewavelengthstypicallyofabout5times10-12mHoweverthe numerical aperture of electron microscopes is considerablysmaller thanthatofanopticalmicroscopetypicallyabout002NonethelessthismeansthataTEMcanresolvetwopointsthatareabout025nmapartThisresolving power is certainly high enough to make out the shape of large molecules

Another type of electron microscope uses a techniqueby which electrons are scattered from the surface of the sample The scattered electrons and then focused as in theTEM to form an image of the surfaceThese so-calledscanningelectronmicroscopes(SEM)havealowerresolving power than TEMrsquos but give very good threedimensional images

The eye

We saw in the last section that the resolving power of the human eye is about 2 times 10-4 rad Suppose that youare looking at car headlights on a dark night and the car

PHYSICS 2015indb 302 150514 337 PM

Wave Phenomena

303

AH

L

is a distance D away If the separation of the headlight is say 15 m then the headlights will subtend an angle θ 15

Dθ =

at your eye Assuming an average wavelength of

500 nm your eye will resolve the headlights into twoseparatesourcesifthisangleequals2times10-4 rad This gives D=75kmInotherwordsifthecarisapproachingyouona straight road then you will be able to distinguish the two headlightsasseparatesourceswhenthecaris75kmawayfrom you Actually because of the structure of the retina and optical defects the resolving power of the average eye is about 6 times 10-4 rad This means that the car is more likely tobe25kmawaybeforeyouresolvetheheadlights

Astronomical telescope

Letusreturntotheexampleofthebinarystarsdiscussedatthe beginning of this section on resolution The stars Kruger A and B form a binary system The average separation of the stars is 14 times 1012mandtheiraveragedistancefromEarthis12times1017mWhenviewedthroughatelescopeonEarththe system will therefore subtend an angle

12

17

14 1012 10

θtimes

=times

= 12times10-5 rad at the objective lens of the telescope Assuming that the average wavelength of the light emitted bythestarsis500nmthenifthetelescopeistoresolvethesystem into two separate images it must have a minimum diameter Dwhere12times10-5 =

7122 500 10D

minustimes times

This gives D=0050m which is about 5 cm So thisparticular system is easily resolved with a small astronomical telescope

Exercise 961 ItissuggestedthatusingtheISASVLBAitwould

bepossibletoldquoseerdquoagrainofriceatadistanceof5000kmEstimatetheresolvingpoweroftheVLBA

2 Thedistancefromtheeyelenstotheretinais20mm The light receptors in the central part of the retinaareabout5times10-6apartDeterminewhetherthe spacing of the receptors will allow for the eye to resolve the headlights in the above discussion whentheyare25kmfromtheeye

3 ThediameterofPlutois23times106 m and its averagedistancefromEarthis60times1012 m EstimatetheminimumdiameteroftheobjectiveofatelescopethatwillenablePlutotobeseenasadisc as opposed to a point source

95 Doppler effect

Essential ideaTheDopplereffectdescribesthephenomenonofwavelengthfrequencyshiftwhenrelativemotionoccurs

Understandingsbull TheDopplereffectforsoundwavesandlightwaves

The Doppler effect for sound waves and light waves

Consider two observers A and B at rest with respect to asoundsourcethatemitsasoundofconstantfrequencyf Clearly both observers will hear a sound of the same frequencyHoweversupposethatthesourcenowmovesat constant speed towards A A will now hear a sound of frequencyfA that is greater than f and B will hear a sound of frequency fB that is less than f This phenomenon is known as the Doppler Effect orDopplerPrinciple after C J Doppler(1803-1853)

The same effect arises for an observer who is either moving towards or away from a stationary source

Figure 930 shows the waves spreading out from a stationary source that emits a sound of constant frequency f The observersAandBhearasoundofthesamefrequency

A B

wavefront

S

Figure 930 Sound waves from a stationary source

Suppose now that the source moves towards A withconstant speed vFigure9310(a)showsasnapshotofthenew wave pattern

NATURE OF SCIENCETechnology Although originally based on physical observations of the pitch of fast moving sources of sound the Doppler effect has an important role in many different areas such as evidence for the expansion of the universe and generating images used in weather reports and in medicine (55)

copy IBO 2014

PHYSICS 2015indb 303 150514 337 PM

Chapter 9

304

AH

L

sourceA B

smallerwavelength

largerwavelength

Figure 931 (a) Sound waves from a moving source

The wavefronts are now crowded together in the direction of travel of the source and stretched out in the opposite direction This is why now the two observers will now hear notesofdifferentfrequenciesHowmuchthewavesbunchtogether and how much they stretch out will depend on the speed vEssentially A

A

cfλ

= and BB

cfλ

= where λA lt λB and v is the speed of sound

IfthesourceisstationaryandAismovingtowardsitthenthe waves from the source incident on A will be bunched up If A is moving away from the stationary source then the waves from the source incident on A will be stretched out

Christian Doppler(1803ndash1853)actuallyappliedtheprinciple(incorrectlyasithappens)totryandexplainthecolourofstars However the Doppler effect does apply to light aswellastosoundIfalightsourceemitsalightoffrequencyf then if it is moving away from an observer the observer willmeasurethelightemittedashavingalowerfrequencythan fSincethesensationofcolourvisionisrelatedtothefrequencyoflight(bluelightisofahigherfrequencythanred light) light emitted by objects moving way from anobserver is often referred to as being red-shifted whereas if the object is moving toward the observer it is referred to as blue-shiftedThisideaisusedinOptionE(Chapter16)

We do not need to consider here the situations where either the source or the observer is accelerating In a situation for example where an accelerating source is approaching astationaryobserverthentheobserverwillhearasoundof ever increasing frequency This sometimes leads toconfusion in describing what is heard when a source approaches passes and then recedes from a stationaryobserver Suppose for example that you are standingon a station platform and a train sounding its whistle is approaching at constant speed What will you hear as the train approaches and then passes through the station As the train approaches you will hear a sound of constant pitch but increasing loudness The pitch of the sound will be greater than if the train were stationary As the train passes through the station you will hear the pitch change at the momentthetrainpassesyoutoasoundagainofconstantpitch The pitch of this sound will be lower than the sound of the approaching train and its intensity will decrease as the train recedes from you What you do not hear is a sound of increasing pitch and then decreasing pitch

The Doppler equations for sound

Although you will not be expected in an IB examination to derivetheequationsassociatedwithaspectsoftheDopplereffectyouwillbeexpectedtoapplythemForcompletenesstherefore thederivationoftheequationsassociatedwiththeDopplereffectasoutlinedaboveisgivenhere

S primeS O

vs

Figure 931 (b)

InFigure931(b)theobserverOisatrestwithrespecttoasourceofsoundSismovingwithconstanspeedvs directly towards O The source is emitting a note of constant frequencyf and the speed of the emitted sound is v

S shows the position of the source ∆t later When the sourceisatresttheninatime∆t the observer will receive f∆t waves and these waves will occupy a distance v∆t ie

v f

λ =f∆tv∆t =

(Because of the motion of the source this number of waves will now occupy a distance (v∆t ndash vs∆t)The lsquonewrsquowavelength is therefore

v ndash vs

fλrsquo =

f∆tv∆t ndash vs∆t

=

If fisthefrequencyheardbyOthen

vfλ

= or

vf

λ = = sv vfminus

From which

s

vf fv v

=minus

Dividingthroughbyv gives

Equation1111

1 sf f v

v=

minus

If the source is moving away from the observer then we have

s

vf fv v

=minus

1

1 sf vv

=+

Equation112

PHYSICS 2015indb 304 150514 337 PM

Wave Phenomena

305

AH

L

We now consider the case where the source is stationary and the observerismoving towardsthesource with speed v0 In this situation the speed of the sound waves as measured by the observer will be v0 + v We therefore have that

v0 + v = ffraslλ = vfλ

times

From which

o1 vf fv

= +

Equation113

If the observerismovingawayfromthesource then

o1 vf fv

= minus

Fromequation(113)wehavethat

o o1 v vf f f f f fv v

∆ = minus = + minus =

Equation114

Thevelocitiesthatwerefertointheaboveequationsarethevelocities with respect to the medium in which the waves fromthesourcetravelHoweverwhenwearedealingwithalight source it is the relative velocity between the source and the observer that we must consider The reason for this is that lightisuniqueintherespectthatthespeedofthelightwavesdoes not depend on the speed of the source All observers irrespective of their speed or the speed of the source will measure the same velocity for the speed of light This is oneof thecornerstonesof theSpecialTheoryofRelativitywhichisdiscussedinmoredetailinOptionH(Chapter18)When applying the Doppler effect to light we aremainlyconcerned with the motion of the source We look here only at the situation where the speed of the source v is much smaller than the speed of light cinfreespace(v ltlt c)Underthesecircumstanceswhenthesourceismovingtowardstheobserverequation111becomes

vf f f fc

minus = ∆ = Equation115

andwhenthesourceismovingawayfromtheobserverequation112becomes vf f f f

cminus = ∆ = minus

Provided that v ltlt c these same equations apply for astationary source and moving observer

We look at the following example and exercise

Example

Asourceemitsasoundoffrequency440HzItmovesinastraight line towards a stationary observer with a speed of 30ms-1Theobserverhearsasoundoffrequency484HzCalculate the speed of sound in air

Solution

We use equation 111 and substitute f= 484 Hz f = 440 Hz

and vs = 30 m s-1

therefore 1

484 440 301v

=minus

such that 30 4401

484vminus = to

give v = 330 m s-1

Example

A particular radio signal from a galaxy is measured as havingafrequencyof139times109HzThesamesignalfromasourceinalaboratoryhasafrequencyof142times109Hz

Suggestwhy thegalaxy ismovingaway fromEarthandcalculateitsrecessionspeed(iethespeedwithwhichitismovingawayfromEarth)

Solution

The fact that the frequency from the moving source is less than that when it is stationary indicates that it is moving away from the stationary observer ie Earth

Using vf fc

∆ = we have

8 96

9

3 10 (142 139) 10 634 10142 10

c fvf∆ times times minus times

= = = timestimes

m s-1

It is usual when dealing with the Doppler effect of light to express speeds as a fraction of c So in this instance we have v = 0021 c

Using the Doppler effectWe have seen in the above example and exercise how the Dopplereffectmaybeusedtomeasuretherecessionspeedof distant galaxies The effect is also used to measure speed inothersituationsHerewewilllookatthegeneralprincipleinvolved in using the Doppler effect to measure speedFigure932showsasource(thetransmitter)thatemitseithersound or em waves of constant frequency f The waves

PHYSICS 2015indb 305 150514 337 PM

Chapter 9

306

AH

L

fromthesourceare incidentonareflectorthat ismovingtowards the transmitter with speed vThereflectedwavesare detected by the receiver placed alongside the transmitter

reector

v

transmitter

receiver

f f

f f

Figure 932 Using the Doppler effect to measure speed

We shall consider the situation where v ltlt c where c is the speed of the waves from the transmitter

For the reflector receiving waves from the transmitterit is effectively an observer moving towards a stationary sourceFromequation(114) itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation116

For the receiver receivingwaves from the reflector it iseffectively a stationary observer receiving waves from a movingsourceFromequation(115)itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation117

Ifweaddequations(116)and(117)wegetthatthetotalDopplershiftatthereceiver∆f is

v vf f f f fc c

minus = ∆ = +

But 1 vf fc

= +

hence

1 v v vf f fc c c

∆ = + +

But since v ltlt cwe can ignore the term2

2

vc when we

expandthebracketintheaboveequation

Therefore we have2vf fc

∆ = Equation118

If v asymp c then we must use the full Doppler equationsHowever for em radiationwewill alwaysonly considersituations in which v ltlt c

Example

Thespeedofsoundinbloodis1500times103 m s-1 Ultrasound offrequency100MHzisreflectedfrombloodflowinginan arteryThe frequencyof the reflectedwaves receivedbackatthetransmitteris105MHzEstimatethespeedofthebloodflowintheartery

Solution

Using equation (118) we have

6 63

2005 10 1015 10

vtimes = times

timesto give v asymp 36 m s-1 (We have assumed that the ultrasound is incident at right angles to the blood flow)

Exercise 971 Judy is standing on the platform of a station A

high speed train is approaching the station in a straight line at constant speed and is sounding its whistleAsthetrainpassesbyJudythefrequencyof the sound emitted by the whistle as heard by Judychangesfrom640Hzto430HzDetermine

(a) thespeedofthetrain

(b) thefrequencyofthesoundemittedbythewhistle as heard by a person on the train (Speedofsound=330ms-1)

2 AgalaxyismovingawayfromEarthwithaspeedof00500cThewavelengthofaparticularspectral line in light emitted by atomic hydrogen inalaboratoryis656times10-7 m Calculate the valueofthewavelengthofthislinemeasuredinthelaboratoryinlightemittedfromasourceofatomic hydrogen in the galaxy

PHYSICS 2015indb 306 150514 337 PM

10 Fields

Contents

101 ndash Describing fields

102 ndash Fields at work

Essential Ideas

Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Similar approaches can be taken in analysing electrical and gravitational potential problems copy IBO 2014

PHYSICS 2015indb 307 150514 337 PM

Chapter 10

308

AH

L

101 Describing fields

Essential idea Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Understandingsbull Gravitationalfieldsbull Electrostaticfieldsbull Electricpotentialandgravitationalpotentialbull Fieldlinesbull Equipotentialsurfaces

Gravitational potentialWehaveseenthatifweliftanobjectofmassm to a height h abovethesurfaceoftheEarththenitsgainingravitationalpotential energy is mghHoweverthisisbynomeansthefullstoryForastartwenowknowthatgvarieswithh and also the expression really gives a difference in potentialenergybetweenthevaluethattheobjecthasattheEarthrsquossurfaceandthevaluethatithasatheighthSowhatwereallyneedisazeropointCanwefindapointwherethepotentialenergyiszeroandusethispointfromwhichtomeasure changes in potential energy

The point that is chosen is in fact infinity At infinity the gravitationalfieldstrengthofanyobjectwillbezeroSoletusseeifwecandeduceanexpressionforthegaininpotentialenergyofanobjectwhenitisldquoliftedrdquofromthesurfaceoftheEarthtoinfinityThisineffectmeansfindingtheworknecessarytoperformthistask

δr

r r δr+

r infin=

mg

AB

Me

Figure 1001 Gravitational forces

Inthediagramweconsidertheworknecessarytomovethe particle of mass m a distance δr in thegravitationalfield of the Earth

The force on the particle at A is FG Mem

r2----------------=

IftheparticleismovedtoB then since δr isverysmallwe can assume that the field remains constant over thedistance ABTheworkδW done againstthegravitationalfieldoftheEarthinmovingthedistanceABis

δWGMem

r2----------------δrndash=

(rememberthatworkdoneagainstaforceisnegative)

TofindthetotalworkdoneW in going from the surface oftheEarthtoinfinitywehavetoaddalltheselittlebitsofworkThisisdonemathematicallybyusingintegralcalculus

WG Me m

r2----------------ndash

rdR

infin

int G Me m 1

r2----- rd

R

infin

intndash G Me m 1r---ndash

R

infinndash= = =

GMe m 0 1R---ndash

ndashndash=

G MemR----------------ndash=

HencewehavewhereRistheradiusoftheEarththattheworkdonebythegravitationalfieldinmovinganobjectofmass m from R(surfaceoftheEarth)toinfinityisgivenby

WGMe m

R---------------- ndash=

We can generalise the result by calculating the worknecessary per unit mass to take a small mass fromthe surface of the Earth to infinity This we call thegravitational potential Vie

Wewouldgetexactlythesameresultifwecalculatedtheworkdonebythefieldtobringthepointmassfrominfinityto the surface of Earth In this respect the formal definition ofgravitationalpotentialatapointinagravitationalfieldis therefore defined as the work done per unit mass in bringing a point mass from infinity to that point

Clearly then the gravitational potential at any point inthe Earthrsquos field distance r from the centre of the Earth (providingr gt R)is

VGMe

r------------ ndash=

Thepotentialisthereforeameasureoftheamountofworkthathastobedonetomoveparticlesbetweenpointsinagravitationalfieldanditsunit is theJkgndash1 We also note thatthepotentialisnegativesothatthepotentialenergy

NATURE OF SCIENCEParadigm shift The move from direct observable actions being responsible for influence on an object to acceptance of a fieldrsquos ldquoaction at a distancerdquo required a paradigm shift in the world of science (23)

copy IBO 2014

PHYSICS 2015indb 308 150514 337 PM

Fields

309

AH

L

aswemoveawayfromtheEarthrsquossurfaceincreasesuntilitreachesthevalueofzeroatinfinity

IfthegravitationalfieldisduetoapointmassofmassmthenwehavethesameexpressionasaboveexceptthatMe is replaced by m andmustalsoexclude thevalueof thepotential at the point mass itself ie at r = 0

WecanexpressthegravitationalpotentialduetotheEarth(orduetoanysphericalmass)intermsofthegravitationalfield strength at its surface

AtthesurfaceoftheEarthwehave

g 0ndash Re GMRe

---------ndash= e

Sothat

g 0Re2 GM= e

Hence at a distance r from the centre of the Earth the gravitationalpotentialVcanbewrittenas

VGMe

r------------ndashg 0 Re

2

r------------ndash= =

The potential at the surface of the Earth

(r = Re)istherefore-g0Re

It is interesting to see how the expression for thegravitational potential ties in with the expression mgh ThepotentialatthesurfaceoftheEarthis-g0Re (see the exampleabove)andataheighthwillbe-g0g 0 Re h+( )ndash ifweassume that g0doesnotchangeover thedistanceh The differenceinpotentialbetweenthesurfaceandtheheighth is therefore g0hSotheworkneededtoraiseanobjectof mass m to a height h is mgh ie m timesdifference ingravitationalpotential

This we have referred to as the gain in gravitationalpotentialenergy(see235)

However this expression can be extended to any twopoints inanygravitationalfieldsuch that ifanobjectofmass mmovesbetween twopointswhosepotentialsareV1 and V2 respectively then the change in gravitationalpotentialenergyoftheobjectism(V1 ndash V2)

Gravitational potential gradient

Let us consider now a region in space where thegravitationalfieldisconstantInFigure912thetwopointsAandBareseparatedbythedistance∆x

A B

direction of uniform gravitational eld of strength I

∆x

Figure 1002 The gravitational potential gradient

The gravitational field is of strength I and is in the direction shownThe gravitational potential at A is V andatBisV +∆V

Theworkdone is takingapointmassmfromAtoB isF∆x = mI∆x

Howeverbydefinitionthisworkisalsoequalto-m∆V

Therefore mI∆x=-m∆V

or VIx

∆= minus

∆Effectivelythissaysthatthemagnitudeofthegravitationalfield strength is equal to the negative gradient of thepotential If I is constant then V is a linear function of x and Iisequaltothenegativegradientofthestraightlinegraph formed by plotting V against x If I is not constant (asusuallythecase)thenthemagnitudeofI at any point in the field can be found by find the gradient of the V-x graph at that point An example of such a calculation can befoundinSection929

For those of you who do HL maths the relationship between field and potential is seen to follow from the expression for the potential of a point mass viz

mV Gr

= minus

2

ddV mG Ir r

minus = + =

Potential due to one or more point masses

Gravitationalpotentialisascalarquantitysocalculatingthepotential due to more than one point mass is a matter of simpleadditionSoforexamplethepotentialV due to the Moon and Earth and a distance x fromthecentreofEarthonastraightlinebetweenthemisgivenbytheexpression

E MM MV Gx r x

= minus + minus

whereME=massofEarthMM= mass of Moon and r = distancebetweencentreofEarthandMoon

PHYSICS 2015indb 309 150514 337 PM

Chapter 10

310

AH

L

Equipotentials and field linesIfthegravitationalpotentialhasthesamevalueatallpointson a surface the surface is said to be an equipotential surfaceSoforexampleifweimagineasphericalshellaboutEarthwhosecentrecoincideswiththecentreofEarththisshellwillbeanequipotentialsurfaceClearlyifwerepresentthegravitationalfieldstrengthbyfieldlinessincethelinesldquoradiaterdquooutfromthecentreofEarththentheselineswillbeatrightanglestothesurfaceIfthefieldlineswerenotnormaltotheequipotentialsurfacethentherewouldbeacomponentof the field parallel to the surfaceThis wouldmean thatpointsonthesurfacewouldbeatdifferentpotentialsandsoitwouldnolongerbeanequipotentialsurfaceThisofcourseholdstrueforanyequipotentialsurface

Figure 1003 shows the field lines and equipotentials fortwopointmassesm

m m

Figure 1003 Equipotentials for two point masses

It is worth noting that we would get exactly the samepattern ifwewere to replace thepointmasseswith twoequalpointcharges(See935)

Escape speedThe potential at the surface of Earth is which meansthattheenergyrequiredtotakeaparticleofmass mfromthesurfacetoinfinityisequalto

ButwhatdoesitactuallymeantotakesomethingtoinfinityWhentheparticleisonthesurfaceoftheEarthwecanthinkofitassittingatthebottomofaldquopotentialwellrdquoasinfigure1004

particle

surface of Earth

innity

GMR---------

FIgure 1004 A potential well

Theldquodepthrdquoofthewellis and if the particle gains an amount of kinetic energy equal to wherem is its massthenitwillhavejustenoughenergytoldquoliftrdquoitoutofthewell

In reality it doesnrsquot actually go to infinity it justmeansthat the particle is effectively free of the gravitationalattractionof theEarthWe say that ithas ldquoescapedrdquo theEarthrsquosgravitationalpullWemeetthisideainconnectionwithmolecular forcesTwomolecules in a solidwill sitat their equilibrium position the separation where therepulsiveforceisequaltotheattractiveforceIfwesupplyjust enough energy to increase the separation of themolecules such that they are an infinite distance apart then intermolecularforcesnolongeraffectthemoleculesandthesolidwillhavebecomealiquidThereisnoincreaseinthekineticenergyofthemoleculesandsothesolidmeltsat constant temperature

Wecancalculatetheescapespeedofanobjectveryeasilyby equating the kinetic energy to the potential energysuch that

12---mvesc ape

2 GMe mRe

----------------=

vesc aperArr2GMe

Re--------------- 2g0 Re = =

Substituting for g0 and Regivesavalueforvescape of about 11kmsndash1 from the surface of the Earth

Youwillnotethattheescapespeeddoesnotdependonthemassoftheobjectsincebothkineticenergyandpotentialenergy are proportional to the mass

IntheoryifyouwanttogetarockettothemoonitcanbedonewithoutreachingtheescapespeedHoweverthiswouldnecessitate anenormousamountof fuel and it islikely that therocketplus fuelwouldbesoheavythat itwouldnevergetoffthegroundItismuchmorepracticalto accelerate the rocket to the escape speed fromEarthorbitandthenintheoryjustlaunchittotheMoon

Example

Use the followingdata todetermine thepotential at thesurface of Mars and the magnitude of the acceleration of free fall

massofMars =64times1023kg

radiusofMars =34times106 m

Determinealsothegravitationalfieldstrengthatadistanceof 68 times 106mabovethesurfaceofMars

PHYSICS 2015indb 310 150514 337 PM

Fields

311

AH

L

Solution

= minus = minus times times = times23

11 76

64 1067 10 ndash13 10

34 10MV GR

minus timestimes

Nkg-1

But V = -g0R

Therefore 7

0 6

13 10 3834 10

VgR

times= minus = =

timesm s-2

To determine the field strength gh at 68 times 106 m above the surface we use the fact that 0 2

Mg GR

= such that GM = g0R2

Therefore 2 2

0h 2 2 2

h h

38 (34) 042(102)

g RGMgR R

times= = = = m s-2

(the distance from the centre is 34 times 106 + 68 times 106 = 102 times 106 m)

Exercise 1011 Thegraphbelowshowshowthegravitational

potentialoutsideoftheEarthvarieswithdistancefrom the centre

0 10 20 30 40 50 60 70

ndash1

ndash2

ndash3

ndash4

ndash5

ndash6

ndash7

r m times 106

V

Jkgndash1

times 10

7

(a) Usethegraphtodeterminethegainingravitationalpotentialenergyofasatelliteofmass200kgasitmovesfromthesurfaceof the Earth to a height of 30 times 107mabovetheEarthrsquossurface

(b) Calculatetheenergyrequiredtotakeit to infinity

(c) DeterminetheslopeofthegraphatthesurfaceoftheEarthmCommenton youranswer

Electric potential energy

The concept of electric potential energy was developedwith gravitational potential energy in mind Just as anobjectnearthesurfaceoftheEarthhaspotentialenergybecause of its gravitational interaction with the Earthsotoothereiselectricalpotentialenergyassociatedwithinteracting charges

Letusfirstlookatacaseoftwopositivepointchargeseachof 1μC that are initially bound together by a thread in a vacuuminspacewithadistancebetweenthemof10cmasshowninFigure1005Whenthethreadiscutthepointchargesinitiallyatrestwouldmoveinoppositedirectionsmovingwithvelocitiesv1 and v2 along the direction of the electrostatic force of repulsion

BEFORE AFTER

v v21

Figure 1005 Interaction of two positive particles

The electric potential energybetweentwopointchargescan be found by simply adding up the energy associated witheachpairofpointchargesForapairofinteractingchargestheelectricpotentialenergyisgivenby

∆U = ∆Ep + ∆Ek = ∆W = ∆Fr = kqQ

____ r2 times r = kqQ

____ r

Becausenoexternalforceisactingonthesystemtheenergyand momentum must be conserved Initially Ek = 0 and Ep =kqQr=9times109 times 1 times 10-1201m=009JWhentheyareagreatdistance fromeachotherEpwillbenegligibleThefinalenergywillbeequaltofrac12mv1

2+frac12mv22=009J

Momentum isalsoconservedand thevelocitieswouldbethe same magnitude but in opposite directions

Electric potential energyismoreoftendefinedintermsofapointchargemovinginanelectricfieldas

lsquothe electricpotential energybetweenany twopoints inan electric field is defined as negativeoftheworkdone by anelectricfieldinmovingapoint electric chargebetweentwolocationsintheelectricfieldrsquo

ΔU = ΔEp = -ΔW = -Fd = qEx

ΔU = ΔEp +ΔEk = ΔW = ΔFr = kqQ r2 x r = kqQ

wherexisthedistancemovedalong(oroppositeto)thedirection of the electric field

Electric potential energyismeasuredinjoule(J)Justasworkisascalarquantitysotooelectricalpotentialenergyis a scalar quantity The negative of the work done by an electric field in moving a unit electric charge between two points is independent of the path taken Inphysicswesaytheelectricfieldisalsquoconservativersquo field

PHYSICS 2015indb 311 150514 337 PM

Chapter 10

312

AH

L

Supposeanexternalforcesuchasyourhandmovesasmallpositive point test charge in the direction of a uniformelectric fieldAs it ismoving itmust be gaining kineticenergyIfthisoccursthentheelectricpotentialenergyofthe unit charge is changing

InFigure1006apointcharge+qismovedbetweenpointsAandBthroughadistancex in a uniform electric field

B

A

x

+ q

Figure 1006 Movement of a positive point charge in a uniform field

InordertomoveapositivepointchargefrompointAtopointBanexternal forcemustbeapplied to thechargeequaltoqE (F = qE)

Since the force is applied through a distance xthennegativeworkhas to bedone tomove the charge because energyis gainedmeaning there is an increaseelectric potential energybetweenthetwopointsRememberthattheworkdoneisequivalenttotheenergygainedorlostinmovingthe charge through the electric field The concept of electric potential energy is only meaningful as the electric field whichgeneratestheforceinquestionisconservative

W F xtimes E q x times= =

θ x cos θ x

Figure 1007 Charge moved at an angle to the field

If a chargemoves at an angle θ to an electric field thecomponent of the displacement parallel to the electric fieldisusedasshowninFigure918

W F x Eq x θcostimes= =

Theelectricpotentialenergyisstoredintheelectricfieldand theelectricfieldwill return theenergy to thepointcharge when required so as not to violate the Law ofconservationofenergy

Electric potential

The electric potential at a point in an electric field is defined as being the work done per unit charge in bringing a small positive point charge from infinity to that point

ΔV = Vinfin ndash Vf = ndash W ___ q

Ifwedesignatethepotentialenergytobezeroatinfinitythen it follows thatelectricpotentialmustalsobezeroat infinity and the electric potential at any point in an electricfieldwillbe

ΔV = ndash W ___ q

NowsupposeweapplyanexternalforcetoasmallpositivetestchargeasitismovedtowardsanisolatedpositivechargeTheexternalforcemustdoworkonthepositivetestchargetomoveittowardstheisolatedpositivechargeandtheworkmustbepositivewhiletheworkdonebytheelectricfieldmustthereforebenegativeSotheelectricpotentialatthatpointmustbepositiveaccordingtotheaboveequationIfanegativeisolatedchargeisusedtheelectricpotentialatapointonthepositivetestchargewouldbenegativePositivepoint chargesof their ownaccordmove fromaplaceofhighelectricpotential toaplaceof lowelectricpotentialNegative point charges move the other way from lowpotentialtohighpotentialInmovingfrompointAtopointBinthediagramthepositivecharge+qismovingfromalowelectricpotentialtoahighelectricpotential

In thedefinitiongiven the termldquoworkperunitchargerdquohas significance If the test charge is +16 times 10-19Cwherethe charge has a potential energy of 32 times 10-17Jthenthepotentialwouldbe32times10-17J+16times10-19C=200JC-1 Now if the charge was doubled the potential wouldbecome 64times10-17 J However the potential per unitchargewouldbethesame

Electric potential is a scalar quantityandithasunitsJC-1 orvoltswhere1voltequalsone joulepercoloumbThevoltallowsustoadoptaunitfortheelectricfieldintermsofthevolt

PreviouslytheunitfortheelectricfieldwasNCndash1

W = qV and F = qEsoW ___ V = F __ E

E = FV ___ W = FV ___ Fm V mndash1

That is the units of the electric field E can also beexpressed as V mndash1

The work done per unit charge in moving a point charge between two points in an electric field is again independent of the path taken

PHYSICS 2015indb 312 150514 337 PM

Fields

313

AH

L

Electric potential due to a point charge

Letus takeapoint rmetres froma chargedobjectThepotentialatthispointcanbecalculatedusingthefollowing

W = ndashFr = ndashqV and F = ndash q1 q2 _____ 4πε0r

2

Therefore

W = ndash q1 q2 _____ 4πε0r

2 times r = ndash q1 q2 _____ 4πε0r

= ndashq1 times q2 _____ 4πε0r

= ndashq1V

That is V q

4πε0r--------------- =

Orsimply

V kqr------

=

Example

Determinehowmuchworkisdonebytheelectricfieldofpointcharge150μCwhenachargeof200μCismovedfrom infinity to apoint 0400m from thepoint charge(Assumenoaccelerationofthecharges)

Solution

The work done by the electric field is W = -qV = -14πε0 times q times (Q rinfin - Q r0400)

W = (- 200 times 10-6 C times 900 times 109 NmC-2 times 150 times 10-6 C) divide 0400 m = - 0675 J

An external force would have to do +0675 J of work

Electric field strength and electric potential gradient

Letus lookbackatFigure1006Supposeagain that thecharge+qismovedasmalldistancebyaforceFfromAtoBsothattheforcecanbeconsideredconstantTheworkdoneisgivenby

∆W F x∆times=

TheforceFandtheelectricfieldEareoppositelydirectedandweknowthat

F = -qE and ΔW = q ΔV

Thereforetheworkdonecanbegivenas

q ΔV = -q E Δx

Therefore E V∆x∆-------

ndash=

The rate of change of potential ΔV at a point with respect to distance Δx in the direction in which the change is maximum is called the potential gradient We say that the electric field = - the potential gradient and the units are Vm-

1 From the equation we can see that in a graph of electric potential versus distance the gradient of the straight line equals the electric field strength

In reality if a charged particle enters a uniform electric field it will be accelerated uniformly by the field and its kinetic energy will increase This is why we had to assume no acceleration in the last worked example

Ek∆ 12---

mv2 q E xsdot sdot q Vx---

xsdot sdot q Vsdot= = = =

Example

Determine how far apart two parallel plates must besituated so that a potential difference of 150 times 102 V produces an electric field strength of 100 times 103 NC-1

Solution

Using E V∆x------- xhArrndash V∆

E-------15 102 Vtimes

100 103 N C 1ndashtimes------------------------------------------= = =

= 150 times 10-1

The plates are 150 times 10-1 m apart

The electric field and the electric potential at a point due to an evenly distributed charge +q on a sphere can be represented graphically as in Figure 1008

r

r

V

E

r0

r0

r0

Charge of +Q evenly distributed over surface

E 14πε0------------ Q

r2----- r r0gt=

V 14πε0------------Q

r----r r0gt=

Figure 1008 Electric field and potential due to a charged sphere

PHYSICS 2015indb 313 150514 337 PM

Chapter 10

314

AH

L

When the sphere becomes charged we know that thechargedistributes itselfevenlyoverthesurfaceThereforeeverypartofthematerialoftheconductorisatthesamepotential As the electric potential at a point is defined as beingnumericallyequaltotheworkdoneinbringingaunitpositivechargefrominfinitytothatpointithasaconstantvalueineverypartofthematerialoftheconductor

Since the potential is the same at all points on the conducting surfacethen∆V∆xiszeroButE=ndash∆V∆xThereforethe electric field inside the conductor is zero There is no electric field inside the conductor

SomefurtherobservationsofthegraphsinFigure1009are

bull OutsidethespherethegraphsobeytherelationshipsgivenasE α 1 r2 and V α 1 r

bull Atthesurfacer = r0Thereforetheelectricfieldandpotentialhavetheminimumvalueforr at this point and this infers a maximum field and potential

bull Insidethespheretheelectricfieldiszerobull Insidethespherenoworkisdonetomoveacharge

fromapointinsidetothesurfaceThereforethereisno potential difference and the potential is the same asitiswhenr = r0

Similargraphscanbedrawnfortheelectricfieldintensityandthe electric potential as a function of distance from conducting parallelplatesandsurfacesandthesearegiveninFigure1009

Potential plot E eld plotE eld

++++

ndashndashndashndash

xx x

+

x x

ndashx x

+ ndashx

x

+ +

x

x

Figure 1009 Electric field and electric potential at a distance from a charged surface

Potential due to one or more point charges

The potential due to one point charge can be determined byusingtheequationformula

V = kq r

Example 1

Determine the electric potential at a point 20 times 10-1 m fromthecentreofan isolatedconducting spherewithapointchargeof40pCinair

Solution

Using the formula

V = kq r we have

V90 10 9times( ) 40 10 12ndashtimes( )times

20 10 1ndashtimes( )------------------------------------------------------------------ 018 V= =

the potential at the point is 180 times 10-1 V

The potential due to a number of point charges can be determined by adding up the potentials due to individual point charges because the electric potential at any point outside a conducting sphere will be the same as if all the charge was concentrated at its centre

Example 2

Threepointchargesofareplacedattheverticesofaright-angledtriangleasshowninthediagrambelowDeterminetheabsolutepotentialat the+20μCchargeduetothetwoothercharges

- 6μC

+3μC +2μC

4 m

3 m

Solution

The electric potential of the +2 μC charge due to the ndash 6 μC charge is

V = (9 times 109 Nm2C-2 times -6 times 10-6 C) divide (radic 32 + 42) m = - 108 times 104 V

PHYSICS 2015indb 314 150514 338 PM

Fields

315

AH

L

Insummarywecanconcludethat

bull No work is done to move a charge along an equipotential

bull Equipotentials are always perpendicular to the electric lines of force

Figure1011and1012showsomeequipotential lines fortwo oppositely charged and identically positive spheresseparated by a distance

+ve ndashve

equipotential lines

Figure 1011 Equipotential lines between two opposite charges

equipotential lines

Figure 1012 Equipotential lines between two charges which are the same

+

ndash ndash ndash ndash

+ + +

40 V

30 V

20 V

10 V

50 V

Figure 1013 Equipotential lines between charged parallel plates

Figure1013showstheequipotentiallinesbetweenchargedparallel plates Throughout this chapter the similarities and differences between gravitational fields and electric fieldshavebeendiscussedTherelationshipsthatexistsbetweengravitationalandelectricquantitiesandtheeffectsofpointmassesandchargesissummarisedinTable1014

The electric potential of the +2 μC charge due to the +3 μC charge is

V = (9 times 109 Nm2C-2 times 3 times 10-6 C) divide 3m = 9 times 103 V

The net absolute potential is the sum of the 2 potentials- 108 times 104 V + 9 times 103 V =- 18 times 103 V

The absolute potential at the point is - 18 times 103 V

Equipotential surfacesRegions in space where the electric potential of a chargedistributionhasaconstantvaluearecalledequipotentials The placeswherethepotentialisconstantinthreedimensionsarecalled equipotential surfacesandwheretheyareconstantintwodimensionstheyarecalledequipotential lines

TheyareinsomewaysanalogoustothecontourlinesontopographicmapsInthiscasethegravitationalpotentialenergy is constant as amassmoves around the contourlines because the mass remains at the same elevationabovetheEarthrsquossurfaceThegravitationalfieldstrengthacts in a direction perpendicular to a contour line

Similarlybecausetheelectricpotentialonanequipotentiallinehas thesamevalueanelectric forcecandonoworkwhena testchargemovesonanequipotentialThereforethe electric field cannot have a component along anequipotentialandthusitmustbeeverywhereperpendiculartotheequipotentialsurfaceorequipotentiallineThisfactmakesiteasytoplotequipotentialsifthelinesofforceorlinesofelectricfluxofanelectricfieldareknown

Forexamplethereareaseriesofequipotentiallinesbetweentwoparallelplateconductorsthatareperpendiculartotheelectric fieldTherewill be a series of concentric circles(eachcirclefurtherapartthanthepreviousone)thatmapouttheequipotentialsaroundanisolatedpositivesphereThe lines of force and some equipotential lines for anisolatedpositivesphereareshowninFigure1010

Lines ofequipotential

Figure 1010 Equipotentials around an isolated positive sphere

PHYSICS 2015indb 315 150514 338 PM

Chapter 10

316

AH

L

Gravitational quantity Electrical quantityQ

uant

ities

g V∆x∆-------ndash= E V∆

x∆-------ndash=

Poin

t mas

ses a

nd ch

arge

s

V Gmr----

ndash= V 14πε0------------q

r---=

g G m

r2-----ndash= E 14πε0------------ q

r2-----=

F Gm1 m2

r2--------------= F 1

4πε0------------

q1q2

r2-----------=

Figure 1014 Formulas

Example

Deduce the electric potential on the surface of a gold nucleus that has a radius of 62 fm

Solution

Using the formula

V = kq r and knowing the atomic number of gold is 79 We will assume the nucleus is spherical and it behaves as if it were a point charge at its centre (relative to outside points)

V = 90 times 109 Nm2C-2 times 79 times 16 times 10-19 C divide 62 times 10-15 m = 18 times 107 V

The potential at the point is 18 MV

Example

Deduce the ionisation energy in electron-volts of theelectron in the hydrogen atom if the electron is in its ground state and it is in a circular orbit at a distance of 53 times 10-11 m from the proton

Solution

This problem is an energy coulombic circular motion question based on Bohrrsquos model of the atom (not the accepted quantum mechanics model) The ionisation energy is the energy required to remove the electron from the ground state to infinity The electron travels in a circular orbit and therefore has a centripetal acceleration The ionisation energy will counteract the coulombic force and the movement of the electron will be in the opposite direction to the centripetal force

Total energy = Ek electron + Ep due to the proton-electron interaction

ΣF = kqQ r2 = mv2 r and as such mv2 = = kqQ r

Therefore Ek electron = frac12 kqQ r

Ep due to the proton-electron interaction = - kqQ r

Total energy = frac12 kqQ r + - kqQ r = -frac12 kqQ r

= - 90 times 109 Nm2C-2 times (16 times 10-19 C)2 divide 53 times 10-11 m = -217 times 10-18 J

= -217 times 10-18 J divide 16 times 10-19 = -136 eV

The ionisation energy is 136 eV

Exercise 1021 ApointchargePisplacedmidwaybetweentwo

identicalnegativechargesWhichoneofthefollowingiscorrectwithregardstoelectricfieldandelectricpotentialatpointP

Electric field Electric potentialA non-zero zeroB zero non-zeroC non-zero non-zeroD zero zero

2 Twopositivechargedspheresaretiedtogetherinavacuumsomewhereinspacewheretherearenoexternal forces A has a mass of 25 g and a charge of20μCandBhasamassof15gandachargeof30μCThedistancebetweenthemis40cmTheyarethenreleasedasshowninthediagram

BEFORE AFTER

v1 v2

A B

(a) Determinetheirinitialelectricpotentialenergy in the before situation

(b) DeterminethespeedofsphereBafterrelease

3 Thediagrambelowrepresentstwoequipotentiallines in separated by a distance of 5 cm in a uniform electric field

+ + + + + + + +

ndash ndash ndash ndash ndash ndash ndash ndash

40 V

20 V 5 cm

PHYSICS 2015indb 316 150514 338 PM

Fields

317

AH

L

Determine the strength of the electric field

4 Thisquestionisabouttheelectricfieldduetoacharged sphere and the motion of electrons in that fieldThediagrambelowshowsanisolatedmetalsphereinavacuumthatcarriesanegativeelectriccharge of 60 μC

ndash

(a) Onthediagramdrawtheconventionalwayto represent the electric field pattern due to the charged sphere and lines to represent threeequipotentialsurfacesintheregionoutside the sphere

(b) Explainhowthelinesrepresentingtheequipotentialsurfacesthatyouhavesketchedindicatethatthestrengthoftheelectricfieldisdecreasingwithdistancefrom the centre of the sphere

(c) Theelectricfieldstrengthatthesurfaceofthe sphere and at points outside the sphere can be determined by assuming that the sphere acts as a point charge of magnitude 60 μC at its centre The radius of the sphere is 25 times 10ndash2 m Deduce that the magnitude of the field strength at the surface of the sphere is 86 times 107 Vmndash1

An electron is initially at rest on the surface of the sphere

(d) (i) Describethepathfollowedbytheelectronasitleavesthesurfaceofthesphere

(ii) Calculatetheinitialaccelerationofthe electron

5 Determinetheamountofworkthatisdoneinmovingachargeof100nCthroughapotentialdifferenceof150times102 V

6 Three identical 20 μC conducting spheres are placedatthecornersofanequilateraltriangleofsides 25 cm The triangle has one apex C pointing upthepageand2baseanglesAandBDeterminetheabsolutepotentialatB

7 Determinehowfaraparttwoparallelplatesmustbesituatedsothatapotentialdifferenceof250 times 102 V produces an electric field strength of 200 times 103 NC-1

8 Thegapbetweentwoparallelplatesis10times10-3mandthereisapotentialdifferenceof10times104 V betweentheplatesCalculate

i theworkdonebyanelectroninmovingfrom one plate to the other

ii thespeedwithwhichtheelectronreachesthe second plate if released from rest

iii theelectricfieldintensitybetweentheplates

9 Anelectronguninapicturetubeisacceleratedby a potential 25 times 103VDeterminethekineticenergygainedbytheelectroninelectron-volts

10 Determine the electric potential 20 times 10-2 m from achargeof-10times10-5 C

11 Determinetheelectricpotentialatapointmid-waybetweenachargeofndash20pCandanotherof+5pConthelinejoiningtheircentresifthecharges are 10 cm apart

12 During a thunderstorm the electric potential differencebetweenacloudandthegroundis10 times 109 V Determine the magnitude of the change in electric potential energy of an electron thatmovesbetweenthesepointsinelectron-volts

13 A charge of 15 μC is placed in a uniform electric fieldoftwooppositelychargedparallelplateswithamagnitudeof14times103 NC-1

(a) Determinetheworkthatmustbedoneagainstthefieldtomovethepointchargeadistance of 55 cm

(b) Calculatethepotentialdifferencebetweenthe final and initial positions of the charge

(c) Determinethepotentialdifferencebetweenthe plates if their separation distance is 15 cm

14 Duringaflashoflightningthepotentialdifferencebetweenacloudandthegroundwas12 times 109 V and the amount of transferred charge was32C

(a) Determinethechangeinenergyofthetransferred charge

(b) Iftheenergyreleasedwasallusedtoacceleratea1tonnecardeduceitsfinalspeed

(c) Iftheenergyreleasedcouldbeusedtomelticeat0degCdeducetheamountoficethatcould be melted

PHYSICS 2015indb 317 150514 338 PM

Chapter 10

318

AH

L

15 SupposethatwhenanelectronmovedfromAtoBin the diagram along an electric field line that the electric field does 36 times 10-19Jofworkonit

Determinethedifferencesinelectricpotential

(a) VB ndash VA(b) VC ndash VA(c) VC ndash VB

16 DeterminethepotentialatpointPthatislocatedatthecentreofthesquareasshowninthediagrambelow

- 6μC

+3μC +2μC

1m

1m

5μC

P

102 Fields at work

Essential idea Similar approaches can be taken inanalysingelectricalandgravitationalpotentialproblems

Understandingsbull Potentialandpotentialenergybull Potentialgradientbull Potentialdifferencebull Escapespeedbull Orbitalmotionorbitalspeedandorbitalenergybull Forcesandinverse-squarelawbehaviour

Orbital motion orbital speed and orbital energy

Although orbital motion may be circular elliptical orparabolic this sub-topic only dealswith circular orbitsThis sub-topic is not fundamentally new physics butan application that synthesizes ideas from gravitationcircularmotiondynamicsandenergy

TheMoonorbitstheEarthandinthissenseitisoftenreferredtoasasatelliteoftheEarthBefore1957itwastheonlyEarthsatelliteHoweverin1957theRussianslaunchedthefirstmanmadesatelliteSputnik1SincethisdatemanymoresatelliteshavebeenlaunchedandtherearenowliterallythousandsofthemorbitingtheEarthSomeareusedtomonitortheweathersome used to enable people to find accurately their position on thesurfaceoftheEarthmanyareusedincommunicationsandnodoubtsomeareusedtospyonothercountriesFigure1015showshowinprincipleasatellitecanbeputintoorbit

Theperson(whosesizeisgreatlyexaggeratedwithrespectto Earth) standing on the surface on the Earth throwssome stonesThe greater the speed with which a stoneis thrown the further it will land from her The pathsfollowedbythethrownstonesareparabolasByastretchof the imaginationwecanvisualisea situation inwhichastoneis thrownwithsuchaspeedthatbecauseof thecurvatureoftheEarthitwillnotlandonthesurfaceoftheEarthbutgointoldquoorbitrdquo(Path4onFigure1015)

NATURE OF SCIENCECommunication of scientific explanations The ability to apply field theory to the unobservable (charges) and the massively scaled (motion of satellites) required scientists to develop new ways to investigate analyse and report findings to a general public used to scientific discoveries based on tangible and discernible evidence (51)

copy IBO 2014

PHYSICS 2015indb 318 150514 338 PM

Fields

319

AH

L

1

2

34Earth

Figure 1015 Throwing a stone into orbit

TheforcethatcausesthestonestofollowaparabolicpathandtofalltoEarthisgravityandsimilarlytheforcethatkeeps the stone in orbit is gravity For circular motionto occur we have seen that a force must act at rightanglestothevelocityofanobjectthatistheremustbeacentripetalforceHenceinthesituationwedescribeherethe centripetal force for circular orbital motion about the EarthisprovidedbygravitationalattractionoftheEarth

We can calculate the speedwithwhich a stonemustbethrowninordertoputitintoorbitjustabovethesurfaceof the Earth

If the stone has mass m and speed v thenwehave fromNewtonrsquos2ndlaw

mv2

RE---------- G

MEm

RE2------------=

whereRE is the radius of the Earth and ME is the mass of the Earth

Bearinginmindthatg 0 GME

RE2--------= then

v g RE 10 64 106timestimes 8 103times= = =

Thatisthestonemustbethrownat8times103m sndash1

Clearly we are not going to get a satellite into orbit soclosetothesurfaceoftheEarthMovingatthisspeedthefrictionduetoairresistancewouldmeltthesatellitebeforeithadtravelledacoupleofkilometresInrealitythereforeasatelliteisputintoorbitabouttheEarthbysendingitattachedtoarocketbeyondtheEarthrsquosatmosphereandthengivingitacomponentofvelocityperpendiculartoaradialvectorfromtheEarthSeeFigure1016

Earth Satellite carried by rocket to hereSatellite orbit

Tangential component of velocity

Figure 1016 Getting a satellite into orbit

Keplerrsquos third law(This work of Kepler and Newtonrsquos synthesis of the work is an excellent example of the scientific method and makes for a good TOK discussion)

In 1627 Johannes Kepler (1571-1630)publishedhis lawsof planetary motion The laws are empirical in natureandwere deduced from the observations of theDanishastronomer Tycho de Brahe (1546-1601)The third lawgives a relationship between the radius of orbit R of a planet and its period TofrevolutionabouttheSunThelawisexpressedmathematicallyas

2

3 constantTR

=

We shall nowuseNewtonrsquos LawofGravitation to showhowitisthattheplanetsmoveinaccordancewithKeplerrsquosthirdlaw

InessenceNewtonwasabletousehislawofgravitytopredict the motion of the planets since all he had to do was factor theF givenby this law intohis second law F = ma to find their accelerations and hence their future positions

InFigure1017theEarthisshownorbitingtheSunandthedistancebetweentheircentresisR

RSun

Earth

F es

F se

Figure 1017 Planets move according to Keplerrsquos third law

Fes is the force that the Earth exerts on the Sun and Fse is the force that the Sun exerts on the Earth The forces are equalandoppositeandtheSunandtheEarthwillactuallyorbitaboutacommoncentreHoweversince theSun issoverymuchmoremassivethantheEarththiscommoncentrewillbeclosetothecentreoftheSunandsowecanregard the Earth as orbiting about the centre of the Sun Theotherthingthatweshallassumeisthatwecanignorethe forces that the other planets exert on the Earth (This wouldnotbeawisethingtodoifyouwereplanningtosend a space ship to theMoon for example)We shallalso assume that we have followed Newtonrsquos exampleandindeedprovedthataspherewillactasapointmasssituated at the centre of the sphere

Kepler had postulated that the orbits of the planets areellipticalbut since theeccentricityof theEarthrsquosorbit issmallweshallassumeacircularorbit

PHYSICS 2015indb 319 150514 338 PM

Chapter 10

320

AH

L

TheaccelerationoftheEarthtowardstheSunisa = Rω2

whereω ω 2πT------=

Hencea R 2π

T------ times

2 4π2 RT 2------------- = =

But the acceleration is given by Newtonrsquos Second Law F = mawhereFisnowgivenbytheLawofGravitationSo in this situation

F maGMs Me

R2-------------------= = butwealsohavethat

a a R 2πT

------ times

2 4π2 RT 2------------- = = and m = Me so that

G MsMe

R2------------------- Me4π2R

T2-------------times

GMs

4π2-----------rArr R3

T2------= =

Butthequantity

isaconstantthathasthesamevalueforeachoftheplanetssowehaveforalltheplanetsnotjustEarththat

R3

T2------ k =

wherekisaconstantWhichisofcourseKeplerrsquosthirdlaw

ThisisindeedanamazingbreakthroughItisdifficulttorefutethe idea that all particles attract each other in accordance withtheLawofGravitationwhenthelawisabletoaccountfortheobservedmotionoftheplanetsabouttheSun

The gravitational effects of the planets upon each othershould produce perturbations in their orbits Such is the predictivepoweroftheUniversalGravitationalLawthatit enabled physicists to compute these perturbations The telescopehadbeeninventedin1608andbythemiddleofthe 18th Century had reached a degree of perfection in design that enabled astronomers to actually measure the orbital perturbations of the planets Their measurements werealways inagreementwith thepredictionsmadebyNewtonrsquoslawHoweverin1781anewplanetUranuswasdiscovered and the orbit of this planet did not fit withthe orbit predicted by Universal Gravitation Such wasthe physicistrsquos faith in theNewtonianmethod that theysuspectedthatthediscrepancywasduetothepresenceofayetundetectedplanetUsingtheLawofGravitationtheFrenchastronomerJLeverrier and the English astronomer J C Adamswereable tocalculate justhowmassive thisnewplanetmustbeandalsowhereitshouldbeIn1846the planetNeptunewas discovered justwhere they hadpredicted Ina similarwaydiscrepancies in theorbitofNeptune ledto thepredictionandsubsequentdiscoveryin1930oftheplanetPlutoNewtonrsquosLawofGravitation

had passed the ultimate test of any theory it is not only abletoexplainexistingdatabutalsotomakepredictions

Satellite energyWhenasatelliteisinorbitaboutaplanetitwillhavebothkineticenergyandgravitationalpotentialenergySupposewe consider a satellite ofmass m that is in an orbit of radius r about a planet of mass M

Thegravitationalpotentialduetotheplanetatdistancer from its centre is

ThegravitationalpotentialenergyofthesatelliteVsat

is therefore GMe m

r----------------ndash

ThatisVsat GMem

r----------------ndash=

Thegravitationalfieldstrengthatthesurfaceoftheplanetisgivenby

g 0GMe

Re2------------=

Hencewecanwrite

Vsat g0 Re

2

rndash=

m

ThekineticenergyofthesatelliteKsatisequaltofrac12mv2wherev is its orbital speed

Byequatingthegravitationalforceactingonthesatellitetoitscentripetalaccelerationwehave

G Mem

r2---------------- mv2

r---------- mv2hArr

GMemr----------------= =

Fromwhich12---

mv2 12---

GMemr----------------times=

= 0 e

2g R

r

2m

Which is actually quite interesting since it shows thatirrespective of the orbital radius the KE is numericallyequal to half the PE Also the total energy E tot K sat Vsat+ 1

2---G Me m

r----------------times G Me m

r----------------ndash + 1

2---G Me m

r----------------ndash= = =of the satelliteisalwaysnegativesince

E tot K sat Vsat+ 12---

G Me mr----------------times

G Me mr----------------ndash

+ 12---

G Me mr----------------ndash= = =

PHYSICS 2015indb 320 150514 338 PM

Fields

321

AH

L

The energies of an orbiting satellite as a function of radial distancefromthecentreofaplanetareshownplottedinFigure1018

12

100806

0402

0ndash02

ndash04

ndash06ndash08ndash10

ndash12

2 4 6 8 10 12

ener

gya

rbita

ry u

nits

distance timesR

kinetic energytotal energy

potential energy

Figure 1018 Energy of an orbiting satellite as a function of distance from the centre of a planet

WeightlessnessSuppose that you are inanelevator(lift)thatisdescendingat constant speedandyou let goof abook thatyouareholdinginyourhandThebookwillfalltothefloorwithaccelerationequaltotheaccelerationduetogravityIfthecablethatsupportstheelevatorweretosnap(asituationthatItrustwillneverhappentoanyofyou)andyounowletgothebookthatyouareholdinginyourotherhandthis book will not fall to the floor - it will stay exactlyin linewithyourhandThis isbecause thebook isnowfallingwith the sameaccelerationas theelevatorandassuch the book cannot ldquocatchrdquo up with the floor of theelevatorFurthermoreifyouhappenedtobestandingonasetofbathroomscalesthescaleswouldnowreadzero-youwouldbeapparentlyweightlessItisthisideaoffreefallthatexplainstheapparentweightlessnessofastronautsin an orbiting satellite These astronauts are in free fall inthesensethattheyareacceleratingtowardsthecentre of the Earth

It is actually possible to define theweight of a body inseveral different waysWe can define it for example asthegravitationalforceexertedonthebodybyaspecifiedobject such as the EarthThiswe have seen thatwe doinlotsofsituationswherewedefinetheweightasbeingequaltomgIfweusethisdefinitionthenanobjectinfreefallcannotbydefinitionbeweightlesssinceitisstillinagravitationalfieldHoweverifwedefinetheweightofanobjectintermsofaldquoweighingrdquoprocesssuchasthereadingonasetofbathroomscaleswhichineffectmeasuresthecontact force between the object and the scales thenclearlyobjectsinfreefallareweightlessOnenowhasto

askthequestionwhetherornotitispossibleForexampletomeasurethegravitationalforceactingonanastronautin orbit about the Earth

WecanalsodefineweightintermsofthenetgravitationalforceactingonabodyduetoseveraldifferentobjectsForexample for an object out in space its weight could bedefined in terms of the resultant of the forces exerted on it bytheSuntheMoontheEarthandalltheotherplanetsin the Solar System If this resultant is zero at a particular pointthenthebodyisweightlessatthispoint

Inviewofthevariousdefinitionsofweightthatareavailabletousitisimportantthatwhenweusethewordldquoweightrdquoweareawareofthecontextinwhichitisbeingused

Example

Calculate the height above the surface of the Earth atwhichageo-stationarysatelliteorbits

Solution

A geo-stationary satellite is one that orbits the Earth in such a way that it is stationary with respect to a point on the surface of the Earth This means that its orbital period must be the same as the time for the Earth to spin once on its axis ie 24 hours

From Keplerrsquos third law we have G Ms

4π2----------- R3

T2------=

That is

RMe

mh

using the fact that the force of attraction between the satellite and the Earth is given by

FG Mem

R2----------------=

and that F = ma

where m is the mass of the satellite and a 4π2R

T2-------------=

we have

GMem

R2---------------- m 4π2R

T2-------------times

GMe

4π2------------rArr R3

T2------= =

PHYSICS 2015indb 321 150514 338 PM

Chapter 10

322

AH

L

Now the mass of the Earth is 60 times 1024 kg and the period T measured in seconds is given by T = 86400 s

So substitution gives R = 42 times 106 m

The radius of the Earth is 64 times 106 m so that the orbital height h is about 36 times 107 m

Example

Calculatetheminimumenergyrequiredtoputasatelliteofmass500kgintoanorbitthatisasaheightequaltotheEarthrsquosradiusabovethesurfaceoftheEarth

Solution

We have seen that when dealing with gravitational fields and potential it is useful to remember that

g 0G M

Re2---------= or g0 Re

2 GM=

The gravitational potential at the surface of the Earth is

g 0ndash Re GMRe

---------ndash=

The gravitational potential at a distance R

from the centre of the Earth is

=

The difference in potential between the surface of the Earth and a point distance R from the centre is therefore

V∆ g 0Re 1ReR------ndash

=

If R 2Re= then V∆g0 Re

2------------=

This means that the work required to ldquoliftrdquo the satellite into orbit is g0Rm where m is the mass of the satellite This is equal to

10 32 106 500timestimestimes = 16000 MJ

However the satellite must also have kinetic energy in order to orbit Earth This will be equal to

g 0 mRe2

2R-----------------g 0mRe

2

4----------------- 8000 MJ= =

The minimum energy required is therefore

24000 MJ

Exercise 1031 The speed needed to put a satellite in orbit does

not depend on

A the radius of the orbitB theshapeoftheorbitC thevalueofgattheorbitD the mass of the satellite

2 EstimatethespeedofanEarthsatellitewhoseorbitis400kmabovetheEarthrsquossurfaceAlsodetermine the period of the orbit

3 Calculatethespeedofa200kgsatelliteorbitingthe Earth at a height of 70 times 106 m

Assume that g = 82 m sndash2 for this orbit

4 TheradiioftwosatellitesXandYorbitingthe Earth are 2r and 8rwherer is the radius of the Earth Calculate the ratio of the periods of revolutionofXtoY

5 A satellite of mass mkgissentfromEarthrsquossurfaceinto an orbit of radius 5RwhereR is the radius of theEarthWritedownanexpressionfor

(a) thepotentialenergyofthesatelliteinorbit

(b) thekineticenergyofthesatelliteinorbit

(c) theminimumworkrequiredtosendthesatellitefromrestattheEarthrsquossurfaceintoits orbit

6 A satellite in an orbit of 10rfallsbacktoEarth(radius r)afteramalfunctionDeterminethespeedwithwhichitwillhittheEarthrsquossurface

7 The radius of the moon is frac14 that of the Earth AssumingEarthandtheMoontohavethesamedensitycomparetheaccelerationsoffreefallatthe surface of Earth to that at the surface of the Moon

8 UsethefollowingdatatodeterminethegravitationalfieldstrengthatthesurfaceoftheMoon and hence determine the escape speed from the surface of the Moon

Mass of the Moon = 73 times 1022kg

RadiusoftheMoon=17times106 m

PHYSICS 2015indb 322 150514 338 PM

Wave Phenomena

293

AH

L

We note that that D (asymp1metre)isverymuchgreaterthaneither y or d(asymp fewmillimetres)Thismeansthatbothθ and θʹ areverysmallanglesandforintentsandpurposesequal

From the diagram we have that

θ = Dy

And

2S Xd

θʹ =

(Remembertheanglesareverysmall)

Sinceθ asymp θʹ then

2S XyD d

=

ButS2X is the path difference between the waves as they

traveltoPWehavethereforethat

path difference = Dyd

TherewillthereforebeabrightfringeatPif

Dyd

= nλ

Supposethatthereisabrightfringeaty = y1 corresponding

to n = n1 then

Ddy1 = n

1λ

If the next bright fringe occurs at y = y2 this will correspond

to n = n1+1Hence

Ddy2 =(n

1+1)λ

This means that the spacing between the fringes y2 ndash y

1 is

given by

y2 ndash y

1 = d

Dλ

Youngactuallyusethisexpressiontomeasurethewavelengthof the light he used and it is a method still used today

We see for instance that if in a given set up we move the slits closer together then the spacing between the fringes will get greaterEffectivelyourinterferencepatternspreadsoutthatis there will be fewer fringes in a given distance We can also increase the fringe spacing by increasing the distance betweentheslitsandthescreenYouwillalsonotethatfora

givensetupusinglightofdifferentwavelengthsthenldquoredrdquofringeswillspacefurtherapartthanldquobluerdquofringes

In this analysis we have assumed that the slits act as point sources and as such the fringes will be uniformly spaced andofequalintensityAmorethoroughanalysisshouldtakeintoaccountthefinitewidthoftheslits

ReturningtoFigure915weseethatwecanwritethepathdifferenceS2X as

S2X = dtan θʹ

But since θʹ is a small angle the sine and tangent will be nearlyequalsothat

S2X = dsinθʹ

And since θʹ asymp θ then

S2X = dsinθ

The condition therefore for a bright fringe to be found at a point of the screen can therefore be written as

dsinθ = nλ

Insummary

for constructive interference and minima

d sinθ = nλ wheren=123hellip

for destructive interference and maxima

d sinθ = ( n + frac12 )λ wheren=123hellip

Figure 916 shows the intensity distribution of the fringes on the screen when the separation of the slits is large comparedtotheirwidthThefringesareofequalintensityandofequalseparation

Intensity

distance along screen

Figure 916 The intensity distribution of the fringes

It isworthnoting that if the slits are close together theintensity of the fringes is modulated by the intensity distribution of the diffraction pattern of one of the slits

PHYSICS 2015indb 293 150514 337 PM

Chapter 9

294

AH

L

Example

Light of wavelength 500nm is incident on two smallparallel slits separated by 10mmDetermine the anglewherethefirstmaximumisformed

If after passing through the slits the light is brought toa focusona screen15m fromtheslitscalculate theobserved fringe spacing on the screen

Solution

Using the small angle approximation we have

3

7

10105minus

minustimes==

dλθ

= 5 times 10-4 rad

The fringe spacing is given by

3

7

1010551

minus

minustimestimes==

dDy λ

= 075 mm

Modulation of two-slit interference pattern by one-slit diffraction effect

In practice the intensity pattern of the maxima as shown in Figure 917 is not constant but fluctuates while decreasingsymmetrically on either side of the central maximum as shown inFigure917Theintensitypatternisacombinationofboththe single-slit diffractionenvelope and the double slit pattern That is the amplitude of the two-slit interference pattern is modulated(ieadjustedto)byasingleslitdiffractionenvelope

2λD

___ λD

D

___λd

d

___

Double slitnite width

Figure 917 Double slit pattern

ForagivenslitseparationdwavelengthoflightandfixedslittoscreendistancethevariationinintensitydependsonthewidthoftheslitDAlthoughincreasingthewidthoftheslitsincreasestheintensityoflightinthefringesitalsomakesthem less sharp and they become blurred As the slit width is widenedthefringesgraduallydisappearbecausenumerouspoint sources along the widened slit give rise to their own dark and bright fringes that then overlap with each other

Exercise 931 InFigure913thedistancebetweenthespeakers

is050mandthedistancebetweenthelineofthespeakersandthescreenis20mAsthemicrophoneismovedalongthelineXYthedistance between successive points of maximum soundintensityis030mThefrequencyofthesoundwavesis44times103HzCalculateavalueforthe speed of sound

2 Laserlightofwavelength610nmfallsontwoslits10x10-5apartDeterminetheseparationofthefirstordermaximumformedonascreen20maway

3 Twoparallelslits012mmapartareilluminatedwith red light with a wavelength of 600 nm An interferencepatternfallsonascreen15maway

Calculate

(a) thedistancefromthecentralmaximumtothefirstbrightfringe

(b) thedistancetotheseconddarkline

Multiple slit and diffraction grating interference patterns

If we examine the interference pattern produced when monochromatic light passes through a different number of slits we notice that as the number of slits increases the number of observed fringes decreases the spacingbetween them increases and the individual fringes become much sharper We can get some idea of how this comes about by looking at the way light behaves when a parallel beam passes through a large number of slits The diagram for this is shown in Figure 918

θθd

1

2

3

telescope

Figure 918 A parallel beam passing through several slits

The slits are very small so that they can be considered to act as point sources They are also very close together such that d issmall(10ndash6 m)EachslitbecomesasourceofcircularwavefrontsandthewavesfromeachslitwillinterfereLetus

PHYSICS 2015indb 294 150514 337 PM

Wave Phenomena

295

AH

L

consider the light that leaves the slit at an angle θ as shown Thepathdifferencebetweenwave1andwave2isdsinθ and ifthisisequaltoanintegralnumberofwavelengthsthenthe two waves will interfere constructively in this direction Similarly wave 2 will interfere constructively with wave3 at this angle andwave 3with 4 etc across thewholegratingHenceifwelookatthelightthroughatelescopethatisbringittoafocusthenwhenthetelescopemakesanangle θ to the grating a bright fringe will be observed The condition for observing a bright fringe is therefore

dsinθ = mλ

Supposeweuselightofwavelength500nmandsupposethat d = 16 times 10ndash6 m

Obviously we will see a bright fringe in the straight on position θ=0(thezero order)

The next position will be when m =1(thefirst order)andsubstitutionintheaboveequationgivesθ = 18deg

The next position will be when m=2(thesecond order)and this give θ = 38deg

For m=3sinθ is greater than 1 so with this set up we only obtain5fringesonezeroorderandtwoeithersideofthezero order

The calculation shows that the separation of the orders is relatively large At any angles other than 18deg or 38deg the light leaving the slits interferes destructively We can see that the fringes will be sharp since if we move just a small angle away from 18deg the light from the slits will interfere destructively

An array of narrow slits is usually made by cutting narrow transparent lines very close together into the emulsion on aphotographicplate(typically200linespermillimetre)Suchanarrangementiscalledadiffraction grating

The diffraction grating is of great use in examining the spectral characteristics of light sources

All elements have their own characteristic spectrum An element can be made to emit light either by heating it until it is incandescent or by causing an electric discharge through itwhen it is in a gaseous state Yourschool probably has some discharge tubes and diffraction gratingsIfithasthenlookattheglowingdischargetubethroughadiffractiongratingIfforexampletheelementthat you are looking has three distinct wavelengths then each wavelength will be diffracted by a different amount inaccordancewiththeequationdsinθ = nλ

Alsoiflaserlightisshonethroughagratingontoascreenyou will see just how sharp and spaced out are the maxima By measuring the line spacing and the distance of the screen fromthelaserthewavelengthofthelasercanbemeasured

If your school has a set of multiple slits say from a single slit toeight slits then it isalsoaworthwhileexercise toexamine how the diffraction pattern changes when laser light is shone through increasing numbers of slits

Consider a grating of NparallelequidistantslitsofwidthaseparatedbyanopaqueregionofwidthD We know that light diffracted from one slit will be superimposed and interfere with light diffracted from all the other slits The intensityprofile of the resultant wave will have a shape determined by

bull theintensityoflightincidentontheslitsbull thewavelengthofthislightbull theslitwidthbull theslitseparation

Typical intensity profiles formed when a plane wavemonochromaticlightatnormalincidenceononetwoandsix slits are shown in Figure 919 In this example the slit widthisonequarterofthegratingspacing

Six slits sharperfringes

d = 4a

I α N 2

(N ndash 2)= 4 subsidiary maxima

order0 4321

I = 36

Two slits

order0

single-slitdiraction

envelope

d = 4a4th principal maximum missing

principa maxima

I = 4 (nominal units)

4321

intensity

order

ISS

One slit

0 1

single-slitdiraction

pattern

N = number of slitsslit width a

Figure 919 Intensity profiles

PHYSICS 2015indb 295 150514 337 PM

Chapter 9

296

AH

L

1 For Nge2thefringepatterncontainsprincipalmaximathat are modulated by the diffraction envelope

2 ForNgt2thefringepatterncontains(Nndash2)subsidiary maxima

3 Where the grating spacing d is four time the slit width a(4xa =d)thenthefourthprincipalmaximumismissing from the fringe pattern Where the grating spacingdisthreetimetheslitwidtha(3xa =d)thenthe third principal maximum is missing from the fringepatternandsoon

4 Increasing N increases the absolute intensities of the diffraction pattern (I = N2)

If white light is shone through a grating then the central image will be white but for the other orders each will be spread out intoacontinuousspectrumcomposedofaninfinitenumberof adjacent images of the slit formed by the wavelength of the different wavelengths present in the white light At any given point in the continuous spectrum the light will be very nearly monochromatic because of the narrowness of the images of the slit formed by the grating This is in contrast to the doubleslitwhereifwhitelightisusedtheimagesarebroadand the spectral colours are not separated

In summary as is the case of interference at a doubleslit there is a path difference between the rays fromadjacent slits of d sinθ Where there is a whole number of wavelengthsconstructiveinterferenceoccurs

For constructive interference and minima

d sinθ = mλ wheren=123hellip

When there is a path difference between adjacent rays of halfawavelengthdestructiveinterferenceoccurs

For and maxima

d sinθ = ( m + frac12 )λ wheren=123hellip

Exercise 941 Lightfromalaserisshonethroughadiffraction

grating on to a screen The screen is a distance of 20mfromthelaserThedistancebetweenthecentraldiffractionmaximumandthefirstprincipalmaximum formed on the screen is 094 m The diffractiongratinghas680linespermmEstimatethe wavelength of the light emitted by the laser

2 Monochromaticlightfromalaserisnormallyincident on a six-slit grating The slit spacing is three times the slit width An interference patternisformedonascreen2montheotherside of the grating

(a) Determineandexplaintheorderoftheprincipal maxima missing from the fringe pattern

(b) Howmanysubsidiarymaximaaretherebetween the principal maxima

(c) Howmanyprincipalmaximaarethere

(i) Betweenthetwofirstorderminimain the diffraction envelope

(ii) Betweenthefirstandsecondorderminima in the diffraction envelope

(d) Drawthepatternformedbythebrightfringesin(c)

(e) Sketchtheintensityprofileofthefringepattern

Thin film interferenceYoumightwellbe familiarwith the colouredpatternoffringes that can be seen when light is reflected off thesurfaceofwateruponwhichathinoilfilmhasbeenspiltorfromlightreflectedfrombubblesWecanseehowthesepatternsarisebylookingatFigure920

Extendedlight source air

A

B

C

12

3

4

Oil lm

Water

Figure 920 Reflection from an oil film

Consider light from an extended source incident on a thinfilmWealsoconsiderawave fromonepointofthe source whose direction is represented by the ray shownSomeofthislightwillbereflectedatAandsometransmitted through the filmwhere somewill again bereflectedatB(somewillalsobetransmittedintotheair)SomeofthelightreflectedatBwillthenbetransmittedatCandsomereflectedandsoonIfweconsiderjustrays1and2thenthesewillbenotbeinphasewhenthearebrought to a focus by the eye since they have travelled

PHYSICS 2015indb 296 150514 337 PM

Wave Phenomena

297

AH

L

different distances If the path difference between them is an integral number of half wavelengths then we might expect the twowaves to be out of phaseHowever ray1 undergoes a phase change of π on reflection but ray2 does not since it is reflected at a boundary betweena more dense and less dense medium (See chapter 4)Henceifthepathdifferenceisanintegralnumberofhalf-wavelengthsrays1and2willreinforceieray1and2areinphaseHowever rays357etcwillbeoutofphasewithrays246etcbutsinceray2ismoreintensethanray3andray4moreintensethanray5thesepairswillnot cancel out so there will be a maximum of intensity

Ifthepathdifferenceissuchthatwave1and2areoutofphasesincewave1ismoreintensethanwave2theywillnotcompletelyannulHoweveritcanbeshownthattheintensitiesofwaves2345hellipaddtoequaltheintensityofwave1Sincewaves345hellipareinphasewithwave2there will be complete cancellation

The path difference will be determined by the angle at which ray1 is incident andalsoon the thickness (andtheactualrefractiveindexaswell)ofthefilmSincethesourceisanextendedsourcethelightwillreachtheeyefrom many different angles and so a system of bright and darkfringeswillbeseenYouwillonlyseefringesifthefilmisverythin(exceptifviewedatnormalincidence)sinceincreasingthethicknessofthefilmwillcausethereflected rays to get so far apart that they will not becollected by the pupil of the eye

From the argument above to find the conditions forconstructive and destructive interferenceweneedonlyfindthepathdifferencebetweenray1andray2Figure921showsthe geometry of the situation

A BE

F

C

D

1 2

φrsquo φrsquo

φ

φ

d

Figure 921 The geometry of interference

Thefilmisofthicknessd and refractive index n and the light has wavelength λ If the line BF is perpendicular to ray 1 then the optical path difference (opd) between ray 1 andray2whenbroughttoafocusis

opd = n(AC+CB)ndashAF

We have to multiple by the refractive index for the path travelledbythelightinthefilmsincethelighttravelsmoreslowly in thefilm If the light travels say adistancex in a material of refractive index n then in the time that it takes to travelthisdistancethelightwouldtraveladistancenx in air

IfthelineCEisatrightanglestoray2thenweseethat

AF = nBE

FromthediagramAC=CDsowecanwrite

opd = n(CD+CB)ndashnBE=nDE

Alsofromthediagramweseethatwhereφ is the angle of refraction

DE=2cosφ

From which opd=2ndcosφ

Bearinginmindthechangeinphaseofray1onreflectionwe have therefore that the condition for constructive interference is

2nd φcos m 12---+

λ m 1 2 hellip = =

Andfordestructiveinterference 2ndcosφ = mλ

Eachfringecorrespondstoaparticularopd for a particular value of the integer m and for any fringe the value of the angle φ is fixedThismeans that itwill be in the formof an arc of a circle with the centre of the circle at the point where the perpendicular drawn from the eye meets the surface of the film Such fringes are called fringesofequal inclinationSince theeyehasa smallaperturethesefringesunlessviewedatneartonormalincidence (φ = 0) will only be observed if the film is very thinThisisbecauseasthethicknessofthefilmincreasesthereflected rayswill get further and further apart and sovery few will enter the eye

Ifwhitelightisshoneontothefilmthenwecanseewhywe get multi-coloured fringes since a series of maxima and minima will be formed for each wavelength present in the white light However when viewed at normalincidenceitispossiblethatonlylightofonecolourwillundergoconstructiveinterferenceandthefilmwilltakeon this colour

PHYSICS 2015indb 297 150514 337 PM

Chapter 9

298

AH

L

Thickness of oil films

The exercise to follow will help explain this use

Non-reflecting films

A very important but simple application of thin filminterferenceisintheproductionofnon-reflectingsurfaces

Athinfilmofthicknessd and refractive index n1 is coated onto glass of refractive index n where n1 lt n Light ofwavelength λ that is reflected at normal incidence willundergo destructive interference if 12

2n d λ

= thatis

14d

nλ

=

(remember that there will now no phase change at the glass interface ie we have a rare to dense reflection)

The use of such films can greatly reduce the loss oflightby reflection at the various surfacesof a systemoflenses or prisms Optical parts of high quality systemsare usually all coatedwithnon-reflecting films in orderto reduce stray reflections The films are usually madebyevaporatingcalciumormagnesiumfluorideonto thesurfacesinvacuumorbychemicaltreatmentwithacidsthat leave a thin layer of silica on the surface of the glass ThecoatedsurfaceshaveapurplishhuebyreflectedlightThis is because the condition for destructive interference fromaparticularfilmthicknesscanonlybeobtainedforone wavelength The wavelength chosen is one that has a value corresponding to light near the middle of the visible spectrumThismeansthatreflectionofredandvioletlightis greater combining to give the purple colour Because of the factor cosφatanglesotherthannormalincidencethepath difference will change butnotsignificantlyuntilsayabout30deg(egcos25deg=090)

It should be borne in mind that no light is actually lost by anon-reflectingfilmthedecreaseofreflectedintensityiscompensated by increase of transmitted intensity

Non-reflecting films can be painted onto aircraft tosuppressreflectionofradarThethicknessofthefilmis determined by

4nd λ

= where λ is the wavelength of

the radar waves and ntherefractiveindexofthefilmat this wavelength

Example

A plane-parallel glass plate of thickness 20 mm isilluminated with light from an extended source The refractiveindexoftheglassis15andthewavelengthofthelight is 600 nm Calculate how many fringes are formed

Solution

We assume that the fringes are formed by light incident at all angles from normal to grazing incidence

At normal incidence we have 2nd = mλ

From which

m 2 15 2 10 3ndashtimestimestimes6 10 7ndashtimes

------------------------------------------- 10 000= =

At grazing incidence the angle of refraction φ is in fact the critical angle

Therefore φφ arcsin 115------- 42 deg= =

ie cosφ = 075

At grazing incidence 2ndcosφ = m λ

From which (and using cos(sinndash1(115) = 075)

m 2 15 2 10 3ndash 075timestimestimestimes6 10 7ndashtimes

------------------------------------------------------------ 7500= =

The total number of fringes seen is equal to m ndash m = 2500

Exercise 95When viewed from above the colour of an oil film onwater appears red in colour Use the data below to estimate theminimumthicknessoftheoilfilm

average wavelength of red light = 630 nm

refractiveindexofoilforredlight=15

refractive index of water for red light = 13

PHYSICS 2015indb 298 150514 337 PM

Wave Phenomena

299

AH

L

94 Resolution

Essential idea Resolution places an absolute limit on the extent to which an optical or other system can separate images of objects

Understandings

bull Thesizeofadiffractingaperturebull Theresolutionofsimplemonochromatic

two-source systems

The size of a diffracting apertureOur discussion concerning diffraction so far has been for rectangular slits When light from a point source enters a smallcircularapertureitdoesnotproduceabrightdotasanimagebutacirculardiscknownasAiryrsquosdiscsurroundedbyfainterconcentriccircularringsasshowninFigure922

Laser

Figure 922 Diffraction at a circular aperture

Diffractionatanapertureisofgreatimportancebecausetheeye and many optical instruments have circular apertures

What is the half-angular width of the central maximum of the diffraction formed by a circular aperture This is not easy to calculate since it involves some advanced mathematics The problemwasfirstsolvedbytheEnglishAstronomerRoyalGeorgeAiryin1835whoshowedthatforcircularapertures

θ 122 λb--------------= where b is the diameter of the aperture

Example 2

Inthefollowingdiagramparallellightfromadistantpointsource(suchasastar)isbroughttofocusonthescreenSbyaconverginglens(the lens is shown as a vertical arrow)

Thefocal length(distance fromlens toscreen) is25cmand the diameter of the lens is 30 cm The wavelength of thelightfromthestaris560nmCalculatethediameterofthe diameter of the image on the screen

Solution

The lens actually acts as a circular aperture of diameter 30 cm The half angular width of central maximum of the diffraction pattern that it forms on the screen is

75

2

122 122 56 10 23 1030 10b

λθminus

minusminus

times times= = = times

times rad

The diameter of the central maxima is therefore

25 times 10-2 times 23 times 10-5

= 57 times 10-6 m

Although this is small it is still finite and is effectively the image of the star as the intensity of the secondary maxima are small compared to that of the central maximum

The resolution of simple monochromatic two-source systemsThe astronomers tell us that many of the stars that we observe with the naked eye are in fact binary stars That iswhatwe see as a single star actually consistsof twostarsinorbitaboutacommoncentreFurthermoretheastronomerstellusthatifweusealdquogoodrdquotelescopethenwewillactuallyseethetwostarsThatiswewillresolve the single point source into its two component parts Sowhatisitthatdetermineswhetherornotweseethetwo stars as a single point source ie what determines whether or not two sources can be resolved It canrsquot just bethatthetelescopemagnifiesthestarssinceiftheyareacting as point sources magnifying them is not going to make a great deal of difference

Ineachofoureyesthereisanaperturethepupilthroughwhich the light enters This light is then focused by the eye lens onto the retina But we have seen that when light passes through an aperture it is diffracted and so when we look at

NATURE OF SCIENCEImproved technology The Rayleigh criterion is the limit of resolution Continuing advancement in technology such as large diameter dishes or lenses or the use of smaller wavelength lasers pushes the limits of what we can resolve (18)

copy IBO 2014

PHYSICS 2015indb 299 150514 337 PM

Chapter 9

300

AH

L

apointsourceadiffractionpatternwillbeformedontheretina If we look at two point sources then two diffraction patterns will be formed on the retina and these patterns will overlap The width of our pupil and the wavelength of the light emitted by the sources will determine the amount by which they overlap But the degree of overlap will also depend on the angular separation of the two point sources WecanseethisfromFigures923

pupilretina

S1

S2

P2

P1

θ

Figure 923

LightfromthesourceS1 enters the eye and is diffracted by the pupil such that the central maximum of the diffraction patternisformedontheretinaatP1SimilarlylightfromS2producesamaximumatP2 If the two central maxima are well separated then there is a fair chance that we will see the two sources as separate sources If they overlap then we will not be able to distinguish one source from another From the diagram we see as the sources are movedclosertotheeyethentheangleθ increases and so does the separation of the central maxima

Figures924925926and927showsthedifferentdiffractionpatternsandtheintensitydistributionthatmightresultonthe retina as a result of light from two point sources

S1 S21 Well resolved

Figure 924 Very well resolved

S1 S22 Well resolved

Figure 925 Well resolved

S1 S23 Just resolved Rayleigh criterion Minimum of S2 coincides with maximum peak of S1

Figure 926 Just resolved

4 Not resolved

Figure 927 Not resolved

We have suggested that if the central maxima of the two diffraction patterns are reasonably separated then we should be able to resolve two point sources In the late 19th century Lord Rayleigh suggested by how much they should be separated in order for the two sources to be just resolved Ifthecentralmaximumofonediffractionpattern coincides with the firstminima of the otherdiffraction pattern then the two sources will just beresolved This is known as the Rayleigh Criterion

Figure926showsjustthissituationThetwosourcesarejust resolved according to the Rayleigh criterion since the peak of the central maximum of one diffraction pattern coincideswiththefirstminimumoftheotherdiffractionpattern This means that the angular separation of the peaks of the two central maxima formed by each source is just the half angular width of one central maximum ie

θ λb---

=

where b is the width of the slit through which the light fromthesourcespassesHoweverweseefromFigure1117that θ is the angle that the two sources subtend at the slit Henceweconcludethattwosourceswillberesolvedbyaslit if the angle that they subtend at the slit is greater than or equaltoλfraslb

PHYSICS 2015indb 300 150514 337 PM

Wave Phenomena

301

AH

L

Sofarwehavebeenassumingthattheeyeisarectangularslit whereas clearly it is a circular aperture and so we must use the formula

θ 122 λb--------------=

As mentioned above the angle θ is sometimes called the resolving power but should more accurately be called the minimum angle of resolution (θmin)

Clearly the smaller θ the greater will be the resolving power

The diffraction grating is a useful for differentiating closely spacedlinesinemissionspectraLikeaprismspectrometerit can disperse a spectrum into its components but it is better suited because it has higher resolution than the prism spectrometer

If λ1 and λ2 are two nearly equal wavelengths that canbarelybedistinguishedtheresolvanceorresolvingpowerofthegratingisdefinedas

R = λ _______ (λ2 ndash λ 1) = λ ___ Δλ

Where λ asymp λ1 asymp λ2

Therefore thediffractiongratingwithahighresolvancewill be better suited in determining small differences in wavelength

If the diffraction grating has Nlinesbeingilluminateditcan be shown that the resolving power of the mth order diffraction is given by

R = mN

Obviouslytheresolvancebecomesgreaterwiththeordernumber m and with a greater number of illuminated slits

Example

A benchmark for the resolving power of a grating is the sodium doublet in the sodium emission spectrum Two yellow lines in itsspectrumhavewavelengthsof58900nmand58959nm

(a) Determinetheresolvanceofagratingifthegivenwavelengths are to be distinguished from each other

(b) Howmanylinesinthegratingmustbeilluminated in order to resolve these lines in the second order spectrum

Solution

(a) R = λ ___ Δλ = 58900nm __________________ (58900ndash58959)nm

= 589nm _______ 059nm = 10 times 103

(b) N = R __ m = 10 times 10 3 ________ 2 =50times102 lines

It has been seen that diffraction effectively limits the resolving power of optical systems This includes such systems as the eye telescopes and microscopes Theresolving power of these systems is dealt with in the next section This section looks at links between technology and resolving power when looking at the very distant and when looking at the very small

Radio telescopes

The average diameter of the pupil of the human eye is about 25 mm This means that the eye will justresolve two point sources emitting light of wavelength 500 nm if their angular separation at the eye is 7

43

50 10122 24 1025 10

θminus

minusminus

times= times = times

times rad

If the eye were to be able to detect radio waves of wavelength 015m then tohave the same resolvingpower thepupilwould have to have a diameter of about 600 m Clearly this is nonsense but it does illustrate a problem facingastronomers who wish to view very distant objects such as quasarsandgalaxies thatemitradiowavesConventionalradio telescopes consist of a large dish typically 25 min diameter Even with such a large diameter the radiowavelength resolving power of the telescope is much less thantheopticalresolvingpowerofthehumaneyeLetuslook at an example

Example

The Galaxy Cygnus A can be resolved optically as an elliptically shaped galaxy However it is also a strongemitterof radiowavesofwavelength015mTheGalaxyisestimatedtobe50times1024mfromEarthUseofaradiotelescope shows that the radio emission is from two sources separated by a distance of 30 times 1021m Estimate thediameterofthedishrequiredtojustresolvethesources

Solution

The angle θ that the sources resolve at the telescope is given by

θ 21

424

30 10 60 1050 10

θ minustimes= = times

times rad

and 4

122 122 015 300060 10

d λθ minus

times= = =

timesm = 30 km

A radio telescope dish of this size would be impossible tomake letalonesupportThisshowsthatasingledishtype radio telescope cannot be used to resolve the sources and yet they were resolved To get round the problemastronomers use two radio telescopes separated by a large

PHYSICS 2015indb 301 150514 337 PM

Chapter 9

302

AH

L

distance The telescopes view the same objects at the same time and the signals that each receive from the objects are simultaneously superimposed The result of the superposition of the two signals is a two-slit interference pattern The pattern has much narrower fringe spacing than that of the diffraction pattern produced by either telescope on its own hence producing a much higherresolvingpowerWhentelescopesareusedlikethistheyare called a stellar interferometer

InSocorroinNewMexicothereisastellarinterferometerthat consists of 27 parabolic dishes each of diameter25 m arranged in a Y-shape that covers an area of 570km2Thisisaso-calledVeryLargeArray(VLA)Evenhigher resolution can be obtained by using an array of radio telescopes in observatories thousands of kilometres apart A system that uses this so-called technique oflsquovery-long-baseline interferometryrsquo (VLBI) is known asalsquovery-long-baselinearrayrsquo(VLBA)WithVLBAaradiowavelength resolving power can be achieved that is 100 timesgreaterthanthebestopticaltelescopesEvenhigherresolving power can be achieved by using a telescope that isinasatelliteorbitingEarthJapanrsquosInstituteofSpaceandAstronautical Science (ISAS) launched sucha system inFebruary 1997TheNationalAstronomicalObservatoryof Japan the National Science Foundationrsquos NationalRadio Astronomy Observatory (NRAO) the CanadianSpaceAgency theAustraliaTelescopeNationalFacilitythe European VLBI Network and the Joint Instituteback this project forVeryLongBaseline Interferometryin EuropeThis project is a very good example of howInternationalismcanoperateinPhysics

Electron microscope

Telescopes are used to look at very distant objects that are very large but because of their distance fromus appearverysmallMicroscopesontheotherhandareusedtolookat objects that are close to us but are physically very small Aswehave seen justmagnifyingobjects that ismakingthemappearlargerisnotsufficientonitsowntogaindetailabouttheobjectfordetailhighresolutionisneeded

Figure928isaschematicofhowanoptical microscope is usedtoviewanobjectandFigure929isaschematicofatransmission electron microscope(TEM)

Optical lens system

bright light source

glass slide containing specimen (object)

eye

Figure 928 The principle of a light microscope

magnetic lens system

electron source

thin wafer of material

ScreenCCD

Figure 929 The principle of an electron microscope

In the optical microscope the resolving power isdetermined by the particular lens system employed and the wavelength λ of the light used For example twopoints in the sample separated by a distance d will just be resolved if

2d mλ=

where m is a property of the lens system know as the numerical aperture In practice the largest value of m obtainable is about 16 Hence if the microscope slideis illuminatedwith light ofwavelength 480 nm a goodmicroscope will resolve two points separated by a distance dasymp15times10-7masymp015micromPointscloser together thanthiswillnotberesolvedHoweverthisisgoodenoughtodistinguishsomevirusessuchastheEbolavirus

Clearlythesmallerλ the higher the resolving power and this is where the electron microscope comes to the fore The electron microscope makes use of the wave nature of electrons(see1315)IntheTEMelectronspassthrougha wafer thin sample and are then focused by a magnetic field onto a fluorescent screen orCCD (charge coupleddevicesee142)Electronsusedinanelectronmicroscopehavewavelengthstypicallyofabout5times10-12mHoweverthe numerical aperture of electron microscopes is considerablysmaller thanthatofanopticalmicroscopetypicallyabout002NonethelessthismeansthataTEMcanresolvetwopointsthatareabout025nmapartThisresolving power is certainly high enough to make out the shape of large molecules

Another type of electron microscope uses a techniqueby which electrons are scattered from the surface of the sample The scattered electrons and then focused as in theTEM to form an image of the surfaceThese so-calledscanningelectronmicroscopes(SEM)havealowerresolving power than TEMrsquos but give very good threedimensional images

The eye

We saw in the last section that the resolving power of the human eye is about 2 times 10-4 rad Suppose that youare looking at car headlights on a dark night and the car

PHYSICS 2015indb 302 150514 337 PM

Wave Phenomena

303

AH

L

is a distance D away If the separation of the headlight is say 15 m then the headlights will subtend an angle θ 15

Dθ =

at your eye Assuming an average wavelength of

500 nm your eye will resolve the headlights into twoseparatesourcesifthisangleequals2times10-4 rad This gives D=75kmInotherwordsifthecarisapproachingyouona straight road then you will be able to distinguish the two headlightsasseparatesourceswhenthecaris75kmawayfrom you Actually because of the structure of the retina and optical defects the resolving power of the average eye is about 6 times 10-4 rad This means that the car is more likely tobe25kmawaybeforeyouresolvetheheadlights

Astronomical telescope

Letusreturntotheexampleofthebinarystarsdiscussedatthe beginning of this section on resolution The stars Kruger A and B form a binary system The average separation of the stars is 14 times 1012mandtheiraveragedistancefromEarthis12times1017mWhenviewedthroughatelescopeonEarththe system will therefore subtend an angle

12

17

14 1012 10

θtimes

=times

= 12times10-5 rad at the objective lens of the telescope Assuming that the average wavelength of the light emitted bythestarsis500nmthenifthetelescopeistoresolvethesystem into two separate images it must have a minimum diameter Dwhere12times10-5 =

7122 500 10D

minustimes times

This gives D=0050m which is about 5 cm So thisparticular system is easily resolved with a small astronomical telescope

Exercise 961 ItissuggestedthatusingtheISASVLBAitwould

bepossibletoldquoseerdquoagrainofriceatadistanceof5000kmEstimatetheresolvingpoweroftheVLBA

2 Thedistancefromtheeyelenstotheretinais20mm The light receptors in the central part of the retinaareabout5times10-6apartDeterminewhetherthe spacing of the receptors will allow for the eye to resolve the headlights in the above discussion whentheyare25kmfromtheeye

3 ThediameterofPlutois23times106 m and its averagedistancefromEarthis60times1012 m EstimatetheminimumdiameteroftheobjectiveofatelescopethatwillenablePlutotobeseenasadisc as opposed to a point source

95 Doppler effect

Essential ideaTheDopplereffectdescribesthephenomenonofwavelengthfrequencyshiftwhenrelativemotionoccurs

Understandingsbull TheDopplereffectforsoundwavesandlightwaves

The Doppler effect for sound waves and light waves

Consider two observers A and B at rest with respect to asoundsourcethatemitsasoundofconstantfrequencyf Clearly both observers will hear a sound of the same frequencyHoweversupposethatthesourcenowmovesat constant speed towards A A will now hear a sound of frequencyfA that is greater than f and B will hear a sound of frequency fB that is less than f This phenomenon is known as the Doppler Effect orDopplerPrinciple after C J Doppler(1803-1853)

The same effect arises for an observer who is either moving towards or away from a stationary source

Figure 930 shows the waves spreading out from a stationary source that emits a sound of constant frequency f The observersAandBhearasoundofthesamefrequency

A B

wavefront

S

Figure 930 Sound waves from a stationary source

Suppose now that the source moves towards A withconstant speed vFigure9310(a)showsasnapshotofthenew wave pattern

NATURE OF SCIENCETechnology Although originally based on physical observations of the pitch of fast moving sources of sound the Doppler effect has an important role in many different areas such as evidence for the expansion of the universe and generating images used in weather reports and in medicine (55)

copy IBO 2014

PHYSICS 2015indb 303 150514 337 PM

Chapter 9

304

AH

L

sourceA B

smallerwavelength

largerwavelength

Figure 931 (a) Sound waves from a moving source

The wavefronts are now crowded together in the direction of travel of the source and stretched out in the opposite direction This is why now the two observers will now hear notesofdifferentfrequenciesHowmuchthewavesbunchtogether and how much they stretch out will depend on the speed vEssentially A

A

cfλ

= and BB

cfλ

= where λA lt λB and v is the speed of sound

IfthesourceisstationaryandAismovingtowardsitthenthe waves from the source incident on A will be bunched up If A is moving away from the stationary source then the waves from the source incident on A will be stretched out

Christian Doppler(1803ndash1853)actuallyappliedtheprinciple(incorrectlyasithappens)totryandexplainthecolourofstars However the Doppler effect does apply to light aswellastosoundIfalightsourceemitsalightoffrequencyf then if it is moving away from an observer the observer willmeasurethelightemittedashavingalowerfrequencythan fSincethesensationofcolourvisionisrelatedtothefrequencyoflight(bluelightisofahigherfrequencythanred light) light emitted by objects moving way from anobserver is often referred to as being red-shifted whereas if the object is moving toward the observer it is referred to as blue-shiftedThisideaisusedinOptionE(Chapter16)

We do not need to consider here the situations where either the source or the observer is accelerating In a situation for example where an accelerating source is approaching astationaryobserverthentheobserverwillhearasoundof ever increasing frequency This sometimes leads toconfusion in describing what is heard when a source approaches passes and then recedes from a stationaryobserver Suppose for example that you are standingon a station platform and a train sounding its whistle is approaching at constant speed What will you hear as the train approaches and then passes through the station As the train approaches you will hear a sound of constant pitch but increasing loudness The pitch of the sound will be greater than if the train were stationary As the train passes through the station you will hear the pitch change at the momentthetrainpassesyoutoasoundagainofconstantpitch The pitch of this sound will be lower than the sound of the approaching train and its intensity will decrease as the train recedes from you What you do not hear is a sound of increasing pitch and then decreasing pitch

The Doppler equations for sound

Although you will not be expected in an IB examination to derivetheequationsassociatedwithaspectsoftheDopplereffectyouwillbeexpectedtoapplythemForcompletenesstherefore thederivationoftheequationsassociatedwiththeDopplereffectasoutlinedaboveisgivenhere

S primeS O

vs

Figure 931 (b)

InFigure931(b)theobserverOisatrestwithrespecttoasourceofsoundSismovingwithconstanspeedvs directly towards O The source is emitting a note of constant frequencyf and the speed of the emitted sound is v

S shows the position of the source ∆t later When the sourceisatresttheninatime∆t the observer will receive f∆t waves and these waves will occupy a distance v∆t ie

v f

λ =f∆tv∆t =

(Because of the motion of the source this number of waves will now occupy a distance (v∆t ndash vs∆t)The lsquonewrsquowavelength is therefore

v ndash vs

fλrsquo =

f∆tv∆t ndash vs∆t

=

If fisthefrequencyheardbyOthen

vfλ

= or

vf

λ = = sv vfminus

From which

s

vf fv v

=minus

Dividingthroughbyv gives

Equation1111

1 sf f v

v=

minus

If the source is moving away from the observer then we have

s

vf fv v

=minus

1

1 sf vv

=+

Equation112

PHYSICS 2015indb 304 150514 337 PM

Wave Phenomena

305

AH

L

We now consider the case where the source is stationary and the observerismoving towardsthesource with speed v0 In this situation the speed of the sound waves as measured by the observer will be v0 + v We therefore have that

v0 + v = ffraslλ = vfλ

times

From which

o1 vf fv

= +

Equation113

If the observerismovingawayfromthesource then

o1 vf fv

= minus

Fromequation(113)wehavethat

o o1 v vf f f f f fv v

∆ = minus = + minus =

Equation114

Thevelocitiesthatwerefertointheaboveequationsarethevelocities with respect to the medium in which the waves fromthesourcetravelHoweverwhenwearedealingwithalight source it is the relative velocity between the source and the observer that we must consider The reason for this is that lightisuniqueintherespectthatthespeedofthelightwavesdoes not depend on the speed of the source All observers irrespective of their speed or the speed of the source will measure the same velocity for the speed of light This is oneof thecornerstonesof theSpecialTheoryofRelativitywhichisdiscussedinmoredetailinOptionH(Chapter18)When applying the Doppler effect to light we aremainlyconcerned with the motion of the source We look here only at the situation where the speed of the source v is much smaller than the speed of light cinfreespace(v ltlt c)Underthesecircumstanceswhenthesourceismovingtowardstheobserverequation111becomes

vf f f fc

minus = ∆ = Equation115

andwhenthesourceismovingawayfromtheobserverequation112becomes vf f f f

cminus = ∆ = minus

Provided that v ltlt c these same equations apply for astationary source and moving observer

We look at the following example and exercise

Example

Asourceemitsasoundoffrequency440HzItmovesinastraight line towards a stationary observer with a speed of 30ms-1Theobserverhearsasoundoffrequency484HzCalculate the speed of sound in air

Solution

We use equation 111 and substitute f= 484 Hz f = 440 Hz

and vs = 30 m s-1

therefore 1

484 440 301v

=minus

such that 30 4401

484vminus = to

give v = 330 m s-1

Example

A particular radio signal from a galaxy is measured as havingafrequencyof139times109HzThesamesignalfromasourceinalaboratoryhasafrequencyof142times109Hz

Suggestwhy thegalaxy ismovingaway fromEarthandcalculateitsrecessionspeed(iethespeedwithwhichitismovingawayfromEarth)

Solution

The fact that the frequency from the moving source is less than that when it is stationary indicates that it is moving away from the stationary observer ie Earth

Using vf fc

∆ = we have

8 96

9

3 10 (142 139) 10 634 10142 10

c fvf∆ times times minus times

= = = timestimes

m s-1

It is usual when dealing with the Doppler effect of light to express speeds as a fraction of c So in this instance we have v = 0021 c

Using the Doppler effectWe have seen in the above example and exercise how the Dopplereffectmaybeusedtomeasuretherecessionspeedof distant galaxies The effect is also used to measure speed inothersituationsHerewewilllookatthegeneralprincipleinvolved in using the Doppler effect to measure speedFigure932showsasource(thetransmitter)thatemitseithersound or em waves of constant frequency f The waves

PHYSICS 2015indb 305 150514 337 PM

Chapter 9

306

AH

L

fromthesourceare incidentonareflectorthat ismovingtowards the transmitter with speed vThereflectedwavesare detected by the receiver placed alongside the transmitter

reector

v

transmitter

receiver

f f

f f

Figure 932 Using the Doppler effect to measure speed

We shall consider the situation where v ltlt c where c is the speed of the waves from the transmitter

For the reflector receiving waves from the transmitterit is effectively an observer moving towards a stationary sourceFromequation(114) itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation116

For the receiver receivingwaves from the reflector it iseffectively a stationary observer receiving waves from a movingsourceFromequation(115)itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation117

Ifweaddequations(116)and(117)wegetthatthetotalDopplershiftatthereceiver∆f is

v vf f f f fc c

minus = ∆ = +

But 1 vf fc

= +

hence

1 v v vf f fc c c

∆ = + +

But since v ltlt cwe can ignore the term2

2

vc when we

expandthebracketintheaboveequation

Therefore we have2vf fc

∆ = Equation118

If v asymp c then we must use the full Doppler equationsHowever for em radiationwewill alwaysonly considersituations in which v ltlt c

Example

Thespeedofsoundinbloodis1500times103 m s-1 Ultrasound offrequency100MHzisreflectedfrombloodflowinginan arteryThe frequencyof the reflectedwaves receivedbackatthetransmitteris105MHzEstimatethespeedofthebloodflowintheartery

Solution

Using equation (118) we have

6 63

2005 10 1015 10

vtimes = times

timesto give v asymp 36 m s-1 (We have assumed that the ultrasound is incident at right angles to the blood flow)

Exercise 971 Judy is standing on the platform of a station A

high speed train is approaching the station in a straight line at constant speed and is sounding its whistleAsthetrainpassesbyJudythefrequencyof the sound emitted by the whistle as heard by Judychangesfrom640Hzto430HzDetermine

(a) thespeedofthetrain

(b) thefrequencyofthesoundemittedbythewhistle as heard by a person on the train (Speedofsound=330ms-1)

2 AgalaxyismovingawayfromEarthwithaspeedof00500cThewavelengthofaparticularspectral line in light emitted by atomic hydrogen inalaboratoryis656times10-7 m Calculate the valueofthewavelengthofthislinemeasuredinthelaboratoryinlightemittedfromasourceofatomic hydrogen in the galaxy

PHYSICS 2015indb 306 150514 337 PM

10 Fields

Contents

101 ndash Describing fields

102 ndash Fields at work

Essential Ideas

Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Similar approaches can be taken in analysing electrical and gravitational potential problems copy IBO 2014

PHYSICS 2015indb 307 150514 337 PM

Chapter 10

308

AH

L

101 Describing fields

Essential idea Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Understandingsbull Gravitationalfieldsbull Electrostaticfieldsbull Electricpotentialandgravitationalpotentialbull Fieldlinesbull Equipotentialsurfaces

Gravitational potentialWehaveseenthatifweliftanobjectofmassm to a height h abovethesurfaceoftheEarththenitsgainingravitationalpotential energy is mghHoweverthisisbynomeansthefullstoryForastartwenowknowthatgvarieswithh and also the expression really gives a difference in potentialenergybetweenthevaluethattheobjecthasattheEarthrsquossurfaceandthevaluethatithasatheighthSowhatwereallyneedisazeropointCanwefindapointwherethepotentialenergyiszeroandusethispointfromwhichtomeasure changes in potential energy

The point that is chosen is in fact infinity At infinity the gravitationalfieldstrengthofanyobjectwillbezeroSoletusseeifwecandeduceanexpressionforthegaininpotentialenergyofanobjectwhenitisldquoliftedrdquofromthesurfaceoftheEarthtoinfinityThisineffectmeansfindingtheworknecessarytoperformthistask

δr

r r δr+

r infin=

mg

AB

Me

Figure 1001 Gravitational forces

Inthediagramweconsidertheworknecessarytomovethe particle of mass m a distance δr in thegravitationalfield of the Earth

The force on the particle at A is FG Mem

r2----------------=

IftheparticleismovedtoB then since δr isverysmallwe can assume that the field remains constant over thedistance ABTheworkδW done againstthegravitationalfieldoftheEarthinmovingthedistanceABis

δWGMem

r2----------------δrndash=

(rememberthatworkdoneagainstaforceisnegative)

TofindthetotalworkdoneW in going from the surface oftheEarthtoinfinitywehavetoaddalltheselittlebitsofworkThisisdonemathematicallybyusingintegralcalculus

WG Me m

r2----------------ndash

rdR

infin

int G Me m 1

r2----- rd

R

infin

intndash G Me m 1r---ndash

R

infinndash= = =

GMe m 0 1R---ndash

ndashndash=

G MemR----------------ndash=

HencewehavewhereRistheradiusoftheEarththattheworkdonebythegravitationalfieldinmovinganobjectofmass m from R(surfaceoftheEarth)toinfinityisgivenby

WGMe m

R---------------- ndash=

We can generalise the result by calculating the worknecessary per unit mass to take a small mass fromthe surface of the Earth to infinity This we call thegravitational potential Vie

Wewouldgetexactlythesameresultifwecalculatedtheworkdonebythefieldtobringthepointmassfrominfinityto the surface of Earth In this respect the formal definition ofgravitationalpotentialatapointinagravitationalfieldis therefore defined as the work done per unit mass in bringing a point mass from infinity to that point

Clearly then the gravitational potential at any point inthe Earthrsquos field distance r from the centre of the Earth (providingr gt R)is

VGMe

r------------ ndash=

Thepotentialisthereforeameasureoftheamountofworkthathastobedonetomoveparticlesbetweenpointsinagravitationalfieldanditsunit is theJkgndash1 We also note thatthepotentialisnegativesothatthepotentialenergy

NATURE OF SCIENCEParadigm shift The move from direct observable actions being responsible for influence on an object to acceptance of a fieldrsquos ldquoaction at a distancerdquo required a paradigm shift in the world of science (23)

copy IBO 2014

PHYSICS 2015indb 308 150514 337 PM

Fields

309

AH

L

aswemoveawayfromtheEarthrsquossurfaceincreasesuntilitreachesthevalueofzeroatinfinity

IfthegravitationalfieldisduetoapointmassofmassmthenwehavethesameexpressionasaboveexceptthatMe is replaced by m andmustalsoexclude thevalueof thepotential at the point mass itself ie at r = 0

WecanexpressthegravitationalpotentialduetotheEarth(orduetoanysphericalmass)intermsofthegravitationalfield strength at its surface

AtthesurfaceoftheEarthwehave

g 0ndash Re GMRe

---------ndash= e

Sothat

g 0Re2 GM= e

Hence at a distance r from the centre of the Earth the gravitationalpotentialVcanbewrittenas

VGMe

r------------ndashg 0 Re

2

r------------ndash= =

The potential at the surface of the Earth

(r = Re)istherefore-g0Re

It is interesting to see how the expression for thegravitational potential ties in with the expression mgh ThepotentialatthesurfaceoftheEarthis-g0Re (see the exampleabove)andataheighthwillbe-g0g 0 Re h+( )ndash ifweassume that g0doesnotchangeover thedistanceh The differenceinpotentialbetweenthesurfaceandtheheighth is therefore g0hSotheworkneededtoraiseanobjectof mass m to a height h is mgh ie m timesdifference ingravitationalpotential

This we have referred to as the gain in gravitationalpotentialenergy(see235)

However this expression can be extended to any twopoints inanygravitationalfieldsuch that ifanobjectofmass mmovesbetween twopointswhosepotentialsareV1 and V2 respectively then the change in gravitationalpotentialenergyoftheobjectism(V1 ndash V2)

Gravitational potential gradient

Let us consider now a region in space where thegravitationalfieldisconstantInFigure912thetwopointsAandBareseparatedbythedistance∆x

A B

direction of uniform gravitational eld of strength I

∆x

Figure 1002 The gravitational potential gradient

The gravitational field is of strength I and is in the direction shownThe gravitational potential at A is V andatBisV +∆V

Theworkdone is takingapointmassmfromAtoB isF∆x = mI∆x

Howeverbydefinitionthisworkisalsoequalto-m∆V

Therefore mI∆x=-m∆V

or VIx

∆= minus

∆Effectivelythissaysthatthemagnitudeofthegravitationalfield strength is equal to the negative gradient of thepotential If I is constant then V is a linear function of x and Iisequaltothenegativegradientofthestraightlinegraph formed by plotting V against x If I is not constant (asusuallythecase)thenthemagnitudeofI at any point in the field can be found by find the gradient of the V-x graph at that point An example of such a calculation can befoundinSection929

For those of you who do HL maths the relationship between field and potential is seen to follow from the expression for the potential of a point mass viz

mV Gr

= minus

2

ddV mG Ir r

minus = + =

Potential due to one or more point masses

Gravitationalpotentialisascalarquantitysocalculatingthepotential due to more than one point mass is a matter of simpleadditionSoforexamplethepotentialV due to the Moon and Earth and a distance x fromthecentreofEarthonastraightlinebetweenthemisgivenbytheexpression

E MM MV Gx r x

= minus + minus

whereME=massofEarthMM= mass of Moon and r = distancebetweencentreofEarthandMoon

PHYSICS 2015indb 309 150514 337 PM

Chapter 10

310

AH

L

Equipotentials and field linesIfthegravitationalpotentialhasthesamevalueatallpointson a surface the surface is said to be an equipotential surfaceSoforexampleifweimagineasphericalshellaboutEarthwhosecentrecoincideswiththecentreofEarththisshellwillbeanequipotentialsurfaceClearlyifwerepresentthegravitationalfieldstrengthbyfieldlinessincethelinesldquoradiaterdquooutfromthecentreofEarththentheselineswillbeatrightanglestothesurfaceIfthefieldlineswerenotnormaltotheequipotentialsurfacethentherewouldbeacomponentof the field parallel to the surfaceThis wouldmean thatpointsonthesurfacewouldbeatdifferentpotentialsandsoitwouldnolongerbeanequipotentialsurfaceThisofcourseholdstrueforanyequipotentialsurface

Figure 1003 shows the field lines and equipotentials fortwopointmassesm

m m

Figure 1003 Equipotentials for two point masses

It is worth noting that we would get exactly the samepattern ifwewere to replace thepointmasseswith twoequalpointcharges(See935)

Escape speedThe potential at the surface of Earth is which meansthattheenergyrequiredtotakeaparticleofmass mfromthesurfacetoinfinityisequalto

ButwhatdoesitactuallymeantotakesomethingtoinfinityWhentheparticleisonthesurfaceoftheEarthwecanthinkofitassittingatthebottomofaldquopotentialwellrdquoasinfigure1004

particle

surface of Earth

innity

GMR---------

FIgure 1004 A potential well

Theldquodepthrdquoofthewellis and if the particle gains an amount of kinetic energy equal to wherem is its massthenitwillhavejustenoughenergytoldquoliftrdquoitoutofthewell

In reality it doesnrsquot actually go to infinity it justmeansthat the particle is effectively free of the gravitationalattractionof theEarthWe say that ithas ldquoescapedrdquo theEarthrsquosgravitationalpullWemeetthisideainconnectionwithmolecular forcesTwomolecules in a solidwill sitat their equilibrium position the separation where therepulsiveforceisequaltotheattractiveforceIfwesupplyjust enough energy to increase the separation of themolecules such that they are an infinite distance apart then intermolecularforcesnolongeraffectthemoleculesandthesolidwillhavebecomealiquidThereisnoincreaseinthekineticenergyofthemoleculesandsothesolidmeltsat constant temperature

Wecancalculatetheescapespeedofanobjectveryeasilyby equating the kinetic energy to the potential energysuch that

12---mvesc ape

2 GMe mRe

----------------=

vesc aperArr2GMe

Re--------------- 2g0 Re = =

Substituting for g0 and Regivesavalueforvescape of about 11kmsndash1 from the surface of the Earth

Youwillnotethattheescapespeeddoesnotdependonthemassoftheobjectsincebothkineticenergyandpotentialenergy are proportional to the mass

IntheoryifyouwanttogetarockettothemoonitcanbedonewithoutreachingtheescapespeedHoweverthiswouldnecessitate anenormousamountof fuel and it islikely that therocketplus fuelwouldbesoheavythat itwouldnevergetoffthegroundItismuchmorepracticalto accelerate the rocket to the escape speed fromEarthorbitandthenintheoryjustlaunchittotheMoon

Example

Use the followingdata todetermine thepotential at thesurface of Mars and the magnitude of the acceleration of free fall

massofMars =64times1023kg

radiusofMars =34times106 m

Determinealsothegravitationalfieldstrengthatadistanceof 68 times 106mabovethesurfaceofMars

PHYSICS 2015indb 310 150514 337 PM

Fields

311

AH

L

Solution

= minus = minus times times = times23

11 76

64 1067 10 ndash13 10

34 10MV GR

minus timestimes

Nkg-1

But V = -g0R

Therefore 7

0 6

13 10 3834 10

VgR

times= minus = =

timesm s-2

To determine the field strength gh at 68 times 106 m above the surface we use the fact that 0 2

Mg GR

= such that GM = g0R2

Therefore 2 2

0h 2 2 2

h h

38 (34) 042(102)

g RGMgR R

times= = = = m s-2

(the distance from the centre is 34 times 106 + 68 times 106 = 102 times 106 m)

Exercise 1011 Thegraphbelowshowshowthegravitational

potentialoutsideoftheEarthvarieswithdistancefrom the centre

0 10 20 30 40 50 60 70

ndash1

ndash2

ndash3

ndash4

ndash5

ndash6

ndash7

r m times 106

V

Jkgndash1

times 10

7

(a) Usethegraphtodeterminethegainingravitationalpotentialenergyofasatelliteofmass200kgasitmovesfromthesurfaceof the Earth to a height of 30 times 107mabovetheEarthrsquossurface

(b) Calculatetheenergyrequiredtotakeit to infinity

(c) DeterminetheslopeofthegraphatthesurfaceoftheEarthmCommenton youranswer

Electric potential energy

The concept of electric potential energy was developedwith gravitational potential energy in mind Just as anobjectnearthesurfaceoftheEarthhaspotentialenergybecause of its gravitational interaction with the Earthsotoothereiselectricalpotentialenergyassociatedwithinteracting charges

Letusfirstlookatacaseoftwopositivepointchargeseachof 1μC that are initially bound together by a thread in a vacuuminspacewithadistancebetweenthemof10cmasshowninFigure1005Whenthethreadiscutthepointchargesinitiallyatrestwouldmoveinoppositedirectionsmovingwithvelocitiesv1 and v2 along the direction of the electrostatic force of repulsion

BEFORE AFTER

v v21

Figure 1005 Interaction of two positive particles

The electric potential energybetweentwopointchargescan be found by simply adding up the energy associated witheachpairofpointchargesForapairofinteractingchargestheelectricpotentialenergyisgivenby

∆U = ∆Ep + ∆Ek = ∆W = ∆Fr = kqQ

____ r2 times r = kqQ

____ r

Becausenoexternalforceisactingonthesystemtheenergyand momentum must be conserved Initially Ek = 0 and Ep =kqQr=9times109 times 1 times 10-1201m=009JWhentheyareagreatdistance fromeachotherEpwillbenegligibleThefinalenergywillbeequaltofrac12mv1

2+frac12mv22=009J

Momentum isalsoconservedand thevelocitieswouldbethe same magnitude but in opposite directions

Electric potential energyismoreoftendefinedintermsofapointchargemovinginanelectricfieldas

lsquothe electricpotential energybetweenany twopoints inan electric field is defined as negativeoftheworkdone by anelectricfieldinmovingapoint electric chargebetweentwolocationsintheelectricfieldrsquo

ΔU = ΔEp = -ΔW = -Fd = qEx

ΔU = ΔEp +ΔEk = ΔW = ΔFr = kqQ r2 x r = kqQ

wherexisthedistancemovedalong(oroppositeto)thedirection of the electric field

Electric potential energyismeasuredinjoule(J)Justasworkisascalarquantitysotooelectricalpotentialenergyis a scalar quantity The negative of the work done by an electric field in moving a unit electric charge between two points is independent of the path taken Inphysicswesaytheelectricfieldisalsquoconservativersquo field

PHYSICS 2015indb 311 150514 337 PM

Chapter 10

312

AH

L

Supposeanexternalforcesuchasyourhandmovesasmallpositive point test charge in the direction of a uniformelectric fieldAs it ismoving itmust be gaining kineticenergyIfthisoccursthentheelectricpotentialenergyofthe unit charge is changing

InFigure1006apointcharge+qismovedbetweenpointsAandBthroughadistancex in a uniform electric field

B

A

x

+ q

Figure 1006 Movement of a positive point charge in a uniform field

InordertomoveapositivepointchargefrompointAtopointBanexternal forcemustbeapplied to thechargeequaltoqE (F = qE)

Since the force is applied through a distance xthennegativeworkhas to bedone tomove the charge because energyis gainedmeaning there is an increaseelectric potential energybetweenthetwopointsRememberthattheworkdoneisequivalenttotheenergygainedorlostinmovingthe charge through the electric field The concept of electric potential energy is only meaningful as the electric field whichgeneratestheforceinquestionisconservative

W F xtimes E q x times= =

θ x cos θ x

Figure 1007 Charge moved at an angle to the field

If a chargemoves at an angle θ to an electric field thecomponent of the displacement parallel to the electric fieldisusedasshowninFigure918

W F x Eq x θcostimes= =

Theelectricpotentialenergyisstoredintheelectricfieldand theelectricfieldwill return theenergy to thepointcharge when required so as not to violate the Law ofconservationofenergy

Electric potential

The electric potential at a point in an electric field is defined as being the work done per unit charge in bringing a small positive point charge from infinity to that point

ΔV = Vinfin ndash Vf = ndash W ___ q

Ifwedesignatethepotentialenergytobezeroatinfinitythen it follows thatelectricpotentialmustalsobezeroat infinity and the electric potential at any point in an electricfieldwillbe

ΔV = ndash W ___ q

NowsupposeweapplyanexternalforcetoasmallpositivetestchargeasitismovedtowardsanisolatedpositivechargeTheexternalforcemustdoworkonthepositivetestchargetomoveittowardstheisolatedpositivechargeandtheworkmustbepositivewhiletheworkdonebytheelectricfieldmustthereforebenegativeSotheelectricpotentialatthatpointmustbepositiveaccordingtotheaboveequationIfanegativeisolatedchargeisusedtheelectricpotentialatapointonthepositivetestchargewouldbenegativePositivepoint chargesof their ownaccordmove fromaplaceofhighelectricpotential toaplaceof lowelectricpotentialNegative point charges move the other way from lowpotentialtohighpotentialInmovingfrompointAtopointBinthediagramthepositivecharge+qismovingfromalowelectricpotentialtoahighelectricpotential

In thedefinitiongiven the termldquoworkperunitchargerdquohas significance If the test charge is +16 times 10-19Cwherethe charge has a potential energy of 32 times 10-17Jthenthepotentialwouldbe32times10-17J+16times10-19C=200JC-1 Now if the charge was doubled the potential wouldbecome 64times10-17 J However the potential per unitchargewouldbethesame

Electric potential is a scalar quantityandithasunitsJC-1 orvoltswhere1voltequalsone joulepercoloumbThevoltallowsustoadoptaunitfortheelectricfieldintermsofthevolt

PreviouslytheunitfortheelectricfieldwasNCndash1

W = qV and F = qEsoW ___ V = F __ E

E = FV ___ W = FV ___ Fm V mndash1

That is the units of the electric field E can also beexpressed as V mndash1

The work done per unit charge in moving a point charge between two points in an electric field is again independent of the path taken

PHYSICS 2015indb 312 150514 337 PM

Fields

313

AH

L

Electric potential due to a point charge

Letus takeapoint rmetres froma chargedobjectThepotentialatthispointcanbecalculatedusingthefollowing

W = ndashFr = ndashqV and F = ndash q1 q2 _____ 4πε0r

2

Therefore

W = ndash q1 q2 _____ 4πε0r

2 times r = ndash q1 q2 _____ 4πε0r

= ndashq1 times q2 _____ 4πε0r

= ndashq1V

That is V q

4πε0r--------------- =

Orsimply

V kqr------

=

Example

Determinehowmuchworkisdonebytheelectricfieldofpointcharge150μCwhenachargeof200μCismovedfrom infinity to apoint 0400m from thepoint charge(Assumenoaccelerationofthecharges)

Solution

The work done by the electric field is W = -qV = -14πε0 times q times (Q rinfin - Q r0400)

W = (- 200 times 10-6 C times 900 times 109 NmC-2 times 150 times 10-6 C) divide 0400 m = - 0675 J

An external force would have to do +0675 J of work

Electric field strength and electric potential gradient

Letus lookbackatFigure1006Supposeagain that thecharge+qismovedasmalldistancebyaforceFfromAtoBsothattheforcecanbeconsideredconstantTheworkdoneisgivenby

∆W F x∆times=

TheforceFandtheelectricfieldEareoppositelydirectedandweknowthat

F = -qE and ΔW = q ΔV

Thereforetheworkdonecanbegivenas

q ΔV = -q E Δx

Therefore E V∆x∆-------

ndash=

The rate of change of potential ΔV at a point with respect to distance Δx in the direction in which the change is maximum is called the potential gradient We say that the electric field = - the potential gradient and the units are Vm-

1 From the equation we can see that in a graph of electric potential versus distance the gradient of the straight line equals the electric field strength

In reality if a charged particle enters a uniform electric field it will be accelerated uniformly by the field and its kinetic energy will increase This is why we had to assume no acceleration in the last worked example

Ek∆ 12---

mv2 q E xsdot sdot q Vx---

xsdot sdot q Vsdot= = = =

Example

Determine how far apart two parallel plates must besituated so that a potential difference of 150 times 102 V produces an electric field strength of 100 times 103 NC-1

Solution

Using E V∆x------- xhArrndash V∆

E-------15 102 Vtimes

100 103 N C 1ndashtimes------------------------------------------= = =

= 150 times 10-1

The plates are 150 times 10-1 m apart

The electric field and the electric potential at a point due to an evenly distributed charge +q on a sphere can be represented graphically as in Figure 1008

r

r

V

E

r0

r0

r0

Charge of +Q evenly distributed over surface

E 14πε0------------ Q

r2----- r r0gt=

V 14πε0------------Q

r----r r0gt=

Figure 1008 Electric field and potential due to a charged sphere

PHYSICS 2015indb 313 150514 337 PM

Chapter 10

314

AH

L

When the sphere becomes charged we know that thechargedistributes itselfevenlyoverthesurfaceThereforeeverypartofthematerialoftheconductorisatthesamepotential As the electric potential at a point is defined as beingnumericallyequaltotheworkdoneinbringingaunitpositivechargefrominfinitytothatpointithasaconstantvalueineverypartofthematerialoftheconductor

Since the potential is the same at all points on the conducting surfacethen∆V∆xiszeroButE=ndash∆V∆xThereforethe electric field inside the conductor is zero There is no electric field inside the conductor

SomefurtherobservationsofthegraphsinFigure1009are

bull OutsidethespherethegraphsobeytherelationshipsgivenasE α 1 r2 and V α 1 r

bull Atthesurfacer = r0Thereforetheelectricfieldandpotentialhavetheminimumvalueforr at this point and this infers a maximum field and potential

bull Insidethespheretheelectricfieldiszerobull Insidethespherenoworkisdonetomoveacharge

fromapointinsidetothesurfaceThereforethereisno potential difference and the potential is the same asitiswhenr = r0

Similargraphscanbedrawnfortheelectricfieldintensityandthe electric potential as a function of distance from conducting parallelplatesandsurfacesandthesearegiveninFigure1009

Potential plot E eld plotE eld

++++

ndashndashndashndash

xx x

+

x x

ndashx x

+ ndashx

x

+ +

x

x

Figure 1009 Electric field and electric potential at a distance from a charged surface

Potential due to one or more point charges

The potential due to one point charge can be determined byusingtheequationformula

V = kq r

Example 1

Determine the electric potential at a point 20 times 10-1 m fromthecentreofan isolatedconducting spherewithapointchargeof40pCinair

Solution

Using the formula

V = kq r we have

V90 10 9times( ) 40 10 12ndashtimes( )times

20 10 1ndashtimes( )------------------------------------------------------------------ 018 V= =

the potential at the point is 180 times 10-1 V

The potential due to a number of point charges can be determined by adding up the potentials due to individual point charges because the electric potential at any point outside a conducting sphere will be the same as if all the charge was concentrated at its centre

Example 2

Threepointchargesofareplacedattheverticesofaright-angledtriangleasshowninthediagrambelowDeterminetheabsolutepotentialat the+20μCchargeduetothetwoothercharges

- 6μC

+3μC +2μC

4 m

3 m

Solution

The electric potential of the +2 μC charge due to the ndash 6 μC charge is

V = (9 times 109 Nm2C-2 times -6 times 10-6 C) divide (radic 32 + 42) m = - 108 times 104 V

PHYSICS 2015indb 314 150514 338 PM

Fields

315

AH

L

Insummarywecanconcludethat

bull No work is done to move a charge along an equipotential

bull Equipotentials are always perpendicular to the electric lines of force

Figure1011and1012showsomeequipotential lines fortwo oppositely charged and identically positive spheresseparated by a distance

+ve ndashve

equipotential lines

Figure 1011 Equipotential lines between two opposite charges

equipotential lines

Figure 1012 Equipotential lines between two charges which are the same

+

ndash ndash ndash ndash

+ + +

40 V

30 V

20 V

10 V

50 V

Figure 1013 Equipotential lines between charged parallel plates

Figure1013showstheequipotentiallinesbetweenchargedparallel plates Throughout this chapter the similarities and differences between gravitational fields and electric fieldshavebeendiscussedTherelationshipsthatexistsbetweengravitationalandelectricquantitiesandtheeffectsofpointmassesandchargesissummarisedinTable1014

The electric potential of the +2 μC charge due to the +3 μC charge is

V = (9 times 109 Nm2C-2 times 3 times 10-6 C) divide 3m = 9 times 103 V

The net absolute potential is the sum of the 2 potentials- 108 times 104 V + 9 times 103 V =- 18 times 103 V

The absolute potential at the point is - 18 times 103 V

Equipotential surfacesRegions in space where the electric potential of a chargedistributionhasaconstantvaluearecalledequipotentials The placeswherethepotentialisconstantinthreedimensionsarecalled equipotential surfacesandwheretheyareconstantintwodimensionstheyarecalledequipotential lines

TheyareinsomewaysanalogoustothecontourlinesontopographicmapsInthiscasethegravitationalpotentialenergy is constant as amassmoves around the contourlines because the mass remains at the same elevationabovetheEarthrsquossurfaceThegravitationalfieldstrengthacts in a direction perpendicular to a contour line

Similarlybecausetheelectricpotentialonanequipotentiallinehas thesamevalueanelectric forcecandonoworkwhena testchargemovesonanequipotentialThereforethe electric field cannot have a component along anequipotentialandthusitmustbeeverywhereperpendiculartotheequipotentialsurfaceorequipotentiallineThisfactmakesiteasytoplotequipotentialsifthelinesofforceorlinesofelectricfluxofanelectricfieldareknown

Forexamplethereareaseriesofequipotentiallinesbetweentwoparallelplateconductorsthatareperpendiculartotheelectric fieldTherewill be a series of concentric circles(eachcirclefurtherapartthanthepreviousone)thatmapouttheequipotentialsaroundanisolatedpositivesphereThe lines of force and some equipotential lines for anisolatedpositivesphereareshowninFigure1010

Lines ofequipotential

Figure 1010 Equipotentials around an isolated positive sphere

PHYSICS 2015indb 315 150514 338 PM

Chapter 10

316

AH

L

Gravitational quantity Electrical quantityQ

uant

ities

g V∆x∆-------ndash= E V∆

x∆-------ndash=

Poin

t mas

ses a

nd ch

arge

s

V Gmr----

ndash= V 14πε0------------q

r---=

g G m

r2-----ndash= E 14πε0------------ q

r2-----=

F Gm1 m2

r2--------------= F 1

4πε0------------

q1q2

r2-----------=

Figure 1014 Formulas

Example

Deduce the electric potential on the surface of a gold nucleus that has a radius of 62 fm

Solution

Using the formula

V = kq r and knowing the atomic number of gold is 79 We will assume the nucleus is spherical and it behaves as if it were a point charge at its centre (relative to outside points)

V = 90 times 109 Nm2C-2 times 79 times 16 times 10-19 C divide 62 times 10-15 m = 18 times 107 V

The potential at the point is 18 MV

Example

Deduce the ionisation energy in electron-volts of theelectron in the hydrogen atom if the electron is in its ground state and it is in a circular orbit at a distance of 53 times 10-11 m from the proton

Solution

This problem is an energy coulombic circular motion question based on Bohrrsquos model of the atom (not the accepted quantum mechanics model) The ionisation energy is the energy required to remove the electron from the ground state to infinity The electron travels in a circular orbit and therefore has a centripetal acceleration The ionisation energy will counteract the coulombic force and the movement of the electron will be in the opposite direction to the centripetal force

Total energy = Ek electron + Ep due to the proton-electron interaction

ΣF = kqQ r2 = mv2 r and as such mv2 = = kqQ r

Therefore Ek electron = frac12 kqQ r

Ep due to the proton-electron interaction = - kqQ r

Total energy = frac12 kqQ r + - kqQ r = -frac12 kqQ r

= - 90 times 109 Nm2C-2 times (16 times 10-19 C)2 divide 53 times 10-11 m = -217 times 10-18 J

= -217 times 10-18 J divide 16 times 10-19 = -136 eV

The ionisation energy is 136 eV

Exercise 1021 ApointchargePisplacedmidwaybetweentwo

identicalnegativechargesWhichoneofthefollowingiscorrectwithregardstoelectricfieldandelectricpotentialatpointP

Electric field Electric potentialA non-zero zeroB zero non-zeroC non-zero non-zeroD zero zero

2 Twopositivechargedspheresaretiedtogetherinavacuumsomewhereinspacewheretherearenoexternal forces A has a mass of 25 g and a charge of20μCandBhasamassof15gandachargeof30μCThedistancebetweenthemis40cmTheyarethenreleasedasshowninthediagram

BEFORE AFTER

v1 v2

A B

(a) Determinetheirinitialelectricpotentialenergy in the before situation

(b) DeterminethespeedofsphereBafterrelease

3 Thediagrambelowrepresentstwoequipotentiallines in separated by a distance of 5 cm in a uniform electric field

+ + + + + + + +

ndash ndash ndash ndash ndash ndash ndash ndash

40 V

20 V 5 cm

PHYSICS 2015indb 316 150514 338 PM

Fields

317

AH

L

Determine the strength of the electric field

4 Thisquestionisabouttheelectricfieldduetoacharged sphere and the motion of electrons in that fieldThediagrambelowshowsanisolatedmetalsphereinavacuumthatcarriesanegativeelectriccharge of 60 μC

ndash

(a) Onthediagramdrawtheconventionalwayto represent the electric field pattern due to the charged sphere and lines to represent threeequipotentialsurfacesintheregionoutside the sphere

(b) Explainhowthelinesrepresentingtheequipotentialsurfacesthatyouhavesketchedindicatethatthestrengthoftheelectricfieldisdecreasingwithdistancefrom the centre of the sphere

(c) Theelectricfieldstrengthatthesurfaceofthe sphere and at points outside the sphere can be determined by assuming that the sphere acts as a point charge of magnitude 60 μC at its centre The radius of the sphere is 25 times 10ndash2 m Deduce that the magnitude of the field strength at the surface of the sphere is 86 times 107 Vmndash1

An electron is initially at rest on the surface of the sphere

(d) (i) Describethepathfollowedbytheelectronasitleavesthesurfaceofthesphere

(ii) Calculatetheinitialaccelerationofthe electron

5 Determinetheamountofworkthatisdoneinmovingachargeof100nCthroughapotentialdifferenceof150times102 V

6 Three identical 20 μC conducting spheres are placedatthecornersofanequilateraltriangleofsides 25 cm The triangle has one apex C pointing upthepageand2baseanglesAandBDeterminetheabsolutepotentialatB

7 Determinehowfaraparttwoparallelplatesmustbesituatedsothatapotentialdifferenceof250 times 102 V produces an electric field strength of 200 times 103 NC-1

8 Thegapbetweentwoparallelplatesis10times10-3mandthereisapotentialdifferenceof10times104 V betweentheplatesCalculate

i theworkdonebyanelectroninmovingfrom one plate to the other

ii thespeedwithwhichtheelectronreachesthe second plate if released from rest

iii theelectricfieldintensitybetweentheplates

9 Anelectronguninapicturetubeisacceleratedby a potential 25 times 103VDeterminethekineticenergygainedbytheelectroninelectron-volts

10 Determine the electric potential 20 times 10-2 m from achargeof-10times10-5 C

11 Determinetheelectricpotentialatapointmid-waybetweenachargeofndash20pCandanotherof+5pConthelinejoiningtheircentresifthecharges are 10 cm apart

12 During a thunderstorm the electric potential differencebetweenacloudandthegroundis10 times 109 V Determine the magnitude of the change in electric potential energy of an electron thatmovesbetweenthesepointsinelectron-volts

13 A charge of 15 μC is placed in a uniform electric fieldoftwooppositelychargedparallelplateswithamagnitudeof14times103 NC-1

(a) Determinetheworkthatmustbedoneagainstthefieldtomovethepointchargeadistance of 55 cm

(b) Calculatethepotentialdifferencebetweenthe final and initial positions of the charge

(c) Determinethepotentialdifferencebetweenthe plates if their separation distance is 15 cm

14 Duringaflashoflightningthepotentialdifferencebetweenacloudandthegroundwas12 times 109 V and the amount of transferred charge was32C

(a) Determinethechangeinenergyofthetransferred charge

(b) Iftheenergyreleasedwasallusedtoacceleratea1tonnecardeduceitsfinalspeed

(c) Iftheenergyreleasedcouldbeusedtomelticeat0degCdeducetheamountoficethatcould be melted

PHYSICS 2015indb 317 150514 338 PM

Chapter 10

318

AH

L

15 SupposethatwhenanelectronmovedfromAtoBin the diagram along an electric field line that the electric field does 36 times 10-19Jofworkonit

Determinethedifferencesinelectricpotential

(a) VB ndash VA(b) VC ndash VA(c) VC ndash VB

16 DeterminethepotentialatpointPthatislocatedatthecentreofthesquareasshowninthediagrambelow

- 6μC

+3μC +2μC

1m

1m

5μC

P

102 Fields at work

Essential idea Similar approaches can be taken inanalysingelectricalandgravitationalpotentialproblems

Understandingsbull Potentialandpotentialenergybull Potentialgradientbull Potentialdifferencebull Escapespeedbull Orbitalmotionorbitalspeedandorbitalenergybull Forcesandinverse-squarelawbehaviour

Orbital motion orbital speed and orbital energy

Although orbital motion may be circular elliptical orparabolic this sub-topic only dealswith circular orbitsThis sub-topic is not fundamentally new physics butan application that synthesizes ideas from gravitationcircularmotiondynamicsandenergy

TheMoonorbitstheEarthandinthissenseitisoftenreferredtoasasatelliteoftheEarthBefore1957itwastheonlyEarthsatelliteHoweverin1957theRussianslaunchedthefirstmanmadesatelliteSputnik1SincethisdatemanymoresatelliteshavebeenlaunchedandtherearenowliterallythousandsofthemorbitingtheEarthSomeareusedtomonitortheweathersome used to enable people to find accurately their position on thesurfaceoftheEarthmanyareusedincommunicationsandnodoubtsomeareusedtospyonothercountriesFigure1015showshowinprincipleasatellitecanbeputintoorbit

Theperson(whosesizeisgreatlyexaggeratedwithrespectto Earth) standing on the surface on the Earth throwssome stonesThe greater the speed with which a stoneis thrown the further it will land from her The pathsfollowedbythethrownstonesareparabolasByastretchof the imaginationwecanvisualisea situation inwhichastoneis thrownwithsuchaspeedthatbecauseof thecurvatureoftheEarthitwillnotlandonthesurfaceoftheEarthbutgointoldquoorbitrdquo(Path4onFigure1015)

NATURE OF SCIENCECommunication of scientific explanations The ability to apply field theory to the unobservable (charges) and the massively scaled (motion of satellites) required scientists to develop new ways to investigate analyse and report findings to a general public used to scientific discoveries based on tangible and discernible evidence (51)

copy IBO 2014

PHYSICS 2015indb 318 150514 338 PM

Fields

319

AH

L

1

2

34Earth

Figure 1015 Throwing a stone into orbit

TheforcethatcausesthestonestofollowaparabolicpathandtofalltoEarthisgravityandsimilarlytheforcethatkeeps the stone in orbit is gravity For circular motionto occur we have seen that a force must act at rightanglestothevelocityofanobjectthatistheremustbeacentripetalforceHenceinthesituationwedescribeherethe centripetal force for circular orbital motion about the EarthisprovidedbygravitationalattractionoftheEarth

We can calculate the speedwithwhich a stonemustbethrowninordertoputitintoorbitjustabovethesurfaceof the Earth

If the stone has mass m and speed v thenwehave fromNewtonrsquos2ndlaw

mv2

RE---------- G

MEm

RE2------------=

whereRE is the radius of the Earth and ME is the mass of the Earth

Bearinginmindthatg 0 GME

RE2--------= then

v g RE 10 64 106timestimes 8 103times= = =

Thatisthestonemustbethrownat8times103m sndash1

Clearly we are not going to get a satellite into orbit soclosetothesurfaceoftheEarthMovingatthisspeedthefrictionduetoairresistancewouldmeltthesatellitebeforeithadtravelledacoupleofkilometresInrealitythereforeasatelliteisputintoorbitabouttheEarthbysendingitattachedtoarocketbeyondtheEarthrsquosatmosphereandthengivingitacomponentofvelocityperpendiculartoaradialvectorfromtheEarthSeeFigure1016

Earth Satellite carried by rocket to hereSatellite orbit

Tangential component of velocity

Figure 1016 Getting a satellite into orbit

Keplerrsquos third law(This work of Kepler and Newtonrsquos synthesis of the work is an excellent example of the scientific method and makes for a good TOK discussion)

In 1627 Johannes Kepler (1571-1630)publishedhis lawsof planetary motion The laws are empirical in natureandwere deduced from the observations of theDanishastronomer Tycho de Brahe (1546-1601)The third lawgives a relationship between the radius of orbit R of a planet and its period TofrevolutionabouttheSunThelawisexpressedmathematicallyas

2

3 constantTR

=

We shall nowuseNewtonrsquos LawofGravitation to showhowitisthattheplanetsmoveinaccordancewithKeplerrsquosthirdlaw

InessenceNewtonwasabletousehislawofgravitytopredict the motion of the planets since all he had to do was factor theF givenby this law intohis second law F = ma to find their accelerations and hence their future positions

InFigure1017theEarthisshownorbitingtheSunandthedistancebetweentheircentresisR

RSun

Earth

F es

F se

Figure 1017 Planets move according to Keplerrsquos third law

Fes is the force that the Earth exerts on the Sun and Fse is the force that the Sun exerts on the Earth The forces are equalandoppositeandtheSunandtheEarthwillactuallyorbitaboutacommoncentreHoweversince theSun issoverymuchmoremassivethantheEarththiscommoncentrewillbeclosetothecentreoftheSunandsowecanregard the Earth as orbiting about the centre of the Sun Theotherthingthatweshallassumeisthatwecanignorethe forces that the other planets exert on the Earth (This wouldnotbeawisethingtodoifyouwereplanningtosend a space ship to theMoon for example)We shallalso assume that we have followed Newtonrsquos exampleandindeedprovedthataspherewillactasapointmasssituated at the centre of the sphere

Kepler had postulated that the orbits of the planets areellipticalbut since theeccentricityof theEarthrsquosorbit issmallweshallassumeacircularorbit

PHYSICS 2015indb 319 150514 338 PM

Chapter 10

320

AH

L

TheaccelerationoftheEarthtowardstheSunisa = Rω2

whereω ω 2πT------=

Hencea R 2π

T------ times

2 4π2 RT 2------------- = =

But the acceleration is given by Newtonrsquos Second Law F = mawhereFisnowgivenbytheLawofGravitationSo in this situation

F maGMs Me

R2-------------------= = butwealsohavethat

a a R 2πT

------ times

2 4π2 RT 2------------- = = and m = Me so that

G MsMe

R2------------------- Me4π2R

T2-------------times

GMs

4π2-----------rArr R3

T2------= =

Butthequantity

isaconstantthathasthesamevalueforeachoftheplanetssowehaveforalltheplanetsnotjustEarththat

R3

T2------ k =

wherekisaconstantWhichisofcourseKeplerrsquosthirdlaw

ThisisindeedanamazingbreakthroughItisdifficulttorefutethe idea that all particles attract each other in accordance withtheLawofGravitationwhenthelawisabletoaccountfortheobservedmotionoftheplanetsabouttheSun

The gravitational effects of the planets upon each othershould produce perturbations in their orbits Such is the predictivepoweroftheUniversalGravitationalLawthatit enabled physicists to compute these perturbations The telescopehadbeeninventedin1608andbythemiddleofthe 18th Century had reached a degree of perfection in design that enabled astronomers to actually measure the orbital perturbations of the planets Their measurements werealways inagreementwith thepredictionsmadebyNewtonrsquoslawHoweverin1781anewplanetUranuswasdiscovered and the orbit of this planet did not fit withthe orbit predicted by Universal Gravitation Such wasthe physicistrsquos faith in theNewtonianmethod that theysuspectedthatthediscrepancywasduetothepresenceofayetundetectedplanetUsingtheLawofGravitationtheFrenchastronomerJLeverrier and the English astronomer J C Adamswereable tocalculate justhowmassive thisnewplanetmustbeandalsowhereitshouldbeIn1846the planetNeptunewas discovered justwhere they hadpredicted Ina similarwaydiscrepancies in theorbitofNeptune ledto thepredictionandsubsequentdiscoveryin1930oftheplanetPlutoNewtonrsquosLawofGravitation

had passed the ultimate test of any theory it is not only abletoexplainexistingdatabutalsotomakepredictions

Satellite energyWhenasatelliteisinorbitaboutaplanetitwillhavebothkineticenergyandgravitationalpotentialenergySupposewe consider a satellite ofmass m that is in an orbit of radius r about a planet of mass M

Thegravitationalpotentialduetotheplanetatdistancer from its centre is

ThegravitationalpotentialenergyofthesatelliteVsat

is therefore GMe m

r----------------ndash

ThatisVsat GMem

r----------------ndash=

Thegravitationalfieldstrengthatthesurfaceoftheplanetisgivenby

g 0GMe

Re2------------=

Hencewecanwrite

Vsat g0 Re

2

rndash=

m

ThekineticenergyofthesatelliteKsatisequaltofrac12mv2wherev is its orbital speed

Byequatingthegravitationalforceactingonthesatellitetoitscentripetalaccelerationwehave

G Mem

r2---------------- mv2

r---------- mv2hArr

GMemr----------------= =

Fromwhich12---

mv2 12---

GMemr----------------times=

= 0 e

2g R

r

2m

Which is actually quite interesting since it shows thatirrespective of the orbital radius the KE is numericallyequal to half the PE Also the total energy E tot K sat Vsat+ 1

2---G Me m

r----------------times G Me m

r----------------ndash + 1

2---G Me m

r----------------ndash= = =of the satelliteisalwaysnegativesince

E tot K sat Vsat+ 12---

G Me mr----------------times

G Me mr----------------ndash

+ 12---

G Me mr----------------ndash= = =

PHYSICS 2015indb 320 150514 338 PM

Fields

321

AH

L

The energies of an orbiting satellite as a function of radial distancefromthecentreofaplanetareshownplottedinFigure1018

12

100806

0402

0ndash02

ndash04

ndash06ndash08ndash10

ndash12

2 4 6 8 10 12

ener

gya

rbita

ry u

nits

distance timesR

kinetic energytotal energy

potential energy

Figure 1018 Energy of an orbiting satellite as a function of distance from the centre of a planet

WeightlessnessSuppose that you are inanelevator(lift)thatisdescendingat constant speedandyou let goof abook thatyouareholdinginyourhandThebookwillfalltothefloorwithaccelerationequaltotheaccelerationduetogravityIfthecablethatsupportstheelevatorweretosnap(asituationthatItrustwillneverhappentoanyofyou)andyounowletgothebookthatyouareholdinginyourotherhandthis book will not fall to the floor - it will stay exactlyin linewithyourhandThis isbecause thebook isnowfallingwith the sameaccelerationas theelevatorandassuch the book cannot ldquocatchrdquo up with the floor of theelevatorFurthermoreifyouhappenedtobestandingonasetofbathroomscalesthescaleswouldnowreadzero-youwouldbeapparentlyweightlessItisthisideaoffreefallthatexplainstheapparentweightlessnessofastronautsin an orbiting satellite These astronauts are in free fall inthesensethattheyareacceleratingtowardsthecentre of the Earth

It is actually possible to define theweight of a body inseveral different waysWe can define it for example asthegravitationalforceexertedonthebodybyaspecifiedobject such as the EarthThiswe have seen thatwe doinlotsofsituationswherewedefinetheweightasbeingequaltomgIfweusethisdefinitionthenanobjectinfreefallcannotbydefinitionbeweightlesssinceitisstillinagravitationalfieldHoweverifwedefinetheweightofanobjectintermsofaldquoweighingrdquoprocesssuchasthereadingonasetofbathroomscaleswhichineffectmeasuresthecontact force between the object and the scales thenclearlyobjectsinfreefallareweightlessOnenowhasto

askthequestionwhetherornotitispossibleForexampletomeasurethegravitationalforceactingonanastronautin orbit about the Earth

WecanalsodefineweightintermsofthenetgravitationalforceactingonabodyduetoseveraldifferentobjectsForexample for an object out in space its weight could bedefined in terms of the resultant of the forces exerted on it bytheSuntheMoontheEarthandalltheotherplanetsin the Solar System If this resultant is zero at a particular pointthenthebodyisweightlessatthispoint

Inviewofthevariousdefinitionsofweightthatareavailabletousitisimportantthatwhenweusethewordldquoweightrdquoweareawareofthecontextinwhichitisbeingused

Example

Calculate the height above the surface of the Earth atwhichageo-stationarysatelliteorbits

Solution

A geo-stationary satellite is one that orbits the Earth in such a way that it is stationary with respect to a point on the surface of the Earth This means that its orbital period must be the same as the time for the Earth to spin once on its axis ie 24 hours

From Keplerrsquos third law we have G Ms

4π2----------- R3

T2------=

That is

RMe

mh

using the fact that the force of attraction between the satellite and the Earth is given by

FG Mem

R2----------------=

and that F = ma

where m is the mass of the satellite and a 4π2R

T2-------------=

we have

GMem

R2---------------- m 4π2R

T2-------------times

GMe

4π2------------rArr R3

T2------= =

PHYSICS 2015indb 321 150514 338 PM

Chapter 10

322

AH

L

Now the mass of the Earth is 60 times 1024 kg and the period T measured in seconds is given by T = 86400 s

So substitution gives R = 42 times 106 m

The radius of the Earth is 64 times 106 m so that the orbital height h is about 36 times 107 m

Example

Calculatetheminimumenergyrequiredtoputasatelliteofmass500kgintoanorbitthatisasaheightequaltotheEarthrsquosradiusabovethesurfaceoftheEarth

Solution

We have seen that when dealing with gravitational fields and potential it is useful to remember that

g 0G M

Re2---------= or g0 Re

2 GM=

The gravitational potential at the surface of the Earth is

g 0ndash Re GMRe

---------ndash=

The gravitational potential at a distance R

from the centre of the Earth is

=

The difference in potential between the surface of the Earth and a point distance R from the centre is therefore

V∆ g 0Re 1ReR------ndash

=

If R 2Re= then V∆g0 Re

2------------=

This means that the work required to ldquoliftrdquo the satellite into orbit is g0Rm where m is the mass of the satellite This is equal to

10 32 106 500timestimestimes = 16000 MJ

However the satellite must also have kinetic energy in order to orbit Earth This will be equal to

g 0 mRe2

2R-----------------g 0mRe

2

4----------------- 8000 MJ= =

The minimum energy required is therefore

24000 MJ

Exercise 1031 The speed needed to put a satellite in orbit does

not depend on

A the radius of the orbitB theshapeoftheorbitC thevalueofgattheorbitD the mass of the satellite

2 EstimatethespeedofanEarthsatellitewhoseorbitis400kmabovetheEarthrsquossurfaceAlsodetermine the period of the orbit

3 Calculatethespeedofa200kgsatelliteorbitingthe Earth at a height of 70 times 106 m

Assume that g = 82 m sndash2 for this orbit

4 TheradiioftwosatellitesXandYorbitingthe Earth are 2r and 8rwherer is the radius of the Earth Calculate the ratio of the periods of revolutionofXtoY

5 A satellite of mass mkgissentfromEarthrsquossurfaceinto an orbit of radius 5RwhereR is the radius of theEarthWritedownanexpressionfor

(a) thepotentialenergyofthesatelliteinorbit

(b) thekineticenergyofthesatelliteinorbit

(c) theminimumworkrequiredtosendthesatellitefromrestattheEarthrsquossurfaceintoits orbit

6 A satellite in an orbit of 10rfallsbacktoEarth(radius r)afteramalfunctionDeterminethespeedwithwhichitwillhittheEarthrsquossurface

7 The radius of the moon is frac14 that of the Earth AssumingEarthandtheMoontohavethesamedensitycomparetheaccelerationsoffreefallatthe surface of Earth to that at the surface of the Moon

8 UsethefollowingdatatodeterminethegravitationalfieldstrengthatthesurfaceoftheMoon and hence determine the escape speed from the surface of the Moon

Mass of the Moon = 73 times 1022kg

RadiusoftheMoon=17times106 m

PHYSICS 2015indb 322 150514 338 PM

Chapter 9

294

AH

L

Example

Light of wavelength 500nm is incident on two smallparallel slits separated by 10mmDetermine the anglewherethefirstmaximumisformed

If after passing through the slits the light is brought toa focusona screen15m fromtheslitscalculate theobserved fringe spacing on the screen

Solution

Using the small angle approximation we have

3

7

10105minus

minustimes==

dλθ

= 5 times 10-4 rad

The fringe spacing is given by

3

7

1010551

minus

minustimestimes==

dDy λ

= 075 mm

Modulation of two-slit interference pattern by one-slit diffraction effect

In practice the intensity pattern of the maxima as shown in Figure 917 is not constant but fluctuates while decreasingsymmetrically on either side of the central maximum as shown inFigure917Theintensitypatternisacombinationofboththe single-slit diffractionenvelope and the double slit pattern That is the amplitude of the two-slit interference pattern is modulated(ieadjustedto)byasingleslitdiffractionenvelope

2λD

___ λD

D

___λd

d

___

Double slitnite width

Figure 917 Double slit pattern

ForagivenslitseparationdwavelengthoflightandfixedslittoscreendistancethevariationinintensitydependsonthewidthoftheslitDAlthoughincreasingthewidthoftheslitsincreasestheintensityoflightinthefringesitalsomakesthem less sharp and they become blurred As the slit width is widenedthefringesgraduallydisappearbecausenumerouspoint sources along the widened slit give rise to their own dark and bright fringes that then overlap with each other

Exercise 931 InFigure913thedistancebetweenthespeakers

is050mandthedistancebetweenthelineofthespeakersandthescreenis20mAsthemicrophoneismovedalongthelineXYthedistance between successive points of maximum soundintensityis030mThefrequencyofthesoundwavesis44times103HzCalculateavalueforthe speed of sound

2 Laserlightofwavelength610nmfallsontwoslits10x10-5apartDeterminetheseparationofthefirstordermaximumformedonascreen20maway

3 Twoparallelslits012mmapartareilluminatedwith red light with a wavelength of 600 nm An interferencepatternfallsonascreen15maway

Calculate

(a) thedistancefromthecentralmaximumtothefirstbrightfringe

(b) thedistancetotheseconddarkline

Multiple slit and diffraction grating interference patterns

If we examine the interference pattern produced when monochromatic light passes through a different number of slits we notice that as the number of slits increases the number of observed fringes decreases the spacingbetween them increases and the individual fringes become much sharper We can get some idea of how this comes about by looking at the way light behaves when a parallel beam passes through a large number of slits The diagram for this is shown in Figure 918

θθd

1

2

3

telescope

Figure 918 A parallel beam passing through several slits

The slits are very small so that they can be considered to act as point sources They are also very close together such that d issmall(10ndash6 m)EachslitbecomesasourceofcircularwavefrontsandthewavesfromeachslitwillinterfereLetus

PHYSICS 2015indb 294 150514 337 PM

Wave Phenomena

295

AH

L

consider the light that leaves the slit at an angle θ as shown Thepathdifferencebetweenwave1andwave2isdsinθ and ifthisisequaltoanintegralnumberofwavelengthsthenthe two waves will interfere constructively in this direction Similarly wave 2 will interfere constructively with wave3 at this angle andwave 3with 4 etc across thewholegratingHenceifwelookatthelightthroughatelescopethatisbringittoafocusthenwhenthetelescopemakesanangle θ to the grating a bright fringe will be observed The condition for observing a bright fringe is therefore

dsinθ = mλ

Supposeweuselightofwavelength500nmandsupposethat d = 16 times 10ndash6 m

Obviously we will see a bright fringe in the straight on position θ=0(thezero order)

The next position will be when m =1(thefirst order)andsubstitutionintheaboveequationgivesθ = 18deg

The next position will be when m=2(thesecond order)and this give θ = 38deg

For m=3sinθ is greater than 1 so with this set up we only obtain5fringesonezeroorderandtwoeithersideofthezero order

The calculation shows that the separation of the orders is relatively large At any angles other than 18deg or 38deg the light leaving the slits interferes destructively We can see that the fringes will be sharp since if we move just a small angle away from 18deg the light from the slits will interfere destructively

An array of narrow slits is usually made by cutting narrow transparent lines very close together into the emulsion on aphotographicplate(typically200linespermillimetre)Suchanarrangementiscalledadiffraction grating

The diffraction grating is of great use in examining the spectral characteristics of light sources

All elements have their own characteristic spectrum An element can be made to emit light either by heating it until it is incandescent or by causing an electric discharge through itwhen it is in a gaseous state Yourschool probably has some discharge tubes and diffraction gratingsIfithasthenlookattheglowingdischargetubethroughadiffractiongratingIfforexampletheelementthat you are looking has three distinct wavelengths then each wavelength will be diffracted by a different amount inaccordancewiththeequationdsinθ = nλ

Alsoiflaserlightisshonethroughagratingontoascreenyou will see just how sharp and spaced out are the maxima By measuring the line spacing and the distance of the screen fromthelaserthewavelengthofthelasercanbemeasured

If your school has a set of multiple slits say from a single slit toeight slits then it isalsoaworthwhileexercise toexamine how the diffraction pattern changes when laser light is shone through increasing numbers of slits

Consider a grating of NparallelequidistantslitsofwidthaseparatedbyanopaqueregionofwidthD We know that light diffracted from one slit will be superimposed and interfere with light diffracted from all the other slits The intensityprofile of the resultant wave will have a shape determined by

bull theintensityoflightincidentontheslitsbull thewavelengthofthislightbull theslitwidthbull theslitseparation

Typical intensity profiles formed when a plane wavemonochromaticlightatnormalincidenceononetwoandsix slits are shown in Figure 919 In this example the slit widthisonequarterofthegratingspacing

Six slits sharperfringes

d = 4a

I α N 2

(N ndash 2)= 4 subsidiary maxima

order0 4321

I = 36

Two slits

order0

single-slitdiraction

envelope

d = 4a4th principal maximum missing

principa maxima

I = 4 (nominal units)

4321

intensity

order

ISS

One slit

0 1

single-slitdiraction

pattern

N = number of slitsslit width a

Figure 919 Intensity profiles

PHYSICS 2015indb 295 150514 337 PM

Chapter 9

296

AH

L

1 For Nge2thefringepatterncontainsprincipalmaximathat are modulated by the diffraction envelope

2 ForNgt2thefringepatterncontains(Nndash2)subsidiary maxima

3 Where the grating spacing d is four time the slit width a(4xa =d)thenthefourthprincipalmaximumismissing from the fringe pattern Where the grating spacingdisthreetimetheslitwidtha(3xa =d)thenthe third principal maximum is missing from the fringepatternandsoon

4 Increasing N increases the absolute intensities of the diffraction pattern (I = N2)

If white light is shone through a grating then the central image will be white but for the other orders each will be spread out intoacontinuousspectrumcomposedofaninfinitenumberof adjacent images of the slit formed by the wavelength of the different wavelengths present in the white light At any given point in the continuous spectrum the light will be very nearly monochromatic because of the narrowness of the images of the slit formed by the grating This is in contrast to the doubleslitwhereifwhitelightisusedtheimagesarebroadand the spectral colours are not separated

In summary as is the case of interference at a doubleslit there is a path difference between the rays fromadjacent slits of d sinθ Where there is a whole number of wavelengthsconstructiveinterferenceoccurs

For constructive interference and minima

d sinθ = mλ wheren=123hellip

When there is a path difference between adjacent rays of halfawavelengthdestructiveinterferenceoccurs

For and maxima

d sinθ = ( m + frac12 )λ wheren=123hellip

Exercise 941 Lightfromalaserisshonethroughadiffraction

grating on to a screen The screen is a distance of 20mfromthelaserThedistancebetweenthecentraldiffractionmaximumandthefirstprincipalmaximum formed on the screen is 094 m The diffractiongratinghas680linespermmEstimatethe wavelength of the light emitted by the laser

2 Monochromaticlightfromalaserisnormallyincident on a six-slit grating The slit spacing is three times the slit width An interference patternisformedonascreen2montheotherside of the grating

(a) Determineandexplaintheorderoftheprincipal maxima missing from the fringe pattern

(b) Howmanysubsidiarymaximaaretherebetween the principal maxima

(c) Howmanyprincipalmaximaarethere

(i) Betweenthetwofirstorderminimain the diffraction envelope

(ii) Betweenthefirstandsecondorderminima in the diffraction envelope

(d) Drawthepatternformedbythebrightfringesin(c)

(e) Sketchtheintensityprofileofthefringepattern

Thin film interferenceYoumightwellbe familiarwith the colouredpatternoffringes that can be seen when light is reflected off thesurfaceofwateruponwhichathinoilfilmhasbeenspiltorfromlightreflectedfrombubblesWecanseehowthesepatternsarisebylookingatFigure920

Extendedlight source air

A

B

C

12

3

4

Oil lm

Water

Figure 920 Reflection from an oil film

Consider light from an extended source incident on a thinfilmWealsoconsiderawave fromonepointofthe source whose direction is represented by the ray shownSomeofthislightwillbereflectedatAandsometransmitted through the filmwhere somewill again bereflectedatB(somewillalsobetransmittedintotheair)SomeofthelightreflectedatBwillthenbetransmittedatCandsomereflectedandsoonIfweconsiderjustrays1and2thenthesewillbenotbeinphasewhenthearebrought to a focus by the eye since they have travelled

PHYSICS 2015indb 296 150514 337 PM

Wave Phenomena

297

AH

L

different distances If the path difference between them is an integral number of half wavelengths then we might expect the twowaves to be out of phaseHowever ray1 undergoes a phase change of π on reflection but ray2 does not since it is reflected at a boundary betweena more dense and less dense medium (See chapter 4)Henceifthepathdifferenceisanintegralnumberofhalf-wavelengthsrays1and2willreinforceieray1and2areinphaseHowever rays357etcwillbeoutofphasewithrays246etcbutsinceray2ismoreintensethanray3andray4moreintensethanray5thesepairswillnot cancel out so there will be a maximum of intensity

Ifthepathdifferenceissuchthatwave1and2areoutofphasesincewave1ismoreintensethanwave2theywillnotcompletelyannulHoweveritcanbeshownthattheintensitiesofwaves2345hellipaddtoequaltheintensityofwave1Sincewaves345hellipareinphasewithwave2there will be complete cancellation

The path difference will be determined by the angle at which ray1 is incident andalsoon the thickness (andtheactualrefractiveindexaswell)ofthefilmSincethesourceisanextendedsourcethelightwillreachtheeyefrom many different angles and so a system of bright and darkfringeswillbeseenYouwillonlyseefringesifthefilmisverythin(exceptifviewedatnormalincidence)sinceincreasingthethicknessofthefilmwillcausethereflected rays to get so far apart that they will not becollected by the pupil of the eye

From the argument above to find the conditions forconstructive and destructive interferenceweneedonlyfindthepathdifferencebetweenray1andray2Figure921showsthe geometry of the situation

A BE

F

C

D

1 2

φrsquo φrsquo

φ

φ

d

Figure 921 The geometry of interference

Thefilmisofthicknessd and refractive index n and the light has wavelength λ If the line BF is perpendicular to ray 1 then the optical path difference (opd) between ray 1 andray2whenbroughttoafocusis

opd = n(AC+CB)ndashAF

We have to multiple by the refractive index for the path travelledbythelightinthefilmsincethelighttravelsmoreslowly in thefilm If the light travels say adistancex in a material of refractive index n then in the time that it takes to travelthisdistancethelightwouldtraveladistancenx in air

IfthelineCEisatrightanglestoray2thenweseethat

AF = nBE

FromthediagramAC=CDsowecanwrite

opd = n(CD+CB)ndashnBE=nDE

Alsofromthediagramweseethatwhereφ is the angle of refraction

DE=2cosφ

From which opd=2ndcosφ

Bearinginmindthechangeinphaseofray1onreflectionwe have therefore that the condition for constructive interference is

2nd φcos m 12---+

λ m 1 2 hellip = =

Andfordestructiveinterference 2ndcosφ = mλ

Eachfringecorrespondstoaparticularopd for a particular value of the integer m and for any fringe the value of the angle φ is fixedThismeans that itwill be in the formof an arc of a circle with the centre of the circle at the point where the perpendicular drawn from the eye meets the surface of the film Such fringes are called fringesofequal inclinationSince theeyehasa smallaperturethesefringesunlessviewedatneartonormalincidence (φ = 0) will only be observed if the film is very thinThisisbecauseasthethicknessofthefilmincreasesthereflected rayswill get further and further apart and sovery few will enter the eye

Ifwhitelightisshoneontothefilmthenwecanseewhywe get multi-coloured fringes since a series of maxima and minima will be formed for each wavelength present in the white light However when viewed at normalincidenceitispossiblethatonlylightofonecolourwillundergoconstructiveinterferenceandthefilmwilltakeon this colour

PHYSICS 2015indb 297 150514 337 PM

Chapter 9

298

AH

L

Thickness of oil films

The exercise to follow will help explain this use

Non-reflecting films

A very important but simple application of thin filminterferenceisintheproductionofnon-reflectingsurfaces

Athinfilmofthicknessd and refractive index n1 is coated onto glass of refractive index n where n1 lt n Light ofwavelength λ that is reflected at normal incidence willundergo destructive interference if 12

2n d λ

= thatis

14d

nλ

=

(remember that there will now no phase change at the glass interface ie we have a rare to dense reflection)

The use of such films can greatly reduce the loss oflightby reflection at the various surfacesof a systemoflenses or prisms Optical parts of high quality systemsare usually all coatedwithnon-reflecting films in orderto reduce stray reflections The films are usually madebyevaporatingcalciumormagnesiumfluorideonto thesurfacesinvacuumorbychemicaltreatmentwithacidsthat leave a thin layer of silica on the surface of the glass ThecoatedsurfaceshaveapurplishhuebyreflectedlightThis is because the condition for destructive interference fromaparticularfilmthicknesscanonlybeobtainedforone wavelength The wavelength chosen is one that has a value corresponding to light near the middle of the visible spectrumThismeansthatreflectionofredandvioletlightis greater combining to give the purple colour Because of the factor cosφatanglesotherthannormalincidencethepath difference will change butnotsignificantlyuntilsayabout30deg(egcos25deg=090)

It should be borne in mind that no light is actually lost by anon-reflectingfilmthedecreaseofreflectedintensityiscompensated by increase of transmitted intensity

Non-reflecting films can be painted onto aircraft tosuppressreflectionofradarThethicknessofthefilmis determined by

4nd λ

= where λ is the wavelength of

the radar waves and ntherefractiveindexofthefilmat this wavelength

Example

A plane-parallel glass plate of thickness 20 mm isilluminated with light from an extended source The refractiveindexoftheglassis15andthewavelengthofthelight is 600 nm Calculate how many fringes are formed

Solution

We assume that the fringes are formed by light incident at all angles from normal to grazing incidence

At normal incidence we have 2nd = mλ

From which

m 2 15 2 10 3ndashtimestimestimes6 10 7ndashtimes

------------------------------------------- 10 000= =

At grazing incidence the angle of refraction φ is in fact the critical angle

Therefore φφ arcsin 115------- 42 deg= =

ie cosφ = 075

At grazing incidence 2ndcosφ = m λ

From which (and using cos(sinndash1(115) = 075)

m 2 15 2 10 3ndash 075timestimestimestimes6 10 7ndashtimes

------------------------------------------------------------ 7500= =

The total number of fringes seen is equal to m ndash m = 2500

Exercise 95When viewed from above the colour of an oil film onwater appears red in colour Use the data below to estimate theminimumthicknessoftheoilfilm

average wavelength of red light = 630 nm

refractiveindexofoilforredlight=15

refractive index of water for red light = 13

PHYSICS 2015indb 298 150514 337 PM

Wave Phenomena

299

AH

L

94 Resolution

Essential idea Resolution places an absolute limit on the extent to which an optical or other system can separate images of objects

Understandings

bull Thesizeofadiffractingaperturebull Theresolutionofsimplemonochromatic

two-source systems

The size of a diffracting apertureOur discussion concerning diffraction so far has been for rectangular slits When light from a point source enters a smallcircularapertureitdoesnotproduceabrightdotasanimagebutacirculardiscknownasAiryrsquosdiscsurroundedbyfainterconcentriccircularringsasshowninFigure922

Laser

Figure 922 Diffraction at a circular aperture

Diffractionatanapertureisofgreatimportancebecausetheeye and many optical instruments have circular apertures

What is the half-angular width of the central maximum of the diffraction formed by a circular aperture This is not easy to calculate since it involves some advanced mathematics The problemwasfirstsolvedbytheEnglishAstronomerRoyalGeorgeAiryin1835whoshowedthatforcircularapertures

θ 122 λb--------------= where b is the diameter of the aperture

Example 2

Inthefollowingdiagramparallellightfromadistantpointsource(suchasastar)isbroughttofocusonthescreenSbyaconverginglens(the lens is shown as a vertical arrow)

Thefocal length(distance fromlens toscreen) is25cmand the diameter of the lens is 30 cm The wavelength of thelightfromthestaris560nmCalculatethediameterofthe diameter of the image on the screen

Solution

The lens actually acts as a circular aperture of diameter 30 cm The half angular width of central maximum of the diffraction pattern that it forms on the screen is

75

2

122 122 56 10 23 1030 10b

λθminus

minusminus

times times= = = times

times rad

The diameter of the central maxima is therefore

25 times 10-2 times 23 times 10-5

= 57 times 10-6 m

Although this is small it is still finite and is effectively the image of the star as the intensity of the secondary maxima are small compared to that of the central maximum

The resolution of simple monochromatic two-source systemsThe astronomers tell us that many of the stars that we observe with the naked eye are in fact binary stars That iswhatwe see as a single star actually consistsof twostarsinorbitaboutacommoncentreFurthermoretheastronomerstellusthatifweusealdquogoodrdquotelescopethenwewillactuallyseethetwostarsThatiswewillresolve the single point source into its two component parts Sowhatisitthatdetermineswhetherornotweseethetwo stars as a single point source ie what determines whether or not two sources can be resolved It canrsquot just bethatthetelescopemagnifiesthestarssinceiftheyareacting as point sources magnifying them is not going to make a great deal of difference

Ineachofoureyesthereisanaperturethepupilthroughwhich the light enters This light is then focused by the eye lens onto the retina But we have seen that when light passes through an aperture it is diffracted and so when we look at

NATURE OF SCIENCEImproved technology The Rayleigh criterion is the limit of resolution Continuing advancement in technology such as large diameter dishes or lenses or the use of smaller wavelength lasers pushes the limits of what we can resolve (18)

copy IBO 2014

PHYSICS 2015indb 299 150514 337 PM

Chapter 9

300

AH

L

apointsourceadiffractionpatternwillbeformedontheretina If we look at two point sources then two diffraction patterns will be formed on the retina and these patterns will overlap The width of our pupil and the wavelength of the light emitted by the sources will determine the amount by which they overlap But the degree of overlap will also depend on the angular separation of the two point sources WecanseethisfromFigures923

pupilretina

S1

S2

P2

P1

θ

Figure 923

LightfromthesourceS1 enters the eye and is diffracted by the pupil such that the central maximum of the diffraction patternisformedontheretinaatP1SimilarlylightfromS2producesamaximumatP2 If the two central maxima are well separated then there is a fair chance that we will see the two sources as separate sources If they overlap then we will not be able to distinguish one source from another From the diagram we see as the sources are movedclosertotheeyethentheangleθ increases and so does the separation of the central maxima

Figures924925926and927showsthedifferentdiffractionpatternsandtheintensitydistributionthatmightresultonthe retina as a result of light from two point sources

S1 S21 Well resolved

Figure 924 Very well resolved

S1 S22 Well resolved

Figure 925 Well resolved

S1 S23 Just resolved Rayleigh criterion Minimum of S2 coincides with maximum peak of S1

Figure 926 Just resolved

4 Not resolved

Figure 927 Not resolved

We have suggested that if the central maxima of the two diffraction patterns are reasonably separated then we should be able to resolve two point sources In the late 19th century Lord Rayleigh suggested by how much they should be separated in order for the two sources to be just resolved Ifthecentralmaximumofonediffractionpattern coincides with the firstminima of the otherdiffraction pattern then the two sources will just beresolved This is known as the Rayleigh Criterion

Figure926showsjustthissituationThetwosourcesarejust resolved according to the Rayleigh criterion since the peak of the central maximum of one diffraction pattern coincideswiththefirstminimumoftheotherdiffractionpattern This means that the angular separation of the peaks of the two central maxima formed by each source is just the half angular width of one central maximum ie

θ λb---

=

where b is the width of the slit through which the light fromthesourcespassesHoweverweseefromFigure1117that θ is the angle that the two sources subtend at the slit Henceweconcludethattwosourceswillberesolvedbyaslit if the angle that they subtend at the slit is greater than or equaltoλfraslb

PHYSICS 2015indb 300 150514 337 PM

Wave Phenomena

301

AH

L

Sofarwehavebeenassumingthattheeyeisarectangularslit whereas clearly it is a circular aperture and so we must use the formula

θ 122 λb--------------=

As mentioned above the angle θ is sometimes called the resolving power but should more accurately be called the minimum angle of resolution (θmin)

Clearly the smaller θ the greater will be the resolving power

The diffraction grating is a useful for differentiating closely spacedlinesinemissionspectraLikeaprismspectrometerit can disperse a spectrum into its components but it is better suited because it has higher resolution than the prism spectrometer

If λ1 and λ2 are two nearly equal wavelengths that canbarelybedistinguishedtheresolvanceorresolvingpowerofthegratingisdefinedas

R = λ _______ (λ2 ndash λ 1) = λ ___ Δλ

Where λ asymp λ1 asymp λ2

Therefore thediffractiongratingwithahighresolvancewill be better suited in determining small differences in wavelength

If the diffraction grating has Nlinesbeingilluminateditcan be shown that the resolving power of the mth order diffraction is given by

R = mN

Obviouslytheresolvancebecomesgreaterwiththeordernumber m and with a greater number of illuminated slits

Example

A benchmark for the resolving power of a grating is the sodium doublet in the sodium emission spectrum Two yellow lines in itsspectrumhavewavelengthsof58900nmand58959nm

(a) Determinetheresolvanceofagratingifthegivenwavelengths are to be distinguished from each other

(b) Howmanylinesinthegratingmustbeilluminated in order to resolve these lines in the second order spectrum

Solution

(a) R = λ ___ Δλ = 58900nm __________________ (58900ndash58959)nm

= 589nm _______ 059nm = 10 times 103

(b) N = R __ m = 10 times 10 3 ________ 2 =50times102 lines

It has been seen that diffraction effectively limits the resolving power of optical systems This includes such systems as the eye telescopes and microscopes Theresolving power of these systems is dealt with in the next section This section looks at links between technology and resolving power when looking at the very distant and when looking at the very small

Radio telescopes

The average diameter of the pupil of the human eye is about 25 mm This means that the eye will justresolve two point sources emitting light of wavelength 500 nm if their angular separation at the eye is 7

43

50 10122 24 1025 10

θminus

minusminus

times= times = times

times rad

If the eye were to be able to detect radio waves of wavelength 015m then tohave the same resolvingpower thepupilwould have to have a diameter of about 600 m Clearly this is nonsense but it does illustrate a problem facingastronomers who wish to view very distant objects such as quasarsandgalaxies thatemitradiowavesConventionalradio telescopes consist of a large dish typically 25 min diameter Even with such a large diameter the radiowavelength resolving power of the telescope is much less thantheopticalresolvingpowerofthehumaneyeLetuslook at an example

Example

The Galaxy Cygnus A can be resolved optically as an elliptically shaped galaxy However it is also a strongemitterof radiowavesofwavelength015mTheGalaxyisestimatedtobe50times1024mfromEarthUseofaradiotelescope shows that the radio emission is from two sources separated by a distance of 30 times 1021m Estimate thediameterofthedishrequiredtojustresolvethesources

Solution

The angle θ that the sources resolve at the telescope is given by

θ 21

424

30 10 60 1050 10

θ minustimes= = times

times rad

and 4

122 122 015 300060 10

d λθ minus

times= = =

timesm = 30 km

A radio telescope dish of this size would be impossible tomake letalonesupportThisshowsthatasingledishtype radio telescope cannot be used to resolve the sources and yet they were resolved To get round the problemastronomers use two radio telescopes separated by a large

PHYSICS 2015indb 301 150514 337 PM

Chapter 9

302

AH

L

distance The telescopes view the same objects at the same time and the signals that each receive from the objects are simultaneously superimposed The result of the superposition of the two signals is a two-slit interference pattern The pattern has much narrower fringe spacing than that of the diffraction pattern produced by either telescope on its own hence producing a much higherresolvingpowerWhentelescopesareusedlikethistheyare called a stellar interferometer

InSocorroinNewMexicothereisastellarinterferometerthat consists of 27 parabolic dishes each of diameter25 m arranged in a Y-shape that covers an area of 570km2Thisisaso-calledVeryLargeArray(VLA)Evenhigher resolution can be obtained by using an array of radio telescopes in observatories thousands of kilometres apart A system that uses this so-called technique oflsquovery-long-baseline interferometryrsquo (VLBI) is known asalsquovery-long-baselinearrayrsquo(VLBA)WithVLBAaradiowavelength resolving power can be achieved that is 100 timesgreaterthanthebestopticaltelescopesEvenhigherresolving power can be achieved by using a telescope that isinasatelliteorbitingEarthJapanrsquosInstituteofSpaceandAstronautical Science (ISAS) launched sucha system inFebruary 1997TheNationalAstronomicalObservatoryof Japan the National Science Foundationrsquos NationalRadio Astronomy Observatory (NRAO) the CanadianSpaceAgency theAustraliaTelescopeNationalFacilitythe European VLBI Network and the Joint Instituteback this project forVeryLongBaseline Interferometryin EuropeThis project is a very good example of howInternationalismcanoperateinPhysics

Electron microscope

Telescopes are used to look at very distant objects that are very large but because of their distance fromus appearverysmallMicroscopesontheotherhandareusedtolookat objects that are close to us but are physically very small Aswehave seen justmagnifyingobjects that ismakingthemappearlargerisnotsufficientonitsowntogaindetailabouttheobjectfordetailhighresolutionisneeded

Figure928isaschematicofhowanoptical microscope is usedtoviewanobjectandFigure929isaschematicofatransmission electron microscope(TEM)

Optical lens system

bright light source

glass slide containing specimen (object)

eye

Figure 928 The principle of a light microscope

magnetic lens system

electron source

thin wafer of material

ScreenCCD

Figure 929 The principle of an electron microscope

In the optical microscope the resolving power isdetermined by the particular lens system employed and the wavelength λ of the light used For example twopoints in the sample separated by a distance d will just be resolved if

2d mλ=

where m is a property of the lens system know as the numerical aperture In practice the largest value of m obtainable is about 16 Hence if the microscope slideis illuminatedwith light ofwavelength 480 nm a goodmicroscope will resolve two points separated by a distance dasymp15times10-7masymp015micromPointscloser together thanthiswillnotberesolvedHoweverthisisgoodenoughtodistinguishsomevirusessuchastheEbolavirus

Clearlythesmallerλ the higher the resolving power and this is where the electron microscope comes to the fore The electron microscope makes use of the wave nature of electrons(see1315)IntheTEMelectronspassthrougha wafer thin sample and are then focused by a magnetic field onto a fluorescent screen orCCD (charge coupleddevicesee142)Electronsusedinanelectronmicroscopehavewavelengthstypicallyofabout5times10-12mHoweverthe numerical aperture of electron microscopes is considerablysmaller thanthatofanopticalmicroscopetypicallyabout002NonethelessthismeansthataTEMcanresolvetwopointsthatareabout025nmapartThisresolving power is certainly high enough to make out the shape of large molecules

Another type of electron microscope uses a techniqueby which electrons are scattered from the surface of the sample The scattered electrons and then focused as in theTEM to form an image of the surfaceThese so-calledscanningelectronmicroscopes(SEM)havealowerresolving power than TEMrsquos but give very good threedimensional images

The eye

We saw in the last section that the resolving power of the human eye is about 2 times 10-4 rad Suppose that youare looking at car headlights on a dark night and the car

PHYSICS 2015indb 302 150514 337 PM

Wave Phenomena

303

AH

L

is a distance D away If the separation of the headlight is say 15 m then the headlights will subtend an angle θ 15

Dθ =

at your eye Assuming an average wavelength of

500 nm your eye will resolve the headlights into twoseparatesourcesifthisangleequals2times10-4 rad This gives D=75kmInotherwordsifthecarisapproachingyouona straight road then you will be able to distinguish the two headlightsasseparatesourceswhenthecaris75kmawayfrom you Actually because of the structure of the retina and optical defects the resolving power of the average eye is about 6 times 10-4 rad This means that the car is more likely tobe25kmawaybeforeyouresolvetheheadlights

Astronomical telescope

Letusreturntotheexampleofthebinarystarsdiscussedatthe beginning of this section on resolution The stars Kruger A and B form a binary system The average separation of the stars is 14 times 1012mandtheiraveragedistancefromEarthis12times1017mWhenviewedthroughatelescopeonEarththe system will therefore subtend an angle

12

17

14 1012 10

θtimes

=times

= 12times10-5 rad at the objective lens of the telescope Assuming that the average wavelength of the light emitted bythestarsis500nmthenifthetelescopeistoresolvethesystem into two separate images it must have a minimum diameter Dwhere12times10-5 =

7122 500 10D

minustimes times

This gives D=0050m which is about 5 cm So thisparticular system is easily resolved with a small astronomical telescope

Exercise 961 ItissuggestedthatusingtheISASVLBAitwould

bepossibletoldquoseerdquoagrainofriceatadistanceof5000kmEstimatetheresolvingpoweroftheVLBA

2 Thedistancefromtheeyelenstotheretinais20mm The light receptors in the central part of the retinaareabout5times10-6apartDeterminewhetherthe spacing of the receptors will allow for the eye to resolve the headlights in the above discussion whentheyare25kmfromtheeye

3 ThediameterofPlutois23times106 m and its averagedistancefromEarthis60times1012 m EstimatetheminimumdiameteroftheobjectiveofatelescopethatwillenablePlutotobeseenasadisc as opposed to a point source

95 Doppler effect

Essential ideaTheDopplereffectdescribesthephenomenonofwavelengthfrequencyshiftwhenrelativemotionoccurs

Understandingsbull TheDopplereffectforsoundwavesandlightwaves

The Doppler effect for sound waves and light waves

Consider two observers A and B at rest with respect to asoundsourcethatemitsasoundofconstantfrequencyf Clearly both observers will hear a sound of the same frequencyHoweversupposethatthesourcenowmovesat constant speed towards A A will now hear a sound of frequencyfA that is greater than f and B will hear a sound of frequency fB that is less than f This phenomenon is known as the Doppler Effect orDopplerPrinciple after C J Doppler(1803-1853)

The same effect arises for an observer who is either moving towards or away from a stationary source

Figure 930 shows the waves spreading out from a stationary source that emits a sound of constant frequency f The observersAandBhearasoundofthesamefrequency

A B

wavefront

S

Figure 930 Sound waves from a stationary source

Suppose now that the source moves towards A withconstant speed vFigure9310(a)showsasnapshotofthenew wave pattern

NATURE OF SCIENCETechnology Although originally based on physical observations of the pitch of fast moving sources of sound the Doppler effect has an important role in many different areas such as evidence for the expansion of the universe and generating images used in weather reports and in medicine (55)

copy IBO 2014

PHYSICS 2015indb 303 150514 337 PM

Chapter 9

304

AH

L

sourceA B

smallerwavelength

largerwavelength

Figure 931 (a) Sound waves from a moving source

The wavefronts are now crowded together in the direction of travel of the source and stretched out in the opposite direction This is why now the two observers will now hear notesofdifferentfrequenciesHowmuchthewavesbunchtogether and how much they stretch out will depend on the speed vEssentially A

A

cfλ

= and BB

cfλ

= where λA lt λB and v is the speed of sound

IfthesourceisstationaryandAismovingtowardsitthenthe waves from the source incident on A will be bunched up If A is moving away from the stationary source then the waves from the source incident on A will be stretched out

Christian Doppler(1803ndash1853)actuallyappliedtheprinciple(incorrectlyasithappens)totryandexplainthecolourofstars However the Doppler effect does apply to light aswellastosoundIfalightsourceemitsalightoffrequencyf then if it is moving away from an observer the observer willmeasurethelightemittedashavingalowerfrequencythan fSincethesensationofcolourvisionisrelatedtothefrequencyoflight(bluelightisofahigherfrequencythanred light) light emitted by objects moving way from anobserver is often referred to as being red-shifted whereas if the object is moving toward the observer it is referred to as blue-shiftedThisideaisusedinOptionE(Chapter16)

We do not need to consider here the situations where either the source or the observer is accelerating In a situation for example where an accelerating source is approaching astationaryobserverthentheobserverwillhearasoundof ever increasing frequency This sometimes leads toconfusion in describing what is heard when a source approaches passes and then recedes from a stationaryobserver Suppose for example that you are standingon a station platform and a train sounding its whistle is approaching at constant speed What will you hear as the train approaches and then passes through the station As the train approaches you will hear a sound of constant pitch but increasing loudness The pitch of the sound will be greater than if the train were stationary As the train passes through the station you will hear the pitch change at the momentthetrainpassesyoutoasoundagainofconstantpitch The pitch of this sound will be lower than the sound of the approaching train and its intensity will decrease as the train recedes from you What you do not hear is a sound of increasing pitch and then decreasing pitch

The Doppler equations for sound

Although you will not be expected in an IB examination to derivetheequationsassociatedwithaspectsoftheDopplereffectyouwillbeexpectedtoapplythemForcompletenesstherefore thederivationoftheequationsassociatedwiththeDopplereffectasoutlinedaboveisgivenhere

S primeS O

vs

Figure 931 (b)

InFigure931(b)theobserverOisatrestwithrespecttoasourceofsoundSismovingwithconstanspeedvs directly towards O The source is emitting a note of constant frequencyf and the speed of the emitted sound is v

S shows the position of the source ∆t later When the sourceisatresttheninatime∆t the observer will receive f∆t waves and these waves will occupy a distance v∆t ie

v f

λ =f∆tv∆t =

(Because of the motion of the source this number of waves will now occupy a distance (v∆t ndash vs∆t)The lsquonewrsquowavelength is therefore

v ndash vs

fλrsquo =

f∆tv∆t ndash vs∆t

=

If fisthefrequencyheardbyOthen

vfλ

= or

vf

λ = = sv vfminus

From which

s

vf fv v

=minus

Dividingthroughbyv gives

Equation1111

1 sf f v

v=

minus

If the source is moving away from the observer then we have

s

vf fv v

=minus

1

1 sf vv

=+

Equation112

PHYSICS 2015indb 304 150514 337 PM

Wave Phenomena

305

AH

L

We now consider the case where the source is stationary and the observerismoving towardsthesource with speed v0 In this situation the speed of the sound waves as measured by the observer will be v0 + v We therefore have that

v0 + v = ffraslλ = vfλ

times

From which

o1 vf fv

= +

Equation113

If the observerismovingawayfromthesource then

o1 vf fv

= minus

Fromequation(113)wehavethat

o o1 v vf f f f f fv v

∆ = minus = + minus =

Equation114

Thevelocitiesthatwerefertointheaboveequationsarethevelocities with respect to the medium in which the waves fromthesourcetravelHoweverwhenwearedealingwithalight source it is the relative velocity between the source and the observer that we must consider The reason for this is that lightisuniqueintherespectthatthespeedofthelightwavesdoes not depend on the speed of the source All observers irrespective of their speed or the speed of the source will measure the same velocity for the speed of light This is oneof thecornerstonesof theSpecialTheoryofRelativitywhichisdiscussedinmoredetailinOptionH(Chapter18)When applying the Doppler effect to light we aremainlyconcerned with the motion of the source We look here only at the situation where the speed of the source v is much smaller than the speed of light cinfreespace(v ltlt c)Underthesecircumstanceswhenthesourceismovingtowardstheobserverequation111becomes

vf f f fc

minus = ∆ = Equation115

andwhenthesourceismovingawayfromtheobserverequation112becomes vf f f f

cminus = ∆ = minus

Provided that v ltlt c these same equations apply for astationary source and moving observer

We look at the following example and exercise

Example

Asourceemitsasoundoffrequency440HzItmovesinastraight line towards a stationary observer with a speed of 30ms-1Theobserverhearsasoundoffrequency484HzCalculate the speed of sound in air

Solution

We use equation 111 and substitute f= 484 Hz f = 440 Hz

and vs = 30 m s-1

therefore 1

484 440 301v

=minus

such that 30 4401

484vminus = to

give v = 330 m s-1

Example

A particular radio signal from a galaxy is measured as havingafrequencyof139times109HzThesamesignalfromasourceinalaboratoryhasafrequencyof142times109Hz

Suggestwhy thegalaxy ismovingaway fromEarthandcalculateitsrecessionspeed(iethespeedwithwhichitismovingawayfromEarth)

Solution

The fact that the frequency from the moving source is less than that when it is stationary indicates that it is moving away from the stationary observer ie Earth

Using vf fc

∆ = we have

8 96

9

3 10 (142 139) 10 634 10142 10

c fvf∆ times times minus times

= = = timestimes

m s-1

It is usual when dealing with the Doppler effect of light to express speeds as a fraction of c So in this instance we have v = 0021 c

Using the Doppler effectWe have seen in the above example and exercise how the Dopplereffectmaybeusedtomeasuretherecessionspeedof distant galaxies The effect is also used to measure speed inothersituationsHerewewilllookatthegeneralprincipleinvolved in using the Doppler effect to measure speedFigure932showsasource(thetransmitter)thatemitseithersound or em waves of constant frequency f The waves

PHYSICS 2015indb 305 150514 337 PM

Chapter 9

306

AH

L

fromthesourceare incidentonareflectorthat ismovingtowards the transmitter with speed vThereflectedwavesare detected by the receiver placed alongside the transmitter

reector

v

transmitter

receiver

f f

f f

Figure 932 Using the Doppler effect to measure speed

We shall consider the situation where v ltlt c where c is the speed of the waves from the transmitter

For the reflector receiving waves from the transmitterit is effectively an observer moving towards a stationary sourceFromequation(114) itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation116

For the receiver receivingwaves from the reflector it iseffectively a stationary observer receiving waves from a movingsourceFromequation(115)itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation117

Ifweaddequations(116)and(117)wegetthatthetotalDopplershiftatthereceiver∆f is

v vf f f f fc c

minus = ∆ = +

But 1 vf fc

= +

hence

1 v v vf f fc c c

∆ = + +

But since v ltlt cwe can ignore the term2

2

vc when we

expandthebracketintheaboveequation

Therefore we have2vf fc

∆ = Equation118

If v asymp c then we must use the full Doppler equationsHowever for em radiationwewill alwaysonly considersituations in which v ltlt c

Example

Thespeedofsoundinbloodis1500times103 m s-1 Ultrasound offrequency100MHzisreflectedfrombloodflowinginan arteryThe frequencyof the reflectedwaves receivedbackatthetransmitteris105MHzEstimatethespeedofthebloodflowintheartery

Solution

Using equation (118) we have

6 63

2005 10 1015 10

vtimes = times

timesto give v asymp 36 m s-1 (We have assumed that the ultrasound is incident at right angles to the blood flow)

Exercise 971 Judy is standing on the platform of a station A

high speed train is approaching the station in a straight line at constant speed and is sounding its whistleAsthetrainpassesbyJudythefrequencyof the sound emitted by the whistle as heard by Judychangesfrom640Hzto430HzDetermine

(a) thespeedofthetrain

(b) thefrequencyofthesoundemittedbythewhistle as heard by a person on the train (Speedofsound=330ms-1)

2 AgalaxyismovingawayfromEarthwithaspeedof00500cThewavelengthofaparticularspectral line in light emitted by atomic hydrogen inalaboratoryis656times10-7 m Calculate the valueofthewavelengthofthislinemeasuredinthelaboratoryinlightemittedfromasourceofatomic hydrogen in the galaxy

PHYSICS 2015indb 306 150514 337 PM

10 Fields

Contents

101 ndash Describing fields

102 ndash Fields at work

Essential Ideas

Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Similar approaches can be taken in analysing electrical and gravitational potential problems copy IBO 2014

PHYSICS 2015indb 307 150514 337 PM

Chapter 10

308

AH

L

101 Describing fields

Essential idea Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Understandingsbull Gravitationalfieldsbull Electrostaticfieldsbull Electricpotentialandgravitationalpotentialbull Fieldlinesbull Equipotentialsurfaces

Gravitational potentialWehaveseenthatifweliftanobjectofmassm to a height h abovethesurfaceoftheEarththenitsgainingravitationalpotential energy is mghHoweverthisisbynomeansthefullstoryForastartwenowknowthatgvarieswithh and also the expression really gives a difference in potentialenergybetweenthevaluethattheobjecthasattheEarthrsquossurfaceandthevaluethatithasatheighthSowhatwereallyneedisazeropointCanwefindapointwherethepotentialenergyiszeroandusethispointfromwhichtomeasure changes in potential energy

The point that is chosen is in fact infinity At infinity the gravitationalfieldstrengthofanyobjectwillbezeroSoletusseeifwecandeduceanexpressionforthegaininpotentialenergyofanobjectwhenitisldquoliftedrdquofromthesurfaceoftheEarthtoinfinityThisineffectmeansfindingtheworknecessarytoperformthistask

δr

r r δr+

r infin=

mg

AB

Me

Figure 1001 Gravitational forces

Inthediagramweconsidertheworknecessarytomovethe particle of mass m a distance δr in thegravitationalfield of the Earth

The force on the particle at A is FG Mem

r2----------------=

IftheparticleismovedtoB then since δr isverysmallwe can assume that the field remains constant over thedistance ABTheworkδW done againstthegravitationalfieldoftheEarthinmovingthedistanceABis

δWGMem

r2----------------δrndash=

(rememberthatworkdoneagainstaforceisnegative)

TofindthetotalworkdoneW in going from the surface oftheEarthtoinfinitywehavetoaddalltheselittlebitsofworkThisisdonemathematicallybyusingintegralcalculus

WG Me m

r2----------------ndash

rdR

infin

int G Me m 1

r2----- rd

R

infin

intndash G Me m 1r---ndash

R

infinndash= = =

GMe m 0 1R---ndash

ndashndash=

G MemR----------------ndash=

HencewehavewhereRistheradiusoftheEarththattheworkdonebythegravitationalfieldinmovinganobjectofmass m from R(surfaceoftheEarth)toinfinityisgivenby

WGMe m

R---------------- ndash=

We can generalise the result by calculating the worknecessary per unit mass to take a small mass fromthe surface of the Earth to infinity This we call thegravitational potential Vie

Wewouldgetexactlythesameresultifwecalculatedtheworkdonebythefieldtobringthepointmassfrominfinityto the surface of Earth In this respect the formal definition ofgravitationalpotentialatapointinagravitationalfieldis therefore defined as the work done per unit mass in bringing a point mass from infinity to that point

Clearly then the gravitational potential at any point inthe Earthrsquos field distance r from the centre of the Earth (providingr gt R)is

VGMe

r------------ ndash=

Thepotentialisthereforeameasureoftheamountofworkthathastobedonetomoveparticlesbetweenpointsinagravitationalfieldanditsunit is theJkgndash1 We also note thatthepotentialisnegativesothatthepotentialenergy

NATURE OF SCIENCEParadigm shift The move from direct observable actions being responsible for influence on an object to acceptance of a fieldrsquos ldquoaction at a distancerdquo required a paradigm shift in the world of science (23)

copy IBO 2014

PHYSICS 2015indb 308 150514 337 PM

Fields

309

AH

L

aswemoveawayfromtheEarthrsquossurfaceincreasesuntilitreachesthevalueofzeroatinfinity

IfthegravitationalfieldisduetoapointmassofmassmthenwehavethesameexpressionasaboveexceptthatMe is replaced by m andmustalsoexclude thevalueof thepotential at the point mass itself ie at r = 0

WecanexpressthegravitationalpotentialduetotheEarth(orduetoanysphericalmass)intermsofthegravitationalfield strength at its surface

AtthesurfaceoftheEarthwehave

g 0ndash Re GMRe

---------ndash= e

Sothat

g 0Re2 GM= e

Hence at a distance r from the centre of the Earth the gravitationalpotentialVcanbewrittenas

VGMe

r------------ndashg 0 Re

2

r------------ndash= =

The potential at the surface of the Earth

(r = Re)istherefore-g0Re

It is interesting to see how the expression for thegravitational potential ties in with the expression mgh ThepotentialatthesurfaceoftheEarthis-g0Re (see the exampleabove)andataheighthwillbe-g0g 0 Re h+( )ndash ifweassume that g0doesnotchangeover thedistanceh The differenceinpotentialbetweenthesurfaceandtheheighth is therefore g0hSotheworkneededtoraiseanobjectof mass m to a height h is mgh ie m timesdifference ingravitationalpotential

This we have referred to as the gain in gravitationalpotentialenergy(see235)

However this expression can be extended to any twopoints inanygravitationalfieldsuch that ifanobjectofmass mmovesbetween twopointswhosepotentialsareV1 and V2 respectively then the change in gravitationalpotentialenergyoftheobjectism(V1 ndash V2)

Gravitational potential gradient

Let us consider now a region in space where thegravitationalfieldisconstantInFigure912thetwopointsAandBareseparatedbythedistance∆x

A B

direction of uniform gravitational eld of strength I

∆x

Figure 1002 The gravitational potential gradient

The gravitational field is of strength I and is in the direction shownThe gravitational potential at A is V andatBisV +∆V

Theworkdone is takingapointmassmfromAtoB isF∆x = mI∆x

Howeverbydefinitionthisworkisalsoequalto-m∆V

Therefore mI∆x=-m∆V

or VIx

∆= minus

∆Effectivelythissaysthatthemagnitudeofthegravitationalfield strength is equal to the negative gradient of thepotential If I is constant then V is a linear function of x and Iisequaltothenegativegradientofthestraightlinegraph formed by plotting V against x If I is not constant (asusuallythecase)thenthemagnitudeofI at any point in the field can be found by find the gradient of the V-x graph at that point An example of such a calculation can befoundinSection929

For those of you who do HL maths the relationship between field and potential is seen to follow from the expression for the potential of a point mass viz

mV Gr

= minus

2

ddV mG Ir r

minus = + =

Potential due to one or more point masses

Gravitationalpotentialisascalarquantitysocalculatingthepotential due to more than one point mass is a matter of simpleadditionSoforexamplethepotentialV due to the Moon and Earth and a distance x fromthecentreofEarthonastraightlinebetweenthemisgivenbytheexpression

E MM MV Gx r x

= minus + minus

whereME=massofEarthMM= mass of Moon and r = distancebetweencentreofEarthandMoon

PHYSICS 2015indb 309 150514 337 PM

Chapter 10

310

AH

L

Equipotentials and field linesIfthegravitationalpotentialhasthesamevalueatallpointson a surface the surface is said to be an equipotential surfaceSoforexampleifweimagineasphericalshellaboutEarthwhosecentrecoincideswiththecentreofEarththisshellwillbeanequipotentialsurfaceClearlyifwerepresentthegravitationalfieldstrengthbyfieldlinessincethelinesldquoradiaterdquooutfromthecentreofEarththentheselineswillbeatrightanglestothesurfaceIfthefieldlineswerenotnormaltotheequipotentialsurfacethentherewouldbeacomponentof the field parallel to the surfaceThis wouldmean thatpointsonthesurfacewouldbeatdifferentpotentialsandsoitwouldnolongerbeanequipotentialsurfaceThisofcourseholdstrueforanyequipotentialsurface

Figure 1003 shows the field lines and equipotentials fortwopointmassesm

m m

Figure 1003 Equipotentials for two point masses

It is worth noting that we would get exactly the samepattern ifwewere to replace thepointmasseswith twoequalpointcharges(See935)

Escape speedThe potential at the surface of Earth is which meansthattheenergyrequiredtotakeaparticleofmass mfromthesurfacetoinfinityisequalto

ButwhatdoesitactuallymeantotakesomethingtoinfinityWhentheparticleisonthesurfaceoftheEarthwecanthinkofitassittingatthebottomofaldquopotentialwellrdquoasinfigure1004

particle

surface of Earth

innity

GMR---------

FIgure 1004 A potential well

Theldquodepthrdquoofthewellis and if the particle gains an amount of kinetic energy equal to wherem is its massthenitwillhavejustenoughenergytoldquoliftrdquoitoutofthewell

In reality it doesnrsquot actually go to infinity it justmeansthat the particle is effectively free of the gravitationalattractionof theEarthWe say that ithas ldquoescapedrdquo theEarthrsquosgravitationalpullWemeetthisideainconnectionwithmolecular forcesTwomolecules in a solidwill sitat their equilibrium position the separation where therepulsiveforceisequaltotheattractiveforceIfwesupplyjust enough energy to increase the separation of themolecules such that they are an infinite distance apart then intermolecularforcesnolongeraffectthemoleculesandthesolidwillhavebecomealiquidThereisnoincreaseinthekineticenergyofthemoleculesandsothesolidmeltsat constant temperature

Wecancalculatetheescapespeedofanobjectveryeasilyby equating the kinetic energy to the potential energysuch that

12---mvesc ape

2 GMe mRe

----------------=

vesc aperArr2GMe

Re--------------- 2g0 Re = =

Substituting for g0 and Regivesavalueforvescape of about 11kmsndash1 from the surface of the Earth

Youwillnotethattheescapespeeddoesnotdependonthemassoftheobjectsincebothkineticenergyandpotentialenergy are proportional to the mass

IntheoryifyouwanttogetarockettothemoonitcanbedonewithoutreachingtheescapespeedHoweverthiswouldnecessitate anenormousamountof fuel and it islikely that therocketplus fuelwouldbesoheavythat itwouldnevergetoffthegroundItismuchmorepracticalto accelerate the rocket to the escape speed fromEarthorbitandthenintheoryjustlaunchittotheMoon

Example

Use the followingdata todetermine thepotential at thesurface of Mars and the magnitude of the acceleration of free fall

massofMars =64times1023kg

radiusofMars =34times106 m

Determinealsothegravitationalfieldstrengthatadistanceof 68 times 106mabovethesurfaceofMars

PHYSICS 2015indb 310 150514 337 PM

Fields

311

AH

L

Solution

= minus = minus times times = times23

11 76

64 1067 10 ndash13 10

34 10MV GR

minus timestimes

Nkg-1

But V = -g0R

Therefore 7

0 6

13 10 3834 10

VgR

times= minus = =

timesm s-2

To determine the field strength gh at 68 times 106 m above the surface we use the fact that 0 2

Mg GR

= such that GM = g0R2

Therefore 2 2

0h 2 2 2

h h

38 (34) 042(102)

g RGMgR R

times= = = = m s-2

(the distance from the centre is 34 times 106 + 68 times 106 = 102 times 106 m)

Exercise 1011 Thegraphbelowshowshowthegravitational

potentialoutsideoftheEarthvarieswithdistancefrom the centre

0 10 20 30 40 50 60 70

ndash1

ndash2

ndash3

ndash4

ndash5

ndash6

ndash7

r m times 106

V

Jkgndash1

times 10

7

(a) Usethegraphtodeterminethegainingravitationalpotentialenergyofasatelliteofmass200kgasitmovesfromthesurfaceof the Earth to a height of 30 times 107mabovetheEarthrsquossurface

(b) Calculatetheenergyrequiredtotakeit to infinity

(c) DeterminetheslopeofthegraphatthesurfaceoftheEarthmCommenton youranswer

Electric potential energy

The concept of electric potential energy was developedwith gravitational potential energy in mind Just as anobjectnearthesurfaceoftheEarthhaspotentialenergybecause of its gravitational interaction with the Earthsotoothereiselectricalpotentialenergyassociatedwithinteracting charges

Letusfirstlookatacaseoftwopositivepointchargeseachof 1μC that are initially bound together by a thread in a vacuuminspacewithadistancebetweenthemof10cmasshowninFigure1005Whenthethreadiscutthepointchargesinitiallyatrestwouldmoveinoppositedirectionsmovingwithvelocitiesv1 and v2 along the direction of the electrostatic force of repulsion

BEFORE AFTER

v v21

Figure 1005 Interaction of two positive particles

The electric potential energybetweentwopointchargescan be found by simply adding up the energy associated witheachpairofpointchargesForapairofinteractingchargestheelectricpotentialenergyisgivenby

∆U = ∆Ep + ∆Ek = ∆W = ∆Fr = kqQ

____ r2 times r = kqQ

____ r

Becausenoexternalforceisactingonthesystemtheenergyand momentum must be conserved Initially Ek = 0 and Ep =kqQr=9times109 times 1 times 10-1201m=009JWhentheyareagreatdistance fromeachotherEpwillbenegligibleThefinalenergywillbeequaltofrac12mv1

2+frac12mv22=009J

Momentum isalsoconservedand thevelocitieswouldbethe same magnitude but in opposite directions

Electric potential energyismoreoftendefinedintermsofapointchargemovinginanelectricfieldas

lsquothe electricpotential energybetweenany twopoints inan electric field is defined as negativeoftheworkdone by anelectricfieldinmovingapoint electric chargebetweentwolocationsintheelectricfieldrsquo

ΔU = ΔEp = -ΔW = -Fd = qEx

ΔU = ΔEp +ΔEk = ΔW = ΔFr = kqQ r2 x r = kqQ

wherexisthedistancemovedalong(oroppositeto)thedirection of the electric field

Electric potential energyismeasuredinjoule(J)Justasworkisascalarquantitysotooelectricalpotentialenergyis a scalar quantity The negative of the work done by an electric field in moving a unit electric charge between two points is independent of the path taken Inphysicswesaytheelectricfieldisalsquoconservativersquo field

PHYSICS 2015indb 311 150514 337 PM

Chapter 10

312

AH

L

Supposeanexternalforcesuchasyourhandmovesasmallpositive point test charge in the direction of a uniformelectric fieldAs it ismoving itmust be gaining kineticenergyIfthisoccursthentheelectricpotentialenergyofthe unit charge is changing

InFigure1006apointcharge+qismovedbetweenpointsAandBthroughadistancex in a uniform electric field

B

A

x

+ q

Figure 1006 Movement of a positive point charge in a uniform field

InordertomoveapositivepointchargefrompointAtopointBanexternal forcemustbeapplied to thechargeequaltoqE (F = qE)

Since the force is applied through a distance xthennegativeworkhas to bedone tomove the charge because energyis gainedmeaning there is an increaseelectric potential energybetweenthetwopointsRememberthattheworkdoneisequivalenttotheenergygainedorlostinmovingthe charge through the electric field The concept of electric potential energy is only meaningful as the electric field whichgeneratestheforceinquestionisconservative

W F xtimes E q x times= =

θ x cos θ x

Figure 1007 Charge moved at an angle to the field

If a chargemoves at an angle θ to an electric field thecomponent of the displacement parallel to the electric fieldisusedasshowninFigure918

W F x Eq x θcostimes= =

Theelectricpotentialenergyisstoredintheelectricfieldand theelectricfieldwill return theenergy to thepointcharge when required so as not to violate the Law ofconservationofenergy

Electric potential

The electric potential at a point in an electric field is defined as being the work done per unit charge in bringing a small positive point charge from infinity to that point

ΔV = Vinfin ndash Vf = ndash W ___ q

Ifwedesignatethepotentialenergytobezeroatinfinitythen it follows thatelectricpotentialmustalsobezeroat infinity and the electric potential at any point in an electricfieldwillbe

ΔV = ndash W ___ q

NowsupposeweapplyanexternalforcetoasmallpositivetestchargeasitismovedtowardsanisolatedpositivechargeTheexternalforcemustdoworkonthepositivetestchargetomoveittowardstheisolatedpositivechargeandtheworkmustbepositivewhiletheworkdonebytheelectricfieldmustthereforebenegativeSotheelectricpotentialatthatpointmustbepositiveaccordingtotheaboveequationIfanegativeisolatedchargeisusedtheelectricpotentialatapointonthepositivetestchargewouldbenegativePositivepoint chargesof their ownaccordmove fromaplaceofhighelectricpotential toaplaceof lowelectricpotentialNegative point charges move the other way from lowpotentialtohighpotentialInmovingfrompointAtopointBinthediagramthepositivecharge+qismovingfromalowelectricpotentialtoahighelectricpotential

In thedefinitiongiven the termldquoworkperunitchargerdquohas significance If the test charge is +16 times 10-19Cwherethe charge has a potential energy of 32 times 10-17Jthenthepotentialwouldbe32times10-17J+16times10-19C=200JC-1 Now if the charge was doubled the potential wouldbecome 64times10-17 J However the potential per unitchargewouldbethesame

Electric potential is a scalar quantityandithasunitsJC-1 orvoltswhere1voltequalsone joulepercoloumbThevoltallowsustoadoptaunitfortheelectricfieldintermsofthevolt

PreviouslytheunitfortheelectricfieldwasNCndash1

W = qV and F = qEsoW ___ V = F __ E

E = FV ___ W = FV ___ Fm V mndash1

That is the units of the electric field E can also beexpressed as V mndash1

The work done per unit charge in moving a point charge between two points in an electric field is again independent of the path taken

PHYSICS 2015indb 312 150514 337 PM

Fields

313

AH

L

Electric potential due to a point charge

Letus takeapoint rmetres froma chargedobjectThepotentialatthispointcanbecalculatedusingthefollowing

W = ndashFr = ndashqV and F = ndash q1 q2 _____ 4πε0r

2

Therefore

W = ndash q1 q2 _____ 4πε0r

2 times r = ndash q1 q2 _____ 4πε0r

= ndashq1 times q2 _____ 4πε0r

= ndashq1V

That is V q

4πε0r--------------- =

Orsimply

V kqr------

=

Example

Determinehowmuchworkisdonebytheelectricfieldofpointcharge150μCwhenachargeof200μCismovedfrom infinity to apoint 0400m from thepoint charge(Assumenoaccelerationofthecharges)

Solution

The work done by the electric field is W = -qV = -14πε0 times q times (Q rinfin - Q r0400)

W = (- 200 times 10-6 C times 900 times 109 NmC-2 times 150 times 10-6 C) divide 0400 m = - 0675 J

An external force would have to do +0675 J of work

Electric field strength and electric potential gradient

Letus lookbackatFigure1006Supposeagain that thecharge+qismovedasmalldistancebyaforceFfromAtoBsothattheforcecanbeconsideredconstantTheworkdoneisgivenby

∆W F x∆times=

TheforceFandtheelectricfieldEareoppositelydirectedandweknowthat

F = -qE and ΔW = q ΔV

Thereforetheworkdonecanbegivenas

q ΔV = -q E Δx

Therefore E V∆x∆-------

ndash=

The rate of change of potential ΔV at a point with respect to distance Δx in the direction in which the change is maximum is called the potential gradient We say that the electric field = - the potential gradient and the units are Vm-

1 From the equation we can see that in a graph of electric potential versus distance the gradient of the straight line equals the electric field strength

In reality if a charged particle enters a uniform electric field it will be accelerated uniformly by the field and its kinetic energy will increase This is why we had to assume no acceleration in the last worked example

Ek∆ 12---

mv2 q E xsdot sdot q Vx---

xsdot sdot q Vsdot= = = =

Example

Determine how far apart two parallel plates must besituated so that a potential difference of 150 times 102 V produces an electric field strength of 100 times 103 NC-1

Solution

Using E V∆x------- xhArrndash V∆

E-------15 102 Vtimes

100 103 N C 1ndashtimes------------------------------------------= = =

= 150 times 10-1

The plates are 150 times 10-1 m apart

The electric field and the electric potential at a point due to an evenly distributed charge +q on a sphere can be represented graphically as in Figure 1008

r

r

V

E

r0

r0

r0

Charge of +Q evenly distributed over surface

E 14πε0------------ Q

r2----- r r0gt=

V 14πε0------------Q

r----r r0gt=

Figure 1008 Electric field and potential due to a charged sphere

PHYSICS 2015indb 313 150514 337 PM

Chapter 10

314

AH

L

When the sphere becomes charged we know that thechargedistributes itselfevenlyoverthesurfaceThereforeeverypartofthematerialoftheconductorisatthesamepotential As the electric potential at a point is defined as beingnumericallyequaltotheworkdoneinbringingaunitpositivechargefrominfinitytothatpointithasaconstantvalueineverypartofthematerialoftheconductor

Since the potential is the same at all points on the conducting surfacethen∆V∆xiszeroButE=ndash∆V∆xThereforethe electric field inside the conductor is zero There is no electric field inside the conductor

SomefurtherobservationsofthegraphsinFigure1009are

bull OutsidethespherethegraphsobeytherelationshipsgivenasE α 1 r2 and V α 1 r

bull Atthesurfacer = r0Thereforetheelectricfieldandpotentialhavetheminimumvalueforr at this point and this infers a maximum field and potential

bull Insidethespheretheelectricfieldiszerobull Insidethespherenoworkisdonetomoveacharge

fromapointinsidetothesurfaceThereforethereisno potential difference and the potential is the same asitiswhenr = r0

Similargraphscanbedrawnfortheelectricfieldintensityandthe electric potential as a function of distance from conducting parallelplatesandsurfacesandthesearegiveninFigure1009

Potential plot E eld plotE eld

++++

ndashndashndashndash

xx x

+

x x

ndashx x

+ ndashx

x

+ +

x

x

Figure 1009 Electric field and electric potential at a distance from a charged surface

Potential due to one or more point charges

The potential due to one point charge can be determined byusingtheequationformula

V = kq r

Example 1

Determine the electric potential at a point 20 times 10-1 m fromthecentreofan isolatedconducting spherewithapointchargeof40pCinair

Solution

Using the formula

V = kq r we have

V90 10 9times( ) 40 10 12ndashtimes( )times

20 10 1ndashtimes( )------------------------------------------------------------------ 018 V= =

the potential at the point is 180 times 10-1 V

The potential due to a number of point charges can be determined by adding up the potentials due to individual point charges because the electric potential at any point outside a conducting sphere will be the same as if all the charge was concentrated at its centre

Example 2

Threepointchargesofareplacedattheverticesofaright-angledtriangleasshowninthediagrambelowDeterminetheabsolutepotentialat the+20μCchargeduetothetwoothercharges

- 6μC

+3μC +2μC

4 m

3 m

Solution

The electric potential of the +2 μC charge due to the ndash 6 μC charge is

V = (9 times 109 Nm2C-2 times -6 times 10-6 C) divide (radic 32 + 42) m = - 108 times 104 V

PHYSICS 2015indb 314 150514 338 PM

Fields

315

AH

L

Insummarywecanconcludethat

bull No work is done to move a charge along an equipotential

bull Equipotentials are always perpendicular to the electric lines of force

Figure1011and1012showsomeequipotential lines fortwo oppositely charged and identically positive spheresseparated by a distance

+ve ndashve

equipotential lines

Figure 1011 Equipotential lines between two opposite charges

equipotential lines

Figure 1012 Equipotential lines between two charges which are the same

+

ndash ndash ndash ndash

+ + +

40 V

30 V

20 V

10 V

50 V

Figure 1013 Equipotential lines between charged parallel plates

Figure1013showstheequipotentiallinesbetweenchargedparallel plates Throughout this chapter the similarities and differences between gravitational fields and electric fieldshavebeendiscussedTherelationshipsthatexistsbetweengravitationalandelectricquantitiesandtheeffectsofpointmassesandchargesissummarisedinTable1014

The electric potential of the +2 μC charge due to the +3 μC charge is

V = (9 times 109 Nm2C-2 times 3 times 10-6 C) divide 3m = 9 times 103 V

The net absolute potential is the sum of the 2 potentials- 108 times 104 V + 9 times 103 V =- 18 times 103 V

The absolute potential at the point is - 18 times 103 V

Equipotential surfacesRegions in space where the electric potential of a chargedistributionhasaconstantvaluearecalledequipotentials The placeswherethepotentialisconstantinthreedimensionsarecalled equipotential surfacesandwheretheyareconstantintwodimensionstheyarecalledequipotential lines

TheyareinsomewaysanalogoustothecontourlinesontopographicmapsInthiscasethegravitationalpotentialenergy is constant as amassmoves around the contourlines because the mass remains at the same elevationabovetheEarthrsquossurfaceThegravitationalfieldstrengthacts in a direction perpendicular to a contour line

Similarlybecausetheelectricpotentialonanequipotentiallinehas thesamevalueanelectric forcecandonoworkwhena testchargemovesonanequipotentialThereforethe electric field cannot have a component along anequipotentialandthusitmustbeeverywhereperpendiculartotheequipotentialsurfaceorequipotentiallineThisfactmakesiteasytoplotequipotentialsifthelinesofforceorlinesofelectricfluxofanelectricfieldareknown

Forexamplethereareaseriesofequipotentiallinesbetweentwoparallelplateconductorsthatareperpendiculartotheelectric fieldTherewill be a series of concentric circles(eachcirclefurtherapartthanthepreviousone)thatmapouttheequipotentialsaroundanisolatedpositivesphereThe lines of force and some equipotential lines for anisolatedpositivesphereareshowninFigure1010

Lines ofequipotential

Figure 1010 Equipotentials around an isolated positive sphere

PHYSICS 2015indb 315 150514 338 PM

Chapter 10

316

AH

L

Gravitational quantity Electrical quantityQ

uant

ities

g V∆x∆-------ndash= E V∆

x∆-------ndash=

Poin

t mas

ses a

nd ch

arge

s

V Gmr----

ndash= V 14πε0------------q

r---=

g G m

r2-----ndash= E 14πε0------------ q

r2-----=

F Gm1 m2

r2--------------= F 1

4πε0------------

q1q2

r2-----------=

Figure 1014 Formulas

Example

Deduce the electric potential on the surface of a gold nucleus that has a radius of 62 fm

Solution

Using the formula

V = kq r and knowing the atomic number of gold is 79 We will assume the nucleus is spherical and it behaves as if it were a point charge at its centre (relative to outside points)

V = 90 times 109 Nm2C-2 times 79 times 16 times 10-19 C divide 62 times 10-15 m = 18 times 107 V

The potential at the point is 18 MV

Example

Deduce the ionisation energy in electron-volts of theelectron in the hydrogen atom if the electron is in its ground state and it is in a circular orbit at a distance of 53 times 10-11 m from the proton

Solution

This problem is an energy coulombic circular motion question based on Bohrrsquos model of the atom (not the accepted quantum mechanics model) The ionisation energy is the energy required to remove the electron from the ground state to infinity The electron travels in a circular orbit and therefore has a centripetal acceleration The ionisation energy will counteract the coulombic force and the movement of the electron will be in the opposite direction to the centripetal force

Total energy = Ek electron + Ep due to the proton-electron interaction

ΣF = kqQ r2 = mv2 r and as such mv2 = = kqQ r

Therefore Ek electron = frac12 kqQ r

Ep due to the proton-electron interaction = - kqQ r

Total energy = frac12 kqQ r + - kqQ r = -frac12 kqQ r

= - 90 times 109 Nm2C-2 times (16 times 10-19 C)2 divide 53 times 10-11 m = -217 times 10-18 J

= -217 times 10-18 J divide 16 times 10-19 = -136 eV

The ionisation energy is 136 eV

Exercise 1021 ApointchargePisplacedmidwaybetweentwo

identicalnegativechargesWhichoneofthefollowingiscorrectwithregardstoelectricfieldandelectricpotentialatpointP

Electric field Electric potentialA non-zero zeroB zero non-zeroC non-zero non-zeroD zero zero

2 Twopositivechargedspheresaretiedtogetherinavacuumsomewhereinspacewheretherearenoexternal forces A has a mass of 25 g and a charge of20μCandBhasamassof15gandachargeof30μCThedistancebetweenthemis40cmTheyarethenreleasedasshowninthediagram

BEFORE AFTER

v1 v2

A B

(a) Determinetheirinitialelectricpotentialenergy in the before situation

(b) DeterminethespeedofsphereBafterrelease

3 Thediagrambelowrepresentstwoequipotentiallines in separated by a distance of 5 cm in a uniform electric field

+ + + + + + + +

ndash ndash ndash ndash ndash ndash ndash ndash

40 V

20 V 5 cm

PHYSICS 2015indb 316 150514 338 PM

Fields

317

AH

L

Determine the strength of the electric field

4 Thisquestionisabouttheelectricfieldduetoacharged sphere and the motion of electrons in that fieldThediagrambelowshowsanisolatedmetalsphereinavacuumthatcarriesanegativeelectriccharge of 60 μC

ndash

(a) Onthediagramdrawtheconventionalwayto represent the electric field pattern due to the charged sphere and lines to represent threeequipotentialsurfacesintheregionoutside the sphere

(b) Explainhowthelinesrepresentingtheequipotentialsurfacesthatyouhavesketchedindicatethatthestrengthoftheelectricfieldisdecreasingwithdistancefrom the centre of the sphere

(c) Theelectricfieldstrengthatthesurfaceofthe sphere and at points outside the sphere can be determined by assuming that the sphere acts as a point charge of magnitude 60 μC at its centre The radius of the sphere is 25 times 10ndash2 m Deduce that the magnitude of the field strength at the surface of the sphere is 86 times 107 Vmndash1

An electron is initially at rest on the surface of the sphere

(d) (i) Describethepathfollowedbytheelectronasitleavesthesurfaceofthesphere

(ii) Calculatetheinitialaccelerationofthe electron

5 Determinetheamountofworkthatisdoneinmovingachargeof100nCthroughapotentialdifferenceof150times102 V

6 Three identical 20 μC conducting spheres are placedatthecornersofanequilateraltriangleofsides 25 cm The triangle has one apex C pointing upthepageand2baseanglesAandBDeterminetheabsolutepotentialatB

7 Determinehowfaraparttwoparallelplatesmustbesituatedsothatapotentialdifferenceof250 times 102 V produces an electric field strength of 200 times 103 NC-1

8 Thegapbetweentwoparallelplatesis10times10-3mandthereisapotentialdifferenceof10times104 V betweentheplatesCalculate

i theworkdonebyanelectroninmovingfrom one plate to the other

ii thespeedwithwhichtheelectronreachesthe second plate if released from rest

iii theelectricfieldintensitybetweentheplates

9 Anelectronguninapicturetubeisacceleratedby a potential 25 times 103VDeterminethekineticenergygainedbytheelectroninelectron-volts

10 Determine the electric potential 20 times 10-2 m from achargeof-10times10-5 C

11 Determinetheelectricpotentialatapointmid-waybetweenachargeofndash20pCandanotherof+5pConthelinejoiningtheircentresifthecharges are 10 cm apart

12 During a thunderstorm the electric potential differencebetweenacloudandthegroundis10 times 109 V Determine the magnitude of the change in electric potential energy of an electron thatmovesbetweenthesepointsinelectron-volts

13 A charge of 15 μC is placed in a uniform electric fieldoftwooppositelychargedparallelplateswithamagnitudeof14times103 NC-1

(a) Determinetheworkthatmustbedoneagainstthefieldtomovethepointchargeadistance of 55 cm

(b) Calculatethepotentialdifferencebetweenthe final and initial positions of the charge

(c) Determinethepotentialdifferencebetweenthe plates if their separation distance is 15 cm

14 Duringaflashoflightningthepotentialdifferencebetweenacloudandthegroundwas12 times 109 V and the amount of transferred charge was32C

(a) Determinethechangeinenergyofthetransferred charge

(b) Iftheenergyreleasedwasallusedtoacceleratea1tonnecardeduceitsfinalspeed

(c) Iftheenergyreleasedcouldbeusedtomelticeat0degCdeducetheamountoficethatcould be melted

PHYSICS 2015indb 317 150514 338 PM

Chapter 10

318

AH

L

15 SupposethatwhenanelectronmovedfromAtoBin the diagram along an electric field line that the electric field does 36 times 10-19Jofworkonit

Determinethedifferencesinelectricpotential

(a) VB ndash VA(b) VC ndash VA(c) VC ndash VB

16 DeterminethepotentialatpointPthatislocatedatthecentreofthesquareasshowninthediagrambelow

- 6μC

+3μC +2μC

1m

1m

5μC

P

102 Fields at work

Essential idea Similar approaches can be taken inanalysingelectricalandgravitationalpotentialproblems

Understandingsbull Potentialandpotentialenergybull Potentialgradientbull Potentialdifferencebull Escapespeedbull Orbitalmotionorbitalspeedandorbitalenergybull Forcesandinverse-squarelawbehaviour

Orbital motion orbital speed and orbital energy

Although orbital motion may be circular elliptical orparabolic this sub-topic only dealswith circular orbitsThis sub-topic is not fundamentally new physics butan application that synthesizes ideas from gravitationcircularmotiondynamicsandenergy

TheMoonorbitstheEarthandinthissenseitisoftenreferredtoasasatelliteoftheEarthBefore1957itwastheonlyEarthsatelliteHoweverin1957theRussianslaunchedthefirstmanmadesatelliteSputnik1SincethisdatemanymoresatelliteshavebeenlaunchedandtherearenowliterallythousandsofthemorbitingtheEarthSomeareusedtomonitortheweathersome used to enable people to find accurately their position on thesurfaceoftheEarthmanyareusedincommunicationsandnodoubtsomeareusedtospyonothercountriesFigure1015showshowinprincipleasatellitecanbeputintoorbit

Theperson(whosesizeisgreatlyexaggeratedwithrespectto Earth) standing on the surface on the Earth throwssome stonesThe greater the speed with which a stoneis thrown the further it will land from her The pathsfollowedbythethrownstonesareparabolasByastretchof the imaginationwecanvisualisea situation inwhichastoneis thrownwithsuchaspeedthatbecauseof thecurvatureoftheEarthitwillnotlandonthesurfaceoftheEarthbutgointoldquoorbitrdquo(Path4onFigure1015)

NATURE OF SCIENCECommunication of scientific explanations The ability to apply field theory to the unobservable (charges) and the massively scaled (motion of satellites) required scientists to develop new ways to investigate analyse and report findings to a general public used to scientific discoveries based on tangible and discernible evidence (51)

copy IBO 2014

PHYSICS 2015indb 318 150514 338 PM

Fields

319

AH

L

1

2

34Earth

Figure 1015 Throwing a stone into orbit

TheforcethatcausesthestonestofollowaparabolicpathandtofalltoEarthisgravityandsimilarlytheforcethatkeeps the stone in orbit is gravity For circular motionto occur we have seen that a force must act at rightanglestothevelocityofanobjectthatistheremustbeacentripetalforceHenceinthesituationwedescribeherethe centripetal force for circular orbital motion about the EarthisprovidedbygravitationalattractionoftheEarth

We can calculate the speedwithwhich a stonemustbethrowninordertoputitintoorbitjustabovethesurfaceof the Earth

If the stone has mass m and speed v thenwehave fromNewtonrsquos2ndlaw

mv2

RE---------- G

MEm

RE2------------=

whereRE is the radius of the Earth and ME is the mass of the Earth

Bearinginmindthatg 0 GME

RE2--------= then

v g RE 10 64 106timestimes 8 103times= = =

Thatisthestonemustbethrownat8times103m sndash1

Clearly we are not going to get a satellite into orbit soclosetothesurfaceoftheEarthMovingatthisspeedthefrictionduetoairresistancewouldmeltthesatellitebeforeithadtravelledacoupleofkilometresInrealitythereforeasatelliteisputintoorbitabouttheEarthbysendingitattachedtoarocketbeyondtheEarthrsquosatmosphereandthengivingitacomponentofvelocityperpendiculartoaradialvectorfromtheEarthSeeFigure1016

Earth Satellite carried by rocket to hereSatellite orbit

Tangential component of velocity

Figure 1016 Getting a satellite into orbit

Keplerrsquos third law(This work of Kepler and Newtonrsquos synthesis of the work is an excellent example of the scientific method and makes for a good TOK discussion)

In 1627 Johannes Kepler (1571-1630)publishedhis lawsof planetary motion The laws are empirical in natureandwere deduced from the observations of theDanishastronomer Tycho de Brahe (1546-1601)The third lawgives a relationship between the radius of orbit R of a planet and its period TofrevolutionabouttheSunThelawisexpressedmathematicallyas

2

3 constantTR

=

We shall nowuseNewtonrsquos LawofGravitation to showhowitisthattheplanetsmoveinaccordancewithKeplerrsquosthirdlaw

InessenceNewtonwasabletousehislawofgravitytopredict the motion of the planets since all he had to do was factor theF givenby this law intohis second law F = ma to find their accelerations and hence their future positions

InFigure1017theEarthisshownorbitingtheSunandthedistancebetweentheircentresisR

RSun

Earth

F es

F se

Figure 1017 Planets move according to Keplerrsquos third law

Fes is the force that the Earth exerts on the Sun and Fse is the force that the Sun exerts on the Earth The forces are equalandoppositeandtheSunandtheEarthwillactuallyorbitaboutacommoncentreHoweversince theSun issoverymuchmoremassivethantheEarththiscommoncentrewillbeclosetothecentreoftheSunandsowecanregard the Earth as orbiting about the centre of the Sun Theotherthingthatweshallassumeisthatwecanignorethe forces that the other planets exert on the Earth (This wouldnotbeawisethingtodoifyouwereplanningtosend a space ship to theMoon for example)We shallalso assume that we have followed Newtonrsquos exampleandindeedprovedthataspherewillactasapointmasssituated at the centre of the sphere

Kepler had postulated that the orbits of the planets areellipticalbut since theeccentricityof theEarthrsquosorbit issmallweshallassumeacircularorbit

PHYSICS 2015indb 319 150514 338 PM

Chapter 10

320

AH

L

TheaccelerationoftheEarthtowardstheSunisa = Rω2

whereω ω 2πT------=

Hencea R 2π

T------ times

2 4π2 RT 2------------- = =

But the acceleration is given by Newtonrsquos Second Law F = mawhereFisnowgivenbytheLawofGravitationSo in this situation

F maGMs Me

R2-------------------= = butwealsohavethat

a a R 2πT

------ times

2 4π2 RT 2------------- = = and m = Me so that

G MsMe

R2------------------- Me4π2R

T2-------------times

GMs

4π2-----------rArr R3

T2------= =

Butthequantity

isaconstantthathasthesamevalueforeachoftheplanetssowehaveforalltheplanetsnotjustEarththat

R3

T2------ k =

wherekisaconstantWhichisofcourseKeplerrsquosthirdlaw

ThisisindeedanamazingbreakthroughItisdifficulttorefutethe idea that all particles attract each other in accordance withtheLawofGravitationwhenthelawisabletoaccountfortheobservedmotionoftheplanetsabouttheSun

The gravitational effects of the planets upon each othershould produce perturbations in their orbits Such is the predictivepoweroftheUniversalGravitationalLawthatit enabled physicists to compute these perturbations The telescopehadbeeninventedin1608andbythemiddleofthe 18th Century had reached a degree of perfection in design that enabled astronomers to actually measure the orbital perturbations of the planets Their measurements werealways inagreementwith thepredictionsmadebyNewtonrsquoslawHoweverin1781anewplanetUranuswasdiscovered and the orbit of this planet did not fit withthe orbit predicted by Universal Gravitation Such wasthe physicistrsquos faith in theNewtonianmethod that theysuspectedthatthediscrepancywasduetothepresenceofayetundetectedplanetUsingtheLawofGravitationtheFrenchastronomerJLeverrier and the English astronomer J C Adamswereable tocalculate justhowmassive thisnewplanetmustbeandalsowhereitshouldbeIn1846the planetNeptunewas discovered justwhere they hadpredicted Ina similarwaydiscrepancies in theorbitofNeptune ledto thepredictionandsubsequentdiscoveryin1930oftheplanetPlutoNewtonrsquosLawofGravitation

had passed the ultimate test of any theory it is not only abletoexplainexistingdatabutalsotomakepredictions

Satellite energyWhenasatelliteisinorbitaboutaplanetitwillhavebothkineticenergyandgravitationalpotentialenergySupposewe consider a satellite ofmass m that is in an orbit of radius r about a planet of mass M

Thegravitationalpotentialduetotheplanetatdistancer from its centre is

ThegravitationalpotentialenergyofthesatelliteVsat

is therefore GMe m

r----------------ndash

ThatisVsat GMem

r----------------ndash=

Thegravitationalfieldstrengthatthesurfaceoftheplanetisgivenby

g 0GMe

Re2------------=

Hencewecanwrite

Vsat g0 Re

2

rndash=

m

ThekineticenergyofthesatelliteKsatisequaltofrac12mv2wherev is its orbital speed

Byequatingthegravitationalforceactingonthesatellitetoitscentripetalaccelerationwehave

G Mem

r2---------------- mv2

r---------- mv2hArr

GMemr----------------= =

Fromwhich12---

mv2 12---

GMemr----------------times=

= 0 e

2g R

r

2m

Which is actually quite interesting since it shows thatirrespective of the orbital radius the KE is numericallyequal to half the PE Also the total energy E tot K sat Vsat+ 1

2---G Me m

r----------------times G Me m

r----------------ndash + 1

2---G Me m

r----------------ndash= = =of the satelliteisalwaysnegativesince

E tot K sat Vsat+ 12---

G Me mr----------------times

G Me mr----------------ndash

+ 12---

G Me mr----------------ndash= = =

PHYSICS 2015indb 320 150514 338 PM

Fields

321

AH

L

The energies of an orbiting satellite as a function of radial distancefromthecentreofaplanetareshownplottedinFigure1018

12

100806

0402

0ndash02

ndash04

ndash06ndash08ndash10

ndash12

2 4 6 8 10 12

ener

gya

rbita

ry u

nits

distance timesR

kinetic energytotal energy

potential energy

Figure 1018 Energy of an orbiting satellite as a function of distance from the centre of a planet

WeightlessnessSuppose that you are inanelevator(lift)thatisdescendingat constant speedandyou let goof abook thatyouareholdinginyourhandThebookwillfalltothefloorwithaccelerationequaltotheaccelerationduetogravityIfthecablethatsupportstheelevatorweretosnap(asituationthatItrustwillneverhappentoanyofyou)andyounowletgothebookthatyouareholdinginyourotherhandthis book will not fall to the floor - it will stay exactlyin linewithyourhandThis isbecause thebook isnowfallingwith the sameaccelerationas theelevatorandassuch the book cannot ldquocatchrdquo up with the floor of theelevatorFurthermoreifyouhappenedtobestandingonasetofbathroomscalesthescaleswouldnowreadzero-youwouldbeapparentlyweightlessItisthisideaoffreefallthatexplainstheapparentweightlessnessofastronautsin an orbiting satellite These astronauts are in free fall inthesensethattheyareacceleratingtowardsthecentre of the Earth

It is actually possible to define theweight of a body inseveral different waysWe can define it for example asthegravitationalforceexertedonthebodybyaspecifiedobject such as the EarthThiswe have seen thatwe doinlotsofsituationswherewedefinetheweightasbeingequaltomgIfweusethisdefinitionthenanobjectinfreefallcannotbydefinitionbeweightlesssinceitisstillinagravitationalfieldHoweverifwedefinetheweightofanobjectintermsofaldquoweighingrdquoprocesssuchasthereadingonasetofbathroomscaleswhichineffectmeasuresthecontact force between the object and the scales thenclearlyobjectsinfreefallareweightlessOnenowhasto

askthequestionwhetherornotitispossibleForexampletomeasurethegravitationalforceactingonanastronautin orbit about the Earth

WecanalsodefineweightintermsofthenetgravitationalforceactingonabodyduetoseveraldifferentobjectsForexample for an object out in space its weight could bedefined in terms of the resultant of the forces exerted on it bytheSuntheMoontheEarthandalltheotherplanetsin the Solar System If this resultant is zero at a particular pointthenthebodyisweightlessatthispoint

Inviewofthevariousdefinitionsofweightthatareavailabletousitisimportantthatwhenweusethewordldquoweightrdquoweareawareofthecontextinwhichitisbeingused

Example

Calculate the height above the surface of the Earth atwhichageo-stationarysatelliteorbits

Solution

A geo-stationary satellite is one that orbits the Earth in such a way that it is stationary with respect to a point on the surface of the Earth This means that its orbital period must be the same as the time for the Earth to spin once on its axis ie 24 hours

From Keplerrsquos third law we have G Ms

4π2----------- R3

T2------=

That is

RMe

mh

using the fact that the force of attraction between the satellite and the Earth is given by

FG Mem

R2----------------=

and that F = ma

where m is the mass of the satellite and a 4π2R

T2-------------=

we have

GMem

R2---------------- m 4π2R

T2-------------times

GMe

4π2------------rArr R3

T2------= =

PHYSICS 2015indb 321 150514 338 PM

Chapter 10

322

AH

L

Now the mass of the Earth is 60 times 1024 kg and the period T measured in seconds is given by T = 86400 s

So substitution gives R = 42 times 106 m

The radius of the Earth is 64 times 106 m so that the orbital height h is about 36 times 107 m

Example

Calculatetheminimumenergyrequiredtoputasatelliteofmass500kgintoanorbitthatisasaheightequaltotheEarthrsquosradiusabovethesurfaceoftheEarth

Solution

We have seen that when dealing with gravitational fields and potential it is useful to remember that

g 0G M

Re2---------= or g0 Re

2 GM=

The gravitational potential at the surface of the Earth is

g 0ndash Re GMRe

---------ndash=

The gravitational potential at a distance R

from the centre of the Earth is

=

The difference in potential between the surface of the Earth and a point distance R from the centre is therefore

V∆ g 0Re 1ReR------ndash

=

If R 2Re= then V∆g0 Re

2------------=

This means that the work required to ldquoliftrdquo the satellite into orbit is g0Rm where m is the mass of the satellite This is equal to

10 32 106 500timestimestimes = 16000 MJ

However the satellite must also have kinetic energy in order to orbit Earth This will be equal to

g 0 mRe2

2R-----------------g 0mRe

2

4----------------- 8000 MJ= =

The minimum energy required is therefore

24000 MJ

Exercise 1031 The speed needed to put a satellite in orbit does

not depend on

A the radius of the orbitB theshapeoftheorbitC thevalueofgattheorbitD the mass of the satellite

2 EstimatethespeedofanEarthsatellitewhoseorbitis400kmabovetheEarthrsquossurfaceAlsodetermine the period of the orbit

3 Calculatethespeedofa200kgsatelliteorbitingthe Earth at a height of 70 times 106 m

Assume that g = 82 m sndash2 for this orbit

4 TheradiioftwosatellitesXandYorbitingthe Earth are 2r and 8rwherer is the radius of the Earth Calculate the ratio of the periods of revolutionofXtoY

5 A satellite of mass mkgissentfromEarthrsquossurfaceinto an orbit of radius 5RwhereR is the radius of theEarthWritedownanexpressionfor

(a) thepotentialenergyofthesatelliteinorbit

(b) thekineticenergyofthesatelliteinorbit

(c) theminimumworkrequiredtosendthesatellitefromrestattheEarthrsquossurfaceintoits orbit

6 A satellite in an orbit of 10rfallsbacktoEarth(radius r)afteramalfunctionDeterminethespeedwithwhichitwillhittheEarthrsquossurface

7 The radius of the moon is frac14 that of the Earth AssumingEarthandtheMoontohavethesamedensitycomparetheaccelerationsoffreefallatthe surface of Earth to that at the surface of the Moon

8 UsethefollowingdatatodeterminethegravitationalfieldstrengthatthesurfaceoftheMoon and hence determine the escape speed from the surface of the Moon

Mass of the Moon = 73 times 1022kg

RadiusoftheMoon=17times106 m

PHYSICS 2015indb 322 150514 338 PM

Wave Phenomena

295

AH

L

consider the light that leaves the slit at an angle θ as shown Thepathdifferencebetweenwave1andwave2isdsinθ and ifthisisequaltoanintegralnumberofwavelengthsthenthe two waves will interfere constructively in this direction Similarly wave 2 will interfere constructively with wave3 at this angle andwave 3with 4 etc across thewholegratingHenceifwelookatthelightthroughatelescopethatisbringittoafocusthenwhenthetelescopemakesanangle θ to the grating a bright fringe will be observed The condition for observing a bright fringe is therefore

dsinθ = mλ

Supposeweuselightofwavelength500nmandsupposethat d = 16 times 10ndash6 m

Obviously we will see a bright fringe in the straight on position θ=0(thezero order)

The next position will be when m =1(thefirst order)andsubstitutionintheaboveequationgivesθ = 18deg

The next position will be when m=2(thesecond order)and this give θ = 38deg

For m=3sinθ is greater than 1 so with this set up we only obtain5fringesonezeroorderandtwoeithersideofthezero order

The calculation shows that the separation of the orders is relatively large At any angles other than 18deg or 38deg the light leaving the slits interferes destructively We can see that the fringes will be sharp since if we move just a small angle away from 18deg the light from the slits will interfere destructively

An array of narrow slits is usually made by cutting narrow transparent lines very close together into the emulsion on aphotographicplate(typically200linespermillimetre)Suchanarrangementiscalledadiffraction grating

The diffraction grating is of great use in examining the spectral characteristics of light sources

All elements have their own characteristic spectrum An element can be made to emit light either by heating it until it is incandescent or by causing an electric discharge through itwhen it is in a gaseous state Yourschool probably has some discharge tubes and diffraction gratingsIfithasthenlookattheglowingdischargetubethroughadiffractiongratingIfforexampletheelementthat you are looking has three distinct wavelengths then each wavelength will be diffracted by a different amount inaccordancewiththeequationdsinθ = nλ

Alsoiflaserlightisshonethroughagratingontoascreenyou will see just how sharp and spaced out are the maxima By measuring the line spacing and the distance of the screen fromthelaserthewavelengthofthelasercanbemeasured

If your school has a set of multiple slits say from a single slit toeight slits then it isalsoaworthwhileexercise toexamine how the diffraction pattern changes when laser light is shone through increasing numbers of slits

Consider a grating of NparallelequidistantslitsofwidthaseparatedbyanopaqueregionofwidthD We know that light diffracted from one slit will be superimposed and interfere with light diffracted from all the other slits The intensityprofile of the resultant wave will have a shape determined by

bull theintensityoflightincidentontheslitsbull thewavelengthofthislightbull theslitwidthbull theslitseparation

Typical intensity profiles formed when a plane wavemonochromaticlightatnormalincidenceononetwoandsix slits are shown in Figure 919 In this example the slit widthisonequarterofthegratingspacing

Six slits sharperfringes

d = 4a

I α N 2

(N ndash 2)= 4 subsidiary maxima

order0 4321

I = 36

Two slits

order0

single-slitdiraction

envelope

d = 4a4th principal maximum missing

principa maxima

I = 4 (nominal units)

4321

intensity

order

ISS

One slit

0 1

single-slitdiraction

pattern

N = number of slitsslit width a

Figure 919 Intensity profiles

PHYSICS 2015indb 295 150514 337 PM

Chapter 9

296

AH

L

1 For Nge2thefringepatterncontainsprincipalmaximathat are modulated by the diffraction envelope

2 ForNgt2thefringepatterncontains(Nndash2)subsidiary maxima

3 Where the grating spacing d is four time the slit width a(4xa =d)thenthefourthprincipalmaximumismissing from the fringe pattern Where the grating spacingdisthreetimetheslitwidtha(3xa =d)thenthe third principal maximum is missing from the fringepatternandsoon

4 Increasing N increases the absolute intensities of the diffraction pattern (I = N2)

If white light is shone through a grating then the central image will be white but for the other orders each will be spread out intoacontinuousspectrumcomposedofaninfinitenumberof adjacent images of the slit formed by the wavelength of the different wavelengths present in the white light At any given point in the continuous spectrum the light will be very nearly monochromatic because of the narrowness of the images of the slit formed by the grating This is in contrast to the doubleslitwhereifwhitelightisusedtheimagesarebroadand the spectral colours are not separated

In summary as is the case of interference at a doubleslit there is a path difference between the rays fromadjacent slits of d sinθ Where there is a whole number of wavelengthsconstructiveinterferenceoccurs

For constructive interference and minima

d sinθ = mλ wheren=123hellip

When there is a path difference between adjacent rays of halfawavelengthdestructiveinterferenceoccurs

For and maxima

d sinθ = ( m + frac12 )λ wheren=123hellip

Exercise 941 Lightfromalaserisshonethroughadiffraction

grating on to a screen The screen is a distance of 20mfromthelaserThedistancebetweenthecentraldiffractionmaximumandthefirstprincipalmaximum formed on the screen is 094 m The diffractiongratinghas680linespermmEstimatethe wavelength of the light emitted by the laser

2 Monochromaticlightfromalaserisnormallyincident on a six-slit grating The slit spacing is three times the slit width An interference patternisformedonascreen2montheotherside of the grating

(a) Determineandexplaintheorderoftheprincipal maxima missing from the fringe pattern

(b) Howmanysubsidiarymaximaaretherebetween the principal maxima

(c) Howmanyprincipalmaximaarethere

(i) Betweenthetwofirstorderminimain the diffraction envelope

(ii) Betweenthefirstandsecondorderminima in the diffraction envelope

(d) Drawthepatternformedbythebrightfringesin(c)

(e) Sketchtheintensityprofileofthefringepattern

Thin film interferenceYoumightwellbe familiarwith the colouredpatternoffringes that can be seen when light is reflected off thesurfaceofwateruponwhichathinoilfilmhasbeenspiltorfromlightreflectedfrombubblesWecanseehowthesepatternsarisebylookingatFigure920

Extendedlight source air

A

B

C

12

3

4

Oil lm

Water

Figure 920 Reflection from an oil film

Consider light from an extended source incident on a thinfilmWealsoconsiderawave fromonepointofthe source whose direction is represented by the ray shownSomeofthislightwillbereflectedatAandsometransmitted through the filmwhere somewill again bereflectedatB(somewillalsobetransmittedintotheair)SomeofthelightreflectedatBwillthenbetransmittedatCandsomereflectedandsoonIfweconsiderjustrays1and2thenthesewillbenotbeinphasewhenthearebrought to a focus by the eye since they have travelled

PHYSICS 2015indb 296 150514 337 PM

Wave Phenomena

297

AH

L

different distances If the path difference between them is an integral number of half wavelengths then we might expect the twowaves to be out of phaseHowever ray1 undergoes a phase change of π on reflection but ray2 does not since it is reflected at a boundary betweena more dense and less dense medium (See chapter 4)Henceifthepathdifferenceisanintegralnumberofhalf-wavelengthsrays1and2willreinforceieray1and2areinphaseHowever rays357etcwillbeoutofphasewithrays246etcbutsinceray2ismoreintensethanray3andray4moreintensethanray5thesepairswillnot cancel out so there will be a maximum of intensity

Ifthepathdifferenceissuchthatwave1and2areoutofphasesincewave1ismoreintensethanwave2theywillnotcompletelyannulHoweveritcanbeshownthattheintensitiesofwaves2345hellipaddtoequaltheintensityofwave1Sincewaves345hellipareinphasewithwave2there will be complete cancellation

The path difference will be determined by the angle at which ray1 is incident andalsoon the thickness (andtheactualrefractiveindexaswell)ofthefilmSincethesourceisanextendedsourcethelightwillreachtheeyefrom many different angles and so a system of bright and darkfringeswillbeseenYouwillonlyseefringesifthefilmisverythin(exceptifviewedatnormalincidence)sinceincreasingthethicknessofthefilmwillcausethereflected rays to get so far apart that they will not becollected by the pupil of the eye

From the argument above to find the conditions forconstructive and destructive interferenceweneedonlyfindthepathdifferencebetweenray1andray2Figure921showsthe geometry of the situation

A BE

F

C

D

1 2

φrsquo φrsquo

φ

φ

d

Figure 921 The geometry of interference

Thefilmisofthicknessd and refractive index n and the light has wavelength λ If the line BF is perpendicular to ray 1 then the optical path difference (opd) between ray 1 andray2whenbroughttoafocusis

opd = n(AC+CB)ndashAF

We have to multiple by the refractive index for the path travelledbythelightinthefilmsincethelighttravelsmoreslowly in thefilm If the light travels say adistancex in a material of refractive index n then in the time that it takes to travelthisdistancethelightwouldtraveladistancenx in air

IfthelineCEisatrightanglestoray2thenweseethat

AF = nBE

FromthediagramAC=CDsowecanwrite

opd = n(CD+CB)ndashnBE=nDE

Alsofromthediagramweseethatwhereφ is the angle of refraction

DE=2cosφ

From which opd=2ndcosφ

Bearinginmindthechangeinphaseofray1onreflectionwe have therefore that the condition for constructive interference is

2nd φcos m 12---+

λ m 1 2 hellip = =

Andfordestructiveinterference 2ndcosφ = mλ

Eachfringecorrespondstoaparticularopd for a particular value of the integer m and for any fringe the value of the angle φ is fixedThismeans that itwill be in the formof an arc of a circle with the centre of the circle at the point where the perpendicular drawn from the eye meets the surface of the film Such fringes are called fringesofequal inclinationSince theeyehasa smallaperturethesefringesunlessviewedatneartonormalincidence (φ = 0) will only be observed if the film is very thinThisisbecauseasthethicknessofthefilmincreasesthereflected rayswill get further and further apart and sovery few will enter the eye

Ifwhitelightisshoneontothefilmthenwecanseewhywe get multi-coloured fringes since a series of maxima and minima will be formed for each wavelength present in the white light However when viewed at normalincidenceitispossiblethatonlylightofonecolourwillundergoconstructiveinterferenceandthefilmwilltakeon this colour

PHYSICS 2015indb 297 150514 337 PM

Chapter 9

298

AH

L

Thickness of oil films

The exercise to follow will help explain this use

Non-reflecting films

A very important but simple application of thin filminterferenceisintheproductionofnon-reflectingsurfaces

Athinfilmofthicknessd and refractive index n1 is coated onto glass of refractive index n where n1 lt n Light ofwavelength λ that is reflected at normal incidence willundergo destructive interference if 12

2n d λ

= thatis

14d

nλ

=

(remember that there will now no phase change at the glass interface ie we have a rare to dense reflection)

The use of such films can greatly reduce the loss oflightby reflection at the various surfacesof a systemoflenses or prisms Optical parts of high quality systemsare usually all coatedwithnon-reflecting films in orderto reduce stray reflections The films are usually madebyevaporatingcalciumormagnesiumfluorideonto thesurfacesinvacuumorbychemicaltreatmentwithacidsthat leave a thin layer of silica on the surface of the glass ThecoatedsurfaceshaveapurplishhuebyreflectedlightThis is because the condition for destructive interference fromaparticularfilmthicknesscanonlybeobtainedforone wavelength The wavelength chosen is one that has a value corresponding to light near the middle of the visible spectrumThismeansthatreflectionofredandvioletlightis greater combining to give the purple colour Because of the factor cosφatanglesotherthannormalincidencethepath difference will change butnotsignificantlyuntilsayabout30deg(egcos25deg=090)

It should be borne in mind that no light is actually lost by anon-reflectingfilmthedecreaseofreflectedintensityiscompensated by increase of transmitted intensity

Non-reflecting films can be painted onto aircraft tosuppressreflectionofradarThethicknessofthefilmis determined by

4nd λ

= where λ is the wavelength of

the radar waves and ntherefractiveindexofthefilmat this wavelength

Example

A plane-parallel glass plate of thickness 20 mm isilluminated with light from an extended source The refractiveindexoftheglassis15andthewavelengthofthelight is 600 nm Calculate how many fringes are formed

Solution

We assume that the fringes are formed by light incident at all angles from normal to grazing incidence

At normal incidence we have 2nd = mλ

From which

m 2 15 2 10 3ndashtimestimestimes6 10 7ndashtimes

------------------------------------------- 10 000= =

At grazing incidence the angle of refraction φ is in fact the critical angle

Therefore φφ arcsin 115------- 42 deg= =

ie cosφ = 075

At grazing incidence 2ndcosφ = m λ

From which (and using cos(sinndash1(115) = 075)

m 2 15 2 10 3ndash 075timestimestimestimes6 10 7ndashtimes

------------------------------------------------------------ 7500= =

The total number of fringes seen is equal to m ndash m = 2500

Exercise 95When viewed from above the colour of an oil film onwater appears red in colour Use the data below to estimate theminimumthicknessoftheoilfilm

average wavelength of red light = 630 nm

refractiveindexofoilforredlight=15

refractive index of water for red light = 13

PHYSICS 2015indb 298 150514 337 PM

Wave Phenomena

299

AH

L

94 Resolution

Essential idea Resolution places an absolute limit on the extent to which an optical or other system can separate images of objects

Understandings

bull Thesizeofadiffractingaperturebull Theresolutionofsimplemonochromatic

two-source systems

The size of a diffracting apertureOur discussion concerning diffraction so far has been for rectangular slits When light from a point source enters a smallcircularapertureitdoesnotproduceabrightdotasanimagebutacirculardiscknownasAiryrsquosdiscsurroundedbyfainterconcentriccircularringsasshowninFigure922

Laser

Figure 922 Diffraction at a circular aperture

Diffractionatanapertureisofgreatimportancebecausetheeye and many optical instruments have circular apertures

What is the half-angular width of the central maximum of the diffraction formed by a circular aperture This is not easy to calculate since it involves some advanced mathematics The problemwasfirstsolvedbytheEnglishAstronomerRoyalGeorgeAiryin1835whoshowedthatforcircularapertures

θ 122 λb--------------= where b is the diameter of the aperture

Example 2

Inthefollowingdiagramparallellightfromadistantpointsource(suchasastar)isbroughttofocusonthescreenSbyaconverginglens(the lens is shown as a vertical arrow)

Thefocal length(distance fromlens toscreen) is25cmand the diameter of the lens is 30 cm The wavelength of thelightfromthestaris560nmCalculatethediameterofthe diameter of the image on the screen

Solution

The lens actually acts as a circular aperture of diameter 30 cm The half angular width of central maximum of the diffraction pattern that it forms on the screen is

75

2

122 122 56 10 23 1030 10b

λθminus

minusminus

times times= = = times

times rad

The diameter of the central maxima is therefore

25 times 10-2 times 23 times 10-5

= 57 times 10-6 m

Although this is small it is still finite and is effectively the image of the star as the intensity of the secondary maxima are small compared to that of the central maximum

The resolution of simple monochromatic two-source systemsThe astronomers tell us that many of the stars that we observe with the naked eye are in fact binary stars That iswhatwe see as a single star actually consistsof twostarsinorbitaboutacommoncentreFurthermoretheastronomerstellusthatifweusealdquogoodrdquotelescopethenwewillactuallyseethetwostarsThatiswewillresolve the single point source into its two component parts Sowhatisitthatdetermineswhetherornotweseethetwo stars as a single point source ie what determines whether or not two sources can be resolved It canrsquot just bethatthetelescopemagnifiesthestarssinceiftheyareacting as point sources magnifying them is not going to make a great deal of difference

Ineachofoureyesthereisanaperturethepupilthroughwhich the light enters This light is then focused by the eye lens onto the retina But we have seen that when light passes through an aperture it is diffracted and so when we look at

NATURE OF SCIENCEImproved technology The Rayleigh criterion is the limit of resolution Continuing advancement in technology such as large diameter dishes or lenses or the use of smaller wavelength lasers pushes the limits of what we can resolve (18)

copy IBO 2014

PHYSICS 2015indb 299 150514 337 PM

Chapter 9

300

AH

L

apointsourceadiffractionpatternwillbeformedontheretina If we look at two point sources then two diffraction patterns will be formed on the retina and these patterns will overlap The width of our pupil and the wavelength of the light emitted by the sources will determine the amount by which they overlap But the degree of overlap will also depend on the angular separation of the two point sources WecanseethisfromFigures923

pupilretina

S1

S2

P2

P1

θ

Figure 923

LightfromthesourceS1 enters the eye and is diffracted by the pupil such that the central maximum of the diffraction patternisformedontheretinaatP1SimilarlylightfromS2producesamaximumatP2 If the two central maxima are well separated then there is a fair chance that we will see the two sources as separate sources If they overlap then we will not be able to distinguish one source from another From the diagram we see as the sources are movedclosertotheeyethentheangleθ increases and so does the separation of the central maxima

Figures924925926and927showsthedifferentdiffractionpatternsandtheintensitydistributionthatmightresultonthe retina as a result of light from two point sources

S1 S21 Well resolved

Figure 924 Very well resolved

S1 S22 Well resolved

Figure 925 Well resolved

S1 S23 Just resolved Rayleigh criterion Minimum of S2 coincides with maximum peak of S1

Figure 926 Just resolved

4 Not resolved

Figure 927 Not resolved

We have suggested that if the central maxima of the two diffraction patterns are reasonably separated then we should be able to resolve two point sources In the late 19th century Lord Rayleigh suggested by how much they should be separated in order for the two sources to be just resolved Ifthecentralmaximumofonediffractionpattern coincides with the firstminima of the otherdiffraction pattern then the two sources will just beresolved This is known as the Rayleigh Criterion

Figure926showsjustthissituationThetwosourcesarejust resolved according to the Rayleigh criterion since the peak of the central maximum of one diffraction pattern coincideswiththefirstminimumoftheotherdiffractionpattern This means that the angular separation of the peaks of the two central maxima formed by each source is just the half angular width of one central maximum ie

θ λb---

=

where b is the width of the slit through which the light fromthesourcespassesHoweverweseefromFigure1117that θ is the angle that the two sources subtend at the slit Henceweconcludethattwosourceswillberesolvedbyaslit if the angle that they subtend at the slit is greater than or equaltoλfraslb

PHYSICS 2015indb 300 150514 337 PM

Wave Phenomena

301

AH

L

Sofarwehavebeenassumingthattheeyeisarectangularslit whereas clearly it is a circular aperture and so we must use the formula

θ 122 λb--------------=

As mentioned above the angle θ is sometimes called the resolving power but should more accurately be called the minimum angle of resolution (θmin)

Clearly the smaller θ the greater will be the resolving power

The diffraction grating is a useful for differentiating closely spacedlinesinemissionspectraLikeaprismspectrometerit can disperse a spectrum into its components but it is better suited because it has higher resolution than the prism spectrometer

If λ1 and λ2 are two nearly equal wavelengths that canbarelybedistinguishedtheresolvanceorresolvingpowerofthegratingisdefinedas

R = λ _______ (λ2 ndash λ 1) = λ ___ Δλ

Where λ asymp λ1 asymp λ2

Therefore thediffractiongratingwithahighresolvancewill be better suited in determining small differences in wavelength

If the diffraction grating has Nlinesbeingilluminateditcan be shown that the resolving power of the mth order diffraction is given by

R = mN

Obviouslytheresolvancebecomesgreaterwiththeordernumber m and with a greater number of illuminated slits

Example

A benchmark for the resolving power of a grating is the sodium doublet in the sodium emission spectrum Two yellow lines in itsspectrumhavewavelengthsof58900nmand58959nm

(a) Determinetheresolvanceofagratingifthegivenwavelengths are to be distinguished from each other

(b) Howmanylinesinthegratingmustbeilluminated in order to resolve these lines in the second order spectrum

Solution

(a) R = λ ___ Δλ = 58900nm __________________ (58900ndash58959)nm

= 589nm _______ 059nm = 10 times 103

(b) N = R __ m = 10 times 10 3 ________ 2 =50times102 lines

It has been seen that diffraction effectively limits the resolving power of optical systems This includes such systems as the eye telescopes and microscopes Theresolving power of these systems is dealt with in the next section This section looks at links between technology and resolving power when looking at the very distant and when looking at the very small

Radio telescopes

The average diameter of the pupil of the human eye is about 25 mm This means that the eye will justresolve two point sources emitting light of wavelength 500 nm if their angular separation at the eye is 7

43

50 10122 24 1025 10

θminus

minusminus

times= times = times

times rad

If the eye were to be able to detect radio waves of wavelength 015m then tohave the same resolvingpower thepupilwould have to have a diameter of about 600 m Clearly this is nonsense but it does illustrate a problem facingastronomers who wish to view very distant objects such as quasarsandgalaxies thatemitradiowavesConventionalradio telescopes consist of a large dish typically 25 min diameter Even with such a large diameter the radiowavelength resolving power of the telescope is much less thantheopticalresolvingpowerofthehumaneyeLetuslook at an example

Example

The Galaxy Cygnus A can be resolved optically as an elliptically shaped galaxy However it is also a strongemitterof radiowavesofwavelength015mTheGalaxyisestimatedtobe50times1024mfromEarthUseofaradiotelescope shows that the radio emission is from two sources separated by a distance of 30 times 1021m Estimate thediameterofthedishrequiredtojustresolvethesources

Solution

The angle θ that the sources resolve at the telescope is given by

θ 21

424

30 10 60 1050 10

θ minustimes= = times

times rad

and 4

122 122 015 300060 10

d λθ minus

times= = =

timesm = 30 km

A radio telescope dish of this size would be impossible tomake letalonesupportThisshowsthatasingledishtype radio telescope cannot be used to resolve the sources and yet they were resolved To get round the problemastronomers use two radio telescopes separated by a large

PHYSICS 2015indb 301 150514 337 PM

Chapter 9

302

AH

L

distance The telescopes view the same objects at the same time and the signals that each receive from the objects are simultaneously superimposed The result of the superposition of the two signals is a two-slit interference pattern The pattern has much narrower fringe spacing than that of the diffraction pattern produced by either telescope on its own hence producing a much higherresolvingpowerWhentelescopesareusedlikethistheyare called a stellar interferometer

InSocorroinNewMexicothereisastellarinterferometerthat consists of 27 parabolic dishes each of diameter25 m arranged in a Y-shape that covers an area of 570km2Thisisaso-calledVeryLargeArray(VLA)Evenhigher resolution can be obtained by using an array of radio telescopes in observatories thousands of kilometres apart A system that uses this so-called technique oflsquovery-long-baseline interferometryrsquo (VLBI) is known asalsquovery-long-baselinearrayrsquo(VLBA)WithVLBAaradiowavelength resolving power can be achieved that is 100 timesgreaterthanthebestopticaltelescopesEvenhigherresolving power can be achieved by using a telescope that isinasatelliteorbitingEarthJapanrsquosInstituteofSpaceandAstronautical Science (ISAS) launched sucha system inFebruary 1997TheNationalAstronomicalObservatoryof Japan the National Science Foundationrsquos NationalRadio Astronomy Observatory (NRAO) the CanadianSpaceAgency theAustraliaTelescopeNationalFacilitythe European VLBI Network and the Joint Instituteback this project forVeryLongBaseline Interferometryin EuropeThis project is a very good example of howInternationalismcanoperateinPhysics

Electron microscope

Telescopes are used to look at very distant objects that are very large but because of their distance fromus appearverysmallMicroscopesontheotherhandareusedtolookat objects that are close to us but are physically very small Aswehave seen justmagnifyingobjects that ismakingthemappearlargerisnotsufficientonitsowntogaindetailabouttheobjectfordetailhighresolutionisneeded

Figure928isaschematicofhowanoptical microscope is usedtoviewanobjectandFigure929isaschematicofatransmission electron microscope(TEM)

Optical lens system

bright light source

glass slide containing specimen (object)

eye

Figure 928 The principle of a light microscope

magnetic lens system

electron source

thin wafer of material

ScreenCCD

Figure 929 The principle of an electron microscope

In the optical microscope the resolving power isdetermined by the particular lens system employed and the wavelength λ of the light used For example twopoints in the sample separated by a distance d will just be resolved if

2d mλ=

where m is a property of the lens system know as the numerical aperture In practice the largest value of m obtainable is about 16 Hence if the microscope slideis illuminatedwith light ofwavelength 480 nm a goodmicroscope will resolve two points separated by a distance dasymp15times10-7masymp015micromPointscloser together thanthiswillnotberesolvedHoweverthisisgoodenoughtodistinguishsomevirusessuchastheEbolavirus

Clearlythesmallerλ the higher the resolving power and this is where the electron microscope comes to the fore The electron microscope makes use of the wave nature of electrons(see1315)IntheTEMelectronspassthrougha wafer thin sample and are then focused by a magnetic field onto a fluorescent screen orCCD (charge coupleddevicesee142)Electronsusedinanelectronmicroscopehavewavelengthstypicallyofabout5times10-12mHoweverthe numerical aperture of electron microscopes is considerablysmaller thanthatofanopticalmicroscopetypicallyabout002NonethelessthismeansthataTEMcanresolvetwopointsthatareabout025nmapartThisresolving power is certainly high enough to make out the shape of large molecules

Another type of electron microscope uses a techniqueby which electrons are scattered from the surface of the sample The scattered electrons and then focused as in theTEM to form an image of the surfaceThese so-calledscanningelectronmicroscopes(SEM)havealowerresolving power than TEMrsquos but give very good threedimensional images

The eye

We saw in the last section that the resolving power of the human eye is about 2 times 10-4 rad Suppose that youare looking at car headlights on a dark night and the car

PHYSICS 2015indb 302 150514 337 PM

Wave Phenomena

303

AH

L

is a distance D away If the separation of the headlight is say 15 m then the headlights will subtend an angle θ 15

Dθ =

at your eye Assuming an average wavelength of

500 nm your eye will resolve the headlights into twoseparatesourcesifthisangleequals2times10-4 rad This gives D=75kmInotherwordsifthecarisapproachingyouona straight road then you will be able to distinguish the two headlightsasseparatesourceswhenthecaris75kmawayfrom you Actually because of the structure of the retina and optical defects the resolving power of the average eye is about 6 times 10-4 rad This means that the car is more likely tobe25kmawaybeforeyouresolvetheheadlights

Astronomical telescope

Letusreturntotheexampleofthebinarystarsdiscussedatthe beginning of this section on resolution The stars Kruger A and B form a binary system The average separation of the stars is 14 times 1012mandtheiraveragedistancefromEarthis12times1017mWhenviewedthroughatelescopeonEarththe system will therefore subtend an angle

12

17

14 1012 10

θtimes

=times

= 12times10-5 rad at the objective lens of the telescope Assuming that the average wavelength of the light emitted bythestarsis500nmthenifthetelescopeistoresolvethesystem into two separate images it must have a minimum diameter Dwhere12times10-5 =

7122 500 10D

minustimes times

This gives D=0050m which is about 5 cm So thisparticular system is easily resolved with a small astronomical telescope

Exercise 961 ItissuggestedthatusingtheISASVLBAitwould

bepossibletoldquoseerdquoagrainofriceatadistanceof5000kmEstimatetheresolvingpoweroftheVLBA

2 Thedistancefromtheeyelenstotheretinais20mm The light receptors in the central part of the retinaareabout5times10-6apartDeterminewhetherthe spacing of the receptors will allow for the eye to resolve the headlights in the above discussion whentheyare25kmfromtheeye

3 ThediameterofPlutois23times106 m and its averagedistancefromEarthis60times1012 m EstimatetheminimumdiameteroftheobjectiveofatelescopethatwillenablePlutotobeseenasadisc as opposed to a point source

95 Doppler effect

Essential ideaTheDopplereffectdescribesthephenomenonofwavelengthfrequencyshiftwhenrelativemotionoccurs

Understandingsbull TheDopplereffectforsoundwavesandlightwaves

The Doppler effect for sound waves and light waves

Consider two observers A and B at rest with respect to asoundsourcethatemitsasoundofconstantfrequencyf Clearly both observers will hear a sound of the same frequencyHoweversupposethatthesourcenowmovesat constant speed towards A A will now hear a sound of frequencyfA that is greater than f and B will hear a sound of frequency fB that is less than f This phenomenon is known as the Doppler Effect orDopplerPrinciple after C J Doppler(1803-1853)

The same effect arises for an observer who is either moving towards or away from a stationary source

Figure 930 shows the waves spreading out from a stationary source that emits a sound of constant frequency f The observersAandBhearasoundofthesamefrequency

A B

wavefront

S

Figure 930 Sound waves from a stationary source

Suppose now that the source moves towards A withconstant speed vFigure9310(a)showsasnapshotofthenew wave pattern

NATURE OF SCIENCETechnology Although originally based on physical observations of the pitch of fast moving sources of sound the Doppler effect has an important role in many different areas such as evidence for the expansion of the universe and generating images used in weather reports and in medicine (55)

copy IBO 2014

PHYSICS 2015indb 303 150514 337 PM

Chapter 9

304

AH

L

sourceA B

smallerwavelength

largerwavelength

Figure 931 (a) Sound waves from a moving source

The wavefronts are now crowded together in the direction of travel of the source and stretched out in the opposite direction This is why now the two observers will now hear notesofdifferentfrequenciesHowmuchthewavesbunchtogether and how much they stretch out will depend on the speed vEssentially A

A

cfλ

= and BB

cfλ

= where λA lt λB and v is the speed of sound

IfthesourceisstationaryandAismovingtowardsitthenthe waves from the source incident on A will be bunched up If A is moving away from the stationary source then the waves from the source incident on A will be stretched out

Christian Doppler(1803ndash1853)actuallyappliedtheprinciple(incorrectlyasithappens)totryandexplainthecolourofstars However the Doppler effect does apply to light aswellastosoundIfalightsourceemitsalightoffrequencyf then if it is moving away from an observer the observer willmeasurethelightemittedashavingalowerfrequencythan fSincethesensationofcolourvisionisrelatedtothefrequencyoflight(bluelightisofahigherfrequencythanred light) light emitted by objects moving way from anobserver is often referred to as being red-shifted whereas if the object is moving toward the observer it is referred to as blue-shiftedThisideaisusedinOptionE(Chapter16)

We do not need to consider here the situations where either the source or the observer is accelerating In a situation for example where an accelerating source is approaching astationaryobserverthentheobserverwillhearasoundof ever increasing frequency This sometimes leads toconfusion in describing what is heard when a source approaches passes and then recedes from a stationaryobserver Suppose for example that you are standingon a station platform and a train sounding its whistle is approaching at constant speed What will you hear as the train approaches and then passes through the station As the train approaches you will hear a sound of constant pitch but increasing loudness The pitch of the sound will be greater than if the train were stationary As the train passes through the station you will hear the pitch change at the momentthetrainpassesyoutoasoundagainofconstantpitch The pitch of this sound will be lower than the sound of the approaching train and its intensity will decrease as the train recedes from you What you do not hear is a sound of increasing pitch and then decreasing pitch

The Doppler equations for sound

Although you will not be expected in an IB examination to derivetheequationsassociatedwithaspectsoftheDopplereffectyouwillbeexpectedtoapplythemForcompletenesstherefore thederivationoftheequationsassociatedwiththeDopplereffectasoutlinedaboveisgivenhere

S primeS O

vs

Figure 931 (b)

InFigure931(b)theobserverOisatrestwithrespecttoasourceofsoundSismovingwithconstanspeedvs directly towards O The source is emitting a note of constant frequencyf and the speed of the emitted sound is v

S shows the position of the source ∆t later When the sourceisatresttheninatime∆t the observer will receive f∆t waves and these waves will occupy a distance v∆t ie

v f

λ =f∆tv∆t =

(Because of the motion of the source this number of waves will now occupy a distance (v∆t ndash vs∆t)The lsquonewrsquowavelength is therefore

v ndash vs

fλrsquo =

f∆tv∆t ndash vs∆t

=

If fisthefrequencyheardbyOthen

vfλ

= or

vf

λ = = sv vfminus

From which

s

vf fv v

=minus

Dividingthroughbyv gives

Equation1111

1 sf f v

v=

minus

If the source is moving away from the observer then we have

s

vf fv v

=minus

1

1 sf vv

=+

Equation112

PHYSICS 2015indb 304 150514 337 PM

Wave Phenomena

305

AH

L

We now consider the case where the source is stationary and the observerismoving towardsthesource with speed v0 In this situation the speed of the sound waves as measured by the observer will be v0 + v We therefore have that

v0 + v = ffraslλ = vfλ

times

From which

o1 vf fv

= +

Equation113

If the observerismovingawayfromthesource then

o1 vf fv

= minus

Fromequation(113)wehavethat

o o1 v vf f f f f fv v

∆ = minus = + minus =

Equation114

Thevelocitiesthatwerefertointheaboveequationsarethevelocities with respect to the medium in which the waves fromthesourcetravelHoweverwhenwearedealingwithalight source it is the relative velocity between the source and the observer that we must consider The reason for this is that lightisuniqueintherespectthatthespeedofthelightwavesdoes not depend on the speed of the source All observers irrespective of their speed or the speed of the source will measure the same velocity for the speed of light This is oneof thecornerstonesof theSpecialTheoryofRelativitywhichisdiscussedinmoredetailinOptionH(Chapter18)When applying the Doppler effect to light we aremainlyconcerned with the motion of the source We look here only at the situation where the speed of the source v is much smaller than the speed of light cinfreespace(v ltlt c)Underthesecircumstanceswhenthesourceismovingtowardstheobserverequation111becomes

vf f f fc

minus = ∆ = Equation115

andwhenthesourceismovingawayfromtheobserverequation112becomes vf f f f

cminus = ∆ = minus

Provided that v ltlt c these same equations apply for astationary source and moving observer

We look at the following example and exercise

Example

Asourceemitsasoundoffrequency440HzItmovesinastraight line towards a stationary observer with a speed of 30ms-1Theobserverhearsasoundoffrequency484HzCalculate the speed of sound in air

Solution

We use equation 111 and substitute f= 484 Hz f = 440 Hz

and vs = 30 m s-1

therefore 1

484 440 301v

=minus

such that 30 4401

484vminus = to

give v = 330 m s-1

Example

A particular radio signal from a galaxy is measured as havingafrequencyof139times109HzThesamesignalfromasourceinalaboratoryhasafrequencyof142times109Hz

Suggestwhy thegalaxy ismovingaway fromEarthandcalculateitsrecessionspeed(iethespeedwithwhichitismovingawayfromEarth)

Solution

The fact that the frequency from the moving source is less than that when it is stationary indicates that it is moving away from the stationary observer ie Earth

Using vf fc

∆ = we have

8 96

9

3 10 (142 139) 10 634 10142 10

c fvf∆ times times minus times

= = = timestimes

m s-1

It is usual when dealing with the Doppler effect of light to express speeds as a fraction of c So in this instance we have v = 0021 c

Using the Doppler effectWe have seen in the above example and exercise how the Dopplereffectmaybeusedtomeasuretherecessionspeedof distant galaxies The effect is also used to measure speed inothersituationsHerewewilllookatthegeneralprincipleinvolved in using the Doppler effect to measure speedFigure932showsasource(thetransmitter)thatemitseithersound or em waves of constant frequency f The waves

PHYSICS 2015indb 305 150514 337 PM

Chapter 9

306

AH

L

fromthesourceare incidentonareflectorthat ismovingtowards the transmitter with speed vThereflectedwavesare detected by the receiver placed alongside the transmitter

reector

v

transmitter

receiver

f f

f f

Figure 932 Using the Doppler effect to measure speed

We shall consider the situation where v ltlt c where c is the speed of the waves from the transmitter

For the reflector receiving waves from the transmitterit is effectively an observer moving towards a stationary sourceFromequation(114) itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation116

For the receiver receivingwaves from the reflector it iseffectively a stationary observer receiving waves from a movingsourceFromequation(115)itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation117

Ifweaddequations(116)and(117)wegetthatthetotalDopplershiftatthereceiver∆f is

v vf f f f fc c

minus = ∆ = +

But 1 vf fc

= +

hence

1 v v vf f fc c c

∆ = + +

But since v ltlt cwe can ignore the term2

2

vc when we

expandthebracketintheaboveequation

Therefore we have2vf fc

∆ = Equation118

If v asymp c then we must use the full Doppler equationsHowever for em radiationwewill alwaysonly considersituations in which v ltlt c

Example

Thespeedofsoundinbloodis1500times103 m s-1 Ultrasound offrequency100MHzisreflectedfrombloodflowinginan arteryThe frequencyof the reflectedwaves receivedbackatthetransmitteris105MHzEstimatethespeedofthebloodflowintheartery

Solution

Using equation (118) we have

6 63

2005 10 1015 10

vtimes = times

timesto give v asymp 36 m s-1 (We have assumed that the ultrasound is incident at right angles to the blood flow)

Exercise 971 Judy is standing on the platform of a station A

high speed train is approaching the station in a straight line at constant speed and is sounding its whistleAsthetrainpassesbyJudythefrequencyof the sound emitted by the whistle as heard by Judychangesfrom640Hzto430HzDetermine

(a) thespeedofthetrain

(b) thefrequencyofthesoundemittedbythewhistle as heard by a person on the train (Speedofsound=330ms-1)

2 AgalaxyismovingawayfromEarthwithaspeedof00500cThewavelengthofaparticularspectral line in light emitted by atomic hydrogen inalaboratoryis656times10-7 m Calculate the valueofthewavelengthofthislinemeasuredinthelaboratoryinlightemittedfromasourceofatomic hydrogen in the galaxy

PHYSICS 2015indb 306 150514 337 PM

10 Fields

Contents

101 ndash Describing fields

102 ndash Fields at work

Essential Ideas

Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Similar approaches can be taken in analysing electrical and gravitational potential problems copy IBO 2014

PHYSICS 2015indb 307 150514 337 PM

Chapter 10

308

AH

L

101 Describing fields

Essential idea Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Understandingsbull Gravitationalfieldsbull Electrostaticfieldsbull Electricpotentialandgravitationalpotentialbull Fieldlinesbull Equipotentialsurfaces

Gravitational potentialWehaveseenthatifweliftanobjectofmassm to a height h abovethesurfaceoftheEarththenitsgainingravitationalpotential energy is mghHoweverthisisbynomeansthefullstoryForastartwenowknowthatgvarieswithh and also the expression really gives a difference in potentialenergybetweenthevaluethattheobjecthasattheEarthrsquossurfaceandthevaluethatithasatheighthSowhatwereallyneedisazeropointCanwefindapointwherethepotentialenergyiszeroandusethispointfromwhichtomeasure changes in potential energy

The point that is chosen is in fact infinity At infinity the gravitationalfieldstrengthofanyobjectwillbezeroSoletusseeifwecandeduceanexpressionforthegaininpotentialenergyofanobjectwhenitisldquoliftedrdquofromthesurfaceoftheEarthtoinfinityThisineffectmeansfindingtheworknecessarytoperformthistask

δr

r r δr+

r infin=

mg

AB

Me

Figure 1001 Gravitational forces

Inthediagramweconsidertheworknecessarytomovethe particle of mass m a distance δr in thegravitationalfield of the Earth

The force on the particle at A is FG Mem

r2----------------=

IftheparticleismovedtoB then since δr isverysmallwe can assume that the field remains constant over thedistance ABTheworkδW done againstthegravitationalfieldoftheEarthinmovingthedistanceABis

δWGMem

r2----------------δrndash=

(rememberthatworkdoneagainstaforceisnegative)

TofindthetotalworkdoneW in going from the surface oftheEarthtoinfinitywehavetoaddalltheselittlebitsofworkThisisdonemathematicallybyusingintegralcalculus

WG Me m

r2----------------ndash

rdR

infin

int G Me m 1

r2----- rd

R

infin

intndash G Me m 1r---ndash

R

infinndash= = =

GMe m 0 1R---ndash

ndashndash=

G MemR----------------ndash=

HencewehavewhereRistheradiusoftheEarththattheworkdonebythegravitationalfieldinmovinganobjectofmass m from R(surfaceoftheEarth)toinfinityisgivenby

WGMe m

R---------------- ndash=

We can generalise the result by calculating the worknecessary per unit mass to take a small mass fromthe surface of the Earth to infinity This we call thegravitational potential Vie

Wewouldgetexactlythesameresultifwecalculatedtheworkdonebythefieldtobringthepointmassfrominfinityto the surface of Earth In this respect the formal definition ofgravitationalpotentialatapointinagravitationalfieldis therefore defined as the work done per unit mass in bringing a point mass from infinity to that point

Clearly then the gravitational potential at any point inthe Earthrsquos field distance r from the centre of the Earth (providingr gt R)is

VGMe

r------------ ndash=

Thepotentialisthereforeameasureoftheamountofworkthathastobedonetomoveparticlesbetweenpointsinagravitationalfieldanditsunit is theJkgndash1 We also note thatthepotentialisnegativesothatthepotentialenergy

NATURE OF SCIENCEParadigm shift The move from direct observable actions being responsible for influence on an object to acceptance of a fieldrsquos ldquoaction at a distancerdquo required a paradigm shift in the world of science (23)

copy IBO 2014

PHYSICS 2015indb 308 150514 337 PM

Fields

309

AH

L

aswemoveawayfromtheEarthrsquossurfaceincreasesuntilitreachesthevalueofzeroatinfinity

IfthegravitationalfieldisduetoapointmassofmassmthenwehavethesameexpressionasaboveexceptthatMe is replaced by m andmustalsoexclude thevalueof thepotential at the point mass itself ie at r = 0

WecanexpressthegravitationalpotentialduetotheEarth(orduetoanysphericalmass)intermsofthegravitationalfield strength at its surface

AtthesurfaceoftheEarthwehave

g 0ndash Re GMRe

---------ndash= e

Sothat

g 0Re2 GM= e

Hence at a distance r from the centre of the Earth the gravitationalpotentialVcanbewrittenas

VGMe

r------------ndashg 0 Re

2

r------------ndash= =

The potential at the surface of the Earth

(r = Re)istherefore-g0Re

It is interesting to see how the expression for thegravitational potential ties in with the expression mgh ThepotentialatthesurfaceoftheEarthis-g0Re (see the exampleabove)andataheighthwillbe-g0g 0 Re h+( )ndash ifweassume that g0doesnotchangeover thedistanceh The differenceinpotentialbetweenthesurfaceandtheheighth is therefore g0hSotheworkneededtoraiseanobjectof mass m to a height h is mgh ie m timesdifference ingravitationalpotential

This we have referred to as the gain in gravitationalpotentialenergy(see235)

However this expression can be extended to any twopoints inanygravitationalfieldsuch that ifanobjectofmass mmovesbetween twopointswhosepotentialsareV1 and V2 respectively then the change in gravitationalpotentialenergyoftheobjectism(V1 ndash V2)

Gravitational potential gradient

Let us consider now a region in space where thegravitationalfieldisconstantInFigure912thetwopointsAandBareseparatedbythedistance∆x

A B

direction of uniform gravitational eld of strength I

∆x

Figure 1002 The gravitational potential gradient

The gravitational field is of strength I and is in the direction shownThe gravitational potential at A is V andatBisV +∆V

Theworkdone is takingapointmassmfromAtoB isF∆x = mI∆x

Howeverbydefinitionthisworkisalsoequalto-m∆V

Therefore mI∆x=-m∆V

or VIx

∆= minus

∆Effectivelythissaysthatthemagnitudeofthegravitationalfield strength is equal to the negative gradient of thepotential If I is constant then V is a linear function of x and Iisequaltothenegativegradientofthestraightlinegraph formed by plotting V against x If I is not constant (asusuallythecase)thenthemagnitudeofI at any point in the field can be found by find the gradient of the V-x graph at that point An example of such a calculation can befoundinSection929

For those of you who do HL maths the relationship between field and potential is seen to follow from the expression for the potential of a point mass viz

mV Gr

= minus

2

ddV mG Ir r

minus = + =

Potential due to one or more point masses

Gravitationalpotentialisascalarquantitysocalculatingthepotential due to more than one point mass is a matter of simpleadditionSoforexamplethepotentialV due to the Moon and Earth and a distance x fromthecentreofEarthonastraightlinebetweenthemisgivenbytheexpression

E MM MV Gx r x

= minus + minus

whereME=massofEarthMM= mass of Moon and r = distancebetweencentreofEarthandMoon

PHYSICS 2015indb 309 150514 337 PM

Chapter 10

310

AH

L

Equipotentials and field linesIfthegravitationalpotentialhasthesamevalueatallpointson a surface the surface is said to be an equipotential surfaceSoforexampleifweimagineasphericalshellaboutEarthwhosecentrecoincideswiththecentreofEarththisshellwillbeanequipotentialsurfaceClearlyifwerepresentthegravitationalfieldstrengthbyfieldlinessincethelinesldquoradiaterdquooutfromthecentreofEarththentheselineswillbeatrightanglestothesurfaceIfthefieldlineswerenotnormaltotheequipotentialsurfacethentherewouldbeacomponentof the field parallel to the surfaceThis wouldmean thatpointsonthesurfacewouldbeatdifferentpotentialsandsoitwouldnolongerbeanequipotentialsurfaceThisofcourseholdstrueforanyequipotentialsurface

Figure 1003 shows the field lines and equipotentials fortwopointmassesm

m m

Figure 1003 Equipotentials for two point masses

It is worth noting that we would get exactly the samepattern ifwewere to replace thepointmasseswith twoequalpointcharges(See935)

Escape speedThe potential at the surface of Earth is which meansthattheenergyrequiredtotakeaparticleofmass mfromthesurfacetoinfinityisequalto

ButwhatdoesitactuallymeantotakesomethingtoinfinityWhentheparticleisonthesurfaceoftheEarthwecanthinkofitassittingatthebottomofaldquopotentialwellrdquoasinfigure1004

particle

surface of Earth

innity

GMR---------

FIgure 1004 A potential well

Theldquodepthrdquoofthewellis and if the particle gains an amount of kinetic energy equal to wherem is its massthenitwillhavejustenoughenergytoldquoliftrdquoitoutofthewell

In reality it doesnrsquot actually go to infinity it justmeansthat the particle is effectively free of the gravitationalattractionof theEarthWe say that ithas ldquoescapedrdquo theEarthrsquosgravitationalpullWemeetthisideainconnectionwithmolecular forcesTwomolecules in a solidwill sitat their equilibrium position the separation where therepulsiveforceisequaltotheattractiveforceIfwesupplyjust enough energy to increase the separation of themolecules such that they are an infinite distance apart then intermolecularforcesnolongeraffectthemoleculesandthesolidwillhavebecomealiquidThereisnoincreaseinthekineticenergyofthemoleculesandsothesolidmeltsat constant temperature

Wecancalculatetheescapespeedofanobjectveryeasilyby equating the kinetic energy to the potential energysuch that

12---mvesc ape

2 GMe mRe

----------------=

vesc aperArr2GMe

Re--------------- 2g0 Re = =

Substituting for g0 and Regivesavalueforvescape of about 11kmsndash1 from the surface of the Earth

Youwillnotethattheescapespeeddoesnotdependonthemassoftheobjectsincebothkineticenergyandpotentialenergy are proportional to the mass

IntheoryifyouwanttogetarockettothemoonitcanbedonewithoutreachingtheescapespeedHoweverthiswouldnecessitate anenormousamountof fuel and it islikely that therocketplus fuelwouldbesoheavythat itwouldnevergetoffthegroundItismuchmorepracticalto accelerate the rocket to the escape speed fromEarthorbitandthenintheoryjustlaunchittotheMoon

Example

Use the followingdata todetermine thepotential at thesurface of Mars and the magnitude of the acceleration of free fall

massofMars =64times1023kg

radiusofMars =34times106 m

Determinealsothegravitationalfieldstrengthatadistanceof 68 times 106mabovethesurfaceofMars

PHYSICS 2015indb 310 150514 337 PM

Fields

311

AH

L

Solution

= minus = minus times times = times23

11 76

64 1067 10 ndash13 10

34 10MV GR

minus timestimes

Nkg-1

But V = -g0R

Therefore 7

0 6

13 10 3834 10

VgR

times= minus = =

timesm s-2

To determine the field strength gh at 68 times 106 m above the surface we use the fact that 0 2

Mg GR

= such that GM = g0R2

Therefore 2 2

0h 2 2 2

h h

38 (34) 042(102)

g RGMgR R

times= = = = m s-2

(the distance from the centre is 34 times 106 + 68 times 106 = 102 times 106 m)

Exercise 1011 Thegraphbelowshowshowthegravitational

potentialoutsideoftheEarthvarieswithdistancefrom the centre

0 10 20 30 40 50 60 70

ndash1

ndash2

ndash3

ndash4

ndash5

ndash6

ndash7

r m times 106

V

Jkgndash1

times 10

7

(a) Usethegraphtodeterminethegainingravitationalpotentialenergyofasatelliteofmass200kgasitmovesfromthesurfaceof the Earth to a height of 30 times 107mabovetheEarthrsquossurface

(b) Calculatetheenergyrequiredtotakeit to infinity

(c) DeterminetheslopeofthegraphatthesurfaceoftheEarthmCommenton youranswer

Electric potential energy

The concept of electric potential energy was developedwith gravitational potential energy in mind Just as anobjectnearthesurfaceoftheEarthhaspotentialenergybecause of its gravitational interaction with the Earthsotoothereiselectricalpotentialenergyassociatedwithinteracting charges

Letusfirstlookatacaseoftwopositivepointchargeseachof 1μC that are initially bound together by a thread in a vacuuminspacewithadistancebetweenthemof10cmasshowninFigure1005Whenthethreadiscutthepointchargesinitiallyatrestwouldmoveinoppositedirectionsmovingwithvelocitiesv1 and v2 along the direction of the electrostatic force of repulsion

BEFORE AFTER

v v21

Figure 1005 Interaction of two positive particles

The electric potential energybetweentwopointchargescan be found by simply adding up the energy associated witheachpairofpointchargesForapairofinteractingchargestheelectricpotentialenergyisgivenby

∆U = ∆Ep + ∆Ek = ∆W = ∆Fr = kqQ

____ r2 times r = kqQ

____ r

Becausenoexternalforceisactingonthesystemtheenergyand momentum must be conserved Initially Ek = 0 and Ep =kqQr=9times109 times 1 times 10-1201m=009JWhentheyareagreatdistance fromeachotherEpwillbenegligibleThefinalenergywillbeequaltofrac12mv1

2+frac12mv22=009J

Momentum isalsoconservedand thevelocitieswouldbethe same magnitude but in opposite directions

Electric potential energyismoreoftendefinedintermsofapointchargemovinginanelectricfieldas

lsquothe electricpotential energybetweenany twopoints inan electric field is defined as negativeoftheworkdone by anelectricfieldinmovingapoint electric chargebetweentwolocationsintheelectricfieldrsquo

ΔU = ΔEp = -ΔW = -Fd = qEx

ΔU = ΔEp +ΔEk = ΔW = ΔFr = kqQ r2 x r = kqQ

wherexisthedistancemovedalong(oroppositeto)thedirection of the electric field

Electric potential energyismeasuredinjoule(J)Justasworkisascalarquantitysotooelectricalpotentialenergyis a scalar quantity The negative of the work done by an electric field in moving a unit electric charge between two points is independent of the path taken Inphysicswesaytheelectricfieldisalsquoconservativersquo field

PHYSICS 2015indb 311 150514 337 PM

Chapter 10

312

AH

L

Supposeanexternalforcesuchasyourhandmovesasmallpositive point test charge in the direction of a uniformelectric fieldAs it ismoving itmust be gaining kineticenergyIfthisoccursthentheelectricpotentialenergyofthe unit charge is changing

InFigure1006apointcharge+qismovedbetweenpointsAandBthroughadistancex in a uniform electric field

B

A

x

+ q

Figure 1006 Movement of a positive point charge in a uniform field

InordertomoveapositivepointchargefrompointAtopointBanexternal forcemustbeapplied to thechargeequaltoqE (F = qE)

Since the force is applied through a distance xthennegativeworkhas to bedone tomove the charge because energyis gainedmeaning there is an increaseelectric potential energybetweenthetwopointsRememberthattheworkdoneisequivalenttotheenergygainedorlostinmovingthe charge through the electric field The concept of electric potential energy is only meaningful as the electric field whichgeneratestheforceinquestionisconservative

W F xtimes E q x times= =

θ x cos θ x

Figure 1007 Charge moved at an angle to the field

If a chargemoves at an angle θ to an electric field thecomponent of the displacement parallel to the electric fieldisusedasshowninFigure918

W F x Eq x θcostimes= =

Theelectricpotentialenergyisstoredintheelectricfieldand theelectricfieldwill return theenergy to thepointcharge when required so as not to violate the Law ofconservationofenergy

Electric potential

The electric potential at a point in an electric field is defined as being the work done per unit charge in bringing a small positive point charge from infinity to that point

ΔV = Vinfin ndash Vf = ndash W ___ q

Ifwedesignatethepotentialenergytobezeroatinfinitythen it follows thatelectricpotentialmustalsobezeroat infinity and the electric potential at any point in an electricfieldwillbe

ΔV = ndash W ___ q

NowsupposeweapplyanexternalforcetoasmallpositivetestchargeasitismovedtowardsanisolatedpositivechargeTheexternalforcemustdoworkonthepositivetestchargetomoveittowardstheisolatedpositivechargeandtheworkmustbepositivewhiletheworkdonebytheelectricfieldmustthereforebenegativeSotheelectricpotentialatthatpointmustbepositiveaccordingtotheaboveequationIfanegativeisolatedchargeisusedtheelectricpotentialatapointonthepositivetestchargewouldbenegativePositivepoint chargesof their ownaccordmove fromaplaceofhighelectricpotential toaplaceof lowelectricpotentialNegative point charges move the other way from lowpotentialtohighpotentialInmovingfrompointAtopointBinthediagramthepositivecharge+qismovingfromalowelectricpotentialtoahighelectricpotential

In thedefinitiongiven the termldquoworkperunitchargerdquohas significance If the test charge is +16 times 10-19Cwherethe charge has a potential energy of 32 times 10-17Jthenthepotentialwouldbe32times10-17J+16times10-19C=200JC-1 Now if the charge was doubled the potential wouldbecome 64times10-17 J However the potential per unitchargewouldbethesame

Electric potential is a scalar quantityandithasunitsJC-1 orvoltswhere1voltequalsone joulepercoloumbThevoltallowsustoadoptaunitfortheelectricfieldintermsofthevolt

PreviouslytheunitfortheelectricfieldwasNCndash1

W = qV and F = qEsoW ___ V = F __ E

E = FV ___ W = FV ___ Fm V mndash1

That is the units of the electric field E can also beexpressed as V mndash1

The work done per unit charge in moving a point charge between two points in an electric field is again independent of the path taken

PHYSICS 2015indb 312 150514 337 PM

Fields

313

AH

L

Electric potential due to a point charge

Letus takeapoint rmetres froma chargedobjectThepotentialatthispointcanbecalculatedusingthefollowing

W = ndashFr = ndashqV and F = ndash q1 q2 _____ 4πε0r

2

Therefore

W = ndash q1 q2 _____ 4πε0r

2 times r = ndash q1 q2 _____ 4πε0r

= ndashq1 times q2 _____ 4πε0r

= ndashq1V

That is V q

4πε0r--------------- =

Orsimply

V kqr------

=

Example

Determinehowmuchworkisdonebytheelectricfieldofpointcharge150μCwhenachargeof200μCismovedfrom infinity to apoint 0400m from thepoint charge(Assumenoaccelerationofthecharges)

Solution

The work done by the electric field is W = -qV = -14πε0 times q times (Q rinfin - Q r0400)

W = (- 200 times 10-6 C times 900 times 109 NmC-2 times 150 times 10-6 C) divide 0400 m = - 0675 J

An external force would have to do +0675 J of work

Electric field strength and electric potential gradient

Letus lookbackatFigure1006Supposeagain that thecharge+qismovedasmalldistancebyaforceFfromAtoBsothattheforcecanbeconsideredconstantTheworkdoneisgivenby

∆W F x∆times=

TheforceFandtheelectricfieldEareoppositelydirectedandweknowthat

F = -qE and ΔW = q ΔV

Thereforetheworkdonecanbegivenas

q ΔV = -q E Δx

Therefore E V∆x∆-------

ndash=

The rate of change of potential ΔV at a point with respect to distance Δx in the direction in which the change is maximum is called the potential gradient We say that the electric field = - the potential gradient and the units are Vm-

1 From the equation we can see that in a graph of electric potential versus distance the gradient of the straight line equals the electric field strength

In reality if a charged particle enters a uniform electric field it will be accelerated uniformly by the field and its kinetic energy will increase This is why we had to assume no acceleration in the last worked example

Ek∆ 12---

mv2 q E xsdot sdot q Vx---

xsdot sdot q Vsdot= = = =

Example

Determine how far apart two parallel plates must besituated so that a potential difference of 150 times 102 V produces an electric field strength of 100 times 103 NC-1

Solution

Using E V∆x------- xhArrndash V∆

E-------15 102 Vtimes

100 103 N C 1ndashtimes------------------------------------------= = =

= 150 times 10-1

The plates are 150 times 10-1 m apart

The electric field and the electric potential at a point due to an evenly distributed charge +q on a sphere can be represented graphically as in Figure 1008

r

r

V

E

r0

r0

r0

Charge of +Q evenly distributed over surface

E 14πε0------------ Q

r2----- r r0gt=

V 14πε0------------Q

r----r r0gt=

Figure 1008 Electric field and potential due to a charged sphere

PHYSICS 2015indb 313 150514 337 PM

Chapter 10

314

AH

L

When the sphere becomes charged we know that thechargedistributes itselfevenlyoverthesurfaceThereforeeverypartofthematerialoftheconductorisatthesamepotential As the electric potential at a point is defined as beingnumericallyequaltotheworkdoneinbringingaunitpositivechargefrominfinitytothatpointithasaconstantvalueineverypartofthematerialoftheconductor

Since the potential is the same at all points on the conducting surfacethen∆V∆xiszeroButE=ndash∆V∆xThereforethe electric field inside the conductor is zero There is no electric field inside the conductor

SomefurtherobservationsofthegraphsinFigure1009are

bull OutsidethespherethegraphsobeytherelationshipsgivenasE α 1 r2 and V α 1 r

bull Atthesurfacer = r0Thereforetheelectricfieldandpotentialhavetheminimumvalueforr at this point and this infers a maximum field and potential

bull Insidethespheretheelectricfieldiszerobull Insidethespherenoworkisdonetomoveacharge

fromapointinsidetothesurfaceThereforethereisno potential difference and the potential is the same asitiswhenr = r0

Similargraphscanbedrawnfortheelectricfieldintensityandthe electric potential as a function of distance from conducting parallelplatesandsurfacesandthesearegiveninFigure1009

Potential plot E eld plotE eld

++++

ndashndashndashndash

xx x

+

x x

ndashx x

+ ndashx

x

+ +

x

x

Figure 1009 Electric field and electric potential at a distance from a charged surface

Potential due to one or more point charges

The potential due to one point charge can be determined byusingtheequationformula

V = kq r

Example 1

Determine the electric potential at a point 20 times 10-1 m fromthecentreofan isolatedconducting spherewithapointchargeof40pCinair

Solution

Using the formula

V = kq r we have

V90 10 9times( ) 40 10 12ndashtimes( )times

20 10 1ndashtimes( )------------------------------------------------------------------ 018 V= =

the potential at the point is 180 times 10-1 V

The potential due to a number of point charges can be determined by adding up the potentials due to individual point charges because the electric potential at any point outside a conducting sphere will be the same as if all the charge was concentrated at its centre

Example 2

Threepointchargesofareplacedattheverticesofaright-angledtriangleasshowninthediagrambelowDeterminetheabsolutepotentialat the+20μCchargeduetothetwoothercharges

- 6μC

+3μC +2μC

4 m

3 m

Solution

The electric potential of the +2 μC charge due to the ndash 6 μC charge is

V = (9 times 109 Nm2C-2 times -6 times 10-6 C) divide (radic 32 + 42) m = - 108 times 104 V

PHYSICS 2015indb 314 150514 338 PM

Fields

315

AH

L

Insummarywecanconcludethat

bull No work is done to move a charge along an equipotential

bull Equipotentials are always perpendicular to the electric lines of force

Figure1011and1012showsomeequipotential lines fortwo oppositely charged and identically positive spheresseparated by a distance

+ve ndashve

equipotential lines

Figure 1011 Equipotential lines between two opposite charges

equipotential lines

Figure 1012 Equipotential lines between two charges which are the same

+

ndash ndash ndash ndash

+ + +

40 V

30 V

20 V

10 V

50 V

Figure 1013 Equipotential lines between charged parallel plates

Figure1013showstheequipotentiallinesbetweenchargedparallel plates Throughout this chapter the similarities and differences between gravitational fields and electric fieldshavebeendiscussedTherelationshipsthatexistsbetweengravitationalandelectricquantitiesandtheeffectsofpointmassesandchargesissummarisedinTable1014

The electric potential of the +2 μC charge due to the +3 μC charge is

V = (9 times 109 Nm2C-2 times 3 times 10-6 C) divide 3m = 9 times 103 V

The net absolute potential is the sum of the 2 potentials- 108 times 104 V + 9 times 103 V =- 18 times 103 V

The absolute potential at the point is - 18 times 103 V

Equipotential surfacesRegions in space where the electric potential of a chargedistributionhasaconstantvaluearecalledequipotentials The placeswherethepotentialisconstantinthreedimensionsarecalled equipotential surfacesandwheretheyareconstantintwodimensionstheyarecalledequipotential lines

TheyareinsomewaysanalogoustothecontourlinesontopographicmapsInthiscasethegravitationalpotentialenergy is constant as amassmoves around the contourlines because the mass remains at the same elevationabovetheEarthrsquossurfaceThegravitationalfieldstrengthacts in a direction perpendicular to a contour line

Similarlybecausetheelectricpotentialonanequipotentiallinehas thesamevalueanelectric forcecandonoworkwhena testchargemovesonanequipotentialThereforethe electric field cannot have a component along anequipotentialandthusitmustbeeverywhereperpendiculartotheequipotentialsurfaceorequipotentiallineThisfactmakesiteasytoplotequipotentialsifthelinesofforceorlinesofelectricfluxofanelectricfieldareknown

Forexamplethereareaseriesofequipotentiallinesbetweentwoparallelplateconductorsthatareperpendiculartotheelectric fieldTherewill be a series of concentric circles(eachcirclefurtherapartthanthepreviousone)thatmapouttheequipotentialsaroundanisolatedpositivesphereThe lines of force and some equipotential lines for anisolatedpositivesphereareshowninFigure1010

Lines ofequipotential

Figure 1010 Equipotentials around an isolated positive sphere

PHYSICS 2015indb 315 150514 338 PM

Chapter 10

316

AH

L

Gravitational quantity Electrical quantityQ

uant

ities

g V∆x∆-------ndash= E V∆

x∆-------ndash=

Poin

t mas

ses a

nd ch

arge

s

V Gmr----

ndash= V 14πε0------------q

r---=

g G m

r2-----ndash= E 14πε0------------ q

r2-----=

F Gm1 m2

r2--------------= F 1

4πε0------------

q1q2

r2-----------=

Figure 1014 Formulas

Example

Deduce the electric potential on the surface of a gold nucleus that has a radius of 62 fm

Solution

Using the formula

V = kq r and knowing the atomic number of gold is 79 We will assume the nucleus is spherical and it behaves as if it were a point charge at its centre (relative to outside points)

V = 90 times 109 Nm2C-2 times 79 times 16 times 10-19 C divide 62 times 10-15 m = 18 times 107 V

The potential at the point is 18 MV

Example

Deduce the ionisation energy in electron-volts of theelectron in the hydrogen atom if the electron is in its ground state and it is in a circular orbit at a distance of 53 times 10-11 m from the proton

Solution

This problem is an energy coulombic circular motion question based on Bohrrsquos model of the atom (not the accepted quantum mechanics model) The ionisation energy is the energy required to remove the electron from the ground state to infinity The electron travels in a circular orbit and therefore has a centripetal acceleration The ionisation energy will counteract the coulombic force and the movement of the electron will be in the opposite direction to the centripetal force

Total energy = Ek electron + Ep due to the proton-electron interaction

ΣF = kqQ r2 = mv2 r and as such mv2 = = kqQ r

Therefore Ek electron = frac12 kqQ r

Ep due to the proton-electron interaction = - kqQ r

Total energy = frac12 kqQ r + - kqQ r = -frac12 kqQ r

= - 90 times 109 Nm2C-2 times (16 times 10-19 C)2 divide 53 times 10-11 m = -217 times 10-18 J

= -217 times 10-18 J divide 16 times 10-19 = -136 eV

The ionisation energy is 136 eV

Exercise 1021 ApointchargePisplacedmidwaybetweentwo

identicalnegativechargesWhichoneofthefollowingiscorrectwithregardstoelectricfieldandelectricpotentialatpointP

Electric field Electric potentialA non-zero zeroB zero non-zeroC non-zero non-zeroD zero zero

2 Twopositivechargedspheresaretiedtogetherinavacuumsomewhereinspacewheretherearenoexternal forces A has a mass of 25 g and a charge of20μCandBhasamassof15gandachargeof30μCThedistancebetweenthemis40cmTheyarethenreleasedasshowninthediagram

BEFORE AFTER

v1 v2

A B

(a) Determinetheirinitialelectricpotentialenergy in the before situation

(b) DeterminethespeedofsphereBafterrelease

3 Thediagrambelowrepresentstwoequipotentiallines in separated by a distance of 5 cm in a uniform electric field

+ + + + + + + +

ndash ndash ndash ndash ndash ndash ndash ndash

40 V

20 V 5 cm

PHYSICS 2015indb 316 150514 338 PM

Fields

317

AH

L

Determine the strength of the electric field

4 Thisquestionisabouttheelectricfieldduetoacharged sphere and the motion of electrons in that fieldThediagrambelowshowsanisolatedmetalsphereinavacuumthatcarriesanegativeelectriccharge of 60 μC

ndash

(a) Onthediagramdrawtheconventionalwayto represent the electric field pattern due to the charged sphere and lines to represent threeequipotentialsurfacesintheregionoutside the sphere

(b) Explainhowthelinesrepresentingtheequipotentialsurfacesthatyouhavesketchedindicatethatthestrengthoftheelectricfieldisdecreasingwithdistancefrom the centre of the sphere

(c) Theelectricfieldstrengthatthesurfaceofthe sphere and at points outside the sphere can be determined by assuming that the sphere acts as a point charge of magnitude 60 μC at its centre The radius of the sphere is 25 times 10ndash2 m Deduce that the magnitude of the field strength at the surface of the sphere is 86 times 107 Vmndash1

An electron is initially at rest on the surface of the sphere

(d) (i) Describethepathfollowedbytheelectronasitleavesthesurfaceofthesphere

(ii) Calculatetheinitialaccelerationofthe electron

5 Determinetheamountofworkthatisdoneinmovingachargeof100nCthroughapotentialdifferenceof150times102 V

6 Three identical 20 μC conducting spheres are placedatthecornersofanequilateraltriangleofsides 25 cm The triangle has one apex C pointing upthepageand2baseanglesAandBDeterminetheabsolutepotentialatB

7 Determinehowfaraparttwoparallelplatesmustbesituatedsothatapotentialdifferenceof250 times 102 V produces an electric field strength of 200 times 103 NC-1

8 Thegapbetweentwoparallelplatesis10times10-3mandthereisapotentialdifferenceof10times104 V betweentheplatesCalculate

i theworkdonebyanelectroninmovingfrom one plate to the other

ii thespeedwithwhichtheelectronreachesthe second plate if released from rest

iii theelectricfieldintensitybetweentheplates

9 Anelectronguninapicturetubeisacceleratedby a potential 25 times 103VDeterminethekineticenergygainedbytheelectroninelectron-volts

10 Determine the electric potential 20 times 10-2 m from achargeof-10times10-5 C

11 Determinetheelectricpotentialatapointmid-waybetweenachargeofndash20pCandanotherof+5pConthelinejoiningtheircentresifthecharges are 10 cm apart

12 During a thunderstorm the electric potential differencebetweenacloudandthegroundis10 times 109 V Determine the magnitude of the change in electric potential energy of an electron thatmovesbetweenthesepointsinelectron-volts

13 A charge of 15 μC is placed in a uniform electric fieldoftwooppositelychargedparallelplateswithamagnitudeof14times103 NC-1

(a) Determinetheworkthatmustbedoneagainstthefieldtomovethepointchargeadistance of 55 cm

(b) Calculatethepotentialdifferencebetweenthe final and initial positions of the charge

(c) Determinethepotentialdifferencebetweenthe plates if their separation distance is 15 cm

14 Duringaflashoflightningthepotentialdifferencebetweenacloudandthegroundwas12 times 109 V and the amount of transferred charge was32C

(a) Determinethechangeinenergyofthetransferred charge

(b) Iftheenergyreleasedwasallusedtoacceleratea1tonnecardeduceitsfinalspeed

(c) Iftheenergyreleasedcouldbeusedtomelticeat0degCdeducetheamountoficethatcould be melted

PHYSICS 2015indb 317 150514 338 PM

Chapter 10

318

AH

L

15 SupposethatwhenanelectronmovedfromAtoBin the diagram along an electric field line that the electric field does 36 times 10-19Jofworkonit

Determinethedifferencesinelectricpotential

(a) VB ndash VA(b) VC ndash VA(c) VC ndash VB

16 DeterminethepotentialatpointPthatislocatedatthecentreofthesquareasshowninthediagrambelow

- 6μC

+3μC +2μC

1m

1m

5μC

P

102 Fields at work

Essential idea Similar approaches can be taken inanalysingelectricalandgravitationalpotentialproblems

Understandingsbull Potentialandpotentialenergybull Potentialgradientbull Potentialdifferencebull Escapespeedbull Orbitalmotionorbitalspeedandorbitalenergybull Forcesandinverse-squarelawbehaviour

Orbital motion orbital speed and orbital energy

Although orbital motion may be circular elliptical orparabolic this sub-topic only dealswith circular orbitsThis sub-topic is not fundamentally new physics butan application that synthesizes ideas from gravitationcircularmotiondynamicsandenergy

TheMoonorbitstheEarthandinthissenseitisoftenreferredtoasasatelliteoftheEarthBefore1957itwastheonlyEarthsatelliteHoweverin1957theRussianslaunchedthefirstmanmadesatelliteSputnik1SincethisdatemanymoresatelliteshavebeenlaunchedandtherearenowliterallythousandsofthemorbitingtheEarthSomeareusedtomonitortheweathersome used to enable people to find accurately their position on thesurfaceoftheEarthmanyareusedincommunicationsandnodoubtsomeareusedtospyonothercountriesFigure1015showshowinprincipleasatellitecanbeputintoorbit

Theperson(whosesizeisgreatlyexaggeratedwithrespectto Earth) standing on the surface on the Earth throwssome stonesThe greater the speed with which a stoneis thrown the further it will land from her The pathsfollowedbythethrownstonesareparabolasByastretchof the imaginationwecanvisualisea situation inwhichastoneis thrownwithsuchaspeedthatbecauseof thecurvatureoftheEarthitwillnotlandonthesurfaceoftheEarthbutgointoldquoorbitrdquo(Path4onFigure1015)

NATURE OF SCIENCECommunication of scientific explanations The ability to apply field theory to the unobservable (charges) and the massively scaled (motion of satellites) required scientists to develop new ways to investigate analyse and report findings to a general public used to scientific discoveries based on tangible and discernible evidence (51)

copy IBO 2014

PHYSICS 2015indb 318 150514 338 PM

Fields

319

AH

L

1

2

34Earth

Figure 1015 Throwing a stone into orbit

TheforcethatcausesthestonestofollowaparabolicpathandtofalltoEarthisgravityandsimilarlytheforcethatkeeps the stone in orbit is gravity For circular motionto occur we have seen that a force must act at rightanglestothevelocityofanobjectthatistheremustbeacentripetalforceHenceinthesituationwedescribeherethe centripetal force for circular orbital motion about the EarthisprovidedbygravitationalattractionoftheEarth

We can calculate the speedwithwhich a stonemustbethrowninordertoputitintoorbitjustabovethesurfaceof the Earth

If the stone has mass m and speed v thenwehave fromNewtonrsquos2ndlaw

mv2

RE---------- G

MEm

RE2------------=

whereRE is the radius of the Earth and ME is the mass of the Earth

Bearinginmindthatg 0 GME

RE2--------= then

v g RE 10 64 106timestimes 8 103times= = =

Thatisthestonemustbethrownat8times103m sndash1

Clearly we are not going to get a satellite into orbit soclosetothesurfaceoftheEarthMovingatthisspeedthefrictionduetoairresistancewouldmeltthesatellitebeforeithadtravelledacoupleofkilometresInrealitythereforeasatelliteisputintoorbitabouttheEarthbysendingitattachedtoarocketbeyondtheEarthrsquosatmosphereandthengivingitacomponentofvelocityperpendiculartoaradialvectorfromtheEarthSeeFigure1016

Earth Satellite carried by rocket to hereSatellite orbit

Tangential component of velocity

Figure 1016 Getting a satellite into orbit

Keplerrsquos third law(This work of Kepler and Newtonrsquos synthesis of the work is an excellent example of the scientific method and makes for a good TOK discussion)

In 1627 Johannes Kepler (1571-1630)publishedhis lawsof planetary motion The laws are empirical in natureandwere deduced from the observations of theDanishastronomer Tycho de Brahe (1546-1601)The third lawgives a relationship between the radius of orbit R of a planet and its period TofrevolutionabouttheSunThelawisexpressedmathematicallyas

2

3 constantTR

=

We shall nowuseNewtonrsquos LawofGravitation to showhowitisthattheplanetsmoveinaccordancewithKeplerrsquosthirdlaw

InessenceNewtonwasabletousehislawofgravitytopredict the motion of the planets since all he had to do was factor theF givenby this law intohis second law F = ma to find their accelerations and hence their future positions

InFigure1017theEarthisshownorbitingtheSunandthedistancebetweentheircentresisR

RSun

Earth

F es

F se

Figure 1017 Planets move according to Keplerrsquos third law

Fes is the force that the Earth exerts on the Sun and Fse is the force that the Sun exerts on the Earth The forces are equalandoppositeandtheSunandtheEarthwillactuallyorbitaboutacommoncentreHoweversince theSun issoverymuchmoremassivethantheEarththiscommoncentrewillbeclosetothecentreoftheSunandsowecanregard the Earth as orbiting about the centre of the Sun Theotherthingthatweshallassumeisthatwecanignorethe forces that the other planets exert on the Earth (This wouldnotbeawisethingtodoifyouwereplanningtosend a space ship to theMoon for example)We shallalso assume that we have followed Newtonrsquos exampleandindeedprovedthataspherewillactasapointmasssituated at the centre of the sphere

Kepler had postulated that the orbits of the planets areellipticalbut since theeccentricityof theEarthrsquosorbit issmallweshallassumeacircularorbit

PHYSICS 2015indb 319 150514 338 PM

Chapter 10

320

AH

L

TheaccelerationoftheEarthtowardstheSunisa = Rω2

whereω ω 2πT------=

Hencea R 2π

T------ times

2 4π2 RT 2------------- = =

But the acceleration is given by Newtonrsquos Second Law F = mawhereFisnowgivenbytheLawofGravitationSo in this situation

F maGMs Me

R2-------------------= = butwealsohavethat

a a R 2πT

------ times

2 4π2 RT 2------------- = = and m = Me so that

G MsMe

R2------------------- Me4π2R

T2-------------times

GMs

4π2-----------rArr R3

T2------= =

Butthequantity

isaconstantthathasthesamevalueforeachoftheplanetssowehaveforalltheplanetsnotjustEarththat

R3

T2------ k =

wherekisaconstantWhichisofcourseKeplerrsquosthirdlaw

ThisisindeedanamazingbreakthroughItisdifficulttorefutethe idea that all particles attract each other in accordance withtheLawofGravitationwhenthelawisabletoaccountfortheobservedmotionoftheplanetsabouttheSun

The gravitational effects of the planets upon each othershould produce perturbations in their orbits Such is the predictivepoweroftheUniversalGravitationalLawthatit enabled physicists to compute these perturbations The telescopehadbeeninventedin1608andbythemiddleofthe 18th Century had reached a degree of perfection in design that enabled astronomers to actually measure the orbital perturbations of the planets Their measurements werealways inagreementwith thepredictionsmadebyNewtonrsquoslawHoweverin1781anewplanetUranuswasdiscovered and the orbit of this planet did not fit withthe orbit predicted by Universal Gravitation Such wasthe physicistrsquos faith in theNewtonianmethod that theysuspectedthatthediscrepancywasduetothepresenceofayetundetectedplanetUsingtheLawofGravitationtheFrenchastronomerJLeverrier and the English astronomer J C Adamswereable tocalculate justhowmassive thisnewplanetmustbeandalsowhereitshouldbeIn1846the planetNeptunewas discovered justwhere they hadpredicted Ina similarwaydiscrepancies in theorbitofNeptune ledto thepredictionandsubsequentdiscoveryin1930oftheplanetPlutoNewtonrsquosLawofGravitation

had passed the ultimate test of any theory it is not only abletoexplainexistingdatabutalsotomakepredictions

Satellite energyWhenasatelliteisinorbitaboutaplanetitwillhavebothkineticenergyandgravitationalpotentialenergySupposewe consider a satellite ofmass m that is in an orbit of radius r about a planet of mass M

Thegravitationalpotentialduetotheplanetatdistancer from its centre is

ThegravitationalpotentialenergyofthesatelliteVsat

is therefore GMe m

r----------------ndash

ThatisVsat GMem

r----------------ndash=

Thegravitationalfieldstrengthatthesurfaceoftheplanetisgivenby

g 0GMe

Re2------------=

Hencewecanwrite

Vsat g0 Re

2

rndash=

m

ThekineticenergyofthesatelliteKsatisequaltofrac12mv2wherev is its orbital speed

Byequatingthegravitationalforceactingonthesatellitetoitscentripetalaccelerationwehave

G Mem

r2---------------- mv2

r---------- mv2hArr

GMemr----------------= =

Fromwhich12---

mv2 12---

GMemr----------------times=

= 0 e

2g R

r

2m

Which is actually quite interesting since it shows thatirrespective of the orbital radius the KE is numericallyequal to half the PE Also the total energy E tot K sat Vsat+ 1

2---G Me m

r----------------times G Me m

r----------------ndash + 1

2---G Me m

r----------------ndash= = =of the satelliteisalwaysnegativesince

E tot K sat Vsat+ 12---

G Me mr----------------times

G Me mr----------------ndash

+ 12---

G Me mr----------------ndash= = =

PHYSICS 2015indb 320 150514 338 PM

Fields

321

AH

L

The energies of an orbiting satellite as a function of radial distancefromthecentreofaplanetareshownplottedinFigure1018

12

100806

0402

0ndash02

ndash04

ndash06ndash08ndash10

ndash12

2 4 6 8 10 12

ener

gya

rbita

ry u

nits

distance timesR

kinetic energytotal energy

potential energy

Figure 1018 Energy of an orbiting satellite as a function of distance from the centre of a planet

WeightlessnessSuppose that you are inanelevator(lift)thatisdescendingat constant speedandyou let goof abook thatyouareholdinginyourhandThebookwillfalltothefloorwithaccelerationequaltotheaccelerationduetogravityIfthecablethatsupportstheelevatorweretosnap(asituationthatItrustwillneverhappentoanyofyou)andyounowletgothebookthatyouareholdinginyourotherhandthis book will not fall to the floor - it will stay exactlyin linewithyourhandThis isbecause thebook isnowfallingwith the sameaccelerationas theelevatorandassuch the book cannot ldquocatchrdquo up with the floor of theelevatorFurthermoreifyouhappenedtobestandingonasetofbathroomscalesthescaleswouldnowreadzero-youwouldbeapparentlyweightlessItisthisideaoffreefallthatexplainstheapparentweightlessnessofastronautsin an orbiting satellite These astronauts are in free fall inthesensethattheyareacceleratingtowardsthecentre of the Earth

It is actually possible to define theweight of a body inseveral different waysWe can define it for example asthegravitationalforceexertedonthebodybyaspecifiedobject such as the EarthThiswe have seen thatwe doinlotsofsituationswherewedefinetheweightasbeingequaltomgIfweusethisdefinitionthenanobjectinfreefallcannotbydefinitionbeweightlesssinceitisstillinagravitationalfieldHoweverifwedefinetheweightofanobjectintermsofaldquoweighingrdquoprocesssuchasthereadingonasetofbathroomscaleswhichineffectmeasuresthecontact force between the object and the scales thenclearlyobjectsinfreefallareweightlessOnenowhasto

askthequestionwhetherornotitispossibleForexampletomeasurethegravitationalforceactingonanastronautin orbit about the Earth

WecanalsodefineweightintermsofthenetgravitationalforceactingonabodyduetoseveraldifferentobjectsForexample for an object out in space its weight could bedefined in terms of the resultant of the forces exerted on it bytheSuntheMoontheEarthandalltheotherplanetsin the Solar System If this resultant is zero at a particular pointthenthebodyisweightlessatthispoint

Inviewofthevariousdefinitionsofweightthatareavailabletousitisimportantthatwhenweusethewordldquoweightrdquoweareawareofthecontextinwhichitisbeingused

Example

Calculate the height above the surface of the Earth atwhichageo-stationarysatelliteorbits

Solution

A geo-stationary satellite is one that orbits the Earth in such a way that it is stationary with respect to a point on the surface of the Earth This means that its orbital period must be the same as the time for the Earth to spin once on its axis ie 24 hours

From Keplerrsquos third law we have G Ms

4π2----------- R3

T2------=

That is

RMe

mh

using the fact that the force of attraction between the satellite and the Earth is given by

FG Mem

R2----------------=

and that F = ma

where m is the mass of the satellite and a 4π2R

T2-------------=

we have

GMem

R2---------------- m 4π2R

T2-------------times

GMe

4π2------------rArr R3

T2------= =

PHYSICS 2015indb 321 150514 338 PM

Chapter 10

322

AH

L

Now the mass of the Earth is 60 times 1024 kg and the period T measured in seconds is given by T = 86400 s

So substitution gives R = 42 times 106 m

The radius of the Earth is 64 times 106 m so that the orbital height h is about 36 times 107 m

Example

Calculatetheminimumenergyrequiredtoputasatelliteofmass500kgintoanorbitthatisasaheightequaltotheEarthrsquosradiusabovethesurfaceoftheEarth

Solution

We have seen that when dealing with gravitational fields and potential it is useful to remember that

g 0G M

Re2---------= or g0 Re

2 GM=

The gravitational potential at the surface of the Earth is

g 0ndash Re GMRe

---------ndash=

The gravitational potential at a distance R

from the centre of the Earth is

=

The difference in potential between the surface of the Earth and a point distance R from the centre is therefore

V∆ g 0Re 1ReR------ndash

=

If R 2Re= then V∆g0 Re

2------------=

This means that the work required to ldquoliftrdquo the satellite into orbit is g0Rm where m is the mass of the satellite This is equal to

10 32 106 500timestimestimes = 16000 MJ

However the satellite must also have kinetic energy in order to orbit Earth This will be equal to

g 0 mRe2

2R-----------------g 0mRe

2

4----------------- 8000 MJ= =

The minimum energy required is therefore

24000 MJ

Exercise 1031 The speed needed to put a satellite in orbit does

not depend on

A the radius of the orbitB theshapeoftheorbitC thevalueofgattheorbitD the mass of the satellite

2 EstimatethespeedofanEarthsatellitewhoseorbitis400kmabovetheEarthrsquossurfaceAlsodetermine the period of the orbit

3 Calculatethespeedofa200kgsatelliteorbitingthe Earth at a height of 70 times 106 m

Assume that g = 82 m sndash2 for this orbit

4 TheradiioftwosatellitesXandYorbitingthe Earth are 2r and 8rwherer is the radius of the Earth Calculate the ratio of the periods of revolutionofXtoY

5 A satellite of mass mkgissentfromEarthrsquossurfaceinto an orbit of radius 5RwhereR is the radius of theEarthWritedownanexpressionfor

(a) thepotentialenergyofthesatelliteinorbit

(b) thekineticenergyofthesatelliteinorbit

(c) theminimumworkrequiredtosendthesatellitefromrestattheEarthrsquossurfaceintoits orbit

6 A satellite in an orbit of 10rfallsbacktoEarth(radius r)afteramalfunctionDeterminethespeedwithwhichitwillhittheEarthrsquossurface

7 The radius of the moon is frac14 that of the Earth AssumingEarthandtheMoontohavethesamedensitycomparetheaccelerationsoffreefallatthe surface of Earth to that at the surface of the Moon

8 UsethefollowingdatatodeterminethegravitationalfieldstrengthatthesurfaceoftheMoon and hence determine the escape speed from the surface of the Moon

Mass of the Moon = 73 times 1022kg

RadiusoftheMoon=17times106 m

PHYSICS 2015indb 322 150514 338 PM

Chapter 9

296

AH

L

1 For Nge2thefringepatterncontainsprincipalmaximathat are modulated by the diffraction envelope

2 ForNgt2thefringepatterncontains(Nndash2)subsidiary maxima

3 Where the grating spacing d is four time the slit width a(4xa =d)thenthefourthprincipalmaximumismissing from the fringe pattern Where the grating spacingdisthreetimetheslitwidtha(3xa =d)thenthe third principal maximum is missing from the fringepatternandsoon

4 Increasing N increases the absolute intensities of the diffraction pattern (I = N2)

If white light is shone through a grating then the central image will be white but for the other orders each will be spread out intoacontinuousspectrumcomposedofaninfinitenumberof adjacent images of the slit formed by the wavelength of the different wavelengths present in the white light At any given point in the continuous spectrum the light will be very nearly monochromatic because of the narrowness of the images of the slit formed by the grating This is in contrast to the doubleslitwhereifwhitelightisusedtheimagesarebroadand the spectral colours are not separated

In summary as is the case of interference at a doubleslit there is a path difference between the rays fromadjacent slits of d sinθ Where there is a whole number of wavelengthsconstructiveinterferenceoccurs

For constructive interference and minima

d sinθ = mλ wheren=123hellip

When there is a path difference between adjacent rays of halfawavelengthdestructiveinterferenceoccurs

For and maxima

d sinθ = ( m + frac12 )λ wheren=123hellip

Exercise 941 Lightfromalaserisshonethroughadiffraction

grating on to a screen The screen is a distance of 20mfromthelaserThedistancebetweenthecentraldiffractionmaximumandthefirstprincipalmaximum formed on the screen is 094 m The diffractiongratinghas680linespermmEstimatethe wavelength of the light emitted by the laser

2 Monochromaticlightfromalaserisnormallyincident on a six-slit grating The slit spacing is three times the slit width An interference patternisformedonascreen2montheotherside of the grating

(a) Determineandexplaintheorderoftheprincipal maxima missing from the fringe pattern

(b) Howmanysubsidiarymaximaaretherebetween the principal maxima

(c) Howmanyprincipalmaximaarethere

(i) Betweenthetwofirstorderminimain the diffraction envelope

(ii) Betweenthefirstandsecondorderminima in the diffraction envelope

(d) Drawthepatternformedbythebrightfringesin(c)

(e) Sketchtheintensityprofileofthefringepattern

Thin film interferenceYoumightwellbe familiarwith the colouredpatternoffringes that can be seen when light is reflected off thesurfaceofwateruponwhichathinoilfilmhasbeenspiltorfromlightreflectedfrombubblesWecanseehowthesepatternsarisebylookingatFigure920

Extendedlight source air

A

B

C

12

3

4

Oil lm

Water

Figure 920 Reflection from an oil film

Consider light from an extended source incident on a thinfilmWealsoconsiderawave fromonepointofthe source whose direction is represented by the ray shownSomeofthislightwillbereflectedatAandsometransmitted through the filmwhere somewill again bereflectedatB(somewillalsobetransmittedintotheair)SomeofthelightreflectedatBwillthenbetransmittedatCandsomereflectedandsoonIfweconsiderjustrays1and2thenthesewillbenotbeinphasewhenthearebrought to a focus by the eye since they have travelled

PHYSICS 2015indb 296 150514 337 PM

Wave Phenomena

297

AH

L

different distances If the path difference between them is an integral number of half wavelengths then we might expect the twowaves to be out of phaseHowever ray1 undergoes a phase change of π on reflection but ray2 does not since it is reflected at a boundary betweena more dense and less dense medium (See chapter 4)Henceifthepathdifferenceisanintegralnumberofhalf-wavelengthsrays1and2willreinforceieray1and2areinphaseHowever rays357etcwillbeoutofphasewithrays246etcbutsinceray2ismoreintensethanray3andray4moreintensethanray5thesepairswillnot cancel out so there will be a maximum of intensity

Ifthepathdifferenceissuchthatwave1and2areoutofphasesincewave1ismoreintensethanwave2theywillnotcompletelyannulHoweveritcanbeshownthattheintensitiesofwaves2345hellipaddtoequaltheintensityofwave1Sincewaves345hellipareinphasewithwave2there will be complete cancellation

The path difference will be determined by the angle at which ray1 is incident andalsoon the thickness (andtheactualrefractiveindexaswell)ofthefilmSincethesourceisanextendedsourcethelightwillreachtheeyefrom many different angles and so a system of bright and darkfringeswillbeseenYouwillonlyseefringesifthefilmisverythin(exceptifviewedatnormalincidence)sinceincreasingthethicknessofthefilmwillcausethereflected rays to get so far apart that they will not becollected by the pupil of the eye

From the argument above to find the conditions forconstructive and destructive interferenceweneedonlyfindthepathdifferencebetweenray1andray2Figure921showsthe geometry of the situation

A BE

F

C

D

1 2

φrsquo φrsquo

φ

φ

d

Figure 921 The geometry of interference

Thefilmisofthicknessd and refractive index n and the light has wavelength λ If the line BF is perpendicular to ray 1 then the optical path difference (opd) between ray 1 andray2whenbroughttoafocusis

opd = n(AC+CB)ndashAF

We have to multiple by the refractive index for the path travelledbythelightinthefilmsincethelighttravelsmoreslowly in thefilm If the light travels say adistancex in a material of refractive index n then in the time that it takes to travelthisdistancethelightwouldtraveladistancenx in air

IfthelineCEisatrightanglestoray2thenweseethat

AF = nBE

FromthediagramAC=CDsowecanwrite

opd = n(CD+CB)ndashnBE=nDE

Alsofromthediagramweseethatwhereφ is the angle of refraction

DE=2cosφ

From which opd=2ndcosφ

Bearinginmindthechangeinphaseofray1onreflectionwe have therefore that the condition for constructive interference is

2nd φcos m 12---+

λ m 1 2 hellip = =

Andfordestructiveinterference 2ndcosφ = mλ

Eachfringecorrespondstoaparticularopd for a particular value of the integer m and for any fringe the value of the angle φ is fixedThismeans that itwill be in the formof an arc of a circle with the centre of the circle at the point where the perpendicular drawn from the eye meets the surface of the film Such fringes are called fringesofequal inclinationSince theeyehasa smallaperturethesefringesunlessviewedatneartonormalincidence (φ = 0) will only be observed if the film is very thinThisisbecauseasthethicknessofthefilmincreasesthereflected rayswill get further and further apart and sovery few will enter the eye

Ifwhitelightisshoneontothefilmthenwecanseewhywe get multi-coloured fringes since a series of maxima and minima will be formed for each wavelength present in the white light However when viewed at normalincidenceitispossiblethatonlylightofonecolourwillundergoconstructiveinterferenceandthefilmwilltakeon this colour

PHYSICS 2015indb 297 150514 337 PM

Chapter 9

298

AH

L

Thickness of oil films

The exercise to follow will help explain this use

Non-reflecting films

A very important but simple application of thin filminterferenceisintheproductionofnon-reflectingsurfaces

Athinfilmofthicknessd and refractive index n1 is coated onto glass of refractive index n where n1 lt n Light ofwavelength λ that is reflected at normal incidence willundergo destructive interference if 12

2n d λ

= thatis

14d

nλ

=

(remember that there will now no phase change at the glass interface ie we have a rare to dense reflection)

The use of such films can greatly reduce the loss oflightby reflection at the various surfacesof a systemoflenses or prisms Optical parts of high quality systemsare usually all coatedwithnon-reflecting films in orderto reduce stray reflections The films are usually madebyevaporatingcalciumormagnesiumfluorideonto thesurfacesinvacuumorbychemicaltreatmentwithacidsthat leave a thin layer of silica on the surface of the glass ThecoatedsurfaceshaveapurplishhuebyreflectedlightThis is because the condition for destructive interference fromaparticularfilmthicknesscanonlybeobtainedforone wavelength The wavelength chosen is one that has a value corresponding to light near the middle of the visible spectrumThismeansthatreflectionofredandvioletlightis greater combining to give the purple colour Because of the factor cosφatanglesotherthannormalincidencethepath difference will change butnotsignificantlyuntilsayabout30deg(egcos25deg=090)

It should be borne in mind that no light is actually lost by anon-reflectingfilmthedecreaseofreflectedintensityiscompensated by increase of transmitted intensity

Non-reflecting films can be painted onto aircraft tosuppressreflectionofradarThethicknessofthefilmis determined by

4nd λ

= where λ is the wavelength of

the radar waves and ntherefractiveindexofthefilmat this wavelength

Example

A plane-parallel glass plate of thickness 20 mm isilluminated with light from an extended source The refractiveindexoftheglassis15andthewavelengthofthelight is 600 nm Calculate how many fringes are formed

Solution

We assume that the fringes are formed by light incident at all angles from normal to grazing incidence

At normal incidence we have 2nd = mλ

From which

m 2 15 2 10 3ndashtimestimestimes6 10 7ndashtimes

------------------------------------------- 10 000= =

At grazing incidence the angle of refraction φ is in fact the critical angle

Therefore φφ arcsin 115------- 42 deg= =

ie cosφ = 075

At grazing incidence 2ndcosφ = m λ

From which (and using cos(sinndash1(115) = 075)

m 2 15 2 10 3ndash 075timestimestimestimes6 10 7ndashtimes

------------------------------------------------------------ 7500= =

The total number of fringes seen is equal to m ndash m = 2500

Exercise 95When viewed from above the colour of an oil film onwater appears red in colour Use the data below to estimate theminimumthicknessoftheoilfilm

average wavelength of red light = 630 nm

refractiveindexofoilforredlight=15

refractive index of water for red light = 13

PHYSICS 2015indb 298 150514 337 PM

Wave Phenomena

299

AH

L

94 Resolution

Essential idea Resolution places an absolute limit on the extent to which an optical or other system can separate images of objects

Understandings

bull Thesizeofadiffractingaperturebull Theresolutionofsimplemonochromatic

two-source systems

The size of a diffracting apertureOur discussion concerning diffraction so far has been for rectangular slits When light from a point source enters a smallcircularapertureitdoesnotproduceabrightdotasanimagebutacirculardiscknownasAiryrsquosdiscsurroundedbyfainterconcentriccircularringsasshowninFigure922

Laser

Figure 922 Diffraction at a circular aperture

Diffractionatanapertureisofgreatimportancebecausetheeye and many optical instruments have circular apertures

What is the half-angular width of the central maximum of the diffraction formed by a circular aperture This is not easy to calculate since it involves some advanced mathematics The problemwasfirstsolvedbytheEnglishAstronomerRoyalGeorgeAiryin1835whoshowedthatforcircularapertures

θ 122 λb--------------= where b is the diameter of the aperture

Example 2

Inthefollowingdiagramparallellightfromadistantpointsource(suchasastar)isbroughttofocusonthescreenSbyaconverginglens(the lens is shown as a vertical arrow)

Thefocal length(distance fromlens toscreen) is25cmand the diameter of the lens is 30 cm The wavelength of thelightfromthestaris560nmCalculatethediameterofthe diameter of the image on the screen

Solution

The lens actually acts as a circular aperture of diameter 30 cm The half angular width of central maximum of the diffraction pattern that it forms on the screen is

75

2

122 122 56 10 23 1030 10b

λθminus

minusminus

times times= = = times

times rad

The diameter of the central maxima is therefore

25 times 10-2 times 23 times 10-5

= 57 times 10-6 m

Although this is small it is still finite and is effectively the image of the star as the intensity of the secondary maxima are small compared to that of the central maximum

The resolution of simple monochromatic two-source systemsThe astronomers tell us that many of the stars that we observe with the naked eye are in fact binary stars That iswhatwe see as a single star actually consistsof twostarsinorbitaboutacommoncentreFurthermoretheastronomerstellusthatifweusealdquogoodrdquotelescopethenwewillactuallyseethetwostarsThatiswewillresolve the single point source into its two component parts Sowhatisitthatdetermineswhetherornotweseethetwo stars as a single point source ie what determines whether or not two sources can be resolved It canrsquot just bethatthetelescopemagnifiesthestarssinceiftheyareacting as point sources magnifying them is not going to make a great deal of difference

Ineachofoureyesthereisanaperturethepupilthroughwhich the light enters This light is then focused by the eye lens onto the retina But we have seen that when light passes through an aperture it is diffracted and so when we look at

NATURE OF SCIENCEImproved technology The Rayleigh criterion is the limit of resolution Continuing advancement in technology such as large diameter dishes or lenses or the use of smaller wavelength lasers pushes the limits of what we can resolve (18)

copy IBO 2014

PHYSICS 2015indb 299 150514 337 PM

Chapter 9

300

AH

L

apointsourceadiffractionpatternwillbeformedontheretina If we look at two point sources then two diffraction patterns will be formed on the retina and these patterns will overlap The width of our pupil and the wavelength of the light emitted by the sources will determine the amount by which they overlap But the degree of overlap will also depend on the angular separation of the two point sources WecanseethisfromFigures923

pupilretina

S1

S2

P2

P1

θ

Figure 923

LightfromthesourceS1 enters the eye and is diffracted by the pupil such that the central maximum of the diffraction patternisformedontheretinaatP1SimilarlylightfromS2producesamaximumatP2 If the two central maxima are well separated then there is a fair chance that we will see the two sources as separate sources If they overlap then we will not be able to distinguish one source from another From the diagram we see as the sources are movedclosertotheeyethentheangleθ increases and so does the separation of the central maxima

Figures924925926and927showsthedifferentdiffractionpatternsandtheintensitydistributionthatmightresultonthe retina as a result of light from two point sources

S1 S21 Well resolved

Figure 924 Very well resolved

S1 S22 Well resolved

Figure 925 Well resolved

S1 S23 Just resolved Rayleigh criterion Minimum of S2 coincides with maximum peak of S1

Figure 926 Just resolved

4 Not resolved

Figure 927 Not resolved

We have suggested that if the central maxima of the two diffraction patterns are reasonably separated then we should be able to resolve two point sources In the late 19th century Lord Rayleigh suggested by how much they should be separated in order for the two sources to be just resolved Ifthecentralmaximumofonediffractionpattern coincides with the firstminima of the otherdiffraction pattern then the two sources will just beresolved This is known as the Rayleigh Criterion

Figure926showsjustthissituationThetwosourcesarejust resolved according to the Rayleigh criterion since the peak of the central maximum of one diffraction pattern coincideswiththefirstminimumoftheotherdiffractionpattern This means that the angular separation of the peaks of the two central maxima formed by each source is just the half angular width of one central maximum ie

θ λb---

=

where b is the width of the slit through which the light fromthesourcespassesHoweverweseefromFigure1117that θ is the angle that the two sources subtend at the slit Henceweconcludethattwosourceswillberesolvedbyaslit if the angle that they subtend at the slit is greater than or equaltoλfraslb

PHYSICS 2015indb 300 150514 337 PM

Wave Phenomena

301

AH

L

Sofarwehavebeenassumingthattheeyeisarectangularslit whereas clearly it is a circular aperture and so we must use the formula

θ 122 λb--------------=

As mentioned above the angle θ is sometimes called the resolving power but should more accurately be called the minimum angle of resolution (θmin)

Clearly the smaller θ the greater will be the resolving power

The diffraction grating is a useful for differentiating closely spacedlinesinemissionspectraLikeaprismspectrometerit can disperse a spectrum into its components but it is better suited because it has higher resolution than the prism spectrometer

If λ1 and λ2 are two nearly equal wavelengths that canbarelybedistinguishedtheresolvanceorresolvingpowerofthegratingisdefinedas

R = λ _______ (λ2 ndash λ 1) = λ ___ Δλ

Where λ asymp λ1 asymp λ2

Therefore thediffractiongratingwithahighresolvancewill be better suited in determining small differences in wavelength

If the diffraction grating has Nlinesbeingilluminateditcan be shown that the resolving power of the mth order diffraction is given by

R = mN

Obviouslytheresolvancebecomesgreaterwiththeordernumber m and with a greater number of illuminated slits

Example

A benchmark for the resolving power of a grating is the sodium doublet in the sodium emission spectrum Two yellow lines in itsspectrumhavewavelengthsof58900nmand58959nm

(a) Determinetheresolvanceofagratingifthegivenwavelengths are to be distinguished from each other

(b) Howmanylinesinthegratingmustbeilluminated in order to resolve these lines in the second order spectrum

Solution

(a) R = λ ___ Δλ = 58900nm __________________ (58900ndash58959)nm

= 589nm _______ 059nm = 10 times 103

(b) N = R __ m = 10 times 10 3 ________ 2 =50times102 lines

It has been seen that diffraction effectively limits the resolving power of optical systems This includes such systems as the eye telescopes and microscopes Theresolving power of these systems is dealt with in the next section This section looks at links between technology and resolving power when looking at the very distant and when looking at the very small

Radio telescopes

The average diameter of the pupil of the human eye is about 25 mm This means that the eye will justresolve two point sources emitting light of wavelength 500 nm if their angular separation at the eye is 7

43

50 10122 24 1025 10

θminus

minusminus

times= times = times

times rad

If the eye were to be able to detect radio waves of wavelength 015m then tohave the same resolvingpower thepupilwould have to have a diameter of about 600 m Clearly this is nonsense but it does illustrate a problem facingastronomers who wish to view very distant objects such as quasarsandgalaxies thatemitradiowavesConventionalradio telescopes consist of a large dish typically 25 min diameter Even with such a large diameter the radiowavelength resolving power of the telescope is much less thantheopticalresolvingpowerofthehumaneyeLetuslook at an example

Example

The Galaxy Cygnus A can be resolved optically as an elliptically shaped galaxy However it is also a strongemitterof radiowavesofwavelength015mTheGalaxyisestimatedtobe50times1024mfromEarthUseofaradiotelescope shows that the radio emission is from two sources separated by a distance of 30 times 1021m Estimate thediameterofthedishrequiredtojustresolvethesources

Solution

The angle θ that the sources resolve at the telescope is given by

θ 21

424

30 10 60 1050 10

θ minustimes= = times

times rad

and 4

122 122 015 300060 10

d λθ minus

times= = =

timesm = 30 km

A radio telescope dish of this size would be impossible tomake letalonesupportThisshowsthatasingledishtype radio telescope cannot be used to resolve the sources and yet they were resolved To get round the problemastronomers use two radio telescopes separated by a large

PHYSICS 2015indb 301 150514 337 PM

Chapter 9

302

AH

L

distance The telescopes view the same objects at the same time and the signals that each receive from the objects are simultaneously superimposed The result of the superposition of the two signals is a two-slit interference pattern The pattern has much narrower fringe spacing than that of the diffraction pattern produced by either telescope on its own hence producing a much higherresolvingpowerWhentelescopesareusedlikethistheyare called a stellar interferometer

InSocorroinNewMexicothereisastellarinterferometerthat consists of 27 parabolic dishes each of diameter25 m arranged in a Y-shape that covers an area of 570km2Thisisaso-calledVeryLargeArray(VLA)Evenhigher resolution can be obtained by using an array of radio telescopes in observatories thousands of kilometres apart A system that uses this so-called technique oflsquovery-long-baseline interferometryrsquo (VLBI) is known asalsquovery-long-baselinearrayrsquo(VLBA)WithVLBAaradiowavelength resolving power can be achieved that is 100 timesgreaterthanthebestopticaltelescopesEvenhigherresolving power can be achieved by using a telescope that isinasatelliteorbitingEarthJapanrsquosInstituteofSpaceandAstronautical Science (ISAS) launched sucha system inFebruary 1997TheNationalAstronomicalObservatoryof Japan the National Science Foundationrsquos NationalRadio Astronomy Observatory (NRAO) the CanadianSpaceAgency theAustraliaTelescopeNationalFacilitythe European VLBI Network and the Joint Instituteback this project forVeryLongBaseline Interferometryin EuropeThis project is a very good example of howInternationalismcanoperateinPhysics

Electron microscope

Telescopes are used to look at very distant objects that are very large but because of their distance fromus appearverysmallMicroscopesontheotherhandareusedtolookat objects that are close to us but are physically very small Aswehave seen justmagnifyingobjects that ismakingthemappearlargerisnotsufficientonitsowntogaindetailabouttheobjectfordetailhighresolutionisneeded

Figure928isaschematicofhowanoptical microscope is usedtoviewanobjectandFigure929isaschematicofatransmission electron microscope(TEM)

Optical lens system

bright light source

glass slide containing specimen (object)

eye

Figure 928 The principle of a light microscope

magnetic lens system

electron source

thin wafer of material

ScreenCCD

Figure 929 The principle of an electron microscope

In the optical microscope the resolving power isdetermined by the particular lens system employed and the wavelength λ of the light used For example twopoints in the sample separated by a distance d will just be resolved if

2d mλ=

where m is a property of the lens system know as the numerical aperture In practice the largest value of m obtainable is about 16 Hence if the microscope slideis illuminatedwith light ofwavelength 480 nm a goodmicroscope will resolve two points separated by a distance dasymp15times10-7masymp015micromPointscloser together thanthiswillnotberesolvedHoweverthisisgoodenoughtodistinguishsomevirusessuchastheEbolavirus

Clearlythesmallerλ the higher the resolving power and this is where the electron microscope comes to the fore The electron microscope makes use of the wave nature of electrons(see1315)IntheTEMelectronspassthrougha wafer thin sample and are then focused by a magnetic field onto a fluorescent screen orCCD (charge coupleddevicesee142)Electronsusedinanelectronmicroscopehavewavelengthstypicallyofabout5times10-12mHoweverthe numerical aperture of electron microscopes is considerablysmaller thanthatofanopticalmicroscopetypicallyabout002NonethelessthismeansthataTEMcanresolvetwopointsthatareabout025nmapartThisresolving power is certainly high enough to make out the shape of large molecules

Another type of electron microscope uses a techniqueby which electrons are scattered from the surface of the sample The scattered electrons and then focused as in theTEM to form an image of the surfaceThese so-calledscanningelectronmicroscopes(SEM)havealowerresolving power than TEMrsquos but give very good threedimensional images

The eye

We saw in the last section that the resolving power of the human eye is about 2 times 10-4 rad Suppose that youare looking at car headlights on a dark night and the car

PHYSICS 2015indb 302 150514 337 PM

Wave Phenomena

303

AH

L

is a distance D away If the separation of the headlight is say 15 m then the headlights will subtend an angle θ 15

Dθ =

at your eye Assuming an average wavelength of

500 nm your eye will resolve the headlights into twoseparatesourcesifthisangleequals2times10-4 rad This gives D=75kmInotherwordsifthecarisapproachingyouona straight road then you will be able to distinguish the two headlightsasseparatesourceswhenthecaris75kmawayfrom you Actually because of the structure of the retina and optical defects the resolving power of the average eye is about 6 times 10-4 rad This means that the car is more likely tobe25kmawaybeforeyouresolvetheheadlights

Astronomical telescope

Letusreturntotheexampleofthebinarystarsdiscussedatthe beginning of this section on resolution The stars Kruger A and B form a binary system The average separation of the stars is 14 times 1012mandtheiraveragedistancefromEarthis12times1017mWhenviewedthroughatelescopeonEarththe system will therefore subtend an angle

12

17

14 1012 10

θtimes

=times

= 12times10-5 rad at the objective lens of the telescope Assuming that the average wavelength of the light emitted bythestarsis500nmthenifthetelescopeistoresolvethesystem into two separate images it must have a minimum diameter Dwhere12times10-5 =

7122 500 10D

minustimes times

This gives D=0050m which is about 5 cm So thisparticular system is easily resolved with a small astronomical telescope

Exercise 961 ItissuggestedthatusingtheISASVLBAitwould

bepossibletoldquoseerdquoagrainofriceatadistanceof5000kmEstimatetheresolvingpoweroftheVLBA

2 Thedistancefromtheeyelenstotheretinais20mm The light receptors in the central part of the retinaareabout5times10-6apartDeterminewhetherthe spacing of the receptors will allow for the eye to resolve the headlights in the above discussion whentheyare25kmfromtheeye

3 ThediameterofPlutois23times106 m and its averagedistancefromEarthis60times1012 m EstimatetheminimumdiameteroftheobjectiveofatelescopethatwillenablePlutotobeseenasadisc as opposed to a point source

95 Doppler effect

Essential ideaTheDopplereffectdescribesthephenomenonofwavelengthfrequencyshiftwhenrelativemotionoccurs

Understandingsbull TheDopplereffectforsoundwavesandlightwaves

The Doppler effect for sound waves and light waves

Consider two observers A and B at rest with respect to asoundsourcethatemitsasoundofconstantfrequencyf Clearly both observers will hear a sound of the same frequencyHoweversupposethatthesourcenowmovesat constant speed towards A A will now hear a sound of frequencyfA that is greater than f and B will hear a sound of frequency fB that is less than f This phenomenon is known as the Doppler Effect orDopplerPrinciple after C J Doppler(1803-1853)

The same effect arises for an observer who is either moving towards or away from a stationary source

Figure 930 shows the waves spreading out from a stationary source that emits a sound of constant frequency f The observersAandBhearasoundofthesamefrequency

A B

wavefront

S

Figure 930 Sound waves from a stationary source

Suppose now that the source moves towards A withconstant speed vFigure9310(a)showsasnapshotofthenew wave pattern

NATURE OF SCIENCETechnology Although originally based on physical observations of the pitch of fast moving sources of sound the Doppler effect has an important role in many different areas such as evidence for the expansion of the universe and generating images used in weather reports and in medicine (55)

copy IBO 2014

PHYSICS 2015indb 303 150514 337 PM

Chapter 9

304

AH

L

sourceA B

smallerwavelength

largerwavelength

Figure 931 (a) Sound waves from a moving source

The wavefronts are now crowded together in the direction of travel of the source and stretched out in the opposite direction This is why now the two observers will now hear notesofdifferentfrequenciesHowmuchthewavesbunchtogether and how much they stretch out will depend on the speed vEssentially A

A

cfλ

= and BB

cfλ

= where λA lt λB and v is the speed of sound

IfthesourceisstationaryandAismovingtowardsitthenthe waves from the source incident on A will be bunched up If A is moving away from the stationary source then the waves from the source incident on A will be stretched out

Christian Doppler(1803ndash1853)actuallyappliedtheprinciple(incorrectlyasithappens)totryandexplainthecolourofstars However the Doppler effect does apply to light aswellastosoundIfalightsourceemitsalightoffrequencyf then if it is moving away from an observer the observer willmeasurethelightemittedashavingalowerfrequencythan fSincethesensationofcolourvisionisrelatedtothefrequencyoflight(bluelightisofahigherfrequencythanred light) light emitted by objects moving way from anobserver is often referred to as being red-shifted whereas if the object is moving toward the observer it is referred to as blue-shiftedThisideaisusedinOptionE(Chapter16)

We do not need to consider here the situations where either the source or the observer is accelerating In a situation for example where an accelerating source is approaching astationaryobserverthentheobserverwillhearasoundof ever increasing frequency This sometimes leads toconfusion in describing what is heard when a source approaches passes and then recedes from a stationaryobserver Suppose for example that you are standingon a station platform and a train sounding its whistle is approaching at constant speed What will you hear as the train approaches and then passes through the station As the train approaches you will hear a sound of constant pitch but increasing loudness The pitch of the sound will be greater than if the train were stationary As the train passes through the station you will hear the pitch change at the momentthetrainpassesyoutoasoundagainofconstantpitch The pitch of this sound will be lower than the sound of the approaching train and its intensity will decrease as the train recedes from you What you do not hear is a sound of increasing pitch and then decreasing pitch

The Doppler equations for sound

Although you will not be expected in an IB examination to derivetheequationsassociatedwithaspectsoftheDopplereffectyouwillbeexpectedtoapplythemForcompletenesstherefore thederivationoftheequationsassociatedwiththeDopplereffectasoutlinedaboveisgivenhere

S primeS O

vs

Figure 931 (b)

InFigure931(b)theobserverOisatrestwithrespecttoasourceofsoundSismovingwithconstanspeedvs directly towards O The source is emitting a note of constant frequencyf and the speed of the emitted sound is v

S shows the position of the source ∆t later When the sourceisatresttheninatime∆t the observer will receive f∆t waves and these waves will occupy a distance v∆t ie

v f

λ =f∆tv∆t =

(Because of the motion of the source this number of waves will now occupy a distance (v∆t ndash vs∆t)The lsquonewrsquowavelength is therefore

v ndash vs

fλrsquo =

f∆tv∆t ndash vs∆t

=

If fisthefrequencyheardbyOthen

vfλ

= or

vf

λ = = sv vfminus

From which

s

vf fv v

=minus

Dividingthroughbyv gives

Equation1111

1 sf f v

v=

minus

If the source is moving away from the observer then we have

s

vf fv v

=minus

1

1 sf vv

=+

Equation112

PHYSICS 2015indb 304 150514 337 PM

Wave Phenomena

305

AH

L

We now consider the case where the source is stationary and the observerismoving towardsthesource with speed v0 In this situation the speed of the sound waves as measured by the observer will be v0 + v We therefore have that

v0 + v = ffraslλ = vfλ

times

From which

o1 vf fv

= +

Equation113

If the observerismovingawayfromthesource then

o1 vf fv

= minus

Fromequation(113)wehavethat

o o1 v vf f f f f fv v

∆ = minus = + minus =

Equation114

Thevelocitiesthatwerefertointheaboveequationsarethevelocities with respect to the medium in which the waves fromthesourcetravelHoweverwhenwearedealingwithalight source it is the relative velocity between the source and the observer that we must consider The reason for this is that lightisuniqueintherespectthatthespeedofthelightwavesdoes not depend on the speed of the source All observers irrespective of their speed or the speed of the source will measure the same velocity for the speed of light This is oneof thecornerstonesof theSpecialTheoryofRelativitywhichisdiscussedinmoredetailinOptionH(Chapter18)When applying the Doppler effect to light we aremainlyconcerned with the motion of the source We look here only at the situation where the speed of the source v is much smaller than the speed of light cinfreespace(v ltlt c)Underthesecircumstanceswhenthesourceismovingtowardstheobserverequation111becomes

vf f f fc

minus = ∆ = Equation115

andwhenthesourceismovingawayfromtheobserverequation112becomes vf f f f

cminus = ∆ = minus

Provided that v ltlt c these same equations apply for astationary source and moving observer

We look at the following example and exercise

Example

Asourceemitsasoundoffrequency440HzItmovesinastraight line towards a stationary observer with a speed of 30ms-1Theobserverhearsasoundoffrequency484HzCalculate the speed of sound in air

Solution

We use equation 111 and substitute f= 484 Hz f = 440 Hz

and vs = 30 m s-1

therefore 1

484 440 301v

=minus

such that 30 4401

484vminus = to

give v = 330 m s-1

Example

A particular radio signal from a galaxy is measured as havingafrequencyof139times109HzThesamesignalfromasourceinalaboratoryhasafrequencyof142times109Hz

Suggestwhy thegalaxy ismovingaway fromEarthandcalculateitsrecessionspeed(iethespeedwithwhichitismovingawayfromEarth)

Solution

The fact that the frequency from the moving source is less than that when it is stationary indicates that it is moving away from the stationary observer ie Earth

Using vf fc

∆ = we have

8 96

9

3 10 (142 139) 10 634 10142 10

c fvf∆ times times minus times

= = = timestimes

m s-1

It is usual when dealing with the Doppler effect of light to express speeds as a fraction of c So in this instance we have v = 0021 c

Using the Doppler effectWe have seen in the above example and exercise how the Dopplereffectmaybeusedtomeasuretherecessionspeedof distant galaxies The effect is also used to measure speed inothersituationsHerewewilllookatthegeneralprincipleinvolved in using the Doppler effect to measure speedFigure932showsasource(thetransmitter)thatemitseithersound or em waves of constant frequency f The waves

PHYSICS 2015indb 305 150514 337 PM

Chapter 9

306

AH

L

fromthesourceare incidentonareflectorthat ismovingtowards the transmitter with speed vThereflectedwavesare detected by the receiver placed alongside the transmitter

reector

v

transmitter

receiver

f f

f f

Figure 932 Using the Doppler effect to measure speed

We shall consider the situation where v ltlt c where c is the speed of the waves from the transmitter

For the reflector receiving waves from the transmitterit is effectively an observer moving towards a stationary sourceFromequation(114) itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation116

For the receiver receivingwaves from the reflector it iseffectively a stationary observer receiving waves from a movingsourceFromequation(115)itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation117

Ifweaddequations(116)and(117)wegetthatthetotalDopplershiftatthereceiver∆f is

v vf f f f fc c

minus = ∆ = +

But 1 vf fc

= +

hence

1 v v vf f fc c c

∆ = + +

But since v ltlt cwe can ignore the term2

2

vc when we

expandthebracketintheaboveequation

Therefore we have2vf fc

∆ = Equation118

If v asymp c then we must use the full Doppler equationsHowever for em radiationwewill alwaysonly considersituations in which v ltlt c

Example

Thespeedofsoundinbloodis1500times103 m s-1 Ultrasound offrequency100MHzisreflectedfrombloodflowinginan arteryThe frequencyof the reflectedwaves receivedbackatthetransmitteris105MHzEstimatethespeedofthebloodflowintheartery

Solution

Using equation (118) we have

6 63

2005 10 1015 10

vtimes = times

timesto give v asymp 36 m s-1 (We have assumed that the ultrasound is incident at right angles to the blood flow)

Exercise 971 Judy is standing on the platform of a station A

high speed train is approaching the station in a straight line at constant speed and is sounding its whistleAsthetrainpassesbyJudythefrequencyof the sound emitted by the whistle as heard by Judychangesfrom640Hzto430HzDetermine

(a) thespeedofthetrain

(b) thefrequencyofthesoundemittedbythewhistle as heard by a person on the train (Speedofsound=330ms-1)

2 AgalaxyismovingawayfromEarthwithaspeedof00500cThewavelengthofaparticularspectral line in light emitted by atomic hydrogen inalaboratoryis656times10-7 m Calculate the valueofthewavelengthofthislinemeasuredinthelaboratoryinlightemittedfromasourceofatomic hydrogen in the galaxy

PHYSICS 2015indb 306 150514 337 PM

10 Fields

Contents

101 ndash Describing fields

102 ndash Fields at work

Essential Ideas

Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Similar approaches can be taken in analysing electrical and gravitational potential problems copy IBO 2014

PHYSICS 2015indb 307 150514 337 PM

Chapter 10

308

AH

L

101 Describing fields

Essential idea Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Understandingsbull Gravitationalfieldsbull Electrostaticfieldsbull Electricpotentialandgravitationalpotentialbull Fieldlinesbull Equipotentialsurfaces

Gravitational potentialWehaveseenthatifweliftanobjectofmassm to a height h abovethesurfaceoftheEarththenitsgainingravitationalpotential energy is mghHoweverthisisbynomeansthefullstoryForastartwenowknowthatgvarieswithh and also the expression really gives a difference in potentialenergybetweenthevaluethattheobjecthasattheEarthrsquossurfaceandthevaluethatithasatheighthSowhatwereallyneedisazeropointCanwefindapointwherethepotentialenergyiszeroandusethispointfromwhichtomeasure changes in potential energy

The point that is chosen is in fact infinity At infinity the gravitationalfieldstrengthofanyobjectwillbezeroSoletusseeifwecandeduceanexpressionforthegaininpotentialenergyofanobjectwhenitisldquoliftedrdquofromthesurfaceoftheEarthtoinfinityThisineffectmeansfindingtheworknecessarytoperformthistask

δr

r r δr+

r infin=

mg

AB

Me

Figure 1001 Gravitational forces

Inthediagramweconsidertheworknecessarytomovethe particle of mass m a distance δr in thegravitationalfield of the Earth

The force on the particle at A is FG Mem

r2----------------=

IftheparticleismovedtoB then since δr isverysmallwe can assume that the field remains constant over thedistance ABTheworkδW done againstthegravitationalfieldoftheEarthinmovingthedistanceABis

δWGMem

r2----------------δrndash=

(rememberthatworkdoneagainstaforceisnegative)

TofindthetotalworkdoneW in going from the surface oftheEarthtoinfinitywehavetoaddalltheselittlebitsofworkThisisdonemathematicallybyusingintegralcalculus

WG Me m

r2----------------ndash

rdR

infin

int G Me m 1

r2----- rd

R

infin

intndash G Me m 1r---ndash

R

infinndash= = =

GMe m 0 1R---ndash

ndashndash=

G MemR----------------ndash=

HencewehavewhereRistheradiusoftheEarththattheworkdonebythegravitationalfieldinmovinganobjectofmass m from R(surfaceoftheEarth)toinfinityisgivenby

WGMe m

R---------------- ndash=

We can generalise the result by calculating the worknecessary per unit mass to take a small mass fromthe surface of the Earth to infinity This we call thegravitational potential Vie

Wewouldgetexactlythesameresultifwecalculatedtheworkdonebythefieldtobringthepointmassfrominfinityto the surface of Earth In this respect the formal definition ofgravitationalpotentialatapointinagravitationalfieldis therefore defined as the work done per unit mass in bringing a point mass from infinity to that point

Clearly then the gravitational potential at any point inthe Earthrsquos field distance r from the centre of the Earth (providingr gt R)is

VGMe

r------------ ndash=

Thepotentialisthereforeameasureoftheamountofworkthathastobedonetomoveparticlesbetweenpointsinagravitationalfieldanditsunit is theJkgndash1 We also note thatthepotentialisnegativesothatthepotentialenergy

NATURE OF SCIENCEParadigm shift The move from direct observable actions being responsible for influence on an object to acceptance of a fieldrsquos ldquoaction at a distancerdquo required a paradigm shift in the world of science (23)

copy IBO 2014

PHYSICS 2015indb 308 150514 337 PM

Fields

309

AH

L

aswemoveawayfromtheEarthrsquossurfaceincreasesuntilitreachesthevalueofzeroatinfinity

IfthegravitationalfieldisduetoapointmassofmassmthenwehavethesameexpressionasaboveexceptthatMe is replaced by m andmustalsoexclude thevalueof thepotential at the point mass itself ie at r = 0

WecanexpressthegravitationalpotentialduetotheEarth(orduetoanysphericalmass)intermsofthegravitationalfield strength at its surface

AtthesurfaceoftheEarthwehave

g 0ndash Re GMRe

---------ndash= e

Sothat

g 0Re2 GM= e

Hence at a distance r from the centre of the Earth the gravitationalpotentialVcanbewrittenas

VGMe

r------------ndashg 0 Re

2

r------------ndash= =

The potential at the surface of the Earth

(r = Re)istherefore-g0Re

It is interesting to see how the expression for thegravitational potential ties in with the expression mgh ThepotentialatthesurfaceoftheEarthis-g0Re (see the exampleabove)andataheighthwillbe-g0g 0 Re h+( )ndash ifweassume that g0doesnotchangeover thedistanceh The differenceinpotentialbetweenthesurfaceandtheheighth is therefore g0hSotheworkneededtoraiseanobjectof mass m to a height h is mgh ie m timesdifference ingravitationalpotential

This we have referred to as the gain in gravitationalpotentialenergy(see235)

However this expression can be extended to any twopoints inanygravitationalfieldsuch that ifanobjectofmass mmovesbetween twopointswhosepotentialsareV1 and V2 respectively then the change in gravitationalpotentialenergyoftheobjectism(V1 ndash V2)

Gravitational potential gradient

Let us consider now a region in space where thegravitationalfieldisconstantInFigure912thetwopointsAandBareseparatedbythedistance∆x

A B

direction of uniform gravitational eld of strength I

∆x

Figure 1002 The gravitational potential gradient

The gravitational field is of strength I and is in the direction shownThe gravitational potential at A is V andatBisV +∆V

Theworkdone is takingapointmassmfromAtoB isF∆x = mI∆x

Howeverbydefinitionthisworkisalsoequalto-m∆V

Therefore mI∆x=-m∆V

or VIx

∆= minus

∆Effectivelythissaysthatthemagnitudeofthegravitationalfield strength is equal to the negative gradient of thepotential If I is constant then V is a linear function of x and Iisequaltothenegativegradientofthestraightlinegraph formed by plotting V against x If I is not constant (asusuallythecase)thenthemagnitudeofI at any point in the field can be found by find the gradient of the V-x graph at that point An example of such a calculation can befoundinSection929

For those of you who do HL maths the relationship between field and potential is seen to follow from the expression for the potential of a point mass viz

mV Gr

= minus

2

ddV mG Ir r

minus = + =

Potential due to one or more point masses

Gravitationalpotentialisascalarquantitysocalculatingthepotential due to more than one point mass is a matter of simpleadditionSoforexamplethepotentialV due to the Moon and Earth and a distance x fromthecentreofEarthonastraightlinebetweenthemisgivenbytheexpression

E MM MV Gx r x

= minus + minus

whereME=massofEarthMM= mass of Moon and r = distancebetweencentreofEarthandMoon

PHYSICS 2015indb 309 150514 337 PM

Chapter 10

310

AH

L

Equipotentials and field linesIfthegravitationalpotentialhasthesamevalueatallpointson a surface the surface is said to be an equipotential surfaceSoforexampleifweimagineasphericalshellaboutEarthwhosecentrecoincideswiththecentreofEarththisshellwillbeanequipotentialsurfaceClearlyifwerepresentthegravitationalfieldstrengthbyfieldlinessincethelinesldquoradiaterdquooutfromthecentreofEarththentheselineswillbeatrightanglestothesurfaceIfthefieldlineswerenotnormaltotheequipotentialsurfacethentherewouldbeacomponentof the field parallel to the surfaceThis wouldmean thatpointsonthesurfacewouldbeatdifferentpotentialsandsoitwouldnolongerbeanequipotentialsurfaceThisofcourseholdstrueforanyequipotentialsurface

Figure 1003 shows the field lines and equipotentials fortwopointmassesm

m m

Figure 1003 Equipotentials for two point masses

It is worth noting that we would get exactly the samepattern ifwewere to replace thepointmasseswith twoequalpointcharges(See935)

Escape speedThe potential at the surface of Earth is which meansthattheenergyrequiredtotakeaparticleofmass mfromthesurfacetoinfinityisequalto

ButwhatdoesitactuallymeantotakesomethingtoinfinityWhentheparticleisonthesurfaceoftheEarthwecanthinkofitassittingatthebottomofaldquopotentialwellrdquoasinfigure1004

particle

surface of Earth

innity

GMR---------

FIgure 1004 A potential well

Theldquodepthrdquoofthewellis and if the particle gains an amount of kinetic energy equal to wherem is its massthenitwillhavejustenoughenergytoldquoliftrdquoitoutofthewell

In reality it doesnrsquot actually go to infinity it justmeansthat the particle is effectively free of the gravitationalattractionof theEarthWe say that ithas ldquoescapedrdquo theEarthrsquosgravitationalpullWemeetthisideainconnectionwithmolecular forcesTwomolecules in a solidwill sitat their equilibrium position the separation where therepulsiveforceisequaltotheattractiveforceIfwesupplyjust enough energy to increase the separation of themolecules such that they are an infinite distance apart then intermolecularforcesnolongeraffectthemoleculesandthesolidwillhavebecomealiquidThereisnoincreaseinthekineticenergyofthemoleculesandsothesolidmeltsat constant temperature

Wecancalculatetheescapespeedofanobjectveryeasilyby equating the kinetic energy to the potential energysuch that

12---mvesc ape

2 GMe mRe

----------------=

vesc aperArr2GMe

Re--------------- 2g0 Re = =

Substituting for g0 and Regivesavalueforvescape of about 11kmsndash1 from the surface of the Earth

Youwillnotethattheescapespeeddoesnotdependonthemassoftheobjectsincebothkineticenergyandpotentialenergy are proportional to the mass

IntheoryifyouwanttogetarockettothemoonitcanbedonewithoutreachingtheescapespeedHoweverthiswouldnecessitate anenormousamountof fuel and it islikely that therocketplus fuelwouldbesoheavythat itwouldnevergetoffthegroundItismuchmorepracticalto accelerate the rocket to the escape speed fromEarthorbitandthenintheoryjustlaunchittotheMoon

Example

Use the followingdata todetermine thepotential at thesurface of Mars and the magnitude of the acceleration of free fall

massofMars =64times1023kg

radiusofMars =34times106 m

Determinealsothegravitationalfieldstrengthatadistanceof 68 times 106mabovethesurfaceofMars

PHYSICS 2015indb 310 150514 337 PM

Fields

311

AH

L

Solution

= minus = minus times times = times23

11 76

64 1067 10 ndash13 10

34 10MV GR

minus timestimes

Nkg-1

But V = -g0R

Therefore 7

0 6

13 10 3834 10

VgR

times= minus = =

timesm s-2

To determine the field strength gh at 68 times 106 m above the surface we use the fact that 0 2

Mg GR

= such that GM = g0R2

Therefore 2 2

0h 2 2 2

h h

38 (34) 042(102)

g RGMgR R

times= = = = m s-2

(the distance from the centre is 34 times 106 + 68 times 106 = 102 times 106 m)

Exercise 1011 Thegraphbelowshowshowthegravitational

potentialoutsideoftheEarthvarieswithdistancefrom the centre

0 10 20 30 40 50 60 70

ndash1

ndash2

ndash3

ndash4

ndash5

ndash6

ndash7

r m times 106

V

Jkgndash1

times 10

7

(a) Usethegraphtodeterminethegainingravitationalpotentialenergyofasatelliteofmass200kgasitmovesfromthesurfaceof the Earth to a height of 30 times 107mabovetheEarthrsquossurface

(b) Calculatetheenergyrequiredtotakeit to infinity

(c) DeterminetheslopeofthegraphatthesurfaceoftheEarthmCommenton youranswer

Electric potential energy

The concept of electric potential energy was developedwith gravitational potential energy in mind Just as anobjectnearthesurfaceoftheEarthhaspotentialenergybecause of its gravitational interaction with the Earthsotoothereiselectricalpotentialenergyassociatedwithinteracting charges

Letusfirstlookatacaseoftwopositivepointchargeseachof 1μC that are initially bound together by a thread in a vacuuminspacewithadistancebetweenthemof10cmasshowninFigure1005Whenthethreadiscutthepointchargesinitiallyatrestwouldmoveinoppositedirectionsmovingwithvelocitiesv1 and v2 along the direction of the electrostatic force of repulsion

BEFORE AFTER

v v21

Figure 1005 Interaction of two positive particles

The electric potential energybetweentwopointchargescan be found by simply adding up the energy associated witheachpairofpointchargesForapairofinteractingchargestheelectricpotentialenergyisgivenby

∆U = ∆Ep + ∆Ek = ∆W = ∆Fr = kqQ

____ r2 times r = kqQ

____ r

Becausenoexternalforceisactingonthesystemtheenergyand momentum must be conserved Initially Ek = 0 and Ep =kqQr=9times109 times 1 times 10-1201m=009JWhentheyareagreatdistance fromeachotherEpwillbenegligibleThefinalenergywillbeequaltofrac12mv1

2+frac12mv22=009J

Momentum isalsoconservedand thevelocitieswouldbethe same magnitude but in opposite directions

Electric potential energyismoreoftendefinedintermsofapointchargemovinginanelectricfieldas

lsquothe electricpotential energybetweenany twopoints inan electric field is defined as negativeoftheworkdone by anelectricfieldinmovingapoint electric chargebetweentwolocationsintheelectricfieldrsquo

ΔU = ΔEp = -ΔW = -Fd = qEx

ΔU = ΔEp +ΔEk = ΔW = ΔFr = kqQ r2 x r = kqQ

wherexisthedistancemovedalong(oroppositeto)thedirection of the electric field

Electric potential energyismeasuredinjoule(J)Justasworkisascalarquantitysotooelectricalpotentialenergyis a scalar quantity The negative of the work done by an electric field in moving a unit electric charge between two points is independent of the path taken Inphysicswesaytheelectricfieldisalsquoconservativersquo field

PHYSICS 2015indb 311 150514 337 PM

Chapter 10

312

AH

L

Supposeanexternalforcesuchasyourhandmovesasmallpositive point test charge in the direction of a uniformelectric fieldAs it ismoving itmust be gaining kineticenergyIfthisoccursthentheelectricpotentialenergyofthe unit charge is changing

InFigure1006apointcharge+qismovedbetweenpointsAandBthroughadistancex in a uniform electric field

B

A

x

+ q

Figure 1006 Movement of a positive point charge in a uniform field

InordertomoveapositivepointchargefrompointAtopointBanexternal forcemustbeapplied to thechargeequaltoqE (F = qE)

Since the force is applied through a distance xthennegativeworkhas to bedone tomove the charge because energyis gainedmeaning there is an increaseelectric potential energybetweenthetwopointsRememberthattheworkdoneisequivalenttotheenergygainedorlostinmovingthe charge through the electric field The concept of electric potential energy is only meaningful as the electric field whichgeneratestheforceinquestionisconservative

W F xtimes E q x times= =

θ x cos θ x

Figure 1007 Charge moved at an angle to the field

If a chargemoves at an angle θ to an electric field thecomponent of the displacement parallel to the electric fieldisusedasshowninFigure918

W F x Eq x θcostimes= =

Theelectricpotentialenergyisstoredintheelectricfieldand theelectricfieldwill return theenergy to thepointcharge when required so as not to violate the Law ofconservationofenergy

Electric potential

The electric potential at a point in an electric field is defined as being the work done per unit charge in bringing a small positive point charge from infinity to that point

ΔV = Vinfin ndash Vf = ndash W ___ q

Ifwedesignatethepotentialenergytobezeroatinfinitythen it follows thatelectricpotentialmustalsobezeroat infinity and the electric potential at any point in an electricfieldwillbe

ΔV = ndash W ___ q

NowsupposeweapplyanexternalforcetoasmallpositivetestchargeasitismovedtowardsanisolatedpositivechargeTheexternalforcemustdoworkonthepositivetestchargetomoveittowardstheisolatedpositivechargeandtheworkmustbepositivewhiletheworkdonebytheelectricfieldmustthereforebenegativeSotheelectricpotentialatthatpointmustbepositiveaccordingtotheaboveequationIfanegativeisolatedchargeisusedtheelectricpotentialatapointonthepositivetestchargewouldbenegativePositivepoint chargesof their ownaccordmove fromaplaceofhighelectricpotential toaplaceof lowelectricpotentialNegative point charges move the other way from lowpotentialtohighpotentialInmovingfrompointAtopointBinthediagramthepositivecharge+qismovingfromalowelectricpotentialtoahighelectricpotential

In thedefinitiongiven the termldquoworkperunitchargerdquohas significance If the test charge is +16 times 10-19Cwherethe charge has a potential energy of 32 times 10-17Jthenthepotentialwouldbe32times10-17J+16times10-19C=200JC-1 Now if the charge was doubled the potential wouldbecome 64times10-17 J However the potential per unitchargewouldbethesame

Electric potential is a scalar quantityandithasunitsJC-1 orvoltswhere1voltequalsone joulepercoloumbThevoltallowsustoadoptaunitfortheelectricfieldintermsofthevolt

PreviouslytheunitfortheelectricfieldwasNCndash1

W = qV and F = qEsoW ___ V = F __ E

E = FV ___ W = FV ___ Fm V mndash1

That is the units of the electric field E can also beexpressed as V mndash1

The work done per unit charge in moving a point charge between two points in an electric field is again independent of the path taken

PHYSICS 2015indb 312 150514 337 PM

Fields

313

AH

L

Electric potential due to a point charge

Letus takeapoint rmetres froma chargedobjectThepotentialatthispointcanbecalculatedusingthefollowing

W = ndashFr = ndashqV and F = ndash q1 q2 _____ 4πε0r

2

Therefore

W = ndash q1 q2 _____ 4πε0r

2 times r = ndash q1 q2 _____ 4πε0r

= ndashq1 times q2 _____ 4πε0r

= ndashq1V

That is V q

4πε0r--------------- =

Orsimply

V kqr------

=

Example

Determinehowmuchworkisdonebytheelectricfieldofpointcharge150μCwhenachargeof200μCismovedfrom infinity to apoint 0400m from thepoint charge(Assumenoaccelerationofthecharges)

Solution

The work done by the electric field is W = -qV = -14πε0 times q times (Q rinfin - Q r0400)

W = (- 200 times 10-6 C times 900 times 109 NmC-2 times 150 times 10-6 C) divide 0400 m = - 0675 J

An external force would have to do +0675 J of work

Electric field strength and electric potential gradient

Letus lookbackatFigure1006Supposeagain that thecharge+qismovedasmalldistancebyaforceFfromAtoBsothattheforcecanbeconsideredconstantTheworkdoneisgivenby

∆W F x∆times=

TheforceFandtheelectricfieldEareoppositelydirectedandweknowthat

F = -qE and ΔW = q ΔV

Thereforetheworkdonecanbegivenas

q ΔV = -q E Δx

Therefore E V∆x∆-------

ndash=

The rate of change of potential ΔV at a point with respect to distance Δx in the direction in which the change is maximum is called the potential gradient We say that the electric field = - the potential gradient and the units are Vm-

1 From the equation we can see that in a graph of electric potential versus distance the gradient of the straight line equals the electric field strength

In reality if a charged particle enters a uniform electric field it will be accelerated uniformly by the field and its kinetic energy will increase This is why we had to assume no acceleration in the last worked example

Ek∆ 12---

mv2 q E xsdot sdot q Vx---

xsdot sdot q Vsdot= = = =

Example

Determine how far apart two parallel plates must besituated so that a potential difference of 150 times 102 V produces an electric field strength of 100 times 103 NC-1

Solution

Using E V∆x------- xhArrndash V∆

E-------15 102 Vtimes

100 103 N C 1ndashtimes------------------------------------------= = =

= 150 times 10-1

The plates are 150 times 10-1 m apart

The electric field and the electric potential at a point due to an evenly distributed charge +q on a sphere can be represented graphically as in Figure 1008

r

r

V

E

r0

r0

r0

Charge of +Q evenly distributed over surface

E 14πε0------------ Q

r2----- r r0gt=

V 14πε0------------Q

r----r r0gt=

Figure 1008 Electric field and potential due to a charged sphere

PHYSICS 2015indb 313 150514 337 PM

Chapter 10

314

AH

L

When the sphere becomes charged we know that thechargedistributes itselfevenlyoverthesurfaceThereforeeverypartofthematerialoftheconductorisatthesamepotential As the electric potential at a point is defined as beingnumericallyequaltotheworkdoneinbringingaunitpositivechargefrominfinitytothatpointithasaconstantvalueineverypartofthematerialoftheconductor

Since the potential is the same at all points on the conducting surfacethen∆V∆xiszeroButE=ndash∆V∆xThereforethe electric field inside the conductor is zero There is no electric field inside the conductor

SomefurtherobservationsofthegraphsinFigure1009are

bull OutsidethespherethegraphsobeytherelationshipsgivenasE α 1 r2 and V α 1 r

bull Atthesurfacer = r0Thereforetheelectricfieldandpotentialhavetheminimumvalueforr at this point and this infers a maximum field and potential

bull Insidethespheretheelectricfieldiszerobull Insidethespherenoworkisdonetomoveacharge

fromapointinsidetothesurfaceThereforethereisno potential difference and the potential is the same asitiswhenr = r0

Similargraphscanbedrawnfortheelectricfieldintensityandthe electric potential as a function of distance from conducting parallelplatesandsurfacesandthesearegiveninFigure1009

Potential plot E eld plotE eld

++++

ndashndashndashndash

xx x

+

x x

ndashx x

+ ndashx

x

+ +

x

x

Figure 1009 Electric field and electric potential at a distance from a charged surface

Potential due to one or more point charges

The potential due to one point charge can be determined byusingtheequationformula

V = kq r

Example 1

Determine the electric potential at a point 20 times 10-1 m fromthecentreofan isolatedconducting spherewithapointchargeof40pCinair

Solution

Using the formula

V = kq r we have

V90 10 9times( ) 40 10 12ndashtimes( )times

20 10 1ndashtimes( )------------------------------------------------------------------ 018 V= =

the potential at the point is 180 times 10-1 V

The potential due to a number of point charges can be determined by adding up the potentials due to individual point charges because the electric potential at any point outside a conducting sphere will be the same as if all the charge was concentrated at its centre

Example 2

Threepointchargesofareplacedattheverticesofaright-angledtriangleasshowninthediagrambelowDeterminetheabsolutepotentialat the+20μCchargeduetothetwoothercharges

- 6μC

+3μC +2μC

4 m

3 m

Solution

The electric potential of the +2 μC charge due to the ndash 6 μC charge is

V = (9 times 109 Nm2C-2 times -6 times 10-6 C) divide (radic 32 + 42) m = - 108 times 104 V

PHYSICS 2015indb 314 150514 338 PM

Fields

315

AH

L

Insummarywecanconcludethat

bull No work is done to move a charge along an equipotential

bull Equipotentials are always perpendicular to the electric lines of force

Figure1011and1012showsomeequipotential lines fortwo oppositely charged and identically positive spheresseparated by a distance

+ve ndashve

equipotential lines

Figure 1011 Equipotential lines between two opposite charges

equipotential lines

Figure 1012 Equipotential lines between two charges which are the same

+

ndash ndash ndash ndash

+ + +

40 V

30 V

20 V

10 V

50 V

Figure 1013 Equipotential lines between charged parallel plates

Figure1013showstheequipotentiallinesbetweenchargedparallel plates Throughout this chapter the similarities and differences between gravitational fields and electric fieldshavebeendiscussedTherelationshipsthatexistsbetweengravitationalandelectricquantitiesandtheeffectsofpointmassesandchargesissummarisedinTable1014

The electric potential of the +2 μC charge due to the +3 μC charge is

V = (9 times 109 Nm2C-2 times 3 times 10-6 C) divide 3m = 9 times 103 V

The net absolute potential is the sum of the 2 potentials- 108 times 104 V + 9 times 103 V =- 18 times 103 V

The absolute potential at the point is - 18 times 103 V

Equipotential surfacesRegions in space where the electric potential of a chargedistributionhasaconstantvaluearecalledequipotentials The placeswherethepotentialisconstantinthreedimensionsarecalled equipotential surfacesandwheretheyareconstantintwodimensionstheyarecalledequipotential lines

TheyareinsomewaysanalogoustothecontourlinesontopographicmapsInthiscasethegravitationalpotentialenergy is constant as amassmoves around the contourlines because the mass remains at the same elevationabovetheEarthrsquossurfaceThegravitationalfieldstrengthacts in a direction perpendicular to a contour line

Similarlybecausetheelectricpotentialonanequipotentiallinehas thesamevalueanelectric forcecandonoworkwhena testchargemovesonanequipotentialThereforethe electric field cannot have a component along anequipotentialandthusitmustbeeverywhereperpendiculartotheequipotentialsurfaceorequipotentiallineThisfactmakesiteasytoplotequipotentialsifthelinesofforceorlinesofelectricfluxofanelectricfieldareknown

Forexamplethereareaseriesofequipotentiallinesbetweentwoparallelplateconductorsthatareperpendiculartotheelectric fieldTherewill be a series of concentric circles(eachcirclefurtherapartthanthepreviousone)thatmapouttheequipotentialsaroundanisolatedpositivesphereThe lines of force and some equipotential lines for anisolatedpositivesphereareshowninFigure1010

Lines ofequipotential

Figure 1010 Equipotentials around an isolated positive sphere

PHYSICS 2015indb 315 150514 338 PM

Chapter 10

316

AH

L

Gravitational quantity Electrical quantityQ

uant

ities

g V∆x∆-------ndash= E V∆

x∆-------ndash=

Poin

t mas

ses a

nd ch

arge

s

V Gmr----

ndash= V 14πε0------------q

r---=

g G m

r2-----ndash= E 14πε0------------ q

r2-----=

F Gm1 m2

r2--------------= F 1

4πε0------------

q1q2

r2-----------=

Figure 1014 Formulas

Example

Deduce the electric potential on the surface of a gold nucleus that has a radius of 62 fm

Solution

Using the formula

V = kq r and knowing the atomic number of gold is 79 We will assume the nucleus is spherical and it behaves as if it were a point charge at its centre (relative to outside points)

V = 90 times 109 Nm2C-2 times 79 times 16 times 10-19 C divide 62 times 10-15 m = 18 times 107 V

The potential at the point is 18 MV

Example

Deduce the ionisation energy in electron-volts of theelectron in the hydrogen atom if the electron is in its ground state and it is in a circular orbit at a distance of 53 times 10-11 m from the proton

Solution

This problem is an energy coulombic circular motion question based on Bohrrsquos model of the atom (not the accepted quantum mechanics model) The ionisation energy is the energy required to remove the electron from the ground state to infinity The electron travels in a circular orbit and therefore has a centripetal acceleration The ionisation energy will counteract the coulombic force and the movement of the electron will be in the opposite direction to the centripetal force

Total energy = Ek electron + Ep due to the proton-electron interaction

ΣF = kqQ r2 = mv2 r and as such mv2 = = kqQ r

Therefore Ek electron = frac12 kqQ r

Ep due to the proton-electron interaction = - kqQ r

Total energy = frac12 kqQ r + - kqQ r = -frac12 kqQ r

= - 90 times 109 Nm2C-2 times (16 times 10-19 C)2 divide 53 times 10-11 m = -217 times 10-18 J

= -217 times 10-18 J divide 16 times 10-19 = -136 eV

The ionisation energy is 136 eV

Exercise 1021 ApointchargePisplacedmidwaybetweentwo

identicalnegativechargesWhichoneofthefollowingiscorrectwithregardstoelectricfieldandelectricpotentialatpointP

Electric field Electric potentialA non-zero zeroB zero non-zeroC non-zero non-zeroD zero zero

2 Twopositivechargedspheresaretiedtogetherinavacuumsomewhereinspacewheretherearenoexternal forces A has a mass of 25 g and a charge of20μCandBhasamassof15gandachargeof30μCThedistancebetweenthemis40cmTheyarethenreleasedasshowninthediagram

BEFORE AFTER

v1 v2

A B

(a) Determinetheirinitialelectricpotentialenergy in the before situation

(b) DeterminethespeedofsphereBafterrelease

3 Thediagrambelowrepresentstwoequipotentiallines in separated by a distance of 5 cm in a uniform electric field

+ + + + + + + +

ndash ndash ndash ndash ndash ndash ndash ndash

40 V

20 V 5 cm

PHYSICS 2015indb 316 150514 338 PM

Fields

317

AH

L

Determine the strength of the electric field

4 Thisquestionisabouttheelectricfieldduetoacharged sphere and the motion of electrons in that fieldThediagrambelowshowsanisolatedmetalsphereinavacuumthatcarriesanegativeelectriccharge of 60 μC

ndash

(a) Onthediagramdrawtheconventionalwayto represent the electric field pattern due to the charged sphere and lines to represent threeequipotentialsurfacesintheregionoutside the sphere

(b) Explainhowthelinesrepresentingtheequipotentialsurfacesthatyouhavesketchedindicatethatthestrengthoftheelectricfieldisdecreasingwithdistancefrom the centre of the sphere

(c) Theelectricfieldstrengthatthesurfaceofthe sphere and at points outside the sphere can be determined by assuming that the sphere acts as a point charge of magnitude 60 μC at its centre The radius of the sphere is 25 times 10ndash2 m Deduce that the magnitude of the field strength at the surface of the sphere is 86 times 107 Vmndash1

An electron is initially at rest on the surface of the sphere

(d) (i) Describethepathfollowedbytheelectronasitleavesthesurfaceofthesphere

(ii) Calculatetheinitialaccelerationofthe electron

5 Determinetheamountofworkthatisdoneinmovingachargeof100nCthroughapotentialdifferenceof150times102 V

6 Three identical 20 μC conducting spheres are placedatthecornersofanequilateraltriangleofsides 25 cm The triangle has one apex C pointing upthepageand2baseanglesAandBDeterminetheabsolutepotentialatB

7 Determinehowfaraparttwoparallelplatesmustbesituatedsothatapotentialdifferenceof250 times 102 V produces an electric field strength of 200 times 103 NC-1

8 Thegapbetweentwoparallelplatesis10times10-3mandthereisapotentialdifferenceof10times104 V betweentheplatesCalculate

i theworkdonebyanelectroninmovingfrom one plate to the other

ii thespeedwithwhichtheelectronreachesthe second plate if released from rest

iii theelectricfieldintensitybetweentheplates

9 Anelectronguninapicturetubeisacceleratedby a potential 25 times 103VDeterminethekineticenergygainedbytheelectroninelectron-volts

10 Determine the electric potential 20 times 10-2 m from achargeof-10times10-5 C

11 Determinetheelectricpotentialatapointmid-waybetweenachargeofndash20pCandanotherof+5pConthelinejoiningtheircentresifthecharges are 10 cm apart

12 During a thunderstorm the electric potential differencebetweenacloudandthegroundis10 times 109 V Determine the magnitude of the change in electric potential energy of an electron thatmovesbetweenthesepointsinelectron-volts

13 A charge of 15 μC is placed in a uniform electric fieldoftwooppositelychargedparallelplateswithamagnitudeof14times103 NC-1

(a) Determinetheworkthatmustbedoneagainstthefieldtomovethepointchargeadistance of 55 cm

(b) Calculatethepotentialdifferencebetweenthe final and initial positions of the charge

(c) Determinethepotentialdifferencebetweenthe plates if their separation distance is 15 cm

14 Duringaflashoflightningthepotentialdifferencebetweenacloudandthegroundwas12 times 109 V and the amount of transferred charge was32C

(a) Determinethechangeinenergyofthetransferred charge

(b) Iftheenergyreleasedwasallusedtoacceleratea1tonnecardeduceitsfinalspeed

(c) Iftheenergyreleasedcouldbeusedtomelticeat0degCdeducetheamountoficethatcould be melted

PHYSICS 2015indb 317 150514 338 PM

Chapter 10

318

AH

L

15 SupposethatwhenanelectronmovedfromAtoBin the diagram along an electric field line that the electric field does 36 times 10-19Jofworkonit

Determinethedifferencesinelectricpotential

(a) VB ndash VA(b) VC ndash VA(c) VC ndash VB

16 DeterminethepotentialatpointPthatislocatedatthecentreofthesquareasshowninthediagrambelow

- 6μC

+3μC +2μC

1m

1m

5μC

P

102 Fields at work

Essential idea Similar approaches can be taken inanalysingelectricalandgravitationalpotentialproblems

Understandingsbull Potentialandpotentialenergybull Potentialgradientbull Potentialdifferencebull Escapespeedbull Orbitalmotionorbitalspeedandorbitalenergybull Forcesandinverse-squarelawbehaviour

Orbital motion orbital speed and orbital energy

Although orbital motion may be circular elliptical orparabolic this sub-topic only dealswith circular orbitsThis sub-topic is not fundamentally new physics butan application that synthesizes ideas from gravitationcircularmotiondynamicsandenergy

TheMoonorbitstheEarthandinthissenseitisoftenreferredtoasasatelliteoftheEarthBefore1957itwastheonlyEarthsatelliteHoweverin1957theRussianslaunchedthefirstmanmadesatelliteSputnik1SincethisdatemanymoresatelliteshavebeenlaunchedandtherearenowliterallythousandsofthemorbitingtheEarthSomeareusedtomonitortheweathersome used to enable people to find accurately their position on thesurfaceoftheEarthmanyareusedincommunicationsandnodoubtsomeareusedtospyonothercountriesFigure1015showshowinprincipleasatellitecanbeputintoorbit

Theperson(whosesizeisgreatlyexaggeratedwithrespectto Earth) standing on the surface on the Earth throwssome stonesThe greater the speed with which a stoneis thrown the further it will land from her The pathsfollowedbythethrownstonesareparabolasByastretchof the imaginationwecanvisualisea situation inwhichastoneis thrownwithsuchaspeedthatbecauseof thecurvatureoftheEarthitwillnotlandonthesurfaceoftheEarthbutgointoldquoorbitrdquo(Path4onFigure1015)

NATURE OF SCIENCECommunication of scientific explanations The ability to apply field theory to the unobservable (charges) and the massively scaled (motion of satellites) required scientists to develop new ways to investigate analyse and report findings to a general public used to scientific discoveries based on tangible and discernible evidence (51)

copy IBO 2014

PHYSICS 2015indb 318 150514 338 PM

Fields

319

AH

L

1

2

34Earth

Figure 1015 Throwing a stone into orbit

TheforcethatcausesthestonestofollowaparabolicpathandtofalltoEarthisgravityandsimilarlytheforcethatkeeps the stone in orbit is gravity For circular motionto occur we have seen that a force must act at rightanglestothevelocityofanobjectthatistheremustbeacentripetalforceHenceinthesituationwedescribeherethe centripetal force for circular orbital motion about the EarthisprovidedbygravitationalattractionoftheEarth

We can calculate the speedwithwhich a stonemustbethrowninordertoputitintoorbitjustabovethesurfaceof the Earth

If the stone has mass m and speed v thenwehave fromNewtonrsquos2ndlaw

mv2

RE---------- G

MEm

RE2------------=

whereRE is the radius of the Earth and ME is the mass of the Earth

Bearinginmindthatg 0 GME

RE2--------= then

v g RE 10 64 106timestimes 8 103times= = =

Thatisthestonemustbethrownat8times103m sndash1

Clearly we are not going to get a satellite into orbit soclosetothesurfaceoftheEarthMovingatthisspeedthefrictionduetoairresistancewouldmeltthesatellitebeforeithadtravelledacoupleofkilometresInrealitythereforeasatelliteisputintoorbitabouttheEarthbysendingitattachedtoarocketbeyondtheEarthrsquosatmosphereandthengivingitacomponentofvelocityperpendiculartoaradialvectorfromtheEarthSeeFigure1016

Earth Satellite carried by rocket to hereSatellite orbit

Tangential component of velocity

Figure 1016 Getting a satellite into orbit

Keplerrsquos third law(This work of Kepler and Newtonrsquos synthesis of the work is an excellent example of the scientific method and makes for a good TOK discussion)

In 1627 Johannes Kepler (1571-1630)publishedhis lawsof planetary motion The laws are empirical in natureandwere deduced from the observations of theDanishastronomer Tycho de Brahe (1546-1601)The third lawgives a relationship between the radius of orbit R of a planet and its period TofrevolutionabouttheSunThelawisexpressedmathematicallyas

2

3 constantTR

=

We shall nowuseNewtonrsquos LawofGravitation to showhowitisthattheplanetsmoveinaccordancewithKeplerrsquosthirdlaw

InessenceNewtonwasabletousehislawofgravitytopredict the motion of the planets since all he had to do was factor theF givenby this law intohis second law F = ma to find their accelerations and hence their future positions

InFigure1017theEarthisshownorbitingtheSunandthedistancebetweentheircentresisR

RSun

Earth

F es

F se

Figure 1017 Planets move according to Keplerrsquos third law

Fes is the force that the Earth exerts on the Sun and Fse is the force that the Sun exerts on the Earth The forces are equalandoppositeandtheSunandtheEarthwillactuallyorbitaboutacommoncentreHoweversince theSun issoverymuchmoremassivethantheEarththiscommoncentrewillbeclosetothecentreoftheSunandsowecanregard the Earth as orbiting about the centre of the Sun Theotherthingthatweshallassumeisthatwecanignorethe forces that the other planets exert on the Earth (This wouldnotbeawisethingtodoifyouwereplanningtosend a space ship to theMoon for example)We shallalso assume that we have followed Newtonrsquos exampleandindeedprovedthataspherewillactasapointmasssituated at the centre of the sphere

Kepler had postulated that the orbits of the planets areellipticalbut since theeccentricityof theEarthrsquosorbit issmallweshallassumeacircularorbit

PHYSICS 2015indb 319 150514 338 PM

Chapter 10

320

AH

L

TheaccelerationoftheEarthtowardstheSunisa = Rω2

whereω ω 2πT------=

Hencea R 2π

T------ times

2 4π2 RT 2------------- = =

But the acceleration is given by Newtonrsquos Second Law F = mawhereFisnowgivenbytheLawofGravitationSo in this situation

F maGMs Me

R2-------------------= = butwealsohavethat

a a R 2πT

------ times

2 4π2 RT 2------------- = = and m = Me so that

G MsMe

R2------------------- Me4π2R

T2-------------times

GMs

4π2-----------rArr R3

T2------= =

Butthequantity

isaconstantthathasthesamevalueforeachoftheplanetssowehaveforalltheplanetsnotjustEarththat

R3

T2------ k =

wherekisaconstantWhichisofcourseKeplerrsquosthirdlaw

ThisisindeedanamazingbreakthroughItisdifficulttorefutethe idea that all particles attract each other in accordance withtheLawofGravitationwhenthelawisabletoaccountfortheobservedmotionoftheplanetsabouttheSun

The gravitational effects of the planets upon each othershould produce perturbations in their orbits Such is the predictivepoweroftheUniversalGravitationalLawthatit enabled physicists to compute these perturbations The telescopehadbeeninventedin1608andbythemiddleofthe 18th Century had reached a degree of perfection in design that enabled astronomers to actually measure the orbital perturbations of the planets Their measurements werealways inagreementwith thepredictionsmadebyNewtonrsquoslawHoweverin1781anewplanetUranuswasdiscovered and the orbit of this planet did not fit withthe orbit predicted by Universal Gravitation Such wasthe physicistrsquos faith in theNewtonianmethod that theysuspectedthatthediscrepancywasduetothepresenceofayetundetectedplanetUsingtheLawofGravitationtheFrenchastronomerJLeverrier and the English astronomer J C Adamswereable tocalculate justhowmassive thisnewplanetmustbeandalsowhereitshouldbeIn1846the planetNeptunewas discovered justwhere they hadpredicted Ina similarwaydiscrepancies in theorbitofNeptune ledto thepredictionandsubsequentdiscoveryin1930oftheplanetPlutoNewtonrsquosLawofGravitation

had passed the ultimate test of any theory it is not only abletoexplainexistingdatabutalsotomakepredictions

Satellite energyWhenasatelliteisinorbitaboutaplanetitwillhavebothkineticenergyandgravitationalpotentialenergySupposewe consider a satellite ofmass m that is in an orbit of radius r about a planet of mass M

Thegravitationalpotentialduetotheplanetatdistancer from its centre is

ThegravitationalpotentialenergyofthesatelliteVsat

is therefore GMe m

r----------------ndash

ThatisVsat GMem

r----------------ndash=

Thegravitationalfieldstrengthatthesurfaceoftheplanetisgivenby

g 0GMe

Re2------------=

Hencewecanwrite

Vsat g0 Re

2

rndash=

m

ThekineticenergyofthesatelliteKsatisequaltofrac12mv2wherev is its orbital speed

Byequatingthegravitationalforceactingonthesatellitetoitscentripetalaccelerationwehave

G Mem

r2---------------- mv2

r---------- mv2hArr

GMemr----------------= =

Fromwhich12---

mv2 12---

GMemr----------------times=

= 0 e

2g R

r

2m

Which is actually quite interesting since it shows thatirrespective of the orbital radius the KE is numericallyequal to half the PE Also the total energy E tot K sat Vsat+ 1

2---G Me m

r----------------times G Me m

r----------------ndash + 1

2---G Me m

r----------------ndash= = =of the satelliteisalwaysnegativesince

E tot K sat Vsat+ 12---

G Me mr----------------times

G Me mr----------------ndash

+ 12---

G Me mr----------------ndash= = =

PHYSICS 2015indb 320 150514 338 PM

Fields

321

AH

L

The energies of an orbiting satellite as a function of radial distancefromthecentreofaplanetareshownplottedinFigure1018

12

100806

0402

0ndash02

ndash04

ndash06ndash08ndash10

ndash12

2 4 6 8 10 12

ener

gya

rbita

ry u

nits

distance timesR

kinetic energytotal energy

potential energy

Figure 1018 Energy of an orbiting satellite as a function of distance from the centre of a planet

WeightlessnessSuppose that you are inanelevator(lift)thatisdescendingat constant speedandyou let goof abook thatyouareholdinginyourhandThebookwillfalltothefloorwithaccelerationequaltotheaccelerationduetogravityIfthecablethatsupportstheelevatorweretosnap(asituationthatItrustwillneverhappentoanyofyou)andyounowletgothebookthatyouareholdinginyourotherhandthis book will not fall to the floor - it will stay exactlyin linewithyourhandThis isbecause thebook isnowfallingwith the sameaccelerationas theelevatorandassuch the book cannot ldquocatchrdquo up with the floor of theelevatorFurthermoreifyouhappenedtobestandingonasetofbathroomscalesthescaleswouldnowreadzero-youwouldbeapparentlyweightlessItisthisideaoffreefallthatexplainstheapparentweightlessnessofastronautsin an orbiting satellite These astronauts are in free fall inthesensethattheyareacceleratingtowardsthecentre of the Earth

It is actually possible to define theweight of a body inseveral different waysWe can define it for example asthegravitationalforceexertedonthebodybyaspecifiedobject such as the EarthThiswe have seen thatwe doinlotsofsituationswherewedefinetheweightasbeingequaltomgIfweusethisdefinitionthenanobjectinfreefallcannotbydefinitionbeweightlesssinceitisstillinagravitationalfieldHoweverifwedefinetheweightofanobjectintermsofaldquoweighingrdquoprocesssuchasthereadingonasetofbathroomscaleswhichineffectmeasuresthecontact force between the object and the scales thenclearlyobjectsinfreefallareweightlessOnenowhasto

askthequestionwhetherornotitispossibleForexampletomeasurethegravitationalforceactingonanastronautin orbit about the Earth

WecanalsodefineweightintermsofthenetgravitationalforceactingonabodyduetoseveraldifferentobjectsForexample for an object out in space its weight could bedefined in terms of the resultant of the forces exerted on it bytheSuntheMoontheEarthandalltheotherplanetsin the Solar System If this resultant is zero at a particular pointthenthebodyisweightlessatthispoint

Inviewofthevariousdefinitionsofweightthatareavailabletousitisimportantthatwhenweusethewordldquoweightrdquoweareawareofthecontextinwhichitisbeingused

Example

Calculate the height above the surface of the Earth atwhichageo-stationarysatelliteorbits

Solution

A geo-stationary satellite is one that orbits the Earth in such a way that it is stationary with respect to a point on the surface of the Earth This means that its orbital period must be the same as the time for the Earth to spin once on its axis ie 24 hours

From Keplerrsquos third law we have G Ms

4π2----------- R3

T2------=

That is

RMe

mh

using the fact that the force of attraction between the satellite and the Earth is given by

FG Mem

R2----------------=

and that F = ma

where m is the mass of the satellite and a 4π2R

T2-------------=

we have

GMem

R2---------------- m 4π2R

T2-------------times

GMe

4π2------------rArr R3

T2------= =

PHYSICS 2015indb 321 150514 338 PM

Chapter 10

322

AH

L

Now the mass of the Earth is 60 times 1024 kg and the period T measured in seconds is given by T = 86400 s

So substitution gives R = 42 times 106 m

The radius of the Earth is 64 times 106 m so that the orbital height h is about 36 times 107 m

Example

Calculatetheminimumenergyrequiredtoputasatelliteofmass500kgintoanorbitthatisasaheightequaltotheEarthrsquosradiusabovethesurfaceoftheEarth

Solution

We have seen that when dealing with gravitational fields and potential it is useful to remember that

g 0G M

Re2---------= or g0 Re

2 GM=

The gravitational potential at the surface of the Earth is

g 0ndash Re GMRe

---------ndash=

The gravitational potential at a distance R

from the centre of the Earth is

=

The difference in potential between the surface of the Earth and a point distance R from the centre is therefore

V∆ g 0Re 1ReR------ndash

=

If R 2Re= then V∆g0 Re

2------------=

This means that the work required to ldquoliftrdquo the satellite into orbit is g0Rm where m is the mass of the satellite This is equal to

10 32 106 500timestimestimes = 16000 MJ

However the satellite must also have kinetic energy in order to orbit Earth This will be equal to

g 0 mRe2

2R-----------------g 0mRe

2

4----------------- 8000 MJ= =

The minimum energy required is therefore

24000 MJ

Exercise 1031 The speed needed to put a satellite in orbit does

not depend on

A the radius of the orbitB theshapeoftheorbitC thevalueofgattheorbitD the mass of the satellite

2 EstimatethespeedofanEarthsatellitewhoseorbitis400kmabovetheEarthrsquossurfaceAlsodetermine the period of the orbit

3 Calculatethespeedofa200kgsatelliteorbitingthe Earth at a height of 70 times 106 m

Assume that g = 82 m sndash2 for this orbit

4 TheradiioftwosatellitesXandYorbitingthe Earth are 2r and 8rwherer is the radius of the Earth Calculate the ratio of the periods of revolutionofXtoY

5 A satellite of mass mkgissentfromEarthrsquossurfaceinto an orbit of radius 5RwhereR is the radius of theEarthWritedownanexpressionfor

(a) thepotentialenergyofthesatelliteinorbit

(b) thekineticenergyofthesatelliteinorbit

(c) theminimumworkrequiredtosendthesatellitefromrestattheEarthrsquossurfaceintoits orbit

6 A satellite in an orbit of 10rfallsbacktoEarth(radius r)afteramalfunctionDeterminethespeedwithwhichitwillhittheEarthrsquossurface

7 The radius of the moon is frac14 that of the Earth AssumingEarthandtheMoontohavethesamedensitycomparetheaccelerationsoffreefallatthe surface of Earth to that at the surface of the Moon

8 UsethefollowingdatatodeterminethegravitationalfieldstrengthatthesurfaceoftheMoon and hence determine the escape speed from the surface of the Moon

Mass of the Moon = 73 times 1022kg

RadiusoftheMoon=17times106 m

PHYSICS 2015indb 322 150514 338 PM

Wave Phenomena

297

AH

L

different distances If the path difference between them is an integral number of half wavelengths then we might expect the twowaves to be out of phaseHowever ray1 undergoes a phase change of π on reflection but ray2 does not since it is reflected at a boundary betweena more dense and less dense medium (See chapter 4)Henceifthepathdifferenceisanintegralnumberofhalf-wavelengthsrays1and2willreinforceieray1and2areinphaseHowever rays357etcwillbeoutofphasewithrays246etcbutsinceray2ismoreintensethanray3andray4moreintensethanray5thesepairswillnot cancel out so there will be a maximum of intensity

Ifthepathdifferenceissuchthatwave1and2areoutofphasesincewave1ismoreintensethanwave2theywillnotcompletelyannulHoweveritcanbeshownthattheintensitiesofwaves2345hellipaddtoequaltheintensityofwave1Sincewaves345hellipareinphasewithwave2there will be complete cancellation

The path difference will be determined by the angle at which ray1 is incident andalsoon the thickness (andtheactualrefractiveindexaswell)ofthefilmSincethesourceisanextendedsourcethelightwillreachtheeyefrom many different angles and so a system of bright and darkfringeswillbeseenYouwillonlyseefringesifthefilmisverythin(exceptifviewedatnormalincidence)sinceincreasingthethicknessofthefilmwillcausethereflected rays to get so far apart that they will not becollected by the pupil of the eye

From the argument above to find the conditions forconstructive and destructive interferenceweneedonlyfindthepathdifferencebetweenray1andray2Figure921showsthe geometry of the situation

A BE

F

C

D

1 2

φrsquo φrsquo

φ

φ

d

Figure 921 The geometry of interference

Thefilmisofthicknessd and refractive index n and the light has wavelength λ If the line BF is perpendicular to ray 1 then the optical path difference (opd) between ray 1 andray2whenbroughttoafocusis

opd = n(AC+CB)ndashAF

We have to multiple by the refractive index for the path travelledbythelightinthefilmsincethelighttravelsmoreslowly in thefilm If the light travels say adistancex in a material of refractive index n then in the time that it takes to travelthisdistancethelightwouldtraveladistancenx in air

IfthelineCEisatrightanglestoray2thenweseethat

AF = nBE

FromthediagramAC=CDsowecanwrite

opd = n(CD+CB)ndashnBE=nDE

Alsofromthediagramweseethatwhereφ is the angle of refraction

DE=2cosφ

From which opd=2ndcosφ

Bearinginmindthechangeinphaseofray1onreflectionwe have therefore that the condition for constructive interference is

2nd φcos m 12---+

λ m 1 2 hellip = =

Andfordestructiveinterference 2ndcosφ = mλ

Eachfringecorrespondstoaparticularopd for a particular value of the integer m and for any fringe the value of the angle φ is fixedThismeans that itwill be in the formof an arc of a circle with the centre of the circle at the point where the perpendicular drawn from the eye meets the surface of the film Such fringes are called fringesofequal inclinationSince theeyehasa smallaperturethesefringesunlessviewedatneartonormalincidence (φ = 0) will only be observed if the film is very thinThisisbecauseasthethicknessofthefilmincreasesthereflected rayswill get further and further apart and sovery few will enter the eye

Ifwhitelightisshoneontothefilmthenwecanseewhywe get multi-coloured fringes since a series of maxima and minima will be formed for each wavelength present in the white light However when viewed at normalincidenceitispossiblethatonlylightofonecolourwillundergoconstructiveinterferenceandthefilmwilltakeon this colour

PHYSICS 2015indb 297 150514 337 PM

Chapter 9

298

AH

L

Thickness of oil films

The exercise to follow will help explain this use

Non-reflecting films

A very important but simple application of thin filminterferenceisintheproductionofnon-reflectingsurfaces

Athinfilmofthicknessd and refractive index n1 is coated onto glass of refractive index n where n1 lt n Light ofwavelength λ that is reflected at normal incidence willundergo destructive interference if 12

2n d λ

= thatis

14d

nλ

=

(remember that there will now no phase change at the glass interface ie we have a rare to dense reflection)

The use of such films can greatly reduce the loss oflightby reflection at the various surfacesof a systemoflenses or prisms Optical parts of high quality systemsare usually all coatedwithnon-reflecting films in orderto reduce stray reflections The films are usually madebyevaporatingcalciumormagnesiumfluorideonto thesurfacesinvacuumorbychemicaltreatmentwithacidsthat leave a thin layer of silica on the surface of the glass ThecoatedsurfaceshaveapurplishhuebyreflectedlightThis is because the condition for destructive interference fromaparticularfilmthicknesscanonlybeobtainedforone wavelength The wavelength chosen is one that has a value corresponding to light near the middle of the visible spectrumThismeansthatreflectionofredandvioletlightis greater combining to give the purple colour Because of the factor cosφatanglesotherthannormalincidencethepath difference will change butnotsignificantlyuntilsayabout30deg(egcos25deg=090)

It should be borne in mind that no light is actually lost by anon-reflectingfilmthedecreaseofreflectedintensityiscompensated by increase of transmitted intensity

Non-reflecting films can be painted onto aircraft tosuppressreflectionofradarThethicknessofthefilmis determined by

4nd λ

= where λ is the wavelength of

the radar waves and ntherefractiveindexofthefilmat this wavelength

Example

A plane-parallel glass plate of thickness 20 mm isilluminated with light from an extended source The refractiveindexoftheglassis15andthewavelengthofthelight is 600 nm Calculate how many fringes are formed

Solution

We assume that the fringes are formed by light incident at all angles from normal to grazing incidence

At normal incidence we have 2nd = mλ

From which

m 2 15 2 10 3ndashtimestimestimes6 10 7ndashtimes

------------------------------------------- 10 000= =

At grazing incidence the angle of refraction φ is in fact the critical angle

Therefore φφ arcsin 115------- 42 deg= =

ie cosφ = 075

At grazing incidence 2ndcosφ = m λ

From which (and using cos(sinndash1(115) = 075)

m 2 15 2 10 3ndash 075timestimestimestimes6 10 7ndashtimes

------------------------------------------------------------ 7500= =

The total number of fringes seen is equal to m ndash m = 2500

Exercise 95When viewed from above the colour of an oil film onwater appears red in colour Use the data below to estimate theminimumthicknessoftheoilfilm

average wavelength of red light = 630 nm

refractiveindexofoilforredlight=15

refractive index of water for red light = 13

PHYSICS 2015indb 298 150514 337 PM

Wave Phenomena

299

AH

L

94 Resolution

Essential idea Resolution places an absolute limit on the extent to which an optical or other system can separate images of objects

Understandings

bull Thesizeofadiffractingaperturebull Theresolutionofsimplemonochromatic

two-source systems

The size of a diffracting apertureOur discussion concerning diffraction so far has been for rectangular slits When light from a point source enters a smallcircularapertureitdoesnotproduceabrightdotasanimagebutacirculardiscknownasAiryrsquosdiscsurroundedbyfainterconcentriccircularringsasshowninFigure922

Laser

Figure 922 Diffraction at a circular aperture

Diffractionatanapertureisofgreatimportancebecausetheeye and many optical instruments have circular apertures

What is the half-angular width of the central maximum of the diffraction formed by a circular aperture This is not easy to calculate since it involves some advanced mathematics The problemwasfirstsolvedbytheEnglishAstronomerRoyalGeorgeAiryin1835whoshowedthatforcircularapertures

θ 122 λb--------------= where b is the diameter of the aperture

Example 2

Inthefollowingdiagramparallellightfromadistantpointsource(suchasastar)isbroughttofocusonthescreenSbyaconverginglens(the lens is shown as a vertical arrow)

Thefocal length(distance fromlens toscreen) is25cmand the diameter of the lens is 30 cm The wavelength of thelightfromthestaris560nmCalculatethediameterofthe diameter of the image on the screen

Solution

The lens actually acts as a circular aperture of diameter 30 cm The half angular width of central maximum of the diffraction pattern that it forms on the screen is

75

2

122 122 56 10 23 1030 10b

λθminus

minusminus

times times= = = times

times rad

The diameter of the central maxima is therefore

25 times 10-2 times 23 times 10-5

= 57 times 10-6 m

Although this is small it is still finite and is effectively the image of the star as the intensity of the secondary maxima are small compared to that of the central maximum

The resolution of simple monochromatic two-source systemsThe astronomers tell us that many of the stars that we observe with the naked eye are in fact binary stars That iswhatwe see as a single star actually consistsof twostarsinorbitaboutacommoncentreFurthermoretheastronomerstellusthatifweusealdquogoodrdquotelescopethenwewillactuallyseethetwostarsThatiswewillresolve the single point source into its two component parts Sowhatisitthatdetermineswhetherornotweseethetwo stars as a single point source ie what determines whether or not two sources can be resolved It canrsquot just bethatthetelescopemagnifiesthestarssinceiftheyareacting as point sources magnifying them is not going to make a great deal of difference

Ineachofoureyesthereisanaperturethepupilthroughwhich the light enters This light is then focused by the eye lens onto the retina But we have seen that when light passes through an aperture it is diffracted and so when we look at

NATURE OF SCIENCEImproved technology The Rayleigh criterion is the limit of resolution Continuing advancement in technology such as large diameter dishes or lenses or the use of smaller wavelength lasers pushes the limits of what we can resolve (18)

copy IBO 2014

PHYSICS 2015indb 299 150514 337 PM

Chapter 9

300

AH

L

apointsourceadiffractionpatternwillbeformedontheretina If we look at two point sources then two diffraction patterns will be formed on the retina and these patterns will overlap The width of our pupil and the wavelength of the light emitted by the sources will determine the amount by which they overlap But the degree of overlap will also depend on the angular separation of the two point sources WecanseethisfromFigures923

pupilretina

S1

S2

P2

P1

θ

Figure 923

LightfromthesourceS1 enters the eye and is diffracted by the pupil such that the central maximum of the diffraction patternisformedontheretinaatP1SimilarlylightfromS2producesamaximumatP2 If the two central maxima are well separated then there is a fair chance that we will see the two sources as separate sources If they overlap then we will not be able to distinguish one source from another From the diagram we see as the sources are movedclosertotheeyethentheangleθ increases and so does the separation of the central maxima

Figures924925926and927showsthedifferentdiffractionpatternsandtheintensitydistributionthatmightresultonthe retina as a result of light from two point sources

S1 S21 Well resolved

Figure 924 Very well resolved

S1 S22 Well resolved

Figure 925 Well resolved

S1 S23 Just resolved Rayleigh criterion Minimum of S2 coincides with maximum peak of S1

Figure 926 Just resolved

4 Not resolved

Figure 927 Not resolved

We have suggested that if the central maxima of the two diffraction patterns are reasonably separated then we should be able to resolve two point sources In the late 19th century Lord Rayleigh suggested by how much they should be separated in order for the two sources to be just resolved Ifthecentralmaximumofonediffractionpattern coincides with the firstminima of the otherdiffraction pattern then the two sources will just beresolved This is known as the Rayleigh Criterion

Figure926showsjustthissituationThetwosourcesarejust resolved according to the Rayleigh criterion since the peak of the central maximum of one diffraction pattern coincideswiththefirstminimumoftheotherdiffractionpattern This means that the angular separation of the peaks of the two central maxima formed by each source is just the half angular width of one central maximum ie

θ λb---

=

where b is the width of the slit through which the light fromthesourcespassesHoweverweseefromFigure1117that θ is the angle that the two sources subtend at the slit Henceweconcludethattwosourceswillberesolvedbyaslit if the angle that they subtend at the slit is greater than or equaltoλfraslb

PHYSICS 2015indb 300 150514 337 PM

Wave Phenomena

301

AH

L

Sofarwehavebeenassumingthattheeyeisarectangularslit whereas clearly it is a circular aperture and so we must use the formula

θ 122 λb--------------=

As mentioned above the angle θ is sometimes called the resolving power but should more accurately be called the minimum angle of resolution (θmin)

Clearly the smaller θ the greater will be the resolving power

The diffraction grating is a useful for differentiating closely spacedlinesinemissionspectraLikeaprismspectrometerit can disperse a spectrum into its components but it is better suited because it has higher resolution than the prism spectrometer

If λ1 and λ2 are two nearly equal wavelengths that canbarelybedistinguishedtheresolvanceorresolvingpowerofthegratingisdefinedas

R = λ _______ (λ2 ndash λ 1) = λ ___ Δλ

Where λ asymp λ1 asymp λ2

Therefore thediffractiongratingwithahighresolvancewill be better suited in determining small differences in wavelength

If the diffraction grating has Nlinesbeingilluminateditcan be shown that the resolving power of the mth order diffraction is given by

R = mN

Obviouslytheresolvancebecomesgreaterwiththeordernumber m and with a greater number of illuminated slits

Example

A benchmark for the resolving power of a grating is the sodium doublet in the sodium emission spectrum Two yellow lines in itsspectrumhavewavelengthsof58900nmand58959nm

(a) Determinetheresolvanceofagratingifthegivenwavelengths are to be distinguished from each other

(b) Howmanylinesinthegratingmustbeilluminated in order to resolve these lines in the second order spectrum

Solution

(a) R = λ ___ Δλ = 58900nm __________________ (58900ndash58959)nm

= 589nm _______ 059nm = 10 times 103

(b) N = R __ m = 10 times 10 3 ________ 2 =50times102 lines

It has been seen that diffraction effectively limits the resolving power of optical systems This includes such systems as the eye telescopes and microscopes Theresolving power of these systems is dealt with in the next section This section looks at links between technology and resolving power when looking at the very distant and when looking at the very small

Radio telescopes

The average diameter of the pupil of the human eye is about 25 mm This means that the eye will justresolve two point sources emitting light of wavelength 500 nm if their angular separation at the eye is 7

43

50 10122 24 1025 10

θminus

minusminus

times= times = times

times rad

If the eye were to be able to detect radio waves of wavelength 015m then tohave the same resolvingpower thepupilwould have to have a diameter of about 600 m Clearly this is nonsense but it does illustrate a problem facingastronomers who wish to view very distant objects such as quasarsandgalaxies thatemitradiowavesConventionalradio telescopes consist of a large dish typically 25 min diameter Even with such a large diameter the radiowavelength resolving power of the telescope is much less thantheopticalresolvingpowerofthehumaneyeLetuslook at an example

Example

The Galaxy Cygnus A can be resolved optically as an elliptically shaped galaxy However it is also a strongemitterof radiowavesofwavelength015mTheGalaxyisestimatedtobe50times1024mfromEarthUseofaradiotelescope shows that the radio emission is from two sources separated by a distance of 30 times 1021m Estimate thediameterofthedishrequiredtojustresolvethesources

Solution

The angle θ that the sources resolve at the telescope is given by

θ 21

424

30 10 60 1050 10

θ minustimes= = times

times rad

and 4

122 122 015 300060 10

d λθ minus

times= = =

timesm = 30 km

A radio telescope dish of this size would be impossible tomake letalonesupportThisshowsthatasingledishtype radio telescope cannot be used to resolve the sources and yet they were resolved To get round the problemastronomers use two radio telescopes separated by a large

PHYSICS 2015indb 301 150514 337 PM

Chapter 9

302

AH

L

distance The telescopes view the same objects at the same time and the signals that each receive from the objects are simultaneously superimposed The result of the superposition of the two signals is a two-slit interference pattern The pattern has much narrower fringe spacing than that of the diffraction pattern produced by either telescope on its own hence producing a much higherresolvingpowerWhentelescopesareusedlikethistheyare called a stellar interferometer

InSocorroinNewMexicothereisastellarinterferometerthat consists of 27 parabolic dishes each of diameter25 m arranged in a Y-shape that covers an area of 570km2Thisisaso-calledVeryLargeArray(VLA)Evenhigher resolution can be obtained by using an array of radio telescopes in observatories thousands of kilometres apart A system that uses this so-called technique oflsquovery-long-baseline interferometryrsquo (VLBI) is known asalsquovery-long-baselinearrayrsquo(VLBA)WithVLBAaradiowavelength resolving power can be achieved that is 100 timesgreaterthanthebestopticaltelescopesEvenhigherresolving power can be achieved by using a telescope that isinasatelliteorbitingEarthJapanrsquosInstituteofSpaceandAstronautical Science (ISAS) launched sucha system inFebruary 1997TheNationalAstronomicalObservatoryof Japan the National Science Foundationrsquos NationalRadio Astronomy Observatory (NRAO) the CanadianSpaceAgency theAustraliaTelescopeNationalFacilitythe European VLBI Network and the Joint Instituteback this project forVeryLongBaseline Interferometryin EuropeThis project is a very good example of howInternationalismcanoperateinPhysics

Electron microscope

Telescopes are used to look at very distant objects that are very large but because of their distance fromus appearverysmallMicroscopesontheotherhandareusedtolookat objects that are close to us but are physically very small Aswehave seen justmagnifyingobjects that ismakingthemappearlargerisnotsufficientonitsowntogaindetailabouttheobjectfordetailhighresolutionisneeded

Figure928isaschematicofhowanoptical microscope is usedtoviewanobjectandFigure929isaschematicofatransmission electron microscope(TEM)

Optical lens system

bright light source

glass slide containing specimen (object)

eye

Figure 928 The principle of a light microscope

magnetic lens system

electron source

thin wafer of material

ScreenCCD

Figure 929 The principle of an electron microscope

In the optical microscope the resolving power isdetermined by the particular lens system employed and the wavelength λ of the light used For example twopoints in the sample separated by a distance d will just be resolved if

2d mλ=

where m is a property of the lens system know as the numerical aperture In practice the largest value of m obtainable is about 16 Hence if the microscope slideis illuminatedwith light ofwavelength 480 nm a goodmicroscope will resolve two points separated by a distance dasymp15times10-7masymp015micromPointscloser together thanthiswillnotberesolvedHoweverthisisgoodenoughtodistinguishsomevirusessuchastheEbolavirus

Clearlythesmallerλ the higher the resolving power and this is where the electron microscope comes to the fore The electron microscope makes use of the wave nature of electrons(see1315)IntheTEMelectronspassthrougha wafer thin sample and are then focused by a magnetic field onto a fluorescent screen orCCD (charge coupleddevicesee142)Electronsusedinanelectronmicroscopehavewavelengthstypicallyofabout5times10-12mHoweverthe numerical aperture of electron microscopes is considerablysmaller thanthatofanopticalmicroscopetypicallyabout002NonethelessthismeansthataTEMcanresolvetwopointsthatareabout025nmapartThisresolving power is certainly high enough to make out the shape of large molecules

Another type of electron microscope uses a techniqueby which electrons are scattered from the surface of the sample The scattered electrons and then focused as in theTEM to form an image of the surfaceThese so-calledscanningelectronmicroscopes(SEM)havealowerresolving power than TEMrsquos but give very good threedimensional images

The eye

We saw in the last section that the resolving power of the human eye is about 2 times 10-4 rad Suppose that youare looking at car headlights on a dark night and the car

PHYSICS 2015indb 302 150514 337 PM

Wave Phenomena

303

AH

L

is a distance D away If the separation of the headlight is say 15 m then the headlights will subtend an angle θ 15

Dθ =

at your eye Assuming an average wavelength of

500 nm your eye will resolve the headlights into twoseparatesourcesifthisangleequals2times10-4 rad This gives D=75kmInotherwordsifthecarisapproachingyouona straight road then you will be able to distinguish the two headlightsasseparatesourceswhenthecaris75kmawayfrom you Actually because of the structure of the retina and optical defects the resolving power of the average eye is about 6 times 10-4 rad This means that the car is more likely tobe25kmawaybeforeyouresolvetheheadlights

Astronomical telescope

Letusreturntotheexampleofthebinarystarsdiscussedatthe beginning of this section on resolution The stars Kruger A and B form a binary system The average separation of the stars is 14 times 1012mandtheiraveragedistancefromEarthis12times1017mWhenviewedthroughatelescopeonEarththe system will therefore subtend an angle

12

17

14 1012 10

θtimes

=times

= 12times10-5 rad at the objective lens of the telescope Assuming that the average wavelength of the light emitted bythestarsis500nmthenifthetelescopeistoresolvethesystem into two separate images it must have a minimum diameter Dwhere12times10-5 =

7122 500 10D

minustimes times

This gives D=0050m which is about 5 cm So thisparticular system is easily resolved with a small astronomical telescope

Exercise 961 ItissuggestedthatusingtheISASVLBAitwould

bepossibletoldquoseerdquoagrainofriceatadistanceof5000kmEstimatetheresolvingpoweroftheVLBA

2 Thedistancefromtheeyelenstotheretinais20mm The light receptors in the central part of the retinaareabout5times10-6apartDeterminewhetherthe spacing of the receptors will allow for the eye to resolve the headlights in the above discussion whentheyare25kmfromtheeye

3 ThediameterofPlutois23times106 m and its averagedistancefromEarthis60times1012 m EstimatetheminimumdiameteroftheobjectiveofatelescopethatwillenablePlutotobeseenasadisc as opposed to a point source

95 Doppler effect

Essential ideaTheDopplereffectdescribesthephenomenonofwavelengthfrequencyshiftwhenrelativemotionoccurs

Understandingsbull TheDopplereffectforsoundwavesandlightwaves

The Doppler effect for sound waves and light waves

Consider two observers A and B at rest with respect to asoundsourcethatemitsasoundofconstantfrequencyf Clearly both observers will hear a sound of the same frequencyHoweversupposethatthesourcenowmovesat constant speed towards A A will now hear a sound of frequencyfA that is greater than f and B will hear a sound of frequency fB that is less than f This phenomenon is known as the Doppler Effect orDopplerPrinciple after C J Doppler(1803-1853)

The same effect arises for an observer who is either moving towards or away from a stationary source

Figure 930 shows the waves spreading out from a stationary source that emits a sound of constant frequency f The observersAandBhearasoundofthesamefrequency

A B

wavefront

S

Figure 930 Sound waves from a stationary source

Suppose now that the source moves towards A withconstant speed vFigure9310(a)showsasnapshotofthenew wave pattern

NATURE OF SCIENCETechnology Although originally based on physical observations of the pitch of fast moving sources of sound the Doppler effect has an important role in many different areas such as evidence for the expansion of the universe and generating images used in weather reports and in medicine (55)

copy IBO 2014

PHYSICS 2015indb 303 150514 337 PM

Chapter 9

304

AH

L

sourceA B

smallerwavelength

largerwavelength

Figure 931 (a) Sound waves from a moving source

The wavefronts are now crowded together in the direction of travel of the source and stretched out in the opposite direction This is why now the two observers will now hear notesofdifferentfrequenciesHowmuchthewavesbunchtogether and how much they stretch out will depend on the speed vEssentially A

A

cfλ

= and BB

cfλ

= where λA lt λB and v is the speed of sound

IfthesourceisstationaryandAismovingtowardsitthenthe waves from the source incident on A will be bunched up If A is moving away from the stationary source then the waves from the source incident on A will be stretched out

Christian Doppler(1803ndash1853)actuallyappliedtheprinciple(incorrectlyasithappens)totryandexplainthecolourofstars However the Doppler effect does apply to light aswellastosoundIfalightsourceemitsalightoffrequencyf then if it is moving away from an observer the observer willmeasurethelightemittedashavingalowerfrequencythan fSincethesensationofcolourvisionisrelatedtothefrequencyoflight(bluelightisofahigherfrequencythanred light) light emitted by objects moving way from anobserver is often referred to as being red-shifted whereas if the object is moving toward the observer it is referred to as blue-shiftedThisideaisusedinOptionE(Chapter16)

We do not need to consider here the situations where either the source or the observer is accelerating In a situation for example where an accelerating source is approaching astationaryobserverthentheobserverwillhearasoundof ever increasing frequency This sometimes leads toconfusion in describing what is heard when a source approaches passes and then recedes from a stationaryobserver Suppose for example that you are standingon a station platform and a train sounding its whistle is approaching at constant speed What will you hear as the train approaches and then passes through the station As the train approaches you will hear a sound of constant pitch but increasing loudness The pitch of the sound will be greater than if the train were stationary As the train passes through the station you will hear the pitch change at the momentthetrainpassesyoutoasoundagainofconstantpitch The pitch of this sound will be lower than the sound of the approaching train and its intensity will decrease as the train recedes from you What you do not hear is a sound of increasing pitch and then decreasing pitch

The Doppler equations for sound

Although you will not be expected in an IB examination to derivetheequationsassociatedwithaspectsoftheDopplereffectyouwillbeexpectedtoapplythemForcompletenesstherefore thederivationoftheequationsassociatedwiththeDopplereffectasoutlinedaboveisgivenhere

S primeS O

vs

Figure 931 (b)

InFigure931(b)theobserverOisatrestwithrespecttoasourceofsoundSismovingwithconstanspeedvs directly towards O The source is emitting a note of constant frequencyf and the speed of the emitted sound is v

S shows the position of the source ∆t later When the sourceisatresttheninatime∆t the observer will receive f∆t waves and these waves will occupy a distance v∆t ie

v f

λ =f∆tv∆t =

(Because of the motion of the source this number of waves will now occupy a distance (v∆t ndash vs∆t)The lsquonewrsquowavelength is therefore

v ndash vs

fλrsquo =

f∆tv∆t ndash vs∆t

=

If fisthefrequencyheardbyOthen

vfλ

= or

vf

λ = = sv vfminus

From which

s

vf fv v

=minus

Dividingthroughbyv gives

Equation1111

1 sf f v

v=

minus

If the source is moving away from the observer then we have

s

vf fv v

=minus

1

1 sf vv

=+

Equation112

PHYSICS 2015indb 304 150514 337 PM

Wave Phenomena

305

AH

L

We now consider the case where the source is stationary and the observerismoving towardsthesource with speed v0 In this situation the speed of the sound waves as measured by the observer will be v0 + v We therefore have that

v0 + v = ffraslλ = vfλ

times

From which

o1 vf fv

= +

Equation113

If the observerismovingawayfromthesource then

o1 vf fv

= minus

Fromequation(113)wehavethat

o o1 v vf f f f f fv v

∆ = minus = + minus =

Equation114

Thevelocitiesthatwerefertointheaboveequationsarethevelocities with respect to the medium in which the waves fromthesourcetravelHoweverwhenwearedealingwithalight source it is the relative velocity between the source and the observer that we must consider The reason for this is that lightisuniqueintherespectthatthespeedofthelightwavesdoes not depend on the speed of the source All observers irrespective of their speed or the speed of the source will measure the same velocity for the speed of light This is oneof thecornerstonesof theSpecialTheoryofRelativitywhichisdiscussedinmoredetailinOptionH(Chapter18)When applying the Doppler effect to light we aremainlyconcerned with the motion of the source We look here only at the situation where the speed of the source v is much smaller than the speed of light cinfreespace(v ltlt c)Underthesecircumstanceswhenthesourceismovingtowardstheobserverequation111becomes

vf f f fc

minus = ∆ = Equation115

andwhenthesourceismovingawayfromtheobserverequation112becomes vf f f f

cminus = ∆ = minus

Provided that v ltlt c these same equations apply for astationary source and moving observer

We look at the following example and exercise

Example

Asourceemitsasoundoffrequency440HzItmovesinastraight line towards a stationary observer with a speed of 30ms-1Theobserverhearsasoundoffrequency484HzCalculate the speed of sound in air

Solution

We use equation 111 and substitute f= 484 Hz f = 440 Hz

and vs = 30 m s-1

therefore 1

484 440 301v

=minus

such that 30 4401

484vminus = to

give v = 330 m s-1

Example

A particular radio signal from a galaxy is measured as havingafrequencyof139times109HzThesamesignalfromasourceinalaboratoryhasafrequencyof142times109Hz

Suggestwhy thegalaxy ismovingaway fromEarthandcalculateitsrecessionspeed(iethespeedwithwhichitismovingawayfromEarth)

Solution

The fact that the frequency from the moving source is less than that when it is stationary indicates that it is moving away from the stationary observer ie Earth

Using vf fc

∆ = we have

8 96

9

3 10 (142 139) 10 634 10142 10

c fvf∆ times times minus times

= = = timestimes

m s-1

It is usual when dealing with the Doppler effect of light to express speeds as a fraction of c So in this instance we have v = 0021 c

Using the Doppler effectWe have seen in the above example and exercise how the Dopplereffectmaybeusedtomeasuretherecessionspeedof distant galaxies The effect is also used to measure speed inothersituationsHerewewilllookatthegeneralprincipleinvolved in using the Doppler effect to measure speedFigure932showsasource(thetransmitter)thatemitseithersound or em waves of constant frequency f The waves

PHYSICS 2015indb 305 150514 337 PM

Chapter 9

306

AH

L

fromthesourceare incidentonareflectorthat ismovingtowards the transmitter with speed vThereflectedwavesare detected by the receiver placed alongside the transmitter

reector

v

transmitter

receiver

f f

f f

Figure 932 Using the Doppler effect to measure speed

We shall consider the situation where v ltlt c where c is the speed of the waves from the transmitter

For the reflector receiving waves from the transmitterit is effectively an observer moving towards a stationary sourceFromequation(114) itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation116

For the receiver receivingwaves from the reflector it iseffectively a stationary observer receiving waves from a movingsourceFromequation(115)itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation117

Ifweaddequations(116)and(117)wegetthatthetotalDopplershiftatthereceiver∆f is

v vf f f f fc c

minus = ∆ = +

But 1 vf fc

= +

hence

1 v v vf f fc c c

∆ = + +

But since v ltlt cwe can ignore the term2

2

vc when we

expandthebracketintheaboveequation

Therefore we have2vf fc

∆ = Equation118

If v asymp c then we must use the full Doppler equationsHowever for em radiationwewill alwaysonly considersituations in which v ltlt c

Example

Thespeedofsoundinbloodis1500times103 m s-1 Ultrasound offrequency100MHzisreflectedfrombloodflowinginan arteryThe frequencyof the reflectedwaves receivedbackatthetransmitteris105MHzEstimatethespeedofthebloodflowintheartery

Solution

Using equation (118) we have

6 63

2005 10 1015 10

vtimes = times

timesto give v asymp 36 m s-1 (We have assumed that the ultrasound is incident at right angles to the blood flow)

Exercise 971 Judy is standing on the platform of a station A

high speed train is approaching the station in a straight line at constant speed and is sounding its whistleAsthetrainpassesbyJudythefrequencyof the sound emitted by the whistle as heard by Judychangesfrom640Hzto430HzDetermine

(a) thespeedofthetrain

(b) thefrequencyofthesoundemittedbythewhistle as heard by a person on the train (Speedofsound=330ms-1)

2 AgalaxyismovingawayfromEarthwithaspeedof00500cThewavelengthofaparticularspectral line in light emitted by atomic hydrogen inalaboratoryis656times10-7 m Calculate the valueofthewavelengthofthislinemeasuredinthelaboratoryinlightemittedfromasourceofatomic hydrogen in the galaxy

PHYSICS 2015indb 306 150514 337 PM

10 Fields

Contents

101 ndash Describing fields

102 ndash Fields at work

Essential Ideas

Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Similar approaches can be taken in analysing electrical and gravitational potential problems copy IBO 2014

PHYSICS 2015indb 307 150514 337 PM

Chapter 10

308

AH

L

101 Describing fields

Essential idea Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Understandingsbull Gravitationalfieldsbull Electrostaticfieldsbull Electricpotentialandgravitationalpotentialbull Fieldlinesbull Equipotentialsurfaces

Gravitational potentialWehaveseenthatifweliftanobjectofmassm to a height h abovethesurfaceoftheEarththenitsgainingravitationalpotential energy is mghHoweverthisisbynomeansthefullstoryForastartwenowknowthatgvarieswithh and also the expression really gives a difference in potentialenergybetweenthevaluethattheobjecthasattheEarthrsquossurfaceandthevaluethatithasatheighthSowhatwereallyneedisazeropointCanwefindapointwherethepotentialenergyiszeroandusethispointfromwhichtomeasure changes in potential energy

The point that is chosen is in fact infinity At infinity the gravitationalfieldstrengthofanyobjectwillbezeroSoletusseeifwecandeduceanexpressionforthegaininpotentialenergyofanobjectwhenitisldquoliftedrdquofromthesurfaceoftheEarthtoinfinityThisineffectmeansfindingtheworknecessarytoperformthistask

δr

r r δr+

r infin=

mg

AB

Me

Figure 1001 Gravitational forces

Inthediagramweconsidertheworknecessarytomovethe particle of mass m a distance δr in thegravitationalfield of the Earth

The force on the particle at A is FG Mem

r2----------------=

IftheparticleismovedtoB then since δr isverysmallwe can assume that the field remains constant over thedistance ABTheworkδW done againstthegravitationalfieldoftheEarthinmovingthedistanceABis

δWGMem

r2----------------δrndash=

(rememberthatworkdoneagainstaforceisnegative)

TofindthetotalworkdoneW in going from the surface oftheEarthtoinfinitywehavetoaddalltheselittlebitsofworkThisisdonemathematicallybyusingintegralcalculus

WG Me m

r2----------------ndash

rdR

infin

int G Me m 1

r2----- rd

R

infin

intndash G Me m 1r---ndash

R

infinndash= = =

GMe m 0 1R---ndash

ndashndash=

G MemR----------------ndash=

HencewehavewhereRistheradiusoftheEarththattheworkdonebythegravitationalfieldinmovinganobjectofmass m from R(surfaceoftheEarth)toinfinityisgivenby

WGMe m

R---------------- ndash=

We can generalise the result by calculating the worknecessary per unit mass to take a small mass fromthe surface of the Earth to infinity This we call thegravitational potential Vie

Wewouldgetexactlythesameresultifwecalculatedtheworkdonebythefieldtobringthepointmassfrominfinityto the surface of Earth In this respect the formal definition ofgravitationalpotentialatapointinagravitationalfieldis therefore defined as the work done per unit mass in bringing a point mass from infinity to that point

Clearly then the gravitational potential at any point inthe Earthrsquos field distance r from the centre of the Earth (providingr gt R)is

VGMe

r------------ ndash=

Thepotentialisthereforeameasureoftheamountofworkthathastobedonetomoveparticlesbetweenpointsinagravitationalfieldanditsunit is theJkgndash1 We also note thatthepotentialisnegativesothatthepotentialenergy

NATURE OF SCIENCEParadigm shift The move from direct observable actions being responsible for influence on an object to acceptance of a fieldrsquos ldquoaction at a distancerdquo required a paradigm shift in the world of science (23)

copy IBO 2014

PHYSICS 2015indb 308 150514 337 PM

Fields

309

AH

L

aswemoveawayfromtheEarthrsquossurfaceincreasesuntilitreachesthevalueofzeroatinfinity

IfthegravitationalfieldisduetoapointmassofmassmthenwehavethesameexpressionasaboveexceptthatMe is replaced by m andmustalsoexclude thevalueof thepotential at the point mass itself ie at r = 0

WecanexpressthegravitationalpotentialduetotheEarth(orduetoanysphericalmass)intermsofthegravitationalfield strength at its surface

AtthesurfaceoftheEarthwehave

g 0ndash Re GMRe

---------ndash= e

Sothat

g 0Re2 GM= e

Hence at a distance r from the centre of the Earth the gravitationalpotentialVcanbewrittenas

VGMe

r------------ndashg 0 Re

2

r------------ndash= =

The potential at the surface of the Earth

(r = Re)istherefore-g0Re

It is interesting to see how the expression for thegravitational potential ties in with the expression mgh ThepotentialatthesurfaceoftheEarthis-g0Re (see the exampleabove)andataheighthwillbe-g0g 0 Re h+( )ndash ifweassume that g0doesnotchangeover thedistanceh The differenceinpotentialbetweenthesurfaceandtheheighth is therefore g0hSotheworkneededtoraiseanobjectof mass m to a height h is mgh ie m timesdifference ingravitationalpotential

This we have referred to as the gain in gravitationalpotentialenergy(see235)

However this expression can be extended to any twopoints inanygravitationalfieldsuch that ifanobjectofmass mmovesbetween twopointswhosepotentialsareV1 and V2 respectively then the change in gravitationalpotentialenergyoftheobjectism(V1 ndash V2)

Gravitational potential gradient

Let us consider now a region in space where thegravitationalfieldisconstantInFigure912thetwopointsAandBareseparatedbythedistance∆x

A B

direction of uniform gravitational eld of strength I

∆x

Figure 1002 The gravitational potential gradient

The gravitational field is of strength I and is in the direction shownThe gravitational potential at A is V andatBisV +∆V

Theworkdone is takingapointmassmfromAtoB isF∆x = mI∆x

Howeverbydefinitionthisworkisalsoequalto-m∆V

Therefore mI∆x=-m∆V

or VIx

∆= minus

∆Effectivelythissaysthatthemagnitudeofthegravitationalfield strength is equal to the negative gradient of thepotential If I is constant then V is a linear function of x and Iisequaltothenegativegradientofthestraightlinegraph formed by plotting V against x If I is not constant (asusuallythecase)thenthemagnitudeofI at any point in the field can be found by find the gradient of the V-x graph at that point An example of such a calculation can befoundinSection929

For those of you who do HL maths the relationship between field and potential is seen to follow from the expression for the potential of a point mass viz

mV Gr

= minus

2

ddV mG Ir r

minus = + =

Potential due to one or more point masses

Gravitationalpotentialisascalarquantitysocalculatingthepotential due to more than one point mass is a matter of simpleadditionSoforexamplethepotentialV due to the Moon and Earth and a distance x fromthecentreofEarthonastraightlinebetweenthemisgivenbytheexpression

E MM MV Gx r x

= minus + minus

whereME=massofEarthMM= mass of Moon and r = distancebetweencentreofEarthandMoon

PHYSICS 2015indb 309 150514 337 PM

Chapter 10

310

AH

L

Equipotentials and field linesIfthegravitationalpotentialhasthesamevalueatallpointson a surface the surface is said to be an equipotential surfaceSoforexampleifweimagineasphericalshellaboutEarthwhosecentrecoincideswiththecentreofEarththisshellwillbeanequipotentialsurfaceClearlyifwerepresentthegravitationalfieldstrengthbyfieldlinessincethelinesldquoradiaterdquooutfromthecentreofEarththentheselineswillbeatrightanglestothesurfaceIfthefieldlineswerenotnormaltotheequipotentialsurfacethentherewouldbeacomponentof the field parallel to the surfaceThis wouldmean thatpointsonthesurfacewouldbeatdifferentpotentialsandsoitwouldnolongerbeanequipotentialsurfaceThisofcourseholdstrueforanyequipotentialsurface

Figure 1003 shows the field lines and equipotentials fortwopointmassesm

m m

Figure 1003 Equipotentials for two point masses

It is worth noting that we would get exactly the samepattern ifwewere to replace thepointmasseswith twoequalpointcharges(See935)

Escape speedThe potential at the surface of Earth is which meansthattheenergyrequiredtotakeaparticleofmass mfromthesurfacetoinfinityisequalto

ButwhatdoesitactuallymeantotakesomethingtoinfinityWhentheparticleisonthesurfaceoftheEarthwecanthinkofitassittingatthebottomofaldquopotentialwellrdquoasinfigure1004

particle

surface of Earth

innity

GMR---------

FIgure 1004 A potential well

Theldquodepthrdquoofthewellis and if the particle gains an amount of kinetic energy equal to wherem is its massthenitwillhavejustenoughenergytoldquoliftrdquoitoutofthewell

In reality it doesnrsquot actually go to infinity it justmeansthat the particle is effectively free of the gravitationalattractionof theEarthWe say that ithas ldquoescapedrdquo theEarthrsquosgravitationalpullWemeetthisideainconnectionwithmolecular forcesTwomolecules in a solidwill sitat their equilibrium position the separation where therepulsiveforceisequaltotheattractiveforceIfwesupplyjust enough energy to increase the separation of themolecules such that they are an infinite distance apart then intermolecularforcesnolongeraffectthemoleculesandthesolidwillhavebecomealiquidThereisnoincreaseinthekineticenergyofthemoleculesandsothesolidmeltsat constant temperature

Wecancalculatetheescapespeedofanobjectveryeasilyby equating the kinetic energy to the potential energysuch that

12---mvesc ape

2 GMe mRe

----------------=

vesc aperArr2GMe

Re--------------- 2g0 Re = =

Substituting for g0 and Regivesavalueforvescape of about 11kmsndash1 from the surface of the Earth

Youwillnotethattheescapespeeddoesnotdependonthemassoftheobjectsincebothkineticenergyandpotentialenergy are proportional to the mass

IntheoryifyouwanttogetarockettothemoonitcanbedonewithoutreachingtheescapespeedHoweverthiswouldnecessitate anenormousamountof fuel and it islikely that therocketplus fuelwouldbesoheavythat itwouldnevergetoffthegroundItismuchmorepracticalto accelerate the rocket to the escape speed fromEarthorbitandthenintheoryjustlaunchittotheMoon

Example

Use the followingdata todetermine thepotential at thesurface of Mars and the magnitude of the acceleration of free fall

massofMars =64times1023kg

radiusofMars =34times106 m

Determinealsothegravitationalfieldstrengthatadistanceof 68 times 106mabovethesurfaceofMars

PHYSICS 2015indb 310 150514 337 PM

Fields

311

AH

L

Solution

= minus = minus times times = times23

11 76

64 1067 10 ndash13 10

34 10MV GR

minus timestimes

Nkg-1

But V = -g0R

Therefore 7

0 6

13 10 3834 10

VgR

times= minus = =

timesm s-2

To determine the field strength gh at 68 times 106 m above the surface we use the fact that 0 2

Mg GR

= such that GM = g0R2

Therefore 2 2

0h 2 2 2

h h

38 (34) 042(102)

g RGMgR R

times= = = = m s-2

(the distance from the centre is 34 times 106 + 68 times 106 = 102 times 106 m)

Exercise 1011 Thegraphbelowshowshowthegravitational

potentialoutsideoftheEarthvarieswithdistancefrom the centre

0 10 20 30 40 50 60 70

ndash1

ndash2

ndash3

ndash4

ndash5

ndash6

ndash7

r m times 106

V

Jkgndash1

times 10

7

(a) Usethegraphtodeterminethegainingravitationalpotentialenergyofasatelliteofmass200kgasitmovesfromthesurfaceof the Earth to a height of 30 times 107mabovetheEarthrsquossurface

(b) Calculatetheenergyrequiredtotakeit to infinity

(c) DeterminetheslopeofthegraphatthesurfaceoftheEarthmCommenton youranswer

Electric potential energy

The concept of electric potential energy was developedwith gravitational potential energy in mind Just as anobjectnearthesurfaceoftheEarthhaspotentialenergybecause of its gravitational interaction with the Earthsotoothereiselectricalpotentialenergyassociatedwithinteracting charges

Letusfirstlookatacaseoftwopositivepointchargeseachof 1μC that are initially bound together by a thread in a vacuuminspacewithadistancebetweenthemof10cmasshowninFigure1005Whenthethreadiscutthepointchargesinitiallyatrestwouldmoveinoppositedirectionsmovingwithvelocitiesv1 and v2 along the direction of the electrostatic force of repulsion

BEFORE AFTER

v v21

Figure 1005 Interaction of two positive particles

The electric potential energybetweentwopointchargescan be found by simply adding up the energy associated witheachpairofpointchargesForapairofinteractingchargestheelectricpotentialenergyisgivenby

∆U = ∆Ep + ∆Ek = ∆W = ∆Fr = kqQ

____ r2 times r = kqQ

____ r

Becausenoexternalforceisactingonthesystemtheenergyand momentum must be conserved Initially Ek = 0 and Ep =kqQr=9times109 times 1 times 10-1201m=009JWhentheyareagreatdistance fromeachotherEpwillbenegligibleThefinalenergywillbeequaltofrac12mv1

2+frac12mv22=009J

Momentum isalsoconservedand thevelocitieswouldbethe same magnitude but in opposite directions

Electric potential energyismoreoftendefinedintermsofapointchargemovinginanelectricfieldas

lsquothe electricpotential energybetweenany twopoints inan electric field is defined as negativeoftheworkdone by anelectricfieldinmovingapoint electric chargebetweentwolocationsintheelectricfieldrsquo

ΔU = ΔEp = -ΔW = -Fd = qEx

ΔU = ΔEp +ΔEk = ΔW = ΔFr = kqQ r2 x r = kqQ

wherexisthedistancemovedalong(oroppositeto)thedirection of the electric field

Electric potential energyismeasuredinjoule(J)Justasworkisascalarquantitysotooelectricalpotentialenergyis a scalar quantity The negative of the work done by an electric field in moving a unit electric charge between two points is independent of the path taken Inphysicswesaytheelectricfieldisalsquoconservativersquo field

PHYSICS 2015indb 311 150514 337 PM

Chapter 10

312

AH

L

Supposeanexternalforcesuchasyourhandmovesasmallpositive point test charge in the direction of a uniformelectric fieldAs it ismoving itmust be gaining kineticenergyIfthisoccursthentheelectricpotentialenergyofthe unit charge is changing

InFigure1006apointcharge+qismovedbetweenpointsAandBthroughadistancex in a uniform electric field

B

A

x

+ q

Figure 1006 Movement of a positive point charge in a uniform field

InordertomoveapositivepointchargefrompointAtopointBanexternal forcemustbeapplied to thechargeequaltoqE (F = qE)

Since the force is applied through a distance xthennegativeworkhas to bedone tomove the charge because energyis gainedmeaning there is an increaseelectric potential energybetweenthetwopointsRememberthattheworkdoneisequivalenttotheenergygainedorlostinmovingthe charge through the electric field The concept of electric potential energy is only meaningful as the electric field whichgeneratestheforceinquestionisconservative

W F xtimes E q x times= =

θ x cos θ x

Figure 1007 Charge moved at an angle to the field

If a chargemoves at an angle θ to an electric field thecomponent of the displacement parallel to the electric fieldisusedasshowninFigure918

W F x Eq x θcostimes= =

Theelectricpotentialenergyisstoredintheelectricfieldand theelectricfieldwill return theenergy to thepointcharge when required so as not to violate the Law ofconservationofenergy

Electric potential

The electric potential at a point in an electric field is defined as being the work done per unit charge in bringing a small positive point charge from infinity to that point

ΔV = Vinfin ndash Vf = ndash W ___ q

Ifwedesignatethepotentialenergytobezeroatinfinitythen it follows thatelectricpotentialmustalsobezeroat infinity and the electric potential at any point in an electricfieldwillbe

ΔV = ndash W ___ q

NowsupposeweapplyanexternalforcetoasmallpositivetestchargeasitismovedtowardsanisolatedpositivechargeTheexternalforcemustdoworkonthepositivetestchargetomoveittowardstheisolatedpositivechargeandtheworkmustbepositivewhiletheworkdonebytheelectricfieldmustthereforebenegativeSotheelectricpotentialatthatpointmustbepositiveaccordingtotheaboveequationIfanegativeisolatedchargeisusedtheelectricpotentialatapointonthepositivetestchargewouldbenegativePositivepoint chargesof their ownaccordmove fromaplaceofhighelectricpotential toaplaceof lowelectricpotentialNegative point charges move the other way from lowpotentialtohighpotentialInmovingfrompointAtopointBinthediagramthepositivecharge+qismovingfromalowelectricpotentialtoahighelectricpotential

In thedefinitiongiven the termldquoworkperunitchargerdquohas significance If the test charge is +16 times 10-19Cwherethe charge has a potential energy of 32 times 10-17Jthenthepotentialwouldbe32times10-17J+16times10-19C=200JC-1 Now if the charge was doubled the potential wouldbecome 64times10-17 J However the potential per unitchargewouldbethesame

Electric potential is a scalar quantityandithasunitsJC-1 orvoltswhere1voltequalsone joulepercoloumbThevoltallowsustoadoptaunitfortheelectricfieldintermsofthevolt

PreviouslytheunitfortheelectricfieldwasNCndash1

W = qV and F = qEsoW ___ V = F __ E

E = FV ___ W = FV ___ Fm V mndash1

That is the units of the electric field E can also beexpressed as V mndash1

The work done per unit charge in moving a point charge between two points in an electric field is again independent of the path taken

PHYSICS 2015indb 312 150514 337 PM

Fields

313

AH

L

Electric potential due to a point charge

Letus takeapoint rmetres froma chargedobjectThepotentialatthispointcanbecalculatedusingthefollowing

W = ndashFr = ndashqV and F = ndash q1 q2 _____ 4πε0r

2

Therefore

W = ndash q1 q2 _____ 4πε0r

2 times r = ndash q1 q2 _____ 4πε0r

= ndashq1 times q2 _____ 4πε0r

= ndashq1V

That is V q

4πε0r--------------- =

Orsimply

V kqr------

=

Example

Determinehowmuchworkisdonebytheelectricfieldofpointcharge150μCwhenachargeof200μCismovedfrom infinity to apoint 0400m from thepoint charge(Assumenoaccelerationofthecharges)

Solution

The work done by the electric field is W = -qV = -14πε0 times q times (Q rinfin - Q r0400)

W = (- 200 times 10-6 C times 900 times 109 NmC-2 times 150 times 10-6 C) divide 0400 m = - 0675 J

An external force would have to do +0675 J of work

Electric field strength and electric potential gradient

Letus lookbackatFigure1006Supposeagain that thecharge+qismovedasmalldistancebyaforceFfromAtoBsothattheforcecanbeconsideredconstantTheworkdoneisgivenby

∆W F x∆times=

TheforceFandtheelectricfieldEareoppositelydirectedandweknowthat

F = -qE and ΔW = q ΔV

Thereforetheworkdonecanbegivenas

q ΔV = -q E Δx

Therefore E V∆x∆-------

ndash=

The rate of change of potential ΔV at a point with respect to distance Δx in the direction in which the change is maximum is called the potential gradient We say that the electric field = - the potential gradient and the units are Vm-

1 From the equation we can see that in a graph of electric potential versus distance the gradient of the straight line equals the electric field strength

In reality if a charged particle enters a uniform electric field it will be accelerated uniformly by the field and its kinetic energy will increase This is why we had to assume no acceleration in the last worked example

Ek∆ 12---

mv2 q E xsdot sdot q Vx---

xsdot sdot q Vsdot= = = =

Example

Determine how far apart two parallel plates must besituated so that a potential difference of 150 times 102 V produces an electric field strength of 100 times 103 NC-1

Solution

Using E V∆x------- xhArrndash V∆

E-------15 102 Vtimes

100 103 N C 1ndashtimes------------------------------------------= = =

= 150 times 10-1

The plates are 150 times 10-1 m apart

The electric field and the electric potential at a point due to an evenly distributed charge +q on a sphere can be represented graphically as in Figure 1008

r

r

V

E

r0

r0

r0

Charge of +Q evenly distributed over surface

E 14πε0------------ Q

r2----- r r0gt=

V 14πε0------------Q

r----r r0gt=

Figure 1008 Electric field and potential due to a charged sphere

PHYSICS 2015indb 313 150514 337 PM

Chapter 10

314

AH

L

When the sphere becomes charged we know that thechargedistributes itselfevenlyoverthesurfaceThereforeeverypartofthematerialoftheconductorisatthesamepotential As the electric potential at a point is defined as beingnumericallyequaltotheworkdoneinbringingaunitpositivechargefrominfinitytothatpointithasaconstantvalueineverypartofthematerialoftheconductor

Since the potential is the same at all points on the conducting surfacethen∆V∆xiszeroButE=ndash∆V∆xThereforethe electric field inside the conductor is zero There is no electric field inside the conductor

SomefurtherobservationsofthegraphsinFigure1009are

bull OutsidethespherethegraphsobeytherelationshipsgivenasE α 1 r2 and V α 1 r

bull Atthesurfacer = r0Thereforetheelectricfieldandpotentialhavetheminimumvalueforr at this point and this infers a maximum field and potential

bull Insidethespheretheelectricfieldiszerobull Insidethespherenoworkisdonetomoveacharge

fromapointinsidetothesurfaceThereforethereisno potential difference and the potential is the same asitiswhenr = r0

Similargraphscanbedrawnfortheelectricfieldintensityandthe electric potential as a function of distance from conducting parallelplatesandsurfacesandthesearegiveninFigure1009

Potential plot E eld plotE eld

++++

ndashndashndashndash

xx x

+

x x

ndashx x

+ ndashx

x

+ +

x

x

Figure 1009 Electric field and electric potential at a distance from a charged surface

Potential due to one or more point charges

The potential due to one point charge can be determined byusingtheequationformula

V = kq r

Example 1

Determine the electric potential at a point 20 times 10-1 m fromthecentreofan isolatedconducting spherewithapointchargeof40pCinair

Solution

Using the formula

V = kq r we have

V90 10 9times( ) 40 10 12ndashtimes( )times

20 10 1ndashtimes( )------------------------------------------------------------------ 018 V= =

the potential at the point is 180 times 10-1 V

The potential due to a number of point charges can be determined by adding up the potentials due to individual point charges because the electric potential at any point outside a conducting sphere will be the same as if all the charge was concentrated at its centre

Example 2

Threepointchargesofareplacedattheverticesofaright-angledtriangleasshowninthediagrambelowDeterminetheabsolutepotentialat the+20μCchargeduetothetwoothercharges

- 6μC

+3μC +2μC

4 m

3 m

Solution

The electric potential of the +2 μC charge due to the ndash 6 μC charge is

V = (9 times 109 Nm2C-2 times -6 times 10-6 C) divide (radic 32 + 42) m = - 108 times 104 V

PHYSICS 2015indb 314 150514 338 PM

Fields

315

AH

L

Insummarywecanconcludethat

bull No work is done to move a charge along an equipotential

bull Equipotentials are always perpendicular to the electric lines of force

Figure1011and1012showsomeequipotential lines fortwo oppositely charged and identically positive spheresseparated by a distance

+ve ndashve

equipotential lines

Figure 1011 Equipotential lines between two opposite charges

equipotential lines

Figure 1012 Equipotential lines between two charges which are the same

+

ndash ndash ndash ndash

+ + +

40 V

30 V

20 V

10 V

50 V

Figure 1013 Equipotential lines between charged parallel plates

Figure1013showstheequipotentiallinesbetweenchargedparallel plates Throughout this chapter the similarities and differences between gravitational fields and electric fieldshavebeendiscussedTherelationshipsthatexistsbetweengravitationalandelectricquantitiesandtheeffectsofpointmassesandchargesissummarisedinTable1014

The electric potential of the +2 μC charge due to the +3 μC charge is

V = (9 times 109 Nm2C-2 times 3 times 10-6 C) divide 3m = 9 times 103 V

The net absolute potential is the sum of the 2 potentials- 108 times 104 V + 9 times 103 V =- 18 times 103 V

The absolute potential at the point is - 18 times 103 V

Equipotential surfacesRegions in space where the electric potential of a chargedistributionhasaconstantvaluearecalledequipotentials The placeswherethepotentialisconstantinthreedimensionsarecalled equipotential surfacesandwheretheyareconstantintwodimensionstheyarecalledequipotential lines

TheyareinsomewaysanalogoustothecontourlinesontopographicmapsInthiscasethegravitationalpotentialenergy is constant as amassmoves around the contourlines because the mass remains at the same elevationabovetheEarthrsquossurfaceThegravitationalfieldstrengthacts in a direction perpendicular to a contour line

Similarlybecausetheelectricpotentialonanequipotentiallinehas thesamevalueanelectric forcecandonoworkwhena testchargemovesonanequipotentialThereforethe electric field cannot have a component along anequipotentialandthusitmustbeeverywhereperpendiculartotheequipotentialsurfaceorequipotentiallineThisfactmakesiteasytoplotequipotentialsifthelinesofforceorlinesofelectricfluxofanelectricfieldareknown

Forexamplethereareaseriesofequipotentiallinesbetweentwoparallelplateconductorsthatareperpendiculartotheelectric fieldTherewill be a series of concentric circles(eachcirclefurtherapartthanthepreviousone)thatmapouttheequipotentialsaroundanisolatedpositivesphereThe lines of force and some equipotential lines for anisolatedpositivesphereareshowninFigure1010

Lines ofequipotential

Figure 1010 Equipotentials around an isolated positive sphere

PHYSICS 2015indb 315 150514 338 PM

Chapter 10

316

AH

L

Gravitational quantity Electrical quantityQ

uant

ities

g V∆x∆-------ndash= E V∆

x∆-------ndash=

Poin

t mas

ses a

nd ch

arge

s

V Gmr----

ndash= V 14πε0------------q

r---=

g G m

r2-----ndash= E 14πε0------------ q

r2-----=

F Gm1 m2

r2--------------= F 1

4πε0------------

q1q2

r2-----------=

Figure 1014 Formulas

Example

Deduce the electric potential on the surface of a gold nucleus that has a radius of 62 fm

Solution

Using the formula

V = kq r and knowing the atomic number of gold is 79 We will assume the nucleus is spherical and it behaves as if it were a point charge at its centre (relative to outside points)

V = 90 times 109 Nm2C-2 times 79 times 16 times 10-19 C divide 62 times 10-15 m = 18 times 107 V

The potential at the point is 18 MV

Example

Deduce the ionisation energy in electron-volts of theelectron in the hydrogen atom if the electron is in its ground state and it is in a circular orbit at a distance of 53 times 10-11 m from the proton

Solution

This problem is an energy coulombic circular motion question based on Bohrrsquos model of the atom (not the accepted quantum mechanics model) The ionisation energy is the energy required to remove the electron from the ground state to infinity The electron travels in a circular orbit and therefore has a centripetal acceleration The ionisation energy will counteract the coulombic force and the movement of the electron will be in the opposite direction to the centripetal force

Total energy = Ek electron + Ep due to the proton-electron interaction

ΣF = kqQ r2 = mv2 r and as such mv2 = = kqQ r

Therefore Ek electron = frac12 kqQ r

Ep due to the proton-electron interaction = - kqQ r

Total energy = frac12 kqQ r + - kqQ r = -frac12 kqQ r

= - 90 times 109 Nm2C-2 times (16 times 10-19 C)2 divide 53 times 10-11 m = -217 times 10-18 J

= -217 times 10-18 J divide 16 times 10-19 = -136 eV

The ionisation energy is 136 eV

Exercise 1021 ApointchargePisplacedmidwaybetweentwo

identicalnegativechargesWhichoneofthefollowingiscorrectwithregardstoelectricfieldandelectricpotentialatpointP

Electric field Electric potentialA non-zero zeroB zero non-zeroC non-zero non-zeroD zero zero

2 Twopositivechargedspheresaretiedtogetherinavacuumsomewhereinspacewheretherearenoexternal forces A has a mass of 25 g and a charge of20μCandBhasamassof15gandachargeof30μCThedistancebetweenthemis40cmTheyarethenreleasedasshowninthediagram

BEFORE AFTER

v1 v2

A B

(a) Determinetheirinitialelectricpotentialenergy in the before situation

(b) DeterminethespeedofsphereBafterrelease

3 Thediagrambelowrepresentstwoequipotentiallines in separated by a distance of 5 cm in a uniform electric field

+ + + + + + + +

ndash ndash ndash ndash ndash ndash ndash ndash

40 V

20 V 5 cm

PHYSICS 2015indb 316 150514 338 PM

Fields

317

AH

L

Determine the strength of the electric field

4 Thisquestionisabouttheelectricfieldduetoacharged sphere and the motion of electrons in that fieldThediagrambelowshowsanisolatedmetalsphereinavacuumthatcarriesanegativeelectriccharge of 60 μC

ndash

(a) Onthediagramdrawtheconventionalwayto represent the electric field pattern due to the charged sphere and lines to represent threeequipotentialsurfacesintheregionoutside the sphere

(b) Explainhowthelinesrepresentingtheequipotentialsurfacesthatyouhavesketchedindicatethatthestrengthoftheelectricfieldisdecreasingwithdistancefrom the centre of the sphere

(c) Theelectricfieldstrengthatthesurfaceofthe sphere and at points outside the sphere can be determined by assuming that the sphere acts as a point charge of magnitude 60 μC at its centre The radius of the sphere is 25 times 10ndash2 m Deduce that the magnitude of the field strength at the surface of the sphere is 86 times 107 Vmndash1

An electron is initially at rest on the surface of the sphere

(d) (i) Describethepathfollowedbytheelectronasitleavesthesurfaceofthesphere

(ii) Calculatetheinitialaccelerationofthe electron

5 Determinetheamountofworkthatisdoneinmovingachargeof100nCthroughapotentialdifferenceof150times102 V

6 Three identical 20 μC conducting spheres are placedatthecornersofanequilateraltriangleofsides 25 cm The triangle has one apex C pointing upthepageand2baseanglesAandBDeterminetheabsolutepotentialatB

7 Determinehowfaraparttwoparallelplatesmustbesituatedsothatapotentialdifferenceof250 times 102 V produces an electric field strength of 200 times 103 NC-1

8 Thegapbetweentwoparallelplatesis10times10-3mandthereisapotentialdifferenceof10times104 V betweentheplatesCalculate

i theworkdonebyanelectroninmovingfrom one plate to the other

ii thespeedwithwhichtheelectronreachesthe second plate if released from rest

iii theelectricfieldintensitybetweentheplates

9 Anelectronguninapicturetubeisacceleratedby a potential 25 times 103VDeterminethekineticenergygainedbytheelectroninelectron-volts

10 Determine the electric potential 20 times 10-2 m from achargeof-10times10-5 C

11 Determinetheelectricpotentialatapointmid-waybetweenachargeofndash20pCandanotherof+5pConthelinejoiningtheircentresifthecharges are 10 cm apart

12 During a thunderstorm the electric potential differencebetweenacloudandthegroundis10 times 109 V Determine the magnitude of the change in electric potential energy of an electron thatmovesbetweenthesepointsinelectron-volts

13 A charge of 15 μC is placed in a uniform electric fieldoftwooppositelychargedparallelplateswithamagnitudeof14times103 NC-1

(a) Determinetheworkthatmustbedoneagainstthefieldtomovethepointchargeadistance of 55 cm

(b) Calculatethepotentialdifferencebetweenthe final and initial positions of the charge

(c) Determinethepotentialdifferencebetweenthe plates if their separation distance is 15 cm

14 Duringaflashoflightningthepotentialdifferencebetweenacloudandthegroundwas12 times 109 V and the amount of transferred charge was32C

(a) Determinethechangeinenergyofthetransferred charge

(b) Iftheenergyreleasedwasallusedtoacceleratea1tonnecardeduceitsfinalspeed

(c) Iftheenergyreleasedcouldbeusedtomelticeat0degCdeducetheamountoficethatcould be melted

PHYSICS 2015indb 317 150514 338 PM

Chapter 10

318

AH

L

15 SupposethatwhenanelectronmovedfromAtoBin the diagram along an electric field line that the electric field does 36 times 10-19Jofworkonit

Determinethedifferencesinelectricpotential

(a) VB ndash VA(b) VC ndash VA(c) VC ndash VB

16 DeterminethepotentialatpointPthatislocatedatthecentreofthesquareasshowninthediagrambelow

- 6μC

+3μC +2μC

1m

1m

5μC

P

102 Fields at work

Essential idea Similar approaches can be taken inanalysingelectricalandgravitationalpotentialproblems

Understandingsbull Potentialandpotentialenergybull Potentialgradientbull Potentialdifferencebull Escapespeedbull Orbitalmotionorbitalspeedandorbitalenergybull Forcesandinverse-squarelawbehaviour

Orbital motion orbital speed and orbital energy

Although orbital motion may be circular elliptical orparabolic this sub-topic only dealswith circular orbitsThis sub-topic is not fundamentally new physics butan application that synthesizes ideas from gravitationcircularmotiondynamicsandenergy

TheMoonorbitstheEarthandinthissenseitisoftenreferredtoasasatelliteoftheEarthBefore1957itwastheonlyEarthsatelliteHoweverin1957theRussianslaunchedthefirstmanmadesatelliteSputnik1SincethisdatemanymoresatelliteshavebeenlaunchedandtherearenowliterallythousandsofthemorbitingtheEarthSomeareusedtomonitortheweathersome used to enable people to find accurately their position on thesurfaceoftheEarthmanyareusedincommunicationsandnodoubtsomeareusedtospyonothercountriesFigure1015showshowinprincipleasatellitecanbeputintoorbit

Theperson(whosesizeisgreatlyexaggeratedwithrespectto Earth) standing on the surface on the Earth throwssome stonesThe greater the speed with which a stoneis thrown the further it will land from her The pathsfollowedbythethrownstonesareparabolasByastretchof the imaginationwecanvisualisea situation inwhichastoneis thrownwithsuchaspeedthatbecauseof thecurvatureoftheEarthitwillnotlandonthesurfaceoftheEarthbutgointoldquoorbitrdquo(Path4onFigure1015)

NATURE OF SCIENCECommunication of scientific explanations The ability to apply field theory to the unobservable (charges) and the massively scaled (motion of satellites) required scientists to develop new ways to investigate analyse and report findings to a general public used to scientific discoveries based on tangible and discernible evidence (51)

copy IBO 2014

PHYSICS 2015indb 318 150514 338 PM

Fields

319

AH

L

1

2

34Earth

Figure 1015 Throwing a stone into orbit

TheforcethatcausesthestonestofollowaparabolicpathandtofalltoEarthisgravityandsimilarlytheforcethatkeeps the stone in orbit is gravity For circular motionto occur we have seen that a force must act at rightanglestothevelocityofanobjectthatistheremustbeacentripetalforceHenceinthesituationwedescribeherethe centripetal force for circular orbital motion about the EarthisprovidedbygravitationalattractionoftheEarth

We can calculate the speedwithwhich a stonemustbethrowninordertoputitintoorbitjustabovethesurfaceof the Earth

If the stone has mass m and speed v thenwehave fromNewtonrsquos2ndlaw

mv2

RE---------- G

MEm

RE2------------=

whereRE is the radius of the Earth and ME is the mass of the Earth

Bearinginmindthatg 0 GME

RE2--------= then

v g RE 10 64 106timestimes 8 103times= = =

Thatisthestonemustbethrownat8times103m sndash1

Clearly we are not going to get a satellite into orbit soclosetothesurfaceoftheEarthMovingatthisspeedthefrictionduetoairresistancewouldmeltthesatellitebeforeithadtravelledacoupleofkilometresInrealitythereforeasatelliteisputintoorbitabouttheEarthbysendingitattachedtoarocketbeyondtheEarthrsquosatmosphereandthengivingitacomponentofvelocityperpendiculartoaradialvectorfromtheEarthSeeFigure1016

Earth Satellite carried by rocket to hereSatellite orbit

Tangential component of velocity

Figure 1016 Getting a satellite into orbit

Keplerrsquos third law(This work of Kepler and Newtonrsquos synthesis of the work is an excellent example of the scientific method and makes for a good TOK discussion)

In 1627 Johannes Kepler (1571-1630)publishedhis lawsof planetary motion The laws are empirical in natureandwere deduced from the observations of theDanishastronomer Tycho de Brahe (1546-1601)The third lawgives a relationship between the radius of orbit R of a planet and its period TofrevolutionabouttheSunThelawisexpressedmathematicallyas

2

3 constantTR

=

We shall nowuseNewtonrsquos LawofGravitation to showhowitisthattheplanetsmoveinaccordancewithKeplerrsquosthirdlaw

InessenceNewtonwasabletousehislawofgravitytopredict the motion of the planets since all he had to do was factor theF givenby this law intohis second law F = ma to find their accelerations and hence their future positions

InFigure1017theEarthisshownorbitingtheSunandthedistancebetweentheircentresisR

RSun

Earth

F es

F se

Figure 1017 Planets move according to Keplerrsquos third law

Fes is the force that the Earth exerts on the Sun and Fse is the force that the Sun exerts on the Earth The forces are equalandoppositeandtheSunandtheEarthwillactuallyorbitaboutacommoncentreHoweversince theSun issoverymuchmoremassivethantheEarththiscommoncentrewillbeclosetothecentreoftheSunandsowecanregard the Earth as orbiting about the centre of the Sun Theotherthingthatweshallassumeisthatwecanignorethe forces that the other planets exert on the Earth (This wouldnotbeawisethingtodoifyouwereplanningtosend a space ship to theMoon for example)We shallalso assume that we have followed Newtonrsquos exampleandindeedprovedthataspherewillactasapointmasssituated at the centre of the sphere

Kepler had postulated that the orbits of the planets areellipticalbut since theeccentricityof theEarthrsquosorbit issmallweshallassumeacircularorbit

PHYSICS 2015indb 319 150514 338 PM

Chapter 10

320

AH

L

TheaccelerationoftheEarthtowardstheSunisa = Rω2

whereω ω 2πT------=

Hencea R 2π

T------ times

2 4π2 RT 2------------- = =

But the acceleration is given by Newtonrsquos Second Law F = mawhereFisnowgivenbytheLawofGravitationSo in this situation

F maGMs Me

R2-------------------= = butwealsohavethat

a a R 2πT

------ times

2 4π2 RT 2------------- = = and m = Me so that

G MsMe

R2------------------- Me4π2R

T2-------------times

GMs

4π2-----------rArr R3

T2------= =

Butthequantity

isaconstantthathasthesamevalueforeachoftheplanetssowehaveforalltheplanetsnotjustEarththat

R3

T2------ k =

wherekisaconstantWhichisofcourseKeplerrsquosthirdlaw

ThisisindeedanamazingbreakthroughItisdifficulttorefutethe idea that all particles attract each other in accordance withtheLawofGravitationwhenthelawisabletoaccountfortheobservedmotionoftheplanetsabouttheSun

The gravitational effects of the planets upon each othershould produce perturbations in their orbits Such is the predictivepoweroftheUniversalGravitationalLawthatit enabled physicists to compute these perturbations The telescopehadbeeninventedin1608andbythemiddleofthe 18th Century had reached a degree of perfection in design that enabled astronomers to actually measure the orbital perturbations of the planets Their measurements werealways inagreementwith thepredictionsmadebyNewtonrsquoslawHoweverin1781anewplanetUranuswasdiscovered and the orbit of this planet did not fit withthe orbit predicted by Universal Gravitation Such wasthe physicistrsquos faith in theNewtonianmethod that theysuspectedthatthediscrepancywasduetothepresenceofayetundetectedplanetUsingtheLawofGravitationtheFrenchastronomerJLeverrier and the English astronomer J C Adamswereable tocalculate justhowmassive thisnewplanetmustbeandalsowhereitshouldbeIn1846the planetNeptunewas discovered justwhere they hadpredicted Ina similarwaydiscrepancies in theorbitofNeptune ledto thepredictionandsubsequentdiscoveryin1930oftheplanetPlutoNewtonrsquosLawofGravitation

had passed the ultimate test of any theory it is not only abletoexplainexistingdatabutalsotomakepredictions

Satellite energyWhenasatelliteisinorbitaboutaplanetitwillhavebothkineticenergyandgravitationalpotentialenergySupposewe consider a satellite ofmass m that is in an orbit of radius r about a planet of mass M

Thegravitationalpotentialduetotheplanetatdistancer from its centre is

ThegravitationalpotentialenergyofthesatelliteVsat

is therefore GMe m

r----------------ndash

ThatisVsat GMem

r----------------ndash=

Thegravitationalfieldstrengthatthesurfaceoftheplanetisgivenby

g 0GMe

Re2------------=

Hencewecanwrite

Vsat g0 Re

2

rndash=

m

ThekineticenergyofthesatelliteKsatisequaltofrac12mv2wherev is its orbital speed

Byequatingthegravitationalforceactingonthesatellitetoitscentripetalaccelerationwehave

G Mem

r2---------------- mv2

r---------- mv2hArr

GMemr----------------= =

Fromwhich12---

mv2 12---

GMemr----------------times=

= 0 e

2g R

r

2m

Which is actually quite interesting since it shows thatirrespective of the orbital radius the KE is numericallyequal to half the PE Also the total energy E tot K sat Vsat+ 1

2---G Me m

r----------------times G Me m

r----------------ndash + 1

2---G Me m

r----------------ndash= = =of the satelliteisalwaysnegativesince

E tot K sat Vsat+ 12---

G Me mr----------------times

G Me mr----------------ndash

+ 12---

G Me mr----------------ndash= = =

PHYSICS 2015indb 320 150514 338 PM

Fields

321

AH

L

The energies of an orbiting satellite as a function of radial distancefromthecentreofaplanetareshownplottedinFigure1018

12

100806

0402

0ndash02

ndash04

ndash06ndash08ndash10

ndash12

2 4 6 8 10 12

ener

gya

rbita

ry u

nits

distance timesR

kinetic energytotal energy

potential energy

Figure 1018 Energy of an orbiting satellite as a function of distance from the centre of a planet

WeightlessnessSuppose that you are inanelevator(lift)thatisdescendingat constant speedandyou let goof abook thatyouareholdinginyourhandThebookwillfalltothefloorwithaccelerationequaltotheaccelerationduetogravityIfthecablethatsupportstheelevatorweretosnap(asituationthatItrustwillneverhappentoanyofyou)andyounowletgothebookthatyouareholdinginyourotherhandthis book will not fall to the floor - it will stay exactlyin linewithyourhandThis isbecause thebook isnowfallingwith the sameaccelerationas theelevatorandassuch the book cannot ldquocatchrdquo up with the floor of theelevatorFurthermoreifyouhappenedtobestandingonasetofbathroomscalesthescaleswouldnowreadzero-youwouldbeapparentlyweightlessItisthisideaoffreefallthatexplainstheapparentweightlessnessofastronautsin an orbiting satellite These astronauts are in free fall inthesensethattheyareacceleratingtowardsthecentre of the Earth

It is actually possible to define theweight of a body inseveral different waysWe can define it for example asthegravitationalforceexertedonthebodybyaspecifiedobject such as the EarthThiswe have seen thatwe doinlotsofsituationswherewedefinetheweightasbeingequaltomgIfweusethisdefinitionthenanobjectinfreefallcannotbydefinitionbeweightlesssinceitisstillinagravitationalfieldHoweverifwedefinetheweightofanobjectintermsofaldquoweighingrdquoprocesssuchasthereadingonasetofbathroomscaleswhichineffectmeasuresthecontact force between the object and the scales thenclearlyobjectsinfreefallareweightlessOnenowhasto

askthequestionwhetherornotitispossibleForexampletomeasurethegravitationalforceactingonanastronautin orbit about the Earth

WecanalsodefineweightintermsofthenetgravitationalforceactingonabodyduetoseveraldifferentobjectsForexample for an object out in space its weight could bedefined in terms of the resultant of the forces exerted on it bytheSuntheMoontheEarthandalltheotherplanetsin the Solar System If this resultant is zero at a particular pointthenthebodyisweightlessatthispoint

Inviewofthevariousdefinitionsofweightthatareavailabletousitisimportantthatwhenweusethewordldquoweightrdquoweareawareofthecontextinwhichitisbeingused

Example

Calculate the height above the surface of the Earth atwhichageo-stationarysatelliteorbits

Solution

A geo-stationary satellite is one that orbits the Earth in such a way that it is stationary with respect to a point on the surface of the Earth This means that its orbital period must be the same as the time for the Earth to spin once on its axis ie 24 hours

From Keplerrsquos third law we have G Ms

4π2----------- R3

T2------=

That is

RMe

mh

using the fact that the force of attraction between the satellite and the Earth is given by

FG Mem

R2----------------=

and that F = ma

where m is the mass of the satellite and a 4π2R

T2-------------=

we have

GMem

R2---------------- m 4π2R

T2-------------times

GMe

4π2------------rArr R3

T2------= =

PHYSICS 2015indb 321 150514 338 PM

Chapter 10

322

AH

L

Now the mass of the Earth is 60 times 1024 kg and the period T measured in seconds is given by T = 86400 s

So substitution gives R = 42 times 106 m

The radius of the Earth is 64 times 106 m so that the orbital height h is about 36 times 107 m

Example

Calculatetheminimumenergyrequiredtoputasatelliteofmass500kgintoanorbitthatisasaheightequaltotheEarthrsquosradiusabovethesurfaceoftheEarth

Solution

We have seen that when dealing with gravitational fields and potential it is useful to remember that

g 0G M

Re2---------= or g0 Re

2 GM=

The gravitational potential at the surface of the Earth is

g 0ndash Re GMRe

---------ndash=

The gravitational potential at a distance R

from the centre of the Earth is

=

The difference in potential between the surface of the Earth and a point distance R from the centre is therefore

V∆ g 0Re 1ReR------ndash

=

If R 2Re= then V∆g0 Re

2------------=

This means that the work required to ldquoliftrdquo the satellite into orbit is g0Rm where m is the mass of the satellite This is equal to

10 32 106 500timestimestimes = 16000 MJ

However the satellite must also have kinetic energy in order to orbit Earth This will be equal to

g 0 mRe2

2R-----------------g 0mRe

2

4----------------- 8000 MJ= =

The minimum energy required is therefore

24000 MJ

Exercise 1031 The speed needed to put a satellite in orbit does

not depend on

A the radius of the orbitB theshapeoftheorbitC thevalueofgattheorbitD the mass of the satellite

2 EstimatethespeedofanEarthsatellitewhoseorbitis400kmabovetheEarthrsquossurfaceAlsodetermine the period of the orbit

3 Calculatethespeedofa200kgsatelliteorbitingthe Earth at a height of 70 times 106 m

Assume that g = 82 m sndash2 for this orbit

4 TheradiioftwosatellitesXandYorbitingthe Earth are 2r and 8rwherer is the radius of the Earth Calculate the ratio of the periods of revolutionofXtoY

5 A satellite of mass mkgissentfromEarthrsquossurfaceinto an orbit of radius 5RwhereR is the radius of theEarthWritedownanexpressionfor

(a) thepotentialenergyofthesatelliteinorbit

(b) thekineticenergyofthesatelliteinorbit

(c) theminimumworkrequiredtosendthesatellitefromrestattheEarthrsquossurfaceintoits orbit

6 A satellite in an orbit of 10rfallsbacktoEarth(radius r)afteramalfunctionDeterminethespeedwithwhichitwillhittheEarthrsquossurface

7 The radius of the moon is frac14 that of the Earth AssumingEarthandtheMoontohavethesamedensitycomparetheaccelerationsoffreefallatthe surface of Earth to that at the surface of the Moon

8 UsethefollowingdatatodeterminethegravitationalfieldstrengthatthesurfaceoftheMoon and hence determine the escape speed from the surface of the Moon

Mass of the Moon = 73 times 1022kg

RadiusoftheMoon=17times106 m

PHYSICS 2015indb 322 150514 338 PM

Chapter 9

298

AH

L

Thickness of oil films

The exercise to follow will help explain this use

Non-reflecting films

A very important but simple application of thin filminterferenceisintheproductionofnon-reflectingsurfaces

Athinfilmofthicknessd and refractive index n1 is coated onto glass of refractive index n where n1 lt n Light ofwavelength λ that is reflected at normal incidence willundergo destructive interference if 12

2n d λ

= thatis

14d

nλ

=

(remember that there will now no phase change at the glass interface ie we have a rare to dense reflection)

The use of such films can greatly reduce the loss oflightby reflection at the various surfacesof a systemoflenses or prisms Optical parts of high quality systemsare usually all coatedwithnon-reflecting films in orderto reduce stray reflections The films are usually madebyevaporatingcalciumormagnesiumfluorideonto thesurfacesinvacuumorbychemicaltreatmentwithacidsthat leave a thin layer of silica on the surface of the glass ThecoatedsurfaceshaveapurplishhuebyreflectedlightThis is because the condition for destructive interference fromaparticularfilmthicknesscanonlybeobtainedforone wavelength The wavelength chosen is one that has a value corresponding to light near the middle of the visible spectrumThismeansthatreflectionofredandvioletlightis greater combining to give the purple colour Because of the factor cosφatanglesotherthannormalincidencethepath difference will change butnotsignificantlyuntilsayabout30deg(egcos25deg=090)

It should be borne in mind that no light is actually lost by anon-reflectingfilmthedecreaseofreflectedintensityiscompensated by increase of transmitted intensity

Non-reflecting films can be painted onto aircraft tosuppressreflectionofradarThethicknessofthefilmis determined by

4nd λ

= where λ is the wavelength of

the radar waves and ntherefractiveindexofthefilmat this wavelength

Example

A plane-parallel glass plate of thickness 20 mm isilluminated with light from an extended source The refractiveindexoftheglassis15andthewavelengthofthelight is 600 nm Calculate how many fringes are formed

Solution

We assume that the fringes are formed by light incident at all angles from normal to grazing incidence

At normal incidence we have 2nd = mλ

From which

m 2 15 2 10 3ndashtimestimestimes6 10 7ndashtimes

------------------------------------------- 10 000= =

At grazing incidence the angle of refraction φ is in fact the critical angle

Therefore φφ arcsin 115------- 42 deg= =

ie cosφ = 075

At grazing incidence 2ndcosφ = m λ

From which (and using cos(sinndash1(115) = 075)

m 2 15 2 10 3ndash 075timestimestimestimes6 10 7ndashtimes

------------------------------------------------------------ 7500= =

The total number of fringes seen is equal to m ndash m = 2500

Exercise 95When viewed from above the colour of an oil film onwater appears red in colour Use the data below to estimate theminimumthicknessoftheoilfilm

average wavelength of red light = 630 nm

refractiveindexofoilforredlight=15

refractive index of water for red light = 13

PHYSICS 2015indb 298 150514 337 PM

Wave Phenomena

299

AH

L

94 Resolution

Essential idea Resolution places an absolute limit on the extent to which an optical or other system can separate images of objects

Understandings

bull Thesizeofadiffractingaperturebull Theresolutionofsimplemonochromatic

two-source systems

The size of a diffracting apertureOur discussion concerning diffraction so far has been for rectangular slits When light from a point source enters a smallcircularapertureitdoesnotproduceabrightdotasanimagebutacirculardiscknownasAiryrsquosdiscsurroundedbyfainterconcentriccircularringsasshowninFigure922

Laser

Figure 922 Diffraction at a circular aperture

Diffractionatanapertureisofgreatimportancebecausetheeye and many optical instruments have circular apertures

What is the half-angular width of the central maximum of the diffraction formed by a circular aperture This is not easy to calculate since it involves some advanced mathematics The problemwasfirstsolvedbytheEnglishAstronomerRoyalGeorgeAiryin1835whoshowedthatforcircularapertures

θ 122 λb--------------= where b is the diameter of the aperture

Example 2

Inthefollowingdiagramparallellightfromadistantpointsource(suchasastar)isbroughttofocusonthescreenSbyaconverginglens(the lens is shown as a vertical arrow)

Thefocal length(distance fromlens toscreen) is25cmand the diameter of the lens is 30 cm The wavelength of thelightfromthestaris560nmCalculatethediameterofthe diameter of the image on the screen

Solution

The lens actually acts as a circular aperture of diameter 30 cm The half angular width of central maximum of the diffraction pattern that it forms on the screen is

75

2

122 122 56 10 23 1030 10b

λθminus

minusminus

times times= = = times

times rad

The diameter of the central maxima is therefore

25 times 10-2 times 23 times 10-5

= 57 times 10-6 m

Although this is small it is still finite and is effectively the image of the star as the intensity of the secondary maxima are small compared to that of the central maximum

The resolution of simple monochromatic two-source systemsThe astronomers tell us that many of the stars that we observe with the naked eye are in fact binary stars That iswhatwe see as a single star actually consistsof twostarsinorbitaboutacommoncentreFurthermoretheastronomerstellusthatifweusealdquogoodrdquotelescopethenwewillactuallyseethetwostarsThatiswewillresolve the single point source into its two component parts Sowhatisitthatdetermineswhetherornotweseethetwo stars as a single point source ie what determines whether or not two sources can be resolved It canrsquot just bethatthetelescopemagnifiesthestarssinceiftheyareacting as point sources magnifying them is not going to make a great deal of difference

Ineachofoureyesthereisanaperturethepupilthroughwhich the light enters This light is then focused by the eye lens onto the retina But we have seen that when light passes through an aperture it is diffracted and so when we look at

NATURE OF SCIENCEImproved technology The Rayleigh criterion is the limit of resolution Continuing advancement in technology such as large diameter dishes or lenses or the use of smaller wavelength lasers pushes the limits of what we can resolve (18)

copy IBO 2014

PHYSICS 2015indb 299 150514 337 PM

Chapter 9

300

AH

L

apointsourceadiffractionpatternwillbeformedontheretina If we look at two point sources then two diffraction patterns will be formed on the retina and these patterns will overlap The width of our pupil and the wavelength of the light emitted by the sources will determine the amount by which they overlap But the degree of overlap will also depend on the angular separation of the two point sources WecanseethisfromFigures923

pupilretina

S1

S2

P2

P1

θ

Figure 923

LightfromthesourceS1 enters the eye and is diffracted by the pupil such that the central maximum of the diffraction patternisformedontheretinaatP1SimilarlylightfromS2producesamaximumatP2 If the two central maxima are well separated then there is a fair chance that we will see the two sources as separate sources If they overlap then we will not be able to distinguish one source from another From the diagram we see as the sources are movedclosertotheeyethentheangleθ increases and so does the separation of the central maxima

Figures924925926and927showsthedifferentdiffractionpatternsandtheintensitydistributionthatmightresultonthe retina as a result of light from two point sources

S1 S21 Well resolved

Figure 924 Very well resolved

S1 S22 Well resolved

Figure 925 Well resolved

S1 S23 Just resolved Rayleigh criterion Minimum of S2 coincides with maximum peak of S1

Figure 926 Just resolved

4 Not resolved

Figure 927 Not resolved

We have suggested that if the central maxima of the two diffraction patterns are reasonably separated then we should be able to resolve two point sources In the late 19th century Lord Rayleigh suggested by how much they should be separated in order for the two sources to be just resolved Ifthecentralmaximumofonediffractionpattern coincides with the firstminima of the otherdiffraction pattern then the two sources will just beresolved This is known as the Rayleigh Criterion

Figure926showsjustthissituationThetwosourcesarejust resolved according to the Rayleigh criterion since the peak of the central maximum of one diffraction pattern coincideswiththefirstminimumoftheotherdiffractionpattern This means that the angular separation of the peaks of the two central maxima formed by each source is just the half angular width of one central maximum ie

θ λb---

=

where b is the width of the slit through which the light fromthesourcespassesHoweverweseefromFigure1117that θ is the angle that the two sources subtend at the slit Henceweconcludethattwosourceswillberesolvedbyaslit if the angle that they subtend at the slit is greater than or equaltoλfraslb

PHYSICS 2015indb 300 150514 337 PM

Wave Phenomena

301

AH

L

Sofarwehavebeenassumingthattheeyeisarectangularslit whereas clearly it is a circular aperture and so we must use the formula

θ 122 λb--------------=

As mentioned above the angle θ is sometimes called the resolving power but should more accurately be called the minimum angle of resolution (θmin)

Clearly the smaller θ the greater will be the resolving power

The diffraction grating is a useful for differentiating closely spacedlinesinemissionspectraLikeaprismspectrometerit can disperse a spectrum into its components but it is better suited because it has higher resolution than the prism spectrometer

If λ1 and λ2 are two nearly equal wavelengths that canbarelybedistinguishedtheresolvanceorresolvingpowerofthegratingisdefinedas

R = λ _______ (λ2 ndash λ 1) = λ ___ Δλ

Where λ asymp λ1 asymp λ2

Therefore thediffractiongratingwithahighresolvancewill be better suited in determining small differences in wavelength

If the diffraction grating has Nlinesbeingilluminateditcan be shown that the resolving power of the mth order diffraction is given by

R = mN

Obviouslytheresolvancebecomesgreaterwiththeordernumber m and with a greater number of illuminated slits

Example

A benchmark for the resolving power of a grating is the sodium doublet in the sodium emission spectrum Two yellow lines in itsspectrumhavewavelengthsof58900nmand58959nm

(a) Determinetheresolvanceofagratingifthegivenwavelengths are to be distinguished from each other

(b) Howmanylinesinthegratingmustbeilluminated in order to resolve these lines in the second order spectrum

Solution

(a) R = λ ___ Δλ = 58900nm __________________ (58900ndash58959)nm

= 589nm _______ 059nm = 10 times 103

(b) N = R __ m = 10 times 10 3 ________ 2 =50times102 lines

It has been seen that diffraction effectively limits the resolving power of optical systems This includes such systems as the eye telescopes and microscopes Theresolving power of these systems is dealt with in the next section This section looks at links between technology and resolving power when looking at the very distant and when looking at the very small

Radio telescopes

The average diameter of the pupil of the human eye is about 25 mm This means that the eye will justresolve two point sources emitting light of wavelength 500 nm if their angular separation at the eye is 7

43

50 10122 24 1025 10

θminus

minusminus

times= times = times

times rad

If the eye were to be able to detect radio waves of wavelength 015m then tohave the same resolvingpower thepupilwould have to have a diameter of about 600 m Clearly this is nonsense but it does illustrate a problem facingastronomers who wish to view very distant objects such as quasarsandgalaxies thatemitradiowavesConventionalradio telescopes consist of a large dish typically 25 min diameter Even with such a large diameter the radiowavelength resolving power of the telescope is much less thantheopticalresolvingpowerofthehumaneyeLetuslook at an example

Example

The Galaxy Cygnus A can be resolved optically as an elliptically shaped galaxy However it is also a strongemitterof radiowavesofwavelength015mTheGalaxyisestimatedtobe50times1024mfromEarthUseofaradiotelescope shows that the radio emission is from two sources separated by a distance of 30 times 1021m Estimate thediameterofthedishrequiredtojustresolvethesources

Solution

The angle θ that the sources resolve at the telescope is given by

θ 21

424

30 10 60 1050 10

θ minustimes= = times

times rad

and 4

122 122 015 300060 10

d λθ minus

times= = =

timesm = 30 km

A radio telescope dish of this size would be impossible tomake letalonesupportThisshowsthatasingledishtype radio telescope cannot be used to resolve the sources and yet they were resolved To get round the problemastronomers use two radio telescopes separated by a large

PHYSICS 2015indb 301 150514 337 PM

Chapter 9

302

AH

L

distance The telescopes view the same objects at the same time and the signals that each receive from the objects are simultaneously superimposed The result of the superposition of the two signals is a two-slit interference pattern The pattern has much narrower fringe spacing than that of the diffraction pattern produced by either telescope on its own hence producing a much higherresolvingpowerWhentelescopesareusedlikethistheyare called a stellar interferometer

InSocorroinNewMexicothereisastellarinterferometerthat consists of 27 parabolic dishes each of diameter25 m arranged in a Y-shape that covers an area of 570km2Thisisaso-calledVeryLargeArray(VLA)Evenhigher resolution can be obtained by using an array of radio telescopes in observatories thousands of kilometres apart A system that uses this so-called technique oflsquovery-long-baseline interferometryrsquo (VLBI) is known asalsquovery-long-baselinearrayrsquo(VLBA)WithVLBAaradiowavelength resolving power can be achieved that is 100 timesgreaterthanthebestopticaltelescopesEvenhigherresolving power can be achieved by using a telescope that isinasatelliteorbitingEarthJapanrsquosInstituteofSpaceandAstronautical Science (ISAS) launched sucha system inFebruary 1997TheNationalAstronomicalObservatoryof Japan the National Science Foundationrsquos NationalRadio Astronomy Observatory (NRAO) the CanadianSpaceAgency theAustraliaTelescopeNationalFacilitythe European VLBI Network and the Joint Instituteback this project forVeryLongBaseline Interferometryin EuropeThis project is a very good example of howInternationalismcanoperateinPhysics

Electron microscope

Telescopes are used to look at very distant objects that are very large but because of their distance fromus appearverysmallMicroscopesontheotherhandareusedtolookat objects that are close to us but are physically very small Aswehave seen justmagnifyingobjects that ismakingthemappearlargerisnotsufficientonitsowntogaindetailabouttheobjectfordetailhighresolutionisneeded

Figure928isaschematicofhowanoptical microscope is usedtoviewanobjectandFigure929isaschematicofatransmission electron microscope(TEM)

Optical lens system

bright light source

glass slide containing specimen (object)

eye

Figure 928 The principle of a light microscope

magnetic lens system

electron source

thin wafer of material

ScreenCCD

Figure 929 The principle of an electron microscope

In the optical microscope the resolving power isdetermined by the particular lens system employed and the wavelength λ of the light used For example twopoints in the sample separated by a distance d will just be resolved if

2d mλ=

where m is a property of the lens system know as the numerical aperture In practice the largest value of m obtainable is about 16 Hence if the microscope slideis illuminatedwith light ofwavelength 480 nm a goodmicroscope will resolve two points separated by a distance dasymp15times10-7masymp015micromPointscloser together thanthiswillnotberesolvedHoweverthisisgoodenoughtodistinguishsomevirusessuchastheEbolavirus

Clearlythesmallerλ the higher the resolving power and this is where the electron microscope comes to the fore The electron microscope makes use of the wave nature of electrons(see1315)IntheTEMelectronspassthrougha wafer thin sample and are then focused by a magnetic field onto a fluorescent screen orCCD (charge coupleddevicesee142)Electronsusedinanelectronmicroscopehavewavelengthstypicallyofabout5times10-12mHoweverthe numerical aperture of electron microscopes is considerablysmaller thanthatofanopticalmicroscopetypicallyabout002NonethelessthismeansthataTEMcanresolvetwopointsthatareabout025nmapartThisresolving power is certainly high enough to make out the shape of large molecules

Another type of electron microscope uses a techniqueby which electrons are scattered from the surface of the sample The scattered electrons and then focused as in theTEM to form an image of the surfaceThese so-calledscanningelectronmicroscopes(SEM)havealowerresolving power than TEMrsquos but give very good threedimensional images

The eye

We saw in the last section that the resolving power of the human eye is about 2 times 10-4 rad Suppose that youare looking at car headlights on a dark night and the car

PHYSICS 2015indb 302 150514 337 PM

Wave Phenomena

303

AH

L

is a distance D away If the separation of the headlight is say 15 m then the headlights will subtend an angle θ 15

Dθ =

at your eye Assuming an average wavelength of

500 nm your eye will resolve the headlights into twoseparatesourcesifthisangleequals2times10-4 rad This gives D=75kmInotherwordsifthecarisapproachingyouona straight road then you will be able to distinguish the two headlightsasseparatesourceswhenthecaris75kmawayfrom you Actually because of the structure of the retina and optical defects the resolving power of the average eye is about 6 times 10-4 rad This means that the car is more likely tobe25kmawaybeforeyouresolvetheheadlights

Astronomical telescope

Letusreturntotheexampleofthebinarystarsdiscussedatthe beginning of this section on resolution The stars Kruger A and B form a binary system The average separation of the stars is 14 times 1012mandtheiraveragedistancefromEarthis12times1017mWhenviewedthroughatelescopeonEarththe system will therefore subtend an angle

12

17

14 1012 10

θtimes

=times

= 12times10-5 rad at the objective lens of the telescope Assuming that the average wavelength of the light emitted bythestarsis500nmthenifthetelescopeistoresolvethesystem into two separate images it must have a minimum diameter Dwhere12times10-5 =

7122 500 10D

minustimes times

This gives D=0050m which is about 5 cm So thisparticular system is easily resolved with a small astronomical telescope

Exercise 961 ItissuggestedthatusingtheISASVLBAitwould

bepossibletoldquoseerdquoagrainofriceatadistanceof5000kmEstimatetheresolvingpoweroftheVLBA

2 Thedistancefromtheeyelenstotheretinais20mm The light receptors in the central part of the retinaareabout5times10-6apartDeterminewhetherthe spacing of the receptors will allow for the eye to resolve the headlights in the above discussion whentheyare25kmfromtheeye

3 ThediameterofPlutois23times106 m and its averagedistancefromEarthis60times1012 m EstimatetheminimumdiameteroftheobjectiveofatelescopethatwillenablePlutotobeseenasadisc as opposed to a point source

95 Doppler effect

Essential ideaTheDopplereffectdescribesthephenomenonofwavelengthfrequencyshiftwhenrelativemotionoccurs

Understandingsbull TheDopplereffectforsoundwavesandlightwaves

The Doppler effect for sound waves and light waves

Consider two observers A and B at rest with respect to asoundsourcethatemitsasoundofconstantfrequencyf Clearly both observers will hear a sound of the same frequencyHoweversupposethatthesourcenowmovesat constant speed towards A A will now hear a sound of frequencyfA that is greater than f and B will hear a sound of frequency fB that is less than f This phenomenon is known as the Doppler Effect orDopplerPrinciple after C J Doppler(1803-1853)

The same effect arises for an observer who is either moving towards or away from a stationary source

Figure 930 shows the waves spreading out from a stationary source that emits a sound of constant frequency f The observersAandBhearasoundofthesamefrequency

A B

wavefront

S

Figure 930 Sound waves from a stationary source

Suppose now that the source moves towards A withconstant speed vFigure9310(a)showsasnapshotofthenew wave pattern

NATURE OF SCIENCETechnology Although originally based on physical observations of the pitch of fast moving sources of sound the Doppler effect has an important role in many different areas such as evidence for the expansion of the universe and generating images used in weather reports and in medicine (55)

copy IBO 2014

PHYSICS 2015indb 303 150514 337 PM

Chapter 9

304

AH

L

sourceA B

smallerwavelength

largerwavelength

Figure 931 (a) Sound waves from a moving source

The wavefronts are now crowded together in the direction of travel of the source and stretched out in the opposite direction This is why now the two observers will now hear notesofdifferentfrequenciesHowmuchthewavesbunchtogether and how much they stretch out will depend on the speed vEssentially A

A

cfλ

= and BB

cfλ

= where λA lt λB and v is the speed of sound

IfthesourceisstationaryandAismovingtowardsitthenthe waves from the source incident on A will be bunched up If A is moving away from the stationary source then the waves from the source incident on A will be stretched out

Christian Doppler(1803ndash1853)actuallyappliedtheprinciple(incorrectlyasithappens)totryandexplainthecolourofstars However the Doppler effect does apply to light aswellastosoundIfalightsourceemitsalightoffrequencyf then if it is moving away from an observer the observer willmeasurethelightemittedashavingalowerfrequencythan fSincethesensationofcolourvisionisrelatedtothefrequencyoflight(bluelightisofahigherfrequencythanred light) light emitted by objects moving way from anobserver is often referred to as being red-shifted whereas if the object is moving toward the observer it is referred to as blue-shiftedThisideaisusedinOptionE(Chapter16)

We do not need to consider here the situations where either the source or the observer is accelerating In a situation for example where an accelerating source is approaching astationaryobserverthentheobserverwillhearasoundof ever increasing frequency This sometimes leads toconfusion in describing what is heard when a source approaches passes and then recedes from a stationaryobserver Suppose for example that you are standingon a station platform and a train sounding its whistle is approaching at constant speed What will you hear as the train approaches and then passes through the station As the train approaches you will hear a sound of constant pitch but increasing loudness The pitch of the sound will be greater than if the train were stationary As the train passes through the station you will hear the pitch change at the momentthetrainpassesyoutoasoundagainofconstantpitch The pitch of this sound will be lower than the sound of the approaching train and its intensity will decrease as the train recedes from you What you do not hear is a sound of increasing pitch and then decreasing pitch

The Doppler equations for sound

Although you will not be expected in an IB examination to derivetheequationsassociatedwithaspectsoftheDopplereffectyouwillbeexpectedtoapplythemForcompletenesstherefore thederivationoftheequationsassociatedwiththeDopplereffectasoutlinedaboveisgivenhere

S primeS O

vs

Figure 931 (b)

InFigure931(b)theobserverOisatrestwithrespecttoasourceofsoundSismovingwithconstanspeedvs directly towards O The source is emitting a note of constant frequencyf and the speed of the emitted sound is v

S shows the position of the source ∆t later When the sourceisatresttheninatime∆t the observer will receive f∆t waves and these waves will occupy a distance v∆t ie

v f

λ =f∆tv∆t =

(Because of the motion of the source this number of waves will now occupy a distance (v∆t ndash vs∆t)The lsquonewrsquowavelength is therefore

v ndash vs

fλrsquo =

f∆tv∆t ndash vs∆t

=

If fisthefrequencyheardbyOthen

vfλ

= or

vf

λ = = sv vfminus

From which

s

vf fv v

=minus

Dividingthroughbyv gives

Equation1111

1 sf f v

v=

minus

If the source is moving away from the observer then we have

s

vf fv v

=minus

1

1 sf vv

=+

Equation112

PHYSICS 2015indb 304 150514 337 PM

Wave Phenomena

305

AH

L

We now consider the case where the source is stationary and the observerismoving towardsthesource with speed v0 In this situation the speed of the sound waves as measured by the observer will be v0 + v We therefore have that

v0 + v = ffraslλ = vfλ

times

From which

o1 vf fv

= +

Equation113

If the observerismovingawayfromthesource then

o1 vf fv

= minus

Fromequation(113)wehavethat

o o1 v vf f f f f fv v

∆ = minus = + minus =

Equation114

Thevelocitiesthatwerefertointheaboveequationsarethevelocities with respect to the medium in which the waves fromthesourcetravelHoweverwhenwearedealingwithalight source it is the relative velocity between the source and the observer that we must consider The reason for this is that lightisuniqueintherespectthatthespeedofthelightwavesdoes not depend on the speed of the source All observers irrespective of their speed or the speed of the source will measure the same velocity for the speed of light This is oneof thecornerstonesof theSpecialTheoryofRelativitywhichisdiscussedinmoredetailinOptionH(Chapter18)When applying the Doppler effect to light we aremainlyconcerned with the motion of the source We look here only at the situation where the speed of the source v is much smaller than the speed of light cinfreespace(v ltlt c)Underthesecircumstanceswhenthesourceismovingtowardstheobserverequation111becomes

vf f f fc

minus = ∆ = Equation115

andwhenthesourceismovingawayfromtheobserverequation112becomes vf f f f

cminus = ∆ = minus

Provided that v ltlt c these same equations apply for astationary source and moving observer

We look at the following example and exercise

Example

Asourceemitsasoundoffrequency440HzItmovesinastraight line towards a stationary observer with a speed of 30ms-1Theobserverhearsasoundoffrequency484HzCalculate the speed of sound in air

Solution

We use equation 111 and substitute f= 484 Hz f = 440 Hz

and vs = 30 m s-1

therefore 1

484 440 301v

=minus

such that 30 4401

484vminus = to

give v = 330 m s-1

Example

A particular radio signal from a galaxy is measured as havingafrequencyof139times109HzThesamesignalfromasourceinalaboratoryhasafrequencyof142times109Hz

Suggestwhy thegalaxy ismovingaway fromEarthandcalculateitsrecessionspeed(iethespeedwithwhichitismovingawayfromEarth)

Solution

The fact that the frequency from the moving source is less than that when it is stationary indicates that it is moving away from the stationary observer ie Earth

Using vf fc

∆ = we have

8 96

9

3 10 (142 139) 10 634 10142 10

c fvf∆ times times minus times

= = = timestimes

m s-1

It is usual when dealing with the Doppler effect of light to express speeds as a fraction of c So in this instance we have v = 0021 c

Using the Doppler effectWe have seen in the above example and exercise how the Dopplereffectmaybeusedtomeasuretherecessionspeedof distant galaxies The effect is also used to measure speed inothersituationsHerewewilllookatthegeneralprincipleinvolved in using the Doppler effect to measure speedFigure932showsasource(thetransmitter)thatemitseithersound or em waves of constant frequency f The waves

PHYSICS 2015indb 305 150514 337 PM

Chapter 9

306

AH

L

fromthesourceare incidentonareflectorthat ismovingtowards the transmitter with speed vThereflectedwavesare detected by the receiver placed alongside the transmitter

reector

v

transmitter

receiver

f f

f f

Figure 932 Using the Doppler effect to measure speed

We shall consider the situation where v ltlt c where c is the speed of the waves from the transmitter

For the reflector receiving waves from the transmitterit is effectively an observer moving towards a stationary sourceFromequation(114) itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation116

For the receiver receivingwaves from the reflector it iseffectively a stationary observer receiving waves from a movingsourceFromequation(115)itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation117

Ifweaddequations(116)and(117)wegetthatthetotalDopplershiftatthereceiver∆f is

v vf f f f fc c

minus = ∆ = +

But 1 vf fc

= +

hence

1 v v vf f fc c c

∆ = + +

But since v ltlt cwe can ignore the term2

2

vc when we

expandthebracketintheaboveequation

Therefore we have2vf fc

∆ = Equation118

If v asymp c then we must use the full Doppler equationsHowever for em radiationwewill alwaysonly considersituations in which v ltlt c

Example

Thespeedofsoundinbloodis1500times103 m s-1 Ultrasound offrequency100MHzisreflectedfrombloodflowinginan arteryThe frequencyof the reflectedwaves receivedbackatthetransmitteris105MHzEstimatethespeedofthebloodflowintheartery

Solution

Using equation (118) we have

6 63

2005 10 1015 10

vtimes = times

timesto give v asymp 36 m s-1 (We have assumed that the ultrasound is incident at right angles to the blood flow)

Exercise 971 Judy is standing on the platform of a station A

high speed train is approaching the station in a straight line at constant speed and is sounding its whistleAsthetrainpassesbyJudythefrequencyof the sound emitted by the whistle as heard by Judychangesfrom640Hzto430HzDetermine

(a) thespeedofthetrain

(b) thefrequencyofthesoundemittedbythewhistle as heard by a person on the train (Speedofsound=330ms-1)

2 AgalaxyismovingawayfromEarthwithaspeedof00500cThewavelengthofaparticularspectral line in light emitted by atomic hydrogen inalaboratoryis656times10-7 m Calculate the valueofthewavelengthofthislinemeasuredinthelaboratoryinlightemittedfromasourceofatomic hydrogen in the galaxy

PHYSICS 2015indb 306 150514 337 PM

10 Fields

Contents

101 ndash Describing fields

102 ndash Fields at work

Essential Ideas

Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Similar approaches can be taken in analysing electrical and gravitational potential problems copy IBO 2014

PHYSICS 2015indb 307 150514 337 PM

Chapter 10

308

AH

L

101 Describing fields

Essential idea Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Understandingsbull Gravitationalfieldsbull Electrostaticfieldsbull Electricpotentialandgravitationalpotentialbull Fieldlinesbull Equipotentialsurfaces

Gravitational potentialWehaveseenthatifweliftanobjectofmassm to a height h abovethesurfaceoftheEarththenitsgainingravitationalpotential energy is mghHoweverthisisbynomeansthefullstoryForastartwenowknowthatgvarieswithh and also the expression really gives a difference in potentialenergybetweenthevaluethattheobjecthasattheEarthrsquossurfaceandthevaluethatithasatheighthSowhatwereallyneedisazeropointCanwefindapointwherethepotentialenergyiszeroandusethispointfromwhichtomeasure changes in potential energy

The point that is chosen is in fact infinity At infinity the gravitationalfieldstrengthofanyobjectwillbezeroSoletusseeifwecandeduceanexpressionforthegaininpotentialenergyofanobjectwhenitisldquoliftedrdquofromthesurfaceoftheEarthtoinfinityThisineffectmeansfindingtheworknecessarytoperformthistask

δr

r r δr+

r infin=

mg

AB

Me

Figure 1001 Gravitational forces

Inthediagramweconsidertheworknecessarytomovethe particle of mass m a distance δr in thegravitationalfield of the Earth

The force on the particle at A is FG Mem

r2----------------=

IftheparticleismovedtoB then since δr isverysmallwe can assume that the field remains constant over thedistance ABTheworkδW done againstthegravitationalfieldoftheEarthinmovingthedistanceABis

δWGMem

r2----------------δrndash=

(rememberthatworkdoneagainstaforceisnegative)

TofindthetotalworkdoneW in going from the surface oftheEarthtoinfinitywehavetoaddalltheselittlebitsofworkThisisdonemathematicallybyusingintegralcalculus

WG Me m

r2----------------ndash

rdR

infin

int G Me m 1

r2----- rd

R

infin

intndash G Me m 1r---ndash

R

infinndash= = =

GMe m 0 1R---ndash

ndashndash=

G MemR----------------ndash=

HencewehavewhereRistheradiusoftheEarththattheworkdonebythegravitationalfieldinmovinganobjectofmass m from R(surfaceoftheEarth)toinfinityisgivenby

WGMe m

R---------------- ndash=

We can generalise the result by calculating the worknecessary per unit mass to take a small mass fromthe surface of the Earth to infinity This we call thegravitational potential Vie

Wewouldgetexactlythesameresultifwecalculatedtheworkdonebythefieldtobringthepointmassfrominfinityto the surface of Earth In this respect the formal definition ofgravitationalpotentialatapointinagravitationalfieldis therefore defined as the work done per unit mass in bringing a point mass from infinity to that point

Clearly then the gravitational potential at any point inthe Earthrsquos field distance r from the centre of the Earth (providingr gt R)is

VGMe

r------------ ndash=

Thepotentialisthereforeameasureoftheamountofworkthathastobedonetomoveparticlesbetweenpointsinagravitationalfieldanditsunit is theJkgndash1 We also note thatthepotentialisnegativesothatthepotentialenergy

NATURE OF SCIENCEParadigm shift The move from direct observable actions being responsible for influence on an object to acceptance of a fieldrsquos ldquoaction at a distancerdquo required a paradigm shift in the world of science (23)

copy IBO 2014

PHYSICS 2015indb 308 150514 337 PM

Fields

309

AH

L

aswemoveawayfromtheEarthrsquossurfaceincreasesuntilitreachesthevalueofzeroatinfinity

IfthegravitationalfieldisduetoapointmassofmassmthenwehavethesameexpressionasaboveexceptthatMe is replaced by m andmustalsoexclude thevalueof thepotential at the point mass itself ie at r = 0

WecanexpressthegravitationalpotentialduetotheEarth(orduetoanysphericalmass)intermsofthegravitationalfield strength at its surface

AtthesurfaceoftheEarthwehave

g 0ndash Re GMRe

---------ndash= e

Sothat

g 0Re2 GM= e

Hence at a distance r from the centre of the Earth the gravitationalpotentialVcanbewrittenas

VGMe

r------------ndashg 0 Re

2

r------------ndash= =

The potential at the surface of the Earth

(r = Re)istherefore-g0Re

It is interesting to see how the expression for thegravitational potential ties in with the expression mgh ThepotentialatthesurfaceoftheEarthis-g0Re (see the exampleabove)andataheighthwillbe-g0g 0 Re h+( )ndash ifweassume that g0doesnotchangeover thedistanceh The differenceinpotentialbetweenthesurfaceandtheheighth is therefore g0hSotheworkneededtoraiseanobjectof mass m to a height h is mgh ie m timesdifference ingravitationalpotential

This we have referred to as the gain in gravitationalpotentialenergy(see235)

However this expression can be extended to any twopoints inanygravitationalfieldsuch that ifanobjectofmass mmovesbetween twopointswhosepotentialsareV1 and V2 respectively then the change in gravitationalpotentialenergyoftheobjectism(V1 ndash V2)

Gravitational potential gradient

Let us consider now a region in space where thegravitationalfieldisconstantInFigure912thetwopointsAandBareseparatedbythedistance∆x

A B

direction of uniform gravitational eld of strength I

∆x

Figure 1002 The gravitational potential gradient

The gravitational field is of strength I and is in the direction shownThe gravitational potential at A is V andatBisV +∆V

Theworkdone is takingapointmassmfromAtoB isF∆x = mI∆x

Howeverbydefinitionthisworkisalsoequalto-m∆V

Therefore mI∆x=-m∆V

or VIx

∆= minus

∆Effectivelythissaysthatthemagnitudeofthegravitationalfield strength is equal to the negative gradient of thepotential If I is constant then V is a linear function of x and Iisequaltothenegativegradientofthestraightlinegraph formed by plotting V against x If I is not constant (asusuallythecase)thenthemagnitudeofI at any point in the field can be found by find the gradient of the V-x graph at that point An example of such a calculation can befoundinSection929

For those of you who do HL maths the relationship between field and potential is seen to follow from the expression for the potential of a point mass viz

mV Gr

= minus

2

ddV mG Ir r

minus = + =

Potential due to one or more point masses

Gravitationalpotentialisascalarquantitysocalculatingthepotential due to more than one point mass is a matter of simpleadditionSoforexamplethepotentialV due to the Moon and Earth and a distance x fromthecentreofEarthonastraightlinebetweenthemisgivenbytheexpression

E MM MV Gx r x

= minus + minus

whereME=massofEarthMM= mass of Moon and r = distancebetweencentreofEarthandMoon

PHYSICS 2015indb 309 150514 337 PM

Chapter 10

310

AH

L

Equipotentials and field linesIfthegravitationalpotentialhasthesamevalueatallpointson a surface the surface is said to be an equipotential surfaceSoforexampleifweimagineasphericalshellaboutEarthwhosecentrecoincideswiththecentreofEarththisshellwillbeanequipotentialsurfaceClearlyifwerepresentthegravitationalfieldstrengthbyfieldlinessincethelinesldquoradiaterdquooutfromthecentreofEarththentheselineswillbeatrightanglestothesurfaceIfthefieldlineswerenotnormaltotheequipotentialsurfacethentherewouldbeacomponentof the field parallel to the surfaceThis wouldmean thatpointsonthesurfacewouldbeatdifferentpotentialsandsoitwouldnolongerbeanequipotentialsurfaceThisofcourseholdstrueforanyequipotentialsurface

Figure 1003 shows the field lines and equipotentials fortwopointmassesm

m m

Figure 1003 Equipotentials for two point masses

It is worth noting that we would get exactly the samepattern ifwewere to replace thepointmasseswith twoequalpointcharges(See935)

Escape speedThe potential at the surface of Earth is which meansthattheenergyrequiredtotakeaparticleofmass mfromthesurfacetoinfinityisequalto

ButwhatdoesitactuallymeantotakesomethingtoinfinityWhentheparticleisonthesurfaceoftheEarthwecanthinkofitassittingatthebottomofaldquopotentialwellrdquoasinfigure1004

particle

surface of Earth

innity

GMR---------

FIgure 1004 A potential well

Theldquodepthrdquoofthewellis and if the particle gains an amount of kinetic energy equal to wherem is its massthenitwillhavejustenoughenergytoldquoliftrdquoitoutofthewell

In reality it doesnrsquot actually go to infinity it justmeansthat the particle is effectively free of the gravitationalattractionof theEarthWe say that ithas ldquoescapedrdquo theEarthrsquosgravitationalpullWemeetthisideainconnectionwithmolecular forcesTwomolecules in a solidwill sitat their equilibrium position the separation where therepulsiveforceisequaltotheattractiveforceIfwesupplyjust enough energy to increase the separation of themolecules such that they are an infinite distance apart then intermolecularforcesnolongeraffectthemoleculesandthesolidwillhavebecomealiquidThereisnoincreaseinthekineticenergyofthemoleculesandsothesolidmeltsat constant temperature

Wecancalculatetheescapespeedofanobjectveryeasilyby equating the kinetic energy to the potential energysuch that

12---mvesc ape

2 GMe mRe

----------------=

vesc aperArr2GMe

Re--------------- 2g0 Re = =

Substituting for g0 and Regivesavalueforvescape of about 11kmsndash1 from the surface of the Earth

Youwillnotethattheescapespeeddoesnotdependonthemassoftheobjectsincebothkineticenergyandpotentialenergy are proportional to the mass

IntheoryifyouwanttogetarockettothemoonitcanbedonewithoutreachingtheescapespeedHoweverthiswouldnecessitate anenormousamountof fuel and it islikely that therocketplus fuelwouldbesoheavythat itwouldnevergetoffthegroundItismuchmorepracticalto accelerate the rocket to the escape speed fromEarthorbitandthenintheoryjustlaunchittotheMoon

Example

Use the followingdata todetermine thepotential at thesurface of Mars and the magnitude of the acceleration of free fall

massofMars =64times1023kg

radiusofMars =34times106 m

Determinealsothegravitationalfieldstrengthatadistanceof 68 times 106mabovethesurfaceofMars

PHYSICS 2015indb 310 150514 337 PM

Fields

311

AH

L

Solution

= minus = minus times times = times23

11 76

64 1067 10 ndash13 10

34 10MV GR

minus timestimes

Nkg-1

But V = -g0R

Therefore 7

0 6

13 10 3834 10

VgR

times= minus = =

timesm s-2

To determine the field strength gh at 68 times 106 m above the surface we use the fact that 0 2

Mg GR

= such that GM = g0R2

Therefore 2 2

0h 2 2 2

h h

38 (34) 042(102)

g RGMgR R

times= = = = m s-2

(the distance from the centre is 34 times 106 + 68 times 106 = 102 times 106 m)

Exercise 1011 Thegraphbelowshowshowthegravitational

potentialoutsideoftheEarthvarieswithdistancefrom the centre

0 10 20 30 40 50 60 70

ndash1

ndash2

ndash3

ndash4

ndash5

ndash6

ndash7

r m times 106

V

Jkgndash1

times 10

7

(a) Usethegraphtodeterminethegainingravitationalpotentialenergyofasatelliteofmass200kgasitmovesfromthesurfaceof the Earth to a height of 30 times 107mabovetheEarthrsquossurface

(b) Calculatetheenergyrequiredtotakeit to infinity

(c) DeterminetheslopeofthegraphatthesurfaceoftheEarthmCommenton youranswer

Electric potential energy

The concept of electric potential energy was developedwith gravitational potential energy in mind Just as anobjectnearthesurfaceoftheEarthhaspotentialenergybecause of its gravitational interaction with the Earthsotoothereiselectricalpotentialenergyassociatedwithinteracting charges

Letusfirstlookatacaseoftwopositivepointchargeseachof 1μC that are initially bound together by a thread in a vacuuminspacewithadistancebetweenthemof10cmasshowninFigure1005Whenthethreadiscutthepointchargesinitiallyatrestwouldmoveinoppositedirectionsmovingwithvelocitiesv1 and v2 along the direction of the electrostatic force of repulsion

BEFORE AFTER

v v21

Figure 1005 Interaction of two positive particles

The electric potential energybetweentwopointchargescan be found by simply adding up the energy associated witheachpairofpointchargesForapairofinteractingchargestheelectricpotentialenergyisgivenby

∆U = ∆Ep + ∆Ek = ∆W = ∆Fr = kqQ

____ r2 times r = kqQ

____ r

Becausenoexternalforceisactingonthesystemtheenergyand momentum must be conserved Initially Ek = 0 and Ep =kqQr=9times109 times 1 times 10-1201m=009JWhentheyareagreatdistance fromeachotherEpwillbenegligibleThefinalenergywillbeequaltofrac12mv1

2+frac12mv22=009J

Momentum isalsoconservedand thevelocitieswouldbethe same magnitude but in opposite directions

Electric potential energyismoreoftendefinedintermsofapointchargemovinginanelectricfieldas

lsquothe electricpotential energybetweenany twopoints inan electric field is defined as negativeoftheworkdone by anelectricfieldinmovingapoint electric chargebetweentwolocationsintheelectricfieldrsquo

ΔU = ΔEp = -ΔW = -Fd = qEx

ΔU = ΔEp +ΔEk = ΔW = ΔFr = kqQ r2 x r = kqQ

wherexisthedistancemovedalong(oroppositeto)thedirection of the electric field

Electric potential energyismeasuredinjoule(J)Justasworkisascalarquantitysotooelectricalpotentialenergyis a scalar quantity The negative of the work done by an electric field in moving a unit electric charge between two points is independent of the path taken Inphysicswesaytheelectricfieldisalsquoconservativersquo field

PHYSICS 2015indb 311 150514 337 PM

Chapter 10

312

AH

L

Supposeanexternalforcesuchasyourhandmovesasmallpositive point test charge in the direction of a uniformelectric fieldAs it ismoving itmust be gaining kineticenergyIfthisoccursthentheelectricpotentialenergyofthe unit charge is changing

InFigure1006apointcharge+qismovedbetweenpointsAandBthroughadistancex in a uniform electric field

B

A

x

+ q

Figure 1006 Movement of a positive point charge in a uniform field

InordertomoveapositivepointchargefrompointAtopointBanexternal forcemustbeapplied to thechargeequaltoqE (F = qE)

Since the force is applied through a distance xthennegativeworkhas to bedone tomove the charge because energyis gainedmeaning there is an increaseelectric potential energybetweenthetwopointsRememberthattheworkdoneisequivalenttotheenergygainedorlostinmovingthe charge through the electric field The concept of electric potential energy is only meaningful as the electric field whichgeneratestheforceinquestionisconservative

W F xtimes E q x times= =

θ x cos θ x

Figure 1007 Charge moved at an angle to the field

If a chargemoves at an angle θ to an electric field thecomponent of the displacement parallel to the electric fieldisusedasshowninFigure918

W F x Eq x θcostimes= =

Theelectricpotentialenergyisstoredintheelectricfieldand theelectricfieldwill return theenergy to thepointcharge when required so as not to violate the Law ofconservationofenergy

Electric potential

The electric potential at a point in an electric field is defined as being the work done per unit charge in bringing a small positive point charge from infinity to that point

ΔV = Vinfin ndash Vf = ndash W ___ q

Ifwedesignatethepotentialenergytobezeroatinfinitythen it follows thatelectricpotentialmustalsobezeroat infinity and the electric potential at any point in an electricfieldwillbe

ΔV = ndash W ___ q

NowsupposeweapplyanexternalforcetoasmallpositivetestchargeasitismovedtowardsanisolatedpositivechargeTheexternalforcemustdoworkonthepositivetestchargetomoveittowardstheisolatedpositivechargeandtheworkmustbepositivewhiletheworkdonebytheelectricfieldmustthereforebenegativeSotheelectricpotentialatthatpointmustbepositiveaccordingtotheaboveequationIfanegativeisolatedchargeisusedtheelectricpotentialatapointonthepositivetestchargewouldbenegativePositivepoint chargesof their ownaccordmove fromaplaceofhighelectricpotential toaplaceof lowelectricpotentialNegative point charges move the other way from lowpotentialtohighpotentialInmovingfrompointAtopointBinthediagramthepositivecharge+qismovingfromalowelectricpotentialtoahighelectricpotential

In thedefinitiongiven the termldquoworkperunitchargerdquohas significance If the test charge is +16 times 10-19Cwherethe charge has a potential energy of 32 times 10-17Jthenthepotentialwouldbe32times10-17J+16times10-19C=200JC-1 Now if the charge was doubled the potential wouldbecome 64times10-17 J However the potential per unitchargewouldbethesame

Electric potential is a scalar quantityandithasunitsJC-1 orvoltswhere1voltequalsone joulepercoloumbThevoltallowsustoadoptaunitfortheelectricfieldintermsofthevolt

PreviouslytheunitfortheelectricfieldwasNCndash1

W = qV and F = qEsoW ___ V = F __ E

E = FV ___ W = FV ___ Fm V mndash1

That is the units of the electric field E can also beexpressed as V mndash1

The work done per unit charge in moving a point charge between two points in an electric field is again independent of the path taken

PHYSICS 2015indb 312 150514 337 PM

Fields

313

AH

L

Electric potential due to a point charge

Letus takeapoint rmetres froma chargedobjectThepotentialatthispointcanbecalculatedusingthefollowing

W = ndashFr = ndashqV and F = ndash q1 q2 _____ 4πε0r

2

Therefore

W = ndash q1 q2 _____ 4πε0r

2 times r = ndash q1 q2 _____ 4πε0r

= ndashq1 times q2 _____ 4πε0r

= ndashq1V

That is V q

4πε0r--------------- =

Orsimply

V kqr------

=

Example

Determinehowmuchworkisdonebytheelectricfieldofpointcharge150μCwhenachargeof200μCismovedfrom infinity to apoint 0400m from thepoint charge(Assumenoaccelerationofthecharges)

Solution

The work done by the electric field is W = -qV = -14πε0 times q times (Q rinfin - Q r0400)

W = (- 200 times 10-6 C times 900 times 109 NmC-2 times 150 times 10-6 C) divide 0400 m = - 0675 J

An external force would have to do +0675 J of work

Electric field strength and electric potential gradient

Letus lookbackatFigure1006Supposeagain that thecharge+qismovedasmalldistancebyaforceFfromAtoBsothattheforcecanbeconsideredconstantTheworkdoneisgivenby

∆W F x∆times=

TheforceFandtheelectricfieldEareoppositelydirectedandweknowthat

F = -qE and ΔW = q ΔV

Thereforetheworkdonecanbegivenas

q ΔV = -q E Δx

Therefore E V∆x∆-------

ndash=

The rate of change of potential ΔV at a point with respect to distance Δx in the direction in which the change is maximum is called the potential gradient We say that the electric field = - the potential gradient and the units are Vm-

1 From the equation we can see that in a graph of electric potential versus distance the gradient of the straight line equals the electric field strength

In reality if a charged particle enters a uniform electric field it will be accelerated uniformly by the field and its kinetic energy will increase This is why we had to assume no acceleration in the last worked example

Ek∆ 12---

mv2 q E xsdot sdot q Vx---

xsdot sdot q Vsdot= = = =

Example

Determine how far apart two parallel plates must besituated so that a potential difference of 150 times 102 V produces an electric field strength of 100 times 103 NC-1

Solution

Using E V∆x------- xhArrndash V∆

E-------15 102 Vtimes

100 103 N C 1ndashtimes------------------------------------------= = =

= 150 times 10-1

The plates are 150 times 10-1 m apart

The electric field and the electric potential at a point due to an evenly distributed charge +q on a sphere can be represented graphically as in Figure 1008

r

r

V

E

r0

r0

r0

Charge of +Q evenly distributed over surface

E 14πε0------------ Q

r2----- r r0gt=

V 14πε0------------Q

r----r r0gt=

Figure 1008 Electric field and potential due to a charged sphere

PHYSICS 2015indb 313 150514 337 PM

Chapter 10

314

AH

L

When the sphere becomes charged we know that thechargedistributes itselfevenlyoverthesurfaceThereforeeverypartofthematerialoftheconductorisatthesamepotential As the electric potential at a point is defined as beingnumericallyequaltotheworkdoneinbringingaunitpositivechargefrominfinitytothatpointithasaconstantvalueineverypartofthematerialoftheconductor

Since the potential is the same at all points on the conducting surfacethen∆V∆xiszeroButE=ndash∆V∆xThereforethe electric field inside the conductor is zero There is no electric field inside the conductor

SomefurtherobservationsofthegraphsinFigure1009are

bull OutsidethespherethegraphsobeytherelationshipsgivenasE α 1 r2 and V α 1 r

bull Atthesurfacer = r0Thereforetheelectricfieldandpotentialhavetheminimumvalueforr at this point and this infers a maximum field and potential

bull Insidethespheretheelectricfieldiszerobull Insidethespherenoworkisdonetomoveacharge

fromapointinsidetothesurfaceThereforethereisno potential difference and the potential is the same asitiswhenr = r0

Similargraphscanbedrawnfortheelectricfieldintensityandthe electric potential as a function of distance from conducting parallelplatesandsurfacesandthesearegiveninFigure1009

Potential plot E eld plotE eld

++++

ndashndashndashndash

xx x

+

x x

ndashx x

+ ndashx

x

+ +

x

x

Figure 1009 Electric field and electric potential at a distance from a charged surface

Potential due to one or more point charges

The potential due to one point charge can be determined byusingtheequationformula

V = kq r

Example 1

Determine the electric potential at a point 20 times 10-1 m fromthecentreofan isolatedconducting spherewithapointchargeof40pCinair

Solution

Using the formula

V = kq r we have

V90 10 9times( ) 40 10 12ndashtimes( )times

20 10 1ndashtimes( )------------------------------------------------------------------ 018 V= =

the potential at the point is 180 times 10-1 V

The potential due to a number of point charges can be determined by adding up the potentials due to individual point charges because the electric potential at any point outside a conducting sphere will be the same as if all the charge was concentrated at its centre

Example 2

Threepointchargesofareplacedattheverticesofaright-angledtriangleasshowninthediagrambelowDeterminetheabsolutepotentialat the+20μCchargeduetothetwoothercharges

- 6μC

+3μC +2μC

4 m

3 m

Solution

The electric potential of the +2 μC charge due to the ndash 6 μC charge is

V = (9 times 109 Nm2C-2 times -6 times 10-6 C) divide (radic 32 + 42) m = - 108 times 104 V

PHYSICS 2015indb 314 150514 338 PM

Fields

315

AH

L

Insummarywecanconcludethat

bull No work is done to move a charge along an equipotential

bull Equipotentials are always perpendicular to the electric lines of force

Figure1011and1012showsomeequipotential lines fortwo oppositely charged and identically positive spheresseparated by a distance

+ve ndashve

equipotential lines

Figure 1011 Equipotential lines between two opposite charges

equipotential lines

Figure 1012 Equipotential lines between two charges which are the same

+

ndash ndash ndash ndash

+ + +

40 V

30 V

20 V

10 V

50 V

Figure 1013 Equipotential lines between charged parallel plates

Figure1013showstheequipotentiallinesbetweenchargedparallel plates Throughout this chapter the similarities and differences between gravitational fields and electric fieldshavebeendiscussedTherelationshipsthatexistsbetweengravitationalandelectricquantitiesandtheeffectsofpointmassesandchargesissummarisedinTable1014

The electric potential of the +2 μC charge due to the +3 μC charge is

V = (9 times 109 Nm2C-2 times 3 times 10-6 C) divide 3m = 9 times 103 V

The net absolute potential is the sum of the 2 potentials- 108 times 104 V + 9 times 103 V =- 18 times 103 V

The absolute potential at the point is - 18 times 103 V

Equipotential surfacesRegions in space where the electric potential of a chargedistributionhasaconstantvaluearecalledequipotentials The placeswherethepotentialisconstantinthreedimensionsarecalled equipotential surfacesandwheretheyareconstantintwodimensionstheyarecalledequipotential lines

TheyareinsomewaysanalogoustothecontourlinesontopographicmapsInthiscasethegravitationalpotentialenergy is constant as amassmoves around the contourlines because the mass remains at the same elevationabovetheEarthrsquossurfaceThegravitationalfieldstrengthacts in a direction perpendicular to a contour line

Similarlybecausetheelectricpotentialonanequipotentiallinehas thesamevalueanelectric forcecandonoworkwhena testchargemovesonanequipotentialThereforethe electric field cannot have a component along anequipotentialandthusitmustbeeverywhereperpendiculartotheequipotentialsurfaceorequipotentiallineThisfactmakesiteasytoplotequipotentialsifthelinesofforceorlinesofelectricfluxofanelectricfieldareknown

Forexamplethereareaseriesofequipotentiallinesbetweentwoparallelplateconductorsthatareperpendiculartotheelectric fieldTherewill be a series of concentric circles(eachcirclefurtherapartthanthepreviousone)thatmapouttheequipotentialsaroundanisolatedpositivesphereThe lines of force and some equipotential lines for anisolatedpositivesphereareshowninFigure1010

Lines ofequipotential

Figure 1010 Equipotentials around an isolated positive sphere

PHYSICS 2015indb 315 150514 338 PM

Chapter 10

316

AH

L

Gravitational quantity Electrical quantityQ

uant

ities

g V∆x∆-------ndash= E V∆

x∆-------ndash=

Poin

t mas

ses a

nd ch

arge

s

V Gmr----

ndash= V 14πε0------------q

r---=

g G m

r2-----ndash= E 14πε0------------ q

r2-----=

F Gm1 m2

r2--------------= F 1

4πε0------------

q1q2

r2-----------=

Figure 1014 Formulas

Example

Deduce the electric potential on the surface of a gold nucleus that has a radius of 62 fm

Solution

Using the formula

V = kq r and knowing the atomic number of gold is 79 We will assume the nucleus is spherical and it behaves as if it were a point charge at its centre (relative to outside points)

V = 90 times 109 Nm2C-2 times 79 times 16 times 10-19 C divide 62 times 10-15 m = 18 times 107 V

The potential at the point is 18 MV

Example

Deduce the ionisation energy in electron-volts of theelectron in the hydrogen atom if the electron is in its ground state and it is in a circular orbit at a distance of 53 times 10-11 m from the proton

Solution

This problem is an energy coulombic circular motion question based on Bohrrsquos model of the atom (not the accepted quantum mechanics model) The ionisation energy is the energy required to remove the electron from the ground state to infinity The electron travels in a circular orbit and therefore has a centripetal acceleration The ionisation energy will counteract the coulombic force and the movement of the electron will be in the opposite direction to the centripetal force

Total energy = Ek electron + Ep due to the proton-electron interaction

ΣF = kqQ r2 = mv2 r and as such mv2 = = kqQ r

Therefore Ek electron = frac12 kqQ r

Ep due to the proton-electron interaction = - kqQ r

Total energy = frac12 kqQ r + - kqQ r = -frac12 kqQ r

= - 90 times 109 Nm2C-2 times (16 times 10-19 C)2 divide 53 times 10-11 m = -217 times 10-18 J

= -217 times 10-18 J divide 16 times 10-19 = -136 eV

The ionisation energy is 136 eV

Exercise 1021 ApointchargePisplacedmidwaybetweentwo

identicalnegativechargesWhichoneofthefollowingiscorrectwithregardstoelectricfieldandelectricpotentialatpointP

Electric field Electric potentialA non-zero zeroB zero non-zeroC non-zero non-zeroD zero zero

2 Twopositivechargedspheresaretiedtogetherinavacuumsomewhereinspacewheretherearenoexternal forces A has a mass of 25 g and a charge of20μCandBhasamassof15gandachargeof30μCThedistancebetweenthemis40cmTheyarethenreleasedasshowninthediagram

BEFORE AFTER

v1 v2

A B

(a) Determinetheirinitialelectricpotentialenergy in the before situation

(b) DeterminethespeedofsphereBafterrelease

3 Thediagrambelowrepresentstwoequipotentiallines in separated by a distance of 5 cm in a uniform electric field

+ + + + + + + +

ndash ndash ndash ndash ndash ndash ndash ndash

40 V

20 V 5 cm

PHYSICS 2015indb 316 150514 338 PM

Fields

317

AH

L

Determine the strength of the electric field

4 Thisquestionisabouttheelectricfieldduetoacharged sphere and the motion of electrons in that fieldThediagrambelowshowsanisolatedmetalsphereinavacuumthatcarriesanegativeelectriccharge of 60 μC

ndash

(a) Onthediagramdrawtheconventionalwayto represent the electric field pattern due to the charged sphere and lines to represent threeequipotentialsurfacesintheregionoutside the sphere

(b) Explainhowthelinesrepresentingtheequipotentialsurfacesthatyouhavesketchedindicatethatthestrengthoftheelectricfieldisdecreasingwithdistancefrom the centre of the sphere

(c) Theelectricfieldstrengthatthesurfaceofthe sphere and at points outside the sphere can be determined by assuming that the sphere acts as a point charge of magnitude 60 μC at its centre The radius of the sphere is 25 times 10ndash2 m Deduce that the magnitude of the field strength at the surface of the sphere is 86 times 107 Vmndash1

An electron is initially at rest on the surface of the sphere

(d) (i) Describethepathfollowedbytheelectronasitleavesthesurfaceofthesphere

(ii) Calculatetheinitialaccelerationofthe electron

5 Determinetheamountofworkthatisdoneinmovingachargeof100nCthroughapotentialdifferenceof150times102 V

6 Three identical 20 μC conducting spheres are placedatthecornersofanequilateraltriangleofsides 25 cm The triangle has one apex C pointing upthepageand2baseanglesAandBDeterminetheabsolutepotentialatB

7 Determinehowfaraparttwoparallelplatesmustbesituatedsothatapotentialdifferenceof250 times 102 V produces an electric field strength of 200 times 103 NC-1

8 Thegapbetweentwoparallelplatesis10times10-3mandthereisapotentialdifferenceof10times104 V betweentheplatesCalculate

i theworkdonebyanelectroninmovingfrom one plate to the other

ii thespeedwithwhichtheelectronreachesthe second plate if released from rest

iii theelectricfieldintensitybetweentheplates

9 Anelectronguninapicturetubeisacceleratedby a potential 25 times 103VDeterminethekineticenergygainedbytheelectroninelectron-volts

10 Determine the electric potential 20 times 10-2 m from achargeof-10times10-5 C

11 Determinetheelectricpotentialatapointmid-waybetweenachargeofndash20pCandanotherof+5pConthelinejoiningtheircentresifthecharges are 10 cm apart

12 During a thunderstorm the electric potential differencebetweenacloudandthegroundis10 times 109 V Determine the magnitude of the change in electric potential energy of an electron thatmovesbetweenthesepointsinelectron-volts

13 A charge of 15 μC is placed in a uniform electric fieldoftwooppositelychargedparallelplateswithamagnitudeof14times103 NC-1

(a) Determinetheworkthatmustbedoneagainstthefieldtomovethepointchargeadistance of 55 cm

(b) Calculatethepotentialdifferencebetweenthe final and initial positions of the charge

(c) Determinethepotentialdifferencebetweenthe plates if their separation distance is 15 cm

14 Duringaflashoflightningthepotentialdifferencebetweenacloudandthegroundwas12 times 109 V and the amount of transferred charge was32C

(a) Determinethechangeinenergyofthetransferred charge

(b) Iftheenergyreleasedwasallusedtoacceleratea1tonnecardeduceitsfinalspeed

(c) Iftheenergyreleasedcouldbeusedtomelticeat0degCdeducetheamountoficethatcould be melted

PHYSICS 2015indb 317 150514 338 PM

Chapter 10

318

AH

L

15 SupposethatwhenanelectronmovedfromAtoBin the diagram along an electric field line that the electric field does 36 times 10-19Jofworkonit

Determinethedifferencesinelectricpotential

(a) VB ndash VA(b) VC ndash VA(c) VC ndash VB

16 DeterminethepotentialatpointPthatislocatedatthecentreofthesquareasshowninthediagrambelow

- 6μC

+3μC +2μC

1m

1m

5μC

P

102 Fields at work

Essential idea Similar approaches can be taken inanalysingelectricalandgravitationalpotentialproblems

Understandingsbull Potentialandpotentialenergybull Potentialgradientbull Potentialdifferencebull Escapespeedbull Orbitalmotionorbitalspeedandorbitalenergybull Forcesandinverse-squarelawbehaviour

Orbital motion orbital speed and orbital energy

Although orbital motion may be circular elliptical orparabolic this sub-topic only dealswith circular orbitsThis sub-topic is not fundamentally new physics butan application that synthesizes ideas from gravitationcircularmotiondynamicsandenergy

TheMoonorbitstheEarthandinthissenseitisoftenreferredtoasasatelliteoftheEarthBefore1957itwastheonlyEarthsatelliteHoweverin1957theRussianslaunchedthefirstmanmadesatelliteSputnik1SincethisdatemanymoresatelliteshavebeenlaunchedandtherearenowliterallythousandsofthemorbitingtheEarthSomeareusedtomonitortheweathersome used to enable people to find accurately their position on thesurfaceoftheEarthmanyareusedincommunicationsandnodoubtsomeareusedtospyonothercountriesFigure1015showshowinprincipleasatellitecanbeputintoorbit

Theperson(whosesizeisgreatlyexaggeratedwithrespectto Earth) standing on the surface on the Earth throwssome stonesThe greater the speed with which a stoneis thrown the further it will land from her The pathsfollowedbythethrownstonesareparabolasByastretchof the imaginationwecanvisualisea situation inwhichastoneis thrownwithsuchaspeedthatbecauseof thecurvatureoftheEarthitwillnotlandonthesurfaceoftheEarthbutgointoldquoorbitrdquo(Path4onFigure1015)

NATURE OF SCIENCECommunication of scientific explanations The ability to apply field theory to the unobservable (charges) and the massively scaled (motion of satellites) required scientists to develop new ways to investigate analyse and report findings to a general public used to scientific discoveries based on tangible and discernible evidence (51)

copy IBO 2014

PHYSICS 2015indb 318 150514 338 PM

Fields

319

AH

L

1

2

34Earth

Figure 1015 Throwing a stone into orbit

TheforcethatcausesthestonestofollowaparabolicpathandtofalltoEarthisgravityandsimilarlytheforcethatkeeps the stone in orbit is gravity For circular motionto occur we have seen that a force must act at rightanglestothevelocityofanobjectthatistheremustbeacentripetalforceHenceinthesituationwedescribeherethe centripetal force for circular orbital motion about the EarthisprovidedbygravitationalattractionoftheEarth

We can calculate the speedwithwhich a stonemustbethrowninordertoputitintoorbitjustabovethesurfaceof the Earth

If the stone has mass m and speed v thenwehave fromNewtonrsquos2ndlaw

mv2

RE---------- G

MEm

RE2------------=

whereRE is the radius of the Earth and ME is the mass of the Earth

Bearinginmindthatg 0 GME

RE2--------= then

v g RE 10 64 106timestimes 8 103times= = =

Thatisthestonemustbethrownat8times103m sndash1

Clearly we are not going to get a satellite into orbit soclosetothesurfaceoftheEarthMovingatthisspeedthefrictionduetoairresistancewouldmeltthesatellitebeforeithadtravelledacoupleofkilometresInrealitythereforeasatelliteisputintoorbitabouttheEarthbysendingitattachedtoarocketbeyondtheEarthrsquosatmosphereandthengivingitacomponentofvelocityperpendiculartoaradialvectorfromtheEarthSeeFigure1016

Earth Satellite carried by rocket to hereSatellite orbit

Tangential component of velocity

Figure 1016 Getting a satellite into orbit

Keplerrsquos third law(This work of Kepler and Newtonrsquos synthesis of the work is an excellent example of the scientific method and makes for a good TOK discussion)

In 1627 Johannes Kepler (1571-1630)publishedhis lawsof planetary motion The laws are empirical in natureandwere deduced from the observations of theDanishastronomer Tycho de Brahe (1546-1601)The third lawgives a relationship between the radius of orbit R of a planet and its period TofrevolutionabouttheSunThelawisexpressedmathematicallyas

2

3 constantTR

=

We shall nowuseNewtonrsquos LawofGravitation to showhowitisthattheplanetsmoveinaccordancewithKeplerrsquosthirdlaw

InessenceNewtonwasabletousehislawofgravitytopredict the motion of the planets since all he had to do was factor theF givenby this law intohis second law F = ma to find their accelerations and hence their future positions

InFigure1017theEarthisshownorbitingtheSunandthedistancebetweentheircentresisR

RSun

Earth

F es

F se

Figure 1017 Planets move according to Keplerrsquos third law

Fes is the force that the Earth exerts on the Sun and Fse is the force that the Sun exerts on the Earth The forces are equalandoppositeandtheSunandtheEarthwillactuallyorbitaboutacommoncentreHoweversince theSun issoverymuchmoremassivethantheEarththiscommoncentrewillbeclosetothecentreoftheSunandsowecanregard the Earth as orbiting about the centre of the Sun Theotherthingthatweshallassumeisthatwecanignorethe forces that the other planets exert on the Earth (This wouldnotbeawisethingtodoifyouwereplanningtosend a space ship to theMoon for example)We shallalso assume that we have followed Newtonrsquos exampleandindeedprovedthataspherewillactasapointmasssituated at the centre of the sphere

Kepler had postulated that the orbits of the planets areellipticalbut since theeccentricityof theEarthrsquosorbit issmallweshallassumeacircularorbit

PHYSICS 2015indb 319 150514 338 PM

Chapter 10

320

AH

L

TheaccelerationoftheEarthtowardstheSunisa = Rω2

whereω ω 2πT------=

Hencea R 2π

T------ times

2 4π2 RT 2------------- = =

But the acceleration is given by Newtonrsquos Second Law F = mawhereFisnowgivenbytheLawofGravitationSo in this situation

F maGMs Me

R2-------------------= = butwealsohavethat

a a R 2πT

------ times

2 4π2 RT 2------------- = = and m = Me so that

G MsMe

R2------------------- Me4π2R

T2-------------times

GMs

4π2-----------rArr R3

T2------= =

Butthequantity

isaconstantthathasthesamevalueforeachoftheplanetssowehaveforalltheplanetsnotjustEarththat

R3

T2------ k =

wherekisaconstantWhichisofcourseKeplerrsquosthirdlaw

ThisisindeedanamazingbreakthroughItisdifficulttorefutethe idea that all particles attract each other in accordance withtheLawofGravitationwhenthelawisabletoaccountfortheobservedmotionoftheplanetsabouttheSun

The gravitational effects of the planets upon each othershould produce perturbations in their orbits Such is the predictivepoweroftheUniversalGravitationalLawthatit enabled physicists to compute these perturbations The telescopehadbeeninventedin1608andbythemiddleofthe 18th Century had reached a degree of perfection in design that enabled astronomers to actually measure the orbital perturbations of the planets Their measurements werealways inagreementwith thepredictionsmadebyNewtonrsquoslawHoweverin1781anewplanetUranuswasdiscovered and the orbit of this planet did not fit withthe orbit predicted by Universal Gravitation Such wasthe physicistrsquos faith in theNewtonianmethod that theysuspectedthatthediscrepancywasduetothepresenceofayetundetectedplanetUsingtheLawofGravitationtheFrenchastronomerJLeverrier and the English astronomer J C Adamswereable tocalculate justhowmassive thisnewplanetmustbeandalsowhereitshouldbeIn1846the planetNeptunewas discovered justwhere they hadpredicted Ina similarwaydiscrepancies in theorbitofNeptune ledto thepredictionandsubsequentdiscoveryin1930oftheplanetPlutoNewtonrsquosLawofGravitation

had passed the ultimate test of any theory it is not only abletoexplainexistingdatabutalsotomakepredictions

Satellite energyWhenasatelliteisinorbitaboutaplanetitwillhavebothkineticenergyandgravitationalpotentialenergySupposewe consider a satellite ofmass m that is in an orbit of radius r about a planet of mass M

Thegravitationalpotentialduetotheplanetatdistancer from its centre is

ThegravitationalpotentialenergyofthesatelliteVsat

is therefore GMe m

r----------------ndash

ThatisVsat GMem

r----------------ndash=

Thegravitationalfieldstrengthatthesurfaceoftheplanetisgivenby

g 0GMe

Re2------------=

Hencewecanwrite

Vsat g0 Re

2

rndash=

m

ThekineticenergyofthesatelliteKsatisequaltofrac12mv2wherev is its orbital speed

Byequatingthegravitationalforceactingonthesatellitetoitscentripetalaccelerationwehave

G Mem

r2---------------- mv2

r---------- mv2hArr

GMemr----------------= =

Fromwhich12---

mv2 12---

GMemr----------------times=

= 0 e

2g R

r

2m

Which is actually quite interesting since it shows thatirrespective of the orbital radius the KE is numericallyequal to half the PE Also the total energy E tot K sat Vsat+ 1

2---G Me m

r----------------times G Me m

r----------------ndash + 1

2---G Me m

r----------------ndash= = =of the satelliteisalwaysnegativesince

E tot K sat Vsat+ 12---

G Me mr----------------times

G Me mr----------------ndash

+ 12---

G Me mr----------------ndash= = =

PHYSICS 2015indb 320 150514 338 PM

Fields

321

AH

L

The energies of an orbiting satellite as a function of radial distancefromthecentreofaplanetareshownplottedinFigure1018

12

100806

0402

0ndash02

ndash04

ndash06ndash08ndash10

ndash12

2 4 6 8 10 12

ener

gya

rbita

ry u

nits

distance timesR

kinetic energytotal energy

potential energy

Figure 1018 Energy of an orbiting satellite as a function of distance from the centre of a planet

WeightlessnessSuppose that you are inanelevator(lift)thatisdescendingat constant speedandyou let goof abook thatyouareholdinginyourhandThebookwillfalltothefloorwithaccelerationequaltotheaccelerationduetogravityIfthecablethatsupportstheelevatorweretosnap(asituationthatItrustwillneverhappentoanyofyou)andyounowletgothebookthatyouareholdinginyourotherhandthis book will not fall to the floor - it will stay exactlyin linewithyourhandThis isbecause thebook isnowfallingwith the sameaccelerationas theelevatorandassuch the book cannot ldquocatchrdquo up with the floor of theelevatorFurthermoreifyouhappenedtobestandingonasetofbathroomscalesthescaleswouldnowreadzero-youwouldbeapparentlyweightlessItisthisideaoffreefallthatexplainstheapparentweightlessnessofastronautsin an orbiting satellite These astronauts are in free fall inthesensethattheyareacceleratingtowardsthecentre of the Earth

It is actually possible to define theweight of a body inseveral different waysWe can define it for example asthegravitationalforceexertedonthebodybyaspecifiedobject such as the EarthThiswe have seen thatwe doinlotsofsituationswherewedefinetheweightasbeingequaltomgIfweusethisdefinitionthenanobjectinfreefallcannotbydefinitionbeweightlesssinceitisstillinagravitationalfieldHoweverifwedefinetheweightofanobjectintermsofaldquoweighingrdquoprocesssuchasthereadingonasetofbathroomscaleswhichineffectmeasuresthecontact force between the object and the scales thenclearlyobjectsinfreefallareweightlessOnenowhasto

askthequestionwhetherornotitispossibleForexampletomeasurethegravitationalforceactingonanastronautin orbit about the Earth

WecanalsodefineweightintermsofthenetgravitationalforceactingonabodyduetoseveraldifferentobjectsForexample for an object out in space its weight could bedefined in terms of the resultant of the forces exerted on it bytheSuntheMoontheEarthandalltheotherplanetsin the Solar System If this resultant is zero at a particular pointthenthebodyisweightlessatthispoint

Inviewofthevariousdefinitionsofweightthatareavailabletousitisimportantthatwhenweusethewordldquoweightrdquoweareawareofthecontextinwhichitisbeingused

Example

Calculate the height above the surface of the Earth atwhichageo-stationarysatelliteorbits

Solution

A geo-stationary satellite is one that orbits the Earth in such a way that it is stationary with respect to a point on the surface of the Earth This means that its orbital period must be the same as the time for the Earth to spin once on its axis ie 24 hours

From Keplerrsquos third law we have G Ms

4π2----------- R3

T2------=

That is

RMe

mh

using the fact that the force of attraction between the satellite and the Earth is given by

FG Mem

R2----------------=

and that F = ma

where m is the mass of the satellite and a 4π2R

T2-------------=

we have

GMem

R2---------------- m 4π2R

T2-------------times

GMe

4π2------------rArr R3

T2------= =

PHYSICS 2015indb 321 150514 338 PM

Chapter 10

322

AH

L

Now the mass of the Earth is 60 times 1024 kg and the period T measured in seconds is given by T = 86400 s

So substitution gives R = 42 times 106 m

The radius of the Earth is 64 times 106 m so that the orbital height h is about 36 times 107 m

Example

Calculatetheminimumenergyrequiredtoputasatelliteofmass500kgintoanorbitthatisasaheightequaltotheEarthrsquosradiusabovethesurfaceoftheEarth

Solution

We have seen that when dealing with gravitational fields and potential it is useful to remember that

g 0G M

Re2---------= or g0 Re

2 GM=

The gravitational potential at the surface of the Earth is

g 0ndash Re GMRe

---------ndash=

The gravitational potential at a distance R

from the centre of the Earth is

=

The difference in potential between the surface of the Earth and a point distance R from the centre is therefore

V∆ g 0Re 1ReR------ndash

=

If R 2Re= then V∆g0 Re

2------------=

This means that the work required to ldquoliftrdquo the satellite into orbit is g0Rm where m is the mass of the satellite This is equal to

10 32 106 500timestimestimes = 16000 MJ

However the satellite must also have kinetic energy in order to orbit Earth This will be equal to

g 0 mRe2

2R-----------------g 0mRe

2

4----------------- 8000 MJ= =

The minimum energy required is therefore

24000 MJ

Exercise 1031 The speed needed to put a satellite in orbit does

not depend on

A the radius of the orbitB theshapeoftheorbitC thevalueofgattheorbitD the mass of the satellite

2 EstimatethespeedofanEarthsatellitewhoseorbitis400kmabovetheEarthrsquossurfaceAlsodetermine the period of the orbit

3 Calculatethespeedofa200kgsatelliteorbitingthe Earth at a height of 70 times 106 m

Assume that g = 82 m sndash2 for this orbit

4 TheradiioftwosatellitesXandYorbitingthe Earth are 2r and 8rwherer is the radius of the Earth Calculate the ratio of the periods of revolutionofXtoY

5 A satellite of mass mkgissentfromEarthrsquossurfaceinto an orbit of radius 5RwhereR is the radius of theEarthWritedownanexpressionfor

(a) thepotentialenergyofthesatelliteinorbit

(b) thekineticenergyofthesatelliteinorbit

(c) theminimumworkrequiredtosendthesatellitefromrestattheEarthrsquossurfaceintoits orbit

6 A satellite in an orbit of 10rfallsbacktoEarth(radius r)afteramalfunctionDeterminethespeedwithwhichitwillhittheEarthrsquossurface

7 The radius of the moon is frac14 that of the Earth AssumingEarthandtheMoontohavethesamedensitycomparetheaccelerationsoffreefallatthe surface of Earth to that at the surface of the Moon

8 UsethefollowingdatatodeterminethegravitationalfieldstrengthatthesurfaceoftheMoon and hence determine the escape speed from the surface of the Moon

Mass of the Moon = 73 times 1022kg

RadiusoftheMoon=17times106 m

PHYSICS 2015indb 322 150514 338 PM

Wave Phenomena

299

AH

L

94 Resolution

Essential idea Resolution places an absolute limit on the extent to which an optical or other system can separate images of objects

Understandings

bull Thesizeofadiffractingaperturebull Theresolutionofsimplemonochromatic

two-source systems

The size of a diffracting apertureOur discussion concerning diffraction so far has been for rectangular slits When light from a point source enters a smallcircularapertureitdoesnotproduceabrightdotasanimagebutacirculardiscknownasAiryrsquosdiscsurroundedbyfainterconcentriccircularringsasshowninFigure922

Laser

Figure 922 Diffraction at a circular aperture

Diffractionatanapertureisofgreatimportancebecausetheeye and many optical instruments have circular apertures

What is the half-angular width of the central maximum of the diffraction formed by a circular aperture This is not easy to calculate since it involves some advanced mathematics The problemwasfirstsolvedbytheEnglishAstronomerRoyalGeorgeAiryin1835whoshowedthatforcircularapertures

θ 122 λb--------------= where b is the diameter of the aperture

Example 2

Inthefollowingdiagramparallellightfromadistantpointsource(suchasastar)isbroughttofocusonthescreenSbyaconverginglens(the lens is shown as a vertical arrow)

Thefocal length(distance fromlens toscreen) is25cmand the diameter of the lens is 30 cm The wavelength of thelightfromthestaris560nmCalculatethediameterofthe diameter of the image on the screen

Solution

The lens actually acts as a circular aperture of diameter 30 cm The half angular width of central maximum of the diffraction pattern that it forms on the screen is

75

2

122 122 56 10 23 1030 10b

λθminus

minusminus

times times= = = times

times rad

The diameter of the central maxima is therefore

25 times 10-2 times 23 times 10-5

= 57 times 10-6 m

Although this is small it is still finite and is effectively the image of the star as the intensity of the secondary maxima are small compared to that of the central maximum

The resolution of simple monochromatic two-source systemsThe astronomers tell us that many of the stars that we observe with the naked eye are in fact binary stars That iswhatwe see as a single star actually consistsof twostarsinorbitaboutacommoncentreFurthermoretheastronomerstellusthatifweusealdquogoodrdquotelescopethenwewillactuallyseethetwostarsThatiswewillresolve the single point source into its two component parts Sowhatisitthatdetermineswhetherornotweseethetwo stars as a single point source ie what determines whether or not two sources can be resolved It canrsquot just bethatthetelescopemagnifiesthestarssinceiftheyareacting as point sources magnifying them is not going to make a great deal of difference

Ineachofoureyesthereisanaperturethepupilthroughwhich the light enters This light is then focused by the eye lens onto the retina But we have seen that when light passes through an aperture it is diffracted and so when we look at

NATURE OF SCIENCEImproved technology The Rayleigh criterion is the limit of resolution Continuing advancement in technology such as large diameter dishes or lenses or the use of smaller wavelength lasers pushes the limits of what we can resolve (18)

copy IBO 2014

PHYSICS 2015indb 299 150514 337 PM

Chapter 9

300

AH

L

apointsourceadiffractionpatternwillbeformedontheretina If we look at two point sources then two diffraction patterns will be formed on the retina and these patterns will overlap The width of our pupil and the wavelength of the light emitted by the sources will determine the amount by which they overlap But the degree of overlap will also depend on the angular separation of the two point sources WecanseethisfromFigures923

pupilretina

S1

S2

P2

P1

θ

Figure 923

LightfromthesourceS1 enters the eye and is diffracted by the pupil such that the central maximum of the diffraction patternisformedontheretinaatP1SimilarlylightfromS2producesamaximumatP2 If the two central maxima are well separated then there is a fair chance that we will see the two sources as separate sources If they overlap then we will not be able to distinguish one source from another From the diagram we see as the sources are movedclosertotheeyethentheangleθ increases and so does the separation of the central maxima

Figures924925926and927showsthedifferentdiffractionpatternsandtheintensitydistributionthatmightresultonthe retina as a result of light from two point sources

S1 S21 Well resolved

Figure 924 Very well resolved

S1 S22 Well resolved

Figure 925 Well resolved

S1 S23 Just resolved Rayleigh criterion Minimum of S2 coincides with maximum peak of S1

Figure 926 Just resolved

4 Not resolved

Figure 927 Not resolved

We have suggested that if the central maxima of the two diffraction patterns are reasonably separated then we should be able to resolve two point sources In the late 19th century Lord Rayleigh suggested by how much they should be separated in order for the two sources to be just resolved Ifthecentralmaximumofonediffractionpattern coincides with the firstminima of the otherdiffraction pattern then the two sources will just beresolved This is known as the Rayleigh Criterion

Figure926showsjustthissituationThetwosourcesarejust resolved according to the Rayleigh criterion since the peak of the central maximum of one diffraction pattern coincideswiththefirstminimumoftheotherdiffractionpattern This means that the angular separation of the peaks of the two central maxima formed by each source is just the half angular width of one central maximum ie

θ λb---

=

where b is the width of the slit through which the light fromthesourcespassesHoweverweseefromFigure1117that θ is the angle that the two sources subtend at the slit Henceweconcludethattwosourceswillberesolvedbyaslit if the angle that they subtend at the slit is greater than or equaltoλfraslb

PHYSICS 2015indb 300 150514 337 PM

Wave Phenomena

301

AH

L

Sofarwehavebeenassumingthattheeyeisarectangularslit whereas clearly it is a circular aperture and so we must use the formula

θ 122 λb--------------=

As mentioned above the angle θ is sometimes called the resolving power but should more accurately be called the minimum angle of resolution (θmin)

Clearly the smaller θ the greater will be the resolving power

The diffraction grating is a useful for differentiating closely spacedlinesinemissionspectraLikeaprismspectrometerit can disperse a spectrum into its components but it is better suited because it has higher resolution than the prism spectrometer

If λ1 and λ2 are two nearly equal wavelengths that canbarelybedistinguishedtheresolvanceorresolvingpowerofthegratingisdefinedas

R = λ _______ (λ2 ndash λ 1) = λ ___ Δλ

Where λ asymp λ1 asymp λ2

Therefore thediffractiongratingwithahighresolvancewill be better suited in determining small differences in wavelength

If the diffraction grating has Nlinesbeingilluminateditcan be shown that the resolving power of the mth order diffraction is given by

R = mN

Obviouslytheresolvancebecomesgreaterwiththeordernumber m and with a greater number of illuminated slits

Example

A benchmark for the resolving power of a grating is the sodium doublet in the sodium emission spectrum Two yellow lines in itsspectrumhavewavelengthsof58900nmand58959nm

(a) Determinetheresolvanceofagratingifthegivenwavelengths are to be distinguished from each other

(b) Howmanylinesinthegratingmustbeilluminated in order to resolve these lines in the second order spectrum

Solution

(a) R = λ ___ Δλ = 58900nm __________________ (58900ndash58959)nm

= 589nm _______ 059nm = 10 times 103

(b) N = R __ m = 10 times 10 3 ________ 2 =50times102 lines

It has been seen that diffraction effectively limits the resolving power of optical systems This includes such systems as the eye telescopes and microscopes Theresolving power of these systems is dealt with in the next section This section looks at links between technology and resolving power when looking at the very distant and when looking at the very small

Radio telescopes

The average diameter of the pupil of the human eye is about 25 mm This means that the eye will justresolve two point sources emitting light of wavelength 500 nm if their angular separation at the eye is 7

43

50 10122 24 1025 10

θminus

minusminus

times= times = times

times rad

If the eye were to be able to detect radio waves of wavelength 015m then tohave the same resolvingpower thepupilwould have to have a diameter of about 600 m Clearly this is nonsense but it does illustrate a problem facingastronomers who wish to view very distant objects such as quasarsandgalaxies thatemitradiowavesConventionalradio telescopes consist of a large dish typically 25 min diameter Even with such a large diameter the radiowavelength resolving power of the telescope is much less thantheopticalresolvingpowerofthehumaneyeLetuslook at an example

Example

The Galaxy Cygnus A can be resolved optically as an elliptically shaped galaxy However it is also a strongemitterof radiowavesofwavelength015mTheGalaxyisestimatedtobe50times1024mfromEarthUseofaradiotelescope shows that the radio emission is from two sources separated by a distance of 30 times 1021m Estimate thediameterofthedishrequiredtojustresolvethesources

Solution

The angle θ that the sources resolve at the telescope is given by

θ 21

424

30 10 60 1050 10

θ minustimes= = times

times rad

and 4

122 122 015 300060 10

d λθ minus

times= = =

timesm = 30 km

A radio telescope dish of this size would be impossible tomake letalonesupportThisshowsthatasingledishtype radio telescope cannot be used to resolve the sources and yet they were resolved To get round the problemastronomers use two radio telescopes separated by a large

PHYSICS 2015indb 301 150514 337 PM

Chapter 9

302

AH

L

distance The telescopes view the same objects at the same time and the signals that each receive from the objects are simultaneously superimposed The result of the superposition of the two signals is a two-slit interference pattern The pattern has much narrower fringe spacing than that of the diffraction pattern produced by either telescope on its own hence producing a much higherresolvingpowerWhentelescopesareusedlikethistheyare called a stellar interferometer

InSocorroinNewMexicothereisastellarinterferometerthat consists of 27 parabolic dishes each of diameter25 m arranged in a Y-shape that covers an area of 570km2Thisisaso-calledVeryLargeArray(VLA)Evenhigher resolution can be obtained by using an array of radio telescopes in observatories thousands of kilometres apart A system that uses this so-called technique oflsquovery-long-baseline interferometryrsquo (VLBI) is known asalsquovery-long-baselinearrayrsquo(VLBA)WithVLBAaradiowavelength resolving power can be achieved that is 100 timesgreaterthanthebestopticaltelescopesEvenhigherresolving power can be achieved by using a telescope that isinasatelliteorbitingEarthJapanrsquosInstituteofSpaceandAstronautical Science (ISAS) launched sucha system inFebruary 1997TheNationalAstronomicalObservatoryof Japan the National Science Foundationrsquos NationalRadio Astronomy Observatory (NRAO) the CanadianSpaceAgency theAustraliaTelescopeNationalFacilitythe European VLBI Network and the Joint Instituteback this project forVeryLongBaseline Interferometryin EuropeThis project is a very good example of howInternationalismcanoperateinPhysics

Electron microscope

Telescopes are used to look at very distant objects that are very large but because of their distance fromus appearverysmallMicroscopesontheotherhandareusedtolookat objects that are close to us but are physically very small Aswehave seen justmagnifyingobjects that ismakingthemappearlargerisnotsufficientonitsowntogaindetailabouttheobjectfordetailhighresolutionisneeded

Figure928isaschematicofhowanoptical microscope is usedtoviewanobjectandFigure929isaschematicofatransmission electron microscope(TEM)

Optical lens system

bright light source

glass slide containing specimen (object)

eye

Figure 928 The principle of a light microscope

magnetic lens system

electron source

thin wafer of material

ScreenCCD

Figure 929 The principle of an electron microscope

In the optical microscope the resolving power isdetermined by the particular lens system employed and the wavelength λ of the light used For example twopoints in the sample separated by a distance d will just be resolved if

2d mλ=

where m is a property of the lens system know as the numerical aperture In practice the largest value of m obtainable is about 16 Hence if the microscope slideis illuminatedwith light ofwavelength 480 nm a goodmicroscope will resolve two points separated by a distance dasymp15times10-7masymp015micromPointscloser together thanthiswillnotberesolvedHoweverthisisgoodenoughtodistinguishsomevirusessuchastheEbolavirus

Clearlythesmallerλ the higher the resolving power and this is where the electron microscope comes to the fore The electron microscope makes use of the wave nature of electrons(see1315)IntheTEMelectronspassthrougha wafer thin sample and are then focused by a magnetic field onto a fluorescent screen orCCD (charge coupleddevicesee142)Electronsusedinanelectronmicroscopehavewavelengthstypicallyofabout5times10-12mHoweverthe numerical aperture of electron microscopes is considerablysmaller thanthatofanopticalmicroscopetypicallyabout002NonethelessthismeansthataTEMcanresolvetwopointsthatareabout025nmapartThisresolving power is certainly high enough to make out the shape of large molecules

Another type of electron microscope uses a techniqueby which electrons are scattered from the surface of the sample The scattered electrons and then focused as in theTEM to form an image of the surfaceThese so-calledscanningelectronmicroscopes(SEM)havealowerresolving power than TEMrsquos but give very good threedimensional images

The eye

We saw in the last section that the resolving power of the human eye is about 2 times 10-4 rad Suppose that youare looking at car headlights on a dark night and the car

PHYSICS 2015indb 302 150514 337 PM

Wave Phenomena

303

AH

L

is a distance D away If the separation of the headlight is say 15 m then the headlights will subtend an angle θ 15

Dθ =

at your eye Assuming an average wavelength of

500 nm your eye will resolve the headlights into twoseparatesourcesifthisangleequals2times10-4 rad This gives D=75kmInotherwordsifthecarisapproachingyouona straight road then you will be able to distinguish the two headlightsasseparatesourceswhenthecaris75kmawayfrom you Actually because of the structure of the retina and optical defects the resolving power of the average eye is about 6 times 10-4 rad This means that the car is more likely tobe25kmawaybeforeyouresolvetheheadlights

Astronomical telescope

Letusreturntotheexampleofthebinarystarsdiscussedatthe beginning of this section on resolution The stars Kruger A and B form a binary system The average separation of the stars is 14 times 1012mandtheiraveragedistancefromEarthis12times1017mWhenviewedthroughatelescopeonEarththe system will therefore subtend an angle

12

17

14 1012 10

θtimes

=times

= 12times10-5 rad at the objective lens of the telescope Assuming that the average wavelength of the light emitted bythestarsis500nmthenifthetelescopeistoresolvethesystem into two separate images it must have a minimum diameter Dwhere12times10-5 =

7122 500 10D

minustimes times

This gives D=0050m which is about 5 cm So thisparticular system is easily resolved with a small astronomical telescope

Exercise 961 ItissuggestedthatusingtheISASVLBAitwould

bepossibletoldquoseerdquoagrainofriceatadistanceof5000kmEstimatetheresolvingpoweroftheVLBA

2 Thedistancefromtheeyelenstotheretinais20mm The light receptors in the central part of the retinaareabout5times10-6apartDeterminewhetherthe spacing of the receptors will allow for the eye to resolve the headlights in the above discussion whentheyare25kmfromtheeye

3 ThediameterofPlutois23times106 m and its averagedistancefromEarthis60times1012 m EstimatetheminimumdiameteroftheobjectiveofatelescopethatwillenablePlutotobeseenasadisc as opposed to a point source

95 Doppler effect

Essential ideaTheDopplereffectdescribesthephenomenonofwavelengthfrequencyshiftwhenrelativemotionoccurs

Understandingsbull TheDopplereffectforsoundwavesandlightwaves

The Doppler effect for sound waves and light waves

Consider two observers A and B at rest with respect to asoundsourcethatemitsasoundofconstantfrequencyf Clearly both observers will hear a sound of the same frequencyHoweversupposethatthesourcenowmovesat constant speed towards A A will now hear a sound of frequencyfA that is greater than f and B will hear a sound of frequency fB that is less than f This phenomenon is known as the Doppler Effect orDopplerPrinciple after C J Doppler(1803-1853)

The same effect arises for an observer who is either moving towards or away from a stationary source

Figure 930 shows the waves spreading out from a stationary source that emits a sound of constant frequency f The observersAandBhearasoundofthesamefrequency

A B

wavefront

S

Figure 930 Sound waves from a stationary source

Suppose now that the source moves towards A withconstant speed vFigure9310(a)showsasnapshotofthenew wave pattern

NATURE OF SCIENCETechnology Although originally based on physical observations of the pitch of fast moving sources of sound the Doppler effect has an important role in many different areas such as evidence for the expansion of the universe and generating images used in weather reports and in medicine (55)

copy IBO 2014

PHYSICS 2015indb 303 150514 337 PM

Chapter 9

304

AH

L

sourceA B

smallerwavelength

largerwavelength

Figure 931 (a) Sound waves from a moving source

The wavefronts are now crowded together in the direction of travel of the source and stretched out in the opposite direction This is why now the two observers will now hear notesofdifferentfrequenciesHowmuchthewavesbunchtogether and how much they stretch out will depend on the speed vEssentially A

A

cfλ

= and BB

cfλ

= where λA lt λB and v is the speed of sound

IfthesourceisstationaryandAismovingtowardsitthenthe waves from the source incident on A will be bunched up If A is moving away from the stationary source then the waves from the source incident on A will be stretched out

Christian Doppler(1803ndash1853)actuallyappliedtheprinciple(incorrectlyasithappens)totryandexplainthecolourofstars However the Doppler effect does apply to light aswellastosoundIfalightsourceemitsalightoffrequencyf then if it is moving away from an observer the observer willmeasurethelightemittedashavingalowerfrequencythan fSincethesensationofcolourvisionisrelatedtothefrequencyoflight(bluelightisofahigherfrequencythanred light) light emitted by objects moving way from anobserver is often referred to as being red-shifted whereas if the object is moving toward the observer it is referred to as blue-shiftedThisideaisusedinOptionE(Chapter16)

We do not need to consider here the situations where either the source or the observer is accelerating In a situation for example where an accelerating source is approaching astationaryobserverthentheobserverwillhearasoundof ever increasing frequency This sometimes leads toconfusion in describing what is heard when a source approaches passes and then recedes from a stationaryobserver Suppose for example that you are standingon a station platform and a train sounding its whistle is approaching at constant speed What will you hear as the train approaches and then passes through the station As the train approaches you will hear a sound of constant pitch but increasing loudness The pitch of the sound will be greater than if the train were stationary As the train passes through the station you will hear the pitch change at the momentthetrainpassesyoutoasoundagainofconstantpitch The pitch of this sound will be lower than the sound of the approaching train and its intensity will decrease as the train recedes from you What you do not hear is a sound of increasing pitch and then decreasing pitch

The Doppler equations for sound

Although you will not be expected in an IB examination to derivetheequationsassociatedwithaspectsoftheDopplereffectyouwillbeexpectedtoapplythemForcompletenesstherefore thederivationoftheequationsassociatedwiththeDopplereffectasoutlinedaboveisgivenhere

S primeS O

vs

Figure 931 (b)

InFigure931(b)theobserverOisatrestwithrespecttoasourceofsoundSismovingwithconstanspeedvs directly towards O The source is emitting a note of constant frequencyf and the speed of the emitted sound is v

S shows the position of the source ∆t later When the sourceisatresttheninatime∆t the observer will receive f∆t waves and these waves will occupy a distance v∆t ie

v f

λ =f∆tv∆t =

(Because of the motion of the source this number of waves will now occupy a distance (v∆t ndash vs∆t)The lsquonewrsquowavelength is therefore

v ndash vs

fλrsquo =

f∆tv∆t ndash vs∆t

=

If fisthefrequencyheardbyOthen

vfλ

= or

vf

λ = = sv vfminus

From which

s

vf fv v

=minus

Dividingthroughbyv gives

Equation1111

1 sf f v

v=

minus

If the source is moving away from the observer then we have

s

vf fv v

=minus

1

1 sf vv

=+

Equation112

PHYSICS 2015indb 304 150514 337 PM

Wave Phenomena

305

AH

L

We now consider the case where the source is stationary and the observerismoving towardsthesource with speed v0 In this situation the speed of the sound waves as measured by the observer will be v0 + v We therefore have that

v0 + v = ffraslλ = vfλ

times

From which

o1 vf fv

= +

Equation113

If the observerismovingawayfromthesource then

o1 vf fv

= minus

Fromequation(113)wehavethat

o o1 v vf f f f f fv v

∆ = minus = + minus =

Equation114

Thevelocitiesthatwerefertointheaboveequationsarethevelocities with respect to the medium in which the waves fromthesourcetravelHoweverwhenwearedealingwithalight source it is the relative velocity between the source and the observer that we must consider The reason for this is that lightisuniqueintherespectthatthespeedofthelightwavesdoes not depend on the speed of the source All observers irrespective of their speed or the speed of the source will measure the same velocity for the speed of light This is oneof thecornerstonesof theSpecialTheoryofRelativitywhichisdiscussedinmoredetailinOptionH(Chapter18)When applying the Doppler effect to light we aremainlyconcerned with the motion of the source We look here only at the situation where the speed of the source v is much smaller than the speed of light cinfreespace(v ltlt c)Underthesecircumstanceswhenthesourceismovingtowardstheobserverequation111becomes

vf f f fc

minus = ∆ = Equation115

andwhenthesourceismovingawayfromtheobserverequation112becomes vf f f f

cminus = ∆ = minus

Provided that v ltlt c these same equations apply for astationary source and moving observer

We look at the following example and exercise

Example

Asourceemitsasoundoffrequency440HzItmovesinastraight line towards a stationary observer with a speed of 30ms-1Theobserverhearsasoundoffrequency484HzCalculate the speed of sound in air

Solution

We use equation 111 and substitute f= 484 Hz f = 440 Hz

and vs = 30 m s-1

therefore 1

484 440 301v

=minus

such that 30 4401

484vminus = to

give v = 330 m s-1

Example

A particular radio signal from a galaxy is measured as havingafrequencyof139times109HzThesamesignalfromasourceinalaboratoryhasafrequencyof142times109Hz

Suggestwhy thegalaxy ismovingaway fromEarthandcalculateitsrecessionspeed(iethespeedwithwhichitismovingawayfromEarth)

Solution

The fact that the frequency from the moving source is less than that when it is stationary indicates that it is moving away from the stationary observer ie Earth

Using vf fc

∆ = we have

8 96

9

3 10 (142 139) 10 634 10142 10

c fvf∆ times times minus times

= = = timestimes

m s-1

It is usual when dealing with the Doppler effect of light to express speeds as a fraction of c So in this instance we have v = 0021 c

Using the Doppler effectWe have seen in the above example and exercise how the Dopplereffectmaybeusedtomeasuretherecessionspeedof distant galaxies The effect is also used to measure speed inothersituationsHerewewilllookatthegeneralprincipleinvolved in using the Doppler effect to measure speedFigure932showsasource(thetransmitter)thatemitseithersound or em waves of constant frequency f The waves

PHYSICS 2015indb 305 150514 337 PM

Chapter 9

306

AH

L

fromthesourceare incidentonareflectorthat ismovingtowards the transmitter with speed vThereflectedwavesare detected by the receiver placed alongside the transmitter

reector

v

transmitter

receiver

f f

f f

Figure 932 Using the Doppler effect to measure speed

We shall consider the situation where v ltlt c where c is the speed of the waves from the transmitter

For the reflector receiving waves from the transmitterit is effectively an observer moving towards a stationary sourceFromequation(114) itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation116

For the receiver receivingwaves from the reflector it iseffectively a stationary observer receiving waves from a movingsourceFromequation(115)itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation117

Ifweaddequations(116)and(117)wegetthatthetotalDopplershiftatthereceiver∆f is

v vf f f f fc c

minus = ∆ = +

But 1 vf fc

= +

hence

1 v v vf f fc c c

∆ = + +

But since v ltlt cwe can ignore the term2

2

vc when we

expandthebracketintheaboveequation

Therefore we have2vf fc

∆ = Equation118

If v asymp c then we must use the full Doppler equationsHowever for em radiationwewill alwaysonly considersituations in which v ltlt c

Example

Thespeedofsoundinbloodis1500times103 m s-1 Ultrasound offrequency100MHzisreflectedfrombloodflowinginan arteryThe frequencyof the reflectedwaves receivedbackatthetransmitteris105MHzEstimatethespeedofthebloodflowintheartery

Solution

Using equation (118) we have

6 63

2005 10 1015 10

vtimes = times

timesto give v asymp 36 m s-1 (We have assumed that the ultrasound is incident at right angles to the blood flow)

Exercise 971 Judy is standing on the platform of a station A

high speed train is approaching the station in a straight line at constant speed and is sounding its whistleAsthetrainpassesbyJudythefrequencyof the sound emitted by the whistle as heard by Judychangesfrom640Hzto430HzDetermine

(a) thespeedofthetrain

(b) thefrequencyofthesoundemittedbythewhistle as heard by a person on the train (Speedofsound=330ms-1)

2 AgalaxyismovingawayfromEarthwithaspeedof00500cThewavelengthofaparticularspectral line in light emitted by atomic hydrogen inalaboratoryis656times10-7 m Calculate the valueofthewavelengthofthislinemeasuredinthelaboratoryinlightemittedfromasourceofatomic hydrogen in the galaxy

PHYSICS 2015indb 306 150514 337 PM

10 Fields

Contents

101 ndash Describing fields

102 ndash Fields at work

Essential Ideas

Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Similar approaches can be taken in analysing electrical and gravitational potential problems copy IBO 2014

PHYSICS 2015indb 307 150514 337 PM

Chapter 10

308

AH

L

101 Describing fields

Essential idea Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Understandingsbull Gravitationalfieldsbull Electrostaticfieldsbull Electricpotentialandgravitationalpotentialbull Fieldlinesbull Equipotentialsurfaces

Gravitational potentialWehaveseenthatifweliftanobjectofmassm to a height h abovethesurfaceoftheEarththenitsgainingravitationalpotential energy is mghHoweverthisisbynomeansthefullstoryForastartwenowknowthatgvarieswithh and also the expression really gives a difference in potentialenergybetweenthevaluethattheobjecthasattheEarthrsquossurfaceandthevaluethatithasatheighthSowhatwereallyneedisazeropointCanwefindapointwherethepotentialenergyiszeroandusethispointfromwhichtomeasure changes in potential energy

The point that is chosen is in fact infinity At infinity the gravitationalfieldstrengthofanyobjectwillbezeroSoletusseeifwecandeduceanexpressionforthegaininpotentialenergyofanobjectwhenitisldquoliftedrdquofromthesurfaceoftheEarthtoinfinityThisineffectmeansfindingtheworknecessarytoperformthistask

δr

r r δr+

r infin=

mg

AB

Me

Figure 1001 Gravitational forces

Inthediagramweconsidertheworknecessarytomovethe particle of mass m a distance δr in thegravitationalfield of the Earth

The force on the particle at A is FG Mem

r2----------------=

IftheparticleismovedtoB then since δr isverysmallwe can assume that the field remains constant over thedistance ABTheworkδW done againstthegravitationalfieldoftheEarthinmovingthedistanceABis

δWGMem

r2----------------δrndash=

(rememberthatworkdoneagainstaforceisnegative)

TofindthetotalworkdoneW in going from the surface oftheEarthtoinfinitywehavetoaddalltheselittlebitsofworkThisisdonemathematicallybyusingintegralcalculus

WG Me m

r2----------------ndash

rdR

infin

int G Me m 1

r2----- rd

R

infin

intndash G Me m 1r---ndash

R

infinndash= = =

GMe m 0 1R---ndash

ndashndash=

G MemR----------------ndash=

HencewehavewhereRistheradiusoftheEarththattheworkdonebythegravitationalfieldinmovinganobjectofmass m from R(surfaceoftheEarth)toinfinityisgivenby

WGMe m

R---------------- ndash=

We can generalise the result by calculating the worknecessary per unit mass to take a small mass fromthe surface of the Earth to infinity This we call thegravitational potential Vie

Wewouldgetexactlythesameresultifwecalculatedtheworkdonebythefieldtobringthepointmassfrominfinityto the surface of Earth In this respect the formal definition ofgravitationalpotentialatapointinagravitationalfieldis therefore defined as the work done per unit mass in bringing a point mass from infinity to that point

Clearly then the gravitational potential at any point inthe Earthrsquos field distance r from the centre of the Earth (providingr gt R)is

VGMe

r------------ ndash=

Thepotentialisthereforeameasureoftheamountofworkthathastobedonetomoveparticlesbetweenpointsinagravitationalfieldanditsunit is theJkgndash1 We also note thatthepotentialisnegativesothatthepotentialenergy

NATURE OF SCIENCEParadigm shift The move from direct observable actions being responsible for influence on an object to acceptance of a fieldrsquos ldquoaction at a distancerdquo required a paradigm shift in the world of science (23)

copy IBO 2014

PHYSICS 2015indb 308 150514 337 PM

Fields

309

AH

L

aswemoveawayfromtheEarthrsquossurfaceincreasesuntilitreachesthevalueofzeroatinfinity

IfthegravitationalfieldisduetoapointmassofmassmthenwehavethesameexpressionasaboveexceptthatMe is replaced by m andmustalsoexclude thevalueof thepotential at the point mass itself ie at r = 0

WecanexpressthegravitationalpotentialduetotheEarth(orduetoanysphericalmass)intermsofthegravitationalfield strength at its surface

AtthesurfaceoftheEarthwehave

g 0ndash Re GMRe

---------ndash= e

Sothat

g 0Re2 GM= e

Hence at a distance r from the centre of the Earth the gravitationalpotentialVcanbewrittenas

VGMe

r------------ndashg 0 Re

2

r------------ndash= =

The potential at the surface of the Earth

(r = Re)istherefore-g0Re

It is interesting to see how the expression for thegravitational potential ties in with the expression mgh ThepotentialatthesurfaceoftheEarthis-g0Re (see the exampleabove)andataheighthwillbe-g0g 0 Re h+( )ndash ifweassume that g0doesnotchangeover thedistanceh The differenceinpotentialbetweenthesurfaceandtheheighth is therefore g0hSotheworkneededtoraiseanobjectof mass m to a height h is mgh ie m timesdifference ingravitationalpotential

This we have referred to as the gain in gravitationalpotentialenergy(see235)

However this expression can be extended to any twopoints inanygravitationalfieldsuch that ifanobjectofmass mmovesbetween twopointswhosepotentialsareV1 and V2 respectively then the change in gravitationalpotentialenergyoftheobjectism(V1 ndash V2)

Gravitational potential gradient

Let us consider now a region in space where thegravitationalfieldisconstantInFigure912thetwopointsAandBareseparatedbythedistance∆x

A B

direction of uniform gravitational eld of strength I

∆x

Figure 1002 The gravitational potential gradient

The gravitational field is of strength I and is in the direction shownThe gravitational potential at A is V andatBisV +∆V

Theworkdone is takingapointmassmfromAtoB isF∆x = mI∆x

Howeverbydefinitionthisworkisalsoequalto-m∆V

Therefore mI∆x=-m∆V

or VIx

∆= minus

∆Effectivelythissaysthatthemagnitudeofthegravitationalfield strength is equal to the negative gradient of thepotential If I is constant then V is a linear function of x and Iisequaltothenegativegradientofthestraightlinegraph formed by plotting V against x If I is not constant (asusuallythecase)thenthemagnitudeofI at any point in the field can be found by find the gradient of the V-x graph at that point An example of such a calculation can befoundinSection929

For those of you who do HL maths the relationship between field and potential is seen to follow from the expression for the potential of a point mass viz

mV Gr

= minus

2

ddV mG Ir r

minus = + =

Potential due to one or more point masses

Gravitationalpotentialisascalarquantitysocalculatingthepotential due to more than one point mass is a matter of simpleadditionSoforexamplethepotentialV due to the Moon and Earth and a distance x fromthecentreofEarthonastraightlinebetweenthemisgivenbytheexpression

E MM MV Gx r x

= minus + minus

whereME=massofEarthMM= mass of Moon and r = distancebetweencentreofEarthandMoon

PHYSICS 2015indb 309 150514 337 PM

Chapter 10

310

AH

L

Equipotentials and field linesIfthegravitationalpotentialhasthesamevalueatallpointson a surface the surface is said to be an equipotential surfaceSoforexampleifweimagineasphericalshellaboutEarthwhosecentrecoincideswiththecentreofEarththisshellwillbeanequipotentialsurfaceClearlyifwerepresentthegravitationalfieldstrengthbyfieldlinessincethelinesldquoradiaterdquooutfromthecentreofEarththentheselineswillbeatrightanglestothesurfaceIfthefieldlineswerenotnormaltotheequipotentialsurfacethentherewouldbeacomponentof the field parallel to the surfaceThis wouldmean thatpointsonthesurfacewouldbeatdifferentpotentialsandsoitwouldnolongerbeanequipotentialsurfaceThisofcourseholdstrueforanyequipotentialsurface

Figure 1003 shows the field lines and equipotentials fortwopointmassesm

m m

Figure 1003 Equipotentials for two point masses

It is worth noting that we would get exactly the samepattern ifwewere to replace thepointmasseswith twoequalpointcharges(See935)

Escape speedThe potential at the surface of Earth is which meansthattheenergyrequiredtotakeaparticleofmass mfromthesurfacetoinfinityisequalto

ButwhatdoesitactuallymeantotakesomethingtoinfinityWhentheparticleisonthesurfaceoftheEarthwecanthinkofitassittingatthebottomofaldquopotentialwellrdquoasinfigure1004

particle

surface of Earth

innity

GMR---------

FIgure 1004 A potential well

Theldquodepthrdquoofthewellis and if the particle gains an amount of kinetic energy equal to wherem is its massthenitwillhavejustenoughenergytoldquoliftrdquoitoutofthewell

In reality it doesnrsquot actually go to infinity it justmeansthat the particle is effectively free of the gravitationalattractionof theEarthWe say that ithas ldquoescapedrdquo theEarthrsquosgravitationalpullWemeetthisideainconnectionwithmolecular forcesTwomolecules in a solidwill sitat their equilibrium position the separation where therepulsiveforceisequaltotheattractiveforceIfwesupplyjust enough energy to increase the separation of themolecules such that they are an infinite distance apart then intermolecularforcesnolongeraffectthemoleculesandthesolidwillhavebecomealiquidThereisnoincreaseinthekineticenergyofthemoleculesandsothesolidmeltsat constant temperature

Wecancalculatetheescapespeedofanobjectveryeasilyby equating the kinetic energy to the potential energysuch that

12---mvesc ape

2 GMe mRe

----------------=

vesc aperArr2GMe

Re--------------- 2g0 Re = =

Substituting for g0 and Regivesavalueforvescape of about 11kmsndash1 from the surface of the Earth

Youwillnotethattheescapespeeddoesnotdependonthemassoftheobjectsincebothkineticenergyandpotentialenergy are proportional to the mass

IntheoryifyouwanttogetarockettothemoonitcanbedonewithoutreachingtheescapespeedHoweverthiswouldnecessitate anenormousamountof fuel and it islikely that therocketplus fuelwouldbesoheavythat itwouldnevergetoffthegroundItismuchmorepracticalto accelerate the rocket to the escape speed fromEarthorbitandthenintheoryjustlaunchittotheMoon

Example

Use the followingdata todetermine thepotential at thesurface of Mars and the magnitude of the acceleration of free fall

massofMars =64times1023kg

radiusofMars =34times106 m

Determinealsothegravitationalfieldstrengthatadistanceof 68 times 106mabovethesurfaceofMars

PHYSICS 2015indb 310 150514 337 PM

Fields

311

AH

L

Solution

= minus = minus times times = times23

11 76

64 1067 10 ndash13 10

34 10MV GR

minus timestimes

Nkg-1

But V = -g0R

Therefore 7

0 6

13 10 3834 10

VgR

times= minus = =

timesm s-2

To determine the field strength gh at 68 times 106 m above the surface we use the fact that 0 2

Mg GR

= such that GM = g0R2

Therefore 2 2

0h 2 2 2

h h

38 (34) 042(102)

g RGMgR R

times= = = = m s-2

(the distance from the centre is 34 times 106 + 68 times 106 = 102 times 106 m)

Exercise 1011 Thegraphbelowshowshowthegravitational

potentialoutsideoftheEarthvarieswithdistancefrom the centre

0 10 20 30 40 50 60 70

ndash1

ndash2

ndash3

ndash4

ndash5

ndash6

ndash7

r m times 106

V

Jkgndash1

times 10

7

(a) Usethegraphtodeterminethegainingravitationalpotentialenergyofasatelliteofmass200kgasitmovesfromthesurfaceof the Earth to a height of 30 times 107mabovetheEarthrsquossurface

(b) Calculatetheenergyrequiredtotakeit to infinity

(c) DeterminetheslopeofthegraphatthesurfaceoftheEarthmCommenton youranswer

Electric potential energy

The concept of electric potential energy was developedwith gravitational potential energy in mind Just as anobjectnearthesurfaceoftheEarthhaspotentialenergybecause of its gravitational interaction with the Earthsotoothereiselectricalpotentialenergyassociatedwithinteracting charges

Letusfirstlookatacaseoftwopositivepointchargeseachof 1μC that are initially bound together by a thread in a vacuuminspacewithadistancebetweenthemof10cmasshowninFigure1005Whenthethreadiscutthepointchargesinitiallyatrestwouldmoveinoppositedirectionsmovingwithvelocitiesv1 and v2 along the direction of the electrostatic force of repulsion

BEFORE AFTER

v v21

Figure 1005 Interaction of two positive particles

The electric potential energybetweentwopointchargescan be found by simply adding up the energy associated witheachpairofpointchargesForapairofinteractingchargestheelectricpotentialenergyisgivenby

∆U = ∆Ep + ∆Ek = ∆W = ∆Fr = kqQ

____ r2 times r = kqQ

____ r

Becausenoexternalforceisactingonthesystemtheenergyand momentum must be conserved Initially Ek = 0 and Ep =kqQr=9times109 times 1 times 10-1201m=009JWhentheyareagreatdistance fromeachotherEpwillbenegligibleThefinalenergywillbeequaltofrac12mv1

2+frac12mv22=009J

Momentum isalsoconservedand thevelocitieswouldbethe same magnitude but in opposite directions

Electric potential energyismoreoftendefinedintermsofapointchargemovinginanelectricfieldas

lsquothe electricpotential energybetweenany twopoints inan electric field is defined as negativeoftheworkdone by anelectricfieldinmovingapoint electric chargebetweentwolocationsintheelectricfieldrsquo

ΔU = ΔEp = -ΔW = -Fd = qEx

ΔU = ΔEp +ΔEk = ΔW = ΔFr = kqQ r2 x r = kqQ

wherexisthedistancemovedalong(oroppositeto)thedirection of the electric field

Electric potential energyismeasuredinjoule(J)Justasworkisascalarquantitysotooelectricalpotentialenergyis a scalar quantity The negative of the work done by an electric field in moving a unit electric charge between two points is independent of the path taken Inphysicswesaytheelectricfieldisalsquoconservativersquo field

PHYSICS 2015indb 311 150514 337 PM

Chapter 10

312

AH

L

Supposeanexternalforcesuchasyourhandmovesasmallpositive point test charge in the direction of a uniformelectric fieldAs it ismoving itmust be gaining kineticenergyIfthisoccursthentheelectricpotentialenergyofthe unit charge is changing

InFigure1006apointcharge+qismovedbetweenpointsAandBthroughadistancex in a uniform electric field

B

A

x

+ q

Figure 1006 Movement of a positive point charge in a uniform field

InordertomoveapositivepointchargefrompointAtopointBanexternal forcemustbeapplied to thechargeequaltoqE (F = qE)

Since the force is applied through a distance xthennegativeworkhas to bedone tomove the charge because energyis gainedmeaning there is an increaseelectric potential energybetweenthetwopointsRememberthattheworkdoneisequivalenttotheenergygainedorlostinmovingthe charge through the electric field The concept of electric potential energy is only meaningful as the electric field whichgeneratestheforceinquestionisconservative

W F xtimes E q x times= =

θ x cos θ x

Figure 1007 Charge moved at an angle to the field

If a chargemoves at an angle θ to an electric field thecomponent of the displacement parallel to the electric fieldisusedasshowninFigure918

W F x Eq x θcostimes= =

Theelectricpotentialenergyisstoredintheelectricfieldand theelectricfieldwill return theenergy to thepointcharge when required so as not to violate the Law ofconservationofenergy

Electric potential

The electric potential at a point in an electric field is defined as being the work done per unit charge in bringing a small positive point charge from infinity to that point

ΔV = Vinfin ndash Vf = ndash W ___ q

Ifwedesignatethepotentialenergytobezeroatinfinitythen it follows thatelectricpotentialmustalsobezeroat infinity and the electric potential at any point in an electricfieldwillbe

ΔV = ndash W ___ q

NowsupposeweapplyanexternalforcetoasmallpositivetestchargeasitismovedtowardsanisolatedpositivechargeTheexternalforcemustdoworkonthepositivetestchargetomoveittowardstheisolatedpositivechargeandtheworkmustbepositivewhiletheworkdonebytheelectricfieldmustthereforebenegativeSotheelectricpotentialatthatpointmustbepositiveaccordingtotheaboveequationIfanegativeisolatedchargeisusedtheelectricpotentialatapointonthepositivetestchargewouldbenegativePositivepoint chargesof their ownaccordmove fromaplaceofhighelectricpotential toaplaceof lowelectricpotentialNegative point charges move the other way from lowpotentialtohighpotentialInmovingfrompointAtopointBinthediagramthepositivecharge+qismovingfromalowelectricpotentialtoahighelectricpotential

In thedefinitiongiven the termldquoworkperunitchargerdquohas significance If the test charge is +16 times 10-19Cwherethe charge has a potential energy of 32 times 10-17Jthenthepotentialwouldbe32times10-17J+16times10-19C=200JC-1 Now if the charge was doubled the potential wouldbecome 64times10-17 J However the potential per unitchargewouldbethesame

Electric potential is a scalar quantityandithasunitsJC-1 orvoltswhere1voltequalsone joulepercoloumbThevoltallowsustoadoptaunitfortheelectricfieldintermsofthevolt

PreviouslytheunitfortheelectricfieldwasNCndash1

W = qV and F = qEsoW ___ V = F __ E

E = FV ___ W = FV ___ Fm V mndash1

That is the units of the electric field E can also beexpressed as V mndash1

The work done per unit charge in moving a point charge between two points in an electric field is again independent of the path taken

PHYSICS 2015indb 312 150514 337 PM

Fields

313

AH

L

Electric potential due to a point charge

Letus takeapoint rmetres froma chargedobjectThepotentialatthispointcanbecalculatedusingthefollowing

W = ndashFr = ndashqV and F = ndash q1 q2 _____ 4πε0r

2

Therefore

W = ndash q1 q2 _____ 4πε0r

2 times r = ndash q1 q2 _____ 4πε0r

= ndashq1 times q2 _____ 4πε0r

= ndashq1V

That is V q

4πε0r--------------- =

Orsimply

V kqr------

=

Example

Determinehowmuchworkisdonebytheelectricfieldofpointcharge150μCwhenachargeof200μCismovedfrom infinity to apoint 0400m from thepoint charge(Assumenoaccelerationofthecharges)

Solution

The work done by the electric field is W = -qV = -14πε0 times q times (Q rinfin - Q r0400)

W = (- 200 times 10-6 C times 900 times 109 NmC-2 times 150 times 10-6 C) divide 0400 m = - 0675 J

An external force would have to do +0675 J of work

Electric field strength and electric potential gradient

Letus lookbackatFigure1006Supposeagain that thecharge+qismovedasmalldistancebyaforceFfromAtoBsothattheforcecanbeconsideredconstantTheworkdoneisgivenby

∆W F x∆times=

TheforceFandtheelectricfieldEareoppositelydirectedandweknowthat

F = -qE and ΔW = q ΔV

Thereforetheworkdonecanbegivenas

q ΔV = -q E Δx

Therefore E V∆x∆-------

ndash=

The rate of change of potential ΔV at a point with respect to distance Δx in the direction in which the change is maximum is called the potential gradient We say that the electric field = - the potential gradient and the units are Vm-

1 From the equation we can see that in a graph of electric potential versus distance the gradient of the straight line equals the electric field strength

In reality if a charged particle enters a uniform electric field it will be accelerated uniformly by the field and its kinetic energy will increase This is why we had to assume no acceleration in the last worked example

Ek∆ 12---

mv2 q E xsdot sdot q Vx---

xsdot sdot q Vsdot= = = =

Example

Determine how far apart two parallel plates must besituated so that a potential difference of 150 times 102 V produces an electric field strength of 100 times 103 NC-1

Solution

Using E V∆x------- xhArrndash V∆

E-------15 102 Vtimes

100 103 N C 1ndashtimes------------------------------------------= = =

= 150 times 10-1

The plates are 150 times 10-1 m apart

The electric field and the electric potential at a point due to an evenly distributed charge +q on a sphere can be represented graphically as in Figure 1008

r

r

V

E

r0

r0

r0

Charge of +Q evenly distributed over surface

E 14πε0------------ Q

r2----- r r0gt=

V 14πε0------------Q

r----r r0gt=

Figure 1008 Electric field and potential due to a charged sphere

PHYSICS 2015indb 313 150514 337 PM

Chapter 10

314

AH

L

When the sphere becomes charged we know that thechargedistributes itselfevenlyoverthesurfaceThereforeeverypartofthematerialoftheconductorisatthesamepotential As the electric potential at a point is defined as beingnumericallyequaltotheworkdoneinbringingaunitpositivechargefrominfinitytothatpointithasaconstantvalueineverypartofthematerialoftheconductor

Since the potential is the same at all points on the conducting surfacethen∆V∆xiszeroButE=ndash∆V∆xThereforethe electric field inside the conductor is zero There is no electric field inside the conductor

SomefurtherobservationsofthegraphsinFigure1009are

bull OutsidethespherethegraphsobeytherelationshipsgivenasE α 1 r2 and V α 1 r

bull Atthesurfacer = r0Thereforetheelectricfieldandpotentialhavetheminimumvalueforr at this point and this infers a maximum field and potential

bull Insidethespheretheelectricfieldiszerobull Insidethespherenoworkisdonetomoveacharge

fromapointinsidetothesurfaceThereforethereisno potential difference and the potential is the same asitiswhenr = r0

Similargraphscanbedrawnfortheelectricfieldintensityandthe electric potential as a function of distance from conducting parallelplatesandsurfacesandthesearegiveninFigure1009

Potential plot E eld plotE eld

++++

ndashndashndashndash

xx x

+

x x

ndashx x

+ ndashx

x

+ +

x

x

Figure 1009 Electric field and electric potential at a distance from a charged surface

Potential due to one or more point charges

The potential due to one point charge can be determined byusingtheequationformula

V = kq r

Example 1

Determine the electric potential at a point 20 times 10-1 m fromthecentreofan isolatedconducting spherewithapointchargeof40pCinair

Solution

Using the formula

V = kq r we have

V90 10 9times( ) 40 10 12ndashtimes( )times

20 10 1ndashtimes( )------------------------------------------------------------------ 018 V= =

the potential at the point is 180 times 10-1 V

The potential due to a number of point charges can be determined by adding up the potentials due to individual point charges because the electric potential at any point outside a conducting sphere will be the same as if all the charge was concentrated at its centre

Example 2

Threepointchargesofareplacedattheverticesofaright-angledtriangleasshowninthediagrambelowDeterminetheabsolutepotentialat the+20μCchargeduetothetwoothercharges

- 6μC

+3μC +2μC

4 m

3 m

Solution

The electric potential of the +2 μC charge due to the ndash 6 μC charge is

V = (9 times 109 Nm2C-2 times -6 times 10-6 C) divide (radic 32 + 42) m = - 108 times 104 V

PHYSICS 2015indb 314 150514 338 PM

Fields

315

AH

L

Insummarywecanconcludethat

bull No work is done to move a charge along an equipotential

bull Equipotentials are always perpendicular to the electric lines of force

Figure1011and1012showsomeequipotential lines fortwo oppositely charged and identically positive spheresseparated by a distance

+ve ndashve

equipotential lines

Figure 1011 Equipotential lines between two opposite charges

equipotential lines

Figure 1012 Equipotential lines between two charges which are the same

+

ndash ndash ndash ndash

+ + +

40 V

30 V

20 V

10 V

50 V

Figure 1013 Equipotential lines between charged parallel plates

Figure1013showstheequipotentiallinesbetweenchargedparallel plates Throughout this chapter the similarities and differences between gravitational fields and electric fieldshavebeendiscussedTherelationshipsthatexistsbetweengravitationalandelectricquantitiesandtheeffectsofpointmassesandchargesissummarisedinTable1014

The electric potential of the +2 μC charge due to the +3 μC charge is

V = (9 times 109 Nm2C-2 times 3 times 10-6 C) divide 3m = 9 times 103 V

The net absolute potential is the sum of the 2 potentials- 108 times 104 V + 9 times 103 V =- 18 times 103 V

The absolute potential at the point is - 18 times 103 V

Equipotential surfacesRegions in space where the electric potential of a chargedistributionhasaconstantvaluearecalledequipotentials The placeswherethepotentialisconstantinthreedimensionsarecalled equipotential surfacesandwheretheyareconstantintwodimensionstheyarecalledequipotential lines

TheyareinsomewaysanalogoustothecontourlinesontopographicmapsInthiscasethegravitationalpotentialenergy is constant as amassmoves around the contourlines because the mass remains at the same elevationabovetheEarthrsquossurfaceThegravitationalfieldstrengthacts in a direction perpendicular to a contour line

Similarlybecausetheelectricpotentialonanequipotentiallinehas thesamevalueanelectric forcecandonoworkwhena testchargemovesonanequipotentialThereforethe electric field cannot have a component along anequipotentialandthusitmustbeeverywhereperpendiculartotheequipotentialsurfaceorequipotentiallineThisfactmakesiteasytoplotequipotentialsifthelinesofforceorlinesofelectricfluxofanelectricfieldareknown

Forexamplethereareaseriesofequipotentiallinesbetweentwoparallelplateconductorsthatareperpendiculartotheelectric fieldTherewill be a series of concentric circles(eachcirclefurtherapartthanthepreviousone)thatmapouttheequipotentialsaroundanisolatedpositivesphereThe lines of force and some equipotential lines for anisolatedpositivesphereareshowninFigure1010

Lines ofequipotential

Figure 1010 Equipotentials around an isolated positive sphere

PHYSICS 2015indb 315 150514 338 PM

Chapter 10

316

AH

L

Gravitational quantity Electrical quantityQ

uant

ities

g V∆x∆-------ndash= E V∆

x∆-------ndash=

Poin

t mas

ses a

nd ch

arge

s

V Gmr----

ndash= V 14πε0------------q

r---=

g G m

r2-----ndash= E 14πε0------------ q

r2-----=

F Gm1 m2

r2--------------= F 1

4πε0------------

q1q2

r2-----------=

Figure 1014 Formulas

Example

Deduce the electric potential on the surface of a gold nucleus that has a radius of 62 fm

Solution

Using the formula

V = kq r and knowing the atomic number of gold is 79 We will assume the nucleus is spherical and it behaves as if it were a point charge at its centre (relative to outside points)

V = 90 times 109 Nm2C-2 times 79 times 16 times 10-19 C divide 62 times 10-15 m = 18 times 107 V

The potential at the point is 18 MV

Example

Deduce the ionisation energy in electron-volts of theelectron in the hydrogen atom if the electron is in its ground state and it is in a circular orbit at a distance of 53 times 10-11 m from the proton

Solution

This problem is an energy coulombic circular motion question based on Bohrrsquos model of the atom (not the accepted quantum mechanics model) The ionisation energy is the energy required to remove the electron from the ground state to infinity The electron travels in a circular orbit and therefore has a centripetal acceleration The ionisation energy will counteract the coulombic force and the movement of the electron will be in the opposite direction to the centripetal force

Total energy = Ek electron + Ep due to the proton-electron interaction

ΣF = kqQ r2 = mv2 r and as such mv2 = = kqQ r

Therefore Ek electron = frac12 kqQ r

Ep due to the proton-electron interaction = - kqQ r

Total energy = frac12 kqQ r + - kqQ r = -frac12 kqQ r

= - 90 times 109 Nm2C-2 times (16 times 10-19 C)2 divide 53 times 10-11 m = -217 times 10-18 J

= -217 times 10-18 J divide 16 times 10-19 = -136 eV

The ionisation energy is 136 eV

Exercise 1021 ApointchargePisplacedmidwaybetweentwo

identicalnegativechargesWhichoneofthefollowingiscorrectwithregardstoelectricfieldandelectricpotentialatpointP

Electric field Electric potentialA non-zero zeroB zero non-zeroC non-zero non-zeroD zero zero

2 Twopositivechargedspheresaretiedtogetherinavacuumsomewhereinspacewheretherearenoexternal forces A has a mass of 25 g and a charge of20μCandBhasamassof15gandachargeof30μCThedistancebetweenthemis40cmTheyarethenreleasedasshowninthediagram

BEFORE AFTER

v1 v2

A B

(a) Determinetheirinitialelectricpotentialenergy in the before situation

(b) DeterminethespeedofsphereBafterrelease

3 Thediagrambelowrepresentstwoequipotentiallines in separated by a distance of 5 cm in a uniform electric field

+ + + + + + + +

ndash ndash ndash ndash ndash ndash ndash ndash

40 V

20 V 5 cm

PHYSICS 2015indb 316 150514 338 PM

Fields

317

AH

L

Determine the strength of the electric field

4 Thisquestionisabouttheelectricfieldduetoacharged sphere and the motion of electrons in that fieldThediagrambelowshowsanisolatedmetalsphereinavacuumthatcarriesanegativeelectriccharge of 60 μC

ndash

(a) Onthediagramdrawtheconventionalwayto represent the electric field pattern due to the charged sphere and lines to represent threeequipotentialsurfacesintheregionoutside the sphere

(b) Explainhowthelinesrepresentingtheequipotentialsurfacesthatyouhavesketchedindicatethatthestrengthoftheelectricfieldisdecreasingwithdistancefrom the centre of the sphere

(c) Theelectricfieldstrengthatthesurfaceofthe sphere and at points outside the sphere can be determined by assuming that the sphere acts as a point charge of magnitude 60 μC at its centre The radius of the sphere is 25 times 10ndash2 m Deduce that the magnitude of the field strength at the surface of the sphere is 86 times 107 Vmndash1

An electron is initially at rest on the surface of the sphere

(d) (i) Describethepathfollowedbytheelectronasitleavesthesurfaceofthesphere

(ii) Calculatetheinitialaccelerationofthe electron

5 Determinetheamountofworkthatisdoneinmovingachargeof100nCthroughapotentialdifferenceof150times102 V

6 Three identical 20 μC conducting spheres are placedatthecornersofanequilateraltriangleofsides 25 cm The triangle has one apex C pointing upthepageand2baseanglesAandBDeterminetheabsolutepotentialatB

7 Determinehowfaraparttwoparallelplatesmustbesituatedsothatapotentialdifferenceof250 times 102 V produces an electric field strength of 200 times 103 NC-1

8 Thegapbetweentwoparallelplatesis10times10-3mandthereisapotentialdifferenceof10times104 V betweentheplatesCalculate

i theworkdonebyanelectroninmovingfrom one plate to the other

ii thespeedwithwhichtheelectronreachesthe second plate if released from rest

iii theelectricfieldintensitybetweentheplates

9 Anelectronguninapicturetubeisacceleratedby a potential 25 times 103VDeterminethekineticenergygainedbytheelectroninelectron-volts

10 Determine the electric potential 20 times 10-2 m from achargeof-10times10-5 C

11 Determinetheelectricpotentialatapointmid-waybetweenachargeofndash20pCandanotherof+5pConthelinejoiningtheircentresifthecharges are 10 cm apart

12 During a thunderstorm the electric potential differencebetweenacloudandthegroundis10 times 109 V Determine the magnitude of the change in electric potential energy of an electron thatmovesbetweenthesepointsinelectron-volts

13 A charge of 15 μC is placed in a uniform electric fieldoftwooppositelychargedparallelplateswithamagnitudeof14times103 NC-1

(a) Determinetheworkthatmustbedoneagainstthefieldtomovethepointchargeadistance of 55 cm

(b) Calculatethepotentialdifferencebetweenthe final and initial positions of the charge

(c) Determinethepotentialdifferencebetweenthe plates if their separation distance is 15 cm

14 Duringaflashoflightningthepotentialdifferencebetweenacloudandthegroundwas12 times 109 V and the amount of transferred charge was32C

(a) Determinethechangeinenergyofthetransferred charge

(b) Iftheenergyreleasedwasallusedtoacceleratea1tonnecardeduceitsfinalspeed

(c) Iftheenergyreleasedcouldbeusedtomelticeat0degCdeducetheamountoficethatcould be melted

PHYSICS 2015indb 317 150514 338 PM

Chapter 10

318

AH

L

15 SupposethatwhenanelectronmovedfromAtoBin the diagram along an electric field line that the electric field does 36 times 10-19Jofworkonit

Determinethedifferencesinelectricpotential

(a) VB ndash VA(b) VC ndash VA(c) VC ndash VB

16 DeterminethepotentialatpointPthatislocatedatthecentreofthesquareasshowninthediagrambelow

- 6μC

+3μC +2μC

1m

1m

5μC

P

102 Fields at work

Essential idea Similar approaches can be taken inanalysingelectricalandgravitationalpotentialproblems

Understandingsbull Potentialandpotentialenergybull Potentialgradientbull Potentialdifferencebull Escapespeedbull Orbitalmotionorbitalspeedandorbitalenergybull Forcesandinverse-squarelawbehaviour

Orbital motion orbital speed and orbital energy

Although orbital motion may be circular elliptical orparabolic this sub-topic only dealswith circular orbitsThis sub-topic is not fundamentally new physics butan application that synthesizes ideas from gravitationcircularmotiondynamicsandenergy

TheMoonorbitstheEarthandinthissenseitisoftenreferredtoasasatelliteoftheEarthBefore1957itwastheonlyEarthsatelliteHoweverin1957theRussianslaunchedthefirstmanmadesatelliteSputnik1SincethisdatemanymoresatelliteshavebeenlaunchedandtherearenowliterallythousandsofthemorbitingtheEarthSomeareusedtomonitortheweathersome used to enable people to find accurately their position on thesurfaceoftheEarthmanyareusedincommunicationsandnodoubtsomeareusedtospyonothercountriesFigure1015showshowinprincipleasatellitecanbeputintoorbit

Theperson(whosesizeisgreatlyexaggeratedwithrespectto Earth) standing on the surface on the Earth throwssome stonesThe greater the speed with which a stoneis thrown the further it will land from her The pathsfollowedbythethrownstonesareparabolasByastretchof the imaginationwecanvisualisea situation inwhichastoneis thrownwithsuchaspeedthatbecauseof thecurvatureoftheEarthitwillnotlandonthesurfaceoftheEarthbutgointoldquoorbitrdquo(Path4onFigure1015)

NATURE OF SCIENCECommunication of scientific explanations The ability to apply field theory to the unobservable (charges) and the massively scaled (motion of satellites) required scientists to develop new ways to investigate analyse and report findings to a general public used to scientific discoveries based on tangible and discernible evidence (51)

copy IBO 2014

PHYSICS 2015indb 318 150514 338 PM

Fields

319

AH

L

1

2

34Earth

Figure 1015 Throwing a stone into orbit

TheforcethatcausesthestonestofollowaparabolicpathandtofalltoEarthisgravityandsimilarlytheforcethatkeeps the stone in orbit is gravity For circular motionto occur we have seen that a force must act at rightanglestothevelocityofanobjectthatistheremustbeacentripetalforceHenceinthesituationwedescribeherethe centripetal force for circular orbital motion about the EarthisprovidedbygravitationalattractionoftheEarth

We can calculate the speedwithwhich a stonemustbethrowninordertoputitintoorbitjustabovethesurfaceof the Earth

If the stone has mass m and speed v thenwehave fromNewtonrsquos2ndlaw

mv2

RE---------- G

MEm

RE2------------=

whereRE is the radius of the Earth and ME is the mass of the Earth

Bearinginmindthatg 0 GME

RE2--------= then

v g RE 10 64 106timestimes 8 103times= = =

Thatisthestonemustbethrownat8times103m sndash1

Clearly we are not going to get a satellite into orbit soclosetothesurfaceoftheEarthMovingatthisspeedthefrictionduetoairresistancewouldmeltthesatellitebeforeithadtravelledacoupleofkilometresInrealitythereforeasatelliteisputintoorbitabouttheEarthbysendingitattachedtoarocketbeyondtheEarthrsquosatmosphereandthengivingitacomponentofvelocityperpendiculartoaradialvectorfromtheEarthSeeFigure1016

Earth Satellite carried by rocket to hereSatellite orbit

Tangential component of velocity

Figure 1016 Getting a satellite into orbit

Keplerrsquos third law(This work of Kepler and Newtonrsquos synthesis of the work is an excellent example of the scientific method and makes for a good TOK discussion)

In 1627 Johannes Kepler (1571-1630)publishedhis lawsof planetary motion The laws are empirical in natureandwere deduced from the observations of theDanishastronomer Tycho de Brahe (1546-1601)The third lawgives a relationship between the radius of orbit R of a planet and its period TofrevolutionabouttheSunThelawisexpressedmathematicallyas

2

3 constantTR

=

We shall nowuseNewtonrsquos LawofGravitation to showhowitisthattheplanetsmoveinaccordancewithKeplerrsquosthirdlaw

InessenceNewtonwasabletousehislawofgravitytopredict the motion of the planets since all he had to do was factor theF givenby this law intohis second law F = ma to find their accelerations and hence their future positions

InFigure1017theEarthisshownorbitingtheSunandthedistancebetweentheircentresisR

RSun

Earth

F es

F se

Figure 1017 Planets move according to Keplerrsquos third law

Fes is the force that the Earth exerts on the Sun and Fse is the force that the Sun exerts on the Earth The forces are equalandoppositeandtheSunandtheEarthwillactuallyorbitaboutacommoncentreHoweversince theSun issoverymuchmoremassivethantheEarththiscommoncentrewillbeclosetothecentreoftheSunandsowecanregard the Earth as orbiting about the centre of the Sun Theotherthingthatweshallassumeisthatwecanignorethe forces that the other planets exert on the Earth (This wouldnotbeawisethingtodoifyouwereplanningtosend a space ship to theMoon for example)We shallalso assume that we have followed Newtonrsquos exampleandindeedprovedthataspherewillactasapointmasssituated at the centre of the sphere

Kepler had postulated that the orbits of the planets areellipticalbut since theeccentricityof theEarthrsquosorbit issmallweshallassumeacircularorbit

PHYSICS 2015indb 319 150514 338 PM

Chapter 10

320

AH

L

TheaccelerationoftheEarthtowardstheSunisa = Rω2

whereω ω 2πT------=

Hencea R 2π

T------ times

2 4π2 RT 2------------- = =

But the acceleration is given by Newtonrsquos Second Law F = mawhereFisnowgivenbytheLawofGravitationSo in this situation

F maGMs Me

R2-------------------= = butwealsohavethat

a a R 2πT

------ times

2 4π2 RT 2------------- = = and m = Me so that

G MsMe

R2------------------- Me4π2R

T2-------------times

GMs

4π2-----------rArr R3

T2------= =

Butthequantity

isaconstantthathasthesamevalueforeachoftheplanetssowehaveforalltheplanetsnotjustEarththat

R3

T2------ k =

wherekisaconstantWhichisofcourseKeplerrsquosthirdlaw

ThisisindeedanamazingbreakthroughItisdifficulttorefutethe idea that all particles attract each other in accordance withtheLawofGravitationwhenthelawisabletoaccountfortheobservedmotionoftheplanetsabouttheSun

The gravitational effects of the planets upon each othershould produce perturbations in their orbits Such is the predictivepoweroftheUniversalGravitationalLawthatit enabled physicists to compute these perturbations The telescopehadbeeninventedin1608andbythemiddleofthe 18th Century had reached a degree of perfection in design that enabled astronomers to actually measure the orbital perturbations of the planets Their measurements werealways inagreementwith thepredictionsmadebyNewtonrsquoslawHoweverin1781anewplanetUranuswasdiscovered and the orbit of this planet did not fit withthe orbit predicted by Universal Gravitation Such wasthe physicistrsquos faith in theNewtonianmethod that theysuspectedthatthediscrepancywasduetothepresenceofayetundetectedplanetUsingtheLawofGravitationtheFrenchastronomerJLeverrier and the English astronomer J C Adamswereable tocalculate justhowmassive thisnewplanetmustbeandalsowhereitshouldbeIn1846the planetNeptunewas discovered justwhere they hadpredicted Ina similarwaydiscrepancies in theorbitofNeptune ledto thepredictionandsubsequentdiscoveryin1930oftheplanetPlutoNewtonrsquosLawofGravitation

had passed the ultimate test of any theory it is not only abletoexplainexistingdatabutalsotomakepredictions

Satellite energyWhenasatelliteisinorbitaboutaplanetitwillhavebothkineticenergyandgravitationalpotentialenergySupposewe consider a satellite ofmass m that is in an orbit of radius r about a planet of mass M

Thegravitationalpotentialduetotheplanetatdistancer from its centre is

ThegravitationalpotentialenergyofthesatelliteVsat

is therefore GMe m

r----------------ndash

ThatisVsat GMem

r----------------ndash=

Thegravitationalfieldstrengthatthesurfaceoftheplanetisgivenby

g 0GMe

Re2------------=

Hencewecanwrite

Vsat g0 Re

2

rndash=

m

ThekineticenergyofthesatelliteKsatisequaltofrac12mv2wherev is its orbital speed

Byequatingthegravitationalforceactingonthesatellitetoitscentripetalaccelerationwehave

G Mem

r2---------------- mv2

r---------- mv2hArr

GMemr----------------= =

Fromwhich12---

mv2 12---

GMemr----------------times=

= 0 e

2g R

r

2m

Which is actually quite interesting since it shows thatirrespective of the orbital radius the KE is numericallyequal to half the PE Also the total energy E tot K sat Vsat+ 1

2---G Me m

r----------------times G Me m

r----------------ndash + 1

2---G Me m

r----------------ndash= = =of the satelliteisalwaysnegativesince

E tot K sat Vsat+ 12---

G Me mr----------------times

G Me mr----------------ndash

+ 12---

G Me mr----------------ndash= = =

PHYSICS 2015indb 320 150514 338 PM

Fields

321

AH

L

The energies of an orbiting satellite as a function of radial distancefromthecentreofaplanetareshownplottedinFigure1018

12

100806

0402

0ndash02

ndash04

ndash06ndash08ndash10

ndash12

2 4 6 8 10 12

ener

gya

rbita

ry u

nits

distance timesR

kinetic energytotal energy

potential energy

Figure 1018 Energy of an orbiting satellite as a function of distance from the centre of a planet

WeightlessnessSuppose that you are inanelevator(lift)thatisdescendingat constant speedandyou let goof abook thatyouareholdinginyourhandThebookwillfalltothefloorwithaccelerationequaltotheaccelerationduetogravityIfthecablethatsupportstheelevatorweretosnap(asituationthatItrustwillneverhappentoanyofyou)andyounowletgothebookthatyouareholdinginyourotherhandthis book will not fall to the floor - it will stay exactlyin linewithyourhandThis isbecause thebook isnowfallingwith the sameaccelerationas theelevatorandassuch the book cannot ldquocatchrdquo up with the floor of theelevatorFurthermoreifyouhappenedtobestandingonasetofbathroomscalesthescaleswouldnowreadzero-youwouldbeapparentlyweightlessItisthisideaoffreefallthatexplainstheapparentweightlessnessofastronautsin an orbiting satellite These astronauts are in free fall inthesensethattheyareacceleratingtowardsthecentre of the Earth

It is actually possible to define theweight of a body inseveral different waysWe can define it for example asthegravitationalforceexertedonthebodybyaspecifiedobject such as the EarthThiswe have seen thatwe doinlotsofsituationswherewedefinetheweightasbeingequaltomgIfweusethisdefinitionthenanobjectinfreefallcannotbydefinitionbeweightlesssinceitisstillinagravitationalfieldHoweverifwedefinetheweightofanobjectintermsofaldquoweighingrdquoprocesssuchasthereadingonasetofbathroomscaleswhichineffectmeasuresthecontact force between the object and the scales thenclearlyobjectsinfreefallareweightlessOnenowhasto

askthequestionwhetherornotitispossibleForexampletomeasurethegravitationalforceactingonanastronautin orbit about the Earth

WecanalsodefineweightintermsofthenetgravitationalforceactingonabodyduetoseveraldifferentobjectsForexample for an object out in space its weight could bedefined in terms of the resultant of the forces exerted on it bytheSuntheMoontheEarthandalltheotherplanetsin the Solar System If this resultant is zero at a particular pointthenthebodyisweightlessatthispoint

Inviewofthevariousdefinitionsofweightthatareavailabletousitisimportantthatwhenweusethewordldquoweightrdquoweareawareofthecontextinwhichitisbeingused

Example

Calculate the height above the surface of the Earth atwhichageo-stationarysatelliteorbits

Solution

A geo-stationary satellite is one that orbits the Earth in such a way that it is stationary with respect to a point on the surface of the Earth This means that its orbital period must be the same as the time for the Earth to spin once on its axis ie 24 hours

From Keplerrsquos third law we have G Ms

4π2----------- R3

T2------=

That is

RMe

mh

using the fact that the force of attraction between the satellite and the Earth is given by

FG Mem

R2----------------=

and that F = ma

where m is the mass of the satellite and a 4π2R

T2-------------=

we have

GMem

R2---------------- m 4π2R

T2-------------times

GMe

4π2------------rArr R3

T2------= =

PHYSICS 2015indb 321 150514 338 PM

Chapter 10

322

AH

L

Now the mass of the Earth is 60 times 1024 kg and the period T measured in seconds is given by T = 86400 s

So substitution gives R = 42 times 106 m

The radius of the Earth is 64 times 106 m so that the orbital height h is about 36 times 107 m

Example

Calculatetheminimumenergyrequiredtoputasatelliteofmass500kgintoanorbitthatisasaheightequaltotheEarthrsquosradiusabovethesurfaceoftheEarth

Solution

We have seen that when dealing with gravitational fields and potential it is useful to remember that

g 0G M

Re2---------= or g0 Re

2 GM=

The gravitational potential at the surface of the Earth is

g 0ndash Re GMRe

---------ndash=

The gravitational potential at a distance R

from the centre of the Earth is

=

The difference in potential between the surface of the Earth and a point distance R from the centre is therefore

V∆ g 0Re 1ReR------ndash

=

If R 2Re= then V∆g0 Re

2------------=

This means that the work required to ldquoliftrdquo the satellite into orbit is g0Rm where m is the mass of the satellite This is equal to

10 32 106 500timestimestimes = 16000 MJ

However the satellite must also have kinetic energy in order to orbit Earth This will be equal to

g 0 mRe2

2R-----------------g 0mRe

2

4----------------- 8000 MJ= =

The minimum energy required is therefore

24000 MJ

Exercise 1031 The speed needed to put a satellite in orbit does

not depend on

A the radius of the orbitB theshapeoftheorbitC thevalueofgattheorbitD the mass of the satellite

2 EstimatethespeedofanEarthsatellitewhoseorbitis400kmabovetheEarthrsquossurfaceAlsodetermine the period of the orbit

3 Calculatethespeedofa200kgsatelliteorbitingthe Earth at a height of 70 times 106 m

Assume that g = 82 m sndash2 for this orbit

4 TheradiioftwosatellitesXandYorbitingthe Earth are 2r and 8rwherer is the radius of the Earth Calculate the ratio of the periods of revolutionofXtoY

5 A satellite of mass mkgissentfromEarthrsquossurfaceinto an orbit of radius 5RwhereR is the radius of theEarthWritedownanexpressionfor

(a) thepotentialenergyofthesatelliteinorbit

(b) thekineticenergyofthesatelliteinorbit

(c) theminimumworkrequiredtosendthesatellitefromrestattheEarthrsquossurfaceintoits orbit

6 A satellite in an orbit of 10rfallsbacktoEarth(radius r)afteramalfunctionDeterminethespeedwithwhichitwillhittheEarthrsquossurface

7 The radius of the moon is frac14 that of the Earth AssumingEarthandtheMoontohavethesamedensitycomparetheaccelerationsoffreefallatthe surface of Earth to that at the surface of the Moon

8 UsethefollowingdatatodeterminethegravitationalfieldstrengthatthesurfaceoftheMoon and hence determine the escape speed from the surface of the Moon

Mass of the Moon = 73 times 1022kg

RadiusoftheMoon=17times106 m

PHYSICS 2015indb 322 150514 338 PM

Chapter 9

300

AH

L

apointsourceadiffractionpatternwillbeformedontheretina If we look at two point sources then two diffraction patterns will be formed on the retina and these patterns will overlap The width of our pupil and the wavelength of the light emitted by the sources will determine the amount by which they overlap But the degree of overlap will also depend on the angular separation of the two point sources WecanseethisfromFigures923

pupilretina

S1

S2

P2

P1

θ

Figure 923

LightfromthesourceS1 enters the eye and is diffracted by the pupil such that the central maximum of the diffraction patternisformedontheretinaatP1SimilarlylightfromS2producesamaximumatP2 If the two central maxima are well separated then there is a fair chance that we will see the two sources as separate sources If they overlap then we will not be able to distinguish one source from another From the diagram we see as the sources are movedclosertotheeyethentheangleθ increases and so does the separation of the central maxima

Figures924925926and927showsthedifferentdiffractionpatternsandtheintensitydistributionthatmightresultonthe retina as a result of light from two point sources

S1 S21 Well resolved

Figure 924 Very well resolved

S1 S22 Well resolved

Figure 925 Well resolved

S1 S23 Just resolved Rayleigh criterion Minimum of S2 coincides with maximum peak of S1

Figure 926 Just resolved

4 Not resolved

Figure 927 Not resolved

We have suggested that if the central maxima of the two diffraction patterns are reasonably separated then we should be able to resolve two point sources In the late 19th century Lord Rayleigh suggested by how much they should be separated in order for the two sources to be just resolved Ifthecentralmaximumofonediffractionpattern coincides with the firstminima of the otherdiffraction pattern then the two sources will just beresolved This is known as the Rayleigh Criterion

Figure926showsjustthissituationThetwosourcesarejust resolved according to the Rayleigh criterion since the peak of the central maximum of one diffraction pattern coincideswiththefirstminimumoftheotherdiffractionpattern This means that the angular separation of the peaks of the two central maxima formed by each source is just the half angular width of one central maximum ie

θ λb---

=

where b is the width of the slit through which the light fromthesourcespassesHoweverweseefromFigure1117that θ is the angle that the two sources subtend at the slit Henceweconcludethattwosourceswillberesolvedbyaslit if the angle that they subtend at the slit is greater than or equaltoλfraslb

PHYSICS 2015indb 300 150514 337 PM

Wave Phenomena

301

AH

L

Sofarwehavebeenassumingthattheeyeisarectangularslit whereas clearly it is a circular aperture and so we must use the formula

θ 122 λb--------------=

As mentioned above the angle θ is sometimes called the resolving power but should more accurately be called the minimum angle of resolution (θmin)

Clearly the smaller θ the greater will be the resolving power

The diffraction grating is a useful for differentiating closely spacedlinesinemissionspectraLikeaprismspectrometerit can disperse a spectrum into its components but it is better suited because it has higher resolution than the prism spectrometer

If λ1 and λ2 are two nearly equal wavelengths that canbarelybedistinguishedtheresolvanceorresolvingpowerofthegratingisdefinedas

R = λ _______ (λ2 ndash λ 1) = λ ___ Δλ

Where λ asymp λ1 asymp λ2

Therefore thediffractiongratingwithahighresolvancewill be better suited in determining small differences in wavelength

If the diffraction grating has Nlinesbeingilluminateditcan be shown that the resolving power of the mth order diffraction is given by

R = mN

Obviouslytheresolvancebecomesgreaterwiththeordernumber m and with a greater number of illuminated slits

Example

A benchmark for the resolving power of a grating is the sodium doublet in the sodium emission spectrum Two yellow lines in itsspectrumhavewavelengthsof58900nmand58959nm

(a) Determinetheresolvanceofagratingifthegivenwavelengths are to be distinguished from each other

(b) Howmanylinesinthegratingmustbeilluminated in order to resolve these lines in the second order spectrum

Solution

(a) R = λ ___ Δλ = 58900nm __________________ (58900ndash58959)nm

= 589nm _______ 059nm = 10 times 103

(b) N = R __ m = 10 times 10 3 ________ 2 =50times102 lines

It has been seen that diffraction effectively limits the resolving power of optical systems This includes such systems as the eye telescopes and microscopes Theresolving power of these systems is dealt with in the next section This section looks at links between technology and resolving power when looking at the very distant and when looking at the very small

Radio telescopes

The average diameter of the pupil of the human eye is about 25 mm This means that the eye will justresolve two point sources emitting light of wavelength 500 nm if their angular separation at the eye is 7

43

50 10122 24 1025 10

θminus

minusminus

times= times = times

times rad

If the eye were to be able to detect radio waves of wavelength 015m then tohave the same resolvingpower thepupilwould have to have a diameter of about 600 m Clearly this is nonsense but it does illustrate a problem facingastronomers who wish to view very distant objects such as quasarsandgalaxies thatemitradiowavesConventionalradio telescopes consist of a large dish typically 25 min diameter Even with such a large diameter the radiowavelength resolving power of the telescope is much less thantheopticalresolvingpowerofthehumaneyeLetuslook at an example

Example

The Galaxy Cygnus A can be resolved optically as an elliptically shaped galaxy However it is also a strongemitterof radiowavesofwavelength015mTheGalaxyisestimatedtobe50times1024mfromEarthUseofaradiotelescope shows that the radio emission is from two sources separated by a distance of 30 times 1021m Estimate thediameterofthedishrequiredtojustresolvethesources

Solution

The angle θ that the sources resolve at the telescope is given by

θ 21

424

30 10 60 1050 10

θ minustimes= = times

times rad

and 4

122 122 015 300060 10

d λθ minus

times= = =

timesm = 30 km

A radio telescope dish of this size would be impossible tomake letalonesupportThisshowsthatasingledishtype radio telescope cannot be used to resolve the sources and yet they were resolved To get round the problemastronomers use two radio telescopes separated by a large

PHYSICS 2015indb 301 150514 337 PM

Chapter 9

302

AH

L

distance The telescopes view the same objects at the same time and the signals that each receive from the objects are simultaneously superimposed The result of the superposition of the two signals is a two-slit interference pattern The pattern has much narrower fringe spacing than that of the diffraction pattern produced by either telescope on its own hence producing a much higherresolvingpowerWhentelescopesareusedlikethistheyare called a stellar interferometer

InSocorroinNewMexicothereisastellarinterferometerthat consists of 27 parabolic dishes each of diameter25 m arranged in a Y-shape that covers an area of 570km2Thisisaso-calledVeryLargeArray(VLA)Evenhigher resolution can be obtained by using an array of radio telescopes in observatories thousands of kilometres apart A system that uses this so-called technique oflsquovery-long-baseline interferometryrsquo (VLBI) is known asalsquovery-long-baselinearrayrsquo(VLBA)WithVLBAaradiowavelength resolving power can be achieved that is 100 timesgreaterthanthebestopticaltelescopesEvenhigherresolving power can be achieved by using a telescope that isinasatelliteorbitingEarthJapanrsquosInstituteofSpaceandAstronautical Science (ISAS) launched sucha system inFebruary 1997TheNationalAstronomicalObservatoryof Japan the National Science Foundationrsquos NationalRadio Astronomy Observatory (NRAO) the CanadianSpaceAgency theAustraliaTelescopeNationalFacilitythe European VLBI Network and the Joint Instituteback this project forVeryLongBaseline Interferometryin EuropeThis project is a very good example of howInternationalismcanoperateinPhysics

Electron microscope

Telescopes are used to look at very distant objects that are very large but because of their distance fromus appearverysmallMicroscopesontheotherhandareusedtolookat objects that are close to us but are physically very small Aswehave seen justmagnifyingobjects that ismakingthemappearlargerisnotsufficientonitsowntogaindetailabouttheobjectfordetailhighresolutionisneeded

Figure928isaschematicofhowanoptical microscope is usedtoviewanobjectandFigure929isaschematicofatransmission electron microscope(TEM)

Optical lens system

bright light source

glass slide containing specimen (object)

eye

Figure 928 The principle of a light microscope

magnetic lens system

electron source

thin wafer of material

ScreenCCD

Figure 929 The principle of an electron microscope

In the optical microscope the resolving power isdetermined by the particular lens system employed and the wavelength λ of the light used For example twopoints in the sample separated by a distance d will just be resolved if

2d mλ=

where m is a property of the lens system know as the numerical aperture In practice the largest value of m obtainable is about 16 Hence if the microscope slideis illuminatedwith light ofwavelength 480 nm a goodmicroscope will resolve two points separated by a distance dasymp15times10-7masymp015micromPointscloser together thanthiswillnotberesolvedHoweverthisisgoodenoughtodistinguishsomevirusessuchastheEbolavirus

Clearlythesmallerλ the higher the resolving power and this is where the electron microscope comes to the fore The electron microscope makes use of the wave nature of electrons(see1315)IntheTEMelectronspassthrougha wafer thin sample and are then focused by a magnetic field onto a fluorescent screen orCCD (charge coupleddevicesee142)Electronsusedinanelectronmicroscopehavewavelengthstypicallyofabout5times10-12mHoweverthe numerical aperture of electron microscopes is considerablysmaller thanthatofanopticalmicroscopetypicallyabout002NonethelessthismeansthataTEMcanresolvetwopointsthatareabout025nmapartThisresolving power is certainly high enough to make out the shape of large molecules

Another type of electron microscope uses a techniqueby which electrons are scattered from the surface of the sample The scattered electrons and then focused as in theTEM to form an image of the surfaceThese so-calledscanningelectronmicroscopes(SEM)havealowerresolving power than TEMrsquos but give very good threedimensional images

The eye

We saw in the last section that the resolving power of the human eye is about 2 times 10-4 rad Suppose that youare looking at car headlights on a dark night and the car

PHYSICS 2015indb 302 150514 337 PM

Wave Phenomena

303

AH

L

is a distance D away If the separation of the headlight is say 15 m then the headlights will subtend an angle θ 15

Dθ =

at your eye Assuming an average wavelength of

500 nm your eye will resolve the headlights into twoseparatesourcesifthisangleequals2times10-4 rad This gives D=75kmInotherwordsifthecarisapproachingyouona straight road then you will be able to distinguish the two headlightsasseparatesourceswhenthecaris75kmawayfrom you Actually because of the structure of the retina and optical defects the resolving power of the average eye is about 6 times 10-4 rad This means that the car is more likely tobe25kmawaybeforeyouresolvetheheadlights

Astronomical telescope

Letusreturntotheexampleofthebinarystarsdiscussedatthe beginning of this section on resolution The stars Kruger A and B form a binary system The average separation of the stars is 14 times 1012mandtheiraveragedistancefromEarthis12times1017mWhenviewedthroughatelescopeonEarththe system will therefore subtend an angle

12

17

14 1012 10

θtimes

=times

= 12times10-5 rad at the objective lens of the telescope Assuming that the average wavelength of the light emitted bythestarsis500nmthenifthetelescopeistoresolvethesystem into two separate images it must have a minimum diameter Dwhere12times10-5 =

7122 500 10D

minustimes times

This gives D=0050m which is about 5 cm So thisparticular system is easily resolved with a small astronomical telescope

Exercise 961 ItissuggestedthatusingtheISASVLBAitwould

bepossibletoldquoseerdquoagrainofriceatadistanceof5000kmEstimatetheresolvingpoweroftheVLBA

2 Thedistancefromtheeyelenstotheretinais20mm The light receptors in the central part of the retinaareabout5times10-6apartDeterminewhetherthe spacing of the receptors will allow for the eye to resolve the headlights in the above discussion whentheyare25kmfromtheeye

3 ThediameterofPlutois23times106 m and its averagedistancefromEarthis60times1012 m EstimatetheminimumdiameteroftheobjectiveofatelescopethatwillenablePlutotobeseenasadisc as opposed to a point source

95 Doppler effect

Essential ideaTheDopplereffectdescribesthephenomenonofwavelengthfrequencyshiftwhenrelativemotionoccurs

Understandingsbull TheDopplereffectforsoundwavesandlightwaves

The Doppler effect for sound waves and light waves

Consider two observers A and B at rest with respect to asoundsourcethatemitsasoundofconstantfrequencyf Clearly both observers will hear a sound of the same frequencyHoweversupposethatthesourcenowmovesat constant speed towards A A will now hear a sound of frequencyfA that is greater than f and B will hear a sound of frequency fB that is less than f This phenomenon is known as the Doppler Effect orDopplerPrinciple after C J Doppler(1803-1853)

The same effect arises for an observer who is either moving towards or away from a stationary source

Figure 930 shows the waves spreading out from a stationary source that emits a sound of constant frequency f The observersAandBhearasoundofthesamefrequency

A B

wavefront

S

Figure 930 Sound waves from a stationary source

Suppose now that the source moves towards A withconstant speed vFigure9310(a)showsasnapshotofthenew wave pattern

NATURE OF SCIENCETechnology Although originally based on physical observations of the pitch of fast moving sources of sound the Doppler effect has an important role in many different areas such as evidence for the expansion of the universe and generating images used in weather reports and in medicine (55)

copy IBO 2014

PHYSICS 2015indb 303 150514 337 PM

Chapter 9

304

AH

L

sourceA B

smallerwavelength

largerwavelength

Figure 931 (a) Sound waves from a moving source

The wavefronts are now crowded together in the direction of travel of the source and stretched out in the opposite direction This is why now the two observers will now hear notesofdifferentfrequenciesHowmuchthewavesbunchtogether and how much they stretch out will depend on the speed vEssentially A

A

cfλ

= and BB

cfλ

= where λA lt λB and v is the speed of sound

IfthesourceisstationaryandAismovingtowardsitthenthe waves from the source incident on A will be bunched up If A is moving away from the stationary source then the waves from the source incident on A will be stretched out

Christian Doppler(1803ndash1853)actuallyappliedtheprinciple(incorrectlyasithappens)totryandexplainthecolourofstars However the Doppler effect does apply to light aswellastosoundIfalightsourceemitsalightoffrequencyf then if it is moving away from an observer the observer willmeasurethelightemittedashavingalowerfrequencythan fSincethesensationofcolourvisionisrelatedtothefrequencyoflight(bluelightisofahigherfrequencythanred light) light emitted by objects moving way from anobserver is often referred to as being red-shifted whereas if the object is moving toward the observer it is referred to as blue-shiftedThisideaisusedinOptionE(Chapter16)

We do not need to consider here the situations where either the source or the observer is accelerating In a situation for example where an accelerating source is approaching astationaryobserverthentheobserverwillhearasoundof ever increasing frequency This sometimes leads toconfusion in describing what is heard when a source approaches passes and then recedes from a stationaryobserver Suppose for example that you are standingon a station platform and a train sounding its whistle is approaching at constant speed What will you hear as the train approaches and then passes through the station As the train approaches you will hear a sound of constant pitch but increasing loudness The pitch of the sound will be greater than if the train were stationary As the train passes through the station you will hear the pitch change at the momentthetrainpassesyoutoasoundagainofconstantpitch The pitch of this sound will be lower than the sound of the approaching train and its intensity will decrease as the train recedes from you What you do not hear is a sound of increasing pitch and then decreasing pitch

The Doppler equations for sound

Although you will not be expected in an IB examination to derivetheequationsassociatedwithaspectsoftheDopplereffectyouwillbeexpectedtoapplythemForcompletenesstherefore thederivationoftheequationsassociatedwiththeDopplereffectasoutlinedaboveisgivenhere

S primeS O

vs

Figure 931 (b)

InFigure931(b)theobserverOisatrestwithrespecttoasourceofsoundSismovingwithconstanspeedvs directly towards O The source is emitting a note of constant frequencyf and the speed of the emitted sound is v

S shows the position of the source ∆t later When the sourceisatresttheninatime∆t the observer will receive f∆t waves and these waves will occupy a distance v∆t ie

v f

λ =f∆tv∆t =

(Because of the motion of the source this number of waves will now occupy a distance (v∆t ndash vs∆t)The lsquonewrsquowavelength is therefore

v ndash vs

fλrsquo =

f∆tv∆t ndash vs∆t

=

If fisthefrequencyheardbyOthen

vfλ

= or

vf

λ = = sv vfminus

From which

s

vf fv v

=minus

Dividingthroughbyv gives

Equation1111

1 sf f v

v=

minus

If the source is moving away from the observer then we have

s

vf fv v

=minus

1

1 sf vv

=+

Equation112

PHYSICS 2015indb 304 150514 337 PM

Wave Phenomena

305

AH

L

We now consider the case where the source is stationary and the observerismoving towardsthesource with speed v0 In this situation the speed of the sound waves as measured by the observer will be v0 + v We therefore have that

v0 + v = ffraslλ = vfλ

times

From which

o1 vf fv

= +

Equation113

If the observerismovingawayfromthesource then

o1 vf fv

= minus

Fromequation(113)wehavethat

o o1 v vf f f f f fv v

∆ = minus = + minus =

Equation114

Thevelocitiesthatwerefertointheaboveequationsarethevelocities with respect to the medium in which the waves fromthesourcetravelHoweverwhenwearedealingwithalight source it is the relative velocity between the source and the observer that we must consider The reason for this is that lightisuniqueintherespectthatthespeedofthelightwavesdoes not depend on the speed of the source All observers irrespective of their speed or the speed of the source will measure the same velocity for the speed of light This is oneof thecornerstonesof theSpecialTheoryofRelativitywhichisdiscussedinmoredetailinOptionH(Chapter18)When applying the Doppler effect to light we aremainlyconcerned with the motion of the source We look here only at the situation where the speed of the source v is much smaller than the speed of light cinfreespace(v ltlt c)Underthesecircumstanceswhenthesourceismovingtowardstheobserverequation111becomes

vf f f fc

minus = ∆ = Equation115

andwhenthesourceismovingawayfromtheobserverequation112becomes vf f f f

cminus = ∆ = minus

Provided that v ltlt c these same equations apply for astationary source and moving observer

We look at the following example and exercise

Example

Asourceemitsasoundoffrequency440HzItmovesinastraight line towards a stationary observer with a speed of 30ms-1Theobserverhearsasoundoffrequency484HzCalculate the speed of sound in air

Solution

We use equation 111 and substitute f= 484 Hz f = 440 Hz

and vs = 30 m s-1

therefore 1

484 440 301v

=minus

such that 30 4401

484vminus = to

give v = 330 m s-1

Example

A particular radio signal from a galaxy is measured as havingafrequencyof139times109HzThesamesignalfromasourceinalaboratoryhasafrequencyof142times109Hz

Suggestwhy thegalaxy ismovingaway fromEarthandcalculateitsrecessionspeed(iethespeedwithwhichitismovingawayfromEarth)

Solution

The fact that the frequency from the moving source is less than that when it is stationary indicates that it is moving away from the stationary observer ie Earth

Using vf fc

∆ = we have

8 96

9

3 10 (142 139) 10 634 10142 10

c fvf∆ times times minus times

= = = timestimes

m s-1

It is usual when dealing with the Doppler effect of light to express speeds as a fraction of c So in this instance we have v = 0021 c

Using the Doppler effectWe have seen in the above example and exercise how the Dopplereffectmaybeusedtomeasuretherecessionspeedof distant galaxies The effect is also used to measure speed inothersituationsHerewewilllookatthegeneralprincipleinvolved in using the Doppler effect to measure speedFigure932showsasource(thetransmitter)thatemitseithersound or em waves of constant frequency f The waves

PHYSICS 2015indb 305 150514 337 PM

Chapter 9

306

AH

L

fromthesourceare incidentonareflectorthat ismovingtowards the transmitter with speed vThereflectedwavesare detected by the receiver placed alongside the transmitter

reector

v

transmitter

receiver

f f

f f

Figure 932 Using the Doppler effect to measure speed

We shall consider the situation where v ltlt c where c is the speed of the waves from the transmitter

For the reflector receiving waves from the transmitterit is effectively an observer moving towards a stationary sourceFromequation(114) itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation116

For the receiver receivingwaves from the reflector it iseffectively a stationary observer receiving waves from a movingsourceFromequation(115)itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation117

Ifweaddequations(116)and(117)wegetthatthetotalDopplershiftatthereceiver∆f is

v vf f f f fc c

minus = ∆ = +

But 1 vf fc

= +

hence

1 v v vf f fc c c

∆ = + +

But since v ltlt cwe can ignore the term2

2

vc when we

expandthebracketintheaboveequation

Therefore we have2vf fc

∆ = Equation118

If v asymp c then we must use the full Doppler equationsHowever for em radiationwewill alwaysonly considersituations in which v ltlt c

Example

Thespeedofsoundinbloodis1500times103 m s-1 Ultrasound offrequency100MHzisreflectedfrombloodflowinginan arteryThe frequencyof the reflectedwaves receivedbackatthetransmitteris105MHzEstimatethespeedofthebloodflowintheartery

Solution

Using equation (118) we have

6 63

2005 10 1015 10

vtimes = times

timesto give v asymp 36 m s-1 (We have assumed that the ultrasound is incident at right angles to the blood flow)

Exercise 971 Judy is standing on the platform of a station A

high speed train is approaching the station in a straight line at constant speed and is sounding its whistleAsthetrainpassesbyJudythefrequencyof the sound emitted by the whistle as heard by Judychangesfrom640Hzto430HzDetermine

(a) thespeedofthetrain

(b) thefrequencyofthesoundemittedbythewhistle as heard by a person on the train (Speedofsound=330ms-1)

2 AgalaxyismovingawayfromEarthwithaspeedof00500cThewavelengthofaparticularspectral line in light emitted by atomic hydrogen inalaboratoryis656times10-7 m Calculate the valueofthewavelengthofthislinemeasuredinthelaboratoryinlightemittedfromasourceofatomic hydrogen in the galaxy

PHYSICS 2015indb 306 150514 337 PM

10 Fields

Contents

101 ndash Describing fields

102 ndash Fields at work

Essential Ideas

Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Similar approaches can be taken in analysing electrical and gravitational potential problems copy IBO 2014

PHYSICS 2015indb 307 150514 337 PM

Chapter 10

308

AH

L

101 Describing fields

Essential idea Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Understandingsbull Gravitationalfieldsbull Electrostaticfieldsbull Electricpotentialandgravitationalpotentialbull Fieldlinesbull Equipotentialsurfaces

Gravitational potentialWehaveseenthatifweliftanobjectofmassm to a height h abovethesurfaceoftheEarththenitsgainingravitationalpotential energy is mghHoweverthisisbynomeansthefullstoryForastartwenowknowthatgvarieswithh and also the expression really gives a difference in potentialenergybetweenthevaluethattheobjecthasattheEarthrsquossurfaceandthevaluethatithasatheighthSowhatwereallyneedisazeropointCanwefindapointwherethepotentialenergyiszeroandusethispointfromwhichtomeasure changes in potential energy

The point that is chosen is in fact infinity At infinity the gravitationalfieldstrengthofanyobjectwillbezeroSoletusseeifwecandeduceanexpressionforthegaininpotentialenergyofanobjectwhenitisldquoliftedrdquofromthesurfaceoftheEarthtoinfinityThisineffectmeansfindingtheworknecessarytoperformthistask

δr

r r δr+

r infin=

mg

AB

Me

Figure 1001 Gravitational forces

Inthediagramweconsidertheworknecessarytomovethe particle of mass m a distance δr in thegravitationalfield of the Earth

The force on the particle at A is FG Mem

r2----------------=

IftheparticleismovedtoB then since δr isverysmallwe can assume that the field remains constant over thedistance ABTheworkδW done againstthegravitationalfieldoftheEarthinmovingthedistanceABis

δWGMem

r2----------------δrndash=

(rememberthatworkdoneagainstaforceisnegative)

TofindthetotalworkdoneW in going from the surface oftheEarthtoinfinitywehavetoaddalltheselittlebitsofworkThisisdonemathematicallybyusingintegralcalculus

WG Me m

r2----------------ndash

rdR

infin

int G Me m 1

r2----- rd

R

infin

intndash G Me m 1r---ndash

R

infinndash= = =

GMe m 0 1R---ndash

ndashndash=

G MemR----------------ndash=

HencewehavewhereRistheradiusoftheEarththattheworkdonebythegravitationalfieldinmovinganobjectofmass m from R(surfaceoftheEarth)toinfinityisgivenby

WGMe m

R---------------- ndash=

We can generalise the result by calculating the worknecessary per unit mass to take a small mass fromthe surface of the Earth to infinity This we call thegravitational potential Vie

Wewouldgetexactlythesameresultifwecalculatedtheworkdonebythefieldtobringthepointmassfrominfinityto the surface of Earth In this respect the formal definition ofgravitationalpotentialatapointinagravitationalfieldis therefore defined as the work done per unit mass in bringing a point mass from infinity to that point

Clearly then the gravitational potential at any point inthe Earthrsquos field distance r from the centre of the Earth (providingr gt R)is

VGMe

r------------ ndash=

Thepotentialisthereforeameasureoftheamountofworkthathastobedonetomoveparticlesbetweenpointsinagravitationalfieldanditsunit is theJkgndash1 We also note thatthepotentialisnegativesothatthepotentialenergy

NATURE OF SCIENCEParadigm shift The move from direct observable actions being responsible for influence on an object to acceptance of a fieldrsquos ldquoaction at a distancerdquo required a paradigm shift in the world of science (23)

copy IBO 2014

PHYSICS 2015indb 308 150514 337 PM

Fields

309

AH

L

aswemoveawayfromtheEarthrsquossurfaceincreasesuntilitreachesthevalueofzeroatinfinity

IfthegravitationalfieldisduetoapointmassofmassmthenwehavethesameexpressionasaboveexceptthatMe is replaced by m andmustalsoexclude thevalueof thepotential at the point mass itself ie at r = 0

WecanexpressthegravitationalpotentialduetotheEarth(orduetoanysphericalmass)intermsofthegravitationalfield strength at its surface

AtthesurfaceoftheEarthwehave

g 0ndash Re GMRe

---------ndash= e

Sothat

g 0Re2 GM= e

Hence at a distance r from the centre of the Earth the gravitationalpotentialVcanbewrittenas

VGMe

r------------ndashg 0 Re

2

r------------ndash= =

The potential at the surface of the Earth

(r = Re)istherefore-g0Re

It is interesting to see how the expression for thegravitational potential ties in with the expression mgh ThepotentialatthesurfaceoftheEarthis-g0Re (see the exampleabove)andataheighthwillbe-g0g 0 Re h+( )ndash ifweassume that g0doesnotchangeover thedistanceh The differenceinpotentialbetweenthesurfaceandtheheighth is therefore g0hSotheworkneededtoraiseanobjectof mass m to a height h is mgh ie m timesdifference ingravitationalpotential

This we have referred to as the gain in gravitationalpotentialenergy(see235)

However this expression can be extended to any twopoints inanygravitationalfieldsuch that ifanobjectofmass mmovesbetween twopointswhosepotentialsareV1 and V2 respectively then the change in gravitationalpotentialenergyoftheobjectism(V1 ndash V2)

Gravitational potential gradient

Let us consider now a region in space where thegravitationalfieldisconstantInFigure912thetwopointsAandBareseparatedbythedistance∆x

A B

direction of uniform gravitational eld of strength I

∆x

Figure 1002 The gravitational potential gradient

The gravitational field is of strength I and is in the direction shownThe gravitational potential at A is V andatBisV +∆V

Theworkdone is takingapointmassmfromAtoB isF∆x = mI∆x

Howeverbydefinitionthisworkisalsoequalto-m∆V

Therefore mI∆x=-m∆V

or VIx

∆= minus

∆Effectivelythissaysthatthemagnitudeofthegravitationalfield strength is equal to the negative gradient of thepotential If I is constant then V is a linear function of x and Iisequaltothenegativegradientofthestraightlinegraph formed by plotting V against x If I is not constant (asusuallythecase)thenthemagnitudeofI at any point in the field can be found by find the gradient of the V-x graph at that point An example of such a calculation can befoundinSection929

For those of you who do HL maths the relationship between field and potential is seen to follow from the expression for the potential of a point mass viz

mV Gr

= minus

2

ddV mG Ir r

minus = + =

Potential due to one or more point masses

Gravitationalpotentialisascalarquantitysocalculatingthepotential due to more than one point mass is a matter of simpleadditionSoforexamplethepotentialV due to the Moon and Earth and a distance x fromthecentreofEarthonastraightlinebetweenthemisgivenbytheexpression

E MM MV Gx r x

= minus + minus

whereME=massofEarthMM= mass of Moon and r = distancebetweencentreofEarthandMoon

PHYSICS 2015indb 309 150514 337 PM

Chapter 10

310

AH

L

Equipotentials and field linesIfthegravitationalpotentialhasthesamevalueatallpointson a surface the surface is said to be an equipotential surfaceSoforexampleifweimagineasphericalshellaboutEarthwhosecentrecoincideswiththecentreofEarththisshellwillbeanequipotentialsurfaceClearlyifwerepresentthegravitationalfieldstrengthbyfieldlinessincethelinesldquoradiaterdquooutfromthecentreofEarththentheselineswillbeatrightanglestothesurfaceIfthefieldlineswerenotnormaltotheequipotentialsurfacethentherewouldbeacomponentof the field parallel to the surfaceThis wouldmean thatpointsonthesurfacewouldbeatdifferentpotentialsandsoitwouldnolongerbeanequipotentialsurfaceThisofcourseholdstrueforanyequipotentialsurface

Figure 1003 shows the field lines and equipotentials fortwopointmassesm

m m

Figure 1003 Equipotentials for two point masses

It is worth noting that we would get exactly the samepattern ifwewere to replace thepointmasseswith twoequalpointcharges(See935)

Escape speedThe potential at the surface of Earth is which meansthattheenergyrequiredtotakeaparticleofmass mfromthesurfacetoinfinityisequalto

ButwhatdoesitactuallymeantotakesomethingtoinfinityWhentheparticleisonthesurfaceoftheEarthwecanthinkofitassittingatthebottomofaldquopotentialwellrdquoasinfigure1004

particle

surface of Earth

innity

GMR---------

FIgure 1004 A potential well

Theldquodepthrdquoofthewellis and if the particle gains an amount of kinetic energy equal to wherem is its massthenitwillhavejustenoughenergytoldquoliftrdquoitoutofthewell

In reality it doesnrsquot actually go to infinity it justmeansthat the particle is effectively free of the gravitationalattractionof theEarthWe say that ithas ldquoescapedrdquo theEarthrsquosgravitationalpullWemeetthisideainconnectionwithmolecular forcesTwomolecules in a solidwill sitat their equilibrium position the separation where therepulsiveforceisequaltotheattractiveforceIfwesupplyjust enough energy to increase the separation of themolecules such that they are an infinite distance apart then intermolecularforcesnolongeraffectthemoleculesandthesolidwillhavebecomealiquidThereisnoincreaseinthekineticenergyofthemoleculesandsothesolidmeltsat constant temperature

Wecancalculatetheescapespeedofanobjectveryeasilyby equating the kinetic energy to the potential energysuch that

12---mvesc ape

2 GMe mRe

----------------=

vesc aperArr2GMe

Re--------------- 2g0 Re = =

Substituting for g0 and Regivesavalueforvescape of about 11kmsndash1 from the surface of the Earth

Youwillnotethattheescapespeeddoesnotdependonthemassoftheobjectsincebothkineticenergyandpotentialenergy are proportional to the mass

IntheoryifyouwanttogetarockettothemoonitcanbedonewithoutreachingtheescapespeedHoweverthiswouldnecessitate anenormousamountof fuel and it islikely that therocketplus fuelwouldbesoheavythat itwouldnevergetoffthegroundItismuchmorepracticalto accelerate the rocket to the escape speed fromEarthorbitandthenintheoryjustlaunchittotheMoon

Example

Use the followingdata todetermine thepotential at thesurface of Mars and the magnitude of the acceleration of free fall

massofMars =64times1023kg

radiusofMars =34times106 m

Determinealsothegravitationalfieldstrengthatadistanceof 68 times 106mabovethesurfaceofMars

PHYSICS 2015indb 310 150514 337 PM

Fields

311

AH

L

Solution

= minus = minus times times = times23

11 76

64 1067 10 ndash13 10

34 10MV GR

minus timestimes

Nkg-1

But V = -g0R

Therefore 7

0 6

13 10 3834 10

VgR

times= minus = =

timesm s-2

To determine the field strength gh at 68 times 106 m above the surface we use the fact that 0 2

Mg GR

= such that GM = g0R2

Therefore 2 2

0h 2 2 2

h h

38 (34) 042(102)

g RGMgR R

times= = = = m s-2

(the distance from the centre is 34 times 106 + 68 times 106 = 102 times 106 m)

Exercise 1011 Thegraphbelowshowshowthegravitational

potentialoutsideoftheEarthvarieswithdistancefrom the centre

0 10 20 30 40 50 60 70

ndash1

ndash2

ndash3

ndash4

ndash5

ndash6

ndash7

r m times 106

V

Jkgndash1

times 10

7

(a) Usethegraphtodeterminethegainingravitationalpotentialenergyofasatelliteofmass200kgasitmovesfromthesurfaceof the Earth to a height of 30 times 107mabovetheEarthrsquossurface

(b) Calculatetheenergyrequiredtotakeit to infinity

(c) DeterminetheslopeofthegraphatthesurfaceoftheEarthmCommenton youranswer

Electric potential energy

The concept of electric potential energy was developedwith gravitational potential energy in mind Just as anobjectnearthesurfaceoftheEarthhaspotentialenergybecause of its gravitational interaction with the Earthsotoothereiselectricalpotentialenergyassociatedwithinteracting charges

Letusfirstlookatacaseoftwopositivepointchargeseachof 1μC that are initially bound together by a thread in a vacuuminspacewithadistancebetweenthemof10cmasshowninFigure1005Whenthethreadiscutthepointchargesinitiallyatrestwouldmoveinoppositedirectionsmovingwithvelocitiesv1 and v2 along the direction of the electrostatic force of repulsion

BEFORE AFTER

v v21

Figure 1005 Interaction of two positive particles

The electric potential energybetweentwopointchargescan be found by simply adding up the energy associated witheachpairofpointchargesForapairofinteractingchargestheelectricpotentialenergyisgivenby

∆U = ∆Ep + ∆Ek = ∆W = ∆Fr = kqQ

____ r2 times r = kqQ

____ r

Becausenoexternalforceisactingonthesystemtheenergyand momentum must be conserved Initially Ek = 0 and Ep =kqQr=9times109 times 1 times 10-1201m=009JWhentheyareagreatdistance fromeachotherEpwillbenegligibleThefinalenergywillbeequaltofrac12mv1

2+frac12mv22=009J

Momentum isalsoconservedand thevelocitieswouldbethe same magnitude but in opposite directions

Electric potential energyismoreoftendefinedintermsofapointchargemovinginanelectricfieldas

lsquothe electricpotential energybetweenany twopoints inan electric field is defined as negativeoftheworkdone by anelectricfieldinmovingapoint electric chargebetweentwolocationsintheelectricfieldrsquo

ΔU = ΔEp = -ΔW = -Fd = qEx

ΔU = ΔEp +ΔEk = ΔW = ΔFr = kqQ r2 x r = kqQ

wherexisthedistancemovedalong(oroppositeto)thedirection of the electric field

Electric potential energyismeasuredinjoule(J)Justasworkisascalarquantitysotooelectricalpotentialenergyis a scalar quantity The negative of the work done by an electric field in moving a unit electric charge between two points is independent of the path taken Inphysicswesaytheelectricfieldisalsquoconservativersquo field

PHYSICS 2015indb 311 150514 337 PM

Chapter 10

312

AH

L

Supposeanexternalforcesuchasyourhandmovesasmallpositive point test charge in the direction of a uniformelectric fieldAs it ismoving itmust be gaining kineticenergyIfthisoccursthentheelectricpotentialenergyofthe unit charge is changing

InFigure1006apointcharge+qismovedbetweenpointsAandBthroughadistancex in a uniform electric field

B

A

x

+ q

Figure 1006 Movement of a positive point charge in a uniform field

InordertomoveapositivepointchargefrompointAtopointBanexternal forcemustbeapplied to thechargeequaltoqE (F = qE)

Since the force is applied through a distance xthennegativeworkhas to bedone tomove the charge because energyis gainedmeaning there is an increaseelectric potential energybetweenthetwopointsRememberthattheworkdoneisequivalenttotheenergygainedorlostinmovingthe charge through the electric field The concept of electric potential energy is only meaningful as the electric field whichgeneratestheforceinquestionisconservative

W F xtimes E q x times= =

θ x cos θ x

Figure 1007 Charge moved at an angle to the field

If a chargemoves at an angle θ to an electric field thecomponent of the displacement parallel to the electric fieldisusedasshowninFigure918

W F x Eq x θcostimes= =

Theelectricpotentialenergyisstoredintheelectricfieldand theelectricfieldwill return theenergy to thepointcharge when required so as not to violate the Law ofconservationofenergy

Electric potential

The electric potential at a point in an electric field is defined as being the work done per unit charge in bringing a small positive point charge from infinity to that point

ΔV = Vinfin ndash Vf = ndash W ___ q

Ifwedesignatethepotentialenergytobezeroatinfinitythen it follows thatelectricpotentialmustalsobezeroat infinity and the electric potential at any point in an electricfieldwillbe

ΔV = ndash W ___ q

NowsupposeweapplyanexternalforcetoasmallpositivetestchargeasitismovedtowardsanisolatedpositivechargeTheexternalforcemustdoworkonthepositivetestchargetomoveittowardstheisolatedpositivechargeandtheworkmustbepositivewhiletheworkdonebytheelectricfieldmustthereforebenegativeSotheelectricpotentialatthatpointmustbepositiveaccordingtotheaboveequationIfanegativeisolatedchargeisusedtheelectricpotentialatapointonthepositivetestchargewouldbenegativePositivepoint chargesof their ownaccordmove fromaplaceofhighelectricpotential toaplaceof lowelectricpotentialNegative point charges move the other way from lowpotentialtohighpotentialInmovingfrompointAtopointBinthediagramthepositivecharge+qismovingfromalowelectricpotentialtoahighelectricpotential

In thedefinitiongiven the termldquoworkperunitchargerdquohas significance If the test charge is +16 times 10-19Cwherethe charge has a potential energy of 32 times 10-17Jthenthepotentialwouldbe32times10-17J+16times10-19C=200JC-1 Now if the charge was doubled the potential wouldbecome 64times10-17 J However the potential per unitchargewouldbethesame

Electric potential is a scalar quantityandithasunitsJC-1 orvoltswhere1voltequalsone joulepercoloumbThevoltallowsustoadoptaunitfortheelectricfieldintermsofthevolt

PreviouslytheunitfortheelectricfieldwasNCndash1

W = qV and F = qEsoW ___ V = F __ E

E = FV ___ W = FV ___ Fm V mndash1

That is the units of the electric field E can also beexpressed as V mndash1

The work done per unit charge in moving a point charge between two points in an electric field is again independent of the path taken

PHYSICS 2015indb 312 150514 337 PM

Fields

313

AH

L

Electric potential due to a point charge

Letus takeapoint rmetres froma chargedobjectThepotentialatthispointcanbecalculatedusingthefollowing

W = ndashFr = ndashqV and F = ndash q1 q2 _____ 4πε0r

2

Therefore

W = ndash q1 q2 _____ 4πε0r

2 times r = ndash q1 q2 _____ 4πε0r

= ndashq1 times q2 _____ 4πε0r

= ndashq1V

That is V q

4πε0r--------------- =

Orsimply

V kqr------

=

Example

Determinehowmuchworkisdonebytheelectricfieldofpointcharge150μCwhenachargeof200μCismovedfrom infinity to apoint 0400m from thepoint charge(Assumenoaccelerationofthecharges)

Solution

The work done by the electric field is W = -qV = -14πε0 times q times (Q rinfin - Q r0400)

W = (- 200 times 10-6 C times 900 times 109 NmC-2 times 150 times 10-6 C) divide 0400 m = - 0675 J

An external force would have to do +0675 J of work

Electric field strength and electric potential gradient

Letus lookbackatFigure1006Supposeagain that thecharge+qismovedasmalldistancebyaforceFfromAtoBsothattheforcecanbeconsideredconstantTheworkdoneisgivenby

∆W F x∆times=

TheforceFandtheelectricfieldEareoppositelydirectedandweknowthat

F = -qE and ΔW = q ΔV

Thereforetheworkdonecanbegivenas

q ΔV = -q E Δx

Therefore E V∆x∆-------

ndash=

The rate of change of potential ΔV at a point with respect to distance Δx in the direction in which the change is maximum is called the potential gradient We say that the electric field = - the potential gradient and the units are Vm-

1 From the equation we can see that in a graph of electric potential versus distance the gradient of the straight line equals the electric field strength

In reality if a charged particle enters a uniform electric field it will be accelerated uniformly by the field and its kinetic energy will increase This is why we had to assume no acceleration in the last worked example

Ek∆ 12---

mv2 q E xsdot sdot q Vx---

xsdot sdot q Vsdot= = = =

Example

Determine how far apart two parallel plates must besituated so that a potential difference of 150 times 102 V produces an electric field strength of 100 times 103 NC-1

Solution

Using E V∆x------- xhArrndash V∆

E-------15 102 Vtimes

100 103 N C 1ndashtimes------------------------------------------= = =

= 150 times 10-1

The plates are 150 times 10-1 m apart

The electric field and the electric potential at a point due to an evenly distributed charge +q on a sphere can be represented graphically as in Figure 1008

r

r

V

E

r0

r0

r0

Charge of +Q evenly distributed over surface

E 14πε0------------ Q

r2----- r r0gt=

V 14πε0------------Q

r----r r0gt=

Figure 1008 Electric field and potential due to a charged sphere

PHYSICS 2015indb 313 150514 337 PM

Chapter 10

314

AH

L

When the sphere becomes charged we know that thechargedistributes itselfevenlyoverthesurfaceThereforeeverypartofthematerialoftheconductorisatthesamepotential As the electric potential at a point is defined as beingnumericallyequaltotheworkdoneinbringingaunitpositivechargefrominfinitytothatpointithasaconstantvalueineverypartofthematerialoftheconductor

Since the potential is the same at all points on the conducting surfacethen∆V∆xiszeroButE=ndash∆V∆xThereforethe electric field inside the conductor is zero There is no electric field inside the conductor

SomefurtherobservationsofthegraphsinFigure1009are

bull OutsidethespherethegraphsobeytherelationshipsgivenasE α 1 r2 and V α 1 r

bull Atthesurfacer = r0Thereforetheelectricfieldandpotentialhavetheminimumvalueforr at this point and this infers a maximum field and potential

bull Insidethespheretheelectricfieldiszerobull Insidethespherenoworkisdonetomoveacharge

fromapointinsidetothesurfaceThereforethereisno potential difference and the potential is the same asitiswhenr = r0

Similargraphscanbedrawnfortheelectricfieldintensityandthe electric potential as a function of distance from conducting parallelplatesandsurfacesandthesearegiveninFigure1009

Potential plot E eld plotE eld

++++

ndashndashndashndash

xx x

+

x x

ndashx x

+ ndashx

x

+ +

x

x

Figure 1009 Electric field and electric potential at a distance from a charged surface

Potential due to one or more point charges

The potential due to one point charge can be determined byusingtheequationformula

V = kq r

Example 1

Determine the electric potential at a point 20 times 10-1 m fromthecentreofan isolatedconducting spherewithapointchargeof40pCinair

Solution

Using the formula

V = kq r we have

V90 10 9times( ) 40 10 12ndashtimes( )times

20 10 1ndashtimes( )------------------------------------------------------------------ 018 V= =

the potential at the point is 180 times 10-1 V

The potential due to a number of point charges can be determined by adding up the potentials due to individual point charges because the electric potential at any point outside a conducting sphere will be the same as if all the charge was concentrated at its centre

Example 2

Threepointchargesofareplacedattheverticesofaright-angledtriangleasshowninthediagrambelowDeterminetheabsolutepotentialat the+20μCchargeduetothetwoothercharges

- 6μC

+3μC +2μC

4 m

3 m

Solution

The electric potential of the +2 μC charge due to the ndash 6 μC charge is

V = (9 times 109 Nm2C-2 times -6 times 10-6 C) divide (radic 32 + 42) m = - 108 times 104 V

PHYSICS 2015indb 314 150514 338 PM

Fields

315

AH

L

Insummarywecanconcludethat

bull No work is done to move a charge along an equipotential

bull Equipotentials are always perpendicular to the electric lines of force

Figure1011and1012showsomeequipotential lines fortwo oppositely charged and identically positive spheresseparated by a distance

+ve ndashve

equipotential lines

Figure 1011 Equipotential lines between two opposite charges

equipotential lines

Figure 1012 Equipotential lines between two charges which are the same

+

ndash ndash ndash ndash

+ + +

40 V

30 V

20 V

10 V

50 V

Figure 1013 Equipotential lines between charged parallel plates

Figure1013showstheequipotentiallinesbetweenchargedparallel plates Throughout this chapter the similarities and differences between gravitational fields and electric fieldshavebeendiscussedTherelationshipsthatexistsbetweengravitationalandelectricquantitiesandtheeffectsofpointmassesandchargesissummarisedinTable1014

The electric potential of the +2 μC charge due to the +3 μC charge is

V = (9 times 109 Nm2C-2 times 3 times 10-6 C) divide 3m = 9 times 103 V

The net absolute potential is the sum of the 2 potentials- 108 times 104 V + 9 times 103 V =- 18 times 103 V

The absolute potential at the point is - 18 times 103 V

Equipotential surfacesRegions in space where the electric potential of a chargedistributionhasaconstantvaluearecalledequipotentials The placeswherethepotentialisconstantinthreedimensionsarecalled equipotential surfacesandwheretheyareconstantintwodimensionstheyarecalledequipotential lines

TheyareinsomewaysanalogoustothecontourlinesontopographicmapsInthiscasethegravitationalpotentialenergy is constant as amassmoves around the contourlines because the mass remains at the same elevationabovetheEarthrsquossurfaceThegravitationalfieldstrengthacts in a direction perpendicular to a contour line

Similarlybecausetheelectricpotentialonanequipotentiallinehas thesamevalueanelectric forcecandonoworkwhena testchargemovesonanequipotentialThereforethe electric field cannot have a component along anequipotentialandthusitmustbeeverywhereperpendiculartotheequipotentialsurfaceorequipotentiallineThisfactmakesiteasytoplotequipotentialsifthelinesofforceorlinesofelectricfluxofanelectricfieldareknown

Forexamplethereareaseriesofequipotentiallinesbetweentwoparallelplateconductorsthatareperpendiculartotheelectric fieldTherewill be a series of concentric circles(eachcirclefurtherapartthanthepreviousone)thatmapouttheequipotentialsaroundanisolatedpositivesphereThe lines of force and some equipotential lines for anisolatedpositivesphereareshowninFigure1010

Lines ofequipotential

Figure 1010 Equipotentials around an isolated positive sphere

PHYSICS 2015indb 315 150514 338 PM

Chapter 10

316

AH

L

Gravitational quantity Electrical quantityQ

uant

ities

g V∆x∆-------ndash= E V∆

x∆-------ndash=

Poin

t mas

ses a

nd ch

arge

s

V Gmr----

ndash= V 14πε0------------q

r---=

g G m

r2-----ndash= E 14πε0------------ q

r2-----=

F Gm1 m2

r2--------------= F 1

4πε0------------

q1q2

r2-----------=

Figure 1014 Formulas

Example

Deduce the electric potential on the surface of a gold nucleus that has a radius of 62 fm

Solution

Using the formula

V = kq r and knowing the atomic number of gold is 79 We will assume the nucleus is spherical and it behaves as if it were a point charge at its centre (relative to outside points)

V = 90 times 109 Nm2C-2 times 79 times 16 times 10-19 C divide 62 times 10-15 m = 18 times 107 V

The potential at the point is 18 MV

Example

Deduce the ionisation energy in electron-volts of theelectron in the hydrogen atom if the electron is in its ground state and it is in a circular orbit at a distance of 53 times 10-11 m from the proton

Solution

This problem is an energy coulombic circular motion question based on Bohrrsquos model of the atom (not the accepted quantum mechanics model) The ionisation energy is the energy required to remove the electron from the ground state to infinity The electron travels in a circular orbit and therefore has a centripetal acceleration The ionisation energy will counteract the coulombic force and the movement of the electron will be in the opposite direction to the centripetal force

Total energy = Ek electron + Ep due to the proton-electron interaction

ΣF = kqQ r2 = mv2 r and as such mv2 = = kqQ r

Therefore Ek electron = frac12 kqQ r

Ep due to the proton-electron interaction = - kqQ r

Total energy = frac12 kqQ r + - kqQ r = -frac12 kqQ r

= - 90 times 109 Nm2C-2 times (16 times 10-19 C)2 divide 53 times 10-11 m = -217 times 10-18 J

= -217 times 10-18 J divide 16 times 10-19 = -136 eV

The ionisation energy is 136 eV

Exercise 1021 ApointchargePisplacedmidwaybetweentwo

identicalnegativechargesWhichoneofthefollowingiscorrectwithregardstoelectricfieldandelectricpotentialatpointP

Electric field Electric potentialA non-zero zeroB zero non-zeroC non-zero non-zeroD zero zero

2 Twopositivechargedspheresaretiedtogetherinavacuumsomewhereinspacewheretherearenoexternal forces A has a mass of 25 g and a charge of20μCandBhasamassof15gandachargeof30μCThedistancebetweenthemis40cmTheyarethenreleasedasshowninthediagram

BEFORE AFTER

v1 v2

A B

(a) Determinetheirinitialelectricpotentialenergy in the before situation

(b) DeterminethespeedofsphereBafterrelease

3 Thediagrambelowrepresentstwoequipotentiallines in separated by a distance of 5 cm in a uniform electric field

+ + + + + + + +

ndash ndash ndash ndash ndash ndash ndash ndash

40 V

20 V 5 cm

PHYSICS 2015indb 316 150514 338 PM

Fields

317

AH

L

Determine the strength of the electric field

4 Thisquestionisabouttheelectricfieldduetoacharged sphere and the motion of electrons in that fieldThediagrambelowshowsanisolatedmetalsphereinavacuumthatcarriesanegativeelectriccharge of 60 μC

ndash

(a) Onthediagramdrawtheconventionalwayto represent the electric field pattern due to the charged sphere and lines to represent threeequipotentialsurfacesintheregionoutside the sphere

(b) Explainhowthelinesrepresentingtheequipotentialsurfacesthatyouhavesketchedindicatethatthestrengthoftheelectricfieldisdecreasingwithdistancefrom the centre of the sphere

(c) Theelectricfieldstrengthatthesurfaceofthe sphere and at points outside the sphere can be determined by assuming that the sphere acts as a point charge of magnitude 60 μC at its centre The radius of the sphere is 25 times 10ndash2 m Deduce that the magnitude of the field strength at the surface of the sphere is 86 times 107 Vmndash1

An electron is initially at rest on the surface of the sphere

(d) (i) Describethepathfollowedbytheelectronasitleavesthesurfaceofthesphere

(ii) Calculatetheinitialaccelerationofthe electron

5 Determinetheamountofworkthatisdoneinmovingachargeof100nCthroughapotentialdifferenceof150times102 V

6 Three identical 20 μC conducting spheres are placedatthecornersofanequilateraltriangleofsides 25 cm The triangle has one apex C pointing upthepageand2baseanglesAandBDeterminetheabsolutepotentialatB

7 Determinehowfaraparttwoparallelplatesmustbesituatedsothatapotentialdifferenceof250 times 102 V produces an electric field strength of 200 times 103 NC-1

8 Thegapbetweentwoparallelplatesis10times10-3mandthereisapotentialdifferenceof10times104 V betweentheplatesCalculate

i theworkdonebyanelectroninmovingfrom one plate to the other

ii thespeedwithwhichtheelectronreachesthe second plate if released from rest

iii theelectricfieldintensitybetweentheplates

9 Anelectronguninapicturetubeisacceleratedby a potential 25 times 103VDeterminethekineticenergygainedbytheelectroninelectron-volts

10 Determine the electric potential 20 times 10-2 m from achargeof-10times10-5 C

11 Determinetheelectricpotentialatapointmid-waybetweenachargeofndash20pCandanotherof+5pConthelinejoiningtheircentresifthecharges are 10 cm apart

12 During a thunderstorm the electric potential differencebetweenacloudandthegroundis10 times 109 V Determine the magnitude of the change in electric potential energy of an electron thatmovesbetweenthesepointsinelectron-volts

13 A charge of 15 μC is placed in a uniform electric fieldoftwooppositelychargedparallelplateswithamagnitudeof14times103 NC-1

(a) Determinetheworkthatmustbedoneagainstthefieldtomovethepointchargeadistance of 55 cm

(b) Calculatethepotentialdifferencebetweenthe final and initial positions of the charge

(c) Determinethepotentialdifferencebetweenthe plates if their separation distance is 15 cm

14 Duringaflashoflightningthepotentialdifferencebetweenacloudandthegroundwas12 times 109 V and the amount of transferred charge was32C

(a) Determinethechangeinenergyofthetransferred charge

(b) Iftheenergyreleasedwasallusedtoacceleratea1tonnecardeduceitsfinalspeed

(c) Iftheenergyreleasedcouldbeusedtomelticeat0degCdeducetheamountoficethatcould be melted

PHYSICS 2015indb 317 150514 338 PM

Chapter 10

318

AH

L

15 SupposethatwhenanelectronmovedfromAtoBin the diagram along an electric field line that the electric field does 36 times 10-19Jofworkonit

Determinethedifferencesinelectricpotential

(a) VB ndash VA(b) VC ndash VA(c) VC ndash VB

16 DeterminethepotentialatpointPthatislocatedatthecentreofthesquareasshowninthediagrambelow

- 6μC

+3μC +2μC

1m

1m

5μC

P

102 Fields at work

Essential idea Similar approaches can be taken inanalysingelectricalandgravitationalpotentialproblems

Understandingsbull Potentialandpotentialenergybull Potentialgradientbull Potentialdifferencebull Escapespeedbull Orbitalmotionorbitalspeedandorbitalenergybull Forcesandinverse-squarelawbehaviour

Orbital motion orbital speed and orbital energy

Although orbital motion may be circular elliptical orparabolic this sub-topic only dealswith circular orbitsThis sub-topic is not fundamentally new physics butan application that synthesizes ideas from gravitationcircularmotiondynamicsandenergy

TheMoonorbitstheEarthandinthissenseitisoftenreferredtoasasatelliteoftheEarthBefore1957itwastheonlyEarthsatelliteHoweverin1957theRussianslaunchedthefirstmanmadesatelliteSputnik1SincethisdatemanymoresatelliteshavebeenlaunchedandtherearenowliterallythousandsofthemorbitingtheEarthSomeareusedtomonitortheweathersome used to enable people to find accurately their position on thesurfaceoftheEarthmanyareusedincommunicationsandnodoubtsomeareusedtospyonothercountriesFigure1015showshowinprincipleasatellitecanbeputintoorbit

Theperson(whosesizeisgreatlyexaggeratedwithrespectto Earth) standing on the surface on the Earth throwssome stonesThe greater the speed with which a stoneis thrown the further it will land from her The pathsfollowedbythethrownstonesareparabolasByastretchof the imaginationwecanvisualisea situation inwhichastoneis thrownwithsuchaspeedthatbecauseof thecurvatureoftheEarthitwillnotlandonthesurfaceoftheEarthbutgointoldquoorbitrdquo(Path4onFigure1015)

NATURE OF SCIENCECommunication of scientific explanations The ability to apply field theory to the unobservable (charges) and the massively scaled (motion of satellites) required scientists to develop new ways to investigate analyse and report findings to a general public used to scientific discoveries based on tangible and discernible evidence (51)

copy IBO 2014

PHYSICS 2015indb 318 150514 338 PM

Fields

319

AH

L

1

2

34Earth

Figure 1015 Throwing a stone into orbit

TheforcethatcausesthestonestofollowaparabolicpathandtofalltoEarthisgravityandsimilarlytheforcethatkeeps the stone in orbit is gravity For circular motionto occur we have seen that a force must act at rightanglestothevelocityofanobjectthatistheremustbeacentripetalforceHenceinthesituationwedescribeherethe centripetal force for circular orbital motion about the EarthisprovidedbygravitationalattractionoftheEarth

We can calculate the speedwithwhich a stonemustbethrowninordertoputitintoorbitjustabovethesurfaceof the Earth

If the stone has mass m and speed v thenwehave fromNewtonrsquos2ndlaw

mv2

RE---------- G

MEm

RE2------------=

whereRE is the radius of the Earth and ME is the mass of the Earth

Bearinginmindthatg 0 GME

RE2--------= then

v g RE 10 64 106timestimes 8 103times= = =

Thatisthestonemustbethrownat8times103m sndash1

Clearly we are not going to get a satellite into orbit soclosetothesurfaceoftheEarthMovingatthisspeedthefrictionduetoairresistancewouldmeltthesatellitebeforeithadtravelledacoupleofkilometresInrealitythereforeasatelliteisputintoorbitabouttheEarthbysendingitattachedtoarocketbeyondtheEarthrsquosatmosphereandthengivingitacomponentofvelocityperpendiculartoaradialvectorfromtheEarthSeeFigure1016

Earth Satellite carried by rocket to hereSatellite orbit

Tangential component of velocity

Figure 1016 Getting a satellite into orbit

Keplerrsquos third law(This work of Kepler and Newtonrsquos synthesis of the work is an excellent example of the scientific method and makes for a good TOK discussion)

In 1627 Johannes Kepler (1571-1630)publishedhis lawsof planetary motion The laws are empirical in natureandwere deduced from the observations of theDanishastronomer Tycho de Brahe (1546-1601)The third lawgives a relationship between the radius of orbit R of a planet and its period TofrevolutionabouttheSunThelawisexpressedmathematicallyas

2

3 constantTR

=

We shall nowuseNewtonrsquos LawofGravitation to showhowitisthattheplanetsmoveinaccordancewithKeplerrsquosthirdlaw

InessenceNewtonwasabletousehislawofgravitytopredict the motion of the planets since all he had to do was factor theF givenby this law intohis second law F = ma to find their accelerations and hence their future positions

InFigure1017theEarthisshownorbitingtheSunandthedistancebetweentheircentresisR

RSun

Earth

F es

F se

Figure 1017 Planets move according to Keplerrsquos third law

Fes is the force that the Earth exerts on the Sun and Fse is the force that the Sun exerts on the Earth The forces are equalandoppositeandtheSunandtheEarthwillactuallyorbitaboutacommoncentreHoweversince theSun issoverymuchmoremassivethantheEarththiscommoncentrewillbeclosetothecentreoftheSunandsowecanregard the Earth as orbiting about the centre of the Sun Theotherthingthatweshallassumeisthatwecanignorethe forces that the other planets exert on the Earth (This wouldnotbeawisethingtodoifyouwereplanningtosend a space ship to theMoon for example)We shallalso assume that we have followed Newtonrsquos exampleandindeedprovedthataspherewillactasapointmasssituated at the centre of the sphere

Kepler had postulated that the orbits of the planets areellipticalbut since theeccentricityof theEarthrsquosorbit issmallweshallassumeacircularorbit

PHYSICS 2015indb 319 150514 338 PM

Chapter 10

320

AH

L

TheaccelerationoftheEarthtowardstheSunisa = Rω2

whereω ω 2πT------=

Hencea R 2π

T------ times

2 4π2 RT 2------------- = =

But the acceleration is given by Newtonrsquos Second Law F = mawhereFisnowgivenbytheLawofGravitationSo in this situation

F maGMs Me

R2-------------------= = butwealsohavethat

a a R 2πT

------ times

2 4π2 RT 2------------- = = and m = Me so that

G MsMe

R2------------------- Me4π2R

T2-------------times

GMs

4π2-----------rArr R3

T2------= =

Butthequantity

isaconstantthathasthesamevalueforeachoftheplanetssowehaveforalltheplanetsnotjustEarththat

R3

T2------ k =

wherekisaconstantWhichisofcourseKeplerrsquosthirdlaw

ThisisindeedanamazingbreakthroughItisdifficulttorefutethe idea that all particles attract each other in accordance withtheLawofGravitationwhenthelawisabletoaccountfortheobservedmotionoftheplanetsabouttheSun

The gravitational effects of the planets upon each othershould produce perturbations in their orbits Such is the predictivepoweroftheUniversalGravitationalLawthatit enabled physicists to compute these perturbations The telescopehadbeeninventedin1608andbythemiddleofthe 18th Century had reached a degree of perfection in design that enabled astronomers to actually measure the orbital perturbations of the planets Their measurements werealways inagreementwith thepredictionsmadebyNewtonrsquoslawHoweverin1781anewplanetUranuswasdiscovered and the orbit of this planet did not fit withthe orbit predicted by Universal Gravitation Such wasthe physicistrsquos faith in theNewtonianmethod that theysuspectedthatthediscrepancywasduetothepresenceofayetundetectedplanetUsingtheLawofGravitationtheFrenchastronomerJLeverrier and the English astronomer J C Adamswereable tocalculate justhowmassive thisnewplanetmustbeandalsowhereitshouldbeIn1846the planetNeptunewas discovered justwhere they hadpredicted Ina similarwaydiscrepancies in theorbitofNeptune ledto thepredictionandsubsequentdiscoveryin1930oftheplanetPlutoNewtonrsquosLawofGravitation

had passed the ultimate test of any theory it is not only abletoexplainexistingdatabutalsotomakepredictions

Satellite energyWhenasatelliteisinorbitaboutaplanetitwillhavebothkineticenergyandgravitationalpotentialenergySupposewe consider a satellite ofmass m that is in an orbit of radius r about a planet of mass M

Thegravitationalpotentialduetotheplanetatdistancer from its centre is

ThegravitationalpotentialenergyofthesatelliteVsat

is therefore GMe m

r----------------ndash

ThatisVsat GMem

r----------------ndash=

Thegravitationalfieldstrengthatthesurfaceoftheplanetisgivenby

g 0GMe

Re2------------=

Hencewecanwrite

Vsat g0 Re

2

rndash=

m

ThekineticenergyofthesatelliteKsatisequaltofrac12mv2wherev is its orbital speed

Byequatingthegravitationalforceactingonthesatellitetoitscentripetalaccelerationwehave

G Mem

r2---------------- mv2

r---------- mv2hArr

GMemr----------------= =

Fromwhich12---

mv2 12---

GMemr----------------times=

= 0 e

2g R

r

2m

Which is actually quite interesting since it shows thatirrespective of the orbital radius the KE is numericallyequal to half the PE Also the total energy E tot K sat Vsat+ 1

2---G Me m

r----------------times G Me m

r----------------ndash + 1

2---G Me m

r----------------ndash= = =of the satelliteisalwaysnegativesince

E tot K sat Vsat+ 12---

G Me mr----------------times

G Me mr----------------ndash

+ 12---

G Me mr----------------ndash= = =

PHYSICS 2015indb 320 150514 338 PM

Fields

321

AH

L

The energies of an orbiting satellite as a function of radial distancefromthecentreofaplanetareshownplottedinFigure1018

12

100806

0402

0ndash02

ndash04

ndash06ndash08ndash10

ndash12

2 4 6 8 10 12

ener

gya

rbita

ry u

nits

distance timesR

kinetic energytotal energy

potential energy

Figure 1018 Energy of an orbiting satellite as a function of distance from the centre of a planet

WeightlessnessSuppose that you are inanelevator(lift)thatisdescendingat constant speedandyou let goof abook thatyouareholdinginyourhandThebookwillfalltothefloorwithaccelerationequaltotheaccelerationduetogravityIfthecablethatsupportstheelevatorweretosnap(asituationthatItrustwillneverhappentoanyofyou)andyounowletgothebookthatyouareholdinginyourotherhandthis book will not fall to the floor - it will stay exactlyin linewithyourhandThis isbecause thebook isnowfallingwith the sameaccelerationas theelevatorandassuch the book cannot ldquocatchrdquo up with the floor of theelevatorFurthermoreifyouhappenedtobestandingonasetofbathroomscalesthescaleswouldnowreadzero-youwouldbeapparentlyweightlessItisthisideaoffreefallthatexplainstheapparentweightlessnessofastronautsin an orbiting satellite These astronauts are in free fall inthesensethattheyareacceleratingtowardsthecentre of the Earth

It is actually possible to define theweight of a body inseveral different waysWe can define it for example asthegravitationalforceexertedonthebodybyaspecifiedobject such as the EarthThiswe have seen thatwe doinlotsofsituationswherewedefinetheweightasbeingequaltomgIfweusethisdefinitionthenanobjectinfreefallcannotbydefinitionbeweightlesssinceitisstillinagravitationalfieldHoweverifwedefinetheweightofanobjectintermsofaldquoweighingrdquoprocesssuchasthereadingonasetofbathroomscaleswhichineffectmeasuresthecontact force between the object and the scales thenclearlyobjectsinfreefallareweightlessOnenowhasto

askthequestionwhetherornotitispossibleForexampletomeasurethegravitationalforceactingonanastronautin orbit about the Earth

WecanalsodefineweightintermsofthenetgravitationalforceactingonabodyduetoseveraldifferentobjectsForexample for an object out in space its weight could bedefined in terms of the resultant of the forces exerted on it bytheSuntheMoontheEarthandalltheotherplanetsin the Solar System If this resultant is zero at a particular pointthenthebodyisweightlessatthispoint

Inviewofthevariousdefinitionsofweightthatareavailabletousitisimportantthatwhenweusethewordldquoweightrdquoweareawareofthecontextinwhichitisbeingused

Example

Calculate the height above the surface of the Earth atwhichageo-stationarysatelliteorbits

Solution

A geo-stationary satellite is one that orbits the Earth in such a way that it is stationary with respect to a point on the surface of the Earth This means that its orbital period must be the same as the time for the Earth to spin once on its axis ie 24 hours

From Keplerrsquos third law we have G Ms

4π2----------- R3

T2------=

That is

RMe

mh

using the fact that the force of attraction between the satellite and the Earth is given by

FG Mem

R2----------------=

and that F = ma

where m is the mass of the satellite and a 4π2R

T2-------------=

we have

GMem

R2---------------- m 4π2R

T2-------------times

GMe

4π2------------rArr R3

T2------= =

PHYSICS 2015indb 321 150514 338 PM

Chapter 10

322

AH

L

Now the mass of the Earth is 60 times 1024 kg and the period T measured in seconds is given by T = 86400 s

So substitution gives R = 42 times 106 m

The radius of the Earth is 64 times 106 m so that the orbital height h is about 36 times 107 m

Example

Calculatetheminimumenergyrequiredtoputasatelliteofmass500kgintoanorbitthatisasaheightequaltotheEarthrsquosradiusabovethesurfaceoftheEarth

Solution

We have seen that when dealing with gravitational fields and potential it is useful to remember that

g 0G M

Re2---------= or g0 Re

2 GM=

The gravitational potential at the surface of the Earth is

g 0ndash Re GMRe

---------ndash=

The gravitational potential at a distance R

from the centre of the Earth is

=

The difference in potential between the surface of the Earth and a point distance R from the centre is therefore

V∆ g 0Re 1ReR------ndash

=

If R 2Re= then V∆g0 Re

2------------=

This means that the work required to ldquoliftrdquo the satellite into orbit is g0Rm where m is the mass of the satellite This is equal to

10 32 106 500timestimestimes = 16000 MJ

However the satellite must also have kinetic energy in order to orbit Earth This will be equal to

g 0 mRe2

2R-----------------g 0mRe

2

4----------------- 8000 MJ= =

The minimum energy required is therefore

24000 MJ

Exercise 1031 The speed needed to put a satellite in orbit does

not depend on

A the radius of the orbitB theshapeoftheorbitC thevalueofgattheorbitD the mass of the satellite

2 EstimatethespeedofanEarthsatellitewhoseorbitis400kmabovetheEarthrsquossurfaceAlsodetermine the period of the orbit

3 Calculatethespeedofa200kgsatelliteorbitingthe Earth at a height of 70 times 106 m

Assume that g = 82 m sndash2 for this orbit

4 TheradiioftwosatellitesXandYorbitingthe Earth are 2r and 8rwherer is the radius of the Earth Calculate the ratio of the periods of revolutionofXtoY

5 A satellite of mass mkgissentfromEarthrsquossurfaceinto an orbit of radius 5RwhereR is the radius of theEarthWritedownanexpressionfor

(a) thepotentialenergyofthesatelliteinorbit

(b) thekineticenergyofthesatelliteinorbit

(c) theminimumworkrequiredtosendthesatellitefromrestattheEarthrsquossurfaceintoits orbit

6 A satellite in an orbit of 10rfallsbacktoEarth(radius r)afteramalfunctionDeterminethespeedwithwhichitwillhittheEarthrsquossurface

7 The radius of the moon is frac14 that of the Earth AssumingEarthandtheMoontohavethesamedensitycomparetheaccelerationsoffreefallatthe surface of Earth to that at the surface of the Moon

8 UsethefollowingdatatodeterminethegravitationalfieldstrengthatthesurfaceoftheMoon and hence determine the escape speed from the surface of the Moon

Mass of the Moon = 73 times 1022kg

RadiusoftheMoon=17times106 m

PHYSICS 2015indb 322 150514 338 PM

Wave Phenomena

301

AH

L

Sofarwehavebeenassumingthattheeyeisarectangularslit whereas clearly it is a circular aperture and so we must use the formula

θ 122 λb--------------=

As mentioned above the angle θ is sometimes called the resolving power but should more accurately be called the minimum angle of resolution (θmin)

Clearly the smaller θ the greater will be the resolving power

The diffraction grating is a useful for differentiating closely spacedlinesinemissionspectraLikeaprismspectrometerit can disperse a spectrum into its components but it is better suited because it has higher resolution than the prism spectrometer

If λ1 and λ2 are two nearly equal wavelengths that canbarelybedistinguishedtheresolvanceorresolvingpowerofthegratingisdefinedas

R = λ _______ (λ2 ndash λ 1) = λ ___ Δλ

Where λ asymp λ1 asymp λ2

Therefore thediffractiongratingwithahighresolvancewill be better suited in determining small differences in wavelength

If the diffraction grating has Nlinesbeingilluminateditcan be shown that the resolving power of the mth order diffraction is given by

R = mN

Obviouslytheresolvancebecomesgreaterwiththeordernumber m and with a greater number of illuminated slits

Example

A benchmark for the resolving power of a grating is the sodium doublet in the sodium emission spectrum Two yellow lines in itsspectrumhavewavelengthsof58900nmand58959nm

(a) Determinetheresolvanceofagratingifthegivenwavelengths are to be distinguished from each other

(b) Howmanylinesinthegratingmustbeilluminated in order to resolve these lines in the second order spectrum

Solution

(a) R = λ ___ Δλ = 58900nm __________________ (58900ndash58959)nm

= 589nm _______ 059nm = 10 times 103

(b) N = R __ m = 10 times 10 3 ________ 2 =50times102 lines

It has been seen that diffraction effectively limits the resolving power of optical systems This includes such systems as the eye telescopes and microscopes Theresolving power of these systems is dealt with in the next section This section looks at links between technology and resolving power when looking at the very distant and when looking at the very small

Radio telescopes

The average diameter of the pupil of the human eye is about 25 mm This means that the eye will justresolve two point sources emitting light of wavelength 500 nm if their angular separation at the eye is 7

43

50 10122 24 1025 10

θminus

minusminus

times= times = times

times rad

If the eye were to be able to detect radio waves of wavelength 015m then tohave the same resolvingpower thepupilwould have to have a diameter of about 600 m Clearly this is nonsense but it does illustrate a problem facingastronomers who wish to view very distant objects such as quasarsandgalaxies thatemitradiowavesConventionalradio telescopes consist of a large dish typically 25 min diameter Even with such a large diameter the radiowavelength resolving power of the telescope is much less thantheopticalresolvingpowerofthehumaneyeLetuslook at an example

Example

The Galaxy Cygnus A can be resolved optically as an elliptically shaped galaxy However it is also a strongemitterof radiowavesofwavelength015mTheGalaxyisestimatedtobe50times1024mfromEarthUseofaradiotelescope shows that the radio emission is from two sources separated by a distance of 30 times 1021m Estimate thediameterofthedishrequiredtojustresolvethesources

Solution

The angle θ that the sources resolve at the telescope is given by

θ 21

424

30 10 60 1050 10

θ minustimes= = times

times rad

and 4

122 122 015 300060 10

d λθ minus

times= = =

timesm = 30 km

A radio telescope dish of this size would be impossible tomake letalonesupportThisshowsthatasingledishtype radio telescope cannot be used to resolve the sources and yet they were resolved To get round the problemastronomers use two radio telescopes separated by a large

PHYSICS 2015indb 301 150514 337 PM

Chapter 9

302

AH

L

distance The telescopes view the same objects at the same time and the signals that each receive from the objects are simultaneously superimposed The result of the superposition of the two signals is a two-slit interference pattern The pattern has much narrower fringe spacing than that of the diffraction pattern produced by either telescope on its own hence producing a much higherresolvingpowerWhentelescopesareusedlikethistheyare called a stellar interferometer

InSocorroinNewMexicothereisastellarinterferometerthat consists of 27 parabolic dishes each of diameter25 m arranged in a Y-shape that covers an area of 570km2Thisisaso-calledVeryLargeArray(VLA)Evenhigher resolution can be obtained by using an array of radio telescopes in observatories thousands of kilometres apart A system that uses this so-called technique oflsquovery-long-baseline interferometryrsquo (VLBI) is known asalsquovery-long-baselinearrayrsquo(VLBA)WithVLBAaradiowavelength resolving power can be achieved that is 100 timesgreaterthanthebestopticaltelescopesEvenhigherresolving power can be achieved by using a telescope that isinasatelliteorbitingEarthJapanrsquosInstituteofSpaceandAstronautical Science (ISAS) launched sucha system inFebruary 1997TheNationalAstronomicalObservatoryof Japan the National Science Foundationrsquos NationalRadio Astronomy Observatory (NRAO) the CanadianSpaceAgency theAustraliaTelescopeNationalFacilitythe European VLBI Network and the Joint Instituteback this project forVeryLongBaseline Interferometryin EuropeThis project is a very good example of howInternationalismcanoperateinPhysics

Electron microscope

Telescopes are used to look at very distant objects that are very large but because of their distance fromus appearverysmallMicroscopesontheotherhandareusedtolookat objects that are close to us but are physically very small Aswehave seen justmagnifyingobjects that ismakingthemappearlargerisnotsufficientonitsowntogaindetailabouttheobjectfordetailhighresolutionisneeded

Figure928isaschematicofhowanoptical microscope is usedtoviewanobjectandFigure929isaschematicofatransmission electron microscope(TEM)

Optical lens system

bright light source

glass slide containing specimen (object)

eye

Figure 928 The principle of a light microscope

magnetic lens system

electron source

thin wafer of material

ScreenCCD

Figure 929 The principle of an electron microscope

In the optical microscope the resolving power isdetermined by the particular lens system employed and the wavelength λ of the light used For example twopoints in the sample separated by a distance d will just be resolved if

2d mλ=

where m is a property of the lens system know as the numerical aperture In practice the largest value of m obtainable is about 16 Hence if the microscope slideis illuminatedwith light ofwavelength 480 nm a goodmicroscope will resolve two points separated by a distance dasymp15times10-7masymp015micromPointscloser together thanthiswillnotberesolvedHoweverthisisgoodenoughtodistinguishsomevirusessuchastheEbolavirus

Clearlythesmallerλ the higher the resolving power and this is where the electron microscope comes to the fore The electron microscope makes use of the wave nature of electrons(see1315)IntheTEMelectronspassthrougha wafer thin sample and are then focused by a magnetic field onto a fluorescent screen orCCD (charge coupleddevicesee142)Electronsusedinanelectronmicroscopehavewavelengthstypicallyofabout5times10-12mHoweverthe numerical aperture of electron microscopes is considerablysmaller thanthatofanopticalmicroscopetypicallyabout002NonethelessthismeansthataTEMcanresolvetwopointsthatareabout025nmapartThisresolving power is certainly high enough to make out the shape of large molecules

Another type of electron microscope uses a techniqueby which electrons are scattered from the surface of the sample The scattered electrons and then focused as in theTEM to form an image of the surfaceThese so-calledscanningelectronmicroscopes(SEM)havealowerresolving power than TEMrsquos but give very good threedimensional images

The eye

We saw in the last section that the resolving power of the human eye is about 2 times 10-4 rad Suppose that youare looking at car headlights on a dark night and the car

PHYSICS 2015indb 302 150514 337 PM

Wave Phenomena

303

AH

L

is a distance D away If the separation of the headlight is say 15 m then the headlights will subtend an angle θ 15

Dθ =

at your eye Assuming an average wavelength of

500 nm your eye will resolve the headlights into twoseparatesourcesifthisangleequals2times10-4 rad This gives D=75kmInotherwordsifthecarisapproachingyouona straight road then you will be able to distinguish the two headlightsasseparatesourceswhenthecaris75kmawayfrom you Actually because of the structure of the retina and optical defects the resolving power of the average eye is about 6 times 10-4 rad This means that the car is more likely tobe25kmawaybeforeyouresolvetheheadlights

Astronomical telescope

Letusreturntotheexampleofthebinarystarsdiscussedatthe beginning of this section on resolution The stars Kruger A and B form a binary system The average separation of the stars is 14 times 1012mandtheiraveragedistancefromEarthis12times1017mWhenviewedthroughatelescopeonEarththe system will therefore subtend an angle

12

17

14 1012 10

θtimes

=times

= 12times10-5 rad at the objective lens of the telescope Assuming that the average wavelength of the light emitted bythestarsis500nmthenifthetelescopeistoresolvethesystem into two separate images it must have a minimum diameter Dwhere12times10-5 =

7122 500 10D

minustimes times

This gives D=0050m which is about 5 cm So thisparticular system is easily resolved with a small astronomical telescope

Exercise 961 ItissuggestedthatusingtheISASVLBAitwould

bepossibletoldquoseerdquoagrainofriceatadistanceof5000kmEstimatetheresolvingpoweroftheVLBA

2 Thedistancefromtheeyelenstotheretinais20mm The light receptors in the central part of the retinaareabout5times10-6apartDeterminewhetherthe spacing of the receptors will allow for the eye to resolve the headlights in the above discussion whentheyare25kmfromtheeye

3 ThediameterofPlutois23times106 m and its averagedistancefromEarthis60times1012 m EstimatetheminimumdiameteroftheobjectiveofatelescopethatwillenablePlutotobeseenasadisc as opposed to a point source

95 Doppler effect

Essential ideaTheDopplereffectdescribesthephenomenonofwavelengthfrequencyshiftwhenrelativemotionoccurs

Understandingsbull TheDopplereffectforsoundwavesandlightwaves

The Doppler effect for sound waves and light waves

Consider two observers A and B at rest with respect to asoundsourcethatemitsasoundofconstantfrequencyf Clearly both observers will hear a sound of the same frequencyHoweversupposethatthesourcenowmovesat constant speed towards A A will now hear a sound of frequencyfA that is greater than f and B will hear a sound of frequency fB that is less than f This phenomenon is known as the Doppler Effect orDopplerPrinciple after C J Doppler(1803-1853)

The same effect arises for an observer who is either moving towards or away from a stationary source

Figure 930 shows the waves spreading out from a stationary source that emits a sound of constant frequency f The observersAandBhearasoundofthesamefrequency

A B

wavefront

S

Figure 930 Sound waves from a stationary source

Suppose now that the source moves towards A withconstant speed vFigure9310(a)showsasnapshotofthenew wave pattern

NATURE OF SCIENCETechnology Although originally based on physical observations of the pitch of fast moving sources of sound the Doppler effect has an important role in many different areas such as evidence for the expansion of the universe and generating images used in weather reports and in medicine (55)

copy IBO 2014

PHYSICS 2015indb 303 150514 337 PM

Chapter 9

304

AH

L

sourceA B

smallerwavelength

largerwavelength

Figure 931 (a) Sound waves from a moving source

The wavefronts are now crowded together in the direction of travel of the source and stretched out in the opposite direction This is why now the two observers will now hear notesofdifferentfrequenciesHowmuchthewavesbunchtogether and how much they stretch out will depend on the speed vEssentially A

A

cfλ

= and BB

cfλ

= where λA lt λB and v is the speed of sound

IfthesourceisstationaryandAismovingtowardsitthenthe waves from the source incident on A will be bunched up If A is moving away from the stationary source then the waves from the source incident on A will be stretched out

Christian Doppler(1803ndash1853)actuallyappliedtheprinciple(incorrectlyasithappens)totryandexplainthecolourofstars However the Doppler effect does apply to light aswellastosoundIfalightsourceemitsalightoffrequencyf then if it is moving away from an observer the observer willmeasurethelightemittedashavingalowerfrequencythan fSincethesensationofcolourvisionisrelatedtothefrequencyoflight(bluelightisofahigherfrequencythanred light) light emitted by objects moving way from anobserver is often referred to as being red-shifted whereas if the object is moving toward the observer it is referred to as blue-shiftedThisideaisusedinOptionE(Chapter16)

We do not need to consider here the situations where either the source or the observer is accelerating In a situation for example where an accelerating source is approaching astationaryobserverthentheobserverwillhearasoundof ever increasing frequency This sometimes leads toconfusion in describing what is heard when a source approaches passes and then recedes from a stationaryobserver Suppose for example that you are standingon a station platform and a train sounding its whistle is approaching at constant speed What will you hear as the train approaches and then passes through the station As the train approaches you will hear a sound of constant pitch but increasing loudness The pitch of the sound will be greater than if the train were stationary As the train passes through the station you will hear the pitch change at the momentthetrainpassesyoutoasoundagainofconstantpitch The pitch of this sound will be lower than the sound of the approaching train and its intensity will decrease as the train recedes from you What you do not hear is a sound of increasing pitch and then decreasing pitch

The Doppler equations for sound

Although you will not be expected in an IB examination to derivetheequationsassociatedwithaspectsoftheDopplereffectyouwillbeexpectedtoapplythemForcompletenesstherefore thederivationoftheequationsassociatedwiththeDopplereffectasoutlinedaboveisgivenhere

S primeS O

vs

Figure 931 (b)

InFigure931(b)theobserverOisatrestwithrespecttoasourceofsoundSismovingwithconstanspeedvs directly towards O The source is emitting a note of constant frequencyf and the speed of the emitted sound is v

S shows the position of the source ∆t later When the sourceisatresttheninatime∆t the observer will receive f∆t waves and these waves will occupy a distance v∆t ie

v f

λ =f∆tv∆t =

(Because of the motion of the source this number of waves will now occupy a distance (v∆t ndash vs∆t)The lsquonewrsquowavelength is therefore

v ndash vs

fλrsquo =

f∆tv∆t ndash vs∆t

=

If fisthefrequencyheardbyOthen

vfλ

= or

vf

λ = = sv vfminus

From which

s

vf fv v

=minus

Dividingthroughbyv gives

Equation1111

1 sf f v

v=

minus

If the source is moving away from the observer then we have

s

vf fv v

=minus

1

1 sf vv

=+

Equation112

PHYSICS 2015indb 304 150514 337 PM

Wave Phenomena

305

AH

L

We now consider the case where the source is stationary and the observerismoving towardsthesource with speed v0 In this situation the speed of the sound waves as measured by the observer will be v0 + v We therefore have that

v0 + v = ffraslλ = vfλ

times

From which

o1 vf fv

= +

Equation113

If the observerismovingawayfromthesource then

o1 vf fv

= minus

Fromequation(113)wehavethat

o o1 v vf f f f f fv v

∆ = minus = + minus =

Equation114

Thevelocitiesthatwerefertointheaboveequationsarethevelocities with respect to the medium in which the waves fromthesourcetravelHoweverwhenwearedealingwithalight source it is the relative velocity between the source and the observer that we must consider The reason for this is that lightisuniqueintherespectthatthespeedofthelightwavesdoes not depend on the speed of the source All observers irrespective of their speed or the speed of the source will measure the same velocity for the speed of light This is oneof thecornerstonesof theSpecialTheoryofRelativitywhichisdiscussedinmoredetailinOptionH(Chapter18)When applying the Doppler effect to light we aremainlyconcerned with the motion of the source We look here only at the situation where the speed of the source v is much smaller than the speed of light cinfreespace(v ltlt c)Underthesecircumstanceswhenthesourceismovingtowardstheobserverequation111becomes

vf f f fc

minus = ∆ = Equation115

andwhenthesourceismovingawayfromtheobserverequation112becomes vf f f f

cminus = ∆ = minus

Provided that v ltlt c these same equations apply for astationary source and moving observer

We look at the following example and exercise

Example

Asourceemitsasoundoffrequency440HzItmovesinastraight line towards a stationary observer with a speed of 30ms-1Theobserverhearsasoundoffrequency484HzCalculate the speed of sound in air

Solution

We use equation 111 and substitute f= 484 Hz f = 440 Hz

and vs = 30 m s-1

therefore 1

484 440 301v

=minus

such that 30 4401

484vminus = to

give v = 330 m s-1

Example

A particular radio signal from a galaxy is measured as havingafrequencyof139times109HzThesamesignalfromasourceinalaboratoryhasafrequencyof142times109Hz

Suggestwhy thegalaxy ismovingaway fromEarthandcalculateitsrecessionspeed(iethespeedwithwhichitismovingawayfromEarth)

Solution

The fact that the frequency from the moving source is less than that when it is stationary indicates that it is moving away from the stationary observer ie Earth

Using vf fc

∆ = we have

8 96

9

3 10 (142 139) 10 634 10142 10

c fvf∆ times times minus times

= = = timestimes

m s-1

It is usual when dealing with the Doppler effect of light to express speeds as a fraction of c So in this instance we have v = 0021 c

Using the Doppler effectWe have seen in the above example and exercise how the Dopplereffectmaybeusedtomeasuretherecessionspeedof distant galaxies The effect is also used to measure speed inothersituationsHerewewilllookatthegeneralprincipleinvolved in using the Doppler effect to measure speedFigure932showsasource(thetransmitter)thatemitseithersound or em waves of constant frequency f The waves

PHYSICS 2015indb 305 150514 337 PM

Chapter 9

306

AH

L

fromthesourceare incidentonareflectorthat ismovingtowards the transmitter with speed vThereflectedwavesare detected by the receiver placed alongside the transmitter

reector

v

transmitter

receiver

f f

f f

Figure 932 Using the Doppler effect to measure speed

We shall consider the situation where v ltlt c where c is the speed of the waves from the transmitter

For the reflector receiving waves from the transmitterit is effectively an observer moving towards a stationary sourceFromequation(114) itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation116

For the receiver receivingwaves from the reflector it iseffectively a stationary observer receiving waves from a movingsourceFromequation(115)itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation117

Ifweaddequations(116)and(117)wegetthatthetotalDopplershiftatthereceiver∆f is

v vf f f f fc c

minus = ∆ = +

But 1 vf fc

= +

hence

1 v v vf f fc c c

∆ = + +

But since v ltlt cwe can ignore the term2

2

vc when we

expandthebracketintheaboveequation

Therefore we have2vf fc

∆ = Equation118

If v asymp c then we must use the full Doppler equationsHowever for em radiationwewill alwaysonly considersituations in which v ltlt c

Example

Thespeedofsoundinbloodis1500times103 m s-1 Ultrasound offrequency100MHzisreflectedfrombloodflowinginan arteryThe frequencyof the reflectedwaves receivedbackatthetransmitteris105MHzEstimatethespeedofthebloodflowintheartery

Solution

Using equation (118) we have

6 63

2005 10 1015 10

vtimes = times

timesto give v asymp 36 m s-1 (We have assumed that the ultrasound is incident at right angles to the blood flow)

Exercise 971 Judy is standing on the platform of a station A

high speed train is approaching the station in a straight line at constant speed and is sounding its whistleAsthetrainpassesbyJudythefrequencyof the sound emitted by the whistle as heard by Judychangesfrom640Hzto430HzDetermine

(a) thespeedofthetrain

(b) thefrequencyofthesoundemittedbythewhistle as heard by a person on the train (Speedofsound=330ms-1)

2 AgalaxyismovingawayfromEarthwithaspeedof00500cThewavelengthofaparticularspectral line in light emitted by atomic hydrogen inalaboratoryis656times10-7 m Calculate the valueofthewavelengthofthislinemeasuredinthelaboratoryinlightemittedfromasourceofatomic hydrogen in the galaxy

PHYSICS 2015indb 306 150514 337 PM

10 Fields

Contents

101 ndash Describing fields

102 ndash Fields at work

Essential Ideas

Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Similar approaches can be taken in analysing electrical and gravitational potential problems copy IBO 2014

PHYSICS 2015indb 307 150514 337 PM

Chapter 10

308

AH

L

101 Describing fields

Essential idea Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Understandingsbull Gravitationalfieldsbull Electrostaticfieldsbull Electricpotentialandgravitationalpotentialbull Fieldlinesbull Equipotentialsurfaces

Gravitational potentialWehaveseenthatifweliftanobjectofmassm to a height h abovethesurfaceoftheEarththenitsgainingravitationalpotential energy is mghHoweverthisisbynomeansthefullstoryForastartwenowknowthatgvarieswithh and also the expression really gives a difference in potentialenergybetweenthevaluethattheobjecthasattheEarthrsquossurfaceandthevaluethatithasatheighthSowhatwereallyneedisazeropointCanwefindapointwherethepotentialenergyiszeroandusethispointfromwhichtomeasure changes in potential energy

The point that is chosen is in fact infinity At infinity the gravitationalfieldstrengthofanyobjectwillbezeroSoletusseeifwecandeduceanexpressionforthegaininpotentialenergyofanobjectwhenitisldquoliftedrdquofromthesurfaceoftheEarthtoinfinityThisineffectmeansfindingtheworknecessarytoperformthistask

δr

r r δr+

r infin=

mg

AB

Me

Figure 1001 Gravitational forces

Inthediagramweconsidertheworknecessarytomovethe particle of mass m a distance δr in thegravitationalfield of the Earth

The force on the particle at A is FG Mem

r2----------------=

IftheparticleismovedtoB then since δr isverysmallwe can assume that the field remains constant over thedistance ABTheworkδW done againstthegravitationalfieldoftheEarthinmovingthedistanceABis

δWGMem

r2----------------δrndash=

(rememberthatworkdoneagainstaforceisnegative)

TofindthetotalworkdoneW in going from the surface oftheEarthtoinfinitywehavetoaddalltheselittlebitsofworkThisisdonemathematicallybyusingintegralcalculus

WG Me m

r2----------------ndash

rdR

infin

int G Me m 1

r2----- rd

R

infin

intndash G Me m 1r---ndash

R

infinndash= = =

GMe m 0 1R---ndash

ndashndash=

G MemR----------------ndash=

HencewehavewhereRistheradiusoftheEarththattheworkdonebythegravitationalfieldinmovinganobjectofmass m from R(surfaceoftheEarth)toinfinityisgivenby

WGMe m

R---------------- ndash=

We can generalise the result by calculating the worknecessary per unit mass to take a small mass fromthe surface of the Earth to infinity This we call thegravitational potential Vie

Wewouldgetexactlythesameresultifwecalculatedtheworkdonebythefieldtobringthepointmassfrominfinityto the surface of Earth In this respect the formal definition ofgravitationalpotentialatapointinagravitationalfieldis therefore defined as the work done per unit mass in bringing a point mass from infinity to that point

Clearly then the gravitational potential at any point inthe Earthrsquos field distance r from the centre of the Earth (providingr gt R)is

VGMe

r------------ ndash=

Thepotentialisthereforeameasureoftheamountofworkthathastobedonetomoveparticlesbetweenpointsinagravitationalfieldanditsunit is theJkgndash1 We also note thatthepotentialisnegativesothatthepotentialenergy

NATURE OF SCIENCEParadigm shift The move from direct observable actions being responsible for influence on an object to acceptance of a fieldrsquos ldquoaction at a distancerdquo required a paradigm shift in the world of science (23)

copy IBO 2014

PHYSICS 2015indb 308 150514 337 PM

Fields

309

AH

L

aswemoveawayfromtheEarthrsquossurfaceincreasesuntilitreachesthevalueofzeroatinfinity

IfthegravitationalfieldisduetoapointmassofmassmthenwehavethesameexpressionasaboveexceptthatMe is replaced by m andmustalsoexclude thevalueof thepotential at the point mass itself ie at r = 0

WecanexpressthegravitationalpotentialduetotheEarth(orduetoanysphericalmass)intermsofthegravitationalfield strength at its surface

AtthesurfaceoftheEarthwehave

g 0ndash Re GMRe

---------ndash= e

Sothat

g 0Re2 GM= e

Hence at a distance r from the centre of the Earth the gravitationalpotentialVcanbewrittenas

VGMe

r------------ndashg 0 Re

2

r------------ndash= =

The potential at the surface of the Earth

(r = Re)istherefore-g0Re

It is interesting to see how the expression for thegravitational potential ties in with the expression mgh ThepotentialatthesurfaceoftheEarthis-g0Re (see the exampleabove)andataheighthwillbe-g0g 0 Re h+( )ndash ifweassume that g0doesnotchangeover thedistanceh The differenceinpotentialbetweenthesurfaceandtheheighth is therefore g0hSotheworkneededtoraiseanobjectof mass m to a height h is mgh ie m timesdifference ingravitationalpotential

This we have referred to as the gain in gravitationalpotentialenergy(see235)

However this expression can be extended to any twopoints inanygravitationalfieldsuch that ifanobjectofmass mmovesbetween twopointswhosepotentialsareV1 and V2 respectively then the change in gravitationalpotentialenergyoftheobjectism(V1 ndash V2)

Gravitational potential gradient

Let us consider now a region in space where thegravitationalfieldisconstantInFigure912thetwopointsAandBareseparatedbythedistance∆x

A B

direction of uniform gravitational eld of strength I

∆x

Figure 1002 The gravitational potential gradient

The gravitational field is of strength I and is in the direction shownThe gravitational potential at A is V andatBisV +∆V

Theworkdone is takingapointmassmfromAtoB isF∆x = mI∆x

Howeverbydefinitionthisworkisalsoequalto-m∆V

Therefore mI∆x=-m∆V

or VIx

∆= minus

∆Effectivelythissaysthatthemagnitudeofthegravitationalfield strength is equal to the negative gradient of thepotential If I is constant then V is a linear function of x and Iisequaltothenegativegradientofthestraightlinegraph formed by plotting V against x If I is not constant (asusuallythecase)thenthemagnitudeofI at any point in the field can be found by find the gradient of the V-x graph at that point An example of such a calculation can befoundinSection929

For those of you who do HL maths the relationship between field and potential is seen to follow from the expression for the potential of a point mass viz

mV Gr

= minus

2

ddV mG Ir r

minus = + =

Potential due to one or more point masses

Gravitationalpotentialisascalarquantitysocalculatingthepotential due to more than one point mass is a matter of simpleadditionSoforexamplethepotentialV due to the Moon and Earth and a distance x fromthecentreofEarthonastraightlinebetweenthemisgivenbytheexpression

E MM MV Gx r x

= minus + minus

whereME=massofEarthMM= mass of Moon and r = distancebetweencentreofEarthandMoon

PHYSICS 2015indb 309 150514 337 PM

Chapter 10

310

AH

L

Equipotentials and field linesIfthegravitationalpotentialhasthesamevalueatallpointson a surface the surface is said to be an equipotential surfaceSoforexampleifweimagineasphericalshellaboutEarthwhosecentrecoincideswiththecentreofEarththisshellwillbeanequipotentialsurfaceClearlyifwerepresentthegravitationalfieldstrengthbyfieldlinessincethelinesldquoradiaterdquooutfromthecentreofEarththentheselineswillbeatrightanglestothesurfaceIfthefieldlineswerenotnormaltotheequipotentialsurfacethentherewouldbeacomponentof the field parallel to the surfaceThis wouldmean thatpointsonthesurfacewouldbeatdifferentpotentialsandsoitwouldnolongerbeanequipotentialsurfaceThisofcourseholdstrueforanyequipotentialsurface

Figure 1003 shows the field lines and equipotentials fortwopointmassesm

m m

Figure 1003 Equipotentials for two point masses

It is worth noting that we would get exactly the samepattern ifwewere to replace thepointmasseswith twoequalpointcharges(See935)

Escape speedThe potential at the surface of Earth is which meansthattheenergyrequiredtotakeaparticleofmass mfromthesurfacetoinfinityisequalto

ButwhatdoesitactuallymeantotakesomethingtoinfinityWhentheparticleisonthesurfaceoftheEarthwecanthinkofitassittingatthebottomofaldquopotentialwellrdquoasinfigure1004

particle

surface of Earth

innity

GMR---------

FIgure 1004 A potential well

Theldquodepthrdquoofthewellis and if the particle gains an amount of kinetic energy equal to wherem is its massthenitwillhavejustenoughenergytoldquoliftrdquoitoutofthewell

In reality it doesnrsquot actually go to infinity it justmeansthat the particle is effectively free of the gravitationalattractionof theEarthWe say that ithas ldquoescapedrdquo theEarthrsquosgravitationalpullWemeetthisideainconnectionwithmolecular forcesTwomolecules in a solidwill sitat their equilibrium position the separation where therepulsiveforceisequaltotheattractiveforceIfwesupplyjust enough energy to increase the separation of themolecules such that they are an infinite distance apart then intermolecularforcesnolongeraffectthemoleculesandthesolidwillhavebecomealiquidThereisnoincreaseinthekineticenergyofthemoleculesandsothesolidmeltsat constant temperature

Wecancalculatetheescapespeedofanobjectveryeasilyby equating the kinetic energy to the potential energysuch that

12---mvesc ape

2 GMe mRe

----------------=

vesc aperArr2GMe

Re--------------- 2g0 Re = =

Substituting for g0 and Regivesavalueforvescape of about 11kmsndash1 from the surface of the Earth

Youwillnotethattheescapespeeddoesnotdependonthemassoftheobjectsincebothkineticenergyandpotentialenergy are proportional to the mass

IntheoryifyouwanttogetarockettothemoonitcanbedonewithoutreachingtheescapespeedHoweverthiswouldnecessitate anenormousamountof fuel and it islikely that therocketplus fuelwouldbesoheavythat itwouldnevergetoffthegroundItismuchmorepracticalto accelerate the rocket to the escape speed fromEarthorbitandthenintheoryjustlaunchittotheMoon

Example

Use the followingdata todetermine thepotential at thesurface of Mars and the magnitude of the acceleration of free fall

massofMars =64times1023kg

radiusofMars =34times106 m

Determinealsothegravitationalfieldstrengthatadistanceof 68 times 106mabovethesurfaceofMars

PHYSICS 2015indb 310 150514 337 PM

Fields

311

AH

L

Solution

= minus = minus times times = times23

11 76

64 1067 10 ndash13 10

34 10MV GR

minus timestimes

Nkg-1

But V = -g0R

Therefore 7

0 6

13 10 3834 10

VgR

times= minus = =

timesm s-2

To determine the field strength gh at 68 times 106 m above the surface we use the fact that 0 2

Mg GR

= such that GM = g0R2

Therefore 2 2

0h 2 2 2

h h

38 (34) 042(102)

g RGMgR R

times= = = = m s-2

(the distance from the centre is 34 times 106 + 68 times 106 = 102 times 106 m)

Exercise 1011 Thegraphbelowshowshowthegravitational

potentialoutsideoftheEarthvarieswithdistancefrom the centre

0 10 20 30 40 50 60 70

ndash1

ndash2

ndash3

ndash4

ndash5

ndash6

ndash7

r m times 106

V

Jkgndash1

times 10

7

(a) Usethegraphtodeterminethegainingravitationalpotentialenergyofasatelliteofmass200kgasitmovesfromthesurfaceof the Earth to a height of 30 times 107mabovetheEarthrsquossurface

(b) Calculatetheenergyrequiredtotakeit to infinity

(c) DeterminetheslopeofthegraphatthesurfaceoftheEarthmCommenton youranswer

Electric potential energy

The concept of electric potential energy was developedwith gravitational potential energy in mind Just as anobjectnearthesurfaceoftheEarthhaspotentialenergybecause of its gravitational interaction with the Earthsotoothereiselectricalpotentialenergyassociatedwithinteracting charges

Letusfirstlookatacaseoftwopositivepointchargeseachof 1μC that are initially bound together by a thread in a vacuuminspacewithadistancebetweenthemof10cmasshowninFigure1005Whenthethreadiscutthepointchargesinitiallyatrestwouldmoveinoppositedirectionsmovingwithvelocitiesv1 and v2 along the direction of the electrostatic force of repulsion

BEFORE AFTER

v v21

Figure 1005 Interaction of two positive particles

The electric potential energybetweentwopointchargescan be found by simply adding up the energy associated witheachpairofpointchargesForapairofinteractingchargestheelectricpotentialenergyisgivenby

∆U = ∆Ep + ∆Ek = ∆W = ∆Fr = kqQ

____ r2 times r = kqQ

____ r

Becausenoexternalforceisactingonthesystemtheenergyand momentum must be conserved Initially Ek = 0 and Ep =kqQr=9times109 times 1 times 10-1201m=009JWhentheyareagreatdistance fromeachotherEpwillbenegligibleThefinalenergywillbeequaltofrac12mv1

2+frac12mv22=009J

Momentum isalsoconservedand thevelocitieswouldbethe same magnitude but in opposite directions

Electric potential energyismoreoftendefinedintermsofapointchargemovinginanelectricfieldas

lsquothe electricpotential energybetweenany twopoints inan electric field is defined as negativeoftheworkdone by anelectricfieldinmovingapoint electric chargebetweentwolocationsintheelectricfieldrsquo

ΔU = ΔEp = -ΔW = -Fd = qEx

ΔU = ΔEp +ΔEk = ΔW = ΔFr = kqQ r2 x r = kqQ

wherexisthedistancemovedalong(oroppositeto)thedirection of the electric field

Electric potential energyismeasuredinjoule(J)Justasworkisascalarquantitysotooelectricalpotentialenergyis a scalar quantity The negative of the work done by an electric field in moving a unit electric charge between two points is independent of the path taken Inphysicswesaytheelectricfieldisalsquoconservativersquo field

PHYSICS 2015indb 311 150514 337 PM

Chapter 10

312

AH

L

Supposeanexternalforcesuchasyourhandmovesasmallpositive point test charge in the direction of a uniformelectric fieldAs it ismoving itmust be gaining kineticenergyIfthisoccursthentheelectricpotentialenergyofthe unit charge is changing

InFigure1006apointcharge+qismovedbetweenpointsAandBthroughadistancex in a uniform electric field

B

A

x

+ q

Figure 1006 Movement of a positive point charge in a uniform field

InordertomoveapositivepointchargefrompointAtopointBanexternal forcemustbeapplied to thechargeequaltoqE (F = qE)

Since the force is applied through a distance xthennegativeworkhas to bedone tomove the charge because energyis gainedmeaning there is an increaseelectric potential energybetweenthetwopointsRememberthattheworkdoneisequivalenttotheenergygainedorlostinmovingthe charge through the electric field The concept of electric potential energy is only meaningful as the electric field whichgeneratestheforceinquestionisconservative

W F xtimes E q x times= =

θ x cos θ x

Figure 1007 Charge moved at an angle to the field

If a chargemoves at an angle θ to an electric field thecomponent of the displacement parallel to the electric fieldisusedasshowninFigure918

W F x Eq x θcostimes= =

Theelectricpotentialenergyisstoredintheelectricfieldand theelectricfieldwill return theenergy to thepointcharge when required so as not to violate the Law ofconservationofenergy

Electric potential

The electric potential at a point in an electric field is defined as being the work done per unit charge in bringing a small positive point charge from infinity to that point

ΔV = Vinfin ndash Vf = ndash W ___ q

Ifwedesignatethepotentialenergytobezeroatinfinitythen it follows thatelectricpotentialmustalsobezeroat infinity and the electric potential at any point in an electricfieldwillbe

ΔV = ndash W ___ q

NowsupposeweapplyanexternalforcetoasmallpositivetestchargeasitismovedtowardsanisolatedpositivechargeTheexternalforcemustdoworkonthepositivetestchargetomoveittowardstheisolatedpositivechargeandtheworkmustbepositivewhiletheworkdonebytheelectricfieldmustthereforebenegativeSotheelectricpotentialatthatpointmustbepositiveaccordingtotheaboveequationIfanegativeisolatedchargeisusedtheelectricpotentialatapointonthepositivetestchargewouldbenegativePositivepoint chargesof their ownaccordmove fromaplaceofhighelectricpotential toaplaceof lowelectricpotentialNegative point charges move the other way from lowpotentialtohighpotentialInmovingfrompointAtopointBinthediagramthepositivecharge+qismovingfromalowelectricpotentialtoahighelectricpotential

In thedefinitiongiven the termldquoworkperunitchargerdquohas significance If the test charge is +16 times 10-19Cwherethe charge has a potential energy of 32 times 10-17Jthenthepotentialwouldbe32times10-17J+16times10-19C=200JC-1 Now if the charge was doubled the potential wouldbecome 64times10-17 J However the potential per unitchargewouldbethesame

Electric potential is a scalar quantityandithasunitsJC-1 orvoltswhere1voltequalsone joulepercoloumbThevoltallowsustoadoptaunitfortheelectricfieldintermsofthevolt

PreviouslytheunitfortheelectricfieldwasNCndash1

W = qV and F = qEsoW ___ V = F __ E

E = FV ___ W = FV ___ Fm V mndash1

That is the units of the electric field E can also beexpressed as V mndash1

The work done per unit charge in moving a point charge between two points in an electric field is again independent of the path taken

PHYSICS 2015indb 312 150514 337 PM

Fields

313

AH

L

Electric potential due to a point charge

Letus takeapoint rmetres froma chargedobjectThepotentialatthispointcanbecalculatedusingthefollowing

W = ndashFr = ndashqV and F = ndash q1 q2 _____ 4πε0r

2

Therefore

W = ndash q1 q2 _____ 4πε0r

2 times r = ndash q1 q2 _____ 4πε0r

= ndashq1 times q2 _____ 4πε0r

= ndashq1V

That is V q

4πε0r--------------- =

Orsimply

V kqr------

=

Example

Determinehowmuchworkisdonebytheelectricfieldofpointcharge150μCwhenachargeof200μCismovedfrom infinity to apoint 0400m from thepoint charge(Assumenoaccelerationofthecharges)

Solution

The work done by the electric field is W = -qV = -14πε0 times q times (Q rinfin - Q r0400)

W = (- 200 times 10-6 C times 900 times 109 NmC-2 times 150 times 10-6 C) divide 0400 m = - 0675 J

An external force would have to do +0675 J of work

Electric field strength and electric potential gradient

Letus lookbackatFigure1006Supposeagain that thecharge+qismovedasmalldistancebyaforceFfromAtoBsothattheforcecanbeconsideredconstantTheworkdoneisgivenby

∆W F x∆times=

TheforceFandtheelectricfieldEareoppositelydirectedandweknowthat

F = -qE and ΔW = q ΔV

Thereforetheworkdonecanbegivenas

q ΔV = -q E Δx

Therefore E V∆x∆-------

ndash=

The rate of change of potential ΔV at a point with respect to distance Δx in the direction in which the change is maximum is called the potential gradient We say that the electric field = - the potential gradient and the units are Vm-

1 From the equation we can see that in a graph of electric potential versus distance the gradient of the straight line equals the electric field strength

In reality if a charged particle enters a uniform electric field it will be accelerated uniformly by the field and its kinetic energy will increase This is why we had to assume no acceleration in the last worked example

Ek∆ 12---

mv2 q E xsdot sdot q Vx---

xsdot sdot q Vsdot= = = =

Example

Determine how far apart two parallel plates must besituated so that a potential difference of 150 times 102 V produces an electric field strength of 100 times 103 NC-1

Solution

Using E V∆x------- xhArrndash V∆

E-------15 102 Vtimes

100 103 N C 1ndashtimes------------------------------------------= = =

= 150 times 10-1

The plates are 150 times 10-1 m apart

The electric field and the electric potential at a point due to an evenly distributed charge +q on a sphere can be represented graphically as in Figure 1008

r

r

V

E

r0

r0

r0

Charge of +Q evenly distributed over surface

E 14πε0------------ Q

r2----- r r0gt=

V 14πε0------------Q

r----r r0gt=

Figure 1008 Electric field and potential due to a charged sphere

PHYSICS 2015indb 313 150514 337 PM

Chapter 10

314

AH

L

When the sphere becomes charged we know that thechargedistributes itselfevenlyoverthesurfaceThereforeeverypartofthematerialoftheconductorisatthesamepotential As the electric potential at a point is defined as beingnumericallyequaltotheworkdoneinbringingaunitpositivechargefrominfinitytothatpointithasaconstantvalueineverypartofthematerialoftheconductor

Since the potential is the same at all points on the conducting surfacethen∆V∆xiszeroButE=ndash∆V∆xThereforethe electric field inside the conductor is zero There is no electric field inside the conductor

SomefurtherobservationsofthegraphsinFigure1009are

bull OutsidethespherethegraphsobeytherelationshipsgivenasE α 1 r2 and V α 1 r

bull Atthesurfacer = r0Thereforetheelectricfieldandpotentialhavetheminimumvalueforr at this point and this infers a maximum field and potential

bull Insidethespheretheelectricfieldiszerobull Insidethespherenoworkisdonetomoveacharge

fromapointinsidetothesurfaceThereforethereisno potential difference and the potential is the same asitiswhenr = r0

Similargraphscanbedrawnfortheelectricfieldintensityandthe electric potential as a function of distance from conducting parallelplatesandsurfacesandthesearegiveninFigure1009

Potential plot E eld plotE eld

++++

ndashndashndashndash

xx x

+

x x

ndashx x

+ ndashx

x

+ +

x

x

Figure 1009 Electric field and electric potential at a distance from a charged surface

Potential due to one or more point charges

The potential due to one point charge can be determined byusingtheequationformula

V = kq r

Example 1

Determine the electric potential at a point 20 times 10-1 m fromthecentreofan isolatedconducting spherewithapointchargeof40pCinair

Solution

Using the formula

V = kq r we have

V90 10 9times( ) 40 10 12ndashtimes( )times

20 10 1ndashtimes( )------------------------------------------------------------------ 018 V= =

the potential at the point is 180 times 10-1 V

The potential due to a number of point charges can be determined by adding up the potentials due to individual point charges because the electric potential at any point outside a conducting sphere will be the same as if all the charge was concentrated at its centre

Example 2

Threepointchargesofareplacedattheverticesofaright-angledtriangleasshowninthediagrambelowDeterminetheabsolutepotentialat the+20μCchargeduetothetwoothercharges

- 6μC

+3μC +2μC

4 m

3 m

Solution

The electric potential of the +2 μC charge due to the ndash 6 μC charge is

V = (9 times 109 Nm2C-2 times -6 times 10-6 C) divide (radic 32 + 42) m = - 108 times 104 V

PHYSICS 2015indb 314 150514 338 PM

Fields

315

AH

L

Insummarywecanconcludethat

bull No work is done to move a charge along an equipotential

bull Equipotentials are always perpendicular to the electric lines of force

Figure1011and1012showsomeequipotential lines fortwo oppositely charged and identically positive spheresseparated by a distance

+ve ndashve

equipotential lines

Figure 1011 Equipotential lines between two opposite charges

equipotential lines

Figure 1012 Equipotential lines between two charges which are the same

+

ndash ndash ndash ndash

+ + +

40 V

30 V

20 V

10 V

50 V

Figure 1013 Equipotential lines between charged parallel plates

Figure1013showstheequipotentiallinesbetweenchargedparallel plates Throughout this chapter the similarities and differences between gravitational fields and electric fieldshavebeendiscussedTherelationshipsthatexistsbetweengravitationalandelectricquantitiesandtheeffectsofpointmassesandchargesissummarisedinTable1014

The electric potential of the +2 μC charge due to the +3 μC charge is

V = (9 times 109 Nm2C-2 times 3 times 10-6 C) divide 3m = 9 times 103 V

The net absolute potential is the sum of the 2 potentials- 108 times 104 V + 9 times 103 V =- 18 times 103 V

The absolute potential at the point is - 18 times 103 V

Equipotential surfacesRegions in space where the electric potential of a chargedistributionhasaconstantvaluearecalledequipotentials The placeswherethepotentialisconstantinthreedimensionsarecalled equipotential surfacesandwheretheyareconstantintwodimensionstheyarecalledequipotential lines

TheyareinsomewaysanalogoustothecontourlinesontopographicmapsInthiscasethegravitationalpotentialenergy is constant as amassmoves around the contourlines because the mass remains at the same elevationabovetheEarthrsquossurfaceThegravitationalfieldstrengthacts in a direction perpendicular to a contour line

Similarlybecausetheelectricpotentialonanequipotentiallinehas thesamevalueanelectric forcecandonoworkwhena testchargemovesonanequipotentialThereforethe electric field cannot have a component along anequipotentialandthusitmustbeeverywhereperpendiculartotheequipotentialsurfaceorequipotentiallineThisfactmakesiteasytoplotequipotentialsifthelinesofforceorlinesofelectricfluxofanelectricfieldareknown

Forexamplethereareaseriesofequipotentiallinesbetweentwoparallelplateconductorsthatareperpendiculartotheelectric fieldTherewill be a series of concentric circles(eachcirclefurtherapartthanthepreviousone)thatmapouttheequipotentialsaroundanisolatedpositivesphereThe lines of force and some equipotential lines for anisolatedpositivesphereareshowninFigure1010

Lines ofequipotential

Figure 1010 Equipotentials around an isolated positive sphere

PHYSICS 2015indb 315 150514 338 PM

Chapter 10

316

AH

L

Gravitational quantity Electrical quantityQ

uant

ities

g V∆x∆-------ndash= E V∆

x∆-------ndash=

Poin

t mas

ses a

nd ch

arge

s

V Gmr----

ndash= V 14πε0------------q

r---=

g G m

r2-----ndash= E 14πε0------------ q

r2-----=

F Gm1 m2

r2--------------= F 1

4πε0------------

q1q2

r2-----------=

Figure 1014 Formulas

Example

Deduce the electric potential on the surface of a gold nucleus that has a radius of 62 fm

Solution

Using the formula

V = kq r and knowing the atomic number of gold is 79 We will assume the nucleus is spherical and it behaves as if it were a point charge at its centre (relative to outside points)

V = 90 times 109 Nm2C-2 times 79 times 16 times 10-19 C divide 62 times 10-15 m = 18 times 107 V

The potential at the point is 18 MV

Example

Deduce the ionisation energy in electron-volts of theelectron in the hydrogen atom if the electron is in its ground state and it is in a circular orbit at a distance of 53 times 10-11 m from the proton

Solution

This problem is an energy coulombic circular motion question based on Bohrrsquos model of the atom (not the accepted quantum mechanics model) The ionisation energy is the energy required to remove the electron from the ground state to infinity The electron travels in a circular orbit and therefore has a centripetal acceleration The ionisation energy will counteract the coulombic force and the movement of the electron will be in the opposite direction to the centripetal force

Total energy = Ek electron + Ep due to the proton-electron interaction

ΣF = kqQ r2 = mv2 r and as such mv2 = = kqQ r

Therefore Ek electron = frac12 kqQ r

Ep due to the proton-electron interaction = - kqQ r

Total energy = frac12 kqQ r + - kqQ r = -frac12 kqQ r

= - 90 times 109 Nm2C-2 times (16 times 10-19 C)2 divide 53 times 10-11 m = -217 times 10-18 J

= -217 times 10-18 J divide 16 times 10-19 = -136 eV

The ionisation energy is 136 eV

Exercise 1021 ApointchargePisplacedmidwaybetweentwo

identicalnegativechargesWhichoneofthefollowingiscorrectwithregardstoelectricfieldandelectricpotentialatpointP

Electric field Electric potentialA non-zero zeroB zero non-zeroC non-zero non-zeroD zero zero

2 Twopositivechargedspheresaretiedtogetherinavacuumsomewhereinspacewheretherearenoexternal forces A has a mass of 25 g and a charge of20μCandBhasamassof15gandachargeof30μCThedistancebetweenthemis40cmTheyarethenreleasedasshowninthediagram

BEFORE AFTER

v1 v2

A B

(a) Determinetheirinitialelectricpotentialenergy in the before situation

(b) DeterminethespeedofsphereBafterrelease

3 Thediagrambelowrepresentstwoequipotentiallines in separated by a distance of 5 cm in a uniform electric field

+ + + + + + + +

ndash ndash ndash ndash ndash ndash ndash ndash

40 V

20 V 5 cm

PHYSICS 2015indb 316 150514 338 PM

Fields

317

AH

L

Determine the strength of the electric field

4 Thisquestionisabouttheelectricfieldduetoacharged sphere and the motion of electrons in that fieldThediagrambelowshowsanisolatedmetalsphereinavacuumthatcarriesanegativeelectriccharge of 60 μC

ndash

(a) Onthediagramdrawtheconventionalwayto represent the electric field pattern due to the charged sphere and lines to represent threeequipotentialsurfacesintheregionoutside the sphere

(b) Explainhowthelinesrepresentingtheequipotentialsurfacesthatyouhavesketchedindicatethatthestrengthoftheelectricfieldisdecreasingwithdistancefrom the centre of the sphere

(c) Theelectricfieldstrengthatthesurfaceofthe sphere and at points outside the sphere can be determined by assuming that the sphere acts as a point charge of magnitude 60 μC at its centre The radius of the sphere is 25 times 10ndash2 m Deduce that the magnitude of the field strength at the surface of the sphere is 86 times 107 Vmndash1

An electron is initially at rest on the surface of the sphere

(d) (i) Describethepathfollowedbytheelectronasitleavesthesurfaceofthesphere

(ii) Calculatetheinitialaccelerationofthe electron

5 Determinetheamountofworkthatisdoneinmovingachargeof100nCthroughapotentialdifferenceof150times102 V

6 Three identical 20 μC conducting spheres are placedatthecornersofanequilateraltriangleofsides 25 cm The triangle has one apex C pointing upthepageand2baseanglesAandBDeterminetheabsolutepotentialatB

7 Determinehowfaraparttwoparallelplatesmustbesituatedsothatapotentialdifferenceof250 times 102 V produces an electric field strength of 200 times 103 NC-1

8 Thegapbetweentwoparallelplatesis10times10-3mandthereisapotentialdifferenceof10times104 V betweentheplatesCalculate

i theworkdonebyanelectroninmovingfrom one plate to the other

ii thespeedwithwhichtheelectronreachesthe second plate if released from rest

iii theelectricfieldintensitybetweentheplates

9 Anelectronguninapicturetubeisacceleratedby a potential 25 times 103VDeterminethekineticenergygainedbytheelectroninelectron-volts

10 Determine the electric potential 20 times 10-2 m from achargeof-10times10-5 C

11 Determinetheelectricpotentialatapointmid-waybetweenachargeofndash20pCandanotherof+5pConthelinejoiningtheircentresifthecharges are 10 cm apart

12 During a thunderstorm the electric potential differencebetweenacloudandthegroundis10 times 109 V Determine the magnitude of the change in electric potential energy of an electron thatmovesbetweenthesepointsinelectron-volts

13 A charge of 15 μC is placed in a uniform electric fieldoftwooppositelychargedparallelplateswithamagnitudeof14times103 NC-1

(a) Determinetheworkthatmustbedoneagainstthefieldtomovethepointchargeadistance of 55 cm

(b) Calculatethepotentialdifferencebetweenthe final and initial positions of the charge

(c) Determinethepotentialdifferencebetweenthe plates if their separation distance is 15 cm

14 Duringaflashoflightningthepotentialdifferencebetweenacloudandthegroundwas12 times 109 V and the amount of transferred charge was32C

(a) Determinethechangeinenergyofthetransferred charge

(b) Iftheenergyreleasedwasallusedtoacceleratea1tonnecardeduceitsfinalspeed

(c) Iftheenergyreleasedcouldbeusedtomelticeat0degCdeducetheamountoficethatcould be melted

PHYSICS 2015indb 317 150514 338 PM

Chapter 10

318

AH

L

15 SupposethatwhenanelectronmovedfromAtoBin the diagram along an electric field line that the electric field does 36 times 10-19Jofworkonit

Determinethedifferencesinelectricpotential

(a) VB ndash VA(b) VC ndash VA(c) VC ndash VB

16 DeterminethepotentialatpointPthatislocatedatthecentreofthesquareasshowninthediagrambelow

- 6μC

+3μC +2μC

1m

1m

5μC

P

102 Fields at work

Essential idea Similar approaches can be taken inanalysingelectricalandgravitationalpotentialproblems

Understandingsbull Potentialandpotentialenergybull Potentialgradientbull Potentialdifferencebull Escapespeedbull Orbitalmotionorbitalspeedandorbitalenergybull Forcesandinverse-squarelawbehaviour

Orbital motion orbital speed and orbital energy

Although orbital motion may be circular elliptical orparabolic this sub-topic only dealswith circular orbitsThis sub-topic is not fundamentally new physics butan application that synthesizes ideas from gravitationcircularmotiondynamicsandenergy

TheMoonorbitstheEarthandinthissenseitisoftenreferredtoasasatelliteoftheEarthBefore1957itwastheonlyEarthsatelliteHoweverin1957theRussianslaunchedthefirstmanmadesatelliteSputnik1SincethisdatemanymoresatelliteshavebeenlaunchedandtherearenowliterallythousandsofthemorbitingtheEarthSomeareusedtomonitortheweathersome used to enable people to find accurately their position on thesurfaceoftheEarthmanyareusedincommunicationsandnodoubtsomeareusedtospyonothercountriesFigure1015showshowinprincipleasatellitecanbeputintoorbit

Theperson(whosesizeisgreatlyexaggeratedwithrespectto Earth) standing on the surface on the Earth throwssome stonesThe greater the speed with which a stoneis thrown the further it will land from her The pathsfollowedbythethrownstonesareparabolasByastretchof the imaginationwecanvisualisea situation inwhichastoneis thrownwithsuchaspeedthatbecauseof thecurvatureoftheEarthitwillnotlandonthesurfaceoftheEarthbutgointoldquoorbitrdquo(Path4onFigure1015)

NATURE OF SCIENCECommunication of scientific explanations The ability to apply field theory to the unobservable (charges) and the massively scaled (motion of satellites) required scientists to develop new ways to investigate analyse and report findings to a general public used to scientific discoveries based on tangible and discernible evidence (51)

copy IBO 2014

PHYSICS 2015indb 318 150514 338 PM

Fields

319

AH

L

1

2

34Earth

Figure 1015 Throwing a stone into orbit

TheforcethatcausesthestonestofollowaparabolicpathandtofalltoEarthisgravityandsimilarlytheforcethatkeeps the stone in orbit is gravity For circular motionto occur we have seen that a force must act at rightanglestothevelocityofanobjectthatistheremustbeacentripetalforceHenceinthesituationwedescribeherethe centripetal force for circular orbital motion about the EarthisprovidedbygravitationalattractionoftheEarth

We can calculate the speedwithwhich a stonemustbethrowninordertoputitintoorbitjustabovethesurfaceof the Earth

If the stone has mass m and speed v thenwehave fromNewtonrsquos2ndlaw

mv2

RE---------- G

MEm

RE2------------=

whereRE is the radius of the Earth and ME is the mass of the Earth

Bearinginmindthatg 0 GME

RE2--------= then

v g RE 10 64 106timestimes 8 103times= = =

Thatisthestonemustbethrownat8times103m sndash1

Clearly we are not going to get a satellite into orbit soclosetothesurfaceoftheEarthMovingatthisspeedthefrictionduetoairresistancewouldmeltthesatellitebeforeithadtravelledacoupleofkilometresInrealitythereforeasatelliteisputintoorbitabouttheEarthbysendingitattachedtoarocketbeyondtheEarthrsquosatmosphereandthengivingitacomponentofvelocityperpendiculartoaradialvectorfromtheEarthSeeFigure1016

Earth Satellite carried by rocket to hereSatellite orbit

Tangential component of velocity

Figure 1016 Getting a satellite into orbit

Keplerrsquos third law(This work of Kepler and Newtonrsquos synthesis of the work is an excellent example of the scientific method and makes for a good TOK discussion)

In 1627 Johannes Kepler (1571-1630)publishedhis lawsof planetary motion The laws are empirical in natureandwere deduced from the observations of theDanishastronomer Tycho de Brahe (1546-1601)The third lawgives a relationship between the radius of orbit R of a planet and its period TofrevolutionabouttheSunThelawisexpressedmathematicallyas

2

3 constantTR

=

We shall nowuseNewtonrsquos LawofGravitation to showhowitisthattheplanetsmoveinaccordancewithKeplerrsquosthirdlaw

InessenceNewtonwasabletousehislawofgravitytopredict the motion of the planets since all he had to do was factor theF givenby this law intohis second law F = ma to find their accelerations and hence their future positions

InFigure1017theEarthisshownorbitingtheSunandthedistancebetweentheircentresisR

RSun

Earth

F es

F se

Figure 1017 Planets move according to Keplerrsquos third law

Fes is the force that the Earth exerts on the Sun and Fse is the force that the Sun exerts on the Earth The forces are equalandoppositeandtheSunandtheEarthwillactuallyorbitaboutacommoncentreHoweversince theSun issoverymuchmoremassivethantheEarththiscommoncentrewillbeclosetothecentreoftheSunandsowecanregard the Earth as orbiting about the centre of the Sun Theotherthingthatweshallassumeisthatwecanignorethe forces that the other planets exert on the Earth (This wouldnotbeawisethingtodoifyouwereplanningtosend a space ship to theMoon for example)We shallalso assume that we have followed Newtonrsquos exampleandindeedprovedthataspherewillactasapointmasssituated at the centre of the sphere

Kepler had postulated that the orbits of the planets areellipticalbut since theeccentricityof theEarthrsquosorbit issmallweshallassumeacircularorbit

PHYSICS 2015indb 319 150514 338 PM

Chapter 10

320

AH

L

TheaccelerationoftheEarthtowardstheSunisa = Rω2

whereω ω 2πT------=

Hencea R 2π

T------ times

2 4π2 RT 2------------- = =

But the acceleration is given by Newtonrsquos Second Law F = mawhereFisnowgivenbytheLawofGravitationSo in this situation

F maGMs Me

R2-------------------= = butwealsohavethat

a a R 2πT

------ times

2 4π2 RT 2------------- = = and m = Me so that

G MsMe

R2------------------- Me4π2R

T2-------------times

GMs

4π2-----------rArr R3

T2------= =

Butthequantity

isaconstantthathasthesamevalueforeachoftheplanetssowehaveforalltheplanetsnotjustEarththat

R3

T2------ k =

wherekisaconstantWhichisofcourseKeplerrsquosthirdlaw

ThisisindeedanamazingbreakthroughItisdifficulttorefutethe idea that all particles attract each other in accordance withtheLawofGravitationwhenthelawisabletoaccountfortheobservedmotionoftheplanetsabouttheSun

The gravitational effects of the planets upon each othershould produce perturbations in their orbits Such is the predictivepoweroftheUniversalGravitationalLawthatit enabled physicists to compute these perturbations The telescopehadbeeninventedin1608andbythemiddleofthe 18th Century had reached a degree of perfection in design that enabled astronomers to actually measure the orbital perturbations of the planets Their measurements werealways inagreementwith thepredictionsmadebyNewtonrsquoslawHoweverin1781anewplanetUranuswasdiscovered and the orbit of this planet did not fit withthe orbit predicted by Universal Gravitation Such wasthe physicistrsquos faith in theNewtonianmethod that theysuspectedthatthediscrepancywasduetothepresenceofayetundetectedplanetUsingtheLawofGravitationtheFrenchastronomerJLeverrier and the English astronomer J C Adamswereable tocalculate justhowmassive thisnewplanetmustbeandalsowhereitshouldbeIn1846the planetNeptunewas discovered justwhere they hadpredicted Ina similarwaydiscrepancies in theorbitofNeptune ledto thepredictionandsubsequentdiscoveryin1930oftheplanetPlutoNewtonrsquosLawofGravitation

had passed the ultimate test of any theory it is not only abletoexplainexistingdatabutalsotomakepredictions

Satellite energyWhenasatelliteisinorbitaboutaplanetitwillhavebothkineticenergyandgravitationalpotentialenergySupposewe consider a satellite ofmass m that is in an orbit of radius r about a planet of mass M

Thegravitationalpotentialduetotheplanetatdistancer from its centre is

ThegravitationalpotentialenergyofthesatelliteVsat

is therefore GMe m

r----------------ndash

ThatisVsat GMem

r----------------ndash=

Thegravitationalfieldstrengthatthesurfaceoftheplanetisgivenby

g 0GMe

Re2------------=

Hencewecanwrite

Vsat g0 Re

2

rndash=

m

ThekineticenergyofthesatelliteKsatisequaltofrac12mv2wherev is its orbital speed

Byequatingthegravitationalforceactingonthesatellitetoitscentripetalaccelerationwehave

G Mem

r2---------------- mv2

r---------- mv2hArr

GMemr----------------= =

Fromwhich12---

mv2 12---

GMemr----------------times=

= 0 e

2g R

r

2m

Which is actually quite interesting since it shows thatirrespective of the orbital radius the KE is numericallyequal to half the PE Also the total energy E tot K sat Vsat+ 1

2---G Me m

r----------------times G Me m

r----------------ndash + 1

2---G Me m

r----------------ndash= = =of the satelliteisalwaysnegativesince

E tot K sat Vsat+ 12---

G Me mr----------------times

G Me mr----------------ndash

+ 12---

G Me mr----------------ndash= = =

PHYSICS 2015indb 320 150514 338 PM

Fields

321

AH

L

The energies of an orbiting satellite as a function of radial distancefromthecentreofaplanetareshownplottedinFigure1018

12

100806

0402

0ndash02

ndash04

ndash06ndash08ndash10

ndash12

2 4 6 8 10 12

ener

gya

rbita

ry u

nits

distance timesR

kinetic energytotal energy

potential energy

Figure 1018 Energy of an orbiting satellite as a function of distance from the centre of a planet

WeightlessnessSuppose that you are inanelevator(lift)thatisdescendingat constant speedandyou let goof abook thatyouareholdinginyourhandThebookwillfalltothefloorwithaccelerationequaltotheaccelerationduetogravityIfthecablethatsupportstheelevatorweretosnap(asituationthatItrustwillneverhappentoanyofyou)andyounowletgothebookthatyouareholdinginyourotherhandthis book will not fall to the floor - it will stay exactlyin linewithyourhandThis isbecause thebook isnowfallingwith the sameaccelerationas theelevatorandassuch the book cannot ldquocatchrdquo up with the floor of theelevatorFurthermoreifyouhappenedtobestandingonasetofbathroomscalesthescaleswouldnowreadzero-youwouldbeapparentlyweightlessItisthisideaoffreefallthatexplainstheapparentweightlessnessofastronautsin an orbiting satellite These astronauts are in free fall inthesensethattheyareacceleratingtowardsthecentre of the Earth

It is actually possible to define theweight of a body inseveral different waysWe can define it for example asthegravitationalforceexertedonthebodybyaspecifiedobject such as the EarthThiswe have seen thatwe doinlotsofsituationswherewedefinetheweightasbeingequaltomgIfweusethisdefinitionthenanobjectinfreefallcannotbydefinitionbeweightlesssinceitisstillinagravitationalfieldHoweverifwedefinetheweightofanobjectintermsofaldquoweighingrdquoprocesssuchasthereadingonasetofbathroomscaleswhichineffectmeasuresthecontact force between the object and the scales thenclearlyobjectsinfreefallareweightlessOnenowhasto

askthequestionwhetherornotitispossibleForexampletomeasurethegravitationalforceactingonanastronautin orbit about the Earth

WecanalsodefineweightintermsofthenetgravitationalforceactingonabodyduetoseveraldifferentobjectsForexample for an object out in space its weight could bedefined in terms of the resultant of the forces exerted on it bytheSuntheMoontheEarthandalltheotherplanetsin the Solar System If this resultant is zero at a particular pointthenthebodyisweightlessatthispoint

Inviewofthevariousdefinitionsofweightthatareavailabletousitisimportantthatwhenweusethewordldquoweightrdquoweareawareofthecontextinwhichitisbeingused

Example

Calculate the height above the surface of the Earth atwhichageo-stationarysatelliteorbits

Solution

A geo-stationary satellite is one that orbits the Earth in such a way that it is stationary with respect to a point on the surface of the Earth This means that its orbital period must be the same as the time for the Earth to spin once on its axis ie 24 hours

From Keplerrsquos third law we have G Ms

4π2----------- R3

T2------=

That is

RMe

mh

using the fact that the force of attraction between the satellite and the Earth is given by

FG Mem

R2----------------=

and that F = ma

where m is the mass of the satellite and a 4π2R

T2-------------=

we have

GMem

R2---------------- m 4π2R

T2-------------times

GMe

4π2------------rArr R3

T2------= =

PHYSICS 2015indb 321 150514 338 PM

Chapter 10

322

AH

L

Now the mass of the Earth is 60 times 1024 kg and the period T measured in seconds is given by T = 86400 s

So substitution gives R = 42 times 106 m

The radius of the Earth is 64 times 106 m so that the orbital height h is about 36 times 107 m

Example

Calculatetheminimumenergyrequiredtoputasatelliteofmass500kgintoanorbitthatisasaheightequaltotheEarthrsquosradiusabovethesurfaceoftheEarth

Solution

We have seen that when dealing with gravitational fields and potential it is useful to remember that

g 0G M

Re2---------= or g0 Re

2 GM=

The gravitational potential at the surface of the Earth is

g 0ndash Re GMRe

---------ndash=

The gravitational potential at a distance R

from the centre of the Earth is

=

The difference in potential between the surface of the Earth and a point distance R from the centre is therefore

V∆ g 0Re 1ReR------ndash

=

If R 2Re= then V∆g0 Re

2------------=

This means that the work required to ldquoliftrdquo the satellite into orbit is g0Rm where m is the mass of the satellite This is equal to

10 32 106 500timestimestimes = 16000 MJ

However the satellite must also have kinetic energy in order to orbit Earth This will be equal to

g 0 mRe2

2R-----------------g 0mRe

2

4----------------- 8000 MJ= =

The minimum energy required is therefore

24000 MJ

Exercise 1031 The speed needed to put a satellite in orbit does

not depend on

A the radius of the orbitB theshapeoftheorbitC thevalueofgattheorbitD the mass of the satellite

2 EstimatethespeedofanEarthsatellitewhoseorbitis400kmabovetheEarthrsquossurfaceAlsodetermine the period of the orbit

3 Calculatethespeedofa200kgsatelliteorbitingthe Earth at a height of 70 times 106 m

Assume that g = 82 m sndash2 for this orbit

4 TheradiioftwosatellitesXandYorbitingthe Earth are 2r and 8rwherer is the radius of the Earth Calculate the ratio of the periods of revolutionofXtoY

5 A satellite of mass mkgissentfromEarthrsquossurfaceinto an orbit of radius 5RwhereR is the radius of theEarthWritedownanexpressionfor

(a) thepotentialenergyofthesatelliteinorbit

(b) thekineticenergyofthesatelliteinorbit

(c) theminimumworkrequiredtosendthesatellitefromrestattheEarthrsquossurfaceintoits orbit

6 A satellite in an orbit of 10rfallsbacktoEarth(radius r)afteramalfunctionDeterminethespeedwithwhichitwillhittheEarthrsquossurface

7 The radius of the moon is frac14 that of the Earth AssumingEarthandtheMoontohavethesamedensitycomparetheaccelerationsoffreefallatthe surface of Earth to that at the surface of the Moon

8 UsethefollowingdatatodeterminethegravitationalfieldstrengthatthesurfaceoftheMoon and hence determine the escape speed from the surface of the Moon

Mass of the Moon = 73 times 1022kg

RadiusoftheMoon=17times106 m

PHYSICS 2015indb 322 150514 338 PM

Chapter 9

302

AH

L

distance The telescopes view the same objects at the same time and the signals that each receive from the objects are simultaneously superimposed The result of the superposition of the two signals is a two-slit interference pattern The pattern has much narrower fringe spacing than that of the diffraction pattern produced by either telescope on its own hence producing a much higherresolvingpowerWhentelescopesareusedlikethistheyare called a stellar interferometer

InSocorroinNewMexicothereisastellarinterferometerthat consists of 27 parabolic dishes each of diameter25 m arranged in a Y-shape that covers an area of 570km2Thisisaso-calledVeryLargeArray(VLA)Evenhigher resolution can be obtained by using an array of radio telescopes in observatories thousands of kilometres apart A system that uses this so-called technique oflsquovery-long-baseline interferometryrsquo (VLBI) is known asalsquovery-long-baselinearrayrsquo(VLBA)WithVLBAaradiowavelength resolving power can be achieved that is 100 timesgreaterthanthebestopticaltelescopesEvenhigherresolving power can be achieved by using a telescope that isinasatelliteorbitingEarthJapanrsquosInstituteofSpaceandAstronautical Science (ISAS) launched sucha system inFebruary 1997TheNationalAstronomicalObservatoryof Japan the National Science Foundationrsquos NationalRadio Astronomy Observatory (NRAO) the CanadianSpaceAgency theAustraliaTelescopeNationalFacilitythe European VLBI Network and the Joint Instituteback this project forVeryLongBaseline Interferometryin EuropeThis project is a very good example of howInternationalismcanoperateinPhysics

Electron microscope

Telescopes are used to look at very distant objects that are very large but because of their distance fromus appearverysmallMicroscopesontheotherhandareusedtolookat objects that are close to us but are physically very small Aswehave seen justmagnifyingobjects that ismakingthemappearlargerisnotsufficientonitsowntogaindetailabouttheobjectfordetailhighresolutionisneeded

Figure928isaschematicofhowanoptical microscope is usedtoviewanobjectandFigure929isaschematicofatransmission electron microscope(TEM)

Optical lens system

bright light source

glass slide containing specimen (object)

eye

Figure 928 The principle of a light microscope

magnetic lens system

electron source

thin wafer of material

ScreenCCD

Figure 929 The principle of an electron microscope

In the optical microscope the resolving power isdetermined by the particular lens system employed and the wavelength λ of the light used For example twopoints in the sample separated by a distance d will just be resolved if

2d mλ=

where m is a property of the lens system know as the numerical aperture In practice the largest value of m obtainable is about 16 Hence if the microscope slideis illuminatedwith light ofwavelength 480 nm a goodmicroscope will resolve two points separated by a distance dasymp15times10-7masymp015micromPointscloser together thanthiswillnotberesolvedHoweverthisisgoodenoughtodistinguishsomevirusessuchastheEbolavirus

Clearlythesmallerλ the higher the resolving power and this is where the electron microscope comes to the fore The electron microscope makes use of the wave nature of electrons(see1315)IntheTEMelectronspassthrougha wafer thin sample and are then focused by a magnetic field onto a fluorescent screen orCCD (charge coupleddevicesee142)Electronsusedinanelectronmicroscopehavewavelengthstypicallyofabout5times10-12mHoweverthe numerical aperture of electron microscopes is considerablysmaller thanthatofanopticalmicroscopetypicallyabout002NonethelessthismeansthataTEMcanresolvetwopointsthatareabout025nmapartThisresolving power is certainly high enough to make out the shape of large molecules

Another type of electron microscope uses a techniqueby which electrons are scattered from the surface of the sample The scattered electrons and then focused as in theTEM to form an image of the surfaceThese so-calledscanningelectronmicroscopes(SEM)havealowerresolving power than TEMrsquos but give very good threedimensional images

The eye

We saw in the last section that the resolving power of the human eye is about 2 times 10-4 rad Suppose that youare looking at car headlights on a dark night and the car

PHYSICS 2015indb 302 150514 337 PM

Wave Phenomena

303

AH

L

is a distance D away If the separation of the headlight is say 15 m then the headlights will subtend an angle θ 15

Dθ =

at your eye Assuming an average wavelength of

500 nm your eye will resolve the headlights into twoseparatesourcesifthisangleequals2times10-4 rad This gives D=75kmInotherwordsifthecarisapproachingyouona straight road then you will be able to distinguish the two headlightsasseparatesourceswhenthecaris75kmawayfrom you Actually because of the structure of the retina and optical defects the resolving power of the average eye is about 6 times 10-4 rad This means that the car is more likely tobe25kmawaybeforeyouresolvetheheadlights

Astronomical telescope

Letusreturntotheexampleofthebinarystarsdiscussedatthe beginning of this section on resolution The stars Kruger A and B form a binary system The average separation of the stars is 14 times 1012mandtheiraveragedistancefromEarthis12times1017mWhenviewedthroughatelescopeonEarththe system will therefore subtend an angle

12

17

14 1012 10

θtimes

=times

= 12times10-5 rad at the objective lens of the telescope Assuming that the average wavelength of the light emitted bythestarsis500nmthenifthetelescopeistoresolvethesystem into two separate images it must have a minimum diameter Dwhere12times10-5 =

7122 500 10D

minustimes times

This gives D=0050m which is about 5 cm So thisparticular system is easily resolved with a small astronomical telescope

Exercise 961 ItissuggestedthatusingtheISASVLBAitwould

bepossibletoldquoseerdquoagrainofriceatadistanceof5000kmEstimatetheresolvingpoweroftheVLBA

2 Thedistancefromtheeyelenstotheretinais20mm The light receptors in the central part of the retinaareabout5times10-6apartDeterminewhetherthe spacing of the receptors will allow for the eye to resolve the headlights in the above discussion whentheyare25kmfromtheeye

3 ThediameterofPlutois23times106 m and its averagedistancefromEarthis60times1012 m EstimatetheminimumdiameteroftheobjectiveofatelescopethatwillenablePlutotobeseenasadisc as opposed to a point source

95 Doppler effect

Essential ideaTheDopplereffectdescribesthephenomenonofwavelengthfrequencyshiftwhenrelativemotionoccurs

Understandingsbull TheDopplereffectforsoundwavesandlightwaves

The Doppler effect for sound waves and light waves

Consider two observers A and B at rest with respect to asoundsourcethatemitsasoundofconstantfrequencyf Clearly both observers will hear a sound of the same frequencyHoweversupposethatthesourcenowmovesat constant speed towards A A will now hear a sound of frequencyfA that is greater than f and B will hear a sound of frequency fB that is less than f This phenomenon is known as the Doppler Effect orDopplerPrinciple after C J Doppler(1803-1853)

The same effect arises for an observer who is either moving towards or away from a stationary source

Figure 930 shows the waves spreading out from a stationary source that emits a sound of constant frequency f The observersAandBhearasoundofthesamefrequency

A B

wavefront

S

Figure 930 Sound waves from a stationary source

Suppose now that the source moves towards A withconstant speed vFigure9310(a)showsasnapshotofthenew wave pattern

NATURE OF SCIENCETechnology Although originally based on physical observations of the pitch of fast moving sources of sound the Doppler effect has an important role in many different areas such as evidence for the expansion of the universe and generating images used in weather reports and in medicine (55)

copy IBO 2014

PHYSICS 2015indb 303 150514 337 PM

Chapter 9

304

AH

L

sourceA B

smallerwavelength

largerwavelength

Figure 931 (a) Sound waves from a moving source

The wavefronts are now crowded together in the direction of travel of the source and stretched out in the opposite direction This is why now the two observers will now hear notesofdifferentfrequenciesHowmuchthewavesbunchtogether and how much they stretch out will depend on the speed vEssentially A

A

cfλ

= and BB

cfλ

= where λA lt λB and v is the speed of sound

IfthesourceisstationaryandAismovingtowardsitthenthe waves from the source incident on A will be bunched up If A is moving away from the stationary source then the waves from the source incident on A will be stretched out

Christian Doppler(1803ndash1853)actuallyappliedtheprinciple(incorrectlyasithappens)totryandexplainthecolourofstars However the Doppler effect does apply to light aswellastosoundIfalightsourceemitsalightoffrequencyf then if it is moving away from an observer the observer willmeasurethelightemittedashavingalowerfrequencythan fSincethesensationofcolourvisionisrelatedtothefrequencyoflight(bluelightisofahigherfrequencythanred light) light emitted by objects moving way from anobserver is often referred to as being red-shifted whereas if the object is moving toward the observer it is referred to as blue-shiftedThisideaisusedinOptionE(Chapter16)

We do not need to consider here the situations where either the source or the observer is accelerating In a situation for example where an accelerating source is approaching astationaryobserverthentheobserverwillhearasoundof ever increasing frequency This sometimes leads toconfusion in describing what is heard when a source approaches passes and then recedes from a stationaryobserver Suppose for example that you are standingon a station platform and a train sounding its whistle is approaching at constant speed What will you hear as the train approaches and then passes through the station As the train approaches you will hear a sound of constant pitch but increasing loudness The pitch of the sound will be greater than if the train were stationary As the train passes through the station you will hear the pitch change at the momentthetrainpassesyoutoasoundagainofconstantpitch The pitch of this sound will be lower than the sound of the approaching train and its intensity will decrease as the train recedes from you What you do not hear is a sound of increasing pitch and then decreasing pitch

The Doppler equations for sound

Although you will not be expected in an IB examination to derivetheequationsassociatedwithaspectsoftheDopplereffectyouwillbeexpectedtoapplythemForcompletenesstherefore thederivationoftheequationsassociatedwiththeDopplereffectasoutlinedaboveisgivenhere

S primeS O

vs

Figure 931 (b)

InFigure931(b)theobserverOisatrestwithrespecttoasourceofsoundSismovingwithconstanspeedvs directly towards O The source is emitting a note of constant frequencyf and the speed of the emitted sound is v

S shows the position of the source ∆t later When the sourceisatresttheninatime∆t the observer will receive f∆t waves and these waves will occupy a distance v∆t ie

v f

λ =f∆tv∆t =

(Because of the motion of the source this number of waves will now occupy a distance (v∆t ndash vs∆t)The lsquonewrsquowavelength is therefore

v ndash vs

fλrsquo =

f∆tv∆t ndash vs∆t

=

If fisthefrequencyheardbyOthen

vfλ

= or

vf

λ = = sv vfminus

From which

s

vf fv v

=minus

Dividingthroughbyv gives

Equation1111

1 sf f v

v=

minus

If the source is moving away from the observer then we have

s

vf fv v

=minus

1

1 sf vv

=+

Equation112

PHYSICS 2015indb 304 150514 337 PM

Wave Phenomena

305

AH

L

We now consider the case where the source is stationary and the observerismoving towardsthesource with speed v0 In this situation the speed of the sound waves as measured by the observer will be v0 + v We therefore have that

v0 + v = ffraslλ = vfλ

times

From which

o1 vf fv

= +

Equation113

If the observerismovingawayfromthesource then

o1 vf fv

= minus

Fromequation(113)wehavethat

o o1 v vf f f f f fv v

∆ = minus = + minus =

Equation114

Thevelocitiesthatwerefertointheaboveequationsarethevelocities with respect to the medium in which the waves fromthesourcetravelHoweverwhenwearedealingwithalight source it is the relative velocity between the source and the observer that we must consider The reason for this is that lightisuniqueintherespectthatthespeedofthelightwavesdoes not depend on the speed of the source All observers irrespective of their speed or the speed of the source will measure the same velocity for the speed of light This is oneof thecornerstonesof theSpecialTheoryofRelativitywhichisdiscussedinmoredetailinOptionH(Chapter18)When applying the Doppler effect to light we aremainlyconcerned with the motion of the source We look here only at the situation where the speed of the source v is much smaller than the speed of light cinfreespace(v ltlt c)Underthesecircumstanceswhenthesourceismovingtowardstheobserverequation111becomes

vf f f fc

minus = ∆ = Equation115

andwhenthesourceismovingawayfromtheobserverequation112becomes vf f f f

cminus = ∆ = minus

Provided that v ltlt c these same equations apply for astationary source and moving observer

We look at the following example and exercise

Example

Asourceemitsasoundoffrequency440HzItmovesinastraight line towards a stationary observer with a speed of 30ms-1Theobserverhearsasoundoffrequency484HzCalculate the speed of sound in air

Solution

We use equation 111 and substitute f= 484 Hz f = 440 Hz

and vs = 30 m s-1

therefore 1

484 440 301v

=minus

such that 30 4401

484vminus = to

give v = 330 m s-1

Example

A particular radio signal from a galaxy is measured as havingafrequencyof139times109HzThesamesignalfromasourceinalaboratoryhasafrequencyof142times109Hz

Suggestwhy thegalaxy ismovingaway fromEarthandcalculateitsrecessionspeed(iethespeedwithwhichitismovingawayfromEarth)

Solution

The fact that the frequency from the moving source is less than that when it is stationary indicates that it is moving away from the stationary observer ie Earth

Using vf fc

∆ = we have

8 96

9

3 10 (142 139) 10 634 10142 10

c fvf∆ times times minus times

= = = timestimes

m s-1

It is usual when dealing with the Doppler effect of light to express speeds as a fraction of c So in this instance we have v = 0021 c

Using the Doppler effectWe have seen in the above example and exercise how the Dopplereffectmaybeusedtomeasuretherecessionspeedof distant galaxies The effect is also used to measure speed inothersituationsHerewewilllookatthegeneralprincipleinvolved in using the Doppler effect to measure speedFigure932showsasource(thetransmitter)thatemitseithersound or em waves of constant frequency f The waves

PHYSICS 2015indb 305 150514 337 PM

Chapter 9

306

AH

L

fromthesourceare incidentonareflectorthat ismovingtowards the transmitter with speed vThereflectedwavesare detected by the receiver placed alongside the transmitter

reector

v

transmitter

receiver

f f

f f

Figure 932 Using the Doppler effect to measure speed

We shall consider the situation where v ltlt c where c is the speed of the waves from the transmitter

For the reflector receiving waves from the transmitterit is effectively an observer moving towards a stationary sourceFromequation(114) itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation116

For the receiver receivingwaves from the reflector it iseffectively a stationary observer receiving waves from a movingsourceFromequation(115)itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation117

Ifweaddequations(116)and(117)wegetthatthetotalDopplershiftatthereceiver∆f is

v vf f f f fc c

minus = ∆ = +

But 1 vf fc

= +

hence

1 v v vf f fc c c

∆ = + +

But since v ltlt cwe can ignore the term2

2

vc when we

expandthebracketintheaboveequation

Therefore we have2vf fc

∆ = Equation118

If v asymp c then we must use the full Doppler equationsHowever for em radiationwewill alwaysonly considersituations in which v ltlt c

Example

Thespeedofsoundinbloodis1500times103 m s-1 Ultrasound offrequency100MHzisreflectedfrombloodflowinginan arteryThe frequencyof the reflectedwaves receivedbackatthetransmitteris105MHzEstimatethespeedofthebloodflowintheartery

Solution

Using equation (118) we have

6 63

2005 10 1015 10

vtimes = times

timesto give v asymp 36 m s-1 (We have assumed that the ultrasound is incident at right angles to the blood flow)

Exercise 971 Judy is standing on the platform of a station A

high speed train is approaching the station in a straight line at constant speed and is sounding its whistleAsthetrainpassesbyJudythefrequencyof the sound emitted by the whistle as heard by Judychangesfrom640Hzto430HzDetermine

(a) thespeedofthetrain

(b) thefrequencyofthesoundemittedbythewhistle as heard by a person on the train (Speedofsound=330ms-1)

2 AgalaxyismovingawayfromEarthwithaspeedof00500cThewavelengthofaparticularspectral line in light emitted by atomic hydrogen inalaboratoryis656times10-7 m Calculate the valueofthewavelengthofthislinemeasuredinthelaboratoryinlightemittedfromasourceofatomic hydrogen in the galaxy

PHYSICS 2015indb 306 150514 337 PM

10 Fields

Contents

101 ndash Describing fields

102 ndash Fields at work

Essential Ideas

Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Similar approaches can be taken in analysing electrical and gravitational potential problems copy IBO 2014

PHYSICS 2015indb 307 150514 337 PM

Chapter 10

308

AH

L

101 Describing fields

Essential idea Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Understandingsbull Gravitationalfieldsbull Electrostaticfieldsbull Electricpotentialandgravitationalpotentialbull Fieldlinesbull Equipotentialsurfaces

Gravitational potentialWehaveseenthatifweliftanobjectofmassm to a height h abovethesurfaceoftheEarththenitsgainingravitationalpotential energy is mghHoweverthisisbynomeansthefullstoryForastartwenowknowthatgvarieswithh and also the expression really gives a difference in potentialenergybetweenthevaluethattheobjecthasattheEarthrsquossurfaceandthevaluethatithasatheighthSowhatwereallyneedisazeropointCanwefindapointwherethepotentialenergyiszeroandusethispointfromwhichtomeasure changes in potential energy

The point that is chosen is in fact infinity At infinity the gravitationalfieldstrengthofanyobjectwillbezeroSoletusseeifwecandeduceanexpressionforthegaininpotentialenergyofanobjectwhenitisldquoliftedrdquofromthesurfaceoftheEarthtoinfinityThisineffectmeansfindingtheworknecessarytoperformthistask

δr

r r δr+

r infin=

mg

AB

Me

Figure 1001 Gravitational forces

Inthediagramweconsidertheworknecessarytomovethe particle of mass m a distance δr in thegravitationalfield of the Earth

The force on the particle at A is FG Mem

r2----------------=

IftheparticleismovedtoB then since δr isverysmallwe can assume that the field remains constant over thedistance ABTheworkδW done againstthegravitationalfieldoftheEarthinmovingthedistanceABis

δWGMem

r2----------------δrndash=

(rememberthatworkdoneagainstaforceisnegative)

TofindthetotalworkdoneW in going from the surface oftheEarthtoinfinitywehavetoaddalltheselittlebitsofworkThisisdonemathematicallybyusingintegralcalculus

WG Me m

r2----------------ndash

rdR

infin

int G Me m 1

r2----- rd

R

infin

intndash G Me m 1r---ndash

R

infinndash= = =

GMe m 0 1R---ndash

ndashndash=

G MemR----------------ndash=

HencewehavewhereRistheradiusoftheEarththattheworkdonebythegravitationalfieldinmovinganobjectofmass m from R(surfaceoftheEarth)toinfinityisgivenby

WGMe m

R---------------- ndash=

We can generalise the result by calculating the worknecessary per unit mass to take a small mass fromthe surface of the Earth to infinity This we call thegravitational potential Vie

Wewouldgetexactlythesameresultifwecalculatedtheworkdonebythefieldtobringthepointmassfrominfinityto the surface of Earth In this respect the formal definition ofgravitationalpotentialatapointinagravitationalfieldis therefore defined as the work done per unit mass in bringing a point mass from infinity to that point

Clearly then the gravitational potential at any point inthe Earthrsquos field distance r from the centre of the Earth (providingr gt R)is

VGMe

r------------ ndash=

Thepotentialisthereforeameasureoftheamountofworkthathastobedonetomoveparticlesbetweenpointsinagravitationalfieldanditsunit is theJkgndash1 We also note thatthepotentialisnegativesothatthepotentialenergy

NATURE OF SCIENCEParadigm shift The move from direct observable actions being responsible for influence on an object to acceptance of a fieldrsquos ldquoaction at a distancerdquo required a paradigm shift in the world of science (23)

copy IBO 2014

PHYSICS 2015indb 308 150514 337 PM

Fields

309

AH

L

aswemoveawayfromtheEarthrsquossurfaceincreasesuntilitreachesthevalueofzeroatinfinity

IfthegravitationalfieldisduetoapointmassofmassmthenwehavethesameexpressionasaboveexceptthatMe is replaced by m andmustalsoexclude thevalueof thepotential at the point mass itself ie at r = 0

WecanexpressthegravitationalpotentialduetotheEarth(orduetoanysphericalmass)intermsofthegravitationalfield strength at its surface

AtthesurfaceoftheEarthwehave

g 0ndash Re GMRe

---------ndash= e

Sothat

g 0Re2 GM= e

Hence at a distance r from the centre of the Earth the gravitationalpotentialVcanbewrittenas

VGMe

r------------ndashg 0 Re

2

r------------ndash= =

The potential at the surface of the Earth

(r = Re)istherefore-g0Re

It is interesting to see how the expression for thegravitational potential ties in with the expression mgh ThepotentialatthesurfaceoftheEarthis-g0Re (see the exampleabove)andataheighthwillbe-g0g 0 Re h+( )ndash ifweassume that g0doesnotchangeover thedistanceh The differenceinpotentialbetweenthesurfaceandtheheighth is therefore g0hSotheworkneededtoraiseanobjectof mass m to a height h is mgh ie m timesdifference ingravitationalpotential

This we have referred to as the gain in gravitationalpotentialenergy(see235)

However this expression can be extended to any twopoints inanygravitationalfieldsuch that ifanobjectofmass mmovesbetween twopointswhosepotentialsareV1 and V2 respectively then the change in gravitationalpotentialenergyoftheobjectism(V1 ndash V2)

Gravitational potential gradient

Let us consider now a region in space where thegravitationalfieldisconstantInFigure912thetwopointsAandBareseparatedbythedistance∆x

A B

direction of uniform gravitational eld of strength I

∆x

Figure 1002 The gravitational potential gradient

The gravitational field is of strength I and is in the direction shownThe gravitational potential at A is V andatBisV +∆V

Theworkdone is takingapointmassmfromAtoB isF∆x = mI∆x

Howeverbydefinitionthisworkisalsoequalto-m∆V

Therefore mI∆x=-m∆V

or VIx

∆= minus

∆Effectivelythissaysthatthemagnitudeofthegravitationalfield strength is equal to the negative gradient of thepotential If I is constant then V is a linear function of x and Iisequaltothenegativegradientofthestraightlinegraph formed by plotting V against x If I is not constant (asusuallythecase)thenthemagnitudeofI at any point in the field can be found by find the gradient of the V-x graph at that point An example of such a calculation can befoundinSection929

For those of you who do HL maths the relationship between field and potential is seen to follow from the expression for the potential of a point mass viz

mV Gr

= minus

2

ddV mG Ir r

minus = + =

Potential due to one or more point masses

Gravitationalpotentialisascalarquantitysocalculatingthepotential due to more than one point mass is a matter of simpleadditionSoforexamplethepotentialV due to the Moon and Earth and a distance x fromthecentreofEarthonastraightlinebetweenthemisgivenbytheexpression

E MM MV Gx r x

= minus + minus

whereME=massofEarthMM= mass of Moon and r = distancebetweencentreofEarthandMoon

PHYSICS 2015indb 309 150514 337 PM

Chapter 10

310

AH

L

Equipotentials and field linesIfthegravitationalpotentialhasthesamevalueatallpointson a surface the surface is said to be an equipotential surfaceSoforexampleifweimagineasphericalshellaboutEarthwhosecentrecoincideswiththecentreofEarththisshellwillbeanequipotentialsurfaceClearlyifwerepresentthegravitationalfieldstrengthbyfieldlinessincethelinesldquoradiaterdquooutfromthecentreofEarththentheselineswillbeatrightanglestothesurfaceIfthefieldlineswerenotnormaltotheequipotentialsurfacethentherewouldbeacomponentof the field parallel to the surfaceThis wouldmean thatpointsonthesurfacewouldbeatdifferentpotentialsandsoitwouldnolongerbeanequipotentialsurfaceThisofcourseholdstrueforanyequipotentialsurface

Figure 1003 shows the field lines and equipotentials fortwopointmassesm

m m

Figure 1003 Equipotentials for two point masses

It is worth noting that we would get exactly the samepattern ifwewere to replace thepointmasseswith twoequalpointcharges(See935)

Escape speedThe potential at the surface of Earth is which meansthattheenergyrequiredtotakeaparticleofmass mfromthesurfacetoinfinityisequalto

ButwhatdoesitactuallymeantotakesomethingtoinfinityWhentheparticleisonthesurfaceoftheEarthwecanthinkofitassittingatthebottomofaldquopotentialwellrdquoasinfigure1004

particle

surface of Earth

innity

GMR---------

FIgure 1004 A potential well

Theldquodepthrdquoofthewellis and if the particle gains an amount of kinetic energy equal to wherem is its massthenitwillhavejustenoughenergytoldquoliftrdquoitoutofthewell

In reality it doesnrsquot actually go to infinity it justmeansthat the particle is effectively free of the gravitationalattractionof theEarthWe say that ithas ldquoescapedrdquo theEarthrsquosgravitationalpullWemeetthisideainconnectionwithmolecular forcesTwomolecules in a solidwill sitat their equilibrium position the separation where therepulsiveforceisequaltotheattractiveforceIfwesupplyjust enough energy to increase the separation of themolecules such that they are an infinite distance apart then intermolecularforcesnolongeraffectthemoleculesandthesolidwillhavebecomealiquidThereisnoincreaseinthekineticenergyofthemoleculesandsothesolidmeltsat constant temperature

Wecancalculatetheescapespeedofanobjectveryeasilyby equating the kinetic energy to the potential energysuch that

12---mvesc ape

2 GMe mRe

----------------=

vesc aperArr2GMe

Re--------------- 2g0 Re = =

Substituting for g0 and Regivesavalueforvescape of about 11kmsndash1 from the surface of the Earth

Youwillnotethattheescapespeeddoesnotdependonthemassoftheobjectsincebothkineticenergyandpotentialenergy are proportional to the mass

IntheoryifyouwanttogetarockettothemoonitcanbedonewithoutreachingtheescapespeedHoweverthiswouldnecessitate anenormousamountof fuel and it islikely that therocketplus fuelwouldbesoheavythat itwouldnevergetoffthegroundItismuchmorepracticalto accelerate the rocket to the escape speed fromEarthorbitandthenintheoryjustlaunchittotheMoon

Example

Use the followingdata todetermine thepotential at thesurface of Mars and the magnitude of the acceleration of free fall

massofMars =64times1023kg

radiusofMars =34times106 m

Determinealsothegravitationalfieldstrengthatadistanceof 68 times 106mabovethesurfaceofMars

PHYSICS 2015indb 310 150514 337 PM

Fields

311

AH

L

Solution

= minus = minus times times = times23

11 76

64 1067 10 ndash13 10

34 10MV GR

minus timestimes

Nkg-1

But V = -g0R

Therefore 7

0 6

13 10 3834 10

VgR

times= minus = =

timesm s-2

To determine the field strength gh at 68 times 106 m above the surface we use the fact that 0 2

Mg GR

= such that GM = g0R2

Therefore 2 2

0h 2 2 2

h h

38 (34) 042(102)

g RGMgR R

times= = = = m s-2

(the distance from the centre is 34 times 106 + 68 times 106 = 102 times 106 m)

Exercise 1011 Thegraphbelowshowshowthegravitational

potentialoutsideoftheEarthvarieswithdistancefrom the centre

0 10 20 30 40 50 60 70

ndash1

ndash2

ndash3

ndash4

ndash5

ndash6

ndash7

r m times 106

V

Jkgndash1

times 10

7

(a) Usethegraphtodeterminethegainingravitationalpotentialenergyofasatelliteofmass200kgasitmovesfromthesurfaceof the Earth to a height of 30 times 107mabovetheEarthrsquossurface

(b) Calculatetheenergyrequiredtotakeit to infinity

(c) DeterminetheslopeofthegraphatthesurfaceoftheEarthmCommenton youranswer

Electric potential energy

The concept of electric potential energy was developedwith gravitational potential energy in mind Just as anobjectnearthesurfaceoftheEarthhaspotentialenergybecause of its gravitational interaction with the Earthsotoothereiselectricalpotentialenergyassociatedwithinteracting charges

Letusfirstlookatacaseoftwopositivepointchargeseachof 1μC that are initially bound together by a thread in a vacuuminspacewithadistancebetweenthemof10cmasshowninFigure1005Whenthethreadiscutthepointchargesinitiallyatrestwouldmoveinoppositedirectionsmovingwithvelocitiesv1 and v2 along the direction of the electrostatic force of repulsion

BEFORE AFTER

v v21

Figure 1005 Interaction of two positive particles

The electric potential energybetweentwopointchargescan be found by simply adding up the energy associated witheachpairofpointchargesForapairofinteractingchargestheelectricpotentialenergyisgivenby

∆U = ∆Ep + ∆Ek = ∆W = ∆Fr = kqQ

____ r2 times r = kqQ

____ r

Becausenoexternalforceisactingonthesystemtheenergyand momentum must be conserved Initially Ek = 0 and Ep =kqQr=9times109 times 1 times 10-1201m=009JWhentheyareagreatdistance fromeachotherEpwillbenegligibleThefinalenergywillbeequaltofrac12mv1

2+frac12mv22=009J

Momentum isalsoconservedand thevelocitieswouldbethe same magnitude but in opposite directions

Electric potential energyismoreoftendefinedintermsofapointchargemovinginanelectricfieldas

lsquothe electricpotential energybetweenany twopoints inan electric field is defined as negativeoftheworkdone by anelectricfieldinmovingapoint electric chargebetweentwolocationsintheelectricfieldrsquo

ΔU = ΔEp = -ΔW = -Fd = qEx

ΔU = ΔEp +ΔEk = ΔW = ΔFr = kqQ r2 x r = kqQ

wherexisthedistancemovedalong(oroppositeto)thedirection of the electric field

Electric potential energyismeasuredinjoule(J)Justasworkisascalarquantitysotooelectricalpotentialenergyis a scalar quantity The negative of the work done by an electric field in moving a unit electric charge between two points is independent of the path taken Inphysicswesaytheelectricfieldisalsquoconservativersquo field

PHYSICS 2015indb 311 150514 337 PM

Chapter 10

312

AH

L

Supposeanexternalforcesuchasyourhandmovesasmallpositive point test charge in the direction of a uniformelectric fieldAs it ismoving itmust be gaining kineticenergyIfthisoccursthentheelectricpotentialenergyofthe unit charge is changing

InFigure1006apointcharge+qismovedbetweenpointsAandBthroughadistancex in a uniform electric field

B

A

x

+ q

Figure 1006 Movement of a positive point charge in a uniform field

InordertomoveapositivepointchargefrompointAtopointBanexternal forcemustbeapplied to thechargeequaltoqE (F = qE)

Since the force is applied through a distance xthennegativeworkhas to bedone tomove the charge because energyis gainedmeaning there is an increaseelectric potential energybetweenthetwopointsRememberthattheworkdoneisequivalenttotheenergygainedorlostinmovingthe charge through the electric field The concept of electric potential energy is only meaningful as the electric field whichgeneratestheforceinquestionisconservative

W F xtimes E q x times= =

θ x cos θ x

Figure 1007 Charge moved at an angle to the field

If a chargemoves at an angle θ to an electric field thecomponent of the displacement parallel to the electric fieldisusedasshowninFigure918

W F x Eq x θcostimes= =

Theelectricpotentialenergyisstoredintheelectricfieldand theelectricfieldwill return theenergy to thepointcharge when required so as not to violate the Law ofconservationofenergy

Electric potential

The electric potential at a point in an electric field is defined as being the work done per unit charge in bringing a small positive point charge from infinity to that point

ΔV = Vinfin ndash Vf = ndash W ___ q

Ifwedesignatethepotentialenergytobezeroatinfinitythen it follows thatelectricpotentialmustalsobezeroat infinity and the electric potential at any point in an electricfieldwillbe

ΔV = ndash W ___ q

NowsupposeweapplyanexternalforcetoasmallpositivetestchargeasitismovedtowardsanisolatedpositivechargeTheexternalforcemustdoworkonthepositivetestchargetomoveittowardstheisolatedpositivechargeandtheworkmustbepositivewhiletheworkdonebytheelectricfieldmustthereforebenegativeSotheelectricpotentialatthatpointmustbepositiveaccordingtotheaboveequationIfanegativeisolatedchargeisusedtheelectricpotentialatapointonthepositivetestchargewouldbenegativePositivepoint chargesof their ownaccordmove fromaplaceofhighelectricpotential toaplaceof lowelectricpotentialNegative point charges move the other way from lowpotentialtohighpotentialInmovingfrompointAtopointBinthediagramthepositivecharge+qismovingfromalowelectricpotentialtoahighelectricpotential

In thedefinitiongiven the termldquoworkperunitchargerdquohas significance If the test charge is +16 times 10-19Cwherethe charge has a potential energy of 32 times 10-17Jthenthepotentialwouldbe32times10-17J+16times10-19C=200JC-1 Now if the charge was doubled the potential wouldbecome 64times10-17 J However the potential per unitchargewouldbethesame

Electric potential is a scalar quantityandithasunitsJC-1 orvoltswhere1voltequalsone joulepercoloumbThevoltallowsustoadoptaunitfortheelectricfieldintermsofthevolt

PreviouslytheunitfortheelectricfieldwasNCndash1

W = qV and F = qEsoW ___ V = F __ E

E = FV ___ W = FV ___ Fm V mndash1

That is the units of the electric field E can also beexpressed as V mndash1

The work done per unit charge in moving a point charge between two points in an electric field is again independent of the path taken

PHYSICS 2015indb 312 150514 337 PM

Fields

313

AH

L

Electric potential due to a point charge

Letus takeapoint rmetres froma chargedobjectThepotentialatthispointcanbecalculatedusingthefollowing

W = ndashFr = ndashqV and F = ndash q1 q2 _____ 4πε0r

2

Therefore

W = ndash q1 q2 _____ 4πε0r

2 times r = ndash q1 q2 _____ 4πε0r

= ndashq1 times q2 _____ 4πε0r

= ndashq1V

That is V q

4πε0r--------------- =

Orsimply

V kqr------

=

Example

Determinehowmuchworkisdonebytheelectricfieldofpointcharge150μCwhenachargeof200μCismovedfrom infinity to apoint 0400m from thepoint charge(Assumenoaccelerationofthecharges)

Solution

The work done by the electric field is W = -qV = -14πε0 times q times (Q rinfin - Q r0400)

W = (- 200 times 10-6 C times 900 times 109 NmC-2 times 150 times 10-6 C) divide 0400 m = - 0675 J

An external force would have to do +0675 J of work

Electric field strength and electric potential gradient

Letus lookbackatFigure1006Supposeagain that thecharge+qismovedasmalldistancebyaforceFfromAtoBsothattheforcecanbeconsideredconstantTheworkdoneisgivenby

∆W F x∆times=

TheforceFandtheelectricfieldEareoppositelydirectedandweknowthat

F = -qE and ΔW = q ΔV

Thereforetheworkdonecanbegivenas

q ΔV = -q E Δx

Therefore E V∆x∆-------

ndash=

The rate of change of potential ΔV at a point with respect to distance Δx in the direction in which the change is maximum is called the potential gradient We say that the electric field = - the potential gradient and the units are Vm-

1 From the equation we can see that in a graph of electric potential versus distance the gradient of the straight line equals the electric field strength

In reality if a charged particle enters a uniform electric field it will be accelerated uniformly by the field and its kinetic energy will increase This is why we had to assume no acceleration in the last worked example

Ek∆ 12---

mv2 q E xsdot sdot q Vx---

xsdot sdot q Vsdot= = = =

Example

Determine how far apart two parallel plates must besituated so that a potential difference of 150 times 102 V produces an electric field strength of 100 times 103 NC-1

Solution

Using E V∆x------- xhArrndash V∆

E-------15 102 Vtimes

100 103 N C 1ndashtimes------------------------------------------= = =

= 150 times 10-1

The plates are 150 times 10-1 m apart

The electric field and the electric potential at a point due to an evenly distributed charge +q on a sphere can be represented graphically as in Figure 1008

r

r

V

E

r0

r0

r0

Charge of +Q evenly distributed over surface

E 14πε0------------ Q

r2----- r r0gt=

V 14πε0------------Q

r----r r0gt=

Figure 1008 Electric field and potential due to a charged sphere

PHYSICS 2015indb 313 150514 337 PM

Chapter 10

314

AH

L

When the sphere becomes charged we know that thechargedistributes itselfevenlyoverthesurfaceThereforeeverypartofthematerialoftheconductorisatthesamepotential As the electric potential at a point is defined as beingnumericallyequaltotheworkdoneinbringingaunitpositivechargefrominfinitytothatpointithasaconstantvalueineverypartofthematerialoftheconductor

Since the potential is the same at all points on the conducting surfacethen∆V∆xiszeroButE=ndash∆V∆xThereforethe electric field inside the conductor is zero There is no electric field inside the conductor

SomefurtherobservationsofthegraphsinFigure1009are

bull OutsidethespherethegraphsobeytherelationshipsgivenasE α 1 r2 and V α 1 r

bull Atthesurfacer = r0Thereforetheelectricfieldandpotentialhavetheminimumvalueforr at this point and this infers a maximum field and potential

bull Insidethespheretheelectricfieldiszerobull Insidethespherenoworkisdonetomoveacharge

fromapointinsidetothesurfaceThereforethereisno potential difference and the potential is the same asitiswhenr = r0

Similargraphscanbedrawnfortheelectricfieldintensityandthe electric potential as a function of distance from conducting parallelplatesandsurfacesandthesearegiveninFigure1009

Potential plot E eld plotE eld

++++

ndashndashndashndash

xx x

+

x x

ndashx x

+ ndashx

x

+ +

x

x

Figure 1009 Electric field and electric potential at a distance from a charged surface

Potential due to one or more point charges

The potential due to one point charge can be determined byusingtheequationformula

V = kq r

Example 1

Determine the electric potential at a point 20 times 10-1 m fromthecentreofan isolatedconducting spherewithapointchargeof40pCinair

Solution

Using the formula

V = kq r we have

V90 10 9times( ) 40 10 12ndashtimes( )times

20 10 1ndashtimes( )------------------------------------------------------------------ 018 V= =

the potential at the point is 180 times 10-1 V

The potential due to a number of point charges can be determined by adding up the potentials due to individual point charges because the electric potential at any point outside a conducting sphere will be the same as if all the charge was concentrated at its centre

Example 2

Threepointchargesofareplacedattheverticesofaright-angledtriangleasshowninthediagrambelowDeterminetheabsolutepotentialat the+20μCchargeduetothetwoothercharges

- 6μC

+3μC +2μC

4 m

3 m

Solution

The electric potential of the +2 μC charge due to the ndash 6 μC charge is

V = (9 times 109 Nm2C-2 times -6 times 10-6 C) divide (radic 32 + 42) m = - 108 times 104 V

PHYSICS 2015indb 314 150514 338 PM

Fields

315

AH

L

Insummarywecanconcludethat

bull No work is done to move a charge along an equipotential

bull Equipotentials are always perpendicular to the electric lines of force

Figure1011and1012showsomeequipotential lines fortwo oppositely charged and identically positive spheresseparated by a distance

+ve ndashve

equipotential lines

Figure 1011 Equipotential lines between two opposite charges

equipotential lines

Figure 1012 Equipotential lines between two charges which are the same

+

ndash ndash ndash ndash

+ + +

40 V

30 V

20 V

10 V

50 V

Figure 1013 Equipotential lines between charged parallel plates

Figure1013showstheequipotentiallinesbetweenchargedparallel plates Throughout this chapter the similarities and differences between gravitational fields and electric fieldshavebeendiscussedTherelationshipsthatexistsbetweengravitationalandelectricquantitiesandtheeffectsofpointmassesandchargesissummarisedinTable1014

The electric potential of the +2 μC charge due to the +3 μC charge is

V = (9 times 109 Nm2C-2 times 3 times 10-6 C) divide 3m = 9 times 103 V

The net absolute potential is the sum of the 2 potentials- 108 times 104 V + 9 times 103 V =- 18 times 103 V

The absolute potential at the point is - 18 times 103 V

Equipotential surfacesRegions in space where the electric potential of a chargedistributionhasaconstantvaluearecalledequipotentials The placeswherethepotentialisconstantinthreedimensionsarecalled equipotential surfacesandwheretheyareconstantintwodimensionstheyarecalledequipotential lines

TheyareinsomewaysanalogoustothecontourlinesontopographicmapsInthiscasethegravitationalpotentialenergy is constant as amassmoves around the contourlines because the mass remains at the same elevationabovetheEarthrsquossurfaceThegravitationalfieldstrengthacts in a direction perpendicular to a contour line

Similarlybecausetheelectricpotentialonanequipotentiallinehas thesamevalueanelectric forcecandonoworkwhena testchargemovesonanequipotentialThereforethe electric field cannot have a component along anequipotentialandthusitmustbeeverywhereperpendiculartotheequipotentialsurfaceorequipotentiallineThisfactmakesiteasytoplotequipotentialsifthelinesofforceorlinesofelectricfluxofanelectricfieldareknown

Forexamplethereareaseriesofequipotentiallinesbetweentwoparallelplateconductorsthatareperpendiculartotheelectric fieldTherewill be a series of concentric circles(eachcirclefurtherapartthanthepreviousone)thatmapouttheequipotentialsaroundanisolatedpositivesphereThe lines of force and some equipotential lines for anisolatedpositivesphereareshowninFigure1010

Lines ofequipotential

Figure 1010 Equipotentials around an isolated positive sphere

PHYSICS 2015indb 315 150514 338 PM

Chapter 10

316

AH

L

Gravitational quantity Electrical quantityQ

uant

ities

g V∆x∆-------ndash= E V∆

x∆-------ndash=

Poin

t mas

ses a

nd ch

arge

s

V Gmr----

ndash= V 14πε0------------q

r---=

g G m

r2-----ndash= E 14πε0------------ q

r2-----=

F Gm1 m2

r2--------------= F 1

4πε0------------

q1q2

r2-----------=

Figure 1014 Formulas

Example

Deduce the electric potential on the surface of a gold nucleus that has a radius of 62 fm

Solution

Using the formula

V = kq r and knowing the atomic number of gold is 79 We will assume the nucleus is spherical and it behaves as if it were a point charge at its centre (relative to outside points)

V = 90 times 109 Nm2C-2 times 79 times 16 times 10-19 C divide 62 times 10-15 m = 18 times 107 V

The potential at the point is 18 MV

Example

Deduce the ionisation energy in electron-volts of theelectron in the hydrogen atom if the electron is in its ground state and it is in a circular orbit at a distance of 53 times 10-11 m from the proton

Solution

This problem is an energy coulombic circular motion question based on Bohrrsquos model of the atom (not the accepted quantum mechanics model) The ionisation energy is the energy required to remove the electron from the ground state to infinity The electron travels in a circular orbit and therefore has a centripetal acceleration The ionisation energy will counteract the coulombic force and the movement of the electron will be in the opposite direction to the centripetal force

Total energy = Ek electron + Ep due to the proton-electron interaction

ΣF = kqQ r2 = mv2 r and as such mv2 = = kqQ r

Therefore Ek electron = frac12 kqQ r

Ep due to the proton-electron interaction = - kqQ r

Total energy = frac12 kqQ r + - kqQ r = -frac12 kqQ r

= - 90 times 109 Nm2C-2 times (16 times 10-19 C)2 divide 53 times 10-11 m = -217 times 10-18 J

= -217 times 10-18 J divide 16 times 10-19 = -136 eV

The ionisation energy is 136 eV

Exercise 1021 ApointchargePisplacedmidwaybetweentwo

identicalnegativechargesWhichoneofthefollowingiscorrectwithregardstoelectricfieldandelectricpotentialatpointP

Electric field Electric potentialA non-zero zeroB zero non-zeroC non-zero non-zeroD zero zero

2 Twopositivechargedspheresaretiedtogetherinavacuumsomewhereinspacewheretherearenoexternal forces A has a mass of 25 g and a charge of20μCandBhasamassof15gandachargeof30μCThedistancebetweenthemis40cmTheyarethenreleasedasshowninthediagram

BEFORE AFTER

v1 v2

A B

(a) Determinetheirinitialelectricpotentialenergy in the before situation

(b) DeterminethespeedofsphereBafterrelease

3 Thediagrambelowrepresentstwoequipotentiallines in separated by a distance of 5 cm in a uniform electric field

+ + + + + + + +

ndash ndash ndash ndash ndash ndash ndash ndash

40 V

20 V 5 cm

PHYSICS 2015indb 316 150514 338 PM

Fields

317

AH

L

Determine the strength of the electric field

4 Thisquestionisabouttheelectricfieldduetoacharged sphere and the motion of electrons in that fieldThediagrambelowshowsanisolatedmetalsphereinavacuumthatcarriesanegativeelectriccharge of 60 μC

ndash

(a) Onthediagramdrawtheconventionalwayto represent the electric field pattern due to the charged sphere and lines to represent threeequipotentialsurfacesintheregionoutside the sphere

(b) Explainhowthelinesrepresentingtheequipotentialsurfacesthatyouhavesketchedindicatethatthestrengthoftheelectricfieldisdecreasingwithdistancefrom the centre of the sphere

(c) Theelectricfieldstrengthatthesurfaceofthe sphere and at points outside the sphere can be determined by assuming that the sphere acts as a point charge of magnitude 60 μC at its centre The radius of the sphere is 25 times 10ndash2 m Deduce that the magnitude of the field strength at the surface of the sphere is 86 times 107 Vmndash1

An electron is initially at rest on the surface of the sphere

(d) (i) Describethepathfollowedbytheelectronasitleavesthesurfaceofthesphere

(ii) Calculatetheinitialaccelerationofthe electron

5 Determinetheamountofworkthatisdoneinmovingachargeof100nCthroughapotentialdifferenceof150times102 V

6 Three identical 20 μC conducting spheres are placedatthecornersofanequilateraltriangleofsides 25 cm The triangle has one apex C pointing upthepageand2baseanglesAandBDeterminetheabsolutepotentialatB

7 Determinehowfaraparttwoparallelplatesmustbesituatedsothatapotentialdifferenceof250 times 102 V produces an electric field strength of 200 times 103 NC-1

8 Thegapbetweentwoparallelplatesis10times10-3mandthereisapotentialdifferenceof10times104 V betweentheplatesCalculate

i theworkdonebyanelectroninmovingfrom one plate to the other

ii thespeedwithwhichtheelectronreachesthe second plate if released from rest

iii theelectricfieldintensitybetweentheplates

9 Anelectronguninapicturetubeisacceleratedby a potential 25 times 103VDeterminethekineticenergygainedbytheelectroninelectron-volts

10 Determine the electric potential 20 times 10-2 m from achargeof-10times10-5 C

11 Determinetheelectricpotentialatapointmid-waybetweenachargeofndash20pCandanotherof+5pConthelinejoiningtheircentresifthecharges are 10 cm apart

12 During a thunderstorm the electric potential differencebetweenacloudandthegroundis10 times 109 V Determine the magnitude of the change in electric potential energy of an electron thatmovesbetweenthesepointsinelectron-volts

13 A charge of 15 μC is placed in a uniform electric fieldoftwooppositelychargedparallelplateswithamagnitudeof14times103 NC-1

(a) Determinetheworkthatmustbedoneagainstthefieldtomovethepointchargeadistance of 55 cm

(b) Calculatethepotentialdifferencebetweenthe final and initial positions of the charge

(c) Determinethepotentialdifferencebetweenthe plates if their separation distance is 15 cm

14 Duringaflashoflightningthepotentialdifferencebetweenacloudandthegroundwas12 times 109 V and the amount of transferred charge was32C

(a) Determinethechangeinenergyofthetransferred charge

(b) Iftheenergyreleasedwasallusedtoacceleratea1tonnecardeduceitsfinalspeed

(c) Iftheenergyreleasedcouldbeusedtomelticeat0degCdeducetheamountoficethatcould be melted

PHYSICS 2015indb 317 150514 338 PM

Chapter 10

318

AH

L

15 SupposethatwhenanelectronmovedfromAtoBin the diagram along an electric field line that the electric field does 36 times 10-19Jofworkonit

Determinethedifferencesinelectricpotential

(a) VB ndash VA(b) VC ndash VA(c) VC ndash VB

16 DeterminethepotentialatpointPthatislocatedatthecentreofthesquareasshowninthediagrambelow

- 6μC

+3μC +2μC

1m

1m

5μC

P

102 Fields at work

Essential idea Similar approaches can be taken inanalysingelectricalandgravitationalpotentialproblems

Understandingsbull Potentialandpotentialenergybull Potentialgradientbull Potentialdifferencebull Escapespeedbull Orbitalmotionorbitalspeedandorbitalenergybull Forcesandinverse-squarelawbehaviour

Orbital motion orbital speed and orbital energy

Although orbital motion may be circular elliptical orparabolic this sub-topic only dealswith circular orbitsThis sub-topic is not fundamentally new physics butan application that synthesizes ideas from gravitationcircularmotiondynamicsandenergy

TheMoonorbitstheEarthandinthissenseitisoftenreferredtoasasatelliteoftheEarthBefore1957itwastheonlyEarthsatelliteHoweverin1957theRussianslaunchedthefirstmanmadesatelliteSputnik1SincethisdatemanymoresatelliteshavebeenlaunchedandtherearenowliterallythousandsofthemorbitingtheEarthSomeareusedtomonitortheweathersome used to enable people to find accurately their position on thesurfaceoftheEarthmanyareusedincommunicationsandnodoubtsomeareusedtospyonothercountriesFigure1015showshowinprincipleasatellitecanbeputintoorbit

Theperson(whosesizeisgreatlyexaggeratedwithrespectto Earth) standing on the surface on the Earth throwssome stonesThe greater the speed with which a stoneis thrown the further it will land from her The pathsfollowedbythethrownstonesareparabolasByastretchof the imaginationwecanvisualisea situation inwhichastoneis thrownwithsuchaspeedthatbecauseof thecurvatureoftheEarthitwillnotlandonthesurfaceoftheEarthbutgointoldquoorbitrdquo(Path4onFigure1015)

NATURE OF SCIENCECommunication of scientific explanations The ability to apply field theory to the unobservable (charges) and the massively scaled (motion of satellites) required scientists to develop new ways to investigate analyse and report findings to a general public used to scientific discoveries based on tangible and discernible evidence (51)

copy IBO 2014

PHYSICS 2015indb 318 150514 338 PM

Fields

319

AH

L

1

2

34Earth

Figure 1015 Throwing a stone into orbit

TheforcethatcausesthestonestofollowaparabolicpathandtofalltoEarthisgravityandsimilarlytheforcethatkeeps the stone in orbit is gravity For circular motionto occur we have seen that a force must act at rightanglestothevelocityofanobjectthatistheremustbeacentripetalforceHenceinthesituationwedescribeherethe centripetal force for circular orbital motion about the EarthisprovidedbygravitationalattractionoftheEarth

We can calculate the speedwithwhich a stonemustbethrowninordertoputitintoorbitjustabovethesurfaceof the Earth

If the stone has mass m and speed v thenwehave fromNewtonrsquos2ndlaw

mv2

RE---------- G

MEm

RE2------------=

whereRE is the radius of the Earth and ME is the mass of the Earth

Bearinginmindthatg 0 GME

RE2--------= then

v g RE 10 64 106timestimes 8 103times= = =

Thatisthestonemustbethrownat8times103m sndash1

Clearly we are not going to get a satellite into orbit soclosetothesurfaceoftheEarthMovingatthisspeedthefrictionduetoairresistancewouldmeltthesatellitebeforeithadtravelledacoupleofkilometresInrealitythereforeasatelliteisputintoorbitabouttheEarthbysendingitattachedtoarocketbeyondtheEarthrsquosatmosphereandthengivingitacomponentofvelocityperpendiculartoaradialvectorfromtheEarthSeeFigure1016

Earth Satellite carried by rocket to hereSatellite orbit

Tangential component of velocity

Figure 1016 Getting a satellite into orbit

Keplerrsquos third law(This work of Kepler and Newtonrsquos synthesis of the work is an excellent example of the scientific method and makes for a good TOK discussion)

In 1627 Johannes Kepler (1571-1630)publishedhis lawsof planetary motion The laws are empirical in natureandwere deduced from the observations of theDanishastronomer Tycho de Brahe (1546-1601)The third lawgives a relationship between the radius of orbit R of a planet and its period TofrevolutionabouttheSunThelawisexpressedmathematicallyas

2

3 constantTR

=

We shall nowuseNewtonrsquos LawofGravitation to showhowitisthattheplanetsmoveinaccordancewithKeplerrsquosthirdlaw

InessenceNewtonwasabletousehislawofgravitytopredict the motion of the planets since all he had to do was factor theF givenby this law intohis second law F = ma to find their accelerations and hence their future positions

InFigure1017theEarthisshownorbitingtheSunandthedistancebetweentheircentresisR

RSun

Earth

F es

F se

Figure 1017 Planets move according to Keplerrsquos third law

Fes is the force that the Earth exerts on the Sun and Fse is the force that the Sun exerts on the Earth The forces are equalandoppositeandtheSunandtheEarthwillactuallyorbitaboutacommoncentreHoweversince theSun issoverymuchmoremassivethantheEarththiscommoncentrewillbeclosetothecentreoftheSunandsowecanregard the Earth as orbiting about the centre of the Sun Theotherthingthatweshallassumeisthatwecanignorethe forces that the other planets exert on the Earth (This wouldnotbeawisethingtodoifyouwereplanningtosend a space ship to theMoon for example)We shallalso assume that we have followed Newtonrsquos exampleandindeedprovedthataspherewillactasapointmasssituated at the centre of the sphere

Kepler had postulated that the orbits of the planets areellipticalbut since theeccentricityof theEarthrsquosorbit issmallweshallassumeacircularorbit

PHYSICS 2015indb 319 150514 338 PM

Chapter 10

320

AH

L

TheaccelerationoftheEarthtowardstheSunisa = Rω2

whereω ω 2πT------=

Hencea R 2π

T------ times

2 4π2 RT 2------------- = =

But the acceleration is given by Newtonrsquos Second Law F = mawhereFisnowgivenbytheLawofGravitationSo in this situation

F maGMs Me

R2-------------------= = butwealsohavethat

a a R 2πT

------ times

2 4π2 RT 2------------- = = and m = Me so that

G MsMe

R2------------------- Me4π2R

T2-------------times

GMs

4π2-----------rArr R3

T2------= =

Butthequantity

isaconstantthathasthesamevalueforeachoftheplanetssowehaveforalltheplanetsnotjustEarththat

R3

T2------ k =

wherekisaconstantWhichisofcourseKeplerrsquosthirdlaw

ThisisindeedanamazingbreakthroughItisdifficulttorefutethe idea that all particles attract each other in accordance withtheLawofGravitationwhenthelawisabletoaccountfortheobservedmotionoftheplanetsabouttheSun

The gravitational effects of the planets upon each othershould produce perturbations in their orbits Such is the predictivepoweroftheUniversalGravitationalLawthatit enabled physicists to compute these perturbations The telescopehadbeeninventedin1608andbythemiddleofthe 18th Century had reached a degree of perfection in design that enabled astronomers to actually measure the orbital perturbations of the planets Their measurements werealways inagreementwith thepredictionsmadebyNewtonrsquoslawHoweverin1781anewplanetUranuswasdiscovered and the orbit of this planet did not fit withthe orbit predicted by Universal Gravitation Such wasthe physicistrsquos faith in theNewtonianmethod that theysuspectedthatthediscrepancywasduetothepresenceofayetundetectedplanetUsingtheLawofGravitationtheFrenchastronomerJLeverrier and the English astronomer J C Adamswereable tocalculate justhowmassive thisnewplanetmustbeandalsowhereitshouldbeIn1846the planetNeptunewas discovered justwhere they hadpredicted Ina similarwaydiscrepancies in theorbitofNeptune ledto thepredictionandsubsequentdiscoveryin1930oftheplanetPlutoNewtonrsquosLawofGravitation

had passed the ultimate test of any theory it is not only abletoexplainexistingdatabutalsotomakepredictions

Satellite energyWhenasatelliteisinorbitaboutaplanetitwillhavebothkineticenergyandgravitationalpotentialenergySupposewe consider a satellite ofmass m that is in an orbit of radius r about a planet of mass M

Thegravitationalpotentialduetotheplanetatdistancer from its centre is

ThegravitationalpotentialenergyofthesatelliteVsat

is therefore GMe m

r----------------ndash

ThatisVsat GMem

r----------------ndash=

Thegravitationalfieldstrengthatthesurfaceoftheplanetisgivenby

g 0GMe

Re2------------=

Hencewecanwrite

Vsat g0 Re

2

rndash=

m

ThekineticenergyofthesatelliteKsatisequaltofrac12mv2wherev is its orbital speed

Byequatingthegravitationalforceactingonthesatellitetoitscentripetalaccelerationwehave

G Mem

r2---------------- mv2

r---------- mv2hArr

GMemr----------------= =

Fromwhich12---

mv2 12---

GMemr----------------times=

= 0 e

2g R

r

2m

Which is actually quite interesting since it shows thatirrespective of the orbital radius the KE is numericallyequal to half the PE Also the total energy E tot K sat Vsat+ 1

2---G Me m

r----------------times G Me m

r----------------ndash + 1

2---G Me m

r----------------ndash= = =of the satelliteisalwaysnegativesince

E tot K sat Vsat+ 12---

G Me mr----------------times

G Me mr----------------ndash

+ 12---

G Me mr----------------ndash= = =

PHYSICS 2015indb 320 150514 338 PM

Fields

321

AH

L

The energies of an orbiting satellite as a function of radial distancefromthecentreofaplanetareshownplottedinFigure1018

12

100806

0402

0ndash02

ndash04

ndash06ndash08ndash10

ndash12

2 4 6 8 10 12

ener

gya

rbita

ry u

nits

distance timesR

kinetic energytotal energy

potential energy

Figure 1018 Energy of an orbiting satellite as a function of distance from the centre of a planet

WeightlessnessSuppose that you are inanelevator(lift)thatisdescendingat constant speedandyou let goof abook thatyouareholdinginyourhandThebookwillfalltothefloorwithaccelerationequaltotheaccelerationduetogravityIfthecablethatsupportstheelevatorweretosnap(asituationthatItrustwillneverhappentoanyofyou)andyounowletgothebookthatyouareholdinginyourotherhandthis book will not fall to the floor - it will stay exactlyin linewithyourhandThis isbecause thebook isnowfallingwith the sameaccelerationas theelevatorandassuch the book cannot ldquocatchrdquo up with the floor of theelevatorFurthermoreifyouhappenedtobestandingonasetofbathroomscalesthescaleswouldnowreadzero-youwouldbeapparentlyweightlessItisthisideaoffreefallthatexplainstheapparentweightlessnessofastronautsin an orbiting satellite These astronauts are in free fall inthesensethattheyareacceleratingtowardsthecentre of the Earth

It is actually possible to define theweight of a body inseveral different waysWe can define it for example asthegravitationalforceexertedonthebodybyaspecifiedobject such as the EarthThiswe have seen thatwe doinlotsofsituationswherewedefinetheweightasbeingequaltomgIfweusethisdefinitionthenanobjectinfreefallcannotbydefinitionbeweightlesssinceitisstillinagravitationalfieldHoweverifwedefinetheweightofanobjectintermsofaldquoweighingrdquoprocesssuchasthereadingonasetofbathroomscaleswhichineffectmeasuresthecontact force between the object and the scales thenclearlyobjectsinfreefallareweightlessOnenowhasto

askthequestionwhetherornotitispossibleForexampletomeasurethegravitationalforceactingonanastronautin orbit about the Earth

WecanalsodefineweightintermsofthenetgravitationalforceactingonabodyduetoseveraldifferentobjectsForexample for an object out in space its weight could bedefined in terms of the resultant of the forces exerted on it bytheSuntheMoontheEarthandalltheotherplanetsin the Solar System If this resultant is zero at a particular pointthenthebodyisweightlessatthispoint

Inviewofthevariousdefinitionsofweightthatareavailabletousitisimportantthatwhenweusethewordldquoweightrdquoweareawareofthecontextinwhichitisbeingused

Example

Calculate the height above the surface of the Earth atwhichageo-stationarysatelliteorbits

Solution

A geo-stationary satellite is one that orbits the Earth in such a way that it is stationary with respect to a point on the surface of the Earth This means that its orbital period must be the same as the time for the Earth to spin once on its axis ie 24 hours

From Keplerrsquos third law we have G Ms

4π2----------- R3

T2------=

That is

RMe

mh

using the fact that the force of attraction between the satellite and the Earth is given by

FG Mem

R2----------------=

and that F = ma

where m is the mass of the satellite and a 4π2R

T2-------------=

we have

GMem

R2---------------- m 4π2R

T2-------------times

GMe

4π2------------rArr R3

T2------= =

PHYSICS 2015indb 321 150514 338 PM

Chapter 10

322

AH

L

Now the mass of the Earth is 60 times 1024 kg and the period T measured in seconds is given by T = 86400 s

So substitution gives R = 42 times 106 m

The radius of the Earth is 64 times 106 m so that the orbital height h is about 36 times 107 m

Example

Calculatetheminimumenergyrequiredtoputasatelliteofmass500kgintoanorbitthatisasaheightequaltotheEarthrsquosradiusabovethesurfaceoftheEarth

Solution

We have seen that when dealing with gravitational fields and potential it is useful to remember that

g 0G M

Re2---------= or g0 Re

2 GM=

The gravitational potential at the surface of the Earth is

g 0ndash Re GMRe

---------ndash=

The gravitational potential at a distance R

from the centre of the Earth is

=

The difference in potential between the surface of the Earth and a point distance R from the centre is therefore

V∆ g 0Re 1ReR------ndash

=

If R 2Re= then V∆g0 Re

2------------=

This means that the work required to ldquoliftrdquo the satellite into orbit is g0Rm where m is the mass of the satellite This is equal to

10 32 106 500timestimestimes = 16000 MJ

However the satellite must also have kinetic energy in order to orbit Earth This will be equal to

g 0 mRe2

2R-----------------g 0mRe

2

4----------------- 8000 MJ= =

The minimum energy required is therefore

24000 MJ

Exercise 1031 The speed needed to put a satellite in orbit does

not depend on

A the radius of the orbitB theshapeoftheorbitC thevalueofgattheorbitD the mass of the satellite

2 EstimatethespeedofanEarthsatellitewhoseorbitis400kmabovetheEarthrsquossurfaceAlsodetermine the period of the orbit

3 Calculatethespeedofa200kgsatelliteorbitingthe Earth at a height of 70 times 106 m

Assume that g = 82 m sndash2 for this orbit

4 TheradiioftwosatellitesXandYorbitingthe Earth are 2r and 8rwherer is the radius of the Earth Calculate the ratio of the periods of revolutionofXtoY

5 A satellite of mass mkgissentfromEarthrsquossurfaceinto an orbit of radius 5RwhereR is the radius of theEarthWritedownanexpressionfor

(a) thepotentialenergyofthesatelliteinorbit

(b) thekineticenergyofthesatelliteinorbit

(c) theminimumworkrequiredtosendthesatellitefromrestattheEarthrsquossurfaceintoits orbit

6 A satellite in an orbit of 10rfallsbacktoEarth(radius r)afteramalfunctionDeterminethespeedwithwhichitwillhittheEarthrsquossurface

7 The radius of the moon is frac14 that of the Earth AssumingEarthandtheMoontohavethesamedensitycomparetheaccelerationsoffreefallatthe surface of Earth to that at the surface of the Moon

8 UsethefollowingdatatodeterminethegravitationalfieldstrengthatthesurfaceoftheMoon and hence determine the escape speed from the surface of the Moon

Mass of the Moon = 73 times 1022kg

RadiusoftheMoon=17times106 m

PHYSICS 2015indb 322 150514 338 PM

Wave Phenomena

303

AH

L

is a distance D away If the separation of the headlight is say 15 m then the headlights will subtend an angle θ 15

Dθ =

at your eye Assuming an average wavelength of

500 nm your eye will resolve the headlights into twoseparatesourcesifthisangleequals2times10-4 rad This gives D=75kmInotherwordsifthecarisapproachingyouona straight road then you will be able to distinguish the two headlightsasseparatesourceswhenthecaris75kmawayfrom you Actually because of the structure of the retina and optical defects the resolving power of the average eye is about 6 times 10-4 rad This means that the car is more likely tobe25kmawaybeforeyouresolvetheheadlights

Astronomical telescope

Letusreturntotheexampleofthebinarystarsdiscussedatthe beginning of this section on resolution The stars Kruger A and B form a binary system The average separation of the stars is 14 times 1012mandtheiraveragedistancefromEarthis12times1017mWhenviewedthroughatelescopeonEarththe system will therefore subtend an angle

12

17

14 1012 10

θtimes

=times

= 12times10-5 rad at the objective lens of the telescope Assuming that the average wavelength of the light emitted bythestarsis500nmthenifthetelescopeistoresolvethesystem into two separate images it must have a minimum diameter Dwhere12times10-5 =

7122 500 10D

minustimes times

This gives D=0050m which is about 5 cm So thisparticular system is easily resolved with a small astronomical telescope

Exercise 961 ItissuggestedthatusingtheISASVLBAitwould

bepossibletoldquoseerdquoagrainofriceatadistanceof5000kmEstimatetheresolvingpoweroftheVLBA

2 Thedistancefromtheeyelenstotheretinais20mm The light receptors in the central part of the retinaareabout5times10-6apartDeterminewhetherthe spacing of the receptors will allow for the eye to resolve the headlights in the above discussion whentheyare25kmfromtheeye

3 ThediameterofPlutois23times106 m and its averagedistancefromEarthis60times1012 m EstimatetheminimumdiameteroftheobjectiveofatelescopethatwillenablePlutotobeseenasadisc as opposed to a point source

95 Doppler effect

Essential ideaTheDopplereffectdescribesthephenomenonofwavelengthfrequencyshiftwhenrelativemotionoccurs

Understandingsbull TheDopplereffectforsoundwavesandlightwaves

The Doppler effect for sound waves and light waves

Consider two observers A and B at rest with respect to asoundsourcethatemitsasoundofconstantfrequencyf Clearly both observers will hear a sound of the same frequencyHoweversupposethatthesourcenowmovesat constant speed towards A A will now hear a sound of frequencyfA that is greater than f and B will hear a sound of frequency fB that is less than f This phenomenon is known as the Doppler Effect orDopplerPrinciple after C J Doppler(1803-1853)

The same effect arises for an observer who is either moving towards or away from a stationary source

Figure 930 shows the waves spreading out from a stationary source that emits a sound of constant frequency f The observersAandBhearasoundofthesamefrequency

A B

wavefront

S

Figure 930 Sound waves from a stationary source

Suppose now that the source moves towards A withconstant speed vFigure9310(a)showsasnapshotofthenew wave pattern

NATURE OF SCIENCETechnology Although originally based on physical observations of the pitch of fast moving sources of sound the Doppler effect has an important role in many different areas such as evidence for the expansion of the universe and generating images used in weather reports and in medicine (55)

copy IBO 2014

PHYSICS 2015indb 303 150514 337 PM

Chapter 9

304

AH

L

sourceA B

smallerwavelength

largerwavelength

Figure 931 (a) Sound waves from a moving source

The wavefronts are now crowded together in the direction of travel of the source and stretched out in the opposite direction This is why now the two observers will now hear notesofdifferentfrequenciesHowmuchthewavesbunchtogether and how much they stretch out will depend on the speed vEssentially A

A

cfλ

= and BB

cfλ

= where λA lt λB and v is the speed of sound

IfthesourceisstationaryandAismovingtowardsitthenthe waves from the source incident on A will be bunched up If A is moving away from the stationary source then the waves from the source incident on A will be stretched out

Christian Doppler(1803ndash1853)actuallyappliedtheprinciple(incorrectlyasithappens)totryandexplainthecolourofstars However the Doppler effect does apply to light aswellastosoundIfalightsourceemitsalightoffrequencyf then if it is moving away from an observer the observer willmeasurethelightemittedashavingalowerfrequencythan fSincethesensationofcolourvisionisrelatedtothefrequencyoflight(bluelightisofahigherfrequencythanred light) light emitted by objects moving way from anobserver is often referred to as being red-shifted whereas if the object is moving toward the observer it is referred to as blue-shiftedThisideaisusedinOptionE(Chapter16)

We do not need to consider here the situations where either the source or the observer is accelerating In a situation for example where an accelerating source is approaching astationaryobserverthentheobserverwillhearasoundof ever increasing frequency This sometimes leads toconfusion in describing what is heard when a source approaches passes and then recedes from a stationaryobserver Suppose for example that you are standingon a station platform and a train sounding its whistle is approaching at constant speed What will you hear as the train approaches and then passes through the station As the train approaches you will hear a sound of constant pitch but increasing loudness The pitch of the sound will be greater than if the train were stationary As the train passes through the station you will hear the pitch change at the momentthetrainpassesyoutoasoundagainofconstantpitch The pitch of this sound will be lower than the sound of the approaching train and its intensity will decrease as the train recedes from you What you do not hear is a sound of increasing pitch and then decreasing pitch

The Doppler equations for sound

Although you will not be expected in an IB examination to derivetheequationsassociatedwithaspectsoftheDopplereffectyouwillbeexpectedtoapplythemForcompletenesstherefore thederivationoftheequationsassociatedwiththeDopplereffectasoutlinedaboveisgivenhere

S primeS O

vs

Figure 931 (b)

InFigure931(b)theobserverOisatrestwithrespecttoasourceofsoundSismovingwithconstanspeedvs directly towards O The source is emitting a note of constant frequencyf and the speed of the emitted sound is v

S shows the position of the source ∆t later When the sourceisatresttheninatime∆t the observer will receive f∆t waves and these waves will occupy a distance v∆t ie

v f

λ =f∆tv∆t =

(Because of the motion of the source this number of waves will now occupy a distance (v∆t ndash vs∆t)The lsquonewrsquowavelength is therefore

v ndash vs

fλrsquo =

f∆tv∆t ndash vs∆t

=

If fisthefrequencyheardbyOthen

vfλ

= or

vf

λ = = sv vfminus

From which

s

vf fv v

=minus

Dividingthroughbyv gives

Equation1111

1 sf f v

v=

minus

If the source is moving away from the observer then we have

s

vf fv v

=minus

1

1 sf vv

=+

Equation112

PHYSICS 2015indb 304 150514 337 PM

Wave Phenomena

305

AH

L

We now consider the case where the source is stationary and the observerismoving towardsthesource with speed v0 In this situation the speed of the sound waves as measured by the observer will be v0 + v We therefore have that

v0 + v = ffraslλ = vfλ

times

From which

o1 vf fv

= +

Equation113

If the observerismovingawayfromthesource then

o1 vf fv

= minus

Fromequation(113)wehavethat

o o1 v vf f f f f fv v

∆ = minus = + minus =

Equation114

Thevelocitiesthatwerefertointheaboveequationsarethevelocities with respect to the medium in which the waves fromthesourcetravelHoweverwhenwearedealingwithalight source it is the relative velocity between the source and the observer that we must consider The reason for this is that lightisuniqueintherespectthatthespeedofthelightwavesdoes not depend on the speed of the source All observers irrespective of their speed or the speed of the source will measure the same velocity for the speed of light This is oneof thecornerstonesof theSpecialTheoryofRelativitywhichisdiscussedinmoredetailinOptionH(Chapter18)When applying the Doppler effect to light we aremainlyconcerned with the motion of the source We look here only at the situation where the speed of the source v is much smaller than the speed of light cinfreespace(v ltlt c)Underthesecircumstanceswhenthesourceismovingtowardstheobserverequation111becomes

vf f f fc

minus = ∆ = Equation115

andwhenthesourceismovingawayfromtheobserverequation112becomes vf f f f

cminus = ∆ = minus

Provided that v ltlt c these same equations apply for astationary source and moving observer

We look at the following example and exercise

Example

Asourceemitsasoundoffrequency440HzItmovesinastraight line towards a stationary observer with a speed of 30ms-1Theobserverhearsasoundoffrequency484HzCalculate the speed of sound in air

Solution

We use equation 111 and substitute f= 484 Hz f = 440 Hz

and vs = 30 m s-1

therefore 1

484 440 301v

=minus

such that 30 4401

484vminus = to

give v = 330 m s-1

Example

A particular radio signal from a galaxy is measured as havingafrequencyof139times109HzThesamesignalfromasourceinalaboratoryhasafrequencyof142times109Hz

Suggestwhy thegalaxy ismovingaway fromEarthandcalculateitsrecessionspeed(iethespeedwithwhichitismovingawayfromEarth)

Solution

The fact that the frequency from the moving source is less than that when it is stationary indicates that it is moving away from the stationary observer ie Earth

Using vf fc

∆ = we have

8 96

9

3 10 (142 139) 10 634 10142 10

c fvf∆ times times minus times

= = = timestimes

m s-1

It is usual when dealing with the Doppler effect of light to express speeds as a fraction of c So in this instance we have v = 0021 c

Using the Doppler effectWe have seen in the above example and exercise how the Dopplereffectmaybeusedtomeasuretherecessionspeedof distant galaxies The effect is also used to measure speed inothersituationsHerewewilllookatthegeneralprincipleinvolved in using the Doppler effect to measure speedFigure932showsasource(thetransmitter)thatemitseithersound or em waves of constant frequency f The waves

PHYSICS 2015indb 305 150514 337 PM

Chapter 9

306

AH

L

fromthesourceare incidentonareflectorthat ismovingtowards the transmitter with speed vThereflectedwavesare detected by the receiver placed alongside the transmitter

reector

v

transmitter

receiver

f f

f f

Figure 932 Using the Doppler effect to measure speed

We shall consider the situation where v ltlt c where c is the speed of the waves from the transmitter

For the reflector receiving waves from the transmitterit is effectively an observer moving towards a stationary sourceFromequation(114) itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation116

For the receiver receivingwaves from the reflector it iseffectively a stationary observer receiving waves from a movingsourceFromequation(115)itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation117

Ifweaddequations(116)and(117)wegetthatthetotalDopplershiftatthereceiver∆f is

v vf f f f fc c

minus = ∆ = +

But 1 vf fc

= +

hence

1 v v vf f fc c c

∆ = + +

But since v ltlt cwe can ignore the term2

2

vc when we

expandthebracketintheaboveequation

Therefore we have2vf fc

∆ = Equation118

If v asymp c then we must use the full Doppler equationsHowever for em radiationwewill alwaysonly considersituations in which v ltlt c

Example

Thespeedofsoundinbloodis1500times103 m s-1 Ultrasound offrequency100MHzisreflectedfrombloodflowinginan arteryThe frequencyof the reflectedwaves receivedbackatthetransmitteris105MHzEstimatethespeedofthebloodflowintheartery

Solution

Using equation (118) we have

6 63

2005 10 1015 10

vtimes = times

timesto give v asymp 36 m s-1 (We have assumed that the ultrasound is incident at right angles to the blood flow)

Exercise 971 Judy is standing on the platform of a station A

high speed train is approaching the station in a straight line at constant speed and is sounding its whistleAsthetrainpassesbyJudythefrequencyof the sound emitted by the whistle as heard by Judychangesfrom640Hzto430HzDetermine

(a) thespeedofthetrain

(b) thefrequencyofthesoundemittedbythewhistle as heard by a person on the train (Speedofsound=330ms-1)

2 AgalaxyismovingawayfromEarthwithaspeedof00500cThewavelengthofaparticularspectral line in light emitted by atomic hydrogen inalaboratoryis656times10-7 m Calculate the valueofthewavelengthofthislinemeasuredinthelaboratoryinlightemittedfromasourceofatomic hydrogen in the galaxy

PHYSICS 2015indb 306 150514 337 PM

10 Fields

Contents

101 ndash Describing fields

102 ndash Fields at work

Essential Ideas

Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Similar approaches can be taken in analysing electrical and gravitational potential problems copy IBO 2014

PHYSICS 2015indb 307 150514 337 PM

Chapter 10

308

AH

L

101 Describing fields

Essential idea Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Understandingsbull Gravitationalfieldsbull Electrostaticfieldsbull Electricpotentialandgravitationalpotentialbull Fieldlinesbull Equipotentialsurfaces

Gravitational potentialWehaveseenthatifweliftanobjectofmassm to a height h abovethesurfaceoftheEarththenitsgainingravitationalpotential energy is mghHoweverthisisbynomeansthefullstoryForastartwenowknowthatgvarieswithh and also the expression really gives a difference in potentialenergybetweenthevaluethattheobjecthasattheEarthrsquossurfaceandthevaluethatithasatheighthSowhatwereallyneedisazeropointCanwefindapointwherethepotentialenergyiszeroandusethispointfromwhichtomeasure changes in potential energy

The point that is chosen is in fact infinity At infinity the gravitationalfieldstrengthofanyobjectwillbezeroSoletusseeifwecandeduceanexpressionforthegaininpotentialenergyofanobjectwhenitisldquoliftedrdquofromthesurfaceoftheEarthtoinfinityThisineffectmeansfindingtheworknecessarytoperformthistask

δr

r r δr+

r infin=

mg

AB

Me

Figure 1001 Gravitational forces

Inthediagramweconsidertheworknecessarytomovethe particle of mass m a distance δr in thegravitationalfield of the Earth

The force on the particle at A is FG Mem

r2----------------=

IftheparticleismovedtoB then since δr isverysmallwe can assume that the field remains constant over thedistance ABTheworkδW done againstthegravitationalfieldoftheEarthinmovingthedistanceABis

δWGMem

r2----------------δrndash=

(rememberthatworkdoneagainstaforceisnegative)

TofindthetotalworkdoneW in going from the surface oftheEarthtoinfinitywehavetoaddalltheselittlebitsofworkThisisdonemathematicallybyusingintegralcalculus

WG Me m

r2----------------ndash

rdR

infin

int G Me m 1

r2----- rd

R

infin

intndash G Me m 1r---ndash

R

infinndash= = =

GMe m 0 1R---ndash

ndashndash=

G MemR----------------ndash=

HencewehavewhereRistheradiusoftheEarththattheworkdonebythegravitationalfieldinmovinganobjectofmass m from R(surfaceoftheEarth)toinfinityisgivenby

WGMe m

R---------------- ndash=

We can generalise the result by calculating the worknecessary per unit mass to take a small mass fromthe surface of the Earth to infinity This we call thegravitational potential Vie

Wewouldgetexactlythesameresultifwecalculatedtheworkdonebythefieldtobringthepointmassfrominfinityto the surface of Earth In this respect the formal definition ofgravitationalpotentialatapointinagravitationalfieldis therefore defined as the work done per unit mass in bringing a point mass from infinity to that point

Clearly then the gravitational potential at any point inthe Earthrsquos field distance r from the centre of the Earth (providingr gt R)is

VGMe

r------------ ndash=

Thepotentialisthereforeameasureoftheamountofworkthathastobedonetomoveparticlesbetweenpointsinagravitationalfieldanditsunit is theJkgndash1 We also note thatthepotentialisnegativesothatthepotentialenergy

NATURE OF SCIENCEParadigm shift The move from direct observable actions being responsible for influence on an object to acceptance of a fieldrsquos ldquoaction at a distancerdquo required a paradigm shift in the world of science (23)

copy IBO 2014

PHYSICS 2015indb 308 150514 337 PM

Fields

309

AH

L

aswemoveawayfromtheEarthrsquossurfaceincreasesuntilitreachesthevalueofzeroatinfinity

IfthegravitationalfieldisduetoapointmassofmassmthenwehavethesameexpressionasaboveexceptthatMe is replaced by m andmustalsoexclude thevalueof thepotential at the point mass itself ie at r = 0

WecanexpressthegravitationalpotentialduetotheEarth(orduetoanysphericalmass)intermsofthegravitationalfield strength at its surface

AtthesurfaceoftheEarthwehave

g 0ndash Re GMRe

---------ndash= e

Sothat

g 0Re2 GM= e

Hence at a distance r from the centre of the Earth the gravitationalpotentialVcanbewrittenas

VGMe

r------------ndashg 0 Re

2

r------------ndash= =

The potential at the surface of the Earth

(r = Re)istherefore-g0Re

It is interesting to see how the expression for thegravitational potential ties in with the expression mgh ThepotentialatthesurfaceoftheEarthis-g0Re (see the exampleabove)andataheighthwillbe-g0g 0 Re h+( )ndash ifweassume that g0doesnotchangeover thedistanceh The differenceinpotentialbetweenthesurfaceandtheheighth is therefore g0hSotheworkneededtoraiseanobjectof mass m to a height h is mgh ie m timesdifference ingravitationalpotential

This we have referred to as the gain in gravitationalpotentialenergy(see235)

However this expression can be extended to any twopoints inanygravitationalfieldsuch that ifanobjectofmass mmovesbetween twopointswhosepotentialsareV1 and V2 respectively then the change in gravitationalpotentialenergyoftheobjectism(V1 ndash V2)

Gravitational potential gradient

Let us consider now a region in space where thegravitationalfieldisconstantInFigure912thetwopointsAandBareseparatedbythedistance∆x

A B

direction of uniform gravitational eld of strength I

∆x

Figure 1002 The gravitational potential gradient

The gravitational field is of strength I and is in the direction shownThe gravitational potential at A is V andatBisV +∆V

Theworkdone is takingapointmassmfromAtoB isF∆x = mI∆x

Howeverbydefinitionthisworkisalsoequalto-m∆V

Therefore mI∆x=-m∆V

or VIx

∆= minus

∆Effectivelythissaysthatthemagnitudeofthegravitationalfield strength is equal to the negative gradient of thepotential If I is constant then V is a linear function of x and Iisequaltothenegativegradientofthestraightlinegraph formed by plotting V against x If I is not constant (asusuallythecase)thenthemagnitudeofI at any point in the field can be found by find the gradient of the V-x graph at that point An example of such a calculation can befoundinSection929

For those of you who do HL maths the relationship between field and potential is seen to follow from the expression for the potential of a point mass viz

mV Gr

= minus

2

ddV mG Ir r

minus = + =

Potential due to one or more point masses

Gravitationalpotentialisascalarquantitysocalculatingthepotential due to more than one point mass is a matter of simpleadditionSoforexamplethepotentialV due to the Moon and Earth and a distance x fromthecentreofEarthonastraightlinebetweenthemisgivenbytheexpression

E MM MV Gx r x

= minus + minus

whereME=massofEarthMM= mass of Moon and r = distancebetweencentreofEarthandMoon

PHYSICS 2015indb 309 150514 337 PM

Chapter 10

310

AH

L

Equipotentials and field linesIfthegravitationalpotentialhasthesamevalueatallpointson a surface the surface is said to be an equipotential surfaceSoforexampleifweimagineasphericalshellaboutEarthwhosecentrecoincideswiththecentreofEarththisshellwillbeanequipotentialsurfaceClearlyifwerepresentthegravitationalfieldstrengthbyfieldlinessincethelinesldquoradiaterdquooutfromthecentreofEarththentheselineswillbeatrightanglestothesurfaceIfthefieldlineswerenotnormaltotheequipotentialsurfacethentherewouldbeacomponentof the field parallel to the surfaceThis wouldmean thatpointsonthesurfacewouldbeatdifferentpotentialsandsoitwouldnolongerbeanequipotentialsurfaceThisofcourseholdstrueforanyequipotentialsurface

Figure 1003 shows the field lines and equipotentials fortwopointmassesm

m m

Figure 1003 Equipotentials for two point masses

It is worth noting that we would get exactly the samepattern ifwewere to replace thepointmasseswith twoequalpointcharges(See935)

Escape speedThe potential at the surface of Earth is which meansthattheenergyrequiredtotakeaparticleofmass mfromthesurfacetoinfinityisequalto

ButwhatdoesitactuallymeantotakesomethingtoinfinityWhentheparticleisonthesurfaceoftheEarthwecanthinkofitassittingatthebottomofaldquopotentialwellrdquoasinfigure1004

particle

surface of Earth

innity

GMR---------

FIgure 1004 A potential well

Theldquodepthrdquoofthewellis and if the particle gains an amount of kinetic energy equal to wherem is its massthenitwillhavejustenoughenergytoldquoliftrdquoitoutofthewell

In reality it doesnrsquot actually go to infinity it justmeansthat the particle is effectively free of the gravitationalattractionof theEarthWe say that ithas ldquoescapedrdquo theEarthrsquosgravitationalpullWemeetthisideainconnectionwithmolecular forcesTwomolecules in a solidwill sitat their equilibrium position the separation where therepulsiveforceisequaltotheattractiveforceIfwesupplyjust enough energy to increase the separation of themolecules such that they are an infinite distance apart then intermolecularforcesnolongeraffectthemoleculesandthesolidwillhavebecomealiquidThereisnoincreaseinthekineticenergyofthemoleculesandsothesolidmeltsat constant temperature

Wecancalculatetheescapespeedofanobjectveryeasilyby equating the kinetic energy to the potential energysuch that

12---mvesc ape

2 GMe mRe

----------------=

vesc aperArr2GMe

Re--------------- 2g0 Re = =

Substituting for g0 and Regivesavalueforvescape of about 11kmsndash1 from the surface of the Earth

Youwillnotethattheescapespeeddoesnotdependonthemassoftheobjectsincebothkineticenergyandpotentialenergy are proportional to the mass

IntheoryifyouwanttogetarockettothemoonitcanbedonewithoutreachingtheescapespeedHoweverthiswouldnecessitate anenormousamountof fuel and it islikely that therocketplus fuelwouldbesoheavythat itwouldnevergetoffthegroundItismuchmorepracticalto accelerate the rocket to the escape speed fromEarthorbitandthenintheoryjustlaunchittotheMoon

Example

Use the followingdata todetermine thepotential at thesurface of Mars and the magnitude of the acceleration of free fall

massofMars =64times1023kg

radiusofMars =34times106 m

Determinealsothegravitationalfieldstrengthatadistanceof 68 times 106mabovethesurfaceofMars

PHYSICS 2015indb 310 150514 337 PM

Fields

311

AH

L

Solution

= minus = minus times times = times23

11 76

64 1067 10 ndash13 10

34 10MV GR

minus timestimes

Nkg-1

But V = -g0R

Therefore 7

0 6

13 10 3834 10

VgR

times= minus = =

timesm s-2

To determine the field strength gh at 68 times 106 m above the surface we use the fact that 0 2

Mg GR

= such that GM = g0R2

Therefore 2 2

0h 2 2 2

h h

38 (34) 042(102)

g RGMgR R

times= = = = m s-2

(the distance from the centre is 34 times 106 + 68 times 106 = 102 times 106 m)

Exercise 1011 Thegraphbelowshowshowthegravitational

potentialoutsideoftheEarthvarieswithdistancefrom the centre

0 10 20 30 40 50 60 70

ndash1

ndash2

ndash3

ndash4

ndash5

ndash6

ndash7

r m times 106

V

Jkgndash1

times 10

7

(a) Usethegraphtodeterminethegainingravitationalpotentialenergyofasatelliteofmass200kgasitmovesfromthesurfaceof the Earth to a height of 30 times 107mabovetheEarthrsquossurface

(b) Calculatetheenergyrequiredtotakeit to infinity

(c) DeterminetheslopeofthegraphatthesurfaceoftheEarthmCommenton youranswer

Electric potential energy

The concept of electric potential energy was developedwith gravitational potential energy in mind Just as anobjectnearthesurfaceoftheEarthhaspotentialenergybecause of its gravitational interaction with the Earthsotoothereiselectricalpotentialenergyassociatedwithinteracting charges

Letusfirstlookatacaseoftwopositivepointchargeseachof 1μC that are initially bound together by a thread in a vacuuminspacewithadistancebetweenthemof10cmasshowninFigure1005Whenthethreadiscutthepointchargesinitiallyatrestwouldmoveinoppositedirectionsmovingwithvelocitiesv1 and v2 along the direction of the electrostatic force of repulsion

BEFORE AFTER

v v21

Figure 1005 Interaction of two positive particles

The electric potential energybetweentwopointchargescan be found by simply adding up the energy associated witheachpairofpointchargesForapairofinteractingchargestheelectricpotentialenergyisgivenby

∆U = ∆Ep + ∆Ek = ∆W = ∆Fr = kqQ

____ r2 times r = kqQ

____ r

Becausenoexternalforceisactingonthesystemtheenergyand momentum must be conserved Initially Ek = 0 and Ep =kqQr=9times109 times 1 times 10-1201m=009JWhentheyareagreatdistance fromeachotherEpwillbenegligibleThefinalenergywillbeequaltofrac12mv1

2+frac12mv22=009J

Momentum isalsoconservedand thevelocitieswouldbethe same magnitude but in opposite directions

Electric potential energyismoreoftendefinedintermsofapointchargemovinginanelectricfieldas

lsquothe electricpotential energybetweenany twopoints inan electric field is defined as negativeoftheworkdone by anelectricfieldinmovingapoint electric chargebetweentwolocationsintheelectricfieldrsquo

ΔU = ΔEp = -ΔW = -Fd = qEx

ΔU = ΔEp +ΔEk = ΔW = ΔFr = kqQ r2 x r = kqQ

wherexisthedistancemovedalong(oroppositeto)thedirection of the electric field

Electric potential energyismeasuredinjoule(J)Justasworkisascalarquantitysotooelectricalpotentialenergyis a scalar quantity The negative of the work done by an electric field in moving a unit electric charge between two points is independent of the path taken Inphysicswesaytheelectricfieldisalsquoconservativersquo field

PHYSICS 2015indb 311 150514 337 PM

Chapter 10

312

AH

L

Supposeanexternalforcesuchasyourhandmovesasmallpositive point test charge in the direction of a uniformelectric fieldAs it ismoving itmust be gaining kineticenergyIfthisoccursthentheelectricpotentialenergyofthe unit charge is changing

InFigure1006apointcharge+qismovedbetweenpointsAandBthroughadistancex in a uniform electric field

B

A

x

+ q

Figure 1006 Movement of a positive point charge in a uniform field

InordertomoveapositivepointchargefrompointAtopointBanexternal forcemustbeapplied to thechargeequaltoqE (F = qE)

Since the force is applied through a distance xthennegativeworkhas to bedone tomove the charge because energyis gainedmeaning there is an increaseelectric potential energybetweenthetwopointsRememberthattheworkdoneisequivalenttotheenergygainedorlostinmovingthe charge through the electric field The concept of electric potential energy is only meaningful as the electric field whichgeneratestheforceinquestionisconservative

W F xtimes E q x times= =

θ x cos θ x

Figure 1007 Charge moved at an angle to the field

If a chargemoves at an angle θ to an electric field thecomponent of the displacement parallel to the electric fieldisusedasshowninFigure918

W F x Eq x θcostimes= =

Theelectricpotentialenergyisstoredintheelectricfieldand theelectricfieldwill return theenergy to thepointcharge when required so as not to violate the Law ofconservationofenergy

Electric potential

The electric potential at a point in an electric field is defined as being the work done per unit charge in bringing a small positive point charge from infinity to that point

ΔV = Vinfin ndash Vf = ndash W ___ q

Ifwedesignatethepotentialenergytobezeroatinfinitythen it follows thatelectricpotentialmustalsobezeroat infinity and the electric potential at any point in an electricfieldwillbe

ΔV = ndash W ___ q

NowsupposeweapplyanexternalforcetoasmallpositivetestchargeasitismovedtowardsanisolatedpositivechargeTheexternalforcemustdoworkonthepositivetestchargetomoveittowardstheisolatedpositivechargeandtheworkmustbepositivewhiletheworkdonebytheelectricfieldmustthereforebenegativeSotheelectricpotentialatthatpointmustbepositiveaccordingtotheaboveequationIfanegativeisolatedchargeisusedtheelectricpotentialatapointonthepositivetestchargewouldbenegativePositivepoint chargesof their ownaccordmove fromaplaceofhighelectricpotential toaplaceof lowelectricpotentialNegative point charges move the other way from lowpotentialtohighpotentialInmovingfrompointAtopointBinthediagramthepositivecharge+qismovingfromalowelectricpotentialtoahighelectricpotential

In thedefinitiongiven the termldquoworkperunitchargerdquohas significance If the test charge is +16 times 10-19Cwherethe charge has a potential energy of 32 times 10-17Jthenthepotentialwouldbe32times10-17J+16times10-19C=200JC-1 Now if the charge was doubled the potential wouldbecome 64times10-17 J However the potential per unitchargewouldbethesame

Electric potential is a scalar quantityandithasunitsJC-1 orvoltswhere1voltequalsone joulepercoloumbThevoltallowsustoadoptaunitfortheelectricfieldintermsofthevolt

PreviouslytheunitfortheelectricfieldwasNCndash1

W = qV and F = qEsoW ___ V = F __ E

E = FV ___ W = FV ___ Fm V mndash1

That is the units of the electric field E can also beexpressed as V mndash1

The work done per unit charge in moving a point charge between two points in an electric field is again independent of the path taken

PHYSICS 2015indb 312 150514 337 PM

Fields

313

AH

L

Electric potential due to a point charge

Letus takeapoint rmetres froma chargedobjectThepotentialatthispointcanbecalculatedusingthefollowing

W = ndashFr = ndashqV and F = ndash q1 q2 _____ 4πε0r

2

Therefore

W = ndash q1 q2 _____ 4πε0r

2 times r = ndash q1 q2 _____ 4πε0r

= ndashq1 times q2 _____ 4πε0r

= ndashq1V

That is V q

4πε0r--------------- =

Orsimply

V kqr------

=

Example

Determinehowmuchworkisdonebytheelectricfieldofpointcharge150μCwhenachargeof200μCismovedfrom infinity to apoint 0400m from thepoint charge(Assumenoaccelerationofthecharges)

Solution

The work done by the electric field is W = -qV = -14πε0 times q times (Q rinfin - Q r0400)

W = (- 200 times 10-6 C times 900 times 109 NmC-2 times 150 times 10-6 C) divide 0400 m = - 0675 J

An external force would have to do +0675 J of work

Electric field strength and electric potential gradient

Letus lookbackatFigure1006Supposeagain that thecharge+qismovedasmalldistancebyaforceFfromAtoBsothattheforcecanbeconsideredconstantTheworkdoneisgivenby

∆W F x∆times=

TheforceFandtheelectricfieldEareoppositelydirectedandweknowthat

F = -qE and ΔW = q ΔV

Thereforetheworkdonecanbegivenas

q ΔV = -q E Δx

Therefore E V∆x∆-------

ndash=

The rate of change of potential ΔV at a point with respect to distance Δx in the direction in which the change is maximum is called the potential gradient We say that the electric field = - the potential gradient and the units are Vm-

1 From the equation we can see that in a graph of electric potential versus distance the gradient of the straight line equals the electric field strength

In reality if a charged particle enters a uniform electric field it will be accelerated uniformly by the field and its kinetic energy will increase This is why we had to assume no acceleration in the last worked example

Ek∆ 12---

mv2 q E xsdot sdot q Vx---

xsdot sdot q Vsdot= = = =

Example

Determine how far apart two parallel plates must besituated so that a potential difference of 150 times 102 V produces an electric field strength of 100 times 103 NC-1

Solution

Using E V∆x------- xhArrndash V∆

E-------15 102 Vtimes

100 103 N C 1ndashtimes------------------------------------------= = =

= 150 times 10-1

The plates are 150 times 10-1 m apart

The electric field and the electric potential at a point due to an evenly distributed charge +q on a sphere can be represented graphically as in Figure 1008

r

r

V

E

r0

r0

r0

Charge of +Q evenly distributed over surface

E 14πε0------------ Q

r2----- r r0gt=

V 14πε0------------Q

r----r r0gt=

Figure 1008 Electric field and potential due to a charged sphere

PHYSICS 2015indb 313 150514 337 PM

Chapter 10

314

AH

L

When the sphere becomes charged we know that thechargedistributes itselfevenlyoverthesurfaceThereforeeverypartofthematerialoftheconductorisatthesamepotential As the electric potential at a point is defined as beingnumericallyequaltotheworkdoneinbringingaunitpositivechargefrominfinitytothatpointithasaconstantvalueineverypartofthematerialoftheconductor

Since the potential is the same at all points on the conducting surfacethen∆V∆xiszeroButE=ndash∆V∆xThereforethe electric field inside the conductor is zero There is no electric field inside the conductor

SomefurtherobservationsofthegraphsinFigure1009are

bull OutsidethespherethegraphsobeytherelationshipsgivenasE α 1 r2 and V α 1 r

bull Atthesurfacer = r0Thereforetheelectricfieldandpotentialhavetheminimumvalueforr at this point and this infers a maximum field and potential

bull Insidethespheretheelectricfieldiszerobull Insidethespherenoworkisdonetomoveacharge

fromapointinsidetothesurfaceThereforethereisno potential difference and the potential is the same asitiswhenr = r0

Similargraphscanbedrawnfortheelectricfieldintensityandthe electric potential as a function of distance from conducting parallelplatesandsurfacesandthesearegiveninFigure1009

Potential plot E eld plotE eld

++++

ndashndashndashndash

xx x

+

x x

ndashx x

+ ndashx

x

+ +

x

x

Figure 1009 Electric field and electric potential at a distance from a charged surface

Potential due to one or more point charges

The potential due to one point charge can be determined byusingtheequationformula

V = kq r

Example 1

Determine the electric potential at a point 20 times 10-1 m fromthecentreofan isolatedconducting spherewithapointchargeof40pCinair

Solution

Using the formula

V = kq r we have

V90 10 9times( ) 40 10 12ndashtimes( )times

20 10 1ndashtimes( )------------------------------------------------------------------ 018 V= =

the potential at the point is 180 times 10-1 V

The potential due to a number of point charges can be determined by adding up the potentials due to individual point charges because the electric potential at any point outside a conducting sphere will be the same as if all the charge was concentrated at its centre

Example 2

Threepointchargesofareplacedattheverticesofaright-angledtriangleasshowninthediagrambelowDeterminetheabsolutepotentialat the+20μCchargeduetothetwoothercharges

- 6μC

+3μC +2μC

4 m

3 m

Solution

The electric potential of the +2 μC charge due to the ndash 6 μC charge is

V = (9 times 109 Nm2C-2 times -6 times 10-6 C) divide (radic 32 + 42) m = - 108 times 104 V

PHYSICS 2015indb 314 150514 338 PM

Fields

315

AH

L

Insummarywecanconcludethat

bull No work is done to move a charge along an equipotential

bull Equipotentials are always perpendicular to the electric lines of force

Figure1011and1012showsomeequipotential lines fortwo oppositely charged and identically positive spheresseparated by a distance

+ve ndashve

equipotential lines

Figure 1011 Equipotential lines between two opposite charges

equipotential lines

Figure 1012 Equipotential lines between two charges which are the same

+

ndash ndash ndash ndash

+ + +

40 V

30 V

20 V

10 V

50 V

Figure 1013 Equipotential lines between charged parallel plates

Figure1013showstheequipotentiallinesbetweenchargedparallel plates Throughout this chapter the similarities and differences between gravitational fields and electric fieldshavebeendiscussedTherelationshipsthatexistsbetweengravitationalandelectricquantitiesandtheeffectsofpointmassesandchargesissummarisedinTable1014

The electric potential of the +2 μC charge due to the +3 μC charge is

V = (9 times 109 Nm2C-2 times 3 times 10-6 C) divide 3m = 9 times 103 V

The net absolute potential is the sum of the 2 potentials- 108 times 104 V + 9 times 103 V =- 18 times 103 V

The absolute potential at the point is - 18 times 103 V

Equipotential surfacesRegions in space where the electric potential of a chargedistributionhasaconstantvaluearecalledequipotentials The placeswherethepotentialisconstantinthreedimensionsarecalled equipotential surfacesandwheretheyareconstantintwodimensionstheyarecalledequipotential lines

TheyareinsomewaysanalogoustothecontourlinesontopographicmapsInthiscasethegravitationalpotentialenergy is constant as amassmoves around the contourlines because the mass remains at the same elevationabovetheEarthrsquossurfaceThegravitationalfieldstrengthacts in a direction perpendicular to a contour line

Similarlybecausetheelectricpotentialonanequipotentiallinehas thesamevalueanelectric forcecandonoworkwhena testchargemovesonanequipotentialThereforethe electric field cannot have a component along anequipotentialandthusitmustbeeverywhereperpendiculartotheequipotentialsurfaceorequipotentiallineThisfactmakesiteasytoplotequipotentialsifthelinesofforceorlinesofelectricfluxofanelectricfieldareknown

Forexamplethereareaseriesofequipotentiallinesbetweentwoparallelplateconductorsthatareperpendiculartotheelectric fieldTherewill be a series of concentric circles(eachcirclefurtherapartthanthepreviousone)thatmapouttheequipotentialsaroundanisolatedpositivesphereThe lines of force and some equipotential lines for anisolatedpositivesphereareshowninFigure1010

Lines ofequipotential

Figure 1010 Equipotentials around an isolated positive sphere

PHYSICS 2015indb 315 150514 338 PM

Chapter 10

316

AH

L

Gravitational quantity Electrical quantityQ

uant

ities

g V∆x∆-------ndash= E V∆

x∆-------ndash=

Poin

t mas

ses a

nd ch

arge

s

V Gmr----

ndash= V 14πε0------------q

r---=

g G m

r2-----ndash= E 14πε0------------ q

r2-----=

F Gm1 m2

r2--------------= F 1

4πε0------------

q1q2

r2-----------=

Figure 1014 Formulas

Example

Deduce the electric potential on the surface of a gold nucleus that has a radius of 62 fm

Solution

Using the formula

V = kq r and knowing the atomic number of gold is 79 We will assume the nucleus is spherical and it behaves as if it were a point charge at its centre (relative to outside points)

V = 90 times 109 Nm2C-2 times 79 times 16 times 10-19 C divide 62 times 10-15 m = 18 times 107 V

The potential at the point is 18 MV

Example

Deduce the ionisation energy in electron-volts of theelectron in the hydrogen atom if the electron is in its ground state and it is in a circular orbit at a distance of 53 times 10-11 m from the proton

Solution

This problem is an energy coulombic circular motion question based on Bohrrsquos model of the atom (not the accepted quantum mechanics model) The ionisation energy is the energy required to remove the electron from the ground state to infinity The electron travels in a circular orbit and therefore has a centripetal acceleration The ionisation energy will counteract the coulombic force and the movement of the electron will be in the opposite direction to the centripetal force

Total energy = Ek electron + Ep due to the proton-electron interaction

ΣF = kqQ r2 = mv2 r and as such mv2 = = kqQ r

Therefore Ek electron = frac12 kqQ r

Ep due to the proton-electron interaction = - kqQ r

Total energy = frac12 kqQ r + - kqQ r = -frac12 kqQ r

= - 90 times 109 Nm2C-2 times (16 times 10-19 C)2 divide 53 times 10-11 m = -217 times 10-18 J

= -217 times 10-18 J divide 16 times 10-19 = -136 eV

The ionisation energy is 136 eV

Exercise 1021 ApointchargePisplacedmidwaybetweentwo

identicalnegativechargesWhichoneofthefollowingiscorrectwithregardstoelectricfieldandelectricpotentialatpointP

Electric field Electric potentialA non-zero zeroB zero non-zeroC non-zero non-zeroD zero zero

2 Twopositivechargedspheresaretiedtogetherinavacuumsomewhereinspacewheretherearenoexternal forces A has a mass of 25 g and a charge of20μCandBhasamassof15gandachargeof30μCThedistancebetweenthemis40cmTheyarethenreleasedasshowninthediagram

BEFORE AFTER

v1 v2

A B

(a) Determinetheirinitialelectricpotentialenergy in the before situation

(b) DeterminethespeedofsphereBafterrelease

3 Thediagrambelowrepresentstwoequipotentiallines in separated by a distance of 5 cm in a uniform electric field

+ + + + + + + +

ndash ndash ndash ndash ndash ndash ndash ndash

40 V

20 V 5 cm

PHYSICS 2015indb 316 150514 338 PM

Fields

317

AH

L

Determine the strength of the electric field

4 Thisquestionisabouttheelectricfieldduetoacharged sphere and the motion of electrons in that fieldThediagrambelowshowsanisolatedmetalsphereinavacuumthatcarriesanegativeelectriccharge of 60 μC

ndash

(a) Onthediagramdrawtheconventionalwayto represent the electric field pattern due to the charged sphere and lines to represent threeequipotentialsurfacesintheregionoutside the sphere

(b) Explainhowthelinesrepresentingtheequipotentialsurfacesthatyouhavesketchedindicatethatthestrengthoftheelectricfieldisdecreasingwithdistancefrom the centre of the sphere

(c) Theelectricfieldstrengthatthesurfaceofthe sphere and at points outside the sphere can be determined by assuming that the sphere acts as a point charge of magnitude 60 μC at its centre The radius of the sphere is 25 times 10ndash2 m Deduce that the magnitude of the field strength at the surface of the sphere is 86 times 107 Vmndash1

An electron is initially at rest on the surface of the sphere

(d) (i) Describethepathfollowedbytheelectronasitleavesthesurfaceofthesphere

(ii) Calculatetheinitialaccelerationofthe electron

5 Determinetheamountofworkthatisdoneinmovingachargeof100nCthroughapotentialdifferenceof150times102 V

6 Three identical 20 μC conducting spheres are placedatthecornersofanequilateraltriangleofsides 25 cm The triangle has one apex C pointing upthepageand2baseanglesAandBDeterminetheabsolutepotentialatB

7 Determinehowfaraparttwoparallelplatesmustbesituatedsothatapotentialdifferenceof250 times 102 V produces an electric field strength of 200 times 103 NC-1

8 Thegapbetweentwoparallelplatesis10times10-3mandthereisapotentialdifferenceof10times104 V betweentheplatesCalculate

i theworkdonebyanelectroninmovingfrom one plate to the other

ii thespeedwithwhichtheelectronreachesthe second plate if released from rest

iii theelectricfieldintensitybetweentheplates

9 Anelectronguninapicturetubeisacceleratedby a potential 25 times 103VDeterminethekineticenergygainedbytheelectroninelectron-volts

10 Determine the electric potential 20 times 10-2 m from achargeof-10times10-5 C

11 Determinetheelectricpotentialatapointmid-waybetweenachargeofndash20pCandanotherof+5pConthelinejoiningtheircentresifthecharges are 10 cm apart

12 During a thunderstorm the electric potential differencebetweenacloudandthegroundis10 times 109 V Determine the magnitude of the change in electric potential energy of an electron thatmovesbetweenthesepointsinelectron-volts

13 A charge of 15 μC is placed in a uniform electric fieldoftwooppositelychargedparallelplateswithamagnitudeof14times103 NC-1

(a) Determinetheworkthatmustbedoneagainstthefieldtomovethepointchargeadistance of 55 cm

(b) Calculatethepotentialdifferencebetweenthe final and initial positions of the charge

(c) Determinethepotentialdifferencebetweenthe plates if their separation distance is 15 cm

14 Duringaflashoflightningthepotentialdifferencebetweenacloudandthegroundwas12 times 109 V and the amount of transferred charge was32C

(a) Determinethechangeinenergyofthetransferred charge

(b) Iftheenergyreleasedwasallusedtoacceleratea1tonnecardeduceitsfinalspeed

(c) Iftheenergyreleasedcouldbeusedtomelticeat0degCdeducetheamountoficethatcould be melted

PHYSICS 2015indb 317 150514 338 PM

Chapter 10

318

AH

L

15 SupposethatwhenanelectronmovedfromAtoBin the diagram along an electric field line that the electric field does 36 times 10-19Jofworkonit

Determinethedifferencesinelectricpotential

(a) VB ndash VA(b) VC ndash VA(c) VC ndash VB

16 DeterminethepotentialatpointPthatislocatedatthecentreofthesquareasshowninthediagrambelow

- 6μC

+3μC +2μC

1m

1m

5μC

P

102 Fields at work

Essential idea Similar approaches can be taken inanalysingelectricalandgravitationalpotentialproblems

Understandingsbull Potentialandpotentialenergybull Potentialgradientbull Potentialdifferencebull Escapespeedbull Orbitalmotionorbitalspeedandorbitalenergybull Forcesandinverse-squarelawbehaviour

Orbital motion orbital speed and orbital energy

Although orbital motion may be circular elliptical orparabolic this sub-topic only dealswith circular orbitsThis sub-topic is not fundamentally new physics butan application that synthesizes ideas from gravitationcircularmotiondynamicsandenergy

TheMoonorbitstheEarthandinthissenseitisoftenreferredtoasasatelliteoftheEarthBefore1957itwastheonlyEarthsatelliteHoweverin1957theRussianslaunchedthefirstmanmadesatelliteSputnik1SincethisdatemanymoresatelliteshavebeenlaunchedandtherearenowliterallythousandsofthemorbitingtheEarthSomeareusedtomonitortheweathersome used to enable people to find accurately their position on thesurfaceoftheEarthmanyareusedincommunicationsandnodoubtsomeareusedtospyonothercountriesFigure1015showshowinprincipleasatellitecanbeputintoorbit

Theperson(whosesizeisgreatlyexaggeratedwithrespectto Earth) standing on the surface on the Earth throwssome stonesThe greater the speed with which a stoneis thrown the further it will land from her The pathsfollowedbythethrownstonesareparabolasByastretchof the imaginationwecanvisualisea situation inwhichastoneis thrownwithsuchaspeedthatbecauseof thecurvatureoftheEarthitwillnotlandonthesurfaceoftheEarthbutgointoldquoorbitrdquo(Path4onFigure1015)

NATURE OF SCIENCECommunication of scientific explanations The ability to apply field theory to the unobservable (charges) and the massively scaled (motion of satellites) required scientists to develop new ways to investigate analyse and report findings to a general public used to scientific discoveries based on tangible and discernible evidence (51)

copy IBO 2014

PHYSICS 2015indb 318 150514 338 PM

Fields

319

AH

L

1

2

34Earth

Figure 1015 Throwing a stone into orbit

TheforcethatcausesthestonestofollowaparabolicpathandtofalltoEarthisgravityandsimilarlytheforcethatkeeps the stone in orbit is gravity For circular motionto occur we have seen that a force must act at rightanglestothevelocityofanobjectthatistheremustbeacentripetalforceHenceinthesituationwedescribeherethe centripetal force for circular orbital motion about the EarthisprovidedbygravitationalattractionoftheEarth

We can calculate the speedwithwhich a stonemustbethrowninordertoputitintoorbitjustabovethesurfaceof the Earth

If the stone has mass m and speed v thenwehave fromNewtonrsquos2ndlaw

mv2

RE---------- G

MEm

RE2------------=

whereRE is the radius of the Earth and ME is the mass of the Earth

Bearinginmindthatg 0 GME

RE2--------= then

v g RE 10 64 106timestimes 8 103times= = =

Thatisthestonemustbethrownat8times103m sndash1

Clearly we are not going to get a satellite into orbit soclosetothesurfaceoftheEarthMovingatthisspeedthefrictionduetoairresistancewouldmeltthesatellitebeforeithadtravelledacoupleofkilometresInrealitythereforeasatelliteisputintoorbitabouttheEarthbysendingitattachedtoarocketbeyondtheEarthrsquosatmosphereandthengivingitacomponentofvelocityperpendiculartoaradialvectorfromtheEarthSeeFigure1016

Earth Satellite carried by rocket to hereSatellite orbit

Tangential component of velocity

Figure 1016 Getting a satellite into orbit

Keplerrsquos third law(This work of Kepler and Newtonrsquos synthesis of the work is an excellent example of the scientific method and makes for a good TOK discussion)

In 1627 Johannes Kepler (1571-1630)publishedhis lawsof planetary motion The laws are empirical in natureandwere deduced from the observations of theDanishastronomer Tycho de Brahe (1546-1601)The third lawgives a relationship between the radius of orbit R of a planet and its period TofrevolutionabouttheSunThelawisexpressedmathematicallyas

2

3 constantTR

=

We shall nowuseNewtonrsquos LawofGravitation to showhowitisthattheplanetsmoveinaccordancewithKeplerrsquosthirdlaw

InessenceNewtonwasabletousehislawofgravitytopredict the motion of the planets since all he had to do was factor theF givenby this law intohis second law F = ma to find their accelerations and hence their future positions

InFigure1017theEarthisshownorbitingtheSunandthedistancebetweentheircentresisR

RSun

Earth

F es

F se

Figure 1017 Planets move according to Keplerrsquos third law

Fes is the force that the Earth exerts on the Sun and Fse is the force that the Sun exerts on the Earth The forces are equalandoppositeandtheSunandtheEarthwillactuallyorbitaboutacommoncentreHoweversince theSun issoverymuchmoremassivethantheEarththiscommoncentrewillbeclosetothecentreoftheSunandsowecanregard the Earth as orbiting about the centre of the Sun Theotherthingthatweshallassumeisthatwecanignorethe forces that the other planets exert on the Earth (This wouldnotbeawisethingtodoifyouwereplanningtosend a space ship to theMoon for example)We shallalso assume that we have followed Newtonrsquos exampleandindeedprovedthataspherewillactasapointmasssituated at the centre of the sphere

Kepler had postulated that the orbits of the planets areellipticalbut since theeccentricityof theEarthrsquosorbit issmallweshallassumeacircularorbit

PHYSICS 2015indb 319 150514 338 PM

Chapter 10

320

AH

L

TheaccelerationoftheEarthtowardstheSunisa = Rω2

whereω ω 2πT------=

Hencea R 2π

T------ times

2 4π2 RT 2------------- = =

But the acceleration is given by Newtonrsquos Second Law F = mawhereFisnowgivenbytheLawofGravitationSo in this situation

F maGMs Me

R2-------------------= = butwealsohavethat

a a R 2πT

------ times

2 4π2 RT 2------------- = = and m = Me so that

G MsMe

R2------------------- Me4π2R

T2-------------times

GMs

4π2-----------rArr R3

T2------= =

Butthequantity

isaconstantthathasthesamevalueforeachoftheplanetssowehaveforalltheplanetsnotjustEarththat

R3

T2------ k =

wherekisaconstantWhichisofcourseKeplerrsquosthirdlaw

ThisisindeedanamazingbreakthroughItisdifficulttorefutethe idea that all particles attract each other in accordance withtheLawofGravitationwhenthelawisabletoaccountfortheobservedmotionoftheplanetsabouttheSun

The gravitational effects of the planets upon each othershould produce perturbations in their orbits Such is the predictivepoweroftheUniversalGravitationalLawthatit enabled physicists to compute these perturbations The telescopehadbeeninventedin1608andbythemiddleofthe 18th Century had reached a degree of perfection in design that enabled astronomers to actually measure the orbital perturbations of the planets Their measurements werealways inagreementwith thepredictionsmadebyNewtonrsquoslawHoweverin1781anewplanetUranuswasdiscovered and the orbit of this planet did not fit withthe orbit predicted by Universal Gravitation Such wasthe physicistrsquos faith in theNewtonianmethod that theysuspectedthatthediscrepancywasduetothepresenceofayetundetectedplanetUsingtheLawofGravitationtheFrenchastronomerJLeverrier and the English astronomer J C Adamswereable tocalculate justhowmassive thisnewplanetmustbeandalsowhereitshouldbeIn1846the planetNeptunewas discovered justwhere they hadpredicted Ina similarwaydiscrepancies in theorbitofNeptune ledto thepredictionandsubsequentdiscoveryin1930oftheplanetPlutoNewtonrsquosLawofGravitation

had passed the ultimate test of any theory it is not only abletoexplainexistingdatabutalsotomakepredictions

Satellite energyWhenasatelliteisinorbitaboutaplanetitwillhavebothkineticenergyandgravitationalpotentialenergySupposewe consider a satellite ofmass m that is in an orbit of radius r about a planet of mass M

Thegravitationalpotentialduetotheplanetatdistancer from its centre is

ThegravitationalpotentialenergyofthesatelliteVsat

is therefore GMe m

r----------------ndash

ThatisVsat GMem

r----------------ndash=

Thegravitationalfieldstrengthatthesurfaceoftheplanetisgivenby

g 0GMe

Re2------------=

Hencewecanwrite

Vsat g0 Re

2

rndash=

m

ThekineticenergyofthesatelliteKsatisequaltofrac12mv2wherev is its orbital speed

Byequatingthegravitationalforceactingonthesatellitetoitscentripetalaccelerationwehave

G Mem

r2---------------- mv2

r---------- mv2hArr

GMemr----------------= =

Fromwhich12---

mv2 12---

GMemr----------------times=

= 0 e

2g R

r

2m

Which is actually quite interesting since it shows thatirrespective of the orbital radius the KE is numericallyequal to half the PE Also the total energy E tot K sat Vsat+ 1

2---G Me m

r----------------times G Me m

r----------------ndash + 1

2---G Me m

r----------------ndash= = =of the satelliteisalwaysnegativesince

E tot K sat Vsat+ 12---

G Me mr----------------times

G Me mr----------------ndash

+ 12---

G Me mr----------------ndash= = =

PHYSICS 2015indb 320 150514 338 PM

Fields

321

AH

L

The energies of an orbiting satellite as a function of radial distancefromthecentreofaplanetareshownplottedinFigure1018

12

100806

0402

0ndash02

ndash04

ndash06ndash08ndash10

ndash12

2 4 6 8 10 12

ener

gya

rbita

ry u

nits

distance timesR

kinetic energytotal energy

potential energy

Figure 1018 Energy of an orbiting satellite as a function of distance from the centre of a planet

WeightlessnessSuppose that you are inanelevator(lift)thatisdescendingat constant speedandyou let goof abook thatyouareholdinginyourhandThebookwillfalltothefloorwithaccelerationequaltotheaccelerationduetogravityIfthecablethatsupportstheelevatorweretosnap(asituationthatItrustwillneverhappentoanyofyou)andyounowletgothebookthatyouareholdinginyourotherhandthis book will not fall to the floor - it will stay exactlyin linewithyourhandThis isbecause thebook isnowfallingwith the sameaccelerationas theelevatorandassuch the book cannot ldquocatchrdquo up with the floor of theelevatorFurthermoreifyouhappenedtobestandingonasetofbathroomscalesthescaleswouldnowreadzero-youwouldbeapparentlyweightlessItisthisideaoffreefallthatexplainstheapparentweightlessnessofastronautsin an orbiting satellite These astronauts are in free fall inthesensethattheyareacceleratingtowardsthecentre of the Earth

It is actually possible to define theweight of a body inseveral different waysWe can define it for example asthegravitationalforceexertedonthebodybyaspecifiedobject such as the EarthThiswe have seen thatwe doinlotsofsituationswherewedefinetheweightasbeingequaltomgIfweusethisdefinitionthenanobjectinfreefallcannotbydefinitionbeweightlesssinceitisstillinagravitationalfieldHoweverifwedefinetheweightofanobjectintermsofaldquoweighingrdquoprocesssuchasthereadingonasetofbathroomscaleswhichineffectmeasuresthecontact force between the object and the scales thenclearlyobjectsinfreefallareweightlessOnenowhasto

askthequestionwhetherornotitispossibleForexampletomeasurethegravitationalforceactingonanastronautin orbit about the Earth

WecanalsodefineweightintermsofthenetgravitationalforceactingonabodyduetoseveraldifferentobjectsForexample for an object out in space its weight could bedefined in terms of the resultant of the forces exerted on it bytheSuntheMoontheEarthandalltheotherplanetsin the Solar System If this resultant is zero at a particular pointthenthebodyisweightlessatthispoint

Inviewofthevariousdefinitionsofweightthatareavailabletousitisimportantthatwhenweusethewordldquoweightrdquoweareawareofthecontextinwhichitisbeingused

Example

Calculate the height above the surface of the Earth atwhichageo-stationarysatelliteorbits

Solution

A geo-stationary satellite is one that orbits the Earth in such a way that it is stationary with respect to a point on the surface of the Earth This means that its orbital period must be the same as the time for the Earth to spin once on its axis ie 24 hours

From Keplerrsquos third law we have G Ms

4π2----------- R3

T2------=

That is

RMe

mh

using the fact that the force of attraction between the satellite and the Earth is given by

FG Mem

R2----------------=

and that F = ma

where m is the mass of the satellite and a 4π2R

T2-------------=

we have

GMem

R2---------------- m 4π2R

T2-------------times

GMe

4π2------------rArr R3

T2------= =

PHYSICS 2015indb 321 150514 338 PM

Chapter 10

322

AH

L

Now the mass of the Earth is 60 times 1024 kg and the period T measured in seconds is given by T = 86400 s

So substitution gives R = 42 times 106 m

The radius of the Earth is 64 times 106 m so that the orbital height h is about 36 times 107 m

Example

Calculatetheminimumenergyrequiredtoputasatelliteofmass500kgintoanorbitthatisasaheightequaltotheEarthrsquosradiusabovethesurfaceoftheEarth

Solution

We have seen that when dealing with gravitational fields and potential it is useful to remember that

g 0G M

Re2---------= or g0 Re

2 GM=

The gravitational potential at the surface of the Earth is

g 0ndash Re GMRe

---------ndash=

The gravitational potential at a distance R

from the centre of the Earth is

=

The difference in potential between the surface of the Earth and a point distance R from the centre is therefore

V∆ g 0Re 1ReR------ndash

=

If R 2Re= then V∆g0 Re

2------------=

This means that the work required to ldquoliftrdquo the satellite into orbit is g0Rm where m is the mass of the satellite This is equal to

10 32 106 500timestimestimes = 16000 MJ

However the satellite must also have kinetic energy in order to orbit Earth This will be equal to

g 0 mRe2

2R-----------------g 0mRe

2

4----------------- 8000 MJ= =

The minimum energy required is therefore

24000 MJ

Exercise 1031 The speed needed to put a satellite in orbit does

not depend on

A the radius of the orbitB theshapeoftheorbitC thevalueofgattheorbitD the mass of the satellite

2 EstimatethespeedofanEarthsatellitewhoseorbitis400kmabovetheEarthrsquossurfaceAlsodetermine the period of the orbit

3 Calculatethespeedofa200kgsatelliteorbitingthe Earth at a height of 70 times 106 m

Assume that g = 82 m sndash2 for this orbit

4 TheradiioftwosatellitesXandYorbitingthe Earth are 2r and 8rwherer is the radius of the Earth Calculate the ratio of the periods of revolutionofXtoY

5 A satellite of mass mkgissentfromEarthrsquossurfaceinto an orbit of radius 5RwhereR is the radius of theEarthWritedownanexpressionfor

(a) thepotentialenergyofthesatelliteinorbit

(b) thekineticenergyofthesatelliteinorbit

(c) theminimumworkrequiredtosendthesatellitefromrestattheEarthrsquossurfaceintoits orbit

6 A satellite in an orbit of 10rfallsbacktoEarth(radius r)afteramalfunctionDeterminethespeedwithwhichitwillhittheEarthrsquossurface

7 The radius of the moon is frac14 that of the Earth AssumingEarthandtheMoontohavethesamedensitycomparetheaccelerationsoffreefallatthe surface of Earth to that at the surface of the Moon

8 UsethefollowingdatatodeterminethegravitationalfieldstrengthatthesurfaceoftheMoon and hence determine the escape speed from the surface of the Moon

Mass of the Moon = 73 times 1022kg

RadiusoftheMoon=17times106 m

PHYSICS 2015indb 322 150514 338 PM

Chapter 9

304

AH

L

sourceA B

smallerwavelength

largerwavelength

Figure 931 (a) Sound waves from a moving source

The wavefronts are now crowded together in the direction of travel of the source and stretched out in the opposite direction This is why now the two observers will now hear notesofdifferentfrequenciesHowmuchthewavesbunchtogether and how much they stretch out will depend on the speed vEssentially A

A

cfλ

= and BB

cfλ

= where λA lt λB and v is the speed of sound

IfthesourceisstationaryandAismovingtowardsitthenthe waves from the source incident on A will be bunched up If A is moving away from the stationary source then the waves from the source incident on A will be stretched out

Christian Doppler(1803ndash1853)actuallyappliedtheprinciple(incorrectlyasithappens)totryandexplainthecolourofstars However the Doppler effect does apply to light aswellastosoundIfalightsourceemitsalightoffrequencyf then if it is moving away from an observer the observer willmeasurethelightemittedashavingalowerfrequencythan fSincethesensationofcolourvisionisrelatedtothefrequencyoflight(bluelightisofahigherfrequencythanred light) light emitted by objects moving way from anobserver is often referred to as being red-shifted whereas if the object is moving toward the observer it is referred to as blue-shiftedThisideaisusedinOptionE(Chapter16)

We do not need to consider here the situations where either the source or the observer is accelerating In a situation for example where an accelerating source is approaching astationaryobserverthentheobserverwillhearasoundof ever increasing frequency This sometimes leads toconfusion in describing what is heard when a source approaches passes and then recedes from a stationaryobserver Suppose for example that you are standingon a station platform and a train sounding its whistle is approaching at constant speed What will you hear as the train approaches and then passes through the station As the train approaches you will hear a sound of constant pitch but increasing loudness The pitch of the sound will be greater than if the train were stationary As the train passes through the station you will hear the pitch change at the momentthetrainpassesyoutoasoundagainofconstantpitch The pitch of this sound will be lower than the sound of the approaching train and its intensity will decrease as the train recedes from you What you do not hear is a sound of increasing pitch and then decreasing pitch

The Doppler equations for sound

Although you will not be expected in an IB examination to derivetheequationsassociatedwithaspectsoftheDopplereffectyouwillbeexpectedtoapplythemForcompletenesstherefore thederivationoftheequationsassociatedwiththeDopplereffectasoutlinedaboveisgivenhere

S primeS O

vs

Figure 931 (b)

InFigure931(b)theobserverOisatrestwithrespecttoasourceofsoundSismovingwithconstanspeedvs directly towards O The source is emitting a note of constant frequencyf and the speed of the emitted sound is v

S shows the position of the source ∆t later When the sourceisatresttheninatime∆t the observer will receive f∆t waves and these waves will occupy a distance v∆t ie

v f

λ =f∆tv∆t =

(Because of the motion of the source this number of waves will now occupy a distance (v∆t ndash vs∆t)The lsquonewrsquowavelength is therefore

v ndash vs

fλrsquo =

f∆tv∆t ndash vs∆t

=

If fisthefrequencyheardbyOthen

vfλ

= or

vf

λ = = sv vfminus

From which

s

vf fv v

=minus

Dividingthroughbyv gives

Equation1111

1 sf f v

v=

minus

If the source is moving away from the observer then we have

s

vf fv v

=minus

1

1 sf vv

=+

Equation112

PHYSICS 2015indb 304 150514 337 PM

Wave Phenomena

305

AH

L

We now consider the case where the source is stationary and the observerismoving towardsthesource with speed v0 In this situation the speed of the sound waves as measured by the observer will be v0 + v We therefore have that

v0 + v = ffraslλ = vfλ

times

From which

o1 vf fv

= +

Equation113

If the observerismovingawayfromthesource then

o1 vf fv

= minus

Fromequation(113)wehavethat

o o1 v vf f f f f fv v

∆ = minus = + minus =

Equation114

Thevelocitiesthatwerefertointheaboveequationsarethevelocities with respect to the medium in which the waves fromthesourcetravelHoweverwhenwearedealingwithalight source it is the relative velocity between the source and the observer that we must consider The reason for this is that lightisuniqueintherespectthatthespeedofthelightwavesdoes not depend on the speed of the source All observers irrespective of their speed or the speed of the source will measure the same velocity for the speed of light This is oneof thecornerstonesof theSpecialTheoryofRelativitywhichisdiscussedinmoredetailinOptionH(Chapter18)When applying the Doppler effect to light we aremainlyconcerned with the motion of the source We look here only at the situation where the speed of the source v is much smaller than the speed of light cinfreespace(v ltlt c)Underthesecircumstanceswhenthesourceismovingtowardstheobserverequation111becomes

vf f f fc

minus = ∆ = Equation115

andwhenthesourceismovingawayfromtheobserverequation112becomes vf f f f

cminus = ∆ = minus

Provided that v ltlt c these same equations apply for astationary source and moving observer

We look at the following example and exercise

Example

Asourceemitsasoundoffrequency440HzItmovesinastraight line towards a stationary observer with a speed of 30ms-1Theobserverhearsasoundoffrequency484HzCalculate the speed of sound in air

Solution

We use equation 111 and substitute f= 484 Hz f = 440 Hz

and vs = 30 m s-1

therefore 1

484 440 301v

=minus

such that 30 4401

484vminus = to

give v = 330 m s-1

Example

A particular radio signal from a galaxy is measured as havingafrequencyof139times109HzThesamesignalfromasourceinalaboratoryhasafrequencyof142times109Hz

Suggestwhy thegalaxy ismovingaway fromEarthandcalculateitsrecessionspeed(iethespeedwithwhichitismovingawayfromEarth)

Solution

The fact that the frequency from the moving source is less than that when it is stationary indicates that it is moving away from the stationary observer ie Earth

Using vf fc

∆ = we have

8 96

9

3 10 (142 139) 10 634 10142 10

c fvf∆ times times minus times

= = = timestimes

m s-1

It is usual when dealing with the Doppler effect of light to express speeds as a fraction of c So in this instance we have v = 0021 c

Using the Doppler effectWe have seen in the above example and exercise how the Dopplereffectmaybeusedtomeasuretherecessionspeedof distant galaxies The effect is also used to measure speed inothersituationsHerewewilllookatthegeneralprincipleinvolved in using the Doppler effect to measure speedFigure932showsasource(thetransmitter)thatemitseithersound or em waves of constant frequency f The waves

PHYSICS 2015indb 305 150514 337 PM

Chapter 9

306

AH

L

fromthesourceare incidentonareflectorthat ismovingtowards the transmitter with speed vThereflectedwavesare detected by the receiver placed alongside the transmitter

reector

v

transmitter

receiver

f f

f f

Figure 932 Using the Doppler effect to measure speed

We shall consider the situation where v ltlt c where c is the speed of the waves from the transmitter

For the reflector receiving waves from the transmitterit is effectively an observer moving towards a stationary sourceFromequation(114) itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation116

For the receiver receivingwaves from the reflector it iseffectively a stationary observer receiving waves from a movingsourceFromequation(115)itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation117

Ifweaddequations(116)and(117)wegetthatthetotalDopplershiftatthereceiver∆f is

v vf f f f fc c

minus = ∆ = +

But 1 vf fc

= +

hence

1 v v vf f fc c c

∆ = + +

But since v ltlt cwe can ignore the term2

2

vc when we

expandthebracketintheaboveequation

Therefore we have2vf fc

∆ = Equation118

If v asymp c then we must use the full Doppler equationsHowever for em radiationwewill alwaysonly considersituations in which v ltlt c

Example

Thespeedofsoundinbloodis1500times103 m s-1 Ultrasound offrequency100MHzisreflectedfrombloodflowinginan arteryThe frequencyof the reflectedwaves receivedbackatthetransmitteris105MHzEstimatethespeedofthebloodflowintheartery

Solution

Using equation (118) we have

6 63

2005 10 1015 10

vtimes = times

timesto give v asymp 36 m s-1 (We have assumed that the ultrasound is incident at right angles to the blood flow)

Exercise 971 Judy is standing on the platform of a station A

high speed train is approaching the station in a straight line at constant speed and is sounding its whistleAsthetrainpassesbyJudythefrequencyof the sound emitted by the whistle as heard by Judychangesfrom640Hzto430HzDetermine

(a) thespeedofthetrain

(b) thefrequencyofthesoundemittedbythewhistle as heard by a person on the train (Speedofsound=330ms-1)

2 AgalaxyismovingawayfromEarthwithaspeedof00500cThewavelengthofaparticularspectral line in light emitted by atomic hydrogen inalaboratoryis656times10-7 m Calculate the valueofthewavelengthofthislinemeasuredinthelaboratoryinlightemittedfromasourceofatomic hydrogen in the galaxy

PHYSICS 2015indb 306 150514 337 PM

10 Fields

Contents

101 ndash Describing fields

102 ndash Fields at work

Essential Ideas

Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Similar approaches can be taken in analysing electrical and gravitational potential problems copy IBO 2014

PHYSICS 2015indb 307 150514 337 PM

Chapter 10

308

AH

L

101 Describing fields

Essential idea Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Understandingsbull Gravitationalfieldsbull Electrostaticfieldsbull Electricpotentialandgravitationalpotentialbull Fieldlinesbull Equipotentialsurfaces

Gravitational potentialWehaveseenthatifweliftanobjectofmassm to a height h abovethesurfaceoftheEarththenitsgainingravitationalpotential energy is mghHoweverthisisbynomeansthefullstoryForastartwenowknowthatgvarieswithh and also the expression really gives a difference in potentialenergybetweenthevaluethattheobjecthasattheEarthrsquossurfaceandthevaluethatithasatheighthSowhatwereallyneedisazeropointCanwefindapointwherethepotentialenergyiszeroandusethispointfromwhichtomeasure changes in potential energy

The point that is chosen is in fact infinity At infinity the gravitationalfieldstrengthofanyobjectwillbezeroSoletusseeifwecandeduceanexpressionforthegaininpotentialenergyofanobjectwhenitisldquoliftedrdquofromthesurfaceoftheEarthtoinfinityThisineffectmeansfindingtheworknecessarytoperformthistask

δr

r r δr+

r infin=

mg

AB

Me

Figure 1001 Gravitational forces

Inthediagramweconsidertheworknecessarytomovethe particle of mass m a distance δr in thegravitationalfield of the Earth

The force on the particle at A is FG Mem

r2----------------=

IftheparticleismovedtoB then since δr isverysmallwe can assume that the field remains constant over thedistance ABTheworkδW done againstthegravitationalfieldoftheEarthinmovingthedistanceABis

δWGMem

r2----------------δrndash=

(rememberthatworkdoneagainstaforceisnegative)

TofindthetotalworkdoneW in going from the surface oftheEarthtoinfinitywehavetoaddalltheselittlebitsofworkThisisdonemathematicallybyusingintegralcalculus

WG Me m

r2----------------ndash

rdR

infin

int G Me m 1

r2----- rd

R

infin

intndash G Me m 1r---ndash

R

infinndash= = =

GMe m 0 1R---ndash

ndashndash=

G MemR----------------ndash=

HencewehavewhereRistheradiusoftheEarththattheworkdonebythegravitationalfieldinmovinganobjectofmass m from R(surfaceoftheEarth)toinfinityisgivenby

WGMe m

R---------------- ndash=

We can generalise the result by calculating the worknecessary per unit mass to take a small mass fromthe surface of the Earth to infinity This we call thegravitational potential Vie

Wewouldgetexactlythesameresultifwecalculatedtheworkdonebythefieldtobringthepointmassfrominfinityto the surface of Earth In this respect the formal definition ofgravitationalpotentialatapointinagravitationalfieldis therefore defined as the work done per unit mass in bringing a point mass from infinity to that point

Clearly then the gravitational potential at any point inthe Earthrsquos field distance r from the centre of the Earth (providingr gt R)is

VGMe

r------------ ndash=

Thepotentialisthereforeameasureoftheamountofworkthathastobedonetomoveparticlesbetweenpointsinagravitationalfieldanditsunit is theJkgndash1 We also note thatthepotentialisnegativesothatthepotentialenergy

NATURE OF SCIENCEParadigm shift The move from direct observable actions being responsible for influence on an object to acceptance of a fieldrsquos ldquoaction at a distancerdquo required a paradigm shift in the world of science (23)

copy IBO 2014

PHYSICS 2015indb 308 150514 337 PM

Fields

309

AH

L

aswemoveawayfromtheEarthrsquossurfaceincreasesuntilitreachesthevalueofzeroatinfinity

IfthegravitationalfieldisduetoapointmassofmassmthenwehavethesameexpressionasaboveexceptthatMe is replaced by m andmustalsoexclude thevalueof thepotential at the point mass itself ie at r = 0

WecanexpressthegravitationalpotentialduetotheEarth(orduetoanysphericalmass)intermsofthegravitationalfield strength at its surface

AtthesurfaceoftheEarthwehave

g 0ndash Re GMRe

---------ndash= e

Sothat

g 0Re2 GM= e

Hence at a distance r from the centre of the Earth the gravitationalpotentialVcanbewrittenas

VGMe

r------------ndashg 0 Re

2

r------------ndash= =

The potential at the surface of the Earth

(r = Re)istherefore-g0Re

It is interesting to see how the expression for thegravitational potential ties in with the expression mgh ThepotentialatthesurfaceoftheEarthis-g0Re (see the exampleabove)andataheighthwillbe-g0g 0 Re h+( )ndash ifweassume that g0doesnotchangeover thedistanceh The differenceinpotentialbetweenthesurfaceandtheheighth is therefore g0hSotheworkneededtoraiseanobjectof mass m to a height h is mgh ie m timesdifference ingravitationalpotential

This we have referred to as the gain in gravitationalpotentialenergy(see235)

However this expression can be extended to any twopoints inanygravitationalfieldsuch that ifanobjectofmass mmovesbetween twopointswhosepotentialsareV1 and V2 respectively then the change in gravitationalpotentialenergyoftheobjectism(V1 ndash V2)

Gravitational potential gradient

Let us consider now a region in space where thegravitationalfieldisconstantInFigure912thetwopointsAandBareseparatedbythedistance∆x

A B

direction of uniform gravitational eld of strength I

∆x

Figure 1002 The gravitational potential gradient

The gravitational field is of strength I and is in the direction shownThe gravitational potential at A is V andatBisV +∆V

Theworkdone is takingapointmassmfromAtoB isF∆x = mI∆x

Howeverbydefinitionthisworkisalsoequalto-m∆V

Therefore mI∆x=-m∆V

or VIx

∆= minus

∆Effectivelythissaysthatthemagnitudeofthegravitationalfield strength is equal to the negative gradient of thepotential If I is constant then V is a linear function of x and Iisequaltothenegativegradientofthestraightlinegraph formed by plotting V against x If I is not constant (asusuallythecase)thenthemagnitudeofI at any point in the field can be found by find the gradient of the V-x graph at that point An example of such a calculation can befoundinSection929

For those of you who do HL maths the relationship between field and potential is seen to follow from the expression for the potential of a point mass viz

mV Gr

= minus

2

ddV mG Ir r

minus = + =

Potential due to one or more point masses

Gravitationalpotentialisascalarquantitysocalculatingthepotential due to more than one point mass is a matter of simpleadditionSoforexamplethepotentialV due to the Moon and Earth and a distance x fromthecentreofEarthonastraightlinebetweenthemisgivenbytheexpression

E MM MV Gx r x

= minus + minus

whereME=massofEarthMM= mass of Moon and r = distancebetweencentreofEarthandMoon

PHYSICS 2015indb 309 150514 337 PM

Chapter 10

310

AH

L

Equipotentials and field linesIfthegravitationalpotentialhasthesamevalueatallpointson a surface the surface is said to be an equipotential surfaceSoforexampleifweimagineasphericalshellaboutEarthwhosecentrecoincideswiththecentreofEarththisshellwillbeanequipotentialsurfaceClearlyifwerepresentthegravitationalfieldstrengthbyfieldlinessincethelinesldquoradiaterdquooutfromthecentreofEarththentheselineswillbeatrightanglestothesurfaceIfthefieldlineswerenotnormaltotheequipotentialsurfacethentherewouldbeacomponentof the field parallel to the surfaceThis wouldmean thatpointsonthesurfacewouldbeatdifferentpotentialsandsoitwouldnolongerbeanequipotentialsurfaceThisofcourseholdstrueforanyequipotentialsurface

Figure 1003 shows the field lines and equipotentials fortwopointmassesm

m m

Figure 1003 Equipotentials for two point masses

It is worth noting that we would get exactly the samepattern ifwewere to replace thepointmasseswith twoequalpointcharges(See935)

Escape speedThe potential at the surface of Earth is which meansthattheenergyrequiredtotakeaparticleofmass mfromthesurfacetoinfinityisequalto

ButwhatdoesitactuallymeantotakesomethingtoinfinityWhentheparticleisonthesurfaceoftheEarthwecanthinkofitassittingatthebottomofaldquopotentialwellrdquoasinfigure1004

particle

surface of Earth

innity

GMR---------

FIgure 1004 A potential well

Theldquodepthrdquoofthewellis and if the particle gains an amount of kinetic energy equal to wherem is its massthenitwillhavejustenoughenergytoldquoliftrdquoitoutofthewell

In reality it doesnrsquot actually go to infinity it justmeansthat the particle is effectively free of the gravitationalattractionof theEarthWe say that ithas ldquoescapedrdquo theEarthrsquosgravitationalpullWemeetthisideainconnectionwithmolecular forcesTwomolecules in a solidwill sitat their equilibrium position the separation where therepulsiveforceisequaltotheattractiveforceIfwesupplyjust enough energy to increase the separation of themolecules such that they are an infinite distance apart then intermolecularforcesnolongeraffectthemoleculesandthesolidwillhavebecomealiquidThereisnoincreaseinthekineticenergyofthemoleculesandsothesolidmeltsat constant temperature

Wecancalculatetheescapespeedofanobjectveryeasilyby equating the kinetic energy to the potential energysuch that

12---mvesc ape

2 GMe mRe

----------------=

vesc aperArr2GMe

Re--------------- 2g0 Re = =

Substituting for g0 and Regivesavalueforvescape of about 11kmsndash1 from the surface of the Earth

Youwillnotethattheescapespeeddoesnotdependonthemassoftheobjectsincebothkineticenergyandpotentialenergy are proportional to the mass

IntheoryifyouwanttogetarockettothemoonitcanbedonewithoutreachingtheescapespeedHoweverthiswouldnecessitate anenormousamountof fuel and it islikely that therocketplus fuelwouldbesoheavythat itwouldnevergetoffthegroundItismuchmorepracticalto accelerate the rocket to the escape speed fromEarthorbitandthenintheoryjustlaunchittotheMoon

Example

Use the followingdata todetermine thepotential at thesurface of Mars and the magnitude of the acceleration of free fall

massofMars =64times1023kg

radiusofMars =34times106 m

Determinealsothegravitationalfieldstrengthatadistanceof 68 times 106mabovethesurfaceofMars

PHYSICS 2015indb 310 150514 337 PM

Fields

311

AH

L

Solution

= minus = minus times times = times23

11 76

64 1067 10 ndash13 10

34 10MV GR

minus timestimes

Nkg-1

But V = -g0R

Therefore 7

0 6

13 10 3834 10

VgR

times= minus = =

timesm s-2

To determine the field strength gh at 68 times 106 m above the surface we use the fact that 0 2

Mg GR

= such that GM = g0R2

Therefore 2 2

0h 2 2 2

h h

38 (34) 042(102)

g RGMgR R

times= = = = m s-2

(the distance from the centre is 34 times 106 + 68 times 106 = 102 times 106 m)

Exercise 1011 Thegraphbelowshowshowthegravitational

potentialoutsideoftheEarthvarieswithdistancefrom the centre

0 10 20 30 40 50 60 70

ndash1

ndash2

ndash3

ndash4

ndash5

ndash6

ndash7

r m times 106

V

Jkgndash1

times 10

7

(a) Usethegraphtodeterminethegainingravitationalpotentialenergyofasatelliteofmass200kgasitmovesfromthesurfaceof the Earth to a height of 30 times 107mabovetheEarthrsquossurface

(b) Calculatetheenergyrequiredtotakeit to infinity

(c) DeterminetheslopeofthegraphatthesurfaceoftheEarthmCommenton youranswer

Electric potential energy

The concept of electric potential energy was developedwith gravitational potential energy in mind Just as anobjectnearthesurfaceoftheEarthhaspotentialenergybecause of its gravitational interaction with the Earthsotoothereiselectricalpotentialenergyassociatedwithinteracting charges

Letusfirstlookatacaseoftwopositivepointchargeseachof 1μC that are initially bound together by a thread in a vacuuminspacewithadistancebetweenthemof10cmasshowninFigure1005Whenthethreadiscutthepointchargesinitiallyatrestwouldmoveinoppositedirectionsmovingwithvelocitiesv1 and v2 along the direction of the electrostatic force of repulsion

BEFORE AFTER

v v21

Figure 1005 Interaction of two positive particles

The electric potential energybetweentwopointchargescan be found by simply adding up the energy associated witheachpairofpointchargesForapairofinteractingchargestheelectricpotentialenergyisgivenby

∆U = ∆Ep + ∆Ek = ∆W = ∆Fr = kqQ

____ r2 times r = kqQ

____ r

Becausenoexternalforceisactingonthesystemtheenergyand momentum must be conserved Initially Ek = 0 and Ep =kqQr=9times109 times 1 times 10-1201m=009JWhentheyareagreatdistance fromeachotherEpwillbenegligibleThefinalenergywillbeequaltofrac12mv1

2+frac12mv22=009J

Momentum isalsoconservedand thevelocitieswouldbethe same magnitude but in opposite directions

Electric potential energyismoreoftendefinedintermsofapointchargemovinginanelectricfieldas

lsquothe electricpotential energybetweenany twopoints inan electric field is defined as negativeoftheworkdone by anelectricfieldinmovingapoint electric chargebetweentwolocationsintheelectricfieldrsquo

ΔU = ΔEp = -ΔW = -Fd = qEx

ΔU = ΔEp +ΔEk = ΔW = ΔFr = kqQ r2 x r = kqQ

wherexisthedistancemovedalong(oroppositeto)thedirection of the electric field

Electric potential energyismeasuredinjoule(J)Justasworkisascalarquantitysotooelectricalpotentialenergyis a scalar quantity The negative of the work done by an electric field in moving a unit electric charge between two points is independent of the path taken Inphysicswesaytheelectricfieldisalsquoconservativersquo field

PHYSICS 2015indb 311 150514 337 PM

Chapter 10

312

AH

L

Supposeanexternalforcesuchasyourhandmovesasmallpositive point test charge in the direction of a uniformelectric fieldAs it ismoving itmust be gaining kineticenergyIfthisoccursthentheelectricpotentialenergyofthe unit charge is changing

InFigure1006apointcharge+qismovedbetweenpointsAandBthroughadistancex in a uniform electric field

B

A

x

+ q

Figure 1006 Movement of a positive point charge in a uniform field

InordertomoveapositivepointchargefrompointAtopointBanexternal forcemustbeapplied to thechargeequaltoqE (F = qE)

Since the force is applied through a distance xthennegativeworkhas to bedone tomove the charge because energyis gainedmeaning there is an increaseelectric potential energybetweenthetwopointsRememberthattheworkdoneisequivalenttotheenergygainedorlostinmovingthe charge through the electric field The concept of electric potential energy is only meaningful as the electric field whichgeneratestheforceinquestionisconservative

W F xtimes E q x times= =

θ x cos θ x

Figure 1007 Charge moved at an angle to the field

If a chargemoves at an angle θ to an electric field thecomponent of the displacement parallel to the electric fieldisusedasshowninFigure918

W F x Eq x θcostimes= =

Theelectricpotentialenergyisstoredintheelectricfieldand theelectricfieldwill return theenergy to thepointcharge when required so as not to violate the Law ofconservationofenergy

Electric potential

The electric potential at a point in an electric field is defined as being the work done per unit charge in bringing a small positive point charge from infinity to that point

ΔV = Vinfin ndash Vf = ndash W ___ q

Ifwedesignatethepotentialenergytobezeroatinfinitythen it follows thatelectricpotentialmustalsobezeroat infinity and the electric potential at any point in an electricfieldwillbe

ΔV = ndash W ___ q

NowsupposeweapplyanexternalforcetoasmallpositivetestchargeasitismovedtowardsanisolatedpositivechargeTheexternalforcemustdoworkonthepositivetestchargetomoveittowardstheisolatedpositivechargeandtheworkmustbepositivewhiletheworkdonebytheelectricfieldmustthereforebenegativeSotheelectricpotentialatthatpointmustbepositiveaccordingtotheaboveequationIfanegativeisolatedchargeisusedtheelectricpotentialatapointonthepositivetestchargewouldbenegativePositivepoint chargesof their ownaccordmove fromaplaceofhighelectricpotential toaplaceof lowelectricpotentialNegative point charges move the other way from lowpotentialtohighpotentialInmovingfrompointAtopointBinthediagramthepositivecharge+qismovingfromalowelectricpotentialtoahighelectricpotential

In thedefinitiongiven the termldquoworkperunitchargerdquohas significance If the test charge is +16 times 10-19Cwherethe charge has a potential energy of 32 times 10-17Jthenthepotentialwouldbe32times10-17J+16times10-19C=200JC-1 Now if the charge was doubled the potential wouldbecome 64times10-17 J However the potential per unitchargewouldbethesame

Electric potential is a scalar quantityandithasunitsJC-1 orvoltswhere1voltequalsone joulepercoloumbThevoltallowsustoadoptaunitfortheelectricfieldintermsofthevolt

PreviouslytheunitfortheelectricfieldwasNCndash1

W = qV and F = qEsoW ___ V = F __ E

E = FV ___ W = FV ___ Fm V mndash1

That is the units of the electric field E can also beexpressed as V mndash1

The work done per unit charge in moving a point charge between two points in an electric field is again independent of the path taken

PHYSICS 2015indb 312 150514 337 PM

Fields

313

AH

L

Electric potential due to a point charge

Letus takeapoint rmetres froma chargedobjectThepotentialatthispointcanbecalculatedusingthefollowing

W = ndashFr = ndashqV and F = ndash q1 q2 _____ 4πε0r

2

Therefore

W = ndash q1 q2 _____ 4πε0r

2 times r = ndash q1 q2 _____ 4πε0r

= ndashq1 times q2 _____ 4πε0r

= ndashq1V

That is V q

4πε0r--------------- =

Orsimply

V kqr------

=

Example

Determinehowmuchworkisdonebytheelectricfieldofpointcharge150μCwhenachargeof200μCismovedfrom infinity to apoint 0400m from thepoint charge(Assumenoaccelerationofthecharges)

Solution

The work done by the electric field is W = -qV = -14πε0 times q times (Q rinfin - Q r0400)

W = (- 200 times 10-6 C times 900 times 109 NmC-2 times 150 times 10-6 C) divide 0400 m = - 0675 J

An external force would have to do +0675 J of work

Electric field strength and electric potential gradient

Letus lookbackatFigure1006Supposeagain that thecharge+qismovedasmalldistancebyaforceFfromAtoBsothattheforcecanbeconsideredconstantTheworkdoneisgivenby

∆W F x∆times=

TheforceFandtheelectricfieldEareoppositelydirectedandweknowthat

F = -qE and ΔW = q ΔV

Thereforetheworkdonecanbegivenas

q ΔV = -q E Δx

Therefore E V∆x∆-------

ndash=

The rate of change of potential ΔV at a point with respect to distance Δx in the direction in which the change is maximum is called the potential gradient We say that the electric field = - the potential gradient and the units are Vm-

1 From the equation we can see that in a graph of electric potential versus distance the gradient of the straight line equals the electric field strength

In reality if a charged particle enters a uniform electric field it will be accelerated uniformly by the field and its kinetic energy will increase This is why we had to assume no acceleration in the last worked example

Ek∆ 12---

mv2 q E xsdot sdot q Vx---

xsdot sdot q Vsdot= = = =

Example

Determine how far apart two parallel plates must besituated so that a potential difference of 150 times 102 V produces an electric field strength of 100 times 103 NC-1

Solution

Using E V∆x------- xhArrndash V∆

E-------15 102 Vtimes

100 103 N C 1ndashtimes------------------------------------------= = =

= 150 times 10-1

The plates are 150 times 10-1 m apart

The electric field and the electric potential at a point due to an evenly distributed charge +q on a sphere can be represented graphically as in Figure 1008

r

r

V

E

r0

r0

r0

Charge of +Q evenly distributed over surface

E 14πε0------------ Q

r2----- r r0gt=

V 14πε0------------Q

r----r r0gt=

Figure 1008 Electric field and potential due to a charged sphere

PHYSICS 2015indb 313 150514 337 PM

Chapter 10

314

AH

L

When the sphere becomes charged we know that thechargedistributes itselfevenlyoverthesurfaceThereforeeverypartofthematerialoftheconductorisatthesamepotential As the electric potential at a point is defined as beingnumericallyequaltotheworkdoneinbringingaunitpositivechargefrominfinitytothatpointithasaconstantvalueineverypartofthematerialoftheconductor

Since the potential is the same at all points on the conducting surfacethen∆V∆xiszeroButE=ndash∆V∆xThereforethe electric field inside the conductor is zero There is no electric field inside the conductor

SomefurtherobservationsofthegraphsinFigure1009are

bull OutsidethespherethegraphsobeytherelationshipsgivenasE α 1 r2 and V α 1 r

bull Atthesurfacer = r0Thereforetheelectricfieldandpotentialhavetheminimumvalueforr at this point and this infers a maximum field and potential

bull Insidethespheretheelectricfieldiszerobull Insidethespherenoworkisdonetomoveacharge

fromapointinsidetothesurfaceThereforethereisno potential difference and the potential is the same asitiswhenr = r0

Similargraphscanbedrawnfortheelectricfieldintensityandthe electric potential as a function of distance from conducting parallelplatesandsurfacesandthesearegiveninFigure1009

Potential plot E eld plotE eld

++++

ndashndashndashndash

xx x

+

x x

ndashx x

+ ndashx

x

+ +

x

x

Figure 1009 Electric field and electric potential at a distance from a charged surface

Potential due to one or more point charges

The potential due to one point charge can be determined byusingtheequationformula

V = kq r

Example 1

Determine the electric potential at a point 20 times 10-1 m fromthecentreofan isolatedconducting spherewithapointchargeof40pCinair

Solution

Using the formula

V = kq r we have

V90 10 9times( ) 40 10 12ndashtimes( )times

20 10 1ndashtimes( )------------------------------------------------------------------ 018 V= =

the potential at the point is 180 times 10-1 V

The potential due to a number of point charges can be determined by adding up the potentials due to individual point charges because the electric potential at any point outside a conducting sphere will be the same as if all the charge was concentrated at its centre

Example 2

Threepointchargesofareplacedattheverticesofaright-angledtriangleasshowninthediagrambelowDeterminetheabsolutepotentialat the+20μCchargeduetothetwoothercharges

- 6μC

+3μC +2μC

4 m

3 m

Solution

The electric potential of the +2 μC charge due to the ndash 6 μC charge is

V = (9 times 109 Nm2C-2 times -6 times 10-6 C) divide (radic 32 + 42) m = - 108 times 104 V

PHYSICS 2015indb 314 150514 338 PM

Fields

315

AH

L

Insummarywecanconcludethat

bull No work is done to move a charge along an equipotential

bull Equipotentials are always perpendicular to the electric lines of force

Figure1011and1012showsomeequipotential lines fortwo oppositely charged and identically positive spheresseparated by a distance

+ve ndashve

equipotential lines

Figure 1011 Equipotential lines between two opposite charges

equipotential lines

Figure 1012 Equipotential lines between two charges which are the same

+

ndash ndash ndash ndash

+ + +

40 V

30 V

20 V

10 V

50 V

Figure 1013 Equipotential lines between charged parallel plates

Figure1013showstheequipotentiallinesbetweenchargedparallel plates Throughout this chapter the similarities and differences between gravitational fields and electric fieldshavebeendiscussedTherelationshipsthatexistsbetweengravitationalandelectricquantitiesandtheeffectsofpointmassesandchargesissummarisedinTable1014

The electric potential of the +2 μC charge due to the +3 μC charge is

V = (9 times 109 Nm2C-2 times 3 times 10-6 C) divide 3m = 9 times 103 V

The net absolute potential is the sum of the 2 potentials- 108 times 104 V + 9 times 103 V =- 18 times 103 V

The absolute potential at the point is - 18 times 103 V

Equipotential surfacesRegions in space where the electric potential of a chargedistributionhasaconstantvaluearecalledequipotentials The placeswherethepotentialisconstantinthreedimensionsarecalled equipotential surfacesandwheretheyareconstantintwodimensionstheyarecalledequipotential lines

TheyareinsomewaysanalogoustothecontourlinesontopographicmapsInthiscasethegravitationalpotentialenergy is constant as amassmoves around the contourlines because the mass remains at the same elevationabovetheEarthrsquossurfaceThegravitationalfieldstrengthacts in a direction perpendicular to a contour line

Similarlybecausetheelectricpotentialonanequipotentiallinehas thesamevalueanelectric forcecandonoworkwhena testchargemovesonanequipotentialThereforethe electric field cannot have a component along anequipotentialandthusitmustbeeverywhereperpendiculartotheequipotentialsurfaceorequipotentiallineThisfactmakesiteasytoplotequipotentialsifthelinesofforceorlinesofelectricfluxofanelectricfieldareknown

Forexamplethereareaseriesofequipotentiallinesbetweentwoparallelplateconductorsthatareperpendiculartotheelectric fieldTherewill be a series of concentric circles(eachcirclefurtherapartthanthepreviousone)thatmapouttheequipotentialsaroundanisolatedpositivesphereThe lines of force and some equipotential lines for anisolatedpositivesphereareshowninFigure1010

Lines ofequipotential

Figure 1010 Equipotentials around an isolated positive sphere

PHYSICS 2015indb 315 150514 338 PM

Chapter 10

316

AH

L

Gravitational quantity Electrical quantityQ

uant

ities

g V∆x∆-------ndash= E V∆

x∆-------ndash=

Poin

t mas

ses a

nd ch

arge

s

V Gmr----

ndash= V 14πε0------------q

r---=

g G m

r2-----ndash= E 14πε0------------ q

r2-----=

F Gm1 m2

r2--------------= F 1

4πε0------------

q1q2

r2-----------=

Figure 1014 Formulas

Example

Deduce the electric potential on the surface of a gold nucleus that has a radius of 62 fm

Solution

Using the formula

V = kq r and knowing the atomic number of gold is 79 We will assume the nucleus is spherical and it behaves as if it were a point charge at its centre (relative to outside points)

V = 90 times 109 Nm2C-2 times 79 times 16 times 10-19 C divide 62 times 10-15 m = 18 times 107 V

The potential at the point is 18 MV

Example

Deduce the ionisation energy in electron-volts of theelectron in the hydrogen atom if the electron is in its ground state and it is in a circular orbit at a distance of 53 times 10-11 m from the proton

Solution

This problem is an energy coulombic circular motion question based on Bohrrsquos model of the atom (not the accepted quantum mechanics model) The ionisation energy is the energy required to remove the electron from the ground state to infinity The electron travels in a circular orbit and therefore has a centripetal acceleration The ionisation energy will counteract the coulombic force and the movement of the electron will be in the opposite direction to the centripetal force

Total energy = Ek electron + Ep due to the proton-electron interaction

ΣF = kqQ r2 = mv2 r and as such mv2 = = kqQ r

Therefore Ek electron = frac12 kqQ r

Ep due to the proton-electron interaction = - kqQ r

Total energy = frac12 kqQ r + - kqQ r = -frac12 kqQ r

= - 90 times 109 Nm2C-2 times (16 times 10-19 C)2 divide 53 times 10-11 m = -217 times 10-18 J

= -217 times 10-18 J divide 16 times 10-19 = -136 eV

The ionisation energy is 136 eV

Exercise 1021 ApointchargePisplacedmidwaybetweentwo

identicalnegativechargesWhichoneofthefollowingiscorrectwithregardstoelectricfieldandelectricpotentialatpointP

Electric field Electric potentialA non-zero zeroB zero non-zeroC non-zero non-zeroD zero zero

2 Twopositivechargedspheresaretiedtogetherinavacuumsomewhereinspacewheretherearenoexternal forces A has a mass of 25 g and a charge of20μCandBhasamassof15gandachargeof30μCThedistancebetweenthemis40cmTheyarethenreleasedasshowninthediagram

BEFORE AFTER

v1 v2

A B

(a) Determinetheirinitialelectricpotentialenergy in the before situation

(b) DeterminethespeedofsphereBafterrelease

3 Thediagrambelowrepresentstwoequipotentiallines in separated by a distance of 5 cm in a uniform electric field

+ + + + + + + +

ndash ndash ndash ndash ndash ndash ndash ndash

40 V

20 V 5 cm

PHYSICS 2015indb 316 150514 338 PM

Fields

317

AH

L

Determine the strength of the electric field

4 Thisquestionisabouttheelectricfieldduetoacharged sphere and the motion of electrons in that fieldThediagrambelowshowsanisolatedmetalsphereinavacuumthatcarriesanegativeelectriccharge of 60 μC

ndash

(a) Onthediagramdrawtheconventionalwayto represent the electric field pattern due to the charged sphere and lines to represent threeequipotentialsurfacesintheregionoutside the sphere

(b) Explainhowthelinesrepresentingtheequipotentialsurfacesthatyouhavesketchedindicatethatthestrengthoftheelectricfieldisdecreasingwithdistancefrom the centre of the sphere

(c) Theelectricfieldstrengthatthesurfaceofthe sphere and at points outside the sphere can be determined by assuming that the sphere acts as a point charge of magnitude 60 μC at its centre The radius of the sphere is 25 times 10ndash2 m Deduce that the magnitude of the field strength at the surface of the sphere is 86 times 107 Vmndash1

An electron is initially at rest on the surface of the sphere

(d) (i) Describethepathfollowedbytheelectronasitleavesthesurfaceofthesphere

(ii) Calculatetheinitialaccelerationofthe electron

5 Determinetheamountofworkthatisdoneinmovingachargeof100nCthroughapotentialdifferenceof150times102 V

6 Three identical 20 μC conducting spheres are placedatthecornersofanequilateraltriangleofsides 25 cm The triangle has one apex C pointing upthepageand2baseanglesAandBDeterminetheabsolutepotentialatB

7 Determinehowfaraparttwoparallelplatesmustbesituatedsothatapotentialdifferenceof250 times 102 V produces an electric field strength of 200 times 103 NC-1

8 Thegapbetweentwoparallelplatesis10times10-3mandthereisapotentialdifferenceof10times104 V betweentheplatesCalculate

i theworkdonebyanelectroninmovingfrom one plate to the other

ii thespeedwithwhichtheelectronreachesthe second plate if released from rest

iii theelectricfieldintensitybetweentheplates

9 Anelectronguninapicturetubeisacceleratedby a potential 25 times 103VDeterminethekineticenergygainedbytheelectroninelectron-volts

10 Determine the electric potential 20 times 10-2 m from achargeof-10times10-5 C

11 Determinetheelectricpotentialatapointmid-waybetweenachargeofndash20pCandanotherof+5pConthelinejoiningtheircentresifthecharges are 10 cm apart

12 During a thunderstorm the electric potential differencebetweenacloudandthegroundis10 times 109 V Determine the magnitude of the change in electric potential energy of an electron thatmovesbetweenthesepointsinelectron-volts

13 A charge of 15 μC is placed in a uniform electric fieldoftwooppositelychargedparallelplateswithamagnitudeof14times103 NC-1

(a) Determinetheworkthatmustbedoneagainstthefieldtomovethepointchargeadistance of 55 cm

(b) Calculatethepotentialdifferencebetweenthe final and initial positions of the charge

(c) Determinethepotentialdifferencebetweenthe plates if their separation distance is 15 cm

14 Duringaflashoflightningthepotentialdifferencebetweenacloudandthegroundwas12 times 109 V and the amount of transferred charge was32C

(a) Determinethechangeinenergyofthetransferred charge

(b) Iftheenergyreleasedwasallusedtoacceleratea1tonnecardeduceitsfinalspeed

(c) Iftheenergyreleasedcouldbeusedtomelticeat0degCdeducetheamountoficethatcould be melted

PHYSICS 2015indb 317 150514 338 PM

Chapter 10

318

AH

L

15 SupposethatwhenanelectronmovedfromAtoBin the diagram along an electric field line that the electric field does 36 times 10-19Jofworkonit

Determinethedifferencesinelectricpotential

(a) VB ndash VA(b) VC ndash VA(c) VC ndash VB

16 DeterminethepotentialatpointPthatislocatedatthecentreofthesquareasshowninthediagrambelow

- 6μC

+3μC +2μC

1m

1m

5μC

P

102 Fields at work

Essential idea Similar approaches can be taken inanalysingelectricalandgravitationalpotentialproblems

Understandingsbull Potentialandpotentialenergybull Potentialgradientbull Potentialdifferencebull Escapespeedbull Orbitalmotionorbitalspeedandorbitalenergybull Forcesandinverse-squarelawbehaviour

Orbital motion orbital speed and orbital energy

Although orbital motion may be circular elliptical orparabolic this sub-topic only dealswith circular orbitsThis sub-topic is not fundamentally new physics butan application that synthesizes ideas from gravitationcircularmotiondynamicsandenergy

TheMoonorbitstheEarthandinthissenseitisoftenreferredtoasasatelliteoftheEarthBefore1957itwastheonlyEarthsatelliteHoweverin1957theRussianslaunchedthefirstmanmadesatelliteSputnik1SincethisdatemanymoresatelliteshavebeenlaunchedandtherearenowliterallythousandsofthemorbitingtheEarthSomeareusedtomonitortheweathersome used to enable people to find accurately their position on thesurfaceoftheEarthmanyareusedincommunicationsandnodoubtsomeareusedtospyonothercountriesFigure1015showshowinprincipleasatellitecanbeputintoorbit

Theperson(whosesizeisgreatlyexaggeratedwithrespectto Earth) standing on the surface on the Earth throwssome stonesThe greater the speed with which a stoneis thrown the further it will land from her The pathsfollowedbythethrownstonesareparabolasByastretchof the imaginationwecanvisualisea situation inwhichastoneis thrownwithsuchaspeedthatbecauseof thecurvatureoftheEarthitwillnotlandonthesurfaceoftheEarthbutgointoldquoorbitrdquo(Path4onFigure1015)

NATURE OF SCIENCECommunication of scientific explanations The ability to apply field theory to the unobservable (charges) and the massively scaled (motion of satellites) required scientists to develop new ways to investigate analyse and report findings to a general public used to scientific discoveries based on tangible and discernible evidence (51)

copy IBO 2014

PHYSICS 2015indb 318 150514 338 PM

Fields

319

AH

L

1

2

34Earth

Figure 1015 Throwing a stone into orbit

TheforcethatcausesthestonestofollowaparabolicpathandtofalltoEarthisgravityandsimilarlytheforcethatkeeps the stone in orbit is gravity For circular motionto occur we have seen that a force must act at rightanglestothevelocityofanobjectthatistheremustbeacentripetalforceHenceinthesituationwedescribeherethe centripetal force for circular orbital motion about the EarthisprovidedbygravitationalattractionoftheEarth

We can calculate the speedwithwhich a stonemustbethrowninordertoputitintoorbitjustabovethesurfaceof the Earth

If the stone has mass m and speed v thenwehave fromNewtonrsquos2ndlaw

mv2

RE---------- G

MEm

RE2------------=

whereRE is the radius of the Earth and ME is the mass of the Earth

Bearinginmindthatg 0 GME

RE2--------= then

v g RE 10 64 106timestimes 8 103times= = =

Thatisthestonemustbethrownat8times103m sndash1

Clearly we are not going to get a satellite into orbit soclosetothesurfaceoftheEarthMovingatthisspeedthefrictionduetoairresistancewouldmeltthesatellitebeforeithadtravelledacoupleofkilometresInrealitythereforeasatelliteisputintoorbitabouttheEarthbysendingitattachedtoarocketbeyondtheEarthrsquosatmosphereandthengivingitacomponentofvelocityperpendiculartoaradialvectorfromtheEarthSeeFigure1016

Earth Satellite carried by rocket to hereSatellite orbit

Tangential component of velocity

Figure 1016 Getting a satellite into orbit

Keplerrsquos third law(This work of Kepler and Newtonrsquos synthesis of the work is an excellent example of the scientific method and makes for a good TOK discussion)

In 1627 Johannes Kepler (1571-1630)publishedhis lawsof planetary motion The laws are empirical in natureandwere deduced from the observations of theDanishastronomer Tycho de Brahe (1546-1601)The third lawgives a relationship between the radius of orbit R of a planet and its period TofrevolutionabouttheSunThelawisexpressedmathematicallyas

2

3 constantTR

=

We shall nowuseNewtonrsquos LawofGravitation to showhowitisthattheplanetsmoveinaccordancewithKeplerrsquosthirdlaw

InessenceNewtonwasabletousehislawofgravitytopredict the motion of the planets since all he had to do was factor theF givenby this law intohis second law F = ma to find their accelerations and hence their future positions

InFigure1017theEarthisshownorbitingtheSunandthedistancebetweentheircentresisR

RSun

Earth

F es

F se

Figure 1017 Planets move according to Keplerrsquos third law

Fes is the force that the Earth exerts on the Sun and Fse is the force that the Sun exerts on the Earth The forces are equalandoppositeandtheSunandtheEarthwillactuallyorbitaboutacommoncentreHoweversince theSun issoverymuchmoremassivethantheEarththiscommoncentrewillbeclosetothecentreoftheSunandsowecanregard the Earth as orbiting about the centre of the Sun Theotherthingthatweshallassumeisthatwecanignorethe forces that the other planets exert on the Earth (This wouldnotbeawisethingtodoifyouwereplanningtosend a space ship to theMoon for example)We shallalso assume that we have followed Newtonrsquos exampleandindeedprovedthataspherewillactasapointmasssituated at the centre of the sphere

Kepler had postulated that the orbits of the planets areellipticalbut since theeccentricityof theEarthrsquosorbit issmallweshallassumeacircularorbit

PHYSICS 2015indb 319 150514 338 PM

Chapter 10

320

AH

L

TheaccelerationoftheEarthtowardstheSunisa = Rω2

whereω ω 2πT------=

Hencea R 2π

T------ times

2 4π2 RT 2------------- = =

But the acceleration is given by Newtonrsquos Second Law F = mawhereFisnowgivenbytheLawofGravitationSo in this situation

F maGMs Me

R2-------------------= = butwealsohavethat

a a R 2πT

------ times

2 4π2 RT 2------------- = = and m = Me so that

G MsMe

R2------------------- Me4π2R

T2-------------times

GMs

4π2-----------rArr R3

T2------= =

Butthequantity

isaconstantthathasthesamevalueforeachoftheplanetssowehaveforalltheplanetsnotjustEarththat

R3

T2------ k =

wherekisaconstantWhichisofcourseKeplerrsquosthirdlaw

ThisisindeedanamazingbreakthroughItisdifficulttorefutethe idea that all particles attract each other in accordance withtheLawofGravitationwhenthelawisabletoaccountfortheobservedmotionoftheplanetsabouttheSun

The gravitational effects of the planets upon each othershould produce perturbations in their orbits Such is the predictivepoweroftheUniversalGravitationalLawthatit enabled physicists to compute these perturbations The telescopehadbeeninventedin1608andbythemiddleofthe 18th Century had reached a degree of perfection in design that enabled astronomers to actually measure the orbital perturbations of the planets Their measurements werealways inagreementwith thepredictionsmadebyNewtonrsquoslawHoweverin1781anewplanetUranuswasdiscovered and the orbit of this planet did not fit withthe orbit predicted by Universal Gravitation Such wasthe physicistrsquos faith in theNewtonianmethod that theysuspectedthatthediscrepancywasduetothepresenceofayetundetectedplanetUsingtheLawofGravitationtheFrenchastronomerJLeverrier and the English astronomer J C Adamswereable tocalculate justhowmassive thisnewplanetmustbeandalsowhereitshouldbeIn1846the planetNeptunewas discovered justwhere they hadpredicted Ina similarwaydiscrepancies in theorbitofNeptune ledto thepredictionandsubsequentdiscoveryin1930oftheplanetPlutoNewtonrsquosLawofGravitation

had passed the ultimate test of any theory it is not only abletoexplainexistingdatabutalsotomakepredictions

Satellite energyWhenasatelliteisinorbitaboutaplanetitwillhavebothkineticenergyandgravitationalpotentialenergySupposewe consider a satellite ofmass m that is in an orbit of radius r about a planet of mass M

Thegravitationalpotentialduetotheplanetatdistancer from its centre is

ThegravitationalpotentialenergyofthesatelliteVsat

is therefore GMe m

r----------------ndash

ThatisVsat GMem

r----------------ndash=

Thegravitationalfieldstrengthatthesurfaceoftheplanetisgivenby

g 0GMe

Re2------------=

Hencewecanwrite

Vsat g0 Re

2

rndash=

m

ThekineticenergyofthesatelliteKsatisequaltofrac12mv2wherev is its orbital speed

Byequatingthegravitationalforceactingonthesatellitetoitscentripetalaccelerationwehave

G Mem

r2---------------- mv2

r---------- mv2hArr

GMemr----------------= =

Fromwhich12---

mv2 12---

GMemr----------------times=

= 0 e

2g R

r

2m

Which is actually quite interesting since it shows thatirrespective of the orbital radius the KE is numericallyequal to half the PE Also the total energy E tot K sat Vsat+ 1

2---G Me m

r----------------times G Me m

r----------------ndash + 1

2---G Me m

r----------------ndash= = =of the satelliteisalwaysnegativesince

E tot K sat Vsat+ 12---

G Me mr----------------times

G Me mr----------------ndash

+ 12---

G Me mr----------------ndash= = =

PHYSICS 2015indb 320 150514 338 PM

Fields

321

AH

L

The energies of an orbiting satellite as a function of radial distancefromthecentreofaplanetareshownplottedinFigure1018

12

100806

0402

0ndash02

ndash04

ndash06ndash08ndash10

ndash12

2 4 6 8 10 12

ener

gya

rbita

ry u

nits

distance timesR

kinetic energytotal energy

potential energy

Figure 1018 Energy of an orbiting satellite as a function of distance from the centre of a planet

WeightlessnessSuppose that you are inanelevator(lift)thatisdescendingat constant speedandyou let goof abook thatyouareholdinginyourhandThebookwillfalltothefloorwithaccelerationequaltotheaccelerationduetogravityIfthecablethatsupportstheelevatorweretosnap(asituationthatItrustwillneverhappentoanyofyou)andyounowletgothebookthatyouareholdinginyourotherhandthis book will not fall to the floor - it will stay exactlyin linewithyourhandThis isbecause thebook isnowfallingwith the sameaccelerationas theelevatorandassuch the book cannot ldquocatchrdquo up with the floor of theelevatorFurthermoreifyouhappenedtobestandingonasetofbathroomscalesthescaleswouldnowreadzero-youwouldbeapparentlyweightlessItisthisideaoffreefallthatexplainstheapparentweightlessnessofastronautsin an orbiting satellite These astronauts are in free fall inthesensethattheyareacceleratingtowardsthecentre of the Earth

It is actually possible to define theweight of a body inseveral different waysWe can define it for example asthegravitationalforceexertedonthebodybyaspecifiedobject such as the EarthThiswe have seen thatwe doinlotsofsituationswherewedefinetheweightasbeingequaltomgIfweusethisdefinitionthenanobjectinfreefallcannotbydefinitionbeweightlesssinceitisstillinagravitationalfieldHoweverifwedefinetheweightofanobjectintermsofaldquoweighingrdquoprocesssuchasthereadingonasetofbathroomscaleswhichineffectmeasuresthecontact force between the object and the scales thenclearlyobjectsinfreefallareweightlessOnenowhasto

askthequestionwhetherornotitispossibleForexampletomeasurethegravitationalforceactingonanastronautin orbit about the Earth

WecanalsodefineweightintermsofthenetgravitationalforceactingonabodyduetoseveraldifferentobjectsForexample for an object out in space its weight could bedefined in terms of the resultant of the forces exerted on it bytheSuntheMoontheEarthandalltheotherplanetsin the Solar System If this resultant is zero at a particular pointthenthebodyisweightlessatthispoint

Inviewofthevariousdefinitionsofweightthatareavailabletousitisimportantthatwhenweusethewordldquoweightrdquoweareawareofthecontextinwhichitisbeingused

Example

Calculate the height above the surface of the Earth atwhichageo-stationarysatelliteorbits

Solution

A geo-stationary satellite is one that orbits the Earth in such a way that it is stationary with respect to a point on the surface of the Earth This means that its orbital period must be the same as the time for the Earth to spin once on its axis ie 24 hours

From Keplerrsquos third law we have G Ms

4π2----------- R3

T2------=

That is

RMe

mh

using the fact that the force of attraction between the satellite and the Earth is given by

FG Mem

R2----------------=

and that F = ma

where m is the mass of the satellite and a 4π2R

T2-------------=

we have

GMem

R2---------------- m 4π2R

T2-------------times

GMe

4π2------------rArr R3

T2------= =

PHYSICS 2015indb 321 150514 338 PM

Chapter 10

322

AH

L

Now the mass of the Earth is 60 times 1024 kg and the period T measured in seconds is given by T = 86400 s

So substitution gives R = 42 times 106 m

The radius of the Earth is 64 times 106 m so that the orbital height h is about 36 times 107 m

Example

Calculatetheminimumenergyrequiredtoputasatelliteofmass500kgintoanorbitthatisasaheightequaltotheEarthrsquosradiusabovethesurfaceoftheEarth

Solution

We have seen that when dealing with gravitational fields and potential it is useful to remember that

g 0G M

Re2---------= or g0 Re

2 GM=

The gravitational potential at the surface of the Earth is

g 0ndash Re GMRe

---------ndash=

The gravitational potential at a distance R

from the centre of the Earth is

=

The difference in potential between the surface of the Earth and a point distance R from the centre is therefore

V∆ g 0Re 1ReR------ndash

=

If R 2Re= then V∆g0 Re

2------------=

This means that the work required to ldquoliftrdquo the satellite into orbit is g0Rm where m is the mass of the satellite This is equal to

10 32 106 500timestimestimes = 16000 MJ

However the satellite must also have kinetic energy in order to orbit Earth This will be equal to

g 0 mRe2

2R-----------------g 0mRe

2

4----------------- 8000 MJ= =

The minimum energy required is therefore

24000 MJ

Exercise 1031 The speed needed to put a satellite in orbit does

not depend on

A the radius of the orbitB theshapeoftheorbitC thevalueofgattheorbitD the mass of the satellite

2 EstimatethespeedofanEarthsatellitewhoseorbitis400kmabovetheEarthrsquossurfaceAlsodetermine the period of the orbit

3 Calculatethespeedofa200kgsatelliteorbitingthe Earth at a height of 70 times 106 m

Assume that g = 82 m sndash2 for this orbit

4 TheradiioftwosatellitesXandYorbitingthe Earth are 2r and 8rwherer is the radius of the Earth Calculate the ratio of the periods of revolutionofXtoY

5 A satellite of mass mkgissentfromEarthrsquossurfaceinto an orbit of radius 5RwhereR is the radius of theEarthWritedownanexpressionfor

(a) thepotentialenergyofthesatelliteinorbit

(b) thekineticenergyofthesatelliteinorbit

(c) theminimumworkrequiredtosendthesatellitefromrestattheEarthrsquossurfaceintoits orbit

6 A satellite in an orbit of 10rfallsbacktoEarth(radius r)afteramalfunctionDeterminethespeedwithwhichitwillhittheEarthrsquossurface

7 The radius of the moon is frac14 that of the Earth AssumingEarthandtheMoontohavethesamedensitycomparetheaccelerationsoffreefallatthe surface of Earth to that at the surface of the Moon

8 UsethefollowingdatatodeterminethegravitationalfieldstrengthatthesurfaceoftheMoon and hence determine the escape speed from the surface of the Moon

Mass of the Moon = 73 times 1022kg

RadiusoftheMoon=17times106 m

PHYSICS 2015indb 322 150514 338 PM

Wave Phenomena

305

AH

L

We now consider the case where the source is stationary and the observerismoving towardsthesource with speed v0 In this situation the speed of the sound waves as measured by the observer will be v0 + v We therefore have that

v0 + v = ffraslλ = vfλ

times

From which

o1 vf fv

= +

Equation113

If the observerismovingawayfromthesource then

o1 vf fv

= minus

Fromequation(113)wehavethat

o o1 v vf f f f f fv v

∆ = minus = + minus =

Equation114

Thevelocitiesthatwerefertointheaboveequationsarethevelocities with respect to the medium in which the waves fromthesourcetravelHoweverwhenwearedealingwithalight source it is the relative velocity between the source and the observer that we must consider The reason for this is that lightisuniqueintherespectthatthespeedofthelightwavesdoes not depend on the speed of the source All observers irrespective of their speed or the speed of the source will measure the same velocity for the speed of light This is oneof thecornerstonesof theSpecialTheoryofRelativitywhichisdiscussedinmoredetailinOptionH(Chapter18)When applying the Doppler effect to light we aremainlyconcerned with the motion of the source We look here only at the situation where the speed of the source v is much smaller than the speed of light cinfreespace(v ltlt c)Underthesecircumstanceswhenthesourceismovingtowardstheobserverequation111becomes

vf f f fc

minus = ∆ = Equation115

andwhenthesourceismovingawayfromtheobserverequation112becomes vf f f f

cminus = ∆ = minus

Provided that v ltlt c these same equations apply for astationary source and moving observer

We look at the following example and exercise

Example

Asourceemitsasoundoffrequency440HzItmovesinastraight line towards a stationary observer with a speed of 30ms-1Theobserverhearsasoundoffrequency484HzCalculate the speed of sound in air

Solution

We use equation 111 and substitute f= 484 Hz f = 440 Hz

and vs = 30 m s-1

therefore 1

484 440 301v

=minus

such that 30 4401

484vminus = to

give v = 330 m s-1

Example

A particular radio signal from a galaxy is measured as havingafrequencyof139times109HzThesamesignalfromasourceinalaboratoryhasafrequencyof142times109Hz

Suggestwhy thegalaxy ismovingaway fromEarthandcalculateitsrecessionspeed(iethespeedwithwhichitismovingawayfromEarth)

Solution

The fact that the frequency from the moving source is less than that when it is stationary indicates that it is moving away from the stationary observer ie Earth

Using vf fc

∆ = we have

8 96

9

3 10 (142 139) 10 634 10142 10

c fvf∆ times times minus times

= = = timestimes

m s-1

It is usual when dealing with the Doppler effect of light to express speeds as a fraction of c So in this instance we have v = 0021 c

Using the Doppler effectWe have seen in the above example and exercise how the Dopplereffectmaybeusedtomeasuretherecessionspeedof distant galaxies The effect is also used to measure speed inothersituationsHerewewilllookatthegeneralprincipleinvolved in using the Doppler effect to measure speedFigure932showsasource(thetransmitter)thatemitseithersound or em waves of constant frequency f The waves

PHYSICS 2015indb 305 150514 337 PM

Chapter 9

306

AH

L

fromthesourceare incidentonareflectorthat ismovingtowards the transmitter with speed vThereflectedwavesare detected by the receiver placed alongside the transmitter

reector

v

transmitter

receiver

f f

f f

Figure 932 Using the Doppler effect to measure speed

We shall consider the situation where v ltlt c where c is the speed of the waves from the transmitter

For the reflector receiving waves from the transmitterit is effectively an observer moving towards a stationary sourceFromequation(114) itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation116

For the receiver receivingwaves from the reflector it iseffectively a stationary observer receiving waves from a movingsourceFromequation(115)itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation117

Ifweaddequations(116)and(117)wegetthatthetotalDopplershiftatthereceiver∆f is

v vf f f f fc c

minus = ∆ = +

But 1 vf fc

= +

hence

1 v v vf f fc c c

∆ = + +

But since v ltlt cwe can ignore the term2

2

vc when we

expandthebracketintheaboveequation

Therefore we have2vf fc

∆ = Equation118

If v asymp c then we must use the full Doppler equationsHowever for em radiationwewill alwaysonly considersituations in which v ltlt c

Example

Thespeedofsoundinbloodis1500times103 m s-1 Ultrasound offrequency100MHzisreflectedfrombloodflowinginan arteryThe frequencyof the reflectedwaves receivedbackatthetransmitteris105MHzEstimatethespeedofthebloodflowintheartery

Solution

Using equation (118) we have

6 63

2005 10 1015 10

vtimes = times

timesto give v asymp 36 m s-1 (We have assumed that the ultrasound is incident at right angles to the blood flow)

Exercise 971 Judy is standing on the platform of a station A

high speed train is approaching the station in a straight line at constant speed and is sounding its whistleAsthetrainpassesbyJudythefrequencyof the sound emitted by the whistle as heard by Judychangesfrom640Hzto430HzDetermine

(a) thespeedofthetrain

(b) thefrequencyofthesoundemittedbythewhistle as heard by a person on the train (Speedofsound=330ms-1)

2 AgalaxyismovingawayfromEarthwithaspeedof00500cThewavelengthofaparticularspectral line in light emitted by atomic hydrogen inalaboratoryis656times10-7 m Calculate the valueofthewavelengthofthislinemeasuredinthelaboratoryinlightemittedfromasourceofatomic hydrogen in the galaxy

PHYSICS 2015indb 306 150514 337 PM

10 Fields

Contents

101 ndash Describing fields

102 ndash Fields at work

Essential Ideas

Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Similar approaches can be taken in analysing electrical and gravitational potential problems copy IBO 2014

PHYSICS 2015indb 307 150514 337 PM

Chapter 10

308

AH

L

101 Describing fields

Essential idea Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Understandingsbull Gravitationalfieldsbull Electrostaticfieldsbull Electricpotentialandgravitationalpotentialbull Fieldlinesbull Equipotentialsurfaces

Gravitational potentialWehaveseenthatifweliftanobjectofmassm to a height h abovethesurfaceoftheEarththenitsgainingravitationalpotential energy is mghHoweverthisisbynomeansthefullstoryForastartwenowknowthatgvarieswithh and also the expression really gives a difference in potentialenergybetweenthevaluethattheobjecthasattheEarthrsquossurfaceandthevaluethatithasatheighthSowhatwereallyneedisazeropointCanwefindapointwherethepotentialenergyiszeroandusethispointfromwhichtomeasure changes in potential energy

The point that is chosen is in fact infinity At infinity the gravitationalfieldstrengthofanyobjectwillbezeroSoletusseeifwecandeduceanexpressionforthegaininpotentialenergyofanobjectwhenitisldquoliftedrdquofromthesurfaceoftheEarthtoinfinityThisineffectmeansfindingtheworknecessarytoperformthistask

δr

r r δr+

r infin=

mg

AB

Me

Figure 1001 Gravitational forces

Inthediagramweconsidertheworknecessarytomovethe particle of mass m a distance δr in thegravitationalfield of the Earth

The force on the particle at A is FG Mem

r2----------------=

IftheparticleismovedtoB then since δr isverysmallwe can assume that the field remains constant over thedistance ABTheworkδW done againstthegravitationalfieldoftheEarthinmovingthedistanceABis

δWGMem

r2----------------δrndash=

(rememberthatworkdoneagainstaforceisnegative)

TofindthetotalworkdoneW in going from the surface oftheEarthtoinfinitywehavetoaddalltheselittlebitsofworkThisisdonemathematicallybyusingintegralcalculus

WG Me m

r2----------------ndash

rdR

infin

int G Me m 1

r2----- rd

R

infin

intndash G Me m 1r---ndash

R

infinndash= = =

GMe m 0 1R---ndash

ndashndash=

G MemR----------------ndash=

HencewehavewhereRistheradiusoftheEarththattheworkdonebythegravitationalfieldinmovinganobjectofmass m from R(surfaceoftheEarth)toinfinityisgivenby

WGMe m

R---------------- ndash=

We can generalise the result by calculating the worknecessary per unit mass to take a small mass fromthe surface of the Earth to infinity This we call thegravitational potential Vie

Wewouldgetexactlythesameresultifwecalculatedtheworkdonebythefieldtobringthepointmassfrominfinityto the surface of Earth In this respect the formal definition ofgravitationalpotentialatapointinagravitationalfieldis therefore defined as the work done per unit mass in bringing a point mass from infinity to that point

Clearly then the gravitational potential at any point inthe Earthrsquos field distance r from the centre of the Earth (providingr gt R)is

VGMe

r------------ ndash=

Thepotentialisthereforeameasureoftheamountofworkthathastobedonetomoveparticlesbetweenpointsinagravitationalfieldanditsunit is theJkgndash1 We also note thatthepotentialisnegativesothatthepotentialenergy

NATURE OF SCIENCEParadigm shift The move from direct observable actions being responsible for influence on an object to acceptance of a fieldrsquos ldquoaction at a distancerdquo required a paradigm shift in the world of science (23)

copy IBO 2014

PHYSICS 2015indb 308 150514 337 PM

Fields

309

AH

L

aswemoveawayfromtheEarthrsquossurfaceincreasesuntilitreachesthevalueofzeroatinfinity

IfthegravitationalfieldisduetoapointmassofmassmthenwehavethesameexpressionasaboveexceptthatMe is replaced by m andmustalsoexclude thevalueof thepotential at the point mass itself ie at r = 0

WecanexpressthegravitationalpotentialduetotheEarth(orduetoanysphericalmass)intermsofthegravitationalfield strength at its surface

AtthesurfaceoftheEarthwehave

g 0ndash Re GMRe

---------ndash= e

Sothat

g 0Re2 GM= e

Hence at a distance r from the centre of the Earth the gravitationalpotentialVcanbewrittenas

VGMe

r------------ndashg 0 Re

2

r------------ndash= =

The potential at the surface of the Earth

(r = Re)istherefore-g0Re

It is interesting to see how the expression for thegravitational potential ties in with the expression mgh ThepotentialatthesurfaceoftheEarthis-g0Re (see the exampleabove)andataheighthwillbe-g0g 0 Re h+( )ndash ifweassume that g0doesnotchangeover thedistanceh The differenceinpotentialbetweenthesurfaceandtheheighth is therefore g0hSotheworkneededtoraiseanobjectof mass m to a height h is mgh ie m timesdifference ingravitationalpotential

This we have referred to as the gain in gravitationalpotentialenergy(see235)

However this expression can be extended to any twopoints inanygravitationalfieldsuch that ifanobjectofmass mmovesbetween twopointswhosepotentialsareV1 and V2 respectively then the change in gravitationalpotentialenergyoftheobjectism(V1 ndash V2)

Gravitational potential gradient

Let us consider now a region in space where thegravitationalfieldisconstantInFigure912thetwopointsAandBareseparatedbythedistance∆x

A B

direction of uniform gravitational eld of strength I

∆x

Figure 1002 The gravitational potential gradient

The gravitational field is of strength I and is in the direction shownThe gravitational potential at A is V andatBisV +∆V

Theworkdone is takingapointmassmfromAtoB isF∆x = mI∆x

Howeverbydefinitionthisworkisalsoequalto-m∆V

Therefore mI∆x=-m∆V

or VIx

∆= minus

∆Effectivelythissaysthatthemagnitudeofthegravitationalfield strength is equal to the negative gradient of thepotential If I is constant then V is a linear function of x and Iisequaltothenegativegradientofthestraightlinegraph formed by plotting V against x If I is not constant (asusuallythecase)thenthemagnitudeofI at any point in the field can be found by find the gradient of the V-x graph at that point An example of such a calculation can befoundinSection929

For those of you who do HL maths the relationship between field and potential is seen to follow from the expression for the potential of a point mass viz

mV Gr

= minus

2

ddV mG Ir r

minus = + =

Potential due to one or more point masses

Gravitationalpotentialisascalarquantitysocalculatingthepotential due to more than one point mass is a matter of simpleadditionSoforexamplethepotentialV due to the Moon and Earth and a distance x fromthecentreofEarthonastraightlinebetweenthemisgivenbytheexpression

E MM MV Gx r x

= minus + minus

whereME=massofEarthMM= mass of Moon and r = distancebetweencentreofEarthandMoon

PHYSICS 2015indb 309 150514 337 PM

Chapter 10

310

AH

L

Equipotentials and field linesIfthegravitationalpotentialhasthesamevalueatallpointson a surface the surface is said to be an equipotential surfaceSoforexampleifweimagineasphericalshellaboutEarthwhosecentrecoincideswiththecentreofEarththisshellwillbeanequipotentialsurfaceClearlyifwerepresentthegravitationalfieldstrengthbyfieldlinessincethelinesldquoradiaterdquooutfromthecentreofEarththentheselineswillbeatrightanglestothesurfaceIfthefieldlineswerenotnormaltotheequipotentialsurfacethentherewouldbeacomponentof the field parallel to the surfaceThis wouldmean thatpointsonthesurfacewouldbeatdifferentpotentialsandsoitwouldnolongerbeanequipotentialsurfaceThisofcourseholdstrueforanyequipotentialsurface

Figure 1003 shows the field lines and equipotentials fortwopointmassesm

m m

Figure 1003 Equipotentials for two point masses

It is worth noting that we would get exactly the samepattern ifwewere to replace thepointmasseswith twoequalpointcharges(See935)

Escape speedThe potential at the surface of Earth is which meansthattheenergyrequiredtotakeaparticleofmass mfromthesurfacetoinfinityisequalto

ButwhatdoesitactuallymeantotakesomethingtoinfinityWhentheparticleisonthesurfaceoftheEarthwecanthinkofitassittingatthebottomofaldquopotentialwellrdquoasinfigure1004

particle

surface of Earth

innity

GMR---------

FIgure 1004 A potential well

Theldquodepthrdquoofthewellis and if the particle gains an amount of kinetic energy equal to wherem is its massthenitwillhavejustenoughenergytoldquoliftrdquoitoutofthewell

In reality it doesnrsquot actually go to infinity it justmeansthat the particle is effectively free of the gravitationalattractionof theEarthWe say that ithas ldquoescapedrdquo theEarthrsquosgravitationalpullWemeetthisideainconnectionwithmolecular forcesTwomolecules in a solidwill sitat their equilibrium position the separation where therepulsiveforceisequaltotheattractiveforceIfwesupplyjust enough energy to increase the separation of themolecules such that they are an infinite distance apart then intermolecularforcesnolongeraffectthemoleculesandthesolidwillhavebecomealiquidThereisnoincreaseinthekineticenergyofthemoleculesandsothesolidmeltsat constant temperature

Wecancalculatetheescapespeedofanobjectveryeasilyby equating the kinetic energy to the potential energysuch that

12---mvesc ape

2 GMe mRe

----------------=

vesc aperArr2GMe

Re--------------- 2g0 Re = =

Substituting for g0 and Regivesavalueforvescape of about 11kmsndash1 from the surface of the Earth

Youwillnotethattheescapespeeddoesnotdependonthemassoftheobjectsincebothkineticenergyandpotentialenergy are proportional to the mass

IntheoryifyouwanttogetarockettothemoonitcanbedonewithoutreachingtheescapespeedHoweverthiswouldnecessitate anenormousamountof fuel and it islikely that therocketplus fuelwouldbesoheavythat itwouldnevergetoffthegroundItismuchmorepracticalto accelerate the rocket to the escape speed fromEarthorbitandthenintheoryjustlaunchittotheMoon

Example

Use the followingdata todetermine thepotential at thesurface of Mars and the magnitude of the acceleration of free fall

massofMars =64times1023kg

radiusofMars =34times106 m

Determinealsothegravitationalfieldstrengthatadistanceof 68 times 106mabovethesurfaceofMars

PHYSICS 2015indb 310 150514 337 PM

Fields

311

AH

L

Solution

= minus = minus times times = times23

11 76

64 1067 10 ndash13 10

34 10MV GR

minus timestimes

Nkg-1

But V = -g0R

Therefore 7

0 6

13 10 3834 10

VgR

times= minus = =

timesm s-2

To determine the field strength gh at 68 times 106 m above the surface we use the fact that 0 2

Mg GR

= such that GM = g0R2

Therefore 2 2

0h 2 2 2

h h

38 (34) 042(102)

g RGMgR R

times= = = = m s-2

(the distance from the centre is 34 times 106 + 68 times 106 = 102 times 106 m)

Exercise 1011 Thegraphbelowshowshowthegravitational

potentialoutsideoftheEarthvarieswithdistancefrom the centre

0 10 20 30 40 50 60 70

ndash1

ndash2

ndash3

ndash4

ndash5

ndash6

ndash7

r m times 106

V

Jkgndash1

times 10

7

(a) Usethegraphtodeterminethegainingravitationalpotentialenergyofasatelliteofmass200kgasitmovesfromthesurfaceof the Earth to a height of 30 times 107mabovetheEarthrsquossurface

(b) Calculatetheenergyrequiredtotakeit to infinity

(c) DeterminetheslopeofthegraphatthesurfaceoftheEarthmCommenton youranswer

Electric potential energy

The concept of electric potential energy was developedwith gravitational potential energy in mind Just as anobjectnearthesurfaceoftheEarthhaspotentialenergybecause of its gravitational interaction with the Earthsotoothereiselectricalpotentialenergyassociatedwithinteracting charges

Letusfirstlookatacaseoftwopositivepointchargeseachof 1μC that are initially bound together by a thread in a vacuuminspacewithadistancebetweenthemof10cmasshowninFigure1005Whenthethreadiscutthepointchargesinitiallyatrestwouldmoveinoppositedirectionsmovingwithvelocitiesv1 and v2 along the direction of the electrostatic force of repulsion

BEFORE AFTER

v v21

Figure 1005 Interaction of two positive particles

The electric potential energybetweentwopointchargescan be found by simply adding up the energy associated witheachpairofpointchargesForapairofinteractingchargestheelectricpotentialenergyisgivenby

∆U = ∆Ep + ∆Ek = ∆W = ∆Fr = kqQ

____ r2 times r = kqQ

____ r

Becausenoexternalforceisactingonthesystemtheenergyand momentum must be conserved Initially Ek = 0 and Ep =kqQr=9times109 times 1 times 10-1201m=009JWhentheyareagreatdistance fromeachotherEpwillbenegligibleThefinalenergywillbeequaltofrac12mv1

2+frac12mv22=009J

Momentum isalsoconservedand thevelocitieswouldbethe same magnitude but in opposite directions

Electric potential energyismoreoftendefinedintermsofapointchargemovinginanelectricfieldas

lsquothe electricpotential energybetweenany twopoints inan electric field is defined as negativeoftheworkdone by anelectricfieldinmovingapoint electric chargebetweentwolocationsintheelectricfieldrsquo

ΔU = ΔEp = -ΔW = -Fd = qEx

ΔU = ΔEp +ΔEk = ΔW = ΔFr = kqQ r2 x r = kqQ

wherexisthedistancemovedalong(oroppositeto)thedirection of the electric field

Electric potential energyismeasuredinjoule(J)Justasworkisascalarquantitysotooelectricalpotentialenergyis a scalar quantity The negative of the work done by an electric field in moving a unit electric charge between two points is independent of the path taken Inphysicswesaytheelectricfieldisalsquoconservativersquo field

PHYSICS 2015indb 311 150514 337 PM

Chapter 10

312

AH

L

Supposeanexternalforcesuchasyourhandmovesasmallpositive point test charge in the direction of a uniformelectric fieldAs it ismoving itmust be gaining kineticenergyIfthisoccursthentheelectricpotentialenergyofthe unit charge is changing

InFigure1006apointcharge+qismovedbetweenpointsAandBthroughadistancex in a uniform electric field

B

A

x

+ q

Figure 1006 Movement of a positive point charge in a uniform field

InordertomoveapositivepointchargefrompointAtopointBanexternal forcemustbeapplied to thechargeequaltoqE (F = qE)

Since the force is applied through a distance xthennegativeworkhas to bedone tomove the charge because energyis gainedmeaning there is an increaseelectric potential energybetweenthetwopointsRememberthattheworkdoneisequivalenttotheenergygainedorlostinmovingthe charge through the electric field The concept of electric potential energy is only meaningful as the electric field whichgeneratestheforceinquestionisconservative

W F xtimes E q x times= =

θ x cos θ x

Figure 1007 Charge moved at an angle to the field

If a chargemoves at an angle θ to an electric field thecomponent of the displacement parallel to the electric fieldisusedasshowninFigure918

W F x Eq x θcostimes= =

Theelectricpotentialenergyisstoredintheelectricfieldand theelectricfieldwill return theenergy to thepointcharge when required so as not to violate the Law ofconservationofenergy

Electric potential

The electric potential at a point in an electric field is defined as being the work done per unit charge in bringing a small positive point charge from infinity to that point

ΔV = Vinfin ndash Vf = ndash W ___ q

Ifwedesignatethepotentialenergytobezeroatinfinitythen it follows thatelectricpotentialmustalsobezeroat infinity and the electric potential at any point in an electricfieldwillbe

ΔV = ndash W ___ q

NowsupposeweapplyanexternalforcetoasmallpositivetestchargeasitismovedtowardsanisolatedpositivechargeTheexternalforcemustdoworkonthepositivetestchargetomoveittowardstheisolatedpositivechargeandtheworkmustbepositivewhiletheworkdonebytheelectricfieldmustthereforebenegativeSotheelectricpotentialatthatpointmustbepositiveaccordingtotheaboveequationIfanegativeisolatedchargeisusedtheelectricpotentialatapointonthepositivetestchargewouldbenegativePositivepoint chargesof their ownaccordmove fromaplaceofhighelectricpotential toaplaceof lowelectricpotentialNegative point charges move the other way from lowpotentialtohighpotentialInmovingfrompointAtopointBinthediagramthepositivecharge+qismovingfromalowelectricpotentialtoahighelectricpotential

In thedefinitiongiven the termldquoworkperunitchargerdquohas significance If the test charge is +16 times 10-19Cwherethe charge has a potential energy of 32 times 10-17Jthenthepotentialwouldbe32times10-17J+16times10-19C=200JC-1 Now if the charge was doubled the potential wouldbecome 64times10-17 J However the potential per unitchargewouldbethesame

Electric potential is a scalar quantityandithasunitsJC-1 orvoltswhere1voltequalsone joulepercoloumbThevoltallowsustoadoptaunitfortheelectricfieldintermsofthevolt

PreviouslytheunitfortheelectricfieldwasNCndash1

W = qV and F = qEsoW ___ V = F __ E

E = FV ___ W = FV ___ Fm V mndash1

That is the units of the electric field E can also beexpressed as V mndash1

The work done per unit charge in moving a point charge between two points in an electric field is again independent of the path taken

PHYSICS 2015indb 312 150514 337 PM

Fields

313

AH

L

Electric potential due to a point charge

Letus takeapoint rmetres froma chargedobjectThepotentialatthispointcanbecalculatedusingthefollowing

W = ndashFr = ndashqV and F = ndash q1 q2 _____ 4πε0r

2

Therefore

W = ndash q1 q2 _____ 4πε0r

2 times r = ndash q1 q2 _____ 4πε0r

= ndashq1 times q2 _____ 4πε0r

= ndashq1V

That is V q

4πε0r--------------- =

Orsimply

V kqr------

=

Example

Determinehowmuchworkisdonebytheelectricfieldofpointcharge150μCwhenachargeof200μCismovedfrom infinity to apoint 0400m from thepoint charge(Assumenoaccelerationofthecharges)

Solution

The work done by the electric field is W = -qV = -14πε0 times q times (Q rinfin - Q r0400)

W = (- 200 times 10-6 C times 900 times 109 NmC-2 times 150 times 10-6 C) divide 0400 m = - 0675 J

An external force would have to do +0675 J of work

Electric field strength and electric potential gradient

Letus lookbackatFigure1006Supposeagain that thecharge+qismovedasmalldistancebyaforceFfromAtoBsothattheforcecanbeconsideredconstantTheworkdoneisgivenby

∆W F x∆times=

TheforceFandtheelectricfieldEareoppositelydirectedandweknowthat

F = -qE and ΔW = q ΔV

Thereforetheworkdonecanbegivenas

q ΔV = -q E Δx

Therefore E V∆x∆-------

ndash=

The rate of change of potential ΔV at a point with respect to distance Δx in the direction in which the change is maximum is called the potential gradient We say that the electric field = - the potential gradient and the units are Vm-

1 From the equation we can see that in a graph of electric potential versus distance the gradient of the straight line equals the electric field strength

In reality if a charged particle enters a uniform electric field it will be accelerated uniformly by the field and its kinetic energy will increase This is why we had to assume no acceleration in the last worked example

Ek∆ 12---

mv2 q E xsdot sdot q Vx---

xsdot sdot q Vsdot= = = =

Example

Determine how far apart two parallel plates must besituated so that a potential difference of 150 times 102 V produces an electric field strength of 100 times 103 NC-1

Solution

Using E V∆x------- xhArrndash V∆

E-------15 102 Vtimes

100 103 N C 1ndashtimes------------------------------------------= = =

= 150 times 10-1

The plates are 150 times 10-1 m apart

The electric field and the electric potential at a point due to an evenly distributed charge +q on a sphere can be represented graphically as in Figure 1008

r

r

V

E

r0

r0

r0

Charge of +Q evenly distributed over surface

E 14πε0------------ Q

r2----- r r0gt=

V 14πε0------------Q

r----r r0gt=

Figure 1008 Electric field and potential due to a charged sphere

PHYSICS 2015indb 313 150514 337 PM

Chapter 10

314

AH

L

When the sphere becomes charged we know that thechargedistributes itselfevenlyoverthesurfaceThereforeeverypartofthematerialoftheconductorisatthesamepotential As the electric potential at a point is defined as beingnumericallyequaltotheworkdoneinbringingaunitpositivechargefrominfinitytothatpointithasaconstantvalueineverypartofthematerialoftheconductor

Since the potential is the same at all points on the conducting surfacethen∆V∆xiszeroButE=ndash∆V∆xThereforethe electric field inside the conductor is zero There is no electric field inside the conductor

SomefurtherobservationsofthegraphsinFigure1009are

bull OutsidethespherethegraphsobeytherelationshipsgivenasE α 1 r2 and V α 1 r

bull Atthesurfacer = r0Thereforetheelectricfieldandpotentialhavetheminimumvalueforr at this point and this infers a maximum field and potential

bull Insidethespheretheelectricfieldiszerobull Insidethespherenoworkisdonetomoveacharge

fromapointinsidetothesurfaceThereforethereisno potential difference and the potential is the same asitiswhenr = r0

Similargraphscanbedrawnfortheelectricfieldintensityandthe electric potential as a function of distance from conducting parallelplatesandsurfacesandthesearegiveninFigure1009

Potential plot E eld plotE eld

++++

ndashndashndashndash

xx x

+

x x

ndashx x

+ ndashx

x

+ +

x

x

Figure 1009 Electric field and electric potential at a distance from a charged surface

Potential due to one or more point charges

The potential due to one point charge can be determined byusingtheequationformula

V = kq r

Example 1

Determine the electric potential at a point 20 times 10-1 m fromthecentreofan isolatedconducting spherewithapointchargeof40pCinair

Solution

Using the formula

V = kq r we have

V90 10 9times( ) 40 10 12ndashtimes( )times

20 10 1ndashtimes( )------------------------------------------------------------------ 018 V= =

the potential at the point is 180 times 10-1 V

The potential due to a number of point charges can be determined by adding up the potentials due to individual point charges because the electric potential at any point outside a conducting sphere will be the same as if all the charge was concentrated at its centre

Example 2

Threepointchargesofareplacedattheverticesofaright-angledtriangleasshowninthediagrambelowDeterminetheabsolutepotentialat the+20μCchargeduetothetwoothercharges

- 6μC

+3μC +2μC

4 m

3 m

Solution

The electric potential of the +2 μC charge due to the ndash 6 μC charge is

V = (9 times 109 Nm2C-2 times -6 times 10-6 C) divide (radic 32 + 42) m = - 108 times 104 V

PHYSICS 2015indb 314 150514 338 PM

Fields

315

AH

L

Insummarywecanconcludethat

bull No work is done to move a charge along an equipotential

bull Equipotentials are always perpendicular to the electric lines of force

Figure1011and1012showsomeequipotential lines fortwo oppositely charged and identically positive spheresseparated by a distance

+ve ndashve

equipotential lines

Figure 1011 Equipotential lines between two opposite charges

equipotential lines

Figure 1012 Equipotential lines between two charges which are the same

+

ndash ndash ndash ndash

+ + +

40 V

30 V

20 V

10 V

50 V

Figure 1013 Equipotential lines between charged parallel plates

Figure1013showstheequipotentiallinesbetweenchargedparallel plates Throughout this chapter the similarities and differences between gravitational fields and electric fieldshavebeendiscussedTherelationshipsthatexistsbetweengravitationalandelectricquantitiesandtheeffectsofpointmassesandchargesissummarisedinTable1014

The electric potential of the +2 μC charge due to the +3 μC charge is

V = (9 times 109 Nm2C-2 times 3 times 10-6 C) divide 3m = 9 times 103 V

The net absolute potential is the sum of the 2 potentials- 108 times 104 V + 9 times 103 V =- 18 times 103 V

The absolute potential at the point is - 18 times 103 V

Equipotential surfacesRegions in space where the electric potential of a chargedistributionhasaconstantvaluearecalledequipotentials The placeswherethepotentialisconstantinthreedimensionsarecalled equipotential surfacesandwheretheyareconstantintwodimensionstheyarecalledequipotential lines

TheyareinsomewaysanalogoustothecontourlinesontopographicmapsInthiscasethegravitationalpotentialenergy is constant as amassmoves around the contourlines because the mass remains at the same elevationabovetheEarthrsquossurfaceThegravitationalfieldstrengthacts in a direction perpendicular to a contour line

Similarlybecausetheelectricpotentialonanequipotentiallinehas thesamevalueanelectric forcecandonoworkwhena testchargemovesonanequipotentialThereforethe electric field cannot have a component along anequipotentialandthusitmustbeeverywhereperpendiculartotheequipotentialsurfaceorequipotentiallineThisfactmakesiteasytoplotequipotentialsifthelinesofforceorlinesofelectricfluxofanelectricfieldareknown

Forexamplethereareaseriesofequipotentiallinesbetweentwoparallelplateconductorsthatareperpendiculartotheelectric fieldTherewill be a series of concentric circles(eachcirclefurtherapartthanthepreviousone)thatmapouttheequipotentialsaroundanisolatedpositivesphereThe lines of force and some equipotential lines for anisolatedpositivesphereareshowninFigure1010

Lines ofequipotential

Figure 1010 Equipotentials around an isolated positive sphere

PHYSICS 2015indb 315 150514 338 PM

Chapter 10

316

AH

L

Gravitational quantity Electrical quantityQ

uant

ities

g V∆x∆-------ndash= E V∆

x∆-------ndash=

Poin

t mas

ses a

nd ch

arge

s

V Gmr----

ndash= V 14πε0------------q

r---=

g G m

r2-----ndash= E 14πε0------------ q

r2-----=

F Gm1 m2

r2--------------= F 1

4πε0------------

q1q2

r2-----------=

Figure 1014 Formulas

Example

Deduce the electric potential on the surface of a gold nucleus that has a radius of 62 fm

Solution

Using the formula

V = kq r and knowing the atomic number of gold is 79 We will assume the nucleus is spherical and it behaves as if it were a point charge at its centre (relative to outside points)

V = 90 times 109 Nm2C-2 times 79 times 16 times 10-19 C divide 62 times 10-15 m = 18 times 107 V

The potential at the point is 18 MV

Example

Deduce the ionisation energy in electron-volts of theelectron in the hydrogen atom if the electron is in its ground state and it is in a circular orbit at a distance of 53 times 10-11 m from the proton

Solution

This problem is an energy coulombic circular motion question based on Bohrrsquos model of the atom (not the accepted quantum mechanics model) The ionisation energy is the energy required to remove the electron from the ground state to infinity The electron travels in a circular orbit and therefore has a centripetal acceleration The ionisation energy will counteract the coulombic force and the movement of the electron will be in the opposite direction to the centripetal force

Total energy = Ek electron + Ep due to the proton-electron interaction

ΣF = kqQ r2 = mv2 r and as such mv2 = = kqQ r

Therefore Ek electron = frac12 kqQ r

Ep due to the proton-electron interaction = - kqQ r

Total energy = frac12 kqQ r + - kqQ r = -frac12 kqQ r

= - 90 times 109 Nm2C-2 times (16 times 10-19 C)2 divide 53 times 10-11 m = -217 times 10-18 J

= -217 times 10-18 J divide 16 times 10-19 = -136 eV

The ionisation energy is 136 eV

Exercise 1021 ApointchargePisplacedmidwaybetweentwo

identicalnegativechargesWhichoneofthefollowingiscorrectwithregardstoelectricfieldandelectricpotentialatpointP

Electric field Electric potentialA non-zero zeroB zero non-zeroC non-zero non-zeroD zero zero

2 Twopositivechargedspheresaretiedtogetherinavacuumsomewhereinspacewheretherearenoexternal forces A has a mass of 25 g and a charge of20μCandBhasamassof15gandachargeof30μCThedistancebetweenthemis40cmTheyarethenreleasedasshowninthediagram

BEFORE AFTER

v1 v2

A B

(a) Determinetheirinitialelectricpotentialenergy in the before situation

(b) DeterminethespeedofsphereBafterrelease

3 Thediagrambelowrepresentstwoequipotentiallines in separated by a distance of 5 cm in a uniform electric field

+ + + + + + + +

ndash ndash ndash ndash ndash ndash ndash ndash

40 V

20 V 5 cm

PHYSICS 2015indb 316 150514 338 PM

Fields

317

AH

L

Determine the strength of the electric field

4 Thisquestionisabouttheelectricfieldduetoacharged sphere and the motion of electrons in that fieldThediagrambelowshowsanisolatedmetalsphereinavacuumthatcarriesanegativeelectriccharge of 60 μC

ndash

(a) Onthediagramdrawtheconventionalwayto represent the electric field pattern due to the charged sphere and lines to represent threeequipotentialsurfacesintheregionoutside the sphere

(b) Explainhowthelinesrepresentingtheequipotentialsurfacesthatyouhavesketchedindicatethatthestrengthoftheelectricfieldisdecreasingwithdistancefrom the centre of the sphere

(c) Theelectricfieldstrengthatthesurfaceofthe sphere and at points outside the sphere can be determined by assuming that the sphere acts as a point charge of magnitude 60 μC at its centre The radius of the sphere is 25 times 10ndash2 m Deduce that the magnitude of the field strength at the surface of the sphere is 86 times 107 Vmndash1

An electron is initially at rest on the surface of the sphere

(d) (i) Describethepathfollowedbytheelectronasitleavesthesurfaceofthesphere

(ii) Calculatetheinitialaccelerationofthe electron

5 Determinetheamountofworkthatisdoneinmovingachargeof100nCthroughapotentialdifferenceof150times102 V

6 Three identical 20 μC conducting spheres are placedatthecornersofanequilateraltriangleofsides 25 cm The triangle has one apex C pointing upthepageand2baseanglesAandBDeterminetheabsolutepotentialatB

7 Determinehowfaraparttwoparallelplatesmustbesituatedsothatapotentialdifferenceof250 times 102 V produces an electric field strength of 200 times 103 NC-1

8 Thegapbetweentwoparallelplatesis10times10-3mandthereisapotentialdifferenceof10times104 V betweentheplatesCalculate

i theworkdonebyanelectroninmovingfrom one plate to the other

ii thespeedwithwhichtheelectronreachesthe second plate if released from rest

iii theelectricfieldintensitybetweentheplates

9 Anelectronguninapicturetubeisacceleratedby a potential 25 times 103VDeterminethekineticenergygainedbytheelectroninelectron-volts

10 Determine the electric potential 20 times 10-2 m from achargeof-10times10-5 C

11 Determinetheelectricpotentialatapointmid-waybetweenachargeofndash20pCandanotherof+5pConthelinejoiningtheircentresifthecharges are 10 cm apart

12 During a thunderstorm the electric potential differencebetweenacloudandthegroundis10 times 109 V Determine the magnitude of the change in electric potential energy of an electron thatmovesbetweenthesepointsinelectron-volts

13 A charge of 15 μC is placed in a uniform electric fieldoftwooppositelychargedparallelplateswithamagnitudeof14times103 NC-1

(a) Determinetheworkthatmustbedoneagainstthefieldtomovethepointchargeadistance of 55 cm

(b) Calculatethepotentialdifferencebetweenthe final and initial positions of the charge

(c) Determinethepotentialdifferencebetweenthe plates if their separation distance is 15 cm

14 Duringaflashoflightningthepotentialdifferencebetweenacloudandthegroundwas12 times 109 V and the amount of transferred charge was32C

(a) Determinethechangeinenergyofthetransferred charge

(b) Iftheenergyreleasedwasallusedtoacceleratea1tonnecardeduceitsfinalspeed

(c) Iftheenergyreleasedcouldbeusedtomelticeat0degCdeducetheamountoficethatcould be melted

PHYSICS 2015indb 317 150514 338 PM

Chapter 10

318

AH

L

15 SupposethatwhenanelectronmovedfromAtoBin the diagram along an electric field line that the electric field does 36 times 10-19Jofworkonit

Determinethedifferencesinelectricpotential

(a) VB ndash VA(b) VC ndash VA(c) VC ndash VB

16 DeterminethepotentialatpointPthatislocatedatthecentreofthesquareasshowninthediagrambelow

- 6μC

+3μC +2μC

1m

1m

5μC

P

102 Fields at work

Essential idea Similar approaches can be taken inanalysingelectricalandgravitationalpotentialproblems

Understandingsbull Potentialandpotentialenergybull Potentialgradientbull Potentialdifferencebull Escapespeedbull Orbitalmotionorbitalspeedandorbitalenergybull Forcesandinverse-squarelawbehaviour

Orbital motion orbital speed and orbital energy

Although orbital motion may be circular elliptical orparabolic this sub-topic only dealswith circular orbitsThis sub-topic is not fundamentally new physics butan application that synthesizes ideas from gravitationcircularmotiondynamicsandenergy

TheMoonorbitstheEarthandinthissenseitisoftenreferredtoasasatelliteoftheEarthBefore1957itwastheonlyEarthsatelliteHoweverin1957theRussianslaunchedthefirstmanmadesatelliteSputnik1SincethisdatemanymoresatelliteshavebeenlaunchedandtherearenowliterallythousandsofthemorbitingtheEarthSomeareusedtomonitortheweathersome used to enable people to find accurately their position on thesurfaceoftheEarthmanyareusedincommunicationsandnodoubtsomeareusedtospyonothercountriesFigure1015showshowinprincipleasatellitecanbeputintoorbit

Theperson(whosesizeisgreatlyexaggeratedwithrespectto Earth) standing on the surface on the Earth throwssome stonesThe greater the speed with which a stoneis thrown the further it will land from her The pathsfollowedbythethrownstonesareparabolasByastretchof the imaginationwecanvisualisea situation inwhichastoneis thrownwithsuchaspeedthatbecauseof thecurvatureoftheEarthitwillnotlandonthesurfaceoftheEarthbutgointoldquoorbitrdquo(Path4onFigure1015)

NATURE OF SCIENCECommunication of scientific explanations The ability to apply field theory to the unobservable (charges) and the massively scaled (motion of satellites) required scientists to develop new ways to investigate analyse and report findings to a general public used to scientific discoveries based on tangible and discernible evidence (51)

copy IBO 2014

PHYSICS 2015indb 318 150514 338 PM

Fields

319

AH

L

1

2

34Earth

Figure 1015 Throwing a stone into orbit

TheforcethatcausesthestonestofollowaparabolicpathandtofalltoEarthisgravityandsimilarlytheforcethatkeeps the stone in orbit is gravity For circular motionto occur we have seen that a force must act at rightanglestothevelocityofanobjectthatistheremustbeacentripetalforceHenceinthesituationwedescribeherethe centripetal force for circular orbital motion about the EarthisprovidedbygravitationalattractionoftheEarth

We can calculate the speedwithwhich a stonemustbethrowninordertoputitintoorbitjustabovethesurfaceof the Earth

If the stone has mass m and speed v thenwehave fromNewtonrsquos2ndlaw

mv2

RE---------- G

MEm

RE2------------=

whereRE is the radius of the Earth and ME is the mass of the Earth

Bearinginmindthatg 0 GME

RE2--------= then

v g RE 10 64 106timestimes 8 103times= = =

Thatisthestonemustbethrownat8times103m sndash1

Clearly we are not going to get a satellite into orbit soclosetothesurfaceoftheEarthMovingatthisspeedthefrictionduetoairresistancewouldmeltthesatellitebeforeithadtravelledacoupleofkilometresInrealitythereforeasatelliteisputintoorbitabouttheEarthbysendingitattachedtoarocketbeyondtheEarthrsquosatmosphereandthengivingitacomponentofvelocityperpendiculartoaradialvectorfromtheEarthSeeFigure1016

Earth Satellite carried by rocket to hereSatellite orbit

Tangential component of velocity

Figure 1016 Getting a satellite into orbit

Keplerrsquos third law(This work of Kepler and Newtonrsquos synthesis of the work is an excellent example of the scientific method and makes for a good TOK discussion)

In 1627 Johannes Kepler (1571-1630)publishedhis lawsof planetary motion The laws are empirical in natureandwere deduced from the observations of theDanishastronomer Tycho de Brahe (1546-1601)The third lawgives a relationship between the radius of orbit R of a planet and its period TofrevolutionabouttheSunThelawisexpressedmathematicallyas

2

3 constantTR

=

We shall nowuseNewtonrsquos LawofGravitation to showhowitisthattheplanetsmoveinaccordancewithKeplerrsquosthirdlaw

InessenceNewtonwasabletousehislawofgravitytopredict the motion of the planets since all he had to do was factor theF givenby this law intohis second law F = ma to find their accelerations and hence their future positions

InFigure1017theEarthisshownorbitingtheSunandthedistancebetweentheircentresisR

RSun

Earth

F es

F se

Figure 1017 Planets move according to Keplerrsquos third law

Fes is the force that the Earth exerts on the Sun and Fse is the force that the Sun exerts on the Earth The forces are equalandoppositeandtheSunandtheEarthwillactuallyorbitaboutacommoncentreHoweversince theSun issoverymuchmoremassivethantheEarththiscommoncentrewillbeclosetothecentreoftheSunandsowecanregard the Earth as orbiting about the centre of the Sun Theotherthingthatweshallassumeisthatwecanignorethe forces that the other planets exert on the Earth (This wouldnotbeawisethingtodoifyouwereplanningtosend a space ship to theMoon for example)We shallalso assume that we have followed Newtonrsquos exampleandindeedprovedthataspherewillactasapointmasssituated at the centre of the sphere

Kepler had postulated that the orbits of the planets areellipticalbut since theeccentricityof theEarthrsquosorbit issmallweshallassumeacircularorbit

PHYSICS 2015indb 319 150514 338 PM

Chapter 10

320

AH

L

TheaccelerationoftheEarthtowardstheSunisa = Rω2

whereω ω 2πT------=

Hencea R 2π

T------ times

2 4π2 RT 2------------- = =

But the acceleration is given by Newtonrsquos Second Law F = mawhereFisnowgivenbytheLawofGravitationSo in this situation

F maGMs Me

R2-------------------= = butwealsohavethat

a a R 2πT

------ times

2 4π2 RT 2------------- = = and m = Me so that

G MsMe

R2------------------- Me4π2R

T2-------------times

GMs

4π2-----------rArr R3

T2------= =

Butthequantity

isaconstantthathasthesamevalueforeachoftheplanetssowehaveforalltheplanetsnotjustEarththat

R3

T2------ k =

wherekisaconstantWhichisofcourseKeplerrsquosthirdlaw

ThisisindeedanamazingbreakthroughItisdifficulttorefutethe idea that all particles attract each other in accordance withtheLawofGravitationwhenthelawisabletoaccountfortheobservedmotionoftheplanetsabouttheSun

The gravitational effects of the planets upon each othershould produce perturbations in their orbits Such is the predictivepoweroftheUniversalGravitationalLawthatit enabled physicists to compute these perturbations The telescopehadbeeninventedin1608andbythemiddleofthe 18th Century had reached a degree of perfection in design that enabled astronomers to actually measure the orbital perturbations of the planets Their measurements werealways inagreementwith thepredictionsmadebyNewtonrsquoslawHoweverin1781anewplanetUranuswasdiscovered and the orbit of this planet did not fit withthe orbit predicted by Universal Gravitation Such wasthe physicistrsquos faith in theNewtonianmethod that theysuspectedthatthediscrepancywasduetothepresenceofayetundetectedplanetUsingtheLawofGravitationtheFrenchastronomerJLeverrier and the English astronomer J C Adamswereable tocalculate justhowmassive thisnewplanetmustbeandalsowhereitshouldbeIn1846the planetNeptunewas discovered justwhere they hadpredicted Ina similarwaydiscrepancies in theorbitofNeptune ledto thepredictionandsubsequentdiscoveryin1930oftheplanetPlutoNewtonrsquosLawofGravitation

had passed the ultimate test of any theory it is not only abletoexplainexistingdatabutalsotomakepredictions

Satellite energyWhenasatelliteisinorbitaboutaplanetitwillhavebothkineticenergyandgravitationalpotentialenergySupposewe consider a satellite ofmass m that is in an orbit of radius r about a planet of mass M

Thegravitationalpotentialduetotheplanetatdistancer from its centre is

ThegravitationalpotentialenergyofthesatelliteVsat

is therefore GMe m

r----------------ndash

ThatisVsat GMem

r----------------ndash=

Thegravitationalfieldstrengthatthesurfaceoftheplanetisgivenby

g 0GMe

Re2------------=

Hencewecanwrite

Vsat g0 Re

2

rndash=

m

ThekineticenergyofthesatelliteKsatisequaltofrac12mv2wherev is its orbital speed

Byequatingthegravitationalforceactingonthesatellitetoitscentripetalaccelerationwehave

G Mem

r2---------------- mv2

r---------- mv2hArr

GMemr----------------= =

Fromwhich12---

mv2 12---

GMemr----------------times=

= 0 e

2g R

r

2m

Which is actually quite interesting since it shows thatirrespective of the orbital radius the KE is numericallyequal to half the PE Also the total energy E tot K sat Vsat+ 1

2---G Me m

r----------------times G Me m

r----------------ndash + 1

2---G Me m

r----------------ndash= = =of the satelliteisalwaysnegativesince

E tot K sat Vsat+ 12---

G Me mr----------------times

G Me mr----------------ndash

+ 12---

G Me mr----------------ndash= = =

PHYSICS 2015indb 320 150514 338 PM

Fields

321

AH

L

The energies of an orbiting satellite as a function of radial distancefromthecentreofaplanetareshownplottedinFigure1018

12

100806

0402

0ndash02

ndash04

ndash06ndash08ndash10

ndash12

2 4 6 8 10 12

ener

gya

rbita

ry u

nits

distance timesR

kinetic energytotal energy

potential energy

Figure 1018 Energy of an orbiting satellite as a function of distance from the centre of a planet

WeightlessnessSuppose that you are inanelevator(lift)thatisdescendingat constant speedandyou let goof abook thatyouareholdinginyourhandThebookwillfalltothefloorwithaccelerationequaltotheaccelerationduetogravityIfthecablethatsupportstheelevatorweretosnap(asituationthatItrustwillneverhappentoanyofyou)andyounowletgothebookthatyouareholdinginyourotherhandthis book will not fall to the floor - it will stay exactlyin linewithyourhandThis isbecause thebook isnowfallingwith the sameaccelerationas theelevatorandassuch the book cannot ldquocatchrdquo up with the floor of theelevatorFurthermoreifyouhappenedtobestandingonasetofbathroomscalesthescaleswouldnowreadzero-youwouldbeapparentlyweightlessItisthisideaoffreefallthatexplainstheapparentweightlessnessofastronautsin an orbiting satellite These astronauts are in free fall inthesensethattheyareacceleratingtowardsthecentre of the Earth

It is actually possible to define theweight of a body inseveral different waysWe can define it for example asthegravitationalforceexertedonthebodybyaspecifiedobject such as the EarthThiswe have seen thatwe doinlotsofsituationswherewedefinetheweightasbeingequaltomgIfweusethisdefinitionthenanobjectinfreefallcannotbydefinitionbeweightlesssinceitisstillinagravitationalfieldHoweverifwedefinetheweightofanobjectintermsofaldquoweighingrdquoprocesssuchasthereadingonasetofbathroomscaleswhichineffectmeasuresthecontact force between the object and the scales thenclearlyobjectsinfreefallareweightlessOnenowhasto

askthequestionwhetherornotitispossibleForexampletomeasurethegravitationalforceactingonanastronautin orbit about the Earth

WecanalsodefineweightintermsofthenetgravitationalforceactingonabodyduetoseveraldifferentobjectsForexample for an object out in space its weight could bedefined in terms of the resultant of the forces exerted on it bytheSuntheMoontheEarthandalltheotherplanetsin the Solar System If this resultant is zero at a particular pointthenthebodyisweightlessatthispoint

Inviewofthevariousdefinitionsofweightthatareavailabletousitisimportantthatwhenweusethewordldquoweightrdquoweareawareofthecontextinwhichitisbeingused

Example

Calculate the height above the surface of the Earth atwhichageo-stationarysatelliteorbits

Solution

A geo-stationary satellite is one that orbits the Earth in such a way that it is stationary with respect to a point on the surface of the Earth This means that its orbital period must be the same as the time for the Earth to spin once on its axis ie 24 hours

From Keplerrsquos third law we have G Ms

4π2----------- R3

T2------=

That is

RMe

mh

using the fact that the force of attraction between the satellite and the Earth is given by

FG Mem

R2----------------=

and that F = ma

where m is the mass of the satellite and a 4π2R

T2-------------=

we have

GMem

R2---------------- m 4π2R

T2-------------times

GMe

4π2------------rArr R3

T2------= =

PHYSICS 2015indb 321 150514 338 PM

Chapter 10

322

AH

L

Now the mass of the Earth is 60 times 1024 kg and the period T measured in seconds is given by T = 86400 s

So substitution gives R = 42 times 106 m

The radius of the Earth is 64 times 106 m so that the orbital height h is about 36 times 107 m

Example

Calculatetheminimumenergyrequiredtoputasatelliteofmass500kgintoanorbitthatisasaheightequaltotheEarthrsquosradiusabovethesurfaceoftheEarth

Solution

We have seen that when dealing with gravitational fields and potential it is useful to remember that

g 0G M

Re2---------= or g0 Re

2 GM=

The gravitational potential at the surface of the Earth is

g 0ndash Re GMRe

---------ndash=

The gravitational potential at a distance R

from the centre of the Earth is

=

The difference in potential between the surface of the Earth and a point distance R from the centre is therefore

V∆ g 0Re 1ReR------ndash

=

If R 2Re= then V∆g0 Re

2------------=

This means that the work required to ldquoliftrdquo the satellite into orbit is g0Rm where m is the mass of the satellite This is equal to

10 32 106 500timestimestimes = 16000 MJ

However the satellite must also have kinetic energy in order to orbit Earth This will be equal to

g 0 mRe2

2R-----------------g 0mRe

2

4----------------- 8000 MJ= =

The minimum energy required is therefore

24000 MJ

Exercise 1031 The speed needed to put a satellite in orbit does

not depend on

A the radius of the orbitB theshapeoftheorbitC thevalueofgattheorbitD the mass of the satellite

2 EstimatethespeedofanEarthsatellitewhoseorbitis400kmabovetheEarthrsquossurfaceAlsodetermine the period of the orbit

3 Calculatethespeedofa200kgsatelliteorbitingthe Earth at a height of 70 times 106 m

Assume that g = 82 m sndash2 for this orbit

4 TheradiioftwosatellitesXandYorbitingthe Earth are 2r and 8rwherer is the radius of the Earth Calculate the ratio of the periods of revolutionofXtoY

5 A satellite of mass mkgissentfromEarthrsquossurfaceinto an orbit of radius 5RwhereR is the radius of theEarthWritedownanexpressionfor

(a) thepotentialenergyofthesatelliteinorbit

(b) thekineticenergyofthesatelliteinorbit

(c) theminimumworkrequiredtosendthesatellitefromrestattheEarthrsquossurfaceintoits orbit

6 A satellite in an orbit of 10rfallsbacktoEarth(radius r)afteramalfunctionDeterminethespeedwithwhichitwillhittheEarthrsquossurface

7 The radius of the moon is frac14 that of the Earth AssumingEarthandtheMoontohavethesamedensitycomparetheaccelerationsoffreefallatthe surface of Earth to that at the surface of the Moon

8 UsethefollowingdatatodeterminethegravitationalfieldstrengthatthesurfaceoftheMoon and hence determine the escape speed from the surface of the Moon

Mass of the Moon = 73 times 1022kg

RadiusoftheMoon=17times106 m

PHYSICS 2015indb 322 150514 338 PM

Chapter 9

306

AH

L

fromthesourceare incidentonareflectorthat ismovingtowards the transmitter with speed vThereflectedwavesare detected by the receiver placed alongside the transmitter

reector

v

transmitter

receiver

f f

f f

Figure 932 Using the Doppler effect to measure speed

We shall consider the situation where v ltlt c where c is the speed of the waves from the transmitter

For the reflector receiving waves from the transmitterit is effectively an observer moving towards a stationary sourceFromequation(114) itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation116

For the receiver receivingwaves from the reflector it iseffectively a stationary observer receiving waves from a movingsourceFromequation(115)itthereforereceiveswavesthathavebeenDopplershiftedbyanamount

vf f fc

minus = Equation117

Ifweaddequations(116)and(117)wegetthatthetotalDopplershiftatthereceiver∆f is

v vf f f f fc c

minus = ∆ = +

But 1 vf fc

= +

hence

1 v v vf f fc c c

∆ = + +

But since v ltlt cwe can ignore the term2

2

vc when we

expandthebracketintheaboveequation

Therefore we have2vf fc

∆ = Equation118

If v asymp c then we must use the full Doppler equationsHowever for em radiationwewill alwaysonly considersituations in which v ltlt c

Example

Thespeedofsoundinbloodis1500times103 m s-1 Ultrasound offrequency100MHzisreflectedfrombloodflowinginan arteryThe frequencyof the reflectedwaves receivedbackatthetransmitteris105MHzEstimatethespeedofthebloodflowintheartery

Solution

Using equation (118) we have

6 63

2005 10 1015 10

vtimes = times

timesto give v asymp 36 m s-1 (We have assumed that the ultrasound is incident at right angles to the blood flow)

Exercise 971 Judy is standing on the platform of a station A

high speed train is approaching the station in a straight line at constant speed and is sounding its whistleAsthetrainpassesbyJudythefrequencyof the sound emitted by the whistle as heard by Judychangesfrom640Hzto430HzDetermine

(a) thespeedofthetrain

(b) thefrequencyofthesoundemittedbythewhistle as heard by a person on the train (Speedofsound=330ms-1)

2 AgalaxyismovingawayfromEarthwithaspeedof00500cThewavelengthofaparticularspectral line in light emitted by atomic hydrogen inalaboratoryis656times10-7 m Calculate the valueofthewavelengthofthislinemeasuredinthelaboratoryinlightemittedfromasourceofatomic hydrogen in the galaxy

PHYSICS 2015indb 306 150514 337 PM

10 Fields

Contents

101 ndash Describing fields

102 ndash Fields at work

Essential Ideas

Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Similar approaches can be taken in analysing electrical and gravitational potential problems copy IBO 2014

PHYSICS 2015indb 307 150514 337 PM

Chapter 10

308

AH

L

101 Describing fields

Essential idea Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Understandingsbull Gravitationalfieldsbull Electrostaticfieldsbull Electricpotentialandgravitationalpotentialbull Fieldlinesbull Equipotentialsurfaces

Gravitational potentialWehaveseenthatifweliftanobjectofmassm to a height h abovethesurfaceoftheEarththenitsgainingravitationalpotential energy is mghHoweverthisisbynomeansthefullstoryForastartwenowknowthatgvarieswithh and also the expression really gives a difference in potentialenergybetweenthevaluethattheobjecthasattheEarthrsquossurfaceandthevaluethatithasatheighthSowhatwereallyneedisazeropointCanwefindapointwherethepotentialenergyiszeroandusethispointfromwhichtomeasure changes in potential energy

The point that is chosen is in fact infinity At infinity the gravitationalfieldstrengthofanyobjectwillbezeroSoletusseeifwecandeduceanexpressionforthegaininpotentialenergyofanobjectwhenitisldquoliftedrdquofromthesurfaceoftheEarthtoinfinityThisineffectmeansfindingtheworknecessarytoperformthistask

δr

r r δr+

r infin=

mg

AB

Me

Figure 1001 Gravitational forces

Inthediagramweconsidertheworknecessarytomovethe particle of mass m a distance δr in thegravitationalfield of the Earth

The force on the particle at A is FG Mem

r2----------------=

IftheparticleismovedtoB then since δr isverysmallwe can assume that the field remains constant over thedistance ABTheworkδW done againstthegravitationalfieldoftheEarthinmovingthedistanceABis

δWGMem

r2----------------δrndash=

(rememberthatworkdoneagainstaforceisnegative)

TofindthetotalworkdoneW in going from the surface oftheEarthtoinfinitywehavetoaddalltheselittlebitsofworkThisisdonemathematicallybyusingintegralcalculus

WG Me m

r2----------------ndash

rdR

infin

int G Me m 1

r2----- rd

R

infin

intndash G Me m 1r---ndash

R

infinndash= = =

GMe m 0 1R---ndash

ndashndash=

G MemR----------------ndash=

HencewehavewhereRistheradiusoftheEarththattheworkdonebythegravitationalfieldinmovinganobjectofmass m from R(surfaceoftheEarth)toinfinityisgivenby

WGMe m

R---------------- ndash=

We can generalise the result by calculating the worknecessary per unit mass to take a small mass fromthe surface of the Earth to infinity This we call thegravitational potential Vie

Wewouldgetexactlythesameresultifwecalculatedtheworkdonebythefieldtobringthepointmassfrominfinityto the surface of Earth In this respect the formal definition ofgravitationalpotentialatapointinagravitationalfieldis therefore defined as the work done per unit mass in bringing a point mass from infinity to that point

Clearly then the gravitational potential at any point inthe Earthrsquos field distance r from the centre of the Earth (providingr gt R)is

VGMe

r------------ ndash=

Thepotentialisthereforeameasureoftheamountofworkthathastobedonetomoveparticlesbetweenpointsinagravitationalfieldanditsunit is theJkgndash1 We also note thatthepotentialisnegativesothatthepotentialenergy

NATURE OF SCIENCEParadigm shift The move from direct observable actions being responsible for influence on an object to acceptance of a fieldrsquos ldquoaction at a distancerdquo required a paradigm shift in the world of science (23)

copy IBO 2014

PHYSICS 2015indb 308 150514 337 PM

Fields

309

AH

L

aswemoveawayfromtheEarthrsquossurfaceincreasesuntilitreachesthevalueofzeroatinfinity

IfthegravitationalfieldisduetoapointmassofmassmthenwehavethesameexpressionasaboveexceptthatMe is replaced by m andmustalsoexclude thevalueof thepotential at the point mass itself ie at r = 0

WecanexpressthegravitationalpotentialduetotheEarth(orduetoanysphericalmass)intermsofthegravitationalfield strength at its surface

AtthesurfaceoftheEarthwehave

g 0ndash Re GMRe

---------ndash= e

Sothat

g 0Re2 GM= e

Hence at a distance r from the centre of the Earth the gravitationalpotentialVcanbewrittenas

VGMe

r------------ndashg 0 Re

2

r------------ndash= =

The potential at the surface of the Earth

(r = Re)istherefore-g0Re

It is interesting to see how the expression for thegravitational potential ties in with the expression mgh ThepotentialatthesurfaceoftheEarthis-g0Re (see the exampleabove)andataheighthwillbe-g0g 0 Re h+( )ndash ifweassume that g0doesnotchangeover thedistanceh The differenceinpotentialbetweenthesurfaceandtheheighth is therefore g0hSotheworkneededtoraiseanobjectof mass m to a height h is mgh ie m timesdifference ingravitationalpotential

This we have referred to as the gain in gravitationalpotentialenergy(see235)

However this expression can be extended to any twopoints inanygravitationalfieldsuch that ifanobjectofmass mmovesbetween twopointswhosepotentialsareV1 and V2 respectively then the change in gravitationalpotentialenergyoftheobjectism(V1 ndash V2)

Gravitational potential gradient

Let us consider now a region in space where thegravitationalfieldisconstantInFigure912thetwopointsAandBareseparatedbythedistance∆x

A B

direction of uniform gravitational eld of strength I

∆x

Figure 1002 The gravitational potential gradient

The gravitational field is of strength I and is in the direction shownThe gravitational potential at A is V andatBisV +∆V

Theworkdone is takingapointmassmfromAtoB isF∆x = mI∆x

Howeverbydefinitionthisworkisalsoequalto-m∆V

Therefore mI∆x=-m∆V

or VIx

∆= minus

∆Effectivelythissaysthatthemagnitudeofthegravitationalfield strength is equal to the negative gradient of thepotential If I is constant then V is a linear function of x and Iisequaltothenegativegradientofthestraightlinegraph formed by plotting V against x If I is not constant (asusuallythecase)thenthemagnitudeofI at any point in the field can be found by find the gradient of the V-x graph at that point An example of such a calculation can befoundinSection929

For those of you who do HL maths the relationship between field and potential is seen to follow from the expression for the potential of a point mass viz

mV Gr

= minus

2

ddV mG Ir r

minus = + =

Potential due to one or more point masses

Gravitationalpotentialisascalarquantitysocalculatingthepotential due to more than one point mass is a matter of simpleadditionSoforexamplethepotentialV due to the Moon and Earth and a distance x fromthecentreofEarthonastraightlinebetweenthemisgivenbytheexpression

E MM MV Gx r x

= minus + minus

whereME=massofEarthMM= mass of Moon and r = distancebetweencentreofEarthandMoon

PHYSICS 2015indb 309 150514 337 PM

Chapter 10

310

AH

L

Equipotentials and field linesIfthegravitationalpotentialhasthesamevalueatallpointson a surface the surface is said to be an equipotential surfaceSoforexampleifweimagineasphericalshellaboutEarthwhosecentrecoincideswiththecentreofEarththisshellwillbeanequipotentialsurfaceClearlyifwerepresentthegravitationalfieldstrengthbyfieldlinessincethelinesldquoradiaterdquooutfromthecentreofEarththentheselineswillbeatrightanglestothesurfaceIfthefieldlineswerenotnormaltotheequipotentialsurfacethentherewouldbeacomponentof the field parallel to the surfaceThis wouldmean thatpointsonthesurfacewouldbeatdifferentpotentialsandsoitwouldnolongerbeanequipotentialsurfaceThisofcourseholdstrueforanyequipotentialsurface

Figure 1003 shows the field lines and equipotentials fortwopointmassesm

m m

Figure 1003 Equipotentials for two point masses

It is worth noting that we would get exactly the samepattern ifwewere to replace thepointmasseswith twoequalpointcharges(See935)

Escape speedThe potential at the surface of Earth is which meansthattheenergyrequiredtotakeaparticleofmass mfromthesurfacetoinfinityisequalto

ButwhatdoesitactuallymeantotakesomethingtoinfinityWhentheparticleisonthesurfaceoftheEarthwecanthinkofitassittingatthebottomofaldquopotentialwellrdquoasinfigure1004

particle

surface of Earth

innity

GMR---------

FIgure 1004 A potential well

Theldquodepthrdquoofthewellis and if the particle gains an amount of kinetic energy equal to wherem is its massthenitwillhavejustenoughenergytoldquoliftrdquoitoutofthewell

In reality it doesnrsquot actually go to infinity it justmeansthat the particle is effectively free of the gravitationalattractionof theEarthWe say that ithas ldquoescapedrdquo theEarthrsquosgravitationalpullWemeetthisideainconnectionwithmolecular forcesTwomolecules in a solidwill sitat their equilibrium position the separation where therepulsiveforceisequaltotheattractiveforceIfwesupplyjust enough energy to increase the separation of themolecules such that they are an infinite distance apart then intermolecularforcesnolongeraffectthemoleculesandthesolidwillhavebecomealiquidThereisnoincreaseinthekineticenergyofthemoleculesandsothesolidmeltsat constant temperature

Wecancalculatetheescapespeedofanobjectveryeasilyby equating the kinetic energy to the potential energysuch that

12---mvesc ape

2 GMe mRe

----------------=

vesc aperArr2GMe

Re--------------- 2g0 Re = =

Substituting for g0 and Regivesavalueforvescape of about 11kmsndash1 from the surface of the Earth

Youwillnotethattheescapespeeddoesnotdependonthemassoftheobjectsincebothkineticenergyandpotentialenergy are proportional to the mass

IntheoryifyouwanttogetarockettothemoonitcanbedonewithoutreachingtheescapespeedHoweverthiswouldnecessitate anenormousamountof fuel and it islikely that therocketplus fuelwouldbesoheavythat itwouldnevergetoffthegroundItismuchmorepracticalto accelerate the rocket to the escape speed fromEarthorbitandthenintheoryjustlaunchittotheMoon

Example

Use the followingdata todetermine thepotential at thesurface of Mars and the magnitude of the acceleration of free fall

massofMars =64times1023kg

radiusofMars =34times106 m

Determinealsothegravitationalfieldstrengthatadistanceof 68 times 106mabovethesurfaceofMars

PHYSICS 2015indb 310 150514 337 PM

Fields

311

AH

L

Solution

= minus = minus times times = times23

11 76

64 1067 10 ndash13 10

34 10MV GR

minus timestimes

Nkg-1

But V = -g0R

Therefore 7

0 6

13 10 3834 10

VgR

times= minus = =

timesm s-2

To determine the field strength gh at 68 times 106 m above the surface we use the fact that 0 2

Mg GR

= such that GM = g0R2

Therefore 2 2

0h 2 2 2

h h

38 (34) 042(102)

g RGMgR R

times= = = = m s-2

(the distance from the centre is 34 times 106 + 68 times 106 = 102 times 106 m)

Exercise 1011 Thegraphbelowshowshowthegravitational

potentialoutsideoftheEarthvarieswithdistancefrom the centre

0 10 20 30 40 50 60 70

ndash1

ndash2

ndash3

ndash4

ndash5

ndash6

ndash7

r m times 106

V

Jkgndash1

times 10

7

(a) Usethegraphtodeterminethegainingravitationalpotentialenergyofasatelliteofmass200kgasitmovesfromthesurfaceof the Earth to a height of 30 times 107mabovetheEarthrsquossurface

(b) Calculatetheenergyrequiredtotakeit to infinity

(c) DeterminetheslopeofthegraphatthesurfaceoftheEarthmCommenton youranswer

Electric potential energy

The concept of electric potential energy was developedwith gravitational potential energy in mind Just as anobjectnearthesurfaceoftheEarthhaspotentialenergybecause of its gravitational interaction with the Earthsotoothereiselectricalpotentialenergyassociatedwithinteracting charges

Letusfirstlookatacaseoftwopositivepointchargeseachof 1μC that are initially bound together by a thread in a vacuuminspacewithadistancebetweenthemof10cmasshowninFigure1005Whenthethreadiscutthepointchargesinitiallyatrestwouldmoveinoppositedirectionsmovingwithvelocitiesv1 and v2 along the direction of the electrostatic force of repulsion

BEFORE AFTER

v v21

Figure 1005 Interaction of two positive particles

The electric potential energybetweentwopointchargescan be found by simply adding up the energy associated witheachpairofpointchargesForapairofinteractingchargestheelectricpotentialenergyisgivenby

∆U = ∆Ep + ∆Ek = ∆W = ∆Fr = kqQ

____ r2 times r = kqQ

____ r

Becausenoexternalforceisactingonthesystemtheenergyand momentum must be conserved Initially Ek = 0 and Ep =kqQr=9times109 times 1 times 10-1201m=009JWhentheyareagreatdistance fromeachotherEpwillbenegligibleThefinalenergywillbeequaltofrac12mv1

2+frac12mv22=009J

Momentum isalsoconservedand thevelocitieswouldbethe same magnitude but in opposite directions

Electric potential energyismoreoftendefinedintermsofapointchargemovinginanelectricfieldas

lsquothe electricpotential energybetweenany twopoints inan electric field is defined as negativeoftheworkdone by anelectricfieldinmovingapoint electric chargebetweentwolocationsintheelectricfieldrsquo

ΔU = ΔEp = -ΔW = -Fd = qEx

ΔU = ΔEp +ΔEk = ΔW = ΔFr = kqQ r2 x r = kqQ

wherexisthedistancemovedalong(oroppositeto)thedirection of the electric field

Electric potential energyismeasuredinjoule(J)Justasworkisascalarquantitysotooelectricalpotentialenergyis a scalar quantity The negative of the work done by an electric field in moving a unit electric charge between two points is independent of the path taken Inphysicswesaytheelectricfieldisalsquoconservativersquo field

PHYSICS 2015indb 311 150514 337 PM

Chapter 10

312

AH

L

Supposeanexternalforcesuchasyourhandmovesasmallpositive point test charge in the direction of a uniformelectric fieldAs it ismoving itmust be gaining kineticenergyIfthisoccursthentheelectricpotentialenergyofthe unit charge is changing

InFigure1006apointcharge+qismovedbetweenpointsAandBthroughadistancex in a uniform electric field

B

A

x

+ q

Figure 1006 Movement of a positive point charge in a uniform field

InordertomoveapositivepointchargefrompointAtopointBanexternal forcemustbeapplied to thechargeequaltoqE (F = qE)

Since the force is applied through a distance xthennegativeworkhas to bedone tomove the charge because energyis gainedmeaning there is an increaseelectric potential energybetweenthetwopointsRememberthattheworkdoneisequivalenttotheenergygainedorlostinmovingthe charge through the electric field The concept of electric potential energy is only meaningful as the electric field whichgeneratestheforceinquestionisconservative

W F xtimes E q x times= =

θ x cos θ x

Figure 1007 Charge moved at an angle to the field

If a chargemoves at an angle θ to an electric field thecomponent of the displacement parallel to the electric fieldisusedasshowninFigure918

W F x Eq x θcostimes= =

Theelectricpotentialenergyisstoredintheelectricfieldand theelectricfieldwill return theenergy to thepointcharge when required so as not to violate the Law ofconservationofenergy

Electric potential

The electric potential at a point in an electric field is defined as being the work done per unit charge in bringing a small positive point charge from infinity to that point

ΔV = Vinfin ndash Vf = ndash W ___ q

Ifwedesignatethepotentialenergytobezeroatinfinitythen it follows thatelectricpotentialmustalsobezeroat infinity and the electric potential at any point in an electricfieldwillbe

ΔV = ndash W ___ q

NowsupposeweapplyanexternalforcetoasmallpositivetestchargeasitismovedtowardsanisolatedpositivechargeTheexternalforcemustdoworkonthepositivetestchargetomoveittowardstheisolatedpositivechargeandtheworkmustbepositivewhiletheworkdonebytheelectricfieldmustthereforebenegativeSotheelectricpotentialatthatpointmustbepositiveaccordingtotheaboveequationIfanegativeisolatedchargeisusedtheelectricpotentialatapointonthepositivetestchargewouldbenegativePositivepoint chargesof their ownaccordmove fromaplaceofhighelectricpotential toaplaceof lowelectricpotentialNegative point charges move the other way from lowpotentialtohighpotentialInmovingfrompointAtopointBinthediagramthepositivecharge+qismovingfromalowelectricpotentialtoahighelectricpotential

In thedefinitiongiven the termldquoworkperunitchargerdquohas significance If the test charge is +16 times 10-19Cwherethe charge has a potential energy of 32 times 10-17Jthenthepotentialwouldbe32times10-17J+16times10-19C=200JC-1 Now if the charge was doubled the potential wouldbecome 64times10-17 J However the potential per unitchargewouldbethesame

Electric potential is a scalar quantityandithasunitsJC-1 orvoltswhere1voltequalsone joulepercoloumbThevoltallowsustoadoptaunitfortheelectricfieldintermsofthevolt

PreviouslytheunitfortheelectricfieldwasNCndash1

W = qV and F = qEsoW ___ V = F __ E

E = FV ___ W = FV ___ Fm V mndash1

That is the units of the electric field E can also beexpressed as V mndash1

The work done per unit charge in moving a point charge between two points in an electric field is again independent of the path taken

PHYSICS 2015indb 312 150514 337 PM

Fields

313

AH

L

Electric potential due to a point charge

Letus takeapoint rmetres froma chargedobjectThepotentialatthispointcanbecalculatedusingthefollowing

W = ndashFr = ndashqV and F = ndash q1 q2 _____ 4πε0r

2

Therefore

W = ndash q1 q2 _____ 4πε0r

2 times r = ndash q1 q2 _____ 4πε0r

= ndashq1 times q2 _____ 4πε0r

= ndashq1V

That is V q

4πε0r--------------- =

Orsimply

V kqr------

=

Example

Determinehowmuchworkisdonebytheelectricfieldofpointcharge150μCwhenachargeof200μCismovedfrom infinity to apoint 0400m from thepoint charge(Assumenoaccelerationofthecharges)

Solution

The work done by the electric field is W = -qV = -14πε0 times q times (Q rinfin - Q r0400)

W = (- 200 times 10-6 C times 900 times 109 NmC-2 times 150 times 10-6 C) divide 0400 m = - 0675 J

An external force would have to do +0675 J of work

Electric field strength and electric potential gradient

Letus lookbackatFigure1006Supposeagain that thecharge+qismovedasmalldistancebyaforceFfromAtoBsothattheforcecanbeconsideredconstantTheworkdoneisgivenby

∆W F x∆times=

TheforceFandtheelectricfieldEareoppositelydirectedandweknowthat

F = -qE and ΔW = q ΔV

Thereforetheworkdonecanbegivenas

q ΔV = -q E Δx

Therefore E V∆x∆-------

ndash=

The rate of change of potential ΔV at a point with respect to distance Δx in the direction in which the change is maximum is called the potential gradient We say that the electric field = - the potential gradient and the units are Vm-

1 From the equation we can see that in a graph of electric potential versus distance the gradient of the straight line equals the electric field strength

In reality if a charged particle enters a uniform electric field it will be accelerated uniformly by the field and its kinetic energy will increase This is why we had to assume no acceleration in the last worked example

Ek∆ 12---

mv2 q E xsdot sdot q Vx---

xsdot sdot q Vsdot= = = =

Example

Determine how far apart two parallel plates must besituated so that a potential difference of 150 times 102 V produces an electric field strength of 100 times 103 NC-1

Solution

Using E V∆x------- xhArrndash V∆

E-------15 102 Vtimes

100 103 N C 1ndashtimes------------------------------------------= = =

= 150 times 10-1

The plates are 150 times 10-1 m apart

The electric field and the electric potential at a point due to an evenly distributed charge +q on a sphere can be represented graphically as in Figure 1008

r

r

V

E

r0

r0

r0

Charge of +Q evenly distributed over surface

E 14πε0------------ Q

r2----- r r0gt=

V 14πε0------------Q

r----r r0gt=

Figure 1008 Electric field and potential due to a charged sphere

PHYSICS 2015indb 313 150514 337 PM

Chapter 10

314

AH

L

When the sphere becomes charged we know that thechargedistributes itselfevenlyoverthesurfaceThereforeeverypartofthematerialoftheconductorisatthesamepotential As the electric potential at a point is defined as beingnumericallyequaltotheworkdoneinbringingaunitpositivechargefrominfinitytothatpointithasaconstantvalueineverypartofthematerialoftheconductor

Since the potential is the same at all points on the conducting surfacethen∆V∆xiszeroButE=ndash∆V∆xThereforethe electric field inside the conductor is zero There is no electric field inside the conductor

SomefurtherobservationsofthegraphsinFigure1009are

bull OutsidethespherethegraphsobeytherelationshipsgivenasE α 1 r2 and V α 1 r

bull Atthesurfacer = r0Thereforetheelectricfieldandpotentialhavetheminimumvalueforr at this point and this infers a maximum field and potential

bull Insidethespheretheelectricfieldiszerobull Insidethespherenoworkisdonetomoveacharge

fromapointinsidetothesurfaceThereforethereisno potential difference and the potential is the same asitiswhenr = r0

Similargraphscanbedrawnfortheelectricfieldintensityandthe electric potential as a function of distance from conducting parallelplatesandsurfacesandthesearegiveninFigure1009

Potential plot E eld plotE eld

++++

ndashndashndashndash

xx x

+

x x

ndashx x

+ ndashx

x

+ +

x

x

Figure 1009 Electric field and electric potential at a distance from a charged surface

Potential due to one or more point charges

The potential due to one point charge can be determined byusingtheequationformula

V = kq r

Example 1

Determine the electric potential at a point 20 times 10-1 m fromthecentreofan isolatedconducting spherewithapointchargeof40pCinair

Solution

Using the formula

V = kq r we have

V90 10 9times( ) 40 10 12ndashtimes( )times

20 10 1ndashtimes( )------------------------------------------------------------------ 018 V= =

the potential at the point is 180 times 10-1 V

The potential due to a number of point charges can be determined by adding up the potentials due to individual point charges because the electric potential at any point outside a conducting sphere will be the same as if all the charge was concentrated at its centre

Example 2

Threepointchargesofareplacedattheverticesofaright-angledtriangleasshowninthediagrambelowDeterminetheabsolutepotentialat the+20μCchargeduetothetwoothercharges

- 6μC

+3μC +2μC

4 m

3 m

Solution

The electric potential of the +2 μC charge due to the ndash 6 μC charge is

V = (9 times 109 Nm2C-2 times -6 times 10-6 C) divide (radic 32 + 42) m = - 108 times 104 V

PHYSICS 2015indb 314 150514 338 PM

Fields

315

AH

L

Insummarywecanconcludethat

bull No work is done to move a charge along an equipotential

bull Equipotentials are always perpendicular to the electric lines of force

Figure1011and1012showsomeequipotential lines fortwo oppositely charged and identically positive spheresseparated by a distance

+ve ndashve

equipotential lines

Figure 1011 Equipotential lines between two opposite charges

equipotential lines

Figure 1012 Equipotential lines between two charges which are the same

+

ndash ndash ndash ndash

+ + +

40 V

30 V

20 V

10 V

50 V

Figure 1013 Equipotential lines between charged parallel plates

Figure1013showstheequipotentiallinesbetweenchargedparallel plates Throughout this chapter the similarities and differences between gravitational fields and electric fieldshavebeendiscussedTherelationshipsthatexistsbetweengravitationalandelectricquantitiesandtheeffectsofpointmassesandchargesissummarisedinTable1014

The electric potential of the +2 μC charge due to the +3 μC charge is

V = (9 times 109 Nm2C-2 times 3 times 10-6 C) divide 3m = 9 times 103 V

The net absolute potential is the sum of the 2 potentials- 108 times 104 V + 9 times 103 V =- 18 times 103 V

The absolute potential at the point is - 18 times 103 V

Equipotential surfacesRegions in space where the electric potential of a chargedistributionhasaconstantvaluearecalledequipotentials The placeswherethepotentialisconstantinthreedimensionsarecalled equipotential surfacesandwheretheyareconstantintwodimensionstheyarecalledequipotential lines

TheyareinsomewaysanalogoustothecontourlinesontopographicmapsInthiscasethegravitationalpotentialenergy is constant as amassmoves around the contourlines because the mass remains at the same elevationabovetheEarthrsquossurfaceThegravitationalfieldstrengthacts in a direction perpendicular to a contour line

Similarlybecausetheelectricpotentialonanequipotentiallinehas thesamevalueanelectric forcecandonoworkwhena testchargemovesonanequipotentialThereforethe electric field cannot have a component along anequipotentialandthusitmustbeeverywhereperpendiculartotheequipotentialsurfaceorequipotentiallineThisfactmakesiteasytoplotequipotentialsifthelinesofforceorlinesofelectricfluxofanelectricfieldareknown

Forexamplethereareaseriesofequipotentiallinesbetweentwoparallelplateconductorsthatareperpendiculartotheelectric fieldTherewill be a series of concentric circles(eachcirclefurtherapartthanthepreviousone)thatmapouttheequipotentialsaroundanisolatedpositivesphereThe lines of force and some equipotential lines for anisolatedpositivesphereareshowninFigure1010

Lines ofequipotential

Figure 1010 Equipotentials around an isolated positive sphere

PHYSICS 2015indb 315 150514 338 PM

Chapter 10

316

AH

L

Gravitational quantity Electrical quantityQ

uant

ities

g V∆x∆-------ndash= E V∆

x∆-------ndash=

Poin

t mas

ses a

nd ch

arge

s

V Gmr----

ndash= V 14πε0------------q

r---=

g G m

r2-----ndash= E 14πε0------------ q

r2-----=

F Gm1 m2

r2--------------= F 1

4πε0------------

q1q2

r2-----------=

Figure 1014 Formulas

Example

Deduce the electric potential on the surface of a gold nucleus that has a radius of 62 fm

Solution

Using the formula

V = kq r and knowing the atomic number of gold is 79 We will assume the nucleus is spherical and it behaves as if it were a point charge at its centre (relative to outside points)

V = 90 times 109 Nm2C-2 times 79 times 16 times 10-19 C divide 62 times 10-15 m = 18 times 107 V

The potential at the point is 18 MV

Example

Deduce the ionisation energy in electron-volts of theelectron in the hydrogen atom if the electron is in its ground state and it is in a circular orbit at a distance of 53 times 10-11 m from the proton

Solution

This problem is an energy coulombic circular motion question based on Bohrrsquos model of the atom (not the accepted quantum mechanics model) The ionisation energy is the energy required to remove the electron from the ground state to infinity The electron travels in a circular orbit and therefore has a centripetal acceleration The ionisation energy will counteract the coulombic force and the movement of the electron will be in the opposite direction to the centripetal force

Total energy = Ek electron + Ep due to the proton-electron interaction

ΣF = kqQ r2 = mv2 r and as such mv2 = = kqQ r

Therefore Ek electron = frac12 kqQ r

Ep due to the proton-electron interaction = - kqQ r

Total energy = frac12 kqQ r + - kqQ r = -frac12 kqQ r

= - 90 times 109 Nm2C-2 times (16 times 10-19 C)2 divide 53 times 10-11 m = -217 times 10-18 J

= -217 times 10-18 J divide 16 times 10-19 = -136 eV

The ionisation energy is 136 eV

Exercise 1021 ApointchargePisplacedmidwaybetweentwo

identicalnegativechargesWhichoneofthefollowingiscorrectwithregardstoelectricfieldandelectricpotentialatpointP

Electric field Electric potentialA non-zero zeroB zero non-zeroC non-zero non-zeroD zero zero

2 Twopositivechargedspheresaretiedtogetherinavacuumsomewhereinspacewheretherearenoexternal forces A has a mass of 25 g and a charge of20μCandBhasamassof15gandachargeof30μCThedistancebetweenthemis40cmTheyarethenreleasedasshowninthediagram

BEFORE AFTER

v1 v2

A B

(a) Determinetheirinitialelectricpotentialenergy in the before situation

(b) DeterminethespeedofsphereBafterrelease

3 Thediagrambelowrepresentstwoequipotentiallines in separated by a distance of 5 cm in a uniform electric field

+ + + + + + + +

ndash ndash ndash ndash ndash ndash ndash ndash

40 V

20 V 5 cm

PHYSICS 2015indb 316 150514 338 PM

Fields

317

AH

L

Determine the strength of the electric field

4 Thisquestionisabouttheelectricfieldduetoacharged sphere and the motion of electrons in that fieldThediagrambelowshowsanisolatedmetalsphereinavacuumthatcarriesanegativeelectriccharge of 60 μC

ndash

(a) Onthediagramdrawtheconventionalwayto represent the electric field pattern due to the charged sphere and lines to represent threeequipotentialsurfacesintheregionoutside the sphere

(b) Explainhowthelinesrepresentingtheequipotentialsurfacesthatyouhavesketchedindicatethatthestrengthoftheelectricfieldisdecreasingwithdistancefrom the centre of the sphere

(c) Theelectricfieldstrengthatthesurfaceofthe sphere and at points outside the sphere can be determined by assuming that the sphere acts as a point charge of magnitude 60 μC at its centre The radius of the sphere is 25 times 10ndash2 m Deduce that the magnitude of the field strength at the surface of the sphere is 86 times 107 Vmndash1

An electron is initially at rest on the surface of the sphere

(d) (i) Describethepathfollowedbytheelectronasitleavesthesurfaceofthesphere

(ii) Calculatetheinitialaccelerationofthe electron

5 Determinetheamountofworkthatisdoneinmovingachargeof100nCthroughapotentialdifferenceof150times102 V

6 Three identical 20 μC conducting spheres are placedatthecornersofanequilateraltriangleofsides 25 cm The triangle has one apex C pointing upthepageand2baseanglesAandBDeterminetheabsolutepotentialatB

7 Determinehowfaraparttwoparallelplatesmustbesituatedsothatapotentialdifferenceof250 times 102 V produces an electric field strength of 200 times 103 NC-1

8 Thegapbetweentwoparallelplatesis10times10-3mandthereisapotentialdifferenceof10times104 V betweentheplatesCalculate

i theworkdonebyanelectroninmovingfrom one plate to the other

ii thespeedwithwhichtheelectronreachesthe second plate if released from rest

iii theelectricfieldintensitybetweentheplates

9 Anelectronguninapicturetubeisacceleratedby a potential 25 times 103VDeterminethekineticenergygainedbytheelectroninelectron-volts

10 Determine the electric potential 20 times 10-2 m from achargeof-10times10-5 C

11 Determinetheelectricpotentialatapointmid-waybetweenachargeofndash20pCandanotherof+5pConthelinejoiningtheircentresifthecharges are 10 cm apart

12 During a thunderstorm the electric potential differencebetweenacloudandthegroundis10 times 109 V Determine the magnitude of the change in electric potential energy of an electron thatmovesbetweenthesepointsinelectron-volts

13 A charge of 15 μC is placed in a uniform electric fieldoftwooppositelychargedparallelplateswithamagnitudeof14times103 NC-1

(a) Determinetheworkthatmustbedoneagainstthefieldtomovethepointchargeadistance of 55 cm

(b) Calculatethepotentialdifferencebetweenthe final and initial positions of the charge

(c) Determinethepotentialdifferencebetweenthe plates if their separation distance is 15 cm

14 Duringaflashoflightningthepotentialdifferencebetweenacloudandthegroundwas12 times 109 V and the amount of transferred charge was32C

(a) Determinethechangeinenergyofthetransferred charge

(b) Iftheenergyreleasedwasallusedtoacceleratea1tonnecardeduceitsfinalspeed

(c) Iftheenergyreleasedcouldbeusedtomelticeat0degCdeducetheamountoficethatcould be melted

PHYSICS 2015indb 317 150514 338 PM

Chapter 10

318

AH

L

15 SupposethatwhenanelectronmovedfromAtoBin the diagram along an electric field line that the electric field does 36 times 10-19Jofworkonit

Determinethedifferencesinelectricpotential

(a) VB ndash VA(b) VC ndash VA(c) VC ndash VB

16 DeterminethepotentialatpointPthatislocatedatthecentreofthesquareasshowninthediagrambelow

- 6μC

+3μC +2μC

1m

1m

5μC

P

102 Fields at work

Essential idea Similar approaches can be taken inanalysingelectricalandgravitationalpotentialproblems

Understandingsbull Potentialandpotentialenergybull Potentialgradientbull Potentialdifferencebull Escapespeedbull Orbitalmotionorbitalspeedandorbitalenergybull Forcesandinverse-squarelawbehaviour

Orbital motion orbital speed and orbital energy

Although orbital motion may be circular elliptical orparabolic this sub-topic only dealswith circular orbitsThis sub-topic is not fundamentally new physics butan application that synthesizes ideas from gravitationcircularmotiondynamicsandenergy

TheMoonorbitstheEarthandinthissenseitisoftenreferredtoasasatelliteoftheEarthBefore1957itwastheonlyEarthsatelliteHoweverin1957theRussianslaunchedthefirstmanmadesatelliteSputnik1SincethisdatemanymoresatelliteshavebeenlaunchedandtherearenowliterallythousandsofthemorbitingtheEarthSomeareusedtomonitortheweathersome used to enable people to find accurately their position on thesurfaceoftheEarthmanyareusedincommunicationsandnodoubtsomeareusedtospyonothercountriesFigure1015showshowinprincipleasatellitecanbeputintoorbit

Theperson(whosesizeisgreatlyexaggeratedwithrespectto Earth) standing on the surface on the Earth throwssome stonesThe greater the speed with which a stoneis thrown the further it will land from her The pathsfollowedbythethrownstonesareparabolasByastretchof the imaginationwecanvisualisea situation inwhichastoneis thrownwithsuchaspeedthatbecauseof thecurvatureoftheEarthitwillnotlandonthesurfaceoftheEarthbutgointoldquoorbitrdquo(Path4onFigure1015)

NATURE OF SCIENCECommunication of scientific explanations The ability to apply field theory to the unobservable (charges) and the massively scaled (motion of satellites) required scientists to develop new ways to investigate analyse and report findings to a general public used to scientific discoveries based on tangible and discernible evidence (51)

copy IBO 2014

PHYSICS 2015indb 318 150514 338 PM

Fields

319

AH

L

1

2

34Earth

Figure 1015 Throwing a stone into orbit

TheforcethatcausesthestonestofollowaparabolicpathandtofalltoEarthisgravityandsimilarlytheforcethatkeeps the stone in orbit is gravity For circular motionto occur we have seen that a force must act at rightanglestothevelocityofanobjectthatistheremustbeacentripetalforceHenceinthesituationwedescribeherethe centripetal force for circular orbital motion about the EarthisprovidedbygravitationalattractionoftheEarth

We can calculate the speedwithwhich a stonemustbethrowninordertoputitintoorbitjustabovethesurfaceof the Earth

If the stone has mass m and speed v thenwehave fromNewtonrsquos2ndlaw

mv2

RE---------- G

MEm

RE2------------=

whereRE is the radius of the Earth and ME is the mass of the Earth

Bearinginmindthatg 0 GME

RE2--------= then

v g RE 10 64 106timestimes 8 103times= = =

Thatisthestonemustbethrownat8times103m sndash1

Clearly we are not going to get a satellite into orbit soclosetothesurfaceoftheEarthMovingatthisspeedthefrictionduetoairresistancewouldmeltthesatellitebeforeithadtravelledacoupleofkilometresInrealitythereforeasatelliteisputintoorbitabouttheEarthbysendingitattachedtoarocketbeyondtheEarthrsquosatmosphereandthengivingitacomponentofvelocityperpendiculartoaradialvectorfromtheEarthSeeFigure1016

Earth Satellite carried by rocket to hereSatellite orbit

Tangential component of velocity

Figure 1016 Getting a satellite into orbit

Keplerrsquos third law(This work of Kepler and Newtonrsquos synthesis of the work is an excellent example of the scientific method and makes for a good TOK discussion)

In 1627 Johannes Kepler (1571-1630)publishedhis lawsof planetary motion The laws are empirical in natureandwere deduced from the observations of theDanishastronomer Tycho de Brahe (1546-1601)The third lawgives a relationship between the radius of orbit R of a planet and its period TofrevolutionabouttheSunThelawisexpressedmathematicallyas

2

3 constantTR

=

We shall nowuseNewtonrsquos LawofGravitation to showhowitisthattheplanetsmoveinaccordancewithKeplerrsquosthirdlaw

InessenceNewtonwasabletousehislawofgravitytopredict the motion of the planets since all he had to do was factor theF givenby this law intohis second law F = ma to find their accelerations and hence their future positions

InFigure1017theEarthisshownorbitingtheSunandthedistancebetweentheircentresisR

RSun

Earth

F es

F se

Figure 1017 Planets move according to Keplerrsquos third law

Fes is the force that the Earth exerts on the Sun and Fse is the force that the Sun exerts on the Earth The forces are equalandoppositeandtheSunandtheEarthwillactuallyorbitaboutacommoncentreHoweversince theSun issoverymuchmoremassivethantheEarththiscommoncentrewillbeclosetothecentreoftheSunandsowecanregard the Earth as orbiting about the centre of the Sun Theotherthingthatweshallassumeisthatwecanignorethe forces that the other planets exert on the Earth (This wouldnotbeawisethingtodoifyouwereplanningtosend a space ship to theMoon for example)We shallalso assume that we have followed Newtonrsquos exampleandindeedprovedthataspherewillactasapointmasssituated at the centre of the sphere

Kepler had postulated that the orbits of the planets areellipticalbut since theeccentricityof theEarthrsquosorbit issmallweshallassumeacircularorbit

PHYSICS 2015indb 319 150514 338 PM

Chapter 10

320

AH

L

TheaccelerationoftheEarthtowardstheSunisa = Rω2

whereω ω 2πT------=

Hencea R 2π

T------ times

2 4π2 RT 2------------- = =

But the acceleration is given by Newtonrsquos Second Law F = mawhereFisnowgivenbytheLawofGravitationSo in this situation

F maGMs Me

R2-------------------= = butwealsohavethat

a a R 2πT

------ times

2 4π2 RT 2------------- = = and m = Me so that

G MsMe

R2------------------- Me4π2R

T2-------------times

GMs

4π2-----------rArr R3

T2------= =

Butthequantity

isaconstantthathasthesamevalueforeachoftheplanetssowehaveforalltheplanetsnotjustEarththat

R3

T2------ k =

wherekisaconstantWhichisofcourseKeplerrsquosthirdlaw

ThisisindeedanamazingbreakthroughItisdifficulttorefutethe idea that all particles attract each other in accordance withtheLawofGravitationwhenthelawisabletoaccountfortheobservedmotionoftheplanetsabouttheSun

The gravitational effects of the planets upon each othershould produce perturbations in their orbits Such is the predictivepoweroftheUniversalGravitationalLawthatit enabled physicists to compute these perturbations The telescopehadbeeninventedin1608andbythemiddleofthe 18th Century had reached a degree of perfection in design that enabled astronomers to actually measure the orbital perturbations of the planets Their measurements werealways inagreementwith thepredictionsmadebyNewtonrsquoslawHoweverin1781anewplanetUranuswasdiscovered and the orbit of this planet did not fit withthe orbit predicted by Universal Gravitation Such wasthe physicistrsquos faith in theNewtonianmethod that theysuspectedthatthediscrepancywasduetothepresenceofayetundetectedplanetUsingtheLawofGravitationtheFrenchastronomerJLeverrier and the English astronomer J C Adamswereable tocalculate justhowmassive thisnewplanetmustbeandalsowhereitshouldbeIn1846the planetNeptunewas discovered justwhere they hadpredicted Ina similarwaydiscrepancies in theorbitofNeptune ledto thepredictionandsubsequentdiscoveryin1930oftheplanetPlutoNewtonrsquosLawofGravitation

had passed the ultimate test of any theory it is not only abletoexplainexistingdatabutalsotomakepredictions

Satellite energyWhenasatelliteisinorbitaboutaplanetitwillhavebothkineticenergyandgravitationalpotentialenergySupposewe consider a satellite ofmass m that is in an orbit of radius r about a planet of mass M

Thegravitationalpotentialduetotheplanetatdistancer from its centre is

ThegravitationalpotentialenergyofthesatelliteVsat

is therefore GMe m

r----------------ndash

ThatisVsat GMem

r----------------ndash=

Thegravitationalfieldstrengthatthesurfaceoftheplanetisgivenby

g 0GMe

Re2------------=

Hencewecanwrite

Vsat g0 Re

2

rndash=

m

ThekineticenergyofthesatelliteKsatisequaltofrac12mv2wherev is its orbital speed

Byequatingthegravitationalforceactingonthesatellitetoitscentripetalaccelerationwehave

G Mem

r2---------------- mv2

r---------- mv2hArr

GMemr----------------= =

Fromwhich12---

mv2 12---

GMemr----------------times=

= 0 e

2g R

r

2m

Which is actually quite interesting since it shows thatirrespective of the orbital radius the KE is numericallyequal to half the PE Also the total energy E tot K sat Vsat+ 1

2---G Me m

r----------------times G Me m

r----------------ndash + 1

2---G Me m

r----------------ndash= = =of the satelliteisalwaysnegativesince

E tot K sat Vsat+ 12---

G Me mr----------------times

G Me mr----------------ndash

+ 12---

G Me mr----------------ndash= = =

PHYSICS 2015indb 320 150514 338 PM

Fields

321

AH

L

The energies of an orbiting satellite as a function of radial distancefromthecentreofaplanetareshownplottedinFigure1018

12

100806

0402

0ndash02

ndash04

ndash06ndash08ndash10

ndash12

2 4 6 8 10 12

ener

gya

rbita

ry u

nits

distance timesR

kinetic energytotal energy

potential energy

Figure 1018 Energy of an orbiting satellite as a function of distance from the centre of a planet

WeightlessnessSuppose that you are inanelevator(lift)thatisdescendingat constant speedandyou let goof abook thatyouareholdinginyourhandThebookwillfalltothefloorwithaccelerationequaltotheaccelerationduetogravityIfthecablethatsupportstheelevatorweretosnap(asituationthatItrustwillneverhappentoanyofyou)andyounowletgothebookthatyouareholdinginyourotherhandthis book will not fall to the floor - it will stay exactlyin linewithyourhandThis isbecause thebook isnowfallingwith the sameaccelerationas theelevatorandassuch the book cannot ldquocatchrdquo up with the floor of theelevatorFurthermoreifyouhappenedtobestandingonasetofbathroomscalesthescaleswouldnowreadzero-youwouldbeapparentlyweightlessItisthisideaoffreefallthatexplainstheapparentweightlessnessofastronautsin an orbiting satellite These astronauts are in free fall inthesensethattheyareacceleratingtowardsthecentre of the Earth

It is actually possible to define theweight of a body inseveral different waysWe can define it for example asthegravitationalforceexertedonthebodybyaspecifiedobject such as the EarthThiswe have seen thatwe doinlotsofsituationswherewedefinetheweightasbeingequaltomgIfweusethisdefinitionthenanobjectinfreefallcannotbydefinitionbeweightlesssinceitisstillinagravitationalfieldHoweverifwedefinetheweightofanobjectintermsofaldquoweighingrdquoprocesssuchasthereadingonasetofbathroomscaleswhichineffectmeasuresthecontact force between the object and the scales thenclearlyobjectsinfreefallareweightlessOnenowhasto

askthequestionwhetherornotitispossibleForexampletomeasurethegravitationalforceactingonanastronautin orbit about the Earth

WecanalsodefineweightintermsofthenetgravitationalforceactingonabodyduetoseveraldifferentobjectsForexample for an object out in space its weight could bedefined in terms of the resultant of the forces exerted on it bytheSuntheMoontheEarthandalltheotherplanetsin the Solar System If this resultant is zero at a particular pointthenthebodyisweightlessatthispoint

Inviewofthevariousdefinitionsofweightthatareavailabletousitisimportantthatwhenweusethewordldquoweightrdquoweareawareofthecontextinwhichitisbeingused

Example

Calculate the height above the surface of the Earth atwhichageo-stationarysatelliteorbits

Solution

A geo-stationary satellite is one that orbits the Earth in such a way that it is stationary with respect to a point on the surface of the Earth This means that its orbital period must be the same as the time for the Earth to spin once on its axis ie 24 hours

From Keplerrsquos third law we have G Ms

4π2----------- R3

T2------=

That is

RMe

mh

using the fact that the force of attraction between the satellite and the Earth is given by

FG Mem

R2----------------=

and that F = ma

where m is the mass of the satellite and a 4π2R

T2-------------=

we have

GMem

R2---------------- m 4π2R

T2-------------times

GMe

4π2------------rArr R3

T2------= =

PHYSICS 2015indb 321 150514 338 PM

Chapter 10

322

AH

L

Now the mass of the Earth is 60 times 1024 kg and the period T measured in seconds is given by T = 86400 s

So substitution gives R = 42 times 106 m

The radius of the Earth is 64 times 106 m so that the orbital height h is about 36 times 107 m

Example

Calculatetheminimumenergyrequiredtoputasatelliteofmass500kgintoanorbitthatisasaheightequaltotheEarthrsquosradiusabovethesurfaceoftheEarth

Solution

We have seen that when dealing with gravitational fields and potential it is useful to remember that

g 0G M

Re2---------= or g0 Re

2 GM=

The gravitational potential at the surface of the Earth is

g 0ndash Re GMRe

---------ndash=

The gravitational potential at a distance R

from the centre of the Earth is

=

The difference in potential between the surface of the Earth and a point distance R from the centre is therefore

V∆ g 0Re 1ReR------ndash

=

If R 2Re= then V∆g0 Re

2------------=

This means that the work required to ldquoliftrdquo the satellite into orbit is g0Rm where m is the mass of the satellite This is equal to

10 32 106 500timestimestimes = 16000 MJ

However the satellite must also have kinetic energy in order to orbit Earth This will be equal to

g 0 mRe2

2R-----------------g 0mRe

2

4----------------- 8000 MJ= =

The minimum energy required is therefore

24000 MJ

Exercise 1031 The speed needed to put a satellite in orbit does

not depend on

A the radius of the orbitB theshapeoftheorbitC thevalueofgattheorbitD the mass of the satellite

2 EstimatethespeedofanEarthsatellitewhoseorbitis400kmabovetheEarthrsquossurfaceAlsodetermine the period of the orbit

3 Calculatethespeedofa200kgsatelliteorbitingthe Earth at a height of 70 times 106 m

Assume that g = 82 m sndash2 for this orbit

4 TheradiioftwosatellitesXandYorbitingthe Earth are 2r and 8rwherer is the radius of the Earth Calculate the ratio of the periods of revolutionofXtoY

5 A satellite of mass mkgissentfromEarthrsquossurfaceinto an orbit of radius 5RwhereR is the radius of theEarthWritedownanexpressionfor

(a) thepotentialenergyofthesatelliteinorbit

(b) thekineticenergyofthesatelliteinorbit

(c) theminimumworkrequiredtosendthesatellitefromrestattheEarthrsquossurfaceintoits orbit

6 A satellite in an orbit of 10rfallsbacktoEarth(radius r)afteramalfunctionDeterminethespeedwithwhichitwillhittheEarthrsquossurface

7 The radius of the moon is frac14 that of the Earth AssumingEarthandtheMoontohavethesamedensitycomparetheaccelerationsoffreefallatthe surface of Earth to that at the surface of the Moon

8 UsethefollowingdatatodeterminethegravitationalfieldstrengthatthesurfaceoftheMoon and hence determine the escape speed from the surface of the Moon

Mass of the Moon = 73 times 1022kg

RadiusoftheMoon=17times106 m

PHYSICS 2015indb 322 150514 338 PM

10 Fields

Contents

101 ndash Describing fields

102 ndash Fields at work

Essential Ideas

Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Similar approaches can be taken in analysing electrical and gravitational potential problems copy IBO 2014

PHYSICS 2015indb 307 150514 337 PM

Chapter 10

308

AH

L

101 Describing fields

Essential idea Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Understandingsbull Gravitationalfieldsbull Electrostaticfieldsbull Electricpotentialandgravitationalpotentialbull Fieldlinesbull Equipotentialsurfaces

Gravitational potentialWehaveseenthatifweliftanobjectofmassm to a height h abovethesurfaceoftheEarththenitsgainingravitationalpotential energy is mghHoweverthisisbynomeansthefullstoryForastartwenowknowthatgvarieswithh and also the expression really gives a difference in potentialenergybetweenthevaluethattheobjecthasattheEarthrsquossurfaceandthevaluethatithasatheighthSowhatwereallyneedisazeropointCanwefindapointwherethepotentialenergyiszeroandusethispointfromwhichtomeasure changes in potential energy

The point that is chosen is in fact infinity At infinity the gravitationalfieldstrengthofanyobjectwillbezeroSoletusseeifwecandeduceanexpressionforthegaininpotentialenergyofanobjectwhenitisldquoliftedrdquofromthesurfaceoftheEarthtoinfinityThisineffectmeansfindingtheworknecessarytoperformthistask

δr

r r δr+

r infin=

mg

AB

Me

Figure 1001 Gravitational forces

Inthediagramweconsidertheworknecessarytomovethe particle of mass m a distance δr in thegravitationalfield of the Earth

The force on the particle at A is FG Mem

r2----------------=

IftheparticleismovedtoB then since δr isverysmallwe can assume that the field remains constant over thedistance ABTheworkδW done againstthegravitationalfieldoftheEarthinmovingthedistanceABis

δWGMem

r2----------------δrndash=

(rememberthatworkdoneagainstaforceisnegative)

TofindthetotalworkdoneW in going from the surface oftheEarthtoinfinitywehavetoaddalltheselittlebitsofworkThisisdonemathematicallybyusingintegralcalculus

WG Me m

r2----------------ndash

rdR

infin

int G Me m 1

r2----- rd

R

infin

intndash G Me m 1r---ndash

R

infinndash= = =

GMe m 0 1R---ndash

ndashndash=

G MemR----------------ndash=

HencewehavewhereRistheradiusoftheEarththattheworkdonebythegravitationalfieldinmovinganobjectofmass m from R(surfaceoftheEarth)toinfinityisgivenby

WGMe m

R---------------- ndash=

We can generalise the result by calculating the worknecessary per unit mass to take a small mass fromthe surface of the Earth to infinity This we call thegravitational potential Vie

Wewouldgetexactlythesameresultifwecalculatedtheworkdonebythefieldtobringthepointmassfrominfinityto the surface of Earth In this respect the formal definition ofgravitationalpotentialatapointinagravitationalfieldis therefore defined as the work done per unit mass in bringing a point mass from infinity to that point

Clearly then the gravitational potential at any point inthe Earthrsquos field distance r from the centre of the Earth (providingr gt R)is

VGMe

r------------ ndash=

Thepotentialisthereforeameasureoftheamountofworkthathastobedonetomoveparticlesbetweenpointsinagravitationalfieldanditsunit is theJkgndash1 We also note thatthepotentialisnegativesothatthepotentialenergy

NATURE OF SCIENCEParadigm shift The move from direct observable actions being responsible for influence on an object to acceptance of a fieldrsquos ldquoaction at a distancerdquo required a paradigm shift in the world of science (23)

copy IBO 2014

PHYSICS 2015indb 308 150514 337 PM

Fields

309

AH

L

aswemoveawayfromtheEarthrsquossurfaceincreasesuntilitreachesthevalueofzeroatinfinity

IfthegravitationalfieldisduetoapointmassofmassmthenwehavethesameexpressionasaboveexceptthatMe is replaced by m andmustalsoexclude thevalueof thepotential at the point mass itself ie at r = 0

WecanexpressthegravitationalpotentialduetotheEarth(orduetoanysphericalmass)intermsofthegravitationalfield strength at its surface

AtthesurfaceoftheEarthwehave

g 0ndash Re GMRe

---------ndash= e

Sothat

g 0Re2 GM= e

Hence at a distance r from the centre of the Earth the gravitationalpotentialVcanbewrittenas

VGMe

r------------ndashg 0 Re

2

r------------ndash= =

The potential at the surface of the Earth

(r = Re)istherefore-g0Re

It is interesting to see how the expression for thegravitational potential ties in with the expression mgh ThepotentialatthesurfaceoftheEarthis-g0Re (see the exampleabove)andataheighthwillbe-g0g 0 Re h+( )ndash ifweassume that g0doesnotchangeover thedistanceh The differenceinpotentialbetweenthesurfaceandtheheighth is therefore g0hSotheworkneededtoraiseanobjectof mass m to a height h is mgh ie m timesdifference ingravitationalpotential

This we have referred to as the gain in gravitationalpotentialenergy(see235)

However this expression can be extended to any twopoints inanygravitationalfieldsuch that ifanobjectofmass mmovesbetween twopointswhosepotentialsareV1 and V2 respectively then the change in gravitationalpotentialenergyoftheobjectism(V1 ndash V2)

Gravitational potential gradient

Let us consider now a region in space where thegravitationalfieldisconstantInFigure912thetwopointsAandBareseparatedbythedistance∆x

A B

direction of uniform gravitational eld of strength I

∆x

Figure 1002 The gravitational potential gradient

The gravitational field is of strength I and is in the direction shownThe gravitational potential at A is V andatBisV +∆V

Theworkdone is takingapointmassmfromAtoB isF∆x = mI∆x

Howeverbydefinitionthisworkisalsoequalto-m∆V

Therefore mI∆x=-m∆V

or VIx

∆= minus

∆Effectivelythissaysthatthemagnitudeofthegravitationalfield strength is equal to the negative gradient of thepotential If I is constant then V is a linear function of x and Iisequaltothenegativegradientofthestraightlinegraph formed by plotting V against x If I is not constant (asusuallythecase)thenthemagnitudeofI at any point in the field can be found by find the gradient of the V-x graph at that point An example of such a calculation can befoundinSection929

For those of you who do HL maths the relationship between field and potential is seen to follow from the expression for the potential of a point mass viz

mV Gr

= minus

2

ddV mG Ir r

minus = + =

Potential due to one or more point masses

Gravitationalpotentialisascalarquantitysocalculatingthepotential due to more than one point mass is a matter of simpleadditionSoforexamplethepotentialV due to the Moon and Earth and a distance x fromthecentreofEarthonastraightlinebetweenthemisgivenbytheexpression

E MM MV Gx r x

= minus + minus

whereME=massofEarthMM= mass of Moon and r = distancebetweencentreofEarthandMoon

PHYSICS 2015indb 309 150514 337 PM

Chapter 10

310

AH

L

Equipotentials and field linesIfthegravitationalpotentialhasthesamevalueatallpointson a surface the surface is said to be an equipotential surfaceSoforexampleifweimagineasphericalshellaboutEarthwhosecentrecoincideswiththecentreofEarththisshellwillbeanequipotentialsurfaceClearlyifwerepresentthegravitationalfieldstrengthbyfieldlinessincethelinesldquoradiaterdquooutfromthecentreofEarththentheselineswillbeatrightanglestothesurfaceIfthefieldlineswerenotnormaltotheequipotentialsurfacethentherewouldbeacomponentof the field parallel to the surfaceThis wouldmean thatpointsonthesurfacewouldbeatdifferentpotentialsandsoitwouldnolongerbeanequipotentialsurfaceThisofcourseholdstrueforanyequipotentialsurface

Figure 1003 shows the field lines and equipotentials fortwopointmassesm

m m

Figure 1003 Equipotentials for two point masses

It is worth noting that we would get exactly the samepattern ifwewere to replace thepointmasseswith twoequalpointcharges(See935)

Escape speedThe potential at the surface of Earth is which meansthattheenergyrequiredtotakeaparticleofmass mfromthesurfacetoinfinityisequalto

ButwhatdoesitactuallymeantotakesomethingtoinfinityWhentheparticleisonthesurfaceoftheEarthwecanthinkofitassittingatthebottomofaldquopotentialwellrdquoasinfigure1004

particle

surface of Earth

innity

GMR---------

FIgure 1004 A potential well

Theldquodepthrdquoofthewellis and if the particle gains an amount of kinetic energy equal to wherem is its massthenitwillhavejustenoughenergytoldquoliftrdquoitoutofthewell

In reality it doesnrsquot actually go to infinity it justmeansthat the particle is effectively free of the gravitationalattractionof theEarthWe say that ithas ldquoescapedrdquo theEarthrsquosgravitationalpullWemeetthisideainconnectionwithmolecular forcesTwomolecules in a solidwill sitat their equilibrium position the separation where therepulsiveforceisequaltotheattractiveforceIfwesupplyjust enough energy to increase the separation of themolecules such that they are an infinite distance apart then intermolecularforcesnolongeraffectthemoleculesandthesolidwillhavebecomealiquidThereisnoincreaseinthekineticenergyofthemoleculesandsothesolidmeltsat constant temperature

Wecancalculatetheescapespeedofanobjectveryeasilyby equating the kinetic energy to the potential energysuch that

12---mvesc ape

2 GMe mRe

----------------=

vesc aperArr2GMe

Re--------------- 2g0 Re = =

Substituting for g0 and Regivesavalueforvescape of about 11kmsndash1 from the surface of the Earth

Youwillnotethattheescapespeeddoesnotdependonthemassoftheobjectsincebothkineticenergyandpotentialenergy are proportional to the mass

IntheoryifyouwanttogetarockettothemoonitcanbedonewithoutreachingtheescapespeedHoweverthiswouldnecessitate anenormousamountof fuel and it islikely that therocketplus fuelwouldbesoheavythat itwouldnevergetoffthegroundItismuchmorepracticalto accelerate the rocket to the escape speed fromEarthorbitandthenintheoryjustlaunchittotheMoon

Example

Use the followingdata todetermine thepotential at thesurface of Mars and the magnitude of the acceleration of free fall

massofMars =64times1023kg

radiusofMars =34times106 m

Determinealsothegravitationalfieldstrengthatadistanceof 68 times 106mabovethesurfaceofMars

PHYSICS 2015indb 310 150514 337 PM

Fields

311

AH

L

Solution

= minus = minus times times = times23

11 76

64 1067 10 ndash13 10

34 10MV GR

minus timestimes

Nkg-1

But V = -g0R

Therefore 7

0 6

13 10 3834 10

VgR

times= minus = =

timesm s-2

To determine the field strength gh at 68 times 106 m above the surface we use the fact that 0 2

Mg GR

= such that GM = g0R2

Therefore 2 2

0h 2 2 2

h h

38 (34) 042(102)

g RGMgR R

times= = = = m s-2

(the distance from the centre is 34 times 106 + 68 times 106 = 102 times 106 m)

Exercise 1011 Thegraphbelowshowshowthegravitational

potentialoutsideoftheEarthvarieswithdistancefrom the centre

0 10 20 30 40 50 60 70

ndash1

ndash2

ndash3

ndash4

ndash5

ndash6

ndash7

r m times 106

V

Jkgndash1

times 10

7

(a) Usethegraphtodeterminethegainingravitationalpotentialenergyofasatelliteofmass200kgasitmovesfromthesurfaceof the Earth to a height of 30 times 107mabovetheEarthrsquossurface

(b) Calculatetheenergyrequiredtotakeit to infinity

(c) DeterminetheslopeofthegraphatthesurfaceoftheEarthmCommenton youranswer

Electric potential energy

The concept of electric potential energy was developedwith gravitational potential energy in mind Just as anobjectnearthesurfaceoftheEarthhaspotentialenergybecause of its gravitational interaction with the Earthsotoothereiselectricalpotentialenergyassociatedwithinteracting charges

Letusfirstlookatacaseoftwopositivepointchargeseachof 1μC that are initially bound together by a thread in a vacuuminspacewithadistancebetweenthemof10cmasshowninFigure1005Whenthethreadiscutthepointchargesinitiallyatrestwouldmoveinoppositedirectionsmovingwithvelocitiesv1 and v2 along the direction of the electrostatic force of repulsion

BEFORE AFTER

v v21

Figure 1005 Interaction of two positive particles

The electric potential energybetweentwopointchargescan be found by simply adding up the energy associated witheachpairofpointchargesForapairofinteractingchargestheelectricpotentialenergyisgivenby

∆U = ∆Ep + ∆Ek = ∆W = ∆Fr = kqQ

____ r2 times r = kqQ

____ r

Becausenoexternalforceisactingonthesystemtheenergyand momentum must be conserved Initially Ek = 0 and Ep =kqQr=9times109 times 1 times 10-1201m=009JWhentheyareagreatdistance fromeachotherEpwillbenegligibleThefinalenergywillbeequaltofrac12mv1

2+frac12mv22=009J

Momentum isalsoconservedand thevelocitieswouldbethe same magnitude but in opposite directions

Electric potential energyismoreoftendefinedintermsofapointchargemovinginanelectricfieldas

lsquothe electricpotential energybetweenany twopoints inan electric field is defined as negativeoftheworkdone by anelectricfieldinmovingapoint electric chargebetweentwolocationsintheelectricfieldrsquo

ΔU = ΔEp = -ΔW = -Fd = qEx

ΔU = ΔEp +ΔEk = ΔW = ΔFr = kqQ r2 x r = kqQ

wherexisthedistancemovedalong(oroppositeto)thedirection of the electric field

Electric potential energyismeasuredinjoule(J)Justasworkisascalarquantitysotooelectricalpotentialenergyis a scalar quantity The negative of the work done by an electric field in moving a unit electric charge between two points is independent of the path taken Inphysicswesaytheelectricfieldisalsquoconservativersquo field

PHYSICS 2015indb 311 150514 337 PM

Chapter 10

312

AH

L

Supposeanexternalforcesuchasyourhandmovesasmallpositive point test charge in the direction of a uniformelectric fieldAs it ismoving itmust be gaining kineticenergyIfthisoccursthentheelectricpotentialenergyofthe unit charge is changing

InFigure1006apointcharge+qismovedbetweenpointsAandBthroughadistancex in a uniform electric field

B

A

x

+ q

Figure 1006 Movement of a positive point charge in a uniform field

InordertomoveapositivepointchargefrompointAtopointBanexternal forcemustbeapplied to thechargeequaltoqE (F = qE)

Since the force is applied through a distance xthennegativeworkhas to bedone tomove the charge because energyis gainedmeaning there is an increaseelectric potential energybetweenthetwopointsRememberthattheworkdoneisequivalenttotheenergygainedorlostinmovingthe charge through the electric field The concept of electric potential energy is only meaningful as the electric field whichgeneratestheforceinquestionisconservative

W F xtimes E q x times= =

θ x cos θ x

Figure 1007 Charge moved at an angle to the field

If a chargemoves at an angle θ to an electric field thecomponent of the displacement parallel to the electric fieldisusedasshowninFigure918

W F x Eq x θcostimes= =

Theelectricpotentialenergyisstoredintheelectricfieldand theelectricfieldwill return theenergy to thepointcharge when required so as not to violate the Law ofconservationofenergy

Electric potential

The electric potential at a point in an electric field is defined as being the work done per unit charge in bringing a small positive point charge from infinity to that point

ΔV = Vinfin ndash Vf = ndash W ___ q

Ifwedesignatethepotentialenergytobezeroatinfinitythen it follows thatelectricpotentialmustalsobezeroat infinity and the electric potential at any point in an electricfieldwillbe

ΔV = ndash W ___ q

NowsupposeweapplyanexternalforcetoasmallpositivetestchargeasitismovedtowardsanisolatedpositivechargeTheexternalforcemustdoworkonthepositivetestchargetomoveittowardstheisolatedpositivechargeandtheworkmustbepositivewhiletheworkdonebytheelectricfieldmustthereforebenegativeSotheelectricpotentialatthatpointmustbepositiveaccordingtotheaboveequationIfanegativeisolatedchargeisusedtheelectricpotentialatapointonthepositivetestchargewouldbenegativePositivepoint chargesof their ownaccordmove fromaplaceofhighelectricpotential toaplaceof lowelectricpotentialNegative point charges move the other way from lowpotentialtohighpotentialInmovingfrompointAtopointBinthediagramthepositivecharge+qismovingfromalowelectricpotentialtoahighelectricpotential

In thedefinitiongiven the termldquoworkperunitchargerdquohas significance If the test charge is +16 times 10-19Cwherethe charge has a potential energy of 32 times 10-17Jthenthepotentialwouldbe32times10-17J+16times10-19C=200JC-1 Now if the charge was doubled the potential wouldbecome 64times10-17 J However the potential per unitchargewouldbethesame

Electric potential is a scalar quantityandithasunitsJC-1 orvoltswhere1voltequalsone joulepercoloumbThevoltallowsustoadoptaunitfortheelectricfieldintermsofthevolt

PreviouslytheunitfortheelectricfieldwasNCndash1

W = qV and F = qEsoW ___ V = F __ E

E = FV ___ W = FV ___ Fm V mndash1

That is the units of the electric field E can also beexpressed as V mndash1

The work done per unit charge in moving a point charge between two points in an electric field is again independent of the path taken

PHYSICS 2015indb 312 150514 337 PM

Fields

313

AH

L

Electric potential due to a point charge

Letus takeapoint rmetres froma chargedobjectThepotentialatthispointcanbecalculatedusingthefollowing

W = ndashFr = ndashqV and F = ndash q1 q2 _____ 4πε0r

2

Therefore

W = ndash q1 q2 _____ 4πε0r

2 times r = ndash q1 q2 _____ 4πε0r

= ndashq1 times q2 _____ 4πε0r

= ndashq1V

That is V q

4πε0r--------------- =

Orsimply

V kqr------

=

Example

Determinehowmuchworkisdonebytheelectricfieldofpointcharge150μCwhenachargeof200μCismovedfrom infinity to apoint 0400m from thepoint charge(Assumenoaccelerationofthecharges)

Solution

The work done by the electric field is W = -qV = -14πε0 times q times (Q rinfin - Q r0400)

W = (- 200 times 10-6 C times 900 times 109 NmC-2 times 150 times 10-6 C) divide 0400 m = - 0675 J

An external force would have to do +0675 J of work

Electric field strength and electric potential gradient

Letus lookbackatFigure1006Supposeagain that thecharge+qismovedasmalldistancebyaforceFfromAtoBsothattheforcecanbeconsideredconstantTheworkdoneisgivenby

∆W F x∆times=

TheforceFandtheelectricfieldEareoppositelydirectedandweknowthat

F = -qE and ΔW = q ΔV

Thereforetheworkdonecanbegivenas

q ΔV = -q E Δx

Therefore E V∆x∆-------

ndash=

The rate of change of potential ΔV at a point with respect to distance Δx in the direction in which the change is maximum is called the potential gradient We say that the electric field = - the potential gradient and the units are Vm-

1 From the equation we can see that in a graph of electric potential versus distance the gradient of the straight line equals the electric field strength

In reality if a charged particle enters a uniform electric field it will be accelerated uniformly by the field and its kinetic energy will increase This is why we had to assume no acceleration in the last worked example

Ek∆ 12---

mv2 q E xsdot sdot q Vx---

xsdot sdot q Vsdot= = = =

Example

Determine how far apart two parallel plates must besituated so that a potential difference of 150 times 102 V produces an electric field strength of 100 times 103 NC-1

Solution

Using E V∆x------- xhArrndash V∆

E-------15 102 Vtimes

100 103 N C 1ndashtimes------------------------------------------= = =

= 150 times 10-1

The plates are 150 times 10-1 m apart

The electric field and the electric potential at a point due to an evenly distributed charge +q on a sphere can be represented graphically as in Figure 1008

r

r

V

E

r0

r0

r0

Charge of +Q evenly distributed over surface

E 14πε0------------ Q

r2----- r r0gt=

V 14πε0------------Q

r----r r0gt=

Figure 1008 Electric field and potential due to a charged sphere

PHYSICS 2015indb 313 150514 337 PM

Chapter 10

314

AH

L

When the sphere becomes charged we know that thechargedistributes itselfevenlyoverthesurfaceThereforeeverypartofthematerialoftheconductorisatthesamepotential As the electric potential at a point is defined as beingnumericallyequaltotheworkdoneinbringingaunitpositivechargefrominfinitytothatpointithasaconstantvalueineverypartofthematerialoftheconductor

Since the potential is the same at all points on the conducting surfacethen∆V∆xiszeroButE=ndash∆V∆xThereforethe electric field inside the conductor is zero There is no electric field inside the conductor

SomefurtherobservationsofthegraphsinFigure1009are

bull OutsidethespherethegraphsobeytherelationshipsgivenasE α 1 r2 and V α 1 r

bull Atthesurfacer = r0Thereforetheelectricfieldandpotentialhavetheminimumvalueforr at this point and this infers a maximum field and potential

bull Insidethespheretheelectricfieldiszerobull Insidethespherenoworkisdonetomoveacharge

fromapointinsidetothesurfaceThereforethereisno potential difference and the potential is the same asitiswhenr = r0

Similargraphscanbedrawnfortheelectricfieldintensityandthe electric potential as a function of distance from conducting parallelplatesandsurfacesandthesearegiveninFigure1009

Potential plot E eld plotE eld

++++

ndashndashndashndash

xx x

+

x x

ndashx x

+ ndashx

x

+ +

x

x

Figure 1009 Electric field and electric potential at a distance from a charged surface

Potential due to one or more point charges

The potential due to one point charge can be determined byusingtheequationformula

V = kq r

Example 1

Determine the electric potential at a point 20 times 10-1 m fromthecentreofan isolatedconducting spherewithapointchargeof40pCinair

Solution

Using the formula

V = kq r we have

V90 10 9times( ) 40 10 12ndashtimes( )times

20 10 1ndashtimes( )------------------------------------------------------------------ 018 V= =

the potential at the point is 180 times 10-1 V

The potential due to a number of point charges can be determined by adding up the potentials due to individual point charges because the electric potential at any point outside a conducting sphere will be the same as if all the charge was concentrated at its centre

Example 2

Threepointchargesofareplacedattheverticesofaright-angledtriangleasshowninthediagrambelowDeterminetheabsolutepotentialat the+20μCchargeduetothetwoothercharges

- 6μC

+3μC +2μC

4 m

3 m

Solution

The electric potential of the +2 μC charge due to the ndash 6 μC charge is

V = (9 times 109 Nm2C-2 times -6 times 10-6 C) divide (radic 32 + 42) m = - 108 times 104 V

PHYSICS 2015indb 314 150514 338 PM

Fields

315

AH

L

Insummarywecanconcludethat

bull No work is done to move a charge along an equipotential

bull Equipotentials are always perpendicular to the electric lines of force

Figure1011and1012showsomeequipotential lines fortwo oppositely charged and identically positive spheresseparated by a distance

+ve ndashve

equipotential lines

Figure 1011 Equipotential lines between two opposite charges

equipotential lines

Figure 1012 Equipotential lines between two charges which are the same

+

ndash ndash ndash ndash

+ + +

40 V

30 V

20 V

10 V

50 V

Figure 1013 Equipotential lines between charged parallel plates

Figure1013showstheequipotentiallinesbetweenchargedparallel plates Throughout this chapter the similarities and differences between gravitational fields and electric fieldshavebeendiscussedTherelationshipsthatexistsbetweengravitationalandelectricquantitiesandtheeffectsofpointmassesandchargesissummarisedinTable1014

The electric potential of the +2 μC charge due to the +3 μC charge is

V = (9 times 109 Nm2C-2 times 3 times 10-6 C) divide 3m = 9 times 103 V

The net absolute potential is the sum of the 2 potentials- 108 times 104 V + 9 times 103 V =- 18 times 103 V

The absolute potential at the point is - 18 times 103 V

Equipotential surfacesRegions in space where the electric potential of a chargedistributionhasaconstantvaluearecalledequipotentials The placeswherethepotentialisconstantinthreedimensionsarecalled equipotential surfacesandwheretheyareconstantintwodimensionstheyarecalledequipotential lines

TheyareinsomewaysanalogoustothecontourlinesontopographicmapsInthiscasethegravitationalpotentialenergy is constant as amassmoves around the contourlines because the mass remains at the same elevationabovetheEarthrsquossurfaceThegravitationalfieldstrengthacts in a direction perpendicular to a contour line

Similarlybecausetheelectricpotentialonanequipotentiallinehas thesamevalueanelectric forcecandonoworkwhena testchargemovesonanequipotentialThereforethe electric field cannot have a component along anequipotentialandthusitmustbeeverywhereperpendiculartotheequipotentialsurfaceorequipotentiallineThisfactmakesiteasytoplotequipotentialsifthelinesofforceorlinesofelectricfluxofanelectricfieldareknown

Forexamplethereareaseriesofequipotentiallinesbetweentwoparallelplateconductorsthatareperpendiculartotheelectric fieldTherewill be a series of concentric circles(eachcirclefurtherapartthanthepreviousone)thatmapouttheequipotentialsaroundanisolatedpositivesphereThe lines of force and some equipotential lines for anisolatedpositivesphereareshowninFigure1010

Lines ofequipotential

Figure 1010 Equipotentials around an isolated positive sphere

PHYSICS 2015indb 315 150514 338 PM

Chapter 10

316

AH

L

Gravitational quantity Electrical quantityQ

uant

ities

g V∆x∆-------ndash= E V∆

x∆-------ndash=

Poin

t mas

ses a

nd ch

arge

s

V Gmr----

ndash= V 14πε0------------q

r---=

g G m

r2-----ndash= E 14πε0------------ q

r2-----=

F Gm1 m2

r2--------------= F 1

4πε0------------

q1q2

r2-----------=

Figure 1014 Formulas

Example

Deduce the electric potential on the surface of a gold nucleus that has a radius of 62 fm

Solution

Using the formula

V = kq r and knowing the atomic number of gold is 79 We will assume the nucleus is spherical and it behaves as if it were a point charge at its centre (relative to outside points)

V = 90 times 109 Nm2C-2 times 79 times 16 times 10-19 C divide 62 times 10-15 m = 18 times 107 V

The potential at the point is 18 MV

Example

Deduce the ionisation energy in electron-volts of theelectron in the hydrogen atom if the electron is in its ground state and it is in a circular orbit at a distance of 53 times 10-11 m from the proton

Solution

This problem is an energy coulombic circular motion question based on Bohrrsquos model of the atom (not the accepted quantum mechanics model) The ionisation energy is the energy required to remove the electron from the ground state to infinity The electron travels in a circular orbit and therefore has a centripetal acceleration The ionisation energy will counteract the coulombic force and the movement of the electron will be in the opposite direction to the centripetal force

Total energy = Ek electron + Ep due to the proton-electron interaction

ΣF = kqQ r2 = mv2 r and as such mv2 = = kqQ r

Therefore Ek electron = frac12 kqQ r

Ep due to the proton-electron interaction = - kqQ r

Total energy = frac12 kqQ r + - kqQ r = -frac12 kqQ r

= - 90 times 109 Nm2C-2 times (16 times 10-19 C)2 divide 53 times 10-11 m = -217 times 10-18 J

= -217 times 10-18 J divide 16 times 10-19 = -136 eV

The ionisation energy is 136 eV

Exercise 1021 ApointchargePisplacedmidwaybetweentwo

identicalnegativechargesWhichoneofthefollowingiscorrectwithregardstoelectricfieldandelectricpotentialatpointP

Electric field Electric potentialA non-zero zeroB zero non-zeroC non-zero non-zeroD zero zero

2 Twopositivechargedspheresaretiedtogetherinavacuumsomewhereinspacewheretherearenoexternal forces A has a mass of 25 g and a charge of20μCandBhasamassof15gandachargeof30μCThedistancebetweenthemis40cmTheyarethenreleasedasshowninthediagram

BEFORE AFTER

v1 v2

A B

(a) Determinetheirinitialelectricpotentialenergy in the before situation

(b) DeterminethespeedofsphereBafterrelease

3 Thediagrambelowrepresentstwoequipotentiallines in separated by a distance of 5 cm in a uniform electric field

+ + + + + + + +

ndash ndash ndash ndash ndash ndash ndash ndash

40 V

20 V 5 cm

PHYSICS 2015indb 316 150514 338 PM

Fields

317

AH

L

Determine the strength of the electric field

4 Thisquestionisabouttheelectricfieldduetoacharged sphere and the motion of electrons in that fieldThediagrambelowshowsanisolatedmetalsphereinavacuumthatcarriesanegativeelectriccharge of 60 μC

ndash

(a) Onthediagramdrawtheconventionalwayto represent the electric field pattern due to the charged sphere and lines to represent threeequipotentialsurfacesintheregionoutside the sphere

(b) Explainhowthelinesrepresentingtheequipotentialsurfacesthatyouhavesketchedindicatethatthestrengthoftheelectricfieldisdecreasingwithdistancefrom the centre of the sphere

(c) Theelectricfieldstrengthatthesurfaceofthe sphere and at points outside the sphere can be determined by assuming that the sphere acts as a point charge of magnitude 60 μC at its centre The radius of the sphere is 25 times 10ndash2 m Deduce that the magnitude of the field strength at the surface of the sphere is 86 times 107 Vmndash1

An electron is initially at rest on the surface of the sphere

(d) (i) Describethepathfollowedbytheelectronasitleavesthesurfaceofthesphere

(ii) Calculatetheinitialaccelerationofthe electron

5 Determinetheamountofworkthatisdoneinmovingachargeof100nCthroughapotentialdifferenceof150times102 V

6 Three identical 20 μC conducting spheres are placedatthecornersofanequilateraltriangleofsides 25 cm The triangle has one apex C pointing upthepageand2baseanglesAandBDeterminetheabsolutepotentialatB

7 Determinehowfaraparttwoparallelplatesmustbesituatedsothatapotentialdifferenceof250 times 102 V produces an electric field strength of 200 times 103 NC-1

8 Thegapbetweentwoparallelplatesis10times10-3mandthereisapotentialdifferenceof10times104 V betweentheplatesCalculate

i theworkdonebyanelectroninmovingfrom one plate to the other

ii thespeedwithwhichtheelectronreachesthe second plate if released from rest

iii theelectricfieldintensitybetweentheplates

9 Anelectronguninapicturetubeisacceleratedby a potential 25 times 103VDeterminethekineticenergygainedbytheelectroninelectron-volts

10 Determine the electric potential 20 times 10-2 m from achargeof-10times10-5 C

11 Determinetheelectricpotentialatapointmid-waybetweenachargeofndash20pCandanotherof+5pConthelinejoiningtheircentresifthecharges are 10 cm apart

12 During a thunderstorm the electric potential differencebetweenacloudandthegroundis10 times 109 V Determine the magnitude of the change in electric potential energy of an electron thatmovesbetweenthesepointsinelectron-volts

13 A charge of 15 μC is placed in a uniform electric fieldoftwooppositelychargedparallelplateswithamagnitudeof14times103 NC-1

(a) Determinetheworkthatmustbedoneagainstthefieldtomovethepointchargeadistance of 55 cm

(b) Calculatethepotentialdifferencebetweenthe final and initial positions of the charge

(c) Determinethepotentialdifferencebetweenthe plates if their separation distance is 15 cm

14 Duringaflashoflightningthepotentialdifferencebetweenacloudandthegroundwas12 times 109 V and the amount of transferred charge was32C

(a) Determinethechangeinenergyofthetransferred charge

(b) Iftheenergyreleasedwasallusedtoacceleratea1tonnecardeduceitsfinalspeed

(c) Iftheenergyreleasedcouldbeusedtomelticeat0degCdeducetheamountoficethatcould be melted

PHYSICS 2015indb 317 150514 338 PM

Chapter 10

318

AH

L

15 SupposethatwhenanelectronmovedfromAtoBin the diagram along an electric field line that the electric field does 36 times 10-19Jofworkonit

Determinethedifferencesinelectricpotential

(a) VB ndash VA(b) VC ndash VA(c) VC ndash VB

16 DeterminethepotentialatpointPthatislocatedatthecentreofthesquareasshowninthediagrambelow

- 6μC

+3μC +2μC

1m

1m

5μC

P

102 Fields at work

Essential idea Similar approaches can be taken inanalysingelectricalandgravitationalpotentialproblems

Understandingsbull Potentialandpotentialenergybull Potentialgradientbull Potentialdifferencebull Escapespeedbull Orbitalmotionorbitalspeedandorbitalenergybull Forcesandinverse-squarelawbehaviour

Orbital motion orbital speed and orbital energy

Although orbital motion may be circular elliptical orparabolic this sub-topic only dealswith circular orbitsThis sub-topic is not fundamentally new physics butan application that synthesizes ideas from gravitationcircularmotiondynamicsandenergy

TheMoonorbitstheEarthandinthissenseitisoftenreferredtoasasatelliteoftheEarthBefore1957itwastheonlyEarthsatelliteHoweverin1957theRussianslaunchedthefirstmanmadesatelliteSputnik1SincethisdatemanymoresatelliteshavebeenlaunchedandtherearenowliterallythousandsofthemorbitingtheEarthSomeareusedtomonitortheweathersome used to enable people to find accurately their position on thesurfaceoftheEarthmanyareusedincommunicationsandnodoubtsomeareusedtospyonothercountriesFigure1015showshowinprincipleasatellitecanbeputintoorbit

Theperson(whosesizeisgreatlyexaggeratedwithrespectto Earth) standing on the surface on the Earth throwssome stonesThe greater the speed with which a stoneis thrown the further it will land from her The pathsfollowedbythethrownstonesareparabolasByastretchof the imaginationwecanvisualisea situation inwhichastoneis thrownwithsuchaspeedthatbecauseof thecurvatureoftheEarthitwillnotlandonthesurfaceoftheEarthbutgointoldquoorbitrdquo(Path4onFigure1015)

NATURE OF SCIENCECommunication of scientific explanations The ability to apply field theory to the unobservable (charges) and the massively scaled (motion of satellites) required scientists to develop new ways to investigate analyse and report findings to a general public used to scientific discoveries based on tangible and discernible evidence (51)

copy IBO 2014

PHYSICS 2015indb 318 150514 338 PM

Fields

319

AH

L

1

2

34Earth

Figure 1015 Throwing a stone into orbit

TheforcethatcausesthestonestofollowaparabolicpathandtofalltoEarthisgravityandsimilarlytheforcethatkeeps the stone in orbit is gravity For circular motionto occur we have seen that a force must act at rightanglestothevelocityofanobjectthatistheremustbeacentripetalforceHenceinthesituationwedescribeherethe centripetal force for circular orbital motion about the EarthisprovidedbygravitationalattractionoftheEarth

We can calculate the speedwithwhich a stonemustbethrowninordertoputitintoorbitjustabovethesurfaceof the Earth

If the stone has mass m and speed v thenwehave fromNewtonrsquos2ndlaw

mv2

RE---------- G

MEm

RE2------------=

whereRE is the radius of the Earth and ME is the mass of the Earth

Bearinginmindthatg 0 GME

RE2--------= then

v g RE 10 64 106timestimes 8 103times= = =

Thatisthestonemustbethrownat8times103m sndash1

Clearly we are not going to get a satellite into orbit soclosetothesurfaceoftheEarthMovingatthisspeedthefrictionduetoairresistancewouldmeltthesatellitebeforeithadtravelledacoupleofkilometresInrealitythereforeasatelliteisputintoorbitabouttheEarthbysendingitattachedtoarocketbeyondtheEarthrsquosatmosphereandthengivingitacomponentofvelocityperpendiculartoaradialvectorfromtheEarthSeeFigure1016

Earth Satellite carried by rocket to hereSatellite orbit

Tangential component of velocity

Figure 1016 Getting a satellite into orbit

Keplerrsquos third law(This work of Kepler and Newtonrsquos synthesis of the work is an excellent example of the scientific method and makes for a good TOK discussion)

In 1627 Johannes Kepler (1571-1630)publishedhis lawsof planetary motion The laws are empirical in natureandwere deduced from the observations of theDanishastronomer Tycho de Brahe (1546-1601)The third lawgives a relationship between the radius of orbit R of a planet and its period TofrevolutionabouttheSunThelawisexpressedmathematicallyas

2

3 constantTR

=

We shall nowuseNewtonrsquos LawofGravitation to showhowitisthattheplanetsmoveinaccordancewithKeplerrsquosthirdlaw

InessenceNewtonwasabletousehislawofgravitytopredict the motion of the planets since all he had to do was factor theF givenby this law intohis second law F = ma to find their accelerations and hence their future positions

InFigure1017theEarthisshownorbitingtheSunandthedistancebetweentheircentresisR

RSun

Earth

F es

F se

Figure 1017 Planets move according to Keplerrsquos third law

Fes is the force that the Earth exerts on the Sun and Fse is the force that the Sun exerts on the Earth The forces are equalandoppositeandtheSunandtheEarthwillactuallyorbitaboutacommoncentreHoweversince theSun issoverymuchmoremassivethantheEarththiscommoncentrewillbeclosetothecentreoftheSunandsowecanregard the Earth as orbiting about the centre of the Sun Theotherthingthatweshallassumeisthatwecanignorethe forces that the other planets exert on the Earth (This wouldnotbeawisethingtodoifyouwereplanningtosend a space ship to theMoon for example)We shallalso assume that we have followed Newtonrsquos exampleandindeedprovedthataspherewillactasapointmasssituated at the centre of the sphere

Kepler had postulated that the orbits of the planets areellipticalbut since theeccentricityof theEarthrsquosorbit issmallweshallassumeacircularorbit

PHYSICS 2015indb 319 150514 338 PM

Chapter 10

320

AH

L

TheaccelerationoftheEarthtowardstheSunisa = Rω2

whereω ω 2πT------=

Hencea R 2π

T------ times

2 4π2 RT 2------------- = =

But the acceleration is given by Newtonrsquos Second Law F = mawhereFisnowgivenbytheLawofGravitationSo in this situation

F maGMs Me

R2-------------------= = butwealsohavethat

a a R 2πT

------ times

2 4π2 RT 2------------- = = and m = Me so that

G MsMe

R2------------------- Me4π2R

T2-------------times

GMs

4π2-----------rArr R3

T2------= =

Butthequantity

isaconstantthathasthesamevalueforeachoftheplanetssowehaveforalltheplanetsnotjustEarththat

R3

T2------ k =

wherekisaconstantWhichisofcourseKeplerrsquosthirdlaw

ThisisindeedanamazingbreakthroughItisdifficulttorefutethe idea that all particles attract each other in accordance withtheLawofGravitationwhenthelawisabletoaccountfortheobservedmotionoftheplanetsabouttheSun

The gravitational effects of the planets upon each othershould produce perturbations in their orbits Such is the predictivepoweroftheUniversalGravitationalLawthatit enabled physicists to compute these perturbations The telescopehadbeeninventedin1608andbythemiddleofthe 18th Century had reached a degree of perfection in design that enabled astronomers to actually measure the orbital perturbations of the planets Their measurements werealways inagreementwith thepredictionsmadebyNewtonrsquoslawHoweverin1781anewplanetUranuswasdiscovered and the orbit of this planet did not fit withthe orbit predicted by Universal Gravitation Such wasthe physicistrsquos faith in theNewtonianmethod that theysuspectedthatthediscrepancywasduetothepresenceofayetundetectedplanetUsingtheLawofGravitationtheFrenchastronomerJLeverrier and the English astronomer J C Adamswereable tocalculate justhowmassive thisnewplanetmustbeandalsowhereitshouldbeIn1846the planetNeptunewas discovered justwhere they hadpredicted Ina similarwaydiscrepancies in theorbitofNeptune ledto thepredictionandsubsequentdiscoveryin1930oftheplanetPlutoNewtonrsquosLawofGravitation

had passed the ultimate test of any theory it is not only abletoexplainexistingdatabutalsotomakepredictions

Satellite energyWhenasatelliteisinorbitaboutaplanetitwillhavebothkineticenergyandgravitationalpotentialenergySupposewe consider a satellite ofmass m that is in an orbit of radius r about a planet of mass M

Thegravitationalpotentialduetotheplanetatdistancer from its centre is

ThegravitationalpotentialenergyofthesatelliteVsat

is therefore GMe m

r----------------ndash

ThatisVsat GMem

r----------------ndash=

Thegravitationalfieldstrengthatthesurfaceoftheplanetisgivenby

g 0GMe

Re2------------=

Hencewecanwrite

Vsat g0 Re

2

rndash=

m

ThekineticenergyofthesatelliteKsatisequaltofrac12mv2wherev is its orbital speed

Byequatingthegravitationalforceactingonthesatellitetoitscentripetalaccelerationwehave

G Mem

r2---------------- mv2

r---------- mv2hArr

GMemr----------------= =

Fromwhich12---

mv2 12---

GMemr----------------times=

= 0 e

2g R

r

2m

Which is actually quite interesting since it shows thatirrespective of the orbital radius the KE is numericallyequal to half the PE Also the total energy E tot K sat Vsat+ 1

2---G Me m

r----------------times G Me m

r----------------ndash + 1

2---G Me m

r----------------ndash= = =of the satelliteisalwaysnegativesince

E tot K sat Vsat+ 12---

G Me mr----------------times

G Me mr----------------ndash

+ 12---

G Me mr----------------ndash= = =

PHYSICS 2015indb 320 150514 338 PM

Fields

321

AH

L

The energies of an orbiting satellite as a function of radial distancefromthecentreofaplanetareshownplottedinFigure1018

12

100806

0402

0ndash02

ndash04

ndash06ndash08ndash10

ndash12

2 4 6 8 10 12

ener

gya

rbita

ry u

nits

distance timesR

kinetic energytotal energy

potential energy

Figure 1018 Energy of an orbiting satellite as a function of distance from the centre of a planet

WeightlessnessSuppose that you are inanelevator(lift)thatisdescendingat constant speedandyou let goof abook thatyouareholdinginyourhandThebookwillfalltothefloorwithaccelerationequaltotheaccelerationduetogravityIfthecablethatsupportstheelevatorweretosnap(asituationthatItrustwillneverhappentoanyofyou)andyounowletgothebookthatyouareholdinginyourotherhandthis book will not fall to the floor - it will stay exactlyin linewithyourhandThis isbecause thebook isnowfallingwith the sameaccelerationas theelevatorandassuch the book cannot ldquocatchrdquo up with the floor of theelevatorFurthermoreifyouhappenedtobestandingonasetofbathroomscalesthescaleswouldnowreadzero-youwouldbeapparentlyweightlessItisthisideaoffreefallthatexplainstheapparentweightlessnessofastronautsin an orbiting satellite These astronauts are in free fall inthesensethattheyareacceleratingtowardsthecentre of the Earth

It is actually possible to define theweight of a body inseveral different waysWe can define it for example asthegravitationalforceexertedonthebodybyaspecifiedobject such as the EarthThiswe have seen thatwe doinlotsofsituationswherewedefinetheweightasbeingequaltomgIfweusethisdefinitionthenanobjectinfreefallcannotbydefinitionbeweightlesssinceitisstillinagravitationalfieldHoweverifwedefinetheweightofanobjectintermsofaldquoweighingrdquoprocesssuchasthereadingonasetofbathroomscaleswhichineffectmeasuresthecontact force between the object and the scales thenclearlyobjectsinfreefallareweightlessOnenowhasto

askthequestionwhetherornotitispossibleForexampletomeasurethegravitationalforceactingonanastronautin orbit about the Earth

WecanalsodefineweightintermsofthenetgravitationalforceactingonabodyduetoseveraldifferentobjectsForexample for an object out in space its weight could bedefined in terms of the resultant of the forces exerted on it bytheSuntheMoontheEarthandalltheotherplanetsin the Solar System If this resultant is zero at a particular pointthenthebodyisweightlessatthispoint

Inviewofthevariousdefinitionsofweightthatareavailabletousitisimportantthatwhenweusethewordldquoweightrdquoweareawareofthecontextinwhichitisbeingused

Example

Calculate the height above the surface of the Earth atwhichageo-stationarysatelliteorbits

Solution

A geo-stationary satellite is one that orbits the Earth in such a way that it is stationary with respect to a point on the surface of the Earth This means that its orbital period must be the same as the time for the Earth to spin once on its axis ie 24 hours

From Keplerrsquos third law we have G Ms

4π2----------- R3

T2------=

That is

RMe

mh

using the fact that the force of attraction between the satellite and the Earth is given by

FG Mem

R2----------------=

and that F = ma

where m is the mass of the satellite and a 4π2R

T2-------------=

we have

GMem

R2---------------- m 4π2R

T2-------------times

GMe

4π2------------rArr R3

T2------= =

PHYSICS 2015indb 321 150514 338 PM

Chapter 10

322

AH

L

Now the mass of the Earth is 60 times 1024 kg and the period T measured in seconds is given by T = 86400 s

So substitution gives R = 42 times 106 m

The radius of the Earth is 64 times 106 m so that the orbital height h is about 36 times 107 m

Example

Calculatetheminimumenergyrequiredtoputasatelliteofmass500kgintoanorbitthatisasaheightequaltotheEarthrsquosradiusabovethesurfaceoftheEarth

Solution

We have seen that when dealing with gravitational fields and potential it is useful to remember that

g 0G M

Re2---------= or g0 Re

2 GM=

The gravitational potential at the surface of the Earth is

g 0ndash Re GMRe

---------ndash=

The gravitational potential at a distance R

from the centre of the Earth is

=

The difference in potential between the surface of the Earth and a point distance R from the centre is therefore

V∆ g 0Re 1ReR------ndash

=

If R 2Re= then V∆g0 Re

2------------=

This means that the work required to ldquoliftrdquo the satellite into orbit is g0Rm where m is the mass of the satellite This is equal to

10 32 106 500timestimestimes = 16000 MJ

However the satellite must also have kinetic energy in order to orbit Earth This will be equal to

g 0 mRe2

2R-----------------g 0mRe

2

4----------------- 8000 MJ= =

The minimum energy required is therefore

24000 MJ

Exercise 1031 The speed needed to put a satellite in orbit does

not depend on

A the radius of the orbitB theshapeoftheorbitC thevalueofgattheorbitD the mass of the satellite

2 EstimatethespeedofanEarthsatellitewhoseorbitis400kmabovetheEarthrsquossurfaceAlsodetermine the period of the orbit

3 Calculatethespeedofa200kgsatelliteorbitingthe Earth at a height of 70 times 106 m

Assume that g = 82 m sndash2 for this orbit

4 TheradiioftwosatellitesXandYorbitingthe Earth are 2r and 8rwherer is the radius of the Earth Calculate the ratio of the periods of revolutionofXtoY

5 A satellite of mass mkgissentfromEarthrsquossurfaceinto an orbit of radius 5RwhereR is the radius of theEarthWritedownanexpressionfor

(a) thepotentialenergyofthesatelliteinorbit

(b) thekineticenergyofthesatelliteinorbit

(c) theminimumworkrequiredtosendthesatellitefromrestattheEarthrsquossurfaceintoits orbit

6 A satellite in an orbit of 10rfallsbacktoEarth(radius r)afteramalfunctionDeterminethespeedwithwhichitwillhittheEarthrsquossurface

7 The radius of the moon is frac14 that of the Earth AssumingEarthandtheMoontohavethesamedensitycomparetheaccelerationsoffreefallatthe surface of Earth to that at the surface of the Moon

8 UsethefollowingdatatodeterminethegravitationalfieldstrengthatthesurfaceoftheMoon and hence determine the escape speed from the surface of the Moon

Mass of the Moon = 73 times 1022kg

RadiusoftheMoon=17times106 m

PHYSICS 2015indb 322 150514 338 PM

Chapter 10

308

AH

L

101 Describing fields

Essential idea Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields

Understandingsbull Gravitationalfieldsbull Electrostaticfieldsbull Electricpotentialandgravitationalpotentialbull Fieldlinesbull Equipotentialsurfaces

Gravitational potentialWehaveseenthatifweliftanobjectofmassm to a height h abovethesurfaceoftheEarththenitsgainingravitationalpotential energy is mghHoweverthisisbynomeansthefullstoryForastartwenowknowthatgvarieswithh and also the expression really gives a difference in potentialenergybetweenthevaluethattheobjecthasattheEarthrsquossurfaceandthevaluethatithasatheighthSowhatwereallyneedisazeropointCanwefindapointwherethepotentialenergyiszeroandusethispointfromwhichtomeasure changes in potential energy

The point that is chosen is in fact infinity At infinity the gravitationalfieldstrengthofanyobjectwillbezeroSoletusseeifwecandeduceanexpressionforthegaininpotentialenergyofanobjectwhenitisldquoliftedrdquofromthesurfaceoftheEarthtoinfinityThisineffectmeansfindingtheworknecessarytoperformthistask

δr

r r δr+

r infin=

mg

AB

Me

Figure 1001 Gravitational forces

Inthediagramweconsidertheworknecessarytomovethe particle of mass m a distance δr in thegravitationalfield of the Earth

The force on the particle at A is FG Mem

r2----------------=

IftheparticleismovedtoB then since δr isverysmallwe can assume that the field remains constant over thedistance ABTheworkδW done againstthegravitationalfieldoftheEarthinmovingthedistanceABis

δWGMem

r2----------------δrndash=

(rememberthatworkdoneagainstaforceisnegative)

TofindthetotalworkdoneW in going from the surface oftheEarthtoinfinitywehavetoaddalltheselittlebitsofworkThisisdonemathematicallybyusingintegralcalculus

WG Me m

r2----------------ndash

rdR

infin

int G Me m 1

r2----- rd

R

infin

intndash G Me m 1r---ndash

R

infinndash= = =

GMe m 0 1R---ndash

ndashndash=

G MemR----------------ndash=

HencewehavewhereRistheradiusoftheEarththattheworkdonebythegravitationalfieldinmovinganobjectofmass m from R(surfaceoftheEarth)toinfinityisgivenby

WGMe m

R---------------- ndash=

We can generalise the result by calculating the worknecessary per unit mass to take a small mass fromthe surface of the Earth to infinity This we call thegravitational potential Vie

Wewouldgetexactlythesameresultifwecalculatedtheworkdonebythefieldtobringthepointmassfrominfinityto the surface of Earth In this respect the formal definition ofgravitationalpotentialatapointinagravitationalfieldis therefore defined as the work done per unit mass in bringing a point mass from infinity to that point

Clearly then the gravitational potential at any point inthe Earthrsquos field distance r from the centre of the Earth (providingr gt R)is

VGMe

r------------ ndash=

Thepotentialisthereforeameasureoftheamountofworkthathastobedonetomoveparticlesbetweenpointsinagravitationalfieldanditsunit is theJkgndash1 We also note thatthepotentialisnegativesothatthepotentialenergy

NATURE OF SCIENCEParadigm shift The move from direct observable actions being responsible for influence on an object to acceptance of a fieldrsquos ldquoaction at a distancerdquo required a paradigm shift in the world of science (23)

copy IBO 2014

PHYSICS 2015indb 308 150514 337 PM

Fields

309

AH

L

aswemoveawayfromtheEarthrsquossurfaceincreasesuntilitreachesthevalueofzeroatinfinity

IfthegravitationalfieldisduetoapointmassofmassmthenwehavethesameexpressionasaboveexceptthatMe is replaced by m andmustalsoexclude thevalueof thepotential at the point mass itself ie at r = 0

WecanexpressthegravitationalpotentialduetotheEarth(orduetoanysphericalmass)intermsofthegravitationalfield strength at its surface

AtthesurfaceoftheEarthwehave

g 0ndash Re GMRe

---------ndash= e

Sothat

g 0Re2 GM= e

Hence at a distance r from the centre of the Earth the gravitationalpotentialVcanbewrittenas

VGMe

r------------ndashg 0 Re

2

r------------ndash= =

The potential at the surface of the Earth

(r = Re)istherefore-g0Re

It is interesting to see how the expression for thegravitational potential ties in with the expression mgh ThepotentialatthesurfaceoftheEarthis-g0Re (see the exampleabove)andataheighthwillbe-g0g 0 Re h+( )ndash ifweassume that g0doesnotchangeover thedistanceh The differenceinpotentialbetweenthesurfaceandtheheighth is therefore g0hSotheworkneededtoraiseanobjectof mass m to a height h is mgh ie m timesdifference ingravitationalpotential

This we have referred to as the gain in gravitationalpotentialenergy(see235)

However this expression can be extended to any twopoints inanygravitationalfieldsuch that ifanobjectofmass mmovesbetween twopointswhosepotentialsareV1 and V2 respectively then the change in gravitationalpotentialenergyoftheobjectism(V1 ndash V2)

Gravitational potential gradient

Let us consider now a region in space where thegravitationalfieldisconstantInFigure912thetwopointsAandBareseparatedbythedistance∆x

A B

direction of uniform gravitational eld of strength I

∆x

Figure 1002 The gravitational potential gradient

The gravitational field is of strength I and is in the direction shownThe gravitational potential at A is V andatBisV +∆V

Theworkdone is takingapointmassmfromAtoB isF∆x = mI∆x

Howeverbydefinitionthisworkisalsoequalto-m∆V

Therefore mI∆x=-m∆V

or VIx

∆= minus

∆Effectivelythissaysthatthemagnitudeofthegravitationalfield strength is equal to the negative gradient of thepotential If I is constant then V is a linear function of x and Iisequaltothenegativegradientofthestraightlinegraph formed by plotting V against x If I is not constant (asusuallythecase)thenthemagnitudeofI at any point in the field can be found by find the gradient of the V-x graph at that point An example of such a calculation can befoundinSection929

For those of you who do HL maths the relationship between field and potential is seen to follow from the expression for the potential of a point mass viz

mV Gr

= minus

2

ddV mG Ir r

minus = + =

Potential due to one or more point masses

Gravitationalpotentialisascalarquantitysocalculatingthepotential due to more than one point mass is a matter of simpleadditionSoforexamplethepotentialV due to the Moon and Earth and a distance x fromthecentreofEarthonastraightlinebetweenthemisgivenbytheexpression

E MM MV Gx r x

= minus + minus

whereME=massofEarthMM= mass of Moon and r = distancebetweencentreofEarthandMoon

PHYSICS 2015indb 309 150514 337 PM

Chapter 10

310

AH

L

Equipotentials and field linesIfthegravitationalpotentialhasthesamevalueatallpointson a surface the surface is said to be an equipotential surfaceSoforexampleifweimagineasphericalshellaboutEarthwhosecentrecoincideswiththecentreofEarththisshellwillbeanequipotentialsurfaceClearlyifwerepresentthegravitationalfieldstrengthbyfieldlinessincethelinesldquoradiaterdquooutfromthecentreofEarththentheselineswillbeatrightanglestothesurfaceIfthefieldlineswerenotnormaltotheequipotentialsurfacethentherewouldbeacomponentof the field parallel to the surfaceThis wouldmean thatpointsonthesurfacewouldbeatdifferentpotentialsandsoitwouldnolongerbeanequipotentialsurfaceThisofcourseholdstrueforanyequipotentialsurface

Figure 1003 shows the field lines and equipotentials fortwopointmassesm

m m

Figure 1003 Equipotentials for two point masses

It is worth noting that we would get exactly the samepattern ifwewere to replace thepointmasseswith twoequalpointcharges(See935)

Escape speedThe potential at the surface of Earth is which meansthattheenergyrequiredtotakeaparticleofmass mfromthesurfacetoinfinityisequalto

ButwhatdoesitactuallymeantotakesomethingtoinfinityWhentheparticleisonthesurfaceoftheEarthwecanthinkofitassittingatthebottomofaldquopotentialwellrdquoasinfigure1004

particle

surface of Earth

innity

GMR---------

FIgure 1004 A potential well

Theldquodepthrdquoofthewellis and if the particle gains an amount of kinetic energy equal to wherem is its massthenitwillhavejustenoughenergytoldquoliftrdquoitoutofthewell

In reality it doesnrsquot actually go to infinity it justmeansthat the particle is effectively free of the gravitationalattractionof theEarthWe say that ithas ldquoescapedrdquo theEarthrsquosgravitationalpullWemeetthisideainconnectionwithmolecular forcesTwomolecules in a solidwill sitat their equilibrium position the separation where therepulsiveforceisequaltotheattractiveforceIfwesupplyjust enough energy to increase the separation of themolecules such that they are an infinite distance apart then intermolecularforcesnolongeraffectthemoleculesandthesolidwillhavebecomealiquidThereisnoincreaseinthekineticenergyofthemoleculesandsothesolidmeltsat constant temperature

Wecancalculatetheescapespeedofanobjectveryeasilyby equating the kinetic energy to the potential energysuch that

12---mvesc ape

2 GMe mRe

----------------=

vesc aperArr2GMe

Re--------------- 2g0 Re = =

Substituting for g0 and Regivesavalueforvescape of about 11kmsndash1 from the surface of the Earth

Youwillnotethattheescapespeeddoesnotdependonthemassoftheobjectsincebothkineticenergyandpotentialenergy are proportional to the mass

IntheoryifyouwanttogetarockettothemoonitcanbedonewithoutreachingtheescapespeedHoweverthiswouldnecessitate anenormousamountof fuel and it islikely that therocketplus fuelwouldbesoheavythat itwouldnevergetoffthegroundItismuchmorepracticalto accelerate the rocket to the escape speed fromEarthorbitandthenintheoryjustlaunchittotheMoon

Example

Use the followingdata todetermine thepotential at thesurface of Mars and the magnitude of the acceleration of free fall

massofMars =64times1023kg

radiusofMars =34times106 m

Determinealsothegravitationalfieldstrengthatadistanceof 68 times 106mabovethesurfaceofMars

PHYSICS 2015indb 310 150514 337 PM

Fields

311

AH

L

Solution

= minus = minus times times = times23

11 76

64 1067 10 ndash13 10

34 10MV GR

minus timestimes

Nkg-1

But V = -g0R

Therefore 7

0 6

13 10 3834 10

VgR

times= minus = =

timesm s-2

To determine the field strength gh at 68 times 106 m above the surface we use the fact that 0 2

Mg GR

= such that GM = g0R2

Therefore 2 2

0h 2 2 2

h h

38 (34) 042(102)

g RGMgR R

times= = = = m s-2

(the distance from the centre is 34 times 106 + 68 times 106 = 102 times 106 m)

Exercise 1011 Thegraphbelowshowshowthegravitational

potentialoutsideoftheEarthvarieswithdistancefrom the centre

0 10 20 30 40 50 60 70

ndash1

ndash2

ndash3

ndash4

ndash5

ndash6

ndash7

r m times 106

V

Jkgndash1

times 10

7

(a) Usethegraphtodeterminethegainingravitationalpotentialenergyofasatelliteofmass200kgasitmovesfromthesurfaceof the Earth to a height of 30 times 107mabovetheEarthrsquossurface

(b) Calculatetheenergyrequiredtotakeit to infinity

(c) DeterminetheslopeofthegraphatthesurfaceoftheEarthmCommenton youranswer

Electric potential energy

The concept of electric potential energy was developedwith gravitational potential energy in mind Just as anobjectnearthesurfaceoftheEarthhaspotentialenergybecause of its gravitational interaction with the Earthsotoothereiselectricalpotentialenergyassociatedwithinteracting charges

Letusfirstlookatacaseoftwopositivepointchargeseachof 1μC that are initially bound together by a thread in a vacuuminspacewithadistancebetweenthemof10cmasshowninFigure1005Whenthethreadiscutthepointchargesinitiallyatrestwouldmoveinoppositedirectionsmovingwithvelocitiesv1 and v2 along the direction of the electrostatic force of repulsion

BEFORE AFTER

v v21

Figure 1005 Interaction of two positive particles

The electric potential energybetweentwopointchargescan be found by simply adding up the energy associated witheachpairofpointchargesForapairofinteractingchargestheelectricpotentialenergyisgivenby

∆U = ∆Ep + ∆Ek = ∆W = ∆Fr = kqQ

____ r2 times r = kqQ

____ r

Becausenoexternalforceisactingonthesystemtheenergyand momentum must be conserved Initially Ek = 0 and Ep =kqQr=9times109 times 1 times 10-1201m=009JWhentheyareagreatdistance fromeachotherEpwillbenegligibleThefinalenergywillbeequaltofrac12mv1

2+frac12mv22=009J

Momentum isalsoconservedand thevelocitieswouldbethe same magnitude but in opposite directions

Electric potential energyismoreoftendefinedintermsofapointchargemovinginanelectricfieldas

lsquothe electricpotential energybetweenany twopoints inan electric field is defined as negativeoftheworkdone by anelectricfieldinmovingapoint electric chargebetweentwolocationsintheelectricfieldrsquo

ΔU = ΔEp = -ΔW = -Fd = qEx

ΔU = ΔEp +ΔEk = ΔW = ΔFr = kqQ r2 x r = kqQ

wherexisthedistancemovedalong(oroppositeto)thedirection of the electric field

Electric potential energyismeasuredinjoule(J)Justasworkisascalarquantitysotooelectricalpotentialenergyis a scalar quantity The negative of the work done by an electric field in moving a unit electric charge between two points is independent of the path taken Inphysicswesaytheelectricfieldisalsquoconservativersquo field

PHYSICS 2015indb 311 150514 337 PM

Chapter 10

312

AH

L

Supposeanexternalforcesuchasyourhandmovesasmallpositive point test charge in the direction of a uniformelectric fieldAs it ismoving itmust be gaining kineticenergyIfthisoccursthentheelectricpotentialenergyofthe unit charge is changing

InFigure1006apointcharge+qismovedbetweenpointsAandBthroughadistancex in a uniform electric field

B

A

x

+ q

Figure 1006 Movement of a positive point charge in a uniform field

InordertomoveapositivepointchargefrompointAtopointBanexternal forcemustbeapplied to thechargeequaltoqE (F = qE)

Since the force is applied through a distance xthennegativeworkhas to bedone tomove the charge because energyis gainedmeaning there is an increaseelectric potential energybetweenthetwopointsRememberthattheworkdoneisequivalenttotheenergygainedorlostinmovingthe charge through the electric field The concept of electric potential energy is only meaningful as the electric field whichgeneratestheforceinquestionisconservative

W F xtimes E q x times= =

θ x cos θ x

Figure 1007 Charge moved at an angle to the field

If a chargemoves at an angle θ to an electric field thecomponent of the displacement parallel to the electric fieldisusedasshowninFigure918

W F x Eq x θcostimes= =

Theelectricpotentialenergyisstoredintheelectricfieldand theelectricfieldwill return theenergy to thepointcharge when required so as not to violate the Law ofconservationofenergy

Electric potential

The electric potential at a point in an electric field is defined as being the work done per unit charge in bringing a small positive point charge from infinity to that point

ΔV = Vinfin ndash Vf = ndash W ___ q

Ifwedesignatethepotentialenergytobezeroatinfinitythen it follows thatelectricpotentialmustalsobezeroat infinity and the electric potential at any point in an electricfieldwillbe

ΔV = ndash W ___ q

NowsupposeweapplyanexternalforcetoasmallpositivetestchargeasitismovedtowardsanisolatedpositivechargeTheexternalforcemustdoworkonthepositivetestchargetomoveittowardstheisolatedpositivechargeandtheworkmustbepositivewhiletheworkdonebytheelectricfieldmustthereforebenegativeSotheelectricpotentialatthatpointmustbepositiveaccordingtotheaboveequationIfanegativeisolatedchargeisusedtheelectricpotentialatapointonthepositivetestchargewouldbenegativePositivepoint chargesof their ownaccordmove fromaplaceofhighelectricpotential toaplaceof lowelectricpotentialNegative point charges move the other way from lowpotentialtohighpotentialInmovingfrompointAtopointBinthediagramthepositivecharge+qismovingfromalowelectricpotentialtoahighelectricpotential

In thedefinitiongiven the termldquoworkperunitchargerdquohas significance If the test charge is +16 times 10-19Cwherethe charge has a potential energy of 32 times 10-17Jthenthepotentialwouldbe32times10-17J+16times10-19C=200JC-1 Now if the charge was doubled the potential wouldbecome 64times10-17 J However the potential per unitchargewouldbethesame

Electric potential is a scalar quantityandithasunitsJC-1 orvoltswhere1voltequalsone joulepercoloumbThevoltallowsustoadoptaunitfortheelectricfieldintermsofthevolt

PreviouslytheunitfortheelectricfieldwasNCndash1

W = qV and F = qEsoW ___ V = F __ E

E = FV ___ W = FV ___ Fm V mndash1

That is the units of the electric field E can also beexpressed as V mndash1

The work done per unit charge in moving a point charge between two points in an electric field is again independent of the path taken

PHYSICS 2015indb 312 150514 337 PM

Fields

313

AH

L

Electric potential due to a point charge

Letus takeapoint rmetres froma chargedobjectThepotentialatthispointcanbecalculatedusingthefollowing

W = ndashFr = ndashqV and F = ndash q1 q2 _____ 4πε0r

2

Therefore

W = ndash q1 q2 _____ 4πε0r

2 times r = ndash q1 q2 _____ 4πε0r

= ndashq1 times q2 _____ 4πε0r

= ndashq1V

That is V q

4πε0r--------------- =

Orsimply

V kqr------

=

Example

Determinehowmuchworkisdonebytheelectricfieldofpointcharge150μCwhenachargeof200μCismovedfrom infinity to apoint 0400m from thepoint charge(Assumenoaccelerationofthecharges)

Solution

The work done by the electric field is W = -qV = -14πε0 times q times (Q rinfin - Q r0400)

W = (- 200 times 10-6 C times 900 times 109 NmC-2 times 150 times 10-6 C) divide 0400 m = - 0675 J

An external force would have to do +0675 J of work

Electric field strength and electric potential gradient

Letus lookbackatFigure1006Supposeagain that thecharge+qismovedasmalldistancebyaforceFfromAtoBsothattheforcecanbeconsideredconstantTheworkdoneisgivenby

∆W F x∆times=

TheforceFandtheelectricfieldEareoppositelydirectedandweknowthat

F = -qE and ΔW = q ΔV

Thereforetheworkdonecanbegivenas

q ΔV = -q E Δx

Therefore E V∆x∆-------

ndash=

The rate of change of potential ΔV at a point with respect to distance Δx in the direction in which the change is maximum is called the potential gradient We say that the electric field = - the potential gradient and the units are Vm-

1 From the equation we can see that in a graph of electric potential versus distance the gradient of the straight line equals the electric field strength

In reality if a charged particle enters a uniform electric field it will be accelerated uniformly by the field and its kinetic energy will increase This is why we had to assume no acceleration in the last worked example

Ek∆ 12---

mv2 q E xsdot sdot q Vx---

xsdot sdot q Vsdot= = = =

Example

Determine how far apart two parallel plates must besituated so that a potential difference of 150 times 102 V produces an electric field strength of 100 times 103 NC-1

Solution

Using E V∆x------- xhArrndash V∆

E-------15 102 Vtimes

100 103 N C 1ndashtimes------------------------------------------= = =

= 150 times 10-1

The plates are 150 times 10-1 m apart

The electric field and the electric potential at a point due to an evenly distributed charge +q on a sphere can be represented graphically as in Figure 1008

r

r

V

E

r0

r0

r0

Charge of +Q evenly distributed over surface

E 14πε0------------ Q

r2----- r r0gt=

V 14πε0------------Q

r----r r0gt=

Figure 1008 Electric field and potential due to a charged sphere

PHYSICS 2015indb 313 150514 337 PM

Chapter 10

314

AH

L

When the sphere becomes charged we know that thechargedistributes itselfevenlyoverthesurfaceThereforeeverypartofthematerialoftheconductorisatthesamepotential As the electric potential at a point is defined as beingnumericallyequaltotheworkdoneinbringingaunitpositivechargefrominfinitytothatpointithasaconstantvalueineverypartofthematerialoftheconductor

Since the potential is the same at all points on the conducting surfacethen∆V∆xiszeroButE=ndash∆V∆xThereforethe electric field inside the conductor is zero There is no electric field inside the conductor

SomefurtherobservationsofthegraphsinFigure1009are

bull OutsidethespherethegraphsobeytherelationshipsgivenasE α 1 r2 and V α 1 r

bull Atthesurfacer = r0Thereforetheelectricfieldandpotentialhavetheminimumvalueforr at this point and this infers a maximum field and potential

bull Insidethespheretheelectricfieldiszerobull Insidethespherenoworkisdonetomoveacharge

fromapointinsidetothesurfaceThereforethereisno potential difference and the potential is the same asitiswhenr = r0

Similargraphscanbedrawnfortheelectricfieldintensityandthe electric potential as a function of distance from conducting parallelplatesandsurfacesandthesearegiveninFigure1009

Potential plot E eld plotE eld

++++

ndashndashndashndash

xx x

+

x x

ndashx x

+ ndashx

x

+ +

x

x

Figure 1009 Electric field and electric potential at a distance from a charged surface

Potential due to one or more point charges

The potential due to one point charge can be determined byusingtheequationformula

V = kq r

Example 1

Determine the electric potential at a point 20 times 10-1 m fromthecentreofan isolatedconducting spherewithapointchargeof40pCinair

Solution

Using the formula

V = kq r we have

V90 10 9times( ) 40 10 12ndashtimes( )times

20 10 1ndashtimes( )------------------------------------------------------------------ 018 V= =

the potential at the point is 180 times 10-1 V

The potential due to a number of point charges can be determined by adding up the potentials due to individual point charges because the electric potential at any point outside a conducting sphere will be the same as if all the charge was concentrated at its centre

Example 2

Threepointchargesofareplacedattheverticesofaright-angledtriangleasshowninthediagrambelowDeterminetheabsolutepotentialat the+20μCchargeduetothetwoothercharges

- 6μC

+3μC +2μC

4 m

3 m

Solution

The electric potential of the +2 μC charge due to the ndash 6 μC charge is

V = (9 times 109 Nm2C-2 times -6 times 10-6 C) divide (radic 32 + 42) m = - 108 times 104 V

PHYSICS 2015indb 314 150514 338 PM

Fields

315

AH

L

Insummarywecanconcludethat

bull No work is done to move a charge along an equipotential

bull Equipotentials are always perpendicular to the electric lines of force

Figure1011and1012showsomeequipotential lines fortwo oppositely charged and identically positive spheresseparated by a distance

+ve ndashve

equipotential lines

Figure 1011 Equipotential lines between two opposite charges

equipotential lines

Figure 1012 Equipotential lines between two charges which are the same

+

ndash ndash ndash ndash

+ + +

40 V

30 V

20 V

10 V

50 V

Figure 1013 Equipotential lines between charged parallel plates

Figure1013showstheequipotentiallinesbetweenchargedparallel plates Throughout this chapter the similarities and differences between gravitational fields and electric fieldshavebeendiscussedTherelationshipsthatexistsbetweengravitationalandelectricquantitiesandtheeffectsofpointmassesandchargesissummarisedinTable1014

The electric potential of the +2 μC charge due to the +3 μC charge is

V = (9 times 109 Nm2C-2 times 3 times 10-6 C) divide 3m = 9 times 103 V

The net absolute potential is the sum of the 2 potentials- 108 times 104 V + 9 times 103 V =- 18 times 103 V

The absolute potential at the point is - 18 times 103 V

Equipotential surfacesRegions in space where the electric potential of a chargedistributionhasaconstantvaluearecalledequipotentials The placeswherethepotentialisconstantinthreedimensionsarecalled equipotential surfacesandwheretheyareconstantintwodimensionstheyarecalledequipotential lines

TheyareinsomewaysanalogoustothecontourlinesontopographicmapsInthiscasethegravitationalpotentialenergy is constant as amassmoves around the contourlines because the mass remains at the same elevationabovetheEarthrsquossurfaceThegravitationalfieldstrengthacts in a direction perpendicular to a contour line

Similarlybecausetheelectricpotentialonanequipotentiallinehas thesamevalueanelectric forcecandonoworkwhena testchargemovesonanequipotentialThereforethe electric field cannot have a component along anequipotentialandthusitmustbeeverywhereperpendiculartotheequipotentialsurfaceorequipotentiallineThisfactmakesiteasytoplotequipotentialsifthelinesofforceorlinesofelectricfluxofanelectricfieldareknown

Forexamplethereareaseriesofequipotentiallinesbetweentwoparallelplateconductorsthatareperpendiculartotheelectric fieldTherewill be a series of concentric circles(eachcirclefurtherapartthanthepreviousone)thatmapouttheequipotentialsaroundanisolatedpositivesphereThe lines of force and some equipotential lines for anisolatedpositivesphereareshowninFigure1010

Lines ofequipotential

Figure 1010 Equipotentials around an isolated positive sphere

PHYSICS 2015indb 315 150514 338 PM

Chapter 10

316

AH

L

Gravitational quantity Electrical quantityQ

uant

ities

g V∆x∆-------ndash= E V∆

x∆-------ndash=

Poin

t mas

ses a

nd ch

arge

s

V Gmr----

ndash= V 14πε0------------q

r---=

g G m

r2-----ndash= E 14πε0------------ q

r2-----=

F Gm1 m2

r2--------------= F 1

4πε0------------

q1q2

r2-----------=

Figure 1014 Formulas

Example

Deduce the electric potential on the surface of a gold nucleus that has a radius of 62 fm

Solution

Using the formula

V = kq r and knowing the atomic number of gold is 79 We will assume the nucleus is spherical and it behaves as if it were a point charge at its centre (relative to outside points)

V = 90 times 109 Nm2C-2 times 79 times 16 times 10-19 C divide 62 times 10-15 m = 18 times 107 V

The potential at the point is 18 MV

Example

Deduce the ionisation energy in electron-volts of theelectron in the hydrogen atom if the electron is in its ground state and it is in a circular orbit at a distance of 53 times 10-11 m from the proton

Solution

This problem is an energy coulombic circular motion question based on Bohrrsquos model of the atom (not the accepted quantum mechanics model) The ionisation energy is the energy required to remove the electron from the ground state to infinity The electron travels in a circular orbit and therefore has a centripetal acceleration The ionisation energy will counteract the coulombic force and the movement of the electron will be in the opposite direction to the centripetal force

Total energy = Ek electron + Ep due to the proton-electron interaction

ΣF = kqQ r2 = mv2 r and as such mv2 = = kqQ r

Therefore Ek electron = frac12 kqQ r

Ep due to the proton-electron interaction = - kqQ r

Total energy = frac12 kqQ r + - kqQ r = -frac12 kqQ r

= - 90 times 109 Nm2C-2 times (16 times 10-19 C)2 divide 53 times 10-11 m = -217 times 10-18 J

= -217 times 10-18 J divide 16 times 10-19 = -136 eV

The ionisation energy is 136 eV

Exercise 1021 ApointchargePisplacedmidwaybetweentwo

identicalnegativechargesWhichoneofthefollowingiscorrectwithregardstoelectricfieldandelectricpotentialatpointP

Electric field Electric potentialA non-zero zeroB zero non-zeroC non-zero non-zeroD zero zero

2 Twopositivechargedspheresaretiedtogetherinavacuumsomewhereinspacewheretherearenoexternal forces A has a mass of 25 g and a charge of20μCandBhasamassof15gandachargeof30μCThedistancebetweenthemis40cmTheyarethenreleasedasshowninthediagram

BEFORE AFTER

v1 v2

A B

(a) Determinetheirinitialelectricpotentialenergy in the before situation

(b) DeterminethespeedofsphereBafterrelease

3 Thediagrambelowrepresentstwoequipotentiallines in separated by a distance of 5 cm in a uniform electric field

+ + + + + + + +

ndash ndash ndash ndash ndash ndash ndash ndash

40 V

20 V 5 cm

PHYSICS 2015indb 316 150514 338 PM

Fields

317

AH

L

Determine the strength of the electric field

4 Thisquestionisabouttheelectricfieldduetoacharged sphere and the motion of electrons in that fieldThediagrambelowshowsanisolatedmetalsphereinavacuumthatcarriesanegativeelectriccharge of 60 μC

ndash

(a) Onthediagramdrawtheconventionalwayto represent the electric field pattern due to the charged sphere and lines to represent threeequipotentialsurfacesintheregionoutside the sphere

(b) Explainhowthelinesrepresentingtheequipotentialsurfacesthatyouhavesketchedindicatethatthestrengthoftheelectricfieldisdecreasingwithdistancefrom the centre of the sphere

(c) Theelectricfieldstrengthatthesurfaceofthe sphere and at points outside the sphere can be determined by assuming that the sphere acts as a point charge of magnitude 60 μC at its centre The radius of the sphere is 25 times 10ndash2 m Deduce that the magnitude of the field strength at the surface of the sphere is 86 times 107 Vmndash1

An electron is initially at rest on the surface of the sphere

(d) (i) Describethepathfollowedbytheelectronasitleavesthesurfaceofthesphere

(ii) Calculatetheinitialaccelerationofthe electron

5 Determinetheamountofworkthatisdoneinmovingachargeof100nCthroughapotentialdifferenceof150times102 V

6 Three identical 20 μC conducting spheres are placedatthecornersofanequilateraltriangleofsides 25 cm The triangle has one apex C pointing upthepageand2baseanglesAandBDeterminetheabsolutepotentialatB

7 Determinehowfaraparttwoparallelplatesmustbesituatedsothatapotentialdifferenceof250 times 102 V produces an electric field strength of 200 times 103 NC-1

8 Thegapbetweentwoparallelplatesis10times10-3mandthereisapotentialdifferenceof10times104 V betweentheplatesCalculate

i theworkdonebyanelectroninmovingfrom one plate to the other

ii thespeedwithwhichtheelectronreachesthe second plate if released from rest

iii theelectricfieldintensitybetweentheplates

9 Anelectronguninapicturetubeisacceleratedby a potential 25 times 103VDeterminethekineticenergygainedbytheelectroninelectron-volts

10 Determine the electric potential 20 times 10-2 m from achargeof-10times10-5 C

11 Determinetheelectricpotentialatapointmid-waybetweenachargeofndash20pCandanotherof+5pConthelinejoiningtheircentresifthecharges are 10 cm apart

12 During a thunderstorm the electric potential differencebetweenacloudandthegroundis10 times 109 V Determine the magnitude of the change in electric potential energy of an electron thatmovesbetweenthesepointsinelectron-volts

13 A charge of 15 μC is placed in a uniform electric fieldoftwooppositelychargedparallelplateswithamagnitudeof14times103 NC-1

(a) Determinetheworkthatmustbedoneagainstthefieldtomovethepointchargeadistance of 55 cm

(b) Calculatethepotentialdifferencebetweenthe final and initial positions of the charge

(c) Determinethepotentialdifferencebetweenthe plates if their separation distance is 15 cm

14 Duringaflashoflightningthepotentialdifferencebetweenacloudandthegroundwas12 times 109 V and the amount of transferred charge was32C

(a) Determinethechangeinenergyofthetransferred charge

(b) Iftheenergyreleasedwasallusedtoacceleratea1tonnecardeduceitsfinalspeed

(c) Iftheenergyreleasedcouldbeusedtomelticeat0degCdeducetheamountoficethatcould be melted

PHYSICS 2015indb 317 150514 338 PM

Chapter 10

318

AH

L

15 SupposethatwhenanelectronmovedfromAtoBin the diagram along an electric field line that the electric field does 36 times 10-19Jofworkonit

Determinethedifferencesinelectricpotential

(a) VB ndash VA(b) VC ndash VA(c) VC ndash VB

16 DeterminethepotentialatpointPthatislocatedatthecentreofthesquareasshowninthediagrambelow

- 6μC

+3μC +2μC

1m

1m

5μC

P

102 Fields at work

Essential idea Similar approaches can be taken inanalysingelectricalandgravitationalpotentialproblems

Understandingsbull Potentialandpotentialenergybull Potentialgradientbull Potentialdifferencebull Escapespeedbull Orbitalmotionorbitalspeedandorbitalenergybull Forcesandinverse-squarelawbehaviour

Orbital motion orbital speed and orbital energy

Although orbital motion may be circular elliptical orparabolic this sub-topic only dealswith circular orbitsThis sub-topic is not fundamentally new physics butan application that synthesizes ideas from gravitationcircularmotiondynamicsandenergy

TheMoonorbitstheEarthandinthissenseitisoftenreferredtoasasatelliteoftheEarthBefore1957itwastheonlyEarthsatelliteHoweverin1957theRussianslaunchedthefirstmanmadesatelliteSputnik1SincethisdatemanymoresatelliteshavebeenlaunchedandtherearenowliterallythousandsofthemorbitingtheEarthSomeareusedtomonitortheweathersome used to enable people to find accurately their position on thesurfaceoftheEarthmanyareusedincommunicationsandnodoubtsomeareusedtospyonothercountriesFigure1015showshowinprincipleasatellitecanbeputintoorbit

Theperson(whosesizeisgreatlyexaggeratedwithrespectto Earth) standing on the surface on the Earth throwssome stonesThe greater the speed with which a stoneis thrown the further it will land from her The pathsfollowedbythethrownstonesareparabolasByastretchof the imaginationwecanvisualisea situation inwhichastoneis thrownwithsuchaspeedthatbecauseof thecurvatureoftheEarthitwillnotlandonthesurfaceoftheEarthbutgointoldquoorbitrdquo(Path4onFigure1015)

NATURE OF SCIENCECommunication of scientific explanations The ability to apply field theory to the unobservable (charges) and the massively scaled (motion of satellites) required scientists to develop new ways to investigate analyse and report findings to a general public used to scientific discoveries based on tangible and discernible evidence (51)

copy IBO 2014

PHYSICS 2015indb 318 150514 338 PM

Fields

319

AH

L

1

2

34Earth

Figure 1015 Throwing a stone into orbit

TheforcethatcausesthestonestofollowaparabolicpathandtofalltoEarthisgravityandsimilarlytheforcethatkeeps the stone in orbit is gravity For circular motionto occur we have seen that a force must act at rightanglestothevelocityofanobjectthatistheremustbeacentripetalforceHenceinthesituationwedescribeherethe centripetal force for circular orbital motion about the EarthisprovidedbygravitationalattractionoftheEarth

We can calculate the speedwithwhich a stonemustbethrowninordertoputitintoorbitjustabovethesurfaceof the Earth

If the stone has mass m and speed v thenwehave fromNewtonrsquos2ndlaw

mv2

RE---------- G

MEm

RE2------------=

whereRE is the radius of the Earth and ME is the mass of the Earth

Bearinginmindthatg 0 GME

RE2--------= then

v g RE 10 64 106timestimes 8 103times= = =

Thatisthestonemustbethrownat8times103m sndash1

Clearly we are not going to get a satellite into orbit soclosetothesurfaceoftheEarthMovingatthisspeedthefrictionduetoairresistancewouldmeltthesatellitebeforeithadtravelledacoupleofkilometresInrealitythereforeasatelliteisputintoorbitabouttheEarthbysendingitattachedtoarocketbeyondtheEarthrsquosatmosphereandthengivingitacomponentofvelocityperpendiculartoaradialvectorfromtheEarthSeeFigure1016

Earth Satellite carried by rocket to hereSatellite orbit

Tangential component of velocity

Figure 1016 Getting a satellite into orbit

Keplerrsquos third law(This work of Kepler and Newtonrsquos synthesis of the work is an excellent example of the scientific method and makes for a good TOK discussion)

In 1627 Johannes Kepler (1571-1630)publishedhis lawsof planetary motion The laws are empirical in natureandwere deduced from the observations of theDanishastronomer Tycho de Brahe (1546-1601)The third lawgives a relationship between the radius of orbit R of a planet and its period TofrevolutionabouttheSunThelawisexpressedmathematicallyas

2

3 constantTR

=

We shall nowuseNewtonrsquos LawofGravitation to showhowitisthattheplanetsmoveinaccordancewithKeplerrsquosthirdlaw

InessenceNewtonwasabletousehislawofgravitytopredict the motion of the planets since all he had to do was factor theF givenby this law intohis second law F = ma to find their accelerations and hence their future positions

InFigure1017theEarthisshownorbitingtheSunandthedistancebetweentheircentresisR

RSun

Earth

F es

F se

Figure 1017 Planets move according to Keplerrsquos third law

Fes is the force that the Earth exerts on the Sun and Fse is the force that the Sun exerts on the Earth The forces are equalandoppositeandtheSunandtheEarthwillactuallyorbitaboutacommoncentreHoweversince theSun issoverymuchmoremassivethantheEarththiscommoncentrewillbeclosetothecentreoftheSunandsowecanregard the Earth as orbiting about the centre of the Sun Theotherthingthatweshallassumeisthatwecanignorethe forces that the other planets exert on the Earth (This wouldnotbeawisethingtodoifyouwereplanningtosend a space ship to theMoon for example)We shallalso assume that we have followed Newtonrsquos exampleandindeedprovedthataspherewillactasapointmasssituated at the centre of the sphere

Kepler had postulated that the orbits of the planets areellipticalbut since theeccentricityof theEarthrsquosorbit issmallweshallassumeacircularorbit

PHYSICS 2015indb 319 150514 338 PM

Chapter 10

320

AH

L

TheaccelerationoftheEarthtowardstheSunisa = Rω2

whereω ω 2πT------=

Hencea R 2π

T------ times

2 4π2 RT 2------------- = =

But the acceleration is given by Newtonrsquos Second Law F = mawhereFisnowgivenbytheLawofGravitationSo in this situation

F maGMs Me

R2-------------------= = butwealsohavethat

a a R 2πT

------ times

2 4π2 RT 2------------- = = and m = Me so that

G MsMe

R2------------------- Me4π2R

T2-------------times

GMs

4π2-----------rArr R3

T2------= =

Butthequantity

isaconstantthathasthesamevalueforeachoftheplanetssowehaveforalltheplanetsnotjustEarththat

R3

T2------ k =

wherekisaconstantWhichisofcourseKeplerrsquosthirdlaw

ThisisindeedanamazingbreakthroughItisdifficulttorefutethe idea that all particles attract each other in accordance withtheLawofGravitationwhenthelawisabletoaccountfortheobservedmotionoftheplanetsabouttheSun

The gravitational effects of the planets upon each othershould produce perturbations in their orbits Such is the predictivepoweroftheUniversalGravitationalLawthatit enabled physicists to compute these perturbations The telescopehadbeeninventedin1608andbythemiddleofthe 18th Century had reached a degree of perfection in design that enabled astronomers to actually measure the orbital perturbations of the planets Their measurements werealways inagreementwith thepredictionsmadebyNewtonrsquoslawHoweverin1781anewplanetUranuswasdiscovered and the orbit of this planet did not fit withthe orbit predicted by Universal Gravitation Such wasthe physicistrsquos faith in theNewtonianmethod that theysuspectedthatthediscrepancywasduetothepresenceofayetundetectedplanetUsingtheLawofGravitationtheFrenchastronomerJLeverrier and the English astronomer J C Adamswereable tocalculate justhowmassive thisnewplanetmustbeandalsowhereitshouldbeIn1846the planetNeptunewas discovered justwhere they hadpredicted Ina similarwaydiscrepancies in theorbitofNeptune ledto thepredictionandsubsequentdiscoveryin1930oftheplanetPlutoNewtonrsquosLawofGravitation

had passed the ultimate test of any theory it is not only abletoexplainexistingdatabutalsotomakepredictions

Satellite energyWhenasatelliteisinorbitaboutaplanetitwillhavebothkineticenergyandgravitationalpotentialenergySupposewe consider a satellite ofmass m that is in an orbit of radius r about a planet of mass M

Thegravitationalpotentialduetotheplanetatdistancer from its centre is

ThegravitationalpotentialenergyofthesatelliteVsat

is therefore GMe m

r----------------ndash

ThatisVsat GMem

r----------------ndash=

Thegravitationalfieldstrengthatthesurfaceoftheplanetisgivenby

g 0GMe

Re2------------=

Hencewecanwrite

Vsat g0 Re

2

rndash=

m

ThekineticenergyofthesatelliteKsatisequaltofrac12mv2wherev is its orbital speed

Byequatingthegravitationalforceactingonthesatellitetoitscentripetalaccelerationwehave

G Mem

r2---------------- mv2

r---------- mv2hArr

GMemr----------------= =

Fromwhich12---

mv2 12---

GMemr----------------times=

= 0 e

2g R

r

2m

Which is actually quite interesting since it shows thatirrespective of the orbital radius the KE is numericallyequal to half the PE Also the total energy E tot K sat Vsat+ 1

2---G Me m

r----------------times G Me m

r----------------ndash + 1

2---G Me m

r----------------ndash= = =of the satelliteisalwaysnegativesince

E tot K sat Vsat+ 12---

G Me mr----------------times

G Me mr----------------ndash

+ 12---

G Me mr----------------ndash= = =

PHYSICS 2015indb 320 150514 338 PM

Fields

321

AH

L

The energies of an orbiting satellite as a function of radial distancefromthecentreofaplanetareshownplottedinFigure1018

12

100806

0402

0ndash02

ndash04

ndash06ndash08ndash10

ndash12

2 4 6 8 10 12

ener

gya

rbita

ry u

nits

distance timesR

kinetic energytotal energy

potential energy

Figure 1018 Energy of an orbiting satellite as a function of distance from the centre of a planet

WeightlessnessSuppose that you are inanelevator(lift)thatisdescendingat constant speedandyou let goof abook thatyouareholdinginyourhandThebookwillfalltothefloorwithaccelerationequaltotheaccelerationduetogravityIfthecablethatsupportstheelevatorweretosnap(asituationthatItrustwillneverhappentoanyofyou)andyounowletgothebookthatyouareholdinginyourotherhandthis book will not fall to the floor - it will stay exactlyin linewithyourhandThis isbecause thebook isnowfallingwith the sameaccelerationas theelevatorandassuch the book cannot ldquocatchrdquo up with the floor of theelevatorFurthermoreifyouhappenedtobestandingonasetofbathroomscalesthescaleswouldnowreadzero-youwouldbeapparentlyweightlessItisthisideaoffreefallthatexplainstheapparentweightlessnessofastronautsin an orbiting satellite These astronauts are in free fall inthesensethattheyareacceleratingtowardsthecentre of the Earth

It is actually possible to define theweight of a body inseveral different waysWe can define it for example asthegravitationalforceexertedonthebodybyaspecifiedobject such as the EarthThiswe have seen thatwe doinlotsofsituationswherewedefinetheweightasbeingequaltomgIfweusethisdefinitionthenanobjectinfreefallcannotbydefinitionbeweightlesssinceitisstillinagravitationalfieldHoweverifwedefinetheweightofanobjectintermsofaldquoweighingrdquoprocesssuchasthereadingonasetofbathroomscaleswhichineffectmeasuresthecontact force between the object and the scales thenclearlyobjectsinfreefallareweightlessOnenowhasto

askthequestionwhetherornotitispossibleForexampletomeasurethegravitationalforceactingonanastronautin orbit about the Earth

WecanalsodefineweightintermsofthenetgravitationalforceactingonabodyduetoseveraldifferentobjectsForexample for an object out in space its weight could bedefined in terms of the resultant of the forces exerted on it bytheSuntheMoontheEarthandalltheotherplanetsin the Solar System If this resultant is zero at a particular pointthenthebodyisweightlessatthispoint

Inviewofthevariousdefinitionsofweightthatareavailabletousitisimportantthatwhenweusethewordldquoweightrdquoweareawareofthecontextinwhichitisbeingused

Example

Calculate the height above the surface of the Earth atwhichageo-stationarysatelliteorbits

Solution

A geo-stationary satellite is one that orbits the Earth in such a way that it is stationary with respect to a point on the surface of the Earth This means that its orbital period must be the same as the time for the Earth to spin once on its axis ie 24 hours

From Keplerrsquos third law we have G Ms

4π2----------- R3

T2------=

That is

RMe

mh

using the fact that the force of attraction between the satellite and the Earth is given by

FG Mem

R2----------------=

and that F = ma

where m is the mass of the satellite and a 4π2R

T2-------------=

we have

GMem

R2---------------- m 4π2R

T2-------------times

GMe

4π2------------rArr R3

T2------= =

PHYSICS 2015indb 321 150514 338 PM

Chapter 10

322

AH

L

Now the mass of the Earth is 60 times 1024 kg and the period T measured in seconds is given by T = 86400 s

So substitution gives R = 42 times 106 m

The radius of the Earth is 64 times 106 m so that the orbital height h is about 36 times 107 m

Example

Calculatetheminimumenergyrequiredtoputasatelliteofmass500kgintoanorbitthatisasaheightequaltotheEarthrsquosradiusabovethesurfaceoftheEarth

Solution

We have seen that when dealing with gravitational fields and potential it is useful to remember that

g 0G M

Re2---------= or g0 Re

2 GM=

The gravitational potential at the surface of the Earth is

g 0ndash Re GMRe

---------ndash=

The gravitational potential at a distance R

from the centre of the Earth is

=

The difference in potential between the surface of the Earth and a point distance R from the centre is therefore

V∆ g 0Re 1ReR------ndash

=

If R 2Re= then V∆g0 Re

2------------=

This means that the work required to ldquoliftrdquo the satellite into orbit is g0Rm where m is the mass of the satellite This is equal to

10 32 106 500timestimestimes = 16000 MJ

However the satellite must also have kinetic energy in order to orbit Earth This will be equal to

g 0 mRe2

2R-----------------g 0mRe

2

4----------------- 8000 MJ= =

The minimum energy required is therefore

24000 MJ

Exercise 1031 The speed needed to put a satellite in orbit does

not depend on

A the radius of the orbitB theshapeoftheorbitC thevalueofgattheorbitD the mass of the satellite

2 EstimatethespeedofanEarthsatellitewhoseorbitis400kmabovetheEarthrsquossurfaceAlsodetermine the period of the orbit

3 Calculatethespeedofa200kgsatelliteorbitingthe Earth at a height of 70 times 106 m

Assume that g = 82 m sndash2 for this orbit

4 TheradiioftwosatellitesXandYorbitingthe Earth are 2r and 8rwherer is the radius of the Earth Calculate the ratio of the periods of revolutionofXtoY

5 A satellite of mass mkgissentfromEarthrsquossurfaceinto an orbit of radius 5RwhereR is the radius of theEarthWritedownanexpressionfor

(a) thepotentialenergyofthesatelliteinorbit

(b) thekineticenergyofthesatelliteinorbit

(c) theminimumworkrequiredtosendthesatellitefromrestattheEarthrsquossurfaceintoits orbit

6 A satellite in an orbit of 10rfallsbacktoEarth(radius r)afteramalfunctionDeterminethespeedwithwhichitwillhittheEarthrsquossurface

7 The radius of the moon is frac14 that of the Earth AssumingEarthandtheMoontohavethesamedensitycomparetheaccelerationsoffreefallatthe surface of Earth to that at the surface of the Moon

8 UsethefollowingdatatodeterminethegravitationalfieldstrengthatthesurfaceoftheMoon and hence determine the escape speed from the surface of the Moon

Mass of the Moon = 73 times 1022kg

RadiusoftheMoon=17times106 m

PHYSICS 2015indb 322 150514 338 PM

Fields

309

AH

L

aswemoveawayfromtheEarthrsquossurfaceincreasesuntilitreachesthevalueofzeroatinfinity

IfthegravitationalfieldisduetoapointmassofmassmthenwehavethesameexpressionasaboveexceptthatMe is replaced by m andmustalsoexclude thevalueof thepotential at the point mass itself ie at r = 0

WecanexpressthegravitationalpotentialduetotheEarth(orduetoanysphericalmass)intermsofthegravitationalfield strength at its surface

AtthesurfaceoftheEarthwehave

g 0ndash Re GMRe

---------ndash= e

Sothat

g 0Re2 GM= e

Hence at a distance r from the centre of the Earth the gravitationalpotentialVcanbewrittenas

VGMe

r------------ndashg 0 Re

2

r------------ndash= =

The potential at the surface of the Earth

(r = Re)istherefore-g0Re

It is interesting to see how the expression for thegravitational potential ties in with the expression mgh ThepotentialatthesurfaceoftheEarthis-g0Re (see the exampleabove)andataheighthwillbe-g0g 0 Re h+( )ndash ifweassume that g0doesnotchangeover thedistanceh The differenceinpotentialbetweenthesurfaceandtheheighth is therefore g0hSotheworkneededtoraiseanobjectof mass m to a height h is mgh ie m timesdifference ingravitationalpotential

This we have referred to as the gain in gravitationalpotentialenergy(see235)

However this expression can be extended to any twopoints inanygravitationalfieldsuch that ifanobjectofmass mmovesbetween twopointswhosepotentialsareV1 and V2 respectively then the change in gravitationalpotentialenergyoftheobjectism(V1 ndash V2)

Gravitational potential gradient

Let us consider now a region in space where thegravitationalfieldisconstantInFigure912thetwopointsAandBareseparatedbythedistance∆x

A B

direction of uniform gravitational eld of strength I

∆x

Figure 1002 The gravitational potential gradient

The gravitational field is of strength I and is in the direction shownThe gravitational potential at A is V andatBisV +∆V

Theworkdone is takingapointmassmfromAtoB isF∆x = mI∆x

Howeverbydefinitionthisworkisalsoequalto-m∆V

Therefore mI∆x=-m∆V

or VIx

∆= minus

∆Effectivelythissaysthatthemagnitudeofthegravitationalfield strength is equal to the negative gradient of thepotential If I is constant then V is a linear function of x and Iisequaltothenegativegradientofthestraightlinegraph formed by plotting V against x If I is not constant (asusuallythecase)thenthemagnitudeofI at any point in the field can be found by find the gradient of the V-x graph at that point An example of such a calculation can befoundinSection929

For those of you who do HL maths the relationship between field and potential is seen to follow from the expression for the potential of a point mass viz

mV Gr

= minus

2

ddV mG Ir r

minus = + =

Potential due to one or more point masses

Gravitationalpotentialisascalarquantitysocalculatingthepotential due to more than one point mass is a matter of simpleadditionSoforexamplethepotentialV due to the Moon and Earth and a distance x fromthecentreofEarthonastraightlinebetweenthemisgivenbytheexpression

E MM MV Gx r x

= minus + minus

whereME=massofEarthMM= mass of Moon and r = distancebetweencentreofEarthandMoon

PHYSICS 2015indb 309 150514 337 PM

Chapter 10

310

AH

L

Equipotentials and field linesIfthegravitationalpotentialhasthesamevalueatallpointson a surface the surface is said to be an equipotential surfaceSoforexampleifweimagineasphericalshellaboutEarthwhosecentrecoincideswiththecentreofEarththisshellwillbeanequipotentialsurfaceClearlyifwerepresentthegravitationalfieldstrengthbyfieldlinessincethelinesldquoradiaterdquooutfromthecentreofEarththentheselineswillbeatrightanglestothesurfaceIfthefieldlineswerenotnormaltotheequipotentialsurfacethentherewouldbeacomponentof the field parallel to the surfaceThis wouldmean thatpointsonthesurfacewouldbeatdifferentpotentialsandsoitwouldnolongerbeanequipotentialsurfaceThisofcourseholdstrueforanyequipotentialsurface

Figure 1003 shows the field lines and equipotentials fortwopointmassesm

m m

Figure 1003 Equipotentials for two point masses

It is worth noting that we would get exactly the samepattern ifwewere to replace thepointmasseswith twoequalpointcharges(See935)

Escape speedThe potential at the surface of Earth is which meansthattheenergyrequiredtotakeaparticleofmass mfromthesurfacetoinfinityisequalto

ButwhatdoesitactuallymeantotakesomethingtoinfinityWhentheparticleisonthesurfaceoftheEarthwecanthinkofitassittingatthebottomofaldquopotentialwellrdquoasinfigure1004

particle

surface of Earth

innity

GMR---------

FIgure 1004 A potential well

Theldquodepthrdquoofthewellis and if the particle gains an amount of kinetic energy equal to wherem is its massthenitwillhavejustenoughenergytoldquoliftrdquoitoutofthewell

In reality it doesnrsquot actually go to infinity it justmeansthat the particle is effectively free of the gravitationalattractionof theEarthWe say that ithas ldquoescapedrdquo theEarthrsquosgravitationalpullWemeetthisideainconnectionwithmolecular forcesTwomolecules in a solidwill sitat their equilibrium position the separation where therepulsiveforceisequaltotheattractiveforceIfwesupplyjust enough energy to increase the separation of themolecules such that they are an infinite distance apart then intermolecularforcesnolongeraffectthemoleculesandthesolidwillhavebecomealiquidThereisnoincreaseinthekineticenergyofthemoleculesandsothesolidmeltsat constant temperature

Wecancalculatetheescapespeedofanobjectveryeasilyby equating the kinetic energy to the potential energysuch that

12---mvesc ape

2 GMe mRe

----------------=

vesc aperArr2GMe

Re--------------- 2g0 Re = =

Substituting for g0 and Regivesavalueforvescape of about 11kmsndash1 from the surface of the Earth

Youwillnotethattheescapespeeddoesnotdependonthemassoftheobjectsincebothkineticenergyandpotentialenergy are proportional to the mass

IntheoryifyouwanttogetarockettothemoonitcanbedonewithoutreachingtheescapespeedHoweverthiswouldnecessitate anenormousamountof fuel and it islikely that therocketplus fuelwouldbesoheavythat itwouldnevergetoffthegroundItismuchmorepracticalto accelerate the rocket to the escape speed fromEarthorbitandthenintheoryjustlaunchittotheMoon

Example

Use the followingdata todetermine thepotential at thesurface of Mars and the magnitude of the acceleration of free fall

massofMars =64times1023kg

radiusofMars =34times106 m

Determinealsothegravitationalfieldstrengthatadistanceof 68 times 106mabovethesurfaceofMars

PHYSICS 2015indb 310 150514 337 PM

Fields

311

AH

L

Solution

= minus = minus times times = times23

11 76

64 1067 10 ndash13 10

34 10MV GR

minus timestimes

Nkg-1

But V = -g0R

Therefore 7

0 6

13 10 3834 10

VgR

times= minus = =

timesm s-2

To determine the field strength gh at 68 times 106 m above the surface we use the fact that 0 2

Mg GR

= such that GM = g0R2

Therefore 2 2

0h 2 2 2

h h

38 (34) 042(102)

g RGMgR R

times= = = = m s-2

(the distance from the centre is 34 times 106 + 68 times 106 = 102 times 106 m)

Exercise 1011 Thegraphbelowshowshowthegravitational

potentialoutsideoftheEarthvarieswithdistancefrom the centre

0 10 20 30 40 50 60 70

ndash1

ndash2

ndash3

ndash4

ndash5

ndash6

ndash7

r m times 106

V

Jkgndash1

times 10

7

(a) Usethegraphtodeterminethegainingravitationalpotentialenergyofasatelliteofmass200kgasitmovesfromthesurfaceof the Earth to a height of 30 times 107mabovetheEarthrsquossurface

(b) Calculatetheenergyrequiredtotakeit to infinity

(c) DeterminetheslopeofthegraphatthesurfaceoftheEarthmCommenton youranswer

Electric potential energy

The concept of electric potential energy was developedwith gravitational potential energy in mind Just as anobjectnearthesurfaceoftheEarthhaspotentialenergybecause of its gravitational interaction with the Earthsotoothereiselectricalpotentialenergyassociatedwithinteracting charges

Letusfirstlookatacaseoftwopositivepointchargeseachof 1μC that are initially bound together by a thread in a vacuuminspacewithadistancebetweenthemof10cmasshowninFigure1005Whenthethreadiscutthepointchargesinitiallyatrestwouldmoveinoppositedirectionsmovingwithvelocitiesv1 and v2 along the direction of the electrostatic force of repulsion

BEFORE AFTER

v v21

Figure 1005 Interaction of two positive particles

The electric potential energybetweentwopointchargescan be found by simply adding up the energy associated witheachpairofpointchargesForapairofinteractingchargestheelectricpotentialenergyisgivenby

∆U = ∆Ep + ∆Ek = ∆W = ∆Fr = kqQ

____ r2 times r = kqQ

____ r

Becausenoexternalforceisactingonthesystemtheenergyand momentum must be conserved Initially Ek = 0 and Ep =kqQr=9times109 times 1 times 10-1201m=009JWhentheyareagreatdistance fromeachotherEpwillbenegligibleThefinalenergywillbeequaltofrac12mv1

2+frac12mv22=009J

Momentum isalsoconservedand thevelocitieswouldbethe same magnitude but in opposite directions

Electric potential energyismoreoftendefinedintermsofapointchargemovinginanelectricfieldas

lsquothe electricpotential energybetweenany twopoints inan electric field is defined as negativeoftheworkdone by anelectricfieldinmovingapoint electric chargebetweentwolocationsintheelectricfieldrsquo

ΔU = ΔEp = -ΔW = -Fd = qEx

ΔU = ΔEp +ΔEk = ΔW = ΔFr = kqQ r2 x r = kqQ

wherexisthedistancemovedalong(oroppositeto)thedirection of the electric field

Electric potential energyismeasuredinjoule(J)Justasworkisascalarquantitysotooelectricalpotentialenergyis a scalar quantity The negative of the work done by an electric field in moving a unit electric charge between two points is independent of the path taken Inphysicswesaytheelectricfieldisalsquoconservativersquo field

PHYSICS 2015indb 311 150514 337 PM

Chapter 10

312

AH

L

Supposeanexternalforcesuchasyourhandmovesasmallpositive point test charge in the direction of a uniformelectric fieldAs it ismoving itmust be gaining kineticenergyIfthisoccursthentheelectricpotentialenergyofthe unit charge is changing

InFigure1006apointcharge+qismovedbetweenpointsAandBthroughadistancex in a uniform electric field

B

A

x

+ q

Figure 1006 Movement of a positive point charge in a uniform field

InordertomoveapositivepointchargefrompointAtopointBanexternal forcemustbeapplied to thechargeequaltoqE (F = qE)

Since the force is applied through a distance xthennegativeworkhas to bedone tomove the charge because energyis gainedmeaning there is an increaseelectric potential energybetweenthetwopointsRememberthattheworkdoneisequivalenttotheenergygainedorlostinmovingthe charge through the electric field The concept of electric potential energy is only meaningful as the electric field whichgeneratestheforceinquestionisconservative

W F xtimes E q x times= =

θ x cos θ x

Figure 1007 Charge moved at an angle to the field

If a chargemoves at an angle θ to an electric field thecomponent of the displacement parallel to the electric fieldisusedasshowninFigure918

W F x Eq x θcostimes= =

Theelectricpotentialenergyisstoredintheelectricfieldand theelectricfieldwill return theenergy to thepointcharge when required so as not to violate the Law ofconservationofenergy

Electric potential

The electric potential at a point in an electric field is defined as being the work done per unit charge in bringing a small positive point charge from infinity to that point

ΔV = Vinfin ndash Vf = ndash W ___ q

Ifwedesignatethepotentialenergytobezeroatinfinitythen it follows thatelectricpotentialmustalsobezeroat infinity and the electric potential at any point in an electricfieldwillbe

ΔV = ndash W ___ q

NowsupposeweapplyanexternalforcetoasmallpositivetestchargeasitismovedtowardsanisolatedpositivechargeTheexternalforcemustdoworkonthepositivetestchargetomoveittowardstheisolatedpositivechargeandtheworkmustbepositivewhiletheworkdonebytheelectricfieldmustthereforebenegativeSotheelectricpotentialatthatpointmustbepositiveaccordingtotheaboveequationIfanegativeisolatedchargeisusedtheelectricpotentialatapointonthepositivetestchargewouldbenegativePositivepoint chargesof their ownaccordmove fromaplaceofhighelectricpotential toaplaceof lowelectricpotentialNegative point charges move the other way from lowpotentialtohighpotentialInmovingfrompointAtopointBinthediagramthepositivecharge+qismovingfromalowelectricpotentialtoahighelectricpotential

In thedefinitiongiven the termldquoworkperunitchargerdquohas significance If the test charge is +16 times 10-19Cwherethe charge has a potential energy of 32 times 10-17Jthenthepotentialwouldbe32times10-17J+16times10-19C=200JC-1 Now if the charge was doubled the potential wouldbecome 64times10-17 J However the potential per unitchargewouldbethesame

Electric potential is a scalar quantityandithasunitsJC-1 orvoltswhere1voltequalsone joulepercoloumbThevoltallowsustoadoptaunitfortheelectricfieldintermsofthevolt

PreviouslytheunitfortheelectricfieldwasNCndash1

W = qV and F = qEsoW ___ V = F __ E

E = FV ___ W = FV ___ Fm V mndash1

That is the units of the electric field E can also beexpressed as V mndash1

The work done per unit charge in moving a point charge between two points in an electric field is again independent of the path taken

PHYSICS 2015indb 312 150514 337 PM

Fields

313

AH

L

Electric potential due to a point charge

Letus takeapoint rmetres froma chargedobjectThepotentialatthispointcanbecalculatedusingthefollowing

W = ndashFr = ndashqV and F = ndash q1 q2 _____ 4πε0r

2

Therefore

W = ndash q1 q2 _____ 4πε0r

2 times r = ndash q1 q2 _____ 4πε0r

= ndashq1 times q2 _____ 4πε0r

= ndashq1V

That is V q

4πε0r--------------- =

Orsimply

V kqr------

=

Example

Determinehowmuchworkisdonebytheelectricfieldofpointcharge150μCwhenachargeof200μCismovedfrom infinity to apoint 0400m from thepoint charge(Assumenoaccelerationofthecharges)

Solution

The work done by the electric field is W = -qV = -14πε0 times q times (Q rinfin - Q r0400)

W = (- 200 times 10-6 C times 900 times 109 NmC-2 times 150 times 10-6 C) divide 0400 m = - 0675 J

An external force would have to do +0675 J of work

Electric field strength and electric potential gradient

Letus lookbackatFigure1006Supposeagain that thecharge+qismovedasmalldistancebyaforceFfromAtoBsothattheforcecanbeconsideredconstantTheworkdoneisgivenby

∆W F x∆times=

TheforceFandtheelectricfieldEareoppositelydirectedandweknowthat

F = -qE and ΔW = q ΔV

Thereforetheworkdonecanbegivenas

q ΔV = -q E Δx

Therefore E V∆x∆-------

ndash=

The rate of change of potential ΔV at a point with respect to distance Δx in the direction in which the change is maximum is called the potential gradient We say that the electric field = - the potential gradient and the units are Vm-

1 From the equation we can see that in a graph of electric potential versus distance the gradient of the straight line equals the electric field strength

In reality if a charged particle enters a uniform electric field it will be accelerated uniformly by the field and its kinetic energy will increase This is why we had to assume no acceleration in the last worked example

Ek∆ 12---

mv2 q E xsdot sdot q Vx---

xsdot sdot q Vsdot= = = =

Example

Determine how far apart two parallel plates must besituated so that a potential difference of 150 times 102 V produces an electric field strength of 100 times 103 NC-1

Solution

Using E V∆x------- xhArrndash V∆

E-------15 102 Vtimes

100 103 N C 1ndashtimes------------------------------------------= = =

= 150 times 10-1

The plates are 150 times 10-1 m apart

The electric field and the electric potential at a point due to an evenly distributed charge +q on a sphere can be represented graphically as in Figure 1008

r

r

V

E

r0

r0

r0

Charge of +Q evenly distributed over surface

E 14πε0------------ Q

r2----- r r0gt=

V 14πε0------------Q

r----r r0gt=

Figure 1008 Electric field and potential due to a charged sphere

PHYSICS 2015indb 313 150514 337 PM

Chapter 10

314

AH

L

When the sphere becomes charged we know that thechargedistributes itselfevenlyoverthesurfaceThereforeeverypartofthematerialoftheconductorisatthesamepotential As the electric potential at a point is defined as beingnumericallyequaltotheworkdoneinbringingaunitpositivechargefrominfinitytothatpointithasaconstantvalueineverypartofthematerialoftheconductor

Since the potential is the same at all points on the conducting surfacethen∆V∆xiszeroButE=ndash∆V∆xThereforethe electric field inside the conductor is zero There is no electric field inside the conductor

SomefurtherobservationsofthegraphsinFigure1009are

bull OutsidethespherethegraphsobeytherelationshipsgivenasE α 1 r2 and V α 1 r

bull Atthesurfacer = r0Thereforetheelectricfieldandpotentialhavetheminimumvalueforr at this point and this infers a maximum field and potential

bull Insidethespheretheelectricfieldiszerobull Insidethespherenoworkisdonetomoveacharge

fromapointinsidetothesurfaceThereforethereisno potential difference and the potential is the same asitiswhenr = r0

Similargraphscanbedrawnfortheelectricfieldintensityandthe electric potential as a function of distance from conducting parallelplatesandsurfacesandthesearegiveninFigure1009

Potential plot E eld plotE eld

++++

ndashndashndashndash

xx x

+

x x

ndashx x

+ ndashx

x

+ +

x

x

Figure 1009 Electric field and electric potential at a distance from a charged surface

Potential due to one or more point charges

The potential due to one point charge can be determined byusingtheequationformula

V = kq r

Example 1

Determine the electric potential at a point 20 times 10-1 m fromthecentreofan isolatedconducting spherewithapointchargeof40pCinair

Solution

Using the formula

V = kq r we have

V90 10 9times( ) 40 10 12ndashtimes( )times

20 10 1ndashtimes( )------------------------------------------------------------------ 018 V= =

the potential at the point is 180 times 10-1 V

The potential due to a number of point charges can be determined by adding up the potentials due to individual point charges because the electric potential at any point outside a conducting sphere will be the same as if all the charge was concentrated at its centre

Example 2

Threepointchargesofareplacedattheverticesofaright-angledtriangleasshowninthediagrambelowDeterminetheabsolutepotentialat the+20μCchargeduetothetwoothercharges

- 6μC

+3μC +2μC

4 m

3 m

Solution

The electric potential of the +2 μC charge due to the ndash 6 μC charge is

V = (9 times 109 Nm2C-2 times -6 times 10-6 C) divide (radic 32 + 42) m = - 108 times 104 V

PHYSICS 2015indb 314 150514 338 PM

Fields

315

AH

L

Insummarywecanconcludethat

bull No work is done to move a charge along an equipotential

bull Equipotentials are always perpendicular to the electric lines of force

Figure1011and1012showsomeequipotential lines fortwo oppositely charged and identically positive spheresseparated by a distance

+ve ndashve

equipotential lines

Figure 1011 Equipotential lines between two opposite charges

equipotential lines

Figure 1012 Equipotential lines between two charges which are the same

+

ndash ndash ndash ndash

+ + +

40 V

30 V

20 V

10 V

50 V

Figure 1013 Equipotential lines between charged parallel plates

Figure1013showstheequipotentiallinesbetweenchargedparallel plates Throughout this chapter the similarities and differences between gravitational fields and electric fieldshavebeendiscussedTherelationshipsthatexistsbetweengravitationalandelectricquantitiesandtheeffectsofpointmassesandchargesissummarisedinTable1014

The electric potential of the +2 μC charge due to the +3 μC charge is

V = (9 times 109 Nm2C-2 times 3 times 10-6 C) divide 3m = 9 times 103 V

The net absolute potential is the sum of the 2 potentials- 108 times 104 V + 9 times 103 V =- 18 times 103 V

The absolute potential at the point is - 18 times 103 V

Equipotential surfacesRegions in space where the electric potential of a chargedistributionhasaconstantvaluearecalledequipotentials The placeswherethepotentialisconstantinthreedimensionsarecalled equipotential surfacesandwheretheyareconstantintwodimensionstheyarecalledequipotential lines

TheyareinsomewaysanalogoustothecontourlinesontopographicmapsInthiscasethegravitationalpotentialenergy is constant as amassmoves around the contourlines because the mass remains at the same elevationabovetheEarthrsquossurfaceThegravitationalfieldstrengthacts in a direction perpendicular to a contour line

Similarlybecausetheelectricpotentialonanequipotentiallinehas thesamevalueanelectric forcecandonoworkwhena testchargemovesonanequipotentialThereforethe electric field cannot have a component along anequipotentialandthusitmustbeeverywhereperpendiculartotheequipotentialsurfaceorequipotentiallineThisfactmakesiteasytoplotequipotentialsifthelinesofforceorlinesofelectricfluxofanelectricfieldareknown

Forexamplethereareaseriesofequipotentiallinesbetweentwoparallelplateconductorsthatareperpendiculartotheelectric fieldTherewill be a series of concentric circles(eachcirclefurtherapartthanthepreviousone)thatmapouttheequipotentialsaroundanisolatedpositivesphereThe lines of force and some equipotential lines for anisolatedpositivesphereareshowninFigure1010

Lines ofequipotential

Figure 1010 Equipotentials around an isolated positive sphere

PHYSICS 2015indb 315 150514 338 PM

Chapter 10

316

AH

L

Gravitational quantity Electrical quantityQ

uant

ities

g V∆x∆-------ndash= E V∆

x∆-------ndash=

Poin

t mas

ses a

nd ch

arge

s

V Gmr----

ndash= V 14πε0------------q

r---=

g G m

r2-----ndash= E 14πε0------------ q

r2-----=

F Gm1 m2

r2--------------= F 1

4πε0------------

q1q2

r2-----------=

Figure 1014 Formulas

Example

Deduce the electric potential on the surface of a gold nucleus that has a radius of 62 fm

Solution

Using the formula

V = kq r and knowing the atomic number of gold is 79 We will assume the nucleus is spherical and it behaves as if it were a point charge at its centre (relative to outside points)

V = 90 times 109 Nm2C-2 times 79 times 16 times 10-19 C divide 62 times 10-15 m = 18 times 107 V

The potential at the point is 18 MV

Example

Deduce the ionisation energy in electron-volts of theelectron in the hydrogen atom if the electron is in its ground state and it is in a circular orbit at a distance of 53 times 10-11 m from the proton

Solution

This problem is an energy coulombic circular motion question based on Bohrrsquos model of the atom (not the accepted quantum mechanics model) The ionisation energy is the energy required to remove the electron from the ground state to infinity The electron travels in a circular orbit and therefore has a centripetal acceleration The ionisation energy will counteract the coulombic force and the movement of the electron will be in the opposite direction to the centripetal force

Total energy = Ek electron + Ep due to the proton-electron interaction

ΣF = kqQ r2 = mv2 r and as such mv2 = = kqQ r

Therefore Ek electron = frac12 kqQ r

Ep due to the proton-electron interaction = - kqQ r

Total energy = frac12 kqQ r + - kqQ r = -frac12 kqQ r

= - 90 times 109 Nm2C-2 times (16 times 10-19 C)2 divide 53 times 10-11 m = -217 times 10-18 J

= -217 times 10-18 J divide 16 times 10-19 = -136 eV

The ionisation energy is 136 eV

Exercise 1021 ApointchargePisplacedmidwaybetweentwo

identicalnegativechargesWhichoneofthefollowingiscorrectwithregardstoelectricfieldandelectricpotentialatpointP

Electric field Electric potentialA non-zero zeroB zero non-zeroC non-zero non-zeroD zero zero

2 Twopositivechargedspheresaretiedtogetherinavacuumsomewhereinspacewheretherearenoexternal forces A has a mass of 25 g and a charge of20μCandBhasamassof15gandachargeof30μCThedistancebetweenthemis40cmTheyarethenreleasedasshowninthediagram

BEFORE AFTER

v1 v2

A B

(a) Determinetheirinitialelectricpotentialenergy in the before situation

(b) DeterminethespeedofsphereBafterrelease

3 Thediagrambelowrepresentstwoequipotentiallines in separated by a distance of 5 cm in a uniform electric field

+ + + + + + + +

ndash ndash ndash ndash ndash ndash ndash ndash

40 V

20 V 5 cm

PHYSICS 2015indb 316 150514 338 PM

Fields

317

AH

L

Determine the strength of the electric field

4 Thisquestionisabouttheelectricfieldduetoacharged sphere and the motion of electrons in that fieldThediagrambelowshowsanisolatedmetalsphereinavacuumthatcarriesanegativeelectriccharge of 60 μC

ndash

(a) Onthediagramdrawtheconventionalwayto represent the electric field pattern due to the charged sphere and lines to represent threeequipotentialsurfacesintheregionoutside the sphere

(b) Explainhowthelinesrepresentingtheequipotentialsurfacesthatyouhavesketchedindicatethatthestrengthoftheelectricfieldisdecreasingwithdistancefrom the centre of the sphere

(c) Theelectricfieldstrengthatthesurfaceofthe sphere and at points outside the sphere can be determined by assuming that the sphere acts as a point charge of magnitude 60 μC at its centre The radius of the sphere is 25 times 10ndash2 m Deduce that the magnitude of the field strength at the surface of the sphere is 86 times 107 Vmndash1

An electron is initially at rest on the surface of the sphere

(d) (i) Describethepathfollowedbytheelectronasitleavesthesurfaceofthesphere

(ii) Calculatetheinitialaccelerationofthe electron

5 Determinetheamountofworkthatisdoneinmovingachargeof100nCthroughapotentialdifferenceof150times102 V

6 Three identical 20 μC conducting spheres are placedatthecornersofanequilateraltriangleofsides 25 cm The triangle has one apex C pointing upthepageand2baseanglesAandBDeterminetheabsolutepotentialatB

7 Determinehowfaraparttwoparallelplatesmustbesituatedsothatapotentialdifferenceof250 times 102 V produces an electric field strength of 200 times 103 NC-1

8 Thegapbetweentwoparallelplatesis10times10-3mandthereisapotentialdifferenceof10times104 V betweentheplatesCalculate

i theworkdonebyanelectroninmovingfrom one plate to the other

ii thespeedwithwhichtheelectronreachesthe second plate if released from rest

iii theelectricfieldintensitybetweentheplates

9 Anelectronguninapicturetubeisacceleratedby a potential 25 times 103VDeterminethekineticenergygainedbytheelectroninelectron-volts

10 Determine the electric potential 20 times 10-2 m from achargeof-10times10-5 C

11 Determinetheelectricpotentialatapointmid-waybetweenachargeofndash20pCandanotherof+5pConthelinejoiningtheircentresifthecharges are 10 cm apart

12 During a thunderstorm the electric potential differencebetweenacloudandthegroundis10 times 109 V Determine the magnitude of the change in electric potential energy of an electron thatmovesbetweenthesepointsinelectron-volts

13 A charge of 15 μC is placed in a uniform electric fieldoftwooppositelychargedparallelplateswithamagnitudeof14times103 NC-1

(a) Determinetheworkthatmustbedoneagainstthefieldtomovethepointchargeadistance of 55 cm

(b) Calculatethepotentialdifferencebetweenthe final and initial positions of the charge

(c) Determinethepotentialdifferencebetweenthe plates if their separation distance is 15 cm

14 Duringaflashoflightningthepotentialdifferencebetweenacloudandthegroundwas12 times 109 V and the amount of transferred charge was32C

(a) Determinethechangeinenergyofthetransferred charge

(b) Iftheenergyreleasedwasallusedtoacceleratea1tonnecardeduceitsfinalspeed

(c) Iftheenergyreleasedcouldbeusedtomelticeat0degCdeducetheamountoficethatcould be melted

PHYSICS 2015indb 317 150514 338 PM

Chapter 10

318

AH

L

15 SupposethatwhenanelectronmovedfromAtoBin the diagram along an electric field line that the electric field does 36 times 10-19Jofworkonit

Determinethedifferencesinelectricpotential

(a) VB ndash VA(b) VC ndash VA(c) VC ndash VB

16 DeterminethepotentialatpointPthatislocatedatthecentreofthesquareasshowninthediagrambelow

- 6μC

+3μC +2μC

1m

1m

5μC

P

102 Fields at work

Essential idea Similar approaches can be taken inanalysingelectricalandgravitationalpotentialproblems

Understandingsbull Potentialandpotentialenergybull Potentialgradientbull Potentialdifferencebull Escapespeedbull Orbitalmotionorbitalspeedandorbitalenergybull Forcesandinverse-squarelawbehaviour

Orbital motion orbital speed and orbital energy

Although orbital motion may be circular elliptical orparabolic this sub-topic only dealswith circular orbitsThis sub-topic is not fundamentally new physics butan application that synthesizes ideas from gravitationcircularmotiondynamicsandenergy

TheMoonorbitstheEarthandinthissenseitisoftenreferredtoasasatelliteoftheEarthBefore1957itwastheonlyEarthsatelliteHoweverin1957theRussianslaunchedthefirstmanmadesatelliteSputnik1SincethisdatemanymoresatelliteshavebeenlaunchedandtherearenowliterallythousandsofthemorbitingtheEarthSomeareusedtomonitortheweathersome used to enable people to find accurately their position on thesurfaceoftheEarthmanyareusedincommunicationsandnodoubtsomeareusedtospyonothercountriesFigure1015showshowinprincipleasatellitecanbeputintoorbit

Theperson(whosesizeisgreatlyexaggeratedwithrespectto Earth) standing on the surface on the Earth throwssome stonesThe greater the speed with which a stoneis thrown the further it will land from her The pathsfollowedbythethrownstonesareparabolasByastretchof the imaginationwecanvisualisea situation inwhichastoneis thrownwithsuchaspeedthatbecauseof thecurvatureoftheEarthitwillnotlandonthesurfaceoftheEarthbutgointoldquoorbitrdquo(Path4onFigure1015)

NATURE OF SCIENCECommunication of scientific explanations The ability to apply field theory to the unobservable (charges) and the massively scaled (motion of satellites) required scientists to develop new ways to investigate analyse and report findings to a general public used to scientific discoveries based on tangible and discernible evidence (51)

copy IBO 2014

PHYSICS 2015indb 318 150514 338 PM

Fields

319

AH

L

1

2

34Earth

Figure 1015 Throwing a stone into orbit

TheforcethatcausesthestonestofollowaparabolicpathandtofalltoEarthisgravityandsimilarlytheforcethatkeeps the stone in orbit is gravity For circular motionto occur we have seen that a force must act at rightanglestothevelocityofanobjectthatistheremustbeacentripetalforceHenceinthesituationwedescribeherethe centripetal force for circular orbital motion about the EarthisprovidedbygravitationalattractionoftheEarth

We can calculate the speedwithwhich a stonemustbethrowninordertoputitintoorbitjustabovethesurfaceof the Earth

If the stone has mass m and speed v thenwehave fromNewtonrsquos2ndlaw

mv2

RE---------- G

MEm

RE2------------=

whereRE is the radius of the Earth and ME is the mass of the Earth

Bearinginmindthatg 0 GME

RE2--------= then

v g RE 10 64 106timestimes 8 103times= = =

Thatisthestonemustbethrownat8times103m sndash1

Clearly we are not going to get a satellite into orbit soclosetothesurfaceoftheEarthMovingatthisspeedthefrictionduetoairresistancewouldmeltthesatellitebeforeithadtravelledacoupleofkilometresInrealitythereforeasatelliteisputintoorbitabouttheEarthbysendingitattachedtoarocketbeyondtheEarthrsquosatmosphereandthengivingitacomponentofvelocityperpendiculartoaradialvectorfromtheEarthSeeFigure1016

Earth Satellite carried by rocket to hereSatellite orbit

Tangential component of velocity

Figure 1016 Getting a satellite into orbit

Keplerrsquos third law(This work of Kepler and Newtonrsquos synthesis of the work is an excellent example of the scientific method and makes for a good TOK discussion)

In 1627 Johannes Kepler (1571-1630)publishedhis lawsof planetary motion The laws are empirical in natureandwere deduced from the observations of theDanishastronomer Tycho de Brahe (1546-1601)The third lawgives a relationship between the radius of orbit R of a planet and its period TofrevolutionabouttheSunThelawisexpressedmathematicallyas

2

3 constantTR

=

We shall nowuseNewtonrsquos LawofGravitation to showhowitisthattheplanetsmoveinaccordancewithKeplerrsquosthirdlaw

InessenceNewtonwasabletousehislawofgravitytopredict the motion of the planets since all he had to do was factor theF givenby this law intohis second law F = ma to find their accelerations and hence their future positions

InFigure1017theEarthisshownorbitingtheSunandthedistancebetweentheircentresisR

RSun

Earth

F es

F se

Figure 1017 Planets move according to Keplerrsquos third law

Fes is the force that the Earth exerts on the Sun and Fse is the force that the Sun exerts on the Earth The forces are equalandoppositeandtheSunandtheEarthwillactuallyorbitaboutacommoncentreHoweversince theSun issoverymuchmoremassivethantheEarththiscommoncentrewillbeclosetothecentreoftheSunandsowecanregard the Earth as orbiting about the centre of the Sun Theotherthingthatweshallassumeisthatwecanignorethe forces that the other planets exert on the Earth (This wouldnotbeawisethingtodoifyouwereplanningtosend a space ship to theMoon for example)We shallalso assume that we have followed Newtonrsquos exampleandindeedprovedthataspherewillactasapointmasssituated at the centre of the sphere

Kepler had postulated that the orbits of the planets areellipticalbut since theeccentricityof theEarthrsquosorbit issmallweshallassumeacircularorbit

PHYSICS 2015indb 319 150514 338 PM

Chapter 10

320

AH

L

TheaccelerationoftheEarthtowardstheSunisa = Rω2

whereω ω 2πT------=

Hencea R 2π

T------ times

2 4π2 RT 2------------- = =

But the acceleration is given by Newtonrsquos Second Law F = mawhereFisnowgivenbytheLawofGravitationSo in this situation

F maGMs Me

R2-------------------= = butwealsohavethat

a a R 2πT

------ times

2 4π2 RT 2------------- = = and m = Me so that

G MsMe

R2------------------- Me4π2R

T2-------------times

GMs

4π2-----------rArr R3

T2------= =

Butthequantity

isaconstantthathasthesamevalueforeachoftheplanetssowehaveforalltheplanetsnotjustEarththat

R3

T2------ k =

wherekisaconstantWhichisofcourseKeplerrsquosthirdlaw

ThisisindeedanamazingbreakthroughItisdifficulttorefutethe idea that all particles attract each other in accordance withtheLawofGravitationwhenthelawisabletoaccountfortheobservedmotionoftheplanetsabouttheSun

The gravitational effects of the planets upon each othershould produce perturbations in their orbits Such is the predictivepoweroftheUniversalGravitationalLawthatit enabled physicists to compute these perturbations The telescopehadbeeninventedin1608andbythemiddleofthe 18th Century had reached a degree of perfection in design that enabled astronomers to actually measure the orbital perturbations of the planets Their measurements werealways inagreementwith thepredictionsmadebyNewtonrsquoslawHoweverin1781anewplanetUranuswasdiscovered and the orbit of this planet did not fit withthe orbit predicted by Universal Gravitation Such wasthe physicistrsquos faith in theNewtonianmethod that theysuspectedthatthediscrepancywasduetothepresenceofayetundetectedplanetUsingtheLawofGravitationtheFrenchastronomerJLeverrier and the English astronomer J C Adamswereable tocalculate justhowmassive thisnewplanetmustbeandalsowhereitshouldbeIn1846the planetNeptunewas discovered justwhere they hadpredicted Ina similarwaydiscrepancies in theorbitofNeptune ledto thepredictionandsubsequentdiscoveryin1930oftheplanetPlutoNewtonrsquosLawofGravitation

had passed the ultimate test of any theory it is not only abletoexplainexistingdatabutalsotomakepredictions

Satellite energyWhenasatelliteisinorbitaboutaplanetitwillhavebothkineticenergyandgravitationalpotentialenergySupposewe consider a satellite ofmass m that is in an orbit of radius r about a planet of mass M

Thegravitationalpotentialduetotheplanetatdistancer from its centre is

ThegravitationalpotentialenergyofthesatelliteVsat

is therefore GMe m

r----------------ndash

ThatisVsat GMem

r----------------ndash=

Thegravitationalfieldstrengthatthesurfaceoftheplanetisgivenby

g 0GMe

Re2------------=

Hencewecanwrite

Vsat g0 Re

2

rndash=

m

ThekineticenergyofthesatelliteKsatisequaltofrac12mv2wherev is its orbital speed

Byequatingthegravitationalforceactingonthesatellitetoitscentripetalaccelerationwehave

G Mem

r2---------------- mv2

r---------- mv2hArr

GMemr----------------= =

Fromwhich12---

mv2 12---

GMemr----------------times=

= 0 e

2g R

r

2m

Which is actually quite interesting since it shows thatirrespective of the orbital radius the KE is numericallyequal to half the PE Also the total energy E tot K sat Vsat+ 1

2---G Me m

r----------------times G Me m

r----------------ndash + 1

2---G Me m

r----------------ndash= = =of the satelliteisalwaysnegativesince

E tot K sat Vsat+ 12---

G Me mr----------------times

G Me mr----------------ndash

+ 12---

G Me mr----------------ndash= = =

PHYSICS 2015indb 320 150514 338 PM

Fields

321

AH

L

The energies of an orbiting satellite as a function of radial distancefromthecentreofaplanetareshownplottedinFigure1018

12

100806

0402

0ndash02

ndash04

ndash06ndash08ndash10

ndash12

2 4 6 8 10 12

ener

gya

rbita

ry u

nits

distance timesR

kinetic energytotal energy

potential energy

Figure 1018 Energy of an orbiting satellite as a function of distance from the centre of a planet

WeightlessnessSuppose that you are inanelevator(lift)thatisdescendingat constant speedandyou let goof abook thatyouareholdinginyourhandThebookwillfalltothefloorwithaccelerationequaltotheaccelerationduetogravityIfthecablethatsupportstheelevatorweretosnap(asituationthatItrustwillneverhappentoanyofyou)andyounowletgothebookthatyouareholdinginyourotherhandthis book will not fall to the floor - it will stay exactlyin linewithyourhandThis isbecause thebook isnowfallingwith the sameaccelerationas theelevatorandassuch the book cannot ldquocatchrdquo up with the floor of theelevatorFurthermoreifyouhappenedtobestandingonasetofbathroomscalesthescaleswouldnowreadzero-youwouldbeapparentlyweightlessItisthisideaoffreefallthatexplainstheapparentweightlessnessofastronautsin an orbiting satellite These astronauts are in free fall inthesensethattheyareacceleratingtowardsthecentre of the Earth

It is actually possible to define theweight of a body inseveral different waysWe can define it for example asthegravitationalforceexertedonthebodybyaspecifiedobject such as the EarthThiswe have seen thatwe doinlotsofsituationswherewedefinetheweightasbeingequaltomgIfweusethisdefinitionthenanobjectinfreefallcannotbydefinitionbeweightlesssinceitisstillinagravitationalfieldHoweverifwedefinetheweightofanobjectintermsofaldquoweighingrdquoprocesssuchasthereadingonasetofbathroomscaleswhichineffectmeasuresthecontact force between the object and the scales thenclearlyobjectsinfreefallareweightlessOnenowhasto

askthequestionwhetherornotitispossibleForexampletomeasurethegravitationalforceactingonanastronautin orbit about the Earth

WecanalsodefineweightintermsofthenetgravitationalforceactingonabodyduetoseveraldifferentobjectsForexample for an object out in space its weight could bedefined in terms of the resultant of the forces exerted on it bytheSuntheMoontheEarthandalltheotherplanetsin the Solar System If this resultant is zero at a particular pointthenthebodyisweightlessatthispoint

Inviewofthevariousdefinitionsofweightthatareavailabletousitisimportantthatwhenweusethewordldquoweightrdquoweareawareofthecontextinwhichitisbeingused

Example

Calculate the height above the surface of the Earth atwhichageo-stationarysatelliteorbits

Solution

A geo-stationary satellite is one that orbits the Earth in such a way that it is stationary with respect to a point on the surface of the Earth This means that its orbital period must be the same as the time for the Earth to spin once on its axis ie 24 hours

From Keplerrsquos third law we have G Ms

4π2----------- R3

T2------=

That is

RMe

mh

using the fact that the force of attraction between the satellite and the Earth is given by

FG Mem

R2----------------=

and that F = ma

where m is the mass of the satellite and a 4π2R

T2-------------=

we have

GMem

R2---------------- m 4π2R

T2-------------times

GMe

4π2------------rArr R3

T2------= =

PHYSICS 2015indb 321 150514 338 PM

Chapter 10

322

AH

L

Now the mass of the Earth is 60 times 1024 kg and the period T measured in seconds is given by T = 86400 s

So substitution gives R = 42 times 106 m

The radius of the Earth is 64 times 106 m so that the orbital height h is about 36 times 107 m

Example

Calculatetheminimumenergyrequiredtoputasatelliteofmass500kgintoanorbitthatisasaheightequaltotheEarthrsquosradiusabovethesurfaceoftheEarth

Solution

We have seen that when dealing with gravitational fields and potential it is useful to remember that

g 0G M

Re2---------= or g0 Re

2 GM=

The gravitational potential at the surface of the Earth is

g 0ndash Re GMRe

---------ndash=

The gravitational potential at a distance R

from the centre of the Earth is

=

The difference in potential between the surface of the Earth and a point distance R from the centre is therefore

V∆ g 0Re 1ReR------ndash

=

If R 2Re= then V∆g0 Re

2------------=

This means that the work required to ldquoliftrdquo the satellite into orbit is g0Rm where m is the mass of the satellite This is equal to

10 32 106 500timestimestimes = 16000 MJ

However the satellite must also have kinetic energy in order to orbit Earth This will be equal to

g 0 mRe2

2R-----------------g 0mRe

2

4----------------- 8000 MJ= =

The minimum energy required is therefore

24000 MJ

Exercise 1031 The speed needed to put a satellite in orbit does

not depend on

A the radius of the orbitB theshapeoftheorbitC thevalueofgattheorbitD the mass of the satellite

2 EstimatethespeedofanEarthsatellitewhoseorbitis400kmabovetheEarthrsquossurfaceAlsodetermine the period of the orbit

3 Calculatethespeedofa200kgsatelliteorbitingthe Earth at a height of 70 times 106 m

Assume that g = 82 m sndash2 for this orbit

4 TheradiioftwosatellitesXandYorbitingthe Earth are 2r and 8rwherer is the radius of the Earth Calculate the ratio of the periods of revolutionofXtoY

5 A satellite of mass mkgissentfromEarthrsquossurfaceinto an orbit of radius 5RwhereR is the radius of theEarthWritedownanexpressionfor

(a) thepotentialenergyofthesatelliteinorbit

(b) thekineticenergyofthesatelliteinorbit

(c) theminimumworkrequiredtosendthesatellitefromrestattheEarthrsquossurfaceintoits orbit

6 A satellite in an orbit of 10rfallsbacktoEarth(radius r)afteramalfunctionDeterminethespeedwithwhichitwillhittheEarthrsquossurface

7 The radius of the moon is frac14 that of the Earth AssumingEarthandtheMoontohavethesamedensitycomparetheaccelerationsoffreefallatthe surface of Earth to that at the surface of the Moon

8 UsethefollowingdatatodeterminethegravitationalfieldstrengthatthesurfaceoftheMoon and hence determine the escape speed from the surface of the Moon

Mass of the Moon = 73 times 1022kg

RadiusoftheMoon=17times106 m

PHYSICS 2015indb 322 150514 338 PM

Chapter 10

310

AH

L

Equipotentials and field linesIfthegravitationalpotentialhasthesamevalueatallpointson a surface the surface is said to be an equipotential surfaceSoforexampleifweimagineasphericalshellaboutEarthwhosecentrecoincideswiththecentreofEarththisshellwillbeanequipotentialsurfaceClearlyifwerepresentthegravitationalfieldstrengthbyfieldlinessincethelinesldquoradiaterdquooutfromthecentreofEarththentheselineswillbeatrightanglestothesurfaceIfthefieldlineswerenotnormaltotheequipotentialsurfacethentherewouldbeacomponentof the field parallel to the surfaceThis wouldmean thatpointsonthesurfacewouldbeatdifferentpotentialsandsoitwouldnolongerbeanequipotentialsurfaceThisofcourseholdstrueforanyequipotentialsurface

Figure 1003 shows the field lines and equipotentials fortwopointmassesm

m m

Figure 1003 Equipotentials for two point masses

It is worth noting that we would get exactly the samepattern ifwewere to replace thepointmasseswith twoequalpointcharges(See935)

Escape speedThe potential at the surface of Earth is which meansthattheenergyrequiredtotakeaparticleofmass mfromthesurfacetoinfinityisequalto

ButwhatdoesitactuallymeantotakesomethingtoinfinityWhentheparticleisonthesurfaceoftheEarthwecanthinkofitassittingatthebottomofaldquopotentialwellrdquoasinfigure1004

particle

surface of Earth

innity

GMR---------

FIgure 1004 A potential well

Theldquodepthrdquoofthewellis and if the particle gains an amount of kinetic energy equal to wherem is its massthenitwillhavejustenoughenergytoldquoliftrdquoitoutofthewell

In reality it doesnrsquot actually go to infinity it justmeansthat the particle is effectively free of the gravitationalattractionof theEarthWe say that ithas ldquoescapedrdquo theEarthrsquosgravitationalpullWemeetthisideainconnectionwithmolecular forcesTwomolecules in a solidwill sitat their equilibrium position the separation where therepulsiveforceisequaltotheattractiveforceIfwesupplyjust enough energy to increase the separation of themolecules such that they are an infinite distance apart then intermolecularforcesnolongeraffectthemoleculesandthesolidwillhavebecomealiquidThereisnoincreaseinthekineticenergyofthemoleculesandsothesolidmeltsat constant temperature

Wecancalculatetheescapespeedofanobjectveryeasilyby equating the kinetic energy to the potential energysuch that

12---mvesc ape

2 GMe mRe

----------------=

vesc aperArr2GMe

Re--------------- 2g0 Re = =

Substituting for g0 and Regivesavalueforvescape of about 11kmsndash1 from the surface of the Earth

Youwillnotethattheescapespeeddoesnotdependonthemassoftheobjectsincebothkineticenergyandpotentialenergy are proportional to the mass

IntheoryifyouwanttogetarockettothemoonitcanbedonewithoutreachingtheescapespeedHoweverthiswouldnecessitate anenormousamountof fuel and it islikely that therocketplus fuelwouldbesoheavythat itwouldnevergetoffthegroundItismuchmorepracticalto accelerate the rocket to the escape speed fromEarthorbitandthenintheoryjustlaunchittotheMoon

Example

Use the followingdata todetermine thepotential at thesurface of Mars and the magnitude of the acceleration of free fall

massofMars =64times1023kg

radiusofMars =34times106 m

Determinealsothegravitationalfieldstrengthatadistanceof 68 times 106mabovethesurfaceofMars

PHYSICS 2015indb 310 150514 337 PM

Fields

311

AH

L

Solution

= minus = minus times times = times23

11 76

64 1067 10 ndash13 10

34 10MV GR

minus timestimes

Nkg-1

But V = -g0R

Therefore 7

0 6

13 10 3834 10

VgR

times= minus = =

timesm s-2

To determine the field strength gh at 68 times 106 m above the surface we use the fact that 0 2

Mg GR

= such that GM = g0R2

Therefore 2 2

0h 2 2 2

h h

38 (34) 042(102)

g RGMgR R

times= = = = m s-2

(the distance from the centre is 34 times 106 + 68 times 106 = 102 times 106 m)

Exercise 1011 Thegraphbelowshowshowthegravitational

potentialoutsideoftheEarthvarieswithdistancefrom the centre

0 10 20 30 40 50 60 70

ndash1

ndash2

ndash3

ndash4

ndash5

ndash6

ndash7

r m times 106

V

Jkgndash1

times 10

7

(a) Usethegraphtodeterminethegainingravitationalpotentialenergyofasatelliteofmass200kgasitmovesfromthesurfaceof the Earth to a height of 30 times 107mabovetheEarthrsquossurface

(b) Calculatetheenergyrequiredtotakeit to infinity

(c) DeterminetheslopeofthegraphatthesurfaceoftheEarthmCommenton youranswer

Electric potential energy

The concept of electric potential energy was developedwith gravitational potential energy in mind Just as anobjectnearthesurfaceoftheEarthhaspotentialenergybecause of its gravitational interaction with the Earthsotoothereiselectricalpotentialenergyassociatedwithinteracting charges

Letusfirstlookatacaseoftwopositivepointchargeseachof 1μC that are initially bound together by a thread in a vacuuminspacewithadistancebetweenthemof10cmasshowninFigure1005Whenthethreadiscutthepointchargesinitiallyatrestwouldmoveinoppositedirectionsmovingwithvelocitiesv1 and v2 along the direction of the electrostatic force of repulsion

BEFORE AFTER

v v21

Figure 1005 Interaction of two positive particles

The electric potential energybetweentwopointchargescan be found by simply adding up the energy associated witheachpairofpointchargesForapairofinteractingchargestheelectricpotentialenergyisgivenby

∆U = ∆Ep + ∆Ek = ∆W = ∆Fr = kqQ

____ r2 times r = kqQ

____ r

Becausenoexternalforceisactingonthesystemtheenergyand momentum must be conserved Initially Ek = 0 and Ep =kqQr=9times109 times 1 times 10-1201m=009JWhentheyareagreatdistance fromeachotherEpwillbenegligibleThefinalenergywillbeequaltofrac12mv1

2+frac12mv22=009J

Momentum isalsoconservedand thevelocitieswouldbethe same magnitude but in opposite directions

Electric potential energyismoreoftendefinedintermsofapointchargemovinginanelectricfieldas

lsquothe electricpotential energybetweenany twopoints inan electric field is defined as negativeoftheworkdone by anelectricfieldinmovingapoint electric chargebetweentwolocationsintheelectricfieldrsquo

ΔU = ΔEp = -ΔW = -Fd = qEx

ΔU = ΔEp +ΔEk = ΔW = ΔFr = kqQ r2 x r = kqQ

wherexisthedistancemovedalong(oroppositeto)thedirection of the electric field

Electric potential energyismeasuredinjoule(J)Justasworkisascalarquantitysotooelectricalpotentialenergyis a scalar quantity The negative of the work done by an electric field in moving a unit electric charge between two points is independent of the path taken Inphysicswesaytheelectricfieldisalsquoconservativersquo field

PHYSICS 2015indb 311 150514 337 PM

Chapter 10

312

AH

L

Supposeanexternalforcesuchasyourhandmovesasmallpositive point test charge in the direction of a uniformelectric fieldAs it ismoving itmust be gaining kineticenergyIfthisoccursthentheelectricpotentialenergyofthe unit charge is changing

InFigure1006apointcharge+qismovedbetweenpointsAandBthroughadistancex in a uniform electric field

B

A

x

+ q

Figure 1006 Movement of a positive point charge in a uniform field

InordertomoveapositivepointchargefrompointAtopointBanexternal forcemustbeapplied to thechargeequaltoqE (F = qE)

Since the force is applied through a distance xthennegativeworkhas to bedone tomove the charge because energyis gainedmeaning there is an increaseelectric potential energybetweenthetwopointsRememberthattheworkdoneisequivalenttotheenergygainedorlostinmovingthe charge through the electric field The concept of electric potential energy is only meaningful as the electric field whichgeneratestheforceinquestionisconservative

W F xtimes E q x times= =

θ x cos θ x

Figure 1007 Charge moved at an angle to the field

If a chargemoves at an angle θ to an electric field thecomponent of the displacement parallel to the electric fieldisusedasshowninFigure918

W F x Eq x θcostimes= =

Theelectricpotentialenergyisstoredintheelectricfieldand theelectricfieldwill return theenergy to thepointcharge when required so as not to violate the Law ofconservationofenergy

Electric potential

The electric potential at a point in an electric field is defined as being the work done per unit charge in bringing a small positive point charge from infinity to that point

ΔV = Vinfin ndash Vf = ndash W ___ q

Ifwedesignatethepotentialenergytobezeroatinfinitythen it follows thatelectricpotentialmustalsobezeroat infinity and the electric potential at any point in an electricfieldwillbe

ΔV = ndash W ___ q

NowsupposeweapplyanexternalforcetoasmallpositivetestchargeasitismovedtowardsanisolatedpositivechargeTheexternalforcemustdoworkonthepositivetestchargetomoveittowardstheisolatedpositivechargeandtheworkmustbepositivewhiletheworkdonebytheelectricfieldmustthereforebenegativeSotheelectricpotentialatthatpointmustbepositiveaccordingtotheaboveequationIfanegativeisolatedchargeisusedtheelectricpotentialatapointonthepositivetestchargewouldbenegativePositivepoint chargesof their ownaccordmove fromaplaceofhighelectricpotential toaplaceof lowelectricpotentialNegative point charges move the other way from lowpotentialtohighpotentialInmovingfrompointAtopointBinthediagramthepositivecharge+qismovingfromalowelectricpotentialtoahighelectricpotential

In thedefinitiongiven the termldquoworkperunitchargerdquohas significance If the test charge is +16 times 10-19Cwherethe charge has a potential energy of 32 times 10-17Jthenthepotentialwouldbe32times10-17J+16times10-19C=200JC-1 Now if the charge was doubled the potential wouldbecome 64times10-17 J However the potential per unitchargewouldbethesame

Electric potential is a scalar quantityandithasunitsJC-1 orvoltswhere1voltequalsone joulepercoloumbThevoltallowsustoadoptaunitfortheelectricfieldintermsofthevolt

PreviouslytheunitfortheelectricfieldwasNCndash1

W = qV and F = qEsoW ___ V = F __ E

E = FV ___ W = FV ___ Fm V mndash1

That is the units of the electric field E can also beexpressed as V mndash1

The work done per unit charge in moving a point charge between two points in an electric field is again independent of the path taken

PHYSICS 2015indb 312 150514 337 PM

Fields

313

AH

L

Electric potential due to a point charge

Letus takeapoint rmetres froma chargedobjectThepotentialatthispointcanbecalculatedusingthefollowing

W = ndashFr = ndashqV and F = ndash q1 q2 _____ 4πε0r

2

Therefore

W = ndash q1 q2 _____ 4πε0r

2 times r = ndash q1 q2 _____ 4πε0r

= ndashq1 times q2 _____ 4πε0r

= ndashq1V

That is V q

4πε0r--------------- =

Orsimply

V kqr------

=

Example

Determinehowmuchworkisdonebytheelectricfieldofpointcharge150μCwhenachargeof200μCismovedfrom infinity to apoint 0400m from thepoint charge(Assumenoaccelerationofthecharges)

Solution

The work done by the electric field is W = -qV = -14πε0 times q times (Q rinfin - Q r0400)

W = (- 200 times 10-6 C times 900 times 109 NmC-2 times 150 times 10-6 C) divide 0400 m = - 0675 J

An external force would have to do +0675 J of work

Electric field strength and electric potential gradient

Letus lookbackatFigure1006Supposeagain that thecharge+qismovedasmalldistancebyaforceFfromAtoBsothattheforcecanbeconsideredconstantTheworkdoneisgivenby

∆W F x∆times=

TheforceFandtheelectricfieldEareoppositelydirectedandweknowthat

F = -qE and ΔW = q ΔV

Thereforetheworkdonecanbegivenas

q ΔV = -q E Δx

Therefore E V∆x∆-------

ndash=

The rate of change of potential ΔV at a point with respect to distance Δx in the direction in which the change is maximum is called the potential gradient We say that the electric field = - the potential gradient and the units are Vm-

1 From the equation we can see that in a graph of electric potential versus distance the gradient of the straight line equals the electric field strength

In reality if a charged particle enters a uniform electric field it will be accelerated uniformly by the field and its kinetic energy will increase This is why we had to assume no acceleration in the last worked example

Ek∆ 12---

mv2 q E xsdot sdot q Vx---

xsdot sdot q Vsdot= = = =

Example

Determine how far apart two parallel plates must besituated so that a potential difference of 150 times 102 V produces an electric field strength of 100 times 103 NC-1

Solution

Using E V∆x------- xhArrndash V∆

E-------15 102 Vtimes

100 103 N C 1ndashtimes------------------------------------------= = =

= 150 times 10-1

The plates are 150 times 10-1 m apart

The electric field and the electric potential at a point due to an evenly distributed charge +q on a sphere can be represented graphically as in Figure 1008

r

r

V

E

r0

r0

r0

Charge of +Q evenly distributed over surface

E 14πε0------------ Q

r2----- r r0gt=

V 14πε0------------Q

r----r r0gt=

Figure 1008 Electric field and potential due to a charged sphere

PHYSICS 2015indb 313 150514 337 PM

Chapter 10

314

AH

L

When the sphere becomes charged we know that thechargedistributes itselfevenlyoverthesurfaceThereforeeverypartofthematerialoftheconductorisatthesamepotential As the electric potential at a point is defined as beingnumericallyequaltotheworkdoneinbringingaunitpositivechargefrominfinitytothatpointithasaconstantvalueineverypartofthematerialoftheconductor

Since the potential is the same at all points on the conducting surfacethen∆V∆xiszeroButE=ndash∆V∆xThereforethe electric field inside the conductor is zero There is no electric field inside the conductor

SomefurtherobservationsofthegraphsinFigure1009are

bull OutsidethespherethegraphsobeytherelationshipsgivenasE α 1 r2 and V α 1 r

bull Atthesurfacer = r0Thereforetheelectricfieldandpotentialhavetheminimumvalueforr at this point and this infers a maximum field and potential

bull Insidethespheretheelectricfieldiszerobull Insidethespherenoworkisdonetomoveacharge

fromapointinsidetothesurfaceThereforethereisno potential difference and the potential is the same asitiswhenr = r0

Similargraphscanbedrawnfortheelectricfieldintensityandthe electric potential as a function of distance from conducting parallelplatesandsurfacesandthesearegiveninFigure1009

Potential plot E eld plotE eld

++++

ndashndashndashndash

xx x

+

x x

ndashx x

+ ndashx

x

+ +

x

x

Figure 1009 Electric field and electric potential at a distance from a charged surface

Potential due to one or more point charges

The potential due to one point charge can be determined byusingtheequationformula

V = kq r

Example 1

Determine the electric potential at a point 20 times 10-1 m fromthecentreofan isolatedconducting spherewithapointchargeof40pCinair

Solution

Using the formula

V = kq r we have

V90 10 9times( ) 40 10 12ndashtimes( )times

20 10 1ndashtimes( )------------------------------------------------------------------ 018 V= =

the potential at the point is 180 times 10-1 V

The potential due to a number of point charges can be determined by adding up the potentials due to individual point charges because the electric potential at any point outside a conducting sphere will be the same as if all the charge was concentrated at its centre

Example 2

Threepointchargesofareplacedattheverticesofaright-angledtriangleasshowninthediagrambelowDeterminetheabsolutepotentialat the+20μCchargeduetothetwoothercharges

- 6μC

+3μC +2μC

4 m

3 m

Solution

The electric potential of the +2 μC charge due to the ndash 6 μC charge is

V = (9 times 109 Nm2C-2 times -6 times 10-6 C) divide (radic 32 + 42) m = - 108 times 104 V

PHYSICS 2015indb 314 150514 338 PM

Fields

315

AH

L

Insummarywecanconcludethat

bull No work is done to move a charge along an equipotential

bull Equipotentials are always perpendicular to the electric lines of force

Figure1011and1012showsomeequipotential lines fortwo oppositely charged and identically positive spheresseparated by a distance

+ve ndashve

equipotential lines

Figure 1011 Equipotential lines between two opposite charges

equipotential lines

Figure 1012 Equipotential lines between two charges which are the same

+

ndash ndash ndash ndash

+ + +

40 V

30 V

20 V

10 V

50 V

Figure 1013 Equipotential lines between charged parallel plates

Figure1013showstheequipotentiallinesbetweenchargedparallel plates Throughout this chapter the similarities and differences between gravitational fields and electric fieldshavebeendiscussedTherelationshipsthatexistsbetweengravitationalandelectricquantitiesandtheeffectsofpointmassesandchargesissummarisedinTable1014

The electric potential of the +2 μC charge due to the +3 μC charge is

V = (9 times 109 Nm2C-2 times 3 times 10-6 C) divide 3m = 9 times 103 V

The net absolute potential is the sum of the 2 potentials- 108 times 104 V + 9 times 103 V =- 18 times 103 V

The absolute potential at the point is - 18 times 103 V

Equipotential surfacesRegions in space where the electric potential of a chargedistributionhasaconstantvaluearecalledequipotentials The placeswherethepotentialisconstantinthreedimensionsarecalled equipotential surfacesandwheretheyareconstantintwodimensionstheyarecalledequipotential lines

TheyareinsomewaysanalogoustothecontourlinesontopographicmapsInthiscasethegravitationalpotentialenergy is constant as amassmoves around the contourlines because the mass remains at the same elevationabovetheEarthrsquossurfaceThegravitationalfieldstrengthacts in a direction perpendicular to a contour line

Similarlybecausetheelectricpotentialonanequipotentiallinehas thesamevalueanelectric forcecandonoworkwhena testchargemovesonanequipotentialThereforethe electric field cannot have a component along anequipotentialandthusitmustbeeverywhereperpendiculartotheequipotentialsurfaceorequipotentiallineThisfactmakesiteasytoplotequipotentialsifthelinesofforceorlinesofelectricfluxofanelectricfieldareknown

Forexamplethereareaseriesofequipotentiallinesbetweentwoparallelplateconductorsthatareperpendiculartotheelectric fieldTherewill be a series of concentric circles(eachcirclefurtherapartthanthepreviousone)thatmapouttheequipotentialsaroundanisolatedpositivesphereThe lines of force and some equipotential lines for anisolatedpositivesphereareshowninFigure1010

Lines ofequipotential

Figure 1010 Equipotentials around an isolated positive sphere

PHYSICS 2015indb 315 150514 338 PM

Chapter 10

316

AH

L

Gravitational quantity Electrical quantityQ

uant

ities

g V∆x∆-------ndash= E V∆

x∆-------ndash=

Poin

t mas

ses a

nd ch

arge

s

V Gmr----

ndash= V 14πε0------------q

r---=

g G m

r2-----ndash= E 14πε0------------ q

r2-----=

F Gm1 m2

r2--------------= F 1

4πε0------------

q1q2

r2-----------=

Figure 1014 Formulas

Example

Deduce the electric potential on the surface of a gold nucleus that has a radius of 62 fm

Solution

Using the formula

V = kq r and knowing the atomic number of gold is 79 We will assume the nucleus is spherical and it behaves as if it were a point charge at its centre (relative to outside points)

V = 90 times 109 Nm2C-2 times 79 times 16 times 10-19 C divide 62 times 10-15 m = 18 times 107 V

The potential at the point is 18 MV

Example

Deduce the ionisation energy in electron-volts of theelectron in the hydrogen atom if the electron is in its ground state and it is in a circular orbit at a distance of 53 times 10-11 m from the proton

Solution

This problem is an energy coulombic circular motion question based on Bohrrsquos model of the atom (not the accepted quantum mechanics model) The ionisation energy is the energy required to remove the electron from the ground state to infinity The electron travels in a circular orbit and therefore has a centripetal acceleration The ionisation energy will counteract the coulombic force and the movement of the electron will be in the opposite direction to the centripetal force

Total energy = Ek electron + Ep due to the proton-electron interaction

ΣF = kqQ r2 = mv2 r and as such mv2 = = kqQ r

Therefore Ek electron = frac12 kqQ r

Ep due to the proton-electron interaction = - kqQ r

Total energy = frac12 kqQ r + - kqQ r = -frac12 kqQ r

= - 90 times 109 Nm2C-2 times (16 times 10-19 C)2 divide 53 times 10-11 m = -217 times 10-18 J

= -217 times 10-18 J divide 16 times 10-19 = -136 eV

The ionisation energy is 136 eV

Exercise 1021 ApointchargePisplacedmidwaybetweentwo

identicalnegativechargesWhichoneofthefollowingiscorrectwithregardstoelectricfieldandelectricpotentialatpointP

Electric field Electric potentialA non-zero zeroB zero non-zeroC non-zero non-zeroD zero zero

2 Twopositivechargedspheresaretiedtogetherinavacuumsomewhereinspacewheretherearenoexternal forces A has a mass of 25 g and a charge of20μCandBhasamassof15gandachargeof30μCThedistancebetweenthemis40cmTheyarethenreleasedasshowninthediagram

BEFORE AFTER

v1 v2

A B

(a) Determinetheirinitialelectricpotentialenergy in the before situation

(b) DeterminethespeedofsphereBafterrelease

3 Thediagrambelowrepresentstwoequipotentiallines in separated by a distance of 5 cm in a uniform electric field

+ + + + + + + +

ndash ndash ndash ndash ndash ndash ndash ndash

40 V

20 V 5 cm

PHYSICS 2015indb 316 150514 338 PM

Fields

317

AH

L

Determine the strength of the electric field

4 Thisquestionisabouttheelectricfieldduetoacharged sphere and the motion of electrons in that fieldThediagrambelowshowsanisolatedmetalsphereinavacuumthatcarriesanegativeelectriccharge of 60 μC

ndash

(a) Onthediagramdrawtheconventionalwayto represent the electric field pattern due to the charged sphere and lines to represent threeequipotentialsurfacesintheregionoutside the sphere

(b) Explainhowthelinesrepresentingtheequipotentialsurfacesthatyouhavesketchedindicatethatthestrengthoftheelectricfieldisdecreasingwithdistancefrom the centre of the sphere

(c) Theelectricfieldstrengthatthesurfaceofthe sphere and at points outside the sphere can be determined by assuming that the sphere acts as a point charge of magnitude 60 μC at its centre The radius of the sphere is 25 times 10ndash2 m Deduce that the magnitude of the field strength at the surface of the sphere is 86 times 107 Vmndash1

An electron is initially at rest on the surface of the sphere

(d) (i) Describethepathfollowedbytheelectronasitleavesthesurfaceofthesphere

(ii) Calculatetheinitialaccelerationofthe electron

5 Determinetheamountofworkthatisdoneinmovingachargeof100nCthroughapotentialdifferenceof150times102 V

6 Three identical 20 μC conducting spheres are placedatthecornersofanequilateraltriangleofsides 25 cm The triangle has one apex C pointing upthepageand2baseanglesAandBDeterminetheabsolutepotentialatB

7 Determinehowfaraparttwoparallelplatesmustbesituatedsothatapotentialdifferenceof250 times 102 V produces an electric field strength of 200 times 103 NC-1

8 Thegapbetweentwoparallelplatesis10times10-3mandthereisapotentialdifferenceof10times104 V betweentheplatesCalculate

i theworkdonebyanelectroninmovingfrom one plate to the other

ii thespeedwithwhichtheelectronreachesthe second plate if released from rest

iii theelectricfieldintensitybetweentheplates

9 Anelectronguninapicturetubeisacceleratedby a potential 25 times 103VDeterminethekineticenergygainedbytheelectroninelectron-volts

10 Determine the electric potential 20 times 10-2 m from achargeof-10times10-5 C

11 Determinetheelectricpotentialatapointmid-waybetweenachargeofndash20pCandanotherof+5pConthelinejoiningtheircentresifthecharges are 10 cm apart

12 During a thunderstorm the electric potential differencebetweenacloudandthegroundis10 times 109 V Determine the magnitude of the change in electric potential energy of an electron thatmovesbetweenthesepointsinelectron-volts

13 A charge of 15 μC is placed in a uniform electric fieldoftwooppositelychargedparallelplateswithamagnitudeof14times103 NC-1

(a) Determinetheworkthatmustbedoneagainstthefieldtomovethepointchargeadistance of 55 cm

(b) Calculatethepotentialdifferencebetweenthe final and initial positions of the charge

(c) Determinethepotentialdifferencebetweenthe plates if their separation distance is 15 cm

14 Duringaflashoflightningthepotentialdifferencebetweenacloudandthegroundwas12 times 109 V and the amount of transferred charge was32C

(a) Determinethechangeinenergyofthetransferred charge

(b) Iftheenergyreleasedwasallusedtoacceleratea1tonnecardeduceitsfinalspeed

(c) Iftheenergyreleasedcouldbeusedtomelticeat0degCdeducetheamountoficethatcould be melted

PHYSICS 2015indb 317 150514 338 PM

Chapter 10

318

AH

L

15 SupposethatwhenanelectronmovedfromAtoBin the diagram along an electric field line that the electric field does 36 times 10-19Jofworkonit

Determinethedifferencesinelectricpotential

(a) VB ndash VA(b) VC ndash VA(c) VC ndash VB

16 DeterminethepotentialatpointPthatislocatedatthecentreofthesquareasshowninthediagrambelow

- 6μC

+3μC +2μC

1m

1m

5μC

P

102 Fields at work

Essential idea Similar approaches can be taken inanalysingelectricalandgravitationalpotentialproblems

Understandingsbull Potentialandpotentialenergybull Potentialgradientbull Potentialdifferencebull Escapespeedbull Orbitalmotionorbitalspeedandorbitalenergybull Forcesandinverse-squarelawbehaviour

Orbital motion orbital speed and orbital energy

Although orbital motion may be circular elliptical orparabolic this sub-topic only dealswith circular orbitsThis sub-topic is not fundamentally new physics butan application that synthesizes ideas from gravitationcircularmotiondynamicsandenergy

TheMoonorbitstheEarthandinthissenseitisoftenreferredtoasasatelliteoftheEarthBefore1957itwastheonlyEarthsatelliteHoweverin1957theRussianslaunchedthefirstmanmadesatelliteSputnik1SincethisdatemanymoresatelliteshavebeenlaunchedandtherearenowliterallythousandsofthemorbitingtheEarthSomeareusedtomonitortheweathersome used to enable people to find accurately their position on thesurfaceoftheEarthmanyareusedincommunicationsandnodoubtsomeareusedtospyonothercountriesFigure1015showshowinprincipleasatellitecanbeputintoorbit

Theperson(whosesizeisgreatlyexaggeratedwithrespectto Earth) standing on the surface on the Earth throwssome stonesThe greater the speed with which a stoneis thrown the further it will land from her The pathsfollowedbythethrownstonesareparabolasByastretchof the imaginationwecanvisualisea situation inwhichastoneis thrownwithsuchaspeedthatbecauseof thecurvatureoftheEarthitwillnotlandonthesurfaceoftheEarthbutgointoldquoorbitrdquo(Path4onFigure1015)

NATURE OF SCIENCECommunication of scientific explanations The ability to apply field theory to the unobservable (charges) and the massively scaled (motion of satellites) required scientists to develop new ways to investigate analyse and report findings to a general public used to scientific discoveries based on tangible and discernible evidence (51)

copy IBO 2014

PHYSICS 2015indb 318 150514 338 PM

Fields

319

AH

L

1

2

34Earth

Figure 1015 Throwing a stone into orbit

TheforcethatcausesthestonestofollowaparabolicpathandtofalltoEarthisgravityandsimilarlytheforcethatkeeps the stone in orbit is gravity For circular motionto occur we have seen that a force must act at rightanglestothevelocityofanobjectthatistheremustbeacentripetalforceHenceinthesituationwedescribeherethe centripetal force for circular orbital motion about the EarthisprovidedbygravitationalattractionoftheEarth

We can calculate the speedwithwhich a stonemustbethrowninordertoputitintoorbitjustabovethesurfaceof the Earth

If the stone has mass m and speed v thenwehave fromNewtonrsquos2ndlaw

mv2

RE---------- G

MEm

RE2------------=

whereRE is the radius of the Earth and ME is the mass of the Earth

Bearinginmindthatg 0 GME

RE2--------= then

v g RE 10 64 106timestimes 8 103times= = =

Thatisthestonemustbethrownat8times103m sndash1

Clearly we are not going to get a satellite into orbit soclosetothesurfaceoftheEarthMovingatthisspeedthefrictionduetoairresistancewouldmeltthesatellitebeforeithadtravelledacoupleofkilometresInrealitythereforeasatelliteisputintoorbitabouttheEarthbysendingitattachedtoarocketbeyondtheEarthrsquosatmosphereandthengivingitacomponentofvelocityperpendiculartoaradialvectorfromtheEarthSeeFigure1016

Earth Satellite carried by rocket to hereSatellite orbit

Tangential component of velocity

Figure 1016 Getting a satellite into orbit

Keplerrsquos third law(This work of Kepler and Newtonrsquos synthesis of the work is an excellent example of the scientific method and makes for a good TOK discussion)

In 1627 Johannes Kepler (1571-1630)publishedhis lawsof planetary motion The laws are empirical in natureandwere deduced from the observations of theDanishastronomer Tycho de Brahe (1546-1601)The third lawgives a relationship between the radius of orbit R of a planet and its period TofrevolutionabouttheSunThelawisexpressedmathematicallyas

2

3 constantTR

=

We shall nowuseNewtonrsquos LawofGravitation to showhowitisthattheplanetsmoveinaccordancewithKeplerrsquosthirdlaw

InessenceNewtonwasabletousehislawofgravitytopredict the motion of the planets since all he had to do was factor theF givenby this law intohis second law F = ma to find their accelerations and hence their future positions

InFigure1017theEarthisshownorbitingtheSunandthedistancebetweentheircentresisR

RSun

Earth

F es

F se

Figure 1017 Planets move according to Keplerrsquos third law

Fes is the force that the Earth exerts on the Sun and Fse is the force that the Sun exerts on the Earth The forces are equalandoppositeandtheSunandtheEarthwillactuallyorbitaboutacommoncentreHoweversince theSun issoverymuchmoremassivethantheEarththiscommoncentrewillbeclosetothecentreoftheSunandsowecanregard the Earth as orbiting about the centre of the Sun Theotherthingthatweshallassumeisthatwecanignorethe forces that the other planets exert on the Earth (This wouldnotbeawisethingtodoifyouwereplanningtosend a space ship to theMoon for example)We shallalso assume that we have followed Newtonrsquos exampleandindeedprovedthataspherewillactasapointmasssituated at the centre of the sphere

Kepler had postulated that the orbits of the planets areellipticalbut since theeccentricityof theEarthrsquosorbit issmallweshallassumeacircularorbit

PHYSICS 2015indb 319 150514 338 PM

Chapter 10

320

AH

L

TheaccelerationoftheEarthtowardstheSunisa = Rω2

whereω ω 2πT------=

Hencea R 2π

T------ times

2 4π2 RT 2------------- = =

But the acceleration is given by Newtonrsquos Second Law F = mawhereFisnowgivenbytheLawofGravitationSo in this situation

F maGMs Me

R2-------------------= = butwealsohavethat

a a R 2πT

------ times

2 4π2 RT 2------------- = = and m = Me so that

G MsMe

R2------------------- Me4π2R

T2-------------times

GMs

4π2-----------rArr R3

T2------= =

Butthequantity

isaconstantthathasthesamevalueforeachoftheplanetssowehaveforalltheplanetsnotjustEarththat

R3

T2------ k =

wherekisaconstantWhichisofcourseKeplerrsquosthirdlaw

ThisisindeedanamazingbreakthroughItisdifficulttorefutethe idea that all particles attract each other in accordance withtheLawofGravitationwhenthelawisabletoaccountfortheobservedmotionoftheplanetsabouttheSun

The gravitational effects of the planets upon each othershould produce perturbations in their orbits Such is the predictivepoweroftheUniversalGravitationalLawthatit enabled physicists to compute these perturbations The telescopehadbeeninventedin1608andbythemiddleofthe 18th Century had reached a degree of perfection in design that enabled astronomers to actually measure the orbital perturbations of the planets Their measurements werealways inagreementwith thepredictionsmadebyNewtonrsquoslawHoweverin1781anewplanetUranuswasdiscovered and the orbit of this planet did not fit withthe orbit predicted by Universal Gravitation Such wasthe physicistrsquos faith in theNewtonianmethod that theysuspectedthatthediscrepancywasduetothepresenceofayetundetectedplanetUsingtheLawofGravitationtheFrenchastronomerJLeverrier and the English astronomer J C Adamswereable tocalculate justhowmassive thisnewplanetmustbeandalsowhereitshouldbeIn1846the planetNeptunewas discovered justwhere they hadpredicted Ina similarwaydiscrepancies in theorbitofNeptune ledto thepredictionandsubsequentdiscoveryin1930oftheplanetPlutoNewtonrsquosLawofGravitation

had passed the ultimate test of any theory it is not only abletoexplainexistingdatabutalsotomakepredictions

Satellite energyWhenasatelliteisinorbitaboutaplanetitwillhavebothkineticenergyandgravitationalpotentialenergySupposewe consider a satellite ofmass m that is in an orbit of radius r about a planet of mass M

Thegravitationalpotentialduetotheplanetatdistancer from its centre is

ThegravitationalpotentialenergyofthesatelliteVsat

is therefore GMe m

r----------------ndash

ThatisVsat GMem

r----------------ndash=

Thegravitationalfieldstrengthatthesurfaceoftheplanetisgivenby

g 0GMe

Re2------------=

Hencewecanwrite

Vsat g0 Re

2

rndash=

m

ThekineticenergyofthesatelliteKsatisequaltofrac12mv2wherev is its orbital speed

Byequatingthegravitationalforceactingonthesatellitetoitscentripetalaccelerationwehave

G Mem

r2---------------- mv2

r---------- mv2hArr

GMemr----------------= =

Fromwhich12---

mv2 12---

GMemr----------------times=

= 0 e

2g R

r

2m

Which is actually quite interesting since it shows thatirrespective of the orbital radius the KE is numericallyequal to half the PE Also the total energy E tot K sat Vsat+ 1

2---G Me m

r----------------times G Me m

r----------------ndash + 1

2---G Me m

r----------------ndash= = =of the satelliteisalwaysnegativesince

E tot K sat Vsat+ 12---

G Me mr----------------times

G Me mr----------------ndash

+ 12---

G Me mr----------------ndash= = =

PHYSICS 2015indb 320 150514 338 PM

Fields

321

AH

L

The energies of an orbiting satellite as a function of radial distancefromthecentreofaplanetareshownplottedinFigure1018

12

100806

0402

0ndash02

ndash04

ndash06ndash08ndash10

ndash12

2 4 6 8 10 12

ener

gya

rbita

ry u

nits

distance timesR

kinetic energytotal energy

potential energy

Figure 1018 Energy of an orbiting satellite as a function of distance from the centre of a planet

WeightlessnessSuppose that you are inanelevator(lift)thatisdescendingat constant speedandyou let goof abook thatyouareholdinginyourhandThebookwillfalltothefloorwithaccelerationequaltotheaccelerationduetogravityIfthecablethatsupportstheelevatorweretosnap(asituationthatItrustwillneverhappentoanyofyou)andyounowletgothebookthatyouareholdinginyourotherhandthis book will not fall to the floor - it will stay exactlyin linewithyourhandThis isbecause thebook isnowfallingwith the sameaccelerationas theelevatorandassuch the book cannot ldquocatchrdquo up with the floor of theelevatorFurthermoreifyouhappenedtobestandingonasetofbathroomscalesthescaleswouldnowreadzero-youwouldbeapparentlyweightlessItisthisideaoffreefallthatexplainstheapparentweightlessnessofastronautsin an orbiting satellite These astronauts are in free fall inthesensethattheyareacceleratingtowardsthecentre of the Earth

It is actually possible to define theweight of a body inseveral different waysWe can define it for example asthegravitationalforceexertedonthebodybyaspecifiedobject such as the EarthThiswe have seen thatwe doinlotsofsituationswherewedefinetheweightasbeingequaltomgIfweusethisdefinitionthenanobjectinfreefallcannotbydefinitionbeweightlesssinceitisstillinagravitationalfieldHoweverifwedefinetheweightofanobjectintermsofaldquoweighingrdquoprocesssuchasthereadingonasetofbathroomscaleswhichineffectmeasuresthecontact force between the object and the scales thenclearlyobjectsinfreefallareweightlessOnenowhasto

askthequestionwhetherornotitispossibleForexampletomeasurethegravitationalforceactingonanastronautin orbit about the Earth

WecanalsodefineweightintermsofthenetgravitationalforceactingonabodyduetoseveraldifferentobjectsForexample for an object out in space its weight could bedefined in terms of the resultant of the forces exerted on it bytheSuntheMoontheEarthandalltheotherplanetsin the Solar System If this resultant is zero at a particular pointthenthebodyisweightlessatthispoint

Inviewofthevariousdefinitionsofweightthatareavailabletousitisimportantthatwhenweusethewordldquoweightrdquoweareawareofthecontextinwhichitisbeingused

Example

Calculate the height above the surface of the Earth atwhichageo-stationarysatelliteorbits

Solution

A geo-stationary satellite is one that orbits the Earth in such a way that it is stationary with respect to a point on the surface of the Earth This means that its orbital period must be the same as the time for the Earth to spin once on its axis ie 24 hours

From Keplerrsquos third law we have G Ms

4π2----------- R3

T2------=

That is

RMe

mh

using the fact that the force of attraction between the satellite and the Earth is given by

FG Mem

R2----------------=

and that F = ma

where m is the mass of the satellite and a 4π2R

T2-------------=

we have

GMem

R2---------------- m 4π2R

T2-------------times

GMe

4π2------------rArr R3

T2------= =

PHYSICS 2015indb 321 150514 338 PM

Chapter 10

322

AH

L

Now the mass of the Earth is 60 times 1024 kg and the period T measured in seconds is given by T = 86400 s

So substitution gives R = 42 times 106 m

The radius of the Earth is 64 times 106 m so that the orbital height h is about 36 times 107 m

Example

Calculatetheminimumenergyrequiredtoputasatelliteofmass500kgintoanorbitthatisasaheightequaltotheEarthrsquosradiusabovethesurfaceoftheEarth

Solution

We have seen that when dealing with gravitational fields and potential it is useful to remember that

g 0G M

Re2---------= or g0 Re

2 GM=

The gravitational potential at the surface of the Earth is

g 0ndash Re GMRe

---------ndash=

The gravitational potential at a distance R

from the centre of the Earth is

=

The difference in potential between the surface of the Earth and a point distance R from the centre is therefore

V∆ g 0Re 1ReR------ndash

=

If R 2Re= then V∆g0 Re

2------------=

This means that the work required to ldquoliftrdquo the satellite into orbit is g0Rm where m is the mass of the satellite This is equal to

10 32 106 500timestimestimes = 16000 MJ

However the satellite must also have kinetic energy in order to orbit Earth This will be equal to

g 0 mRe2

2R-----------------g 0mRe

2

4----------------- 8000 MJ= =

The minimum energy required is therefore

24000 MJ

Exercise 1031 The speed needed to put a satellite in orbit does

not depend on

A the radius of the orbitB theshapeoftheorbitC thevalueofgattheorbitD the mass of the satellite

2 EstimatethespeedofanEarthsatellitewhoseorbitis400kmabovetheEarthrsquossurfaceAlsodetermine the period of the orbit

3 Calculatethespeedofa200kgsatelliteorbitingthe Earth at a height of 70 times 106 m

Assume that g = 82 m sndash2 for this orbit

4 TheradiioftwosatellitesXandYorbitingthe Earth are 2r and 8rwherer is the radius of the Earth Calculate the ratio of the periods of revolutionofXtoY

5 A satellite of mass mkgissentfromEarthrsquossurfaceinto an orbit of radius 5RwhereR is the radius of theEarthWritedownanexpressionfor

(a) thepotentialenergyofthesatelliteinorbit

(b) thekineticenergyofthesatelliteinorbit

(c) theminimumworkrequiredtosendthesatellitefromrestattheEarthrsquossurfaceintoits orbit

6 A satellite in an orbit of 10rfallsbacktoEarth(radius r)afteramalfunctionDeterminethespeedwithwhichitwillhittheEarthrsquossurface

7 The radius of the moon is frac14 that of the Earth AssumingEarthandtheMoontohavethesamedensitycomparetheaccelerationsoffreefallatthe surface of Earth to that at the surface of the Moon

8 UsethefollowingdatatodeterminethegravitationalfieldstrengthatthesurfaceoftheMoon and hence determine the escape speed from the surface of the Moon

Mass of the Moon = 73 times 1022kg

RadiusoftheMoon=17times106 m

PHYSICS 2015indb 322 150514 338 PM

Fields

311

AH

L

Solution

= minus = minus times times = times23

11 76

64 1067 10 ndash13 10

34 10MV GR

minus timestimes

Nkg-1

But V = -g0R

Therefore 7

0 6

13 10 3834 10

VgR

times= minus = =

timesm s-2

To determine the field strength gh at 68 times 106 m above the surface we use the fact that 0 2

Mg GR

= such that GM = g0R2

Therefore 2 2

0h 2 2 2

h h

38 (34) 042(102)

g RGMgR R

times= = = = m s-2

(the distance from the centre is 34 times 106 + 68 times 106 = 102 times 106 m)

Exercise 1011 Thegraphbelowshowshowthegravitational

potentialoutsideoftheEarthvarieswithdistancefrom the centre

0 10 20 30 40 50 60 70

ndash1

ndash2

ndash3

ndash4

ndash5

ndash6

ndash7

r m times 106

V

Jkgndash1

times 10

7

(a) Usethegraphtodeterminethegainingravitationalpotentialenergyofasatelliteofmass200kgasitmovesfromthesurfaceof the Earth to a height of 30 times 107mabovetheEarthrsquossurface

(b) Calculatetheenergyrequiredtotakeit to infinity

(c) DeterminetheslopeofthegraphatthesurfaceoftheEarthmCommenton youranswer

Electric potential energy

The concept of electric potential energy was developedwith gravitational potential energy in mind Just as anobjectnearthesurfaceoftheEarthhaspotentialenergybecause of its gravitational interaction with the Earthsotoothereiselectricalpotentialenergyassociatedwithinteracting charges

Letusfirstlookatacaseoftwopositivepointchargeseachof 1μC that are initially bound together by a thread in a vacuuminspacewithadistancebetweenthemof10cmasshowninFigure1005Whenthethreadiscutthepointchargesinitiallyatrestwouldmoveinoppositedirectionsmovingwithvelocitiesv1 and v2 along the direction of the electrostatic force of repulsion

BEFORE AFTER

v v21

Figure 1005 Interaction of two positive particles

The electric potential energybetweentwopointchargescan be found by simply adding up the energy associated witheachpairofpointchargesForapairofinteractingchargestheelectricpotentialenergyisgivenby

∆U = ∆Ep + ∆Ek = ∆W = ∆Fr = kqQ

____ r2 times r = kqQ

____ r

Becausenoexternalforceisactingonthesystemtheenergyand momentum must be conserved Initially Ek = 0 and Ep =kqQr=9times109 times 1 times 10-1201m=009JWhentheyareagreatdistance fromeachotherEpwillbenegligibleThefinalenergywillbeequaltofrac12mv1

2+frac12mv22=009J

Momentum isalsoconservedand thevelocitieswouldbethe same magnitude but in opposite directions

Electric potential energyismoreoftendefinedintermsofapointchargemovinginanelectricfieldas

lsquothe electricpotential energybetweenany twopoints inan electric field is defined as negativeoftheworkdone by anelectricfieldinmovingapoint electric chargebetweentwolocationsintheelectricfieldrsquo

ΔU = ΔEp = -ΔW = -Fd = qEx

ΔU = ΔEp +ΔEk = ΔW = ΔFr = kqQ r2 x r = kqQ

wherexisthedistancemovedalong(oroppositeto)thedirection of the electric field

Electric potential energyismeasuredinjoule(J)Justasworkisascalarquantitysotooelectricalpotentialenergyis a scalar quantity The negative of the work done by an electric field in moving a unit electric charge between two points is independent of the path taken Inphysicswesaytheelectricfieldisalsquoconservativersquo field

PHYSICS 2015indb 311 150514 337 PM

Chapter 10

312

AH

L

Supposeanexternalforcesuchasyourhandmovesasmallpositive point test charge in the direction of a uniformelectric fieldAs it ismoving itmust be gaining kineticenergyIfthisoccursthentheelectricpotentialenergyofthe unit charge is changing

InFigure1006apointcharge+qismovedbetweenpointsAandBthroughadistancex in a uniform electric field

B

A

x

+ q

Figure 1006 Movement of a positive point charge in a uniform field

InordertomoveapositivepointchargefrompointAtopointBanexternal forcemustbeapplied to thechargeequaltoqE (F = qE)

Since the force is applied through a distance xthennegativeworkhas to bedone tomove the charge because energyis gainedmeaning there is an increaseelectric potential energybetweenthetwopointsRememberthattheworkdoneisequivalenttotheenergygainedorlostinmovingthe charge through the electric field The concept of electric potential energy is only meaningful as the electric field whichgeneratestheforceinquestionisconservative

W F xtimes E q x times= =

θ x cos θ x

Figure 1007 Charge moved at an angle to the field

If a chargemoves at an angle θ to an electric field thecomponent of the displacement parallel to the electric fieldisusedasshowninFigure918

W F x Eq x θcostimes= =

Theelectricpotentialenergyisstoredintheelectricfieldand theelectricfieldwill return theenergy to thepointcharge when required so as not to violate the Law ofconservationofenergy

Electric potential

The electric potential at a point in an electric field is defined as being the work done per unit charge in bringing a small positive point charge from infinity to that point

ΔV = Vinfin ndash Vf = ndash W ___ q

Ifwedesignatethepotentialenergytobezeroatinfinitythen it follows thatelectricpotentialmustalsobezeroat infinity and the electric potential at any point in an electricfieldwillbe

ΔV = ndash W ___ q

NowsupposeweapplyanexternalforcetoasmallpositivetestchargeasitismovedtowardsanisolatedpositivechargeTheexternalforcemustdoworkonthepositivetestchargetomoveittowardstheisolatedpositivechargeandtheworkmustbepositivewhiletheworkdonebytheelectricfieldmustthereforebenegativeSotheelectricpotentialatthatpointmustbepositiveaccordingtotheaboveequationIfanegativeisolatedchargeisusedtheelectricpotentialatapointonthepositivetestchargewouldbenegativePositivepoint chargesof their ownaccordmove fromaplaceofhighelectricpotential toaplaceof lowelectricpotentialNegative point charges move the other way from lowpotentialtohighpotentialInmovingfrompointAtopointBinthediagramthepositivecharge+qismovingfromalowelectricpotentialtoahighelectricpotential

In thedefinitiongiven the termldquoworkperunitchargerdquohas significance If the test charge is +16 times 10-19Cwherethe charge has a potential energy of 32 times 10-17Jthenthepotentialwouldbe32times10-17J+16times10-19C=200JC-1 Now if the charge was doubled the potential wouldbecome 64times10-17 J However the potential per unitchargewouldbethesame

Electric potential is a scalar quantityandithasunitsJC-1 orvoltswhere1voltequalsone joulepercoloumbThevoltallowsustoadoptaunitfortheelectricfieldintermsofthevolt

PreviouslytheunitfortheelectricfieldwasNCndash1

W = qV and F = qEsoW ___ V = F __ E

E = FV ___ W = FV ___ Fm V mndash1

That is the units of the electric field E can also beexpressed as V mndash1

The work done per unit charge in moving a point charge between two points in an electric field is again independent of the path taken

PHYSICS 2015indb 312 150514 337 PM

Fields

313

AH

L

Electric potential due to a point charge

Letus takeapoint rmetres froma chargedobjectThepotentialatthispointcanbecalculatedusingthefollowing

W = ndashFr = ndashqV and F = ndash q1 q2 _____ 4πε0r

2

Therefore

W = ndash q1 q2 _____ 4πε0r

2 times r = ndash q1 q2 _____ 4πε0r

= ndashq1 times q2 _____ 4πε0r

= ndashq1V

That is V q

4πε0r--------------- =

Orsimply

V kqr------

=

Example

Determinehowmuchworkisdonebytheelectricfieldofpointcharge150μCwhenachargeof200μCismovedfrom infinity to apoint 0400m from thepoint charge(Assumenoaccelerationofthecharges)

Solution

The work done by the electric field is W = -qV = -14πε0 times q times (Q rinfin - Q r0400)

W = (- 200 times 10-6 C times 900 times 109 NmC-2 times 150 times 10-6 C) divide 0400 m = - 0675 J

An external force would have to do +0675 J of work

Electric field strength and electric potential gradient

Letus lookbackatFigure1006Supposeagain that thecharge+qismovedasmalldistancebyaforceFfromAtoBsothattheforcecanbeconsideredconstantTheworkdoneisgivenby

∆W F x∆times=

TheforceFandtheelectricfieldEareoppositelydirectedandweknowthat

F = -qE and ΔW = q ΔV

Thereforetheworkdonecanbegivenas

q ΔV = -q E Δx

Therefore E V∆x∆-------

ndash=

The rate of change of potential ΔV at a point with respect to distance Δx in the direction in which the change is maximum is called the potential gradient We say that the electric field = - the potential gradient and the units are Vm-

1 From the equation we can see that in a graph of electric potential versus distance the gradient of the straight line equals the electric field strength

In reality if a charged particle enters a uniform electric field it will be accelerated uniformly by the field and its kinetic energy will increase This is why we had to assume no acceleration in the last worked example

Ek∆ 12---

mv2 q E xsdot sdot q Vx---

xsdot sdot q Vsdot= = = =

Example

Determine how far apart two parallel plates must besituated so that a potential difference of 150 times 102 V produces an electric field strength of 100 times 103 NC-1

Solution

Using E V∆x------- xhArrndash V∆

E-------15 102 Vtimes

100 103 N C 1ndashtimes------------------------------------------= = =

= 150 times 10-1

The plates are 150 times 10-1 m apart

The electric field and the electric potential at a point due to an evenly distributed charge +q on a sphere can be represented graphically as in Figure 1008

r

r

V

E

r0

r0

r0

Charge of +Q evenly distributed over surface

E 14πε0------------ Q

r2----- r r0gt=

V 14πε0------------Q

r----r r0gt=

Figure 1008 Electric field and potential due to a charged sphere

PHYSICS 2015indb 313 150514 337 PM

Chapter 10

314

AH

L

When the sphere becomes charged we know that thechargedistributes itselfevenlyoverthesurfaceThereforeeverypartofthematerialoftheconductorisatthesamepotential As the electric potential at a point is defined as beingnumericallyequaltotheworkdoneinbringingaunitpositivechargefrominfinitytothatpointithasaconstantvalueineverypartofthematerialoftheconductor

Since the potential is the same at all points on the conducting surfacethen∆V∆xiszeroButE=ndash∆V∆xThereforethe electric field inside the conductor is zero There is no electric field inside the conductor

SomefurtherobservationsofthegraphsinFigure1009are

bull OutsidethespherethegraphsobeytherelationshipsgivenasE α 1 r2 and V α 1 r

bull Atthesurfacer = r0Thereforetheelectricfieldandpotentialhavetheminimumvalueforr at this point and this infers a maximum field and potential

bull Insidethespheretheelectricfieldiszerobull Insidethespherenoworkisdonetomoveacharge

fromapointinsidetothesurfaceThereforethereisno potential difference and the potential is the same asitiswhenr = r0

Similargraphscanbedrawnfortheelectricfieldintensityandthe electric potential as a function of distance from conducting parallelplatesandsurfacesandthesearegiveninFigure1009

Potential plot E eld plotE eld

++++

ndashndashndashndash

xx x

+

x x

ndashx x

+ ndashx

x

+ +

x

x

Figure 1009 Electric field and electric potential at a distance from a charged surface

Potential due to one or more point charges

The potential due to one point charge can be determined byusingtheequationformula

V = kq r

Example 1

Determine the electric potential at a point 20 times 10-1 m fromthecentreofan isolatedconducting spherewithapointchargeof40pCinair

Solution

Using the formula

V = kq r we have

V90 10 9times( ) 40 10 12ndashtimes( )times

20 10 1ndashtimes( )------------------------------------------------------------------ 018 V= =

the potential at the point is 180 times 10-1 V

The potential due to a number of point charges can be determined by adding up the potentials due to individual point charges because the electric potential at any point outside a conducting sphere will be the same as if all the charge was concentrated at its centre

Example 2

Threepointchargesofareplacedattheverticesofaright-angledtriangleasshowninthediagrambelowDeterminetheabsolutepotentialat the+20μCchargeduetothetwoothercharges

- 6μC

+3μC +2μC

4 m

3 m

Solution

The electric potential of the +2 μC charge due to the ndash 6 μC charge is

V = (9 times 109 Nm2C-2 times -6 times 10-6 C) divide (radic 32 + 42) m = - 108 times 104 V

PHYSICS 2015indb 314 150514 338 PM

Fields

315

AH

L

Insummarywecanconcludethat

bull No work is done to move a charge along an equipotential

bull Equipotentials are always perpendicular to the electric lines of force

Figure1011and1012showsomeequipotential lines fortwo oppositely charged and identically positive spheresseparated by a distance

+ve ndashve

equipotential lines

Figure 1011 Equipotential lines between two opposite charges

equipotential lines

Figure 1012 Equipotential lines between two charges which are the same

+

ndash ndash ndash ndash

+ + +

40 V

30 V

20 V

10 V

50 V

Figure 1013 Equipotential lines between charged parallel plates

Figure1013showstheequipotentiallinesbetweenchargedparallel plates Throughout this chapter the similarities and differences between gravitational fields and electric fieldshavebeendiscussedTherelationshipsthatexistsbetweengravitationalandelectricquantitiesandtheeffectsofpointmassesandchargesissummarisedinTable1014

The electric potential of the +2 μC charge due to the +3 μC charge is

V = (9 times 109 Nm2C-2 times 3 times 10-6 C) divide 3m = 9 times 103 V

The net absolute potential is the sum of the 2 potentials- 108 times 104 V + 9 times 103 V =- 18 times 103 V

The absolute potential at the point is - 18 times 103 V

Equipotential surfacesRegions in space where the electric potential of a chargedistributionhasaconstantvaluearecalledequipotentials The placeswherethepotentialisconstantinthreedimensionsarecalled equipotential surfacesandwheretheyareconstantintwodimensionstheyarecalledequipotential lines

TheyareinsomewaysanalogoustothecontourlinesontopographicmapsInthiscasethegravitationalpotentialenergy is constant as amassmoves around the contourlines because the mass remains at the same elevationabovetheEarthrsquossurfaceThegravitationalfieldstrengthacts in a direction perpendicular to a contour line

Similarlybecausetheelectricpotentialonanequipotentiallinehas thesamevalueanelectric forcecandonoworkwhena testchargemovesonanequipotentialThereforethe electric field cannot have a component along anequipotentialandthusitmustbeeverywhereperpendiculartotheequipotentialsurfaceorequipotentiallineThisfactmakesiteasytoplotequipotentialsifthelinesofforceorlinesofelectricfluxofanelectricfieldareknown

Forexamplethereareaseriesofequipotentiallinesbetweentwoparallelplateconductorsthatareperpendiculartotheelectric fieldTherewill be a series of concentric circles(eachcirclefurtherapartthanthepreviousone)thatmapouttheequipotentialsaroundanisolatedpositivesphereThe lines of force and some equipotential lines for anisolatedpositivesphereareshowninFigure1010

Lines ofequipotential

Figure 1010 Equipotentials around an isolated positive sphere

PHYSICS 2015indb 315 150514 338 PM

Chapter 10

316

AH

L

Gravitational quantity Electrical quantityQ

uant

ities

g V∆x∆-------ndash= E V∆

x∆-------ndash=

Poin

t mas

ses a

nd ch

arge

s

V Gmr----

ndash= V 14πε0------------q

r---=

g G m

r2-----ndash= E 14πε0------------ q

r2-----=

F Gm1 m2

r2--------------= F 1

4πε0------------

q1q2

r2-----------=

Figure 1014 Formulas

Example

Deduce the electric potential on the surface of a gold nucleus that has a radius of 62 fm

Solution

Using the formula

V = kq r and knowing the atomic number of gold is 79 We will assume the nucleus is spherical and it behaves as if it were a point charge at its centre (relative to outside points)

V = 90 times 109 Nm2C-2 times 79 times 16 times 10-19 C divide 62 times 10-15 m = 18 times 107 V

The potential at the point is 18 MV

Example

Deduce the ionisation energy in electron-volts of theelectron in the hydrogen atom if the electron is in its ground state and it is in a circular orbit at a distance of 53 times 10-11 m from the proton

Solution

This problem is an energy coulombic circular motion question based on Bohrrsquos model of the atom (not the accepted quantum mechanics model) The ionisation energy is the energy required to remove the electron from the ground state to infinity The electron travels in a circular orbit and therefore has a centripetal acceleration The ionisation energy will counteract the coulombic force and the movement of the electron will be in the opposite direction to the centripetal force

Total energy = Ek electron + Ep due to the proton-electron interaction

ΣF = kqQ r2 = mv2 r and as such mv2 = = kqQ r

Therefore Ek electron = frac12 kqQ r

Ep due to the proton-electron interaction = - kqQ r

Total energy = frac12 kqQ r + - kqQ r = -frac12 kqQ r

= - 90 times 109 Nm2C-2 times (16 times 10-19 C)2 divide 53 times 10-11 m = -217 times 10-18 J

= -217 times 10-18 J divide 16 times 10-19 = -136 eV

The ionisation energy is 136 eV

Exercise 1021 ApointchargePisplacedmidwaybetweentwo

identicalnegativechargesWhichoneofthefollowingiscorrectwithregardstoelectricfieldandelectricpotentialatpointP

Electric field Electric potentialA non-zero zeroB zero non-zeroC non-zero non-zeroD zero zero

2 Twopositivechargedspheresaretiedtogetherinavacuumsomewhereinspacewheretherearenoexternal forces A has a mass of 25 g and a charge of20μCandBhasamassof15gandachargeof30μCThedistancebetweenthemis40cmTheyarethenreleasedasshowninthediagram

BEFORE AFTER

v1 v2

A B

(a) Determinetheirinitialelectricpotentialenergy in the before situation

(b) DeterminethespeedofsphereBafterrelease

3 Thediagrambelowrepresentstwoequipotentiallines in separated by a distance of 5 cm in a uniform electric field

+ + + + + + + +

ndash ndash ndash ndash ndash ndash ndash ndash

40 V

20 V 5 cm

PHYSICS 2015indb 316 150514 338 PM

Fields

317

AH

L

Determine the strength of the electric field

4 Thisquestionisabouttheelectricfieldduetoacharged sphere and the motion of electrons in that fieldThediagrambelowshowsanisolatedmetalsphereinavacuumthatcarriesanegativeelectriccharge of 60 μC

ndash

(a) Onthediagramdrawtheconventionalwayto represent the electric field pattern due to the charged sphere and lines to represent threeequipotentialsurfacesintheregionoutside the sphere

(b) Explainhowthelinesrepresentingtheequipotentialsurfacesthatyouhavesketchedindicatethatthestrengthoftheelectricfieldisdecreasingwithdistancefrom the centre of the sphere

(c) Theelectricfieldstrengthatthesurfaceofthe sphere and at points outside the sphere can be determined by assuming that the sphere acts as a point charge of magnitude 60 μC at its centre The radius of the sphere is 25 times 10ndash2 m Deduce that the magnitude of the field strength at the surface of the sphere is 86 times 107 Vmndash1

An electron is initially at rest on the surface of the sphere

(d) (i) Describethepathfollowedbytheelectronasitleavesthesurfaceofthesphere

(ii) Calculatetheinitialaccelerationofthe electron

5 Determinetheamountofworkthatisdoneinmovingachargeof100nCthroughapotentialdifferenceof150times102 V

6 Three identical 20 μC conducting spheres are placedatthecornersofanequilateraltriangleofsides 25 cm The triangle has one apex C pointing upthepageand2baseanglesAandBDeterminetheabsolutepotentialatB

7 Determinehowfaraparttwoparallelplatesmustbesituatedsothatapotentialdifferenceof250 times 102 V produces an electric field strength of 200 times 103 NC-1

8 Thegapbetweentwoparallelplatesis10times10-3mandthereisapotentialdifferenceof10times104 V betweentheplatesCalculate

i theworkdonebyanelectroninmovingfrom one plate to the other

ii thespeedwithwhichtheelectronreachesthe second plate if released from rest

iii theelectricfieldintensitybetweentheplates

9 Anelectronguninapicturetubeisacceleratedby a potential 25 times 103VDeterminethekineticenergygainedbytheelectroninelectron-volts

10 Determine the electric potential 20 times 10-2 m from achargeof-10times10-5 C

11 Determinetheelectricpotentialatapointmid-waybetweenachargeofndash20pCandanotherof+5pConthelinejoiningtheircentresifthecharges are 10 cm apart

12 During a thunderstorm the electric potential differencebetweenacloudandthegroundis10 times 109 V Determine the magnitude of the change in electric potential energy of an electron thatmovesbetweenthesepointsinelectron-volts

13 A charge of 15 μC is placed in a uniform electric fieldoftwooppositelychargedparallelplateswithamagnitudeof14times103 NC-1

(a) Determinetheworkthatmustbedoneagainstthefieldtomovethepointchargeadistance of 55 cm

(b) Calculatethepotentialdifferencebetweenthe final and initial positions of the charge

(c) Determinethepotentialdifferencebetweenthe plates if their separation distance is 15 cm

14 Duringaflashoflightningthepotentialdifferencebetweenacloudandthegroundwas12 times 109 V and the amount of transferred charge was32C

(a) Determinethechangeinenergyofthetransferred charge

(b) Iftheenergyreleasedwasallusedtoacceleratea1tonnecardeduceitsfinalspeed

(c) Iftheenergyreleasedcouldbeusedtomelticeat0degCdeducetheamountoficethatcould be melted

PHYSICS 2015indb 317 150514 338 PM

Chapter 10

318

AH

L

15 SupposethatwhenanelectronmovedfromAtoBin the diagram along an electric field line that the electric field does 36 times 10-19Jofworkonit

Determinethedifferencesinelectricpotential

(a) VB ndash VA(b) VC ndash VA(c) VC ndash VB

16 DeterminethepotentialatpointPthatislocatedatthecentreofthesquareasshowninthediagrambelow

- 6μC

+3μC +2μC

1m

1m

5μC

P

102 Fields at work

Essential idea Similar approaches can be taken inanalysingelectricalandgravitationalpotentialproblems

Understandingsbull Potentialandpotentialenergybull Potentialgradientbull Potentialdifferencebull Escapespeedbull Orbitalmotionorbitalspeedandorbitalenergybull Forcesandinverse-squarelawbehaviour

Orbital motion orbital speed and orbital energy

Although orbital motion may be circular elliptical orparabolic this sub-topic only dealswith circular orbitsThis sub-topic is not fundamentally new physics butan application that synthesizes ideas from gravitationcircularmotiondynamicsandenergy

TheMoonorbitstheEarthandinthissenseitisoftenreferredtoasasatelliteoftheEarthBefore1957itwastheonlyEarthsatelliteHoweverin1957theRussianslaunchedthefirstmanmadesatelliteSputnik1SincethisdatemanymoresatelliteshavebeenlaunchedandtherearenowliterallythousandsofthemorbitingtheEarthSomeareusedtomonitortheweathersome used to enable people to find accurately their position on thesurfaceoftheEarthmanyareusedincommunicationsandnodoubtsomeareusedtospyonothercountriesFigure1015showshowinprincipleasatellitecanbeputintoorbit

Theperson(whosesizeisgreatlyexaggeratedwithrespectto Earth) standing on the surface on the Earth throwssome stonesThe greater the speed with which a stoneis thrown the further it will land from her The pathsfollowedbythethrownstonesareparabolasByastretchof the imaginationwecanvisualisea situation inwhichastoneis thrownwithsuchaspeedthatbecauseof thecurvatureoftheEarthitwillnotlandonthesurfaceoftheEarthbutgointoldquoorbitrdquo(Path4onFigure1015)

NATURE OF SCIENCECommunication of scientific explanations The ability to apply field theory to the unobservable (charges) and the massively scaled (motion of satellites) required scientists to develop new ways to investigate analyse and report findings to a general public used to scientific discoveries based on tangible and discernible evidence (51)

copy IBO 2014

PHYSICS 2015indb 318 150514 338 PM

Fields

319

AH

L

1

2

34Earth

Figure 1015 Throwing a stone into orbit

TheforcethatcausesthestonestofollowaparabolicpathandtofalltoEarthisgravityandsimilarlytheforcethatkeeps the stone in orbit is gravity For circular motionto occur we have seen that a force must act at rightanglestothevelocityofanobjectthatistheremustbeacentripetalforceHenceinthesituationwedescribeherethe centripetal force for circular orbital motion about the EarthisprovidedbygravitationalattractionoftheEarth

We can calculate the speedwithwhich a stonemustbethrowninordertoputitintoorbitjustabovethesurfaceof the Earth

If the stone has mass m and speed v thenwehave fromNewtonrsquos2ndlaw

mv2

RE---------- G

MEm

RE2------------=

whereRE is the radius of the Earth and ME is the mass of the Earth

Bearinginmindthatg 0 GME

RE2--------= then

v g RE 10 64 106timestimes 8 103times= = =

Thatisthestonemustbethrownat8times103m sndash1

Clearly we are not going to get a satellite into orbit soclosetothesurfaceoftheEarthMovingatthisspeedthefrictionduetoairresistancewouldmeltthesatellitebeforeithadtravelledacoupleofkilometresInrealitythereforeasatelliteisputintoorbitabouttheEarthbysendingitattachedtoarocketbeyondtheEarthrsquosatmosphereandthengivingitacomponentofvelocityperpendiculartoaradialvectorfromtheEarthSeeFigure1016

Earth Satellite carried by rocket to hereSatellite orbit

Tangential component of velocity

Figure 1016 Getting a satellite into orbit

Keplerrsquos third law(This work of Kepler and Newtonrsquos synthesis of the work is an excellent example of the scientific method and makes for a good TOK discussion)

In 1627 Johannes Kepler (1571-1630)publishedhis lawsof planetary motion The laws are empirical in natureandwere deduced from the observations of theDanishastronomer Tycho de Brahe (1546-1601)The third lawgives a relationship between the radius of orbit R of a planet and its period TofrevolutionabouttheSunThelawisexpressedmathematicallyas

2

3 constantTR

=

We shall nowuseNewtonrsquos LawofGravitation to showhowitisthattheplanetsmoveinaccordancewithKeplerrsquosthirdlaw

InessenceNewtonwasabletousehislawofgravitytopredict the motion of the planets since all he had to do was factor theF givenby this law intohis second law F = ma to find their accelerations and hence their future positions

InFigure1017theEarthisshownorbitingtheSunandthedistancebetweentheircentresisR

RSun

Earth

F es

F se

Figure 1017 Planets move according to Keplerrsquos third law

Fes is the force that the Earth exerts on the Sun and Fse is the force that the Sun exerts on the Earth The forces are equalandoppositeandtheSunandtheEarthwillactuallyorbitaboutacommoncentreHoweversince theSun issoverymuchmoremassivethantheEarththiscommoncentrewillbeclosetothecentreoftheSunandsowecanregard the Earth as orbiting about the centre of the Sun Theotherthingthatweshallassumeisthatwecanignorethe forces that the other planets exert on the Earth (This wouldnotbeawisethingtodoifyouwereplanningtosend a space ship to theMoon for example)We shallalso assume that we have followed Newtonrsquos exampleandindeedprovedthataspherewillactasapointmasssituated at the centre of the sphere

Kepler had postulated that the orbits of the planets areellipticalbut since theeccentricityof theEarthrsquosorbit issmallweshallassumeacircularorbit

PHYSICS 2015indb 319 150514 338 PM

Chapter 10

320

AH

L

TheaccelerationoftheEarthtowardstheSunisa = Rω2

whereω ω 2πT------=

Hencea R 2π

T------ times

2 4π2 RT 2------------- = =

But the acceleration is given by Newtonrsquos Second Law F = mawhereFisnowgivenbytheLawofGravitationSo in this situation

F maGMs Me

R2-------------------= = butwealsohavethat

a a R 2πT

------ times

2 4π2 RT 2------------- = = and m = Me so that

G MsMe

R2------------------- Me4π2R

T2-------------times

GMs

4π2-----------rArr R3

T2------= =

Butthequantity

isaconstantthathasthesamevalueforeachoftheplanetssowehaveforalltheplanetsnotjustEarththat

R3

T2------ k =

wherekisaconstantWhichisofcourseKeplerrsquosthirdlaw

ThisisindeedanamazingbreakthroughItisdifficulttorefutethe idea that all particles attract each other in accordance withtheLawofGravitationwhenthelawisabletoaccountfortheobservedmotionoftheplanetsabouttheSun

The gravitational effects of the planets upon each othershould produce perturbations in their orbits Such is the predictivepoweroftheUniversalGravitationalLawthatit enabled physicists to compute these perturbations The telescopehadbeeninventedin1608andbythemiddleofthe 18th Century had reached a degree of perfection in design that enabled astronomers to actually measure the orbital perturbations of the planets Their measurements werealways inagreementwith thepredictionsmadebyNewtonrsquoslawHoweverin1781anewplanetUranuswasdiscovered and the orbit of this planet did not fit withthe orbit predicted by Universal Gravitation Such wasthe physicistrsquos faith in theNewtonianmethod that theysuspectedthatthediscrepancywasduetothepresenceofayetundetectedplanetUsingtheLawofGravitationtheFrenchastronomerJLeverrier and the English astronomer J C Adamswereable tocalculate justhowmassive thisnewplanetmustbeandalsowhereitshouldbeIn1846the planetNeptunewas discovered justwhere they hadpredicted Ina similarwaydiscrepancies in theorbitofNeptune ledto thepredictionandsubsequentdiscoveryin1930oftheplanetPlutoNewtonrsquosLawofGravitation

had passed the ultimate test of any theory it is not only abletoexplainexistingdatabutalsotomakepredictions

Satellite energyWhenasatelliteisinorbitaboutaplanetitwillhavebothkineticenergyandgravitationalpotentialenergySupposewe consider a satellite ofmass m that is in an orbit of radius r about a planet of mass M

Thegravitationalpotentialduetotheplanetatdistancer from its centre is

ThegravitationalpotentialenergyofthesatelliteVsat

is therefore GMe m

r----------------ndash

ThatisVsat GMem

r----------------ndash=

Thegravitationalfieldstrengthatthesurfaceoftheplanetisgivenby

g 0GMe

Re2------------=

Hencewecanwrite

Vsat g0 Re

2

rndash=

m

ThekineticenergyofthesatelliteKsatisequaltofrac12mv2wherev is its orbital speed

Byequatingthegravitationalforceactingonthesatellitetoitscentripetalaccelerationwehave

G Mem

r2---------------- mv2

r---------- mv2hArr

GMemr----------------= =

Fromwhich12---

mv2 12---

GMemr----------------times=

= 0 e

2g R

r

2m

Which is actually quite interesting since it shows thatirrespective of the orbital radius the KE is numericallyequal to half the PE Also the total energy E tot K sat Vsat+ 1

2---G Me m

r----------------times G Me m

r----------------ndash + 1

2---G Me m

r----------------ndash= = =of the satelliteisalwaysnegativesince

E tot K sat Vsat+ 12---

G Me mr----------------times

G Me mr----------------ndash

+ 12---

G Me mr----------------ndash= = =

PHYSICS 2015indb 320 150514 338 PM

Fields

321

AH

L

The energies of an orbiting satellite as a function of radial distancefromthecentreofaplanetareshownplottedinFigure1018

12

100806

0402

0ndash02

ndash04

ndash06ndash08ndash10

ndash12

2 4 6 8 10 12

ener

gya

rbita

ry u

nits

distance timesR

kinetic energytotal energy

potential energy

Figure 1018 Energy of an orbiting satellite as a function of distance from the centre of a planet

WeightlessnessSuppose that you are inanelevator(lift)thatisdescendingat constant speedandyou let goof abook thatyouareholdinginyourhandThebookwillfalltothefloorwithaccelerationequaltotheaccelerationduetogravityIfthecablethatsupportstheelevatorweretosnap(asituationthatItrustwillneverhappentoanyofyou)andyounowletgothebookthatyouareholdinginyourotherhandthis book will not fall to the floor - it will stay exactlyin linewithyourhandThis isbecause thebook isnowfallingwith the sameaccelerationas theelevatorandassuch the book cannot ldquocatchrdquo up with the floor of theelevatorFurthermoreifyouhappenedtobestandingonasetofbathroomscalesthescaleswouldnowreadzero-youwouldbeapparentlyweightlessItisthisideaoffreefallthatexplainstheapparentweightlessnessofastronautsin an orbiting satellite These astronauts are in free fall inthesensethattheyareacceleratingtowardsthecentre of the Earth

It is actually possible to define theweight of a body inseveral different waysWe can define it for example asthegravitationalforceexertedonthebodybyaspecifiedobject such as the EarthThiswe have seen thatwe doinlotsofsituationswherewedefinetheweightasbeingequaltomgIfweusethisdefinitionthenanobjectinfreefallcannotbydefinitionbeweightlesssinceitisstillinagravitationalfieldHoweverifwedefinetheweightofanobjectintermsofaldquoweighingrdquoprocesssuchasthereadingonasetofbathroomscaleswhichineffectmeasuresthecontact force between the object and the scales thenclearlyobjectsinfreefallareweightlessOnenowhasto

askthequestionwhetherornotitispossibleForexampletomeasurethegravitationalforceactingonanastronautin orbit about the Earth

WecanalsodefineweightintermsofthenetgravitationalforceactingonabodyduetoseveraldifferentobjectsForexample for an object out in space its weight could bedefined in terms of the resultant of the forces exerted on it bytheSuntheMoontheEarthandalltheotherplanetsin the Solar System If this resultant is zero at a particular pointthenthebodyisweightlessatthispoint

Inviewofthevariousdefinitionsofweightthatareavailabletousitisimportantthatwhenweusethewordldquoweightrdquoweareawareofthecontextinwhichitisbeingused

Example

Calculate the height above the surface of the Earth atwhichageo-stationarysatelliteorbits

Solution

A geo-stationary satellite is one that orbits the Earth in such a way that it is stationary with respect to a point on the surface of the Earth This means that its orbital period must be the same as the time for the Earth to spin once on its axis ie 24 hours

From Keplerrsquos third law we have G Ms

4π2----------- R3

T2------=

That is

RMe

mh

using the fact that the force of attraction between the satellite and the Earth is given by

FG Mem

R2----------------=

and that F = ma

where m is the mass of the satellite and a 4π2R

T2-------------=

we have

GMem

R2---------------- m 4π2R

T2-------------times

GMe

4π2------------rArr R3

T2------= =

PHYSICS 2015indb 321 150514 338 PM

Chapter 10

322

AH

L

Now the mass of the Earth is 60 times 1024 kg and the period T measured in seconds is given by T = 86400 s

So substitution gives R = 42 times 106 m

The radius of the Earth is 64 times 106 m so that the orbital height h is about 36 times 107 m

Example

Calculatetheminimumenergyrequiredtoputasatelliteofmass500kgintoanorbitthatisasaheightequaltotheEarthrsquosradiusabovethesurfaceoftheEarth

Solution

We have seen that when dealing with gravitational fields and potential it is useful to remember that

g 0G M

Re2---------= or g0 Re

2 GM=

The gravitational potential at the surface of the Earth is

g 0ndash Re GMRe

---------ndash=

The gravitational potential at a distance R

from the centre of the Earth is

=

The difference in potential between the surface of the Earth and a point distance R from the centre is therefore

V∆ g 0Re 1ReR------ndash

=

If R 2Re= then V∆g0 Re

2------------=

This means that the work required to ldquoliftrdquo the satellite into orbit is g0Rm where m is the mass of the satellite This is equal to

10 32 106 500timestimestimes = 16000 MJ

However the satellite must also have kinetic energy in order to orbit Earth This will be equal to

g 0 mRe2

2R-----------------g 0mRe

2

4----------------- 8000 MJ= =

The minimum energy required is therefore

24000 MJ

Exercise 1031 The speed needed to put a satellite in orbit does

not depend on

A the radius of the orbitB theshapeoftheorbitC thevalueofgattheorbitD the mass of the satellite

2 EstimatethespeedofanEarthsatellitewhoseorbitis400kmabovetheEarthrsquossurfaceAlsodetermine the period of the orbit

3 Calculatethespeedofa200kgsatelliteorbitingthe Earth at a height of 70 times 106 m

Assume that g = 82 m sndash2 for this orbit

4 TheradiioftwosatellitesXandYorbitingthe Earth are 2r and 8rwherer is the radius of the Earth Calculate the ratio of the periods of revolutionofXtoY

5 A satellite of mass mkgissentfromEarthrsquossurfaceinto an orbit of radius 5RwhereR is the radius of theEarthWritedownanexpressionfor

(a) thepotentialenergyofthesatelliteinorbit

(b) thekineticenergyofthesatelliteinorbit

(c) theminimumworkrequiredtosendthesatellitefromrestattheEarthrsquossurfaceintoits orbit

6 A satellite in an orbit of 10rfallsbacktoEarth(radius r)afteramalfunctionDeterminethespeedwithwhichitwillhittheEarthrsquossurface

7 The radius of the moon is frac14 that of the Earth AssumingEarthandtheMoontohavethesamedensitycomparetheaccelerationsoffreefallatthe surface of Earth to that at the surface of the Moon

8 UsethefollowingdatatodeterminethegravitationalfieldstrengthatthesurfaceoftheMoon and hence determine the escape speed from the surface of the Moon

Mass of the Moon = 73 times 1022kg

RadiusoftheMoon=17times106 m

PHYSICS 2015indb 322 150514 338 PM

Chapter 10

312

AH

L

Supposeanexternalforcesuchasyourhandmovesasmallpositive point test charge in the direction of a uniformelectric fieldAs it ismoving itmust be gaining kineticenergyIfthisoccursthentheelectricpotentialenergyofthe unit charge is changing

InFigure1006apointcharge+qismovedbetweenpointsAandBthroughadistancex in a uniform electric field

B

A

x

+ q

Figure 1006 Movement of a positive point charge in a uniform field

InordertomoveapositivepointchargefrompointAtopointBanexternal forcemustbeapplied to thechargeequaltoqE (F = qE)

Since the force is applied through a distance xthennegativeworkhas to bedone tomove the charge because energyis gainedmeaning there is an increaseelectric potential energybetweenthetwopointsRememberthattheworkdoneisequivalenttotheenergygainedorlostinmovingthe charge through the electric field The concept of electric potential energy is only meaningful as the electric field whichgeneratestheforceinquestionisconservative

W F xtimes E q x times= =

θ x cos θ x

Figure 1007 Charge moved at an angle to the field

If a chargemoves at an angle θ to an electric field thecomponent of the displacement parallel to the electric fieldisusedasshowninFigure918

W F x Eq x θcostimes= =

Theelectricpotentialenergyisstoredintheelectricfieldand theelectricfieldwill return theenergy to thepointcharge when required so as not to violate the Law ofconservationofenergy

Electric potential

The electric potential at a point in an electric field is defined as being the work done per unit charge in bringing a small positive point charge from infinity to that point

ΔV = Vinfin ndash Vf = ndash W ___ q

Ifwedesignatethepotentialenergytobezeroatinfinitythen it follows thatelectricpotentialmustalsobezeroat infinity and the electric potential at any point in an electricfieldwillbe

ΔV = ndash W ___ q

NowsupposeweapplyanexternalforcetoasmallpositivetestchargeasitismovedtowardsanisolatedpositivechargeTheexternalforcemustdoworkonthepositivetestchargetomoveittowardstheisolatedpositivechargeandtheworkmustbepositivewhiletheworkdonebytheelectricfieldmustthereforebenegativeSotheelectricpotentialatthatpointmustbepositiveaccordingtotheaboveequationIfanegativeisolatedchargeisusedtheelectricpotentialatapointonthepositivetestchargewouldbenegativePositivepoint chargesof their ownaccordmove fromaplaceofhighelectricpotential toaplaceof lowelectricpotentialNegative point charges move the other way from lowpotentialtohighpotentialInmovingfrompointAtopointBinthediagramthepositivecharge+qismovingfromalowelectricpotentialtoahighelectricpotential

In thedefinitiongiven the termldquoworkperunitchargerdquohas significance If the test charge is +16 times 10-19Cwherethe charge has a potential energy of 32 times 10-17Jthenthepotentialwouldbe32times10-17J+16times10-19C=200JC-1 Now if the charge was doubled the potential wouldbecome 64times10-17 J However the potential per unitchargewouldbethesame

Electric potential is a scalar quantityandithasunitsJC-1 orvoltswhere1voltequalsone joulepercoloumbThevoltallowsustoadoptaunitfortheelectricfieldintermsofthevolt

PreviouslytheunitfortheelectricfieldwasNCndash1

W = qV and F = qEsoW ___ V = F __ E

E = FV ___ W = FV ___ Fm V mndash1

That is the units of the electric field E can also beexpressed as V mndash1

The work done per unit charge in moving a point charge between two points in an electric field is again independent of the path taken

PHYSICS 2015indb 312 150514 337 PM

Fields

313

AH

L

Electric potential due to a point charge

Letus takeapoint rmetres froma chargedobjectThepotentialatthispointcanbecalculatedusingthefollowing

W = ndashFr = ndashqV and F = ndash q1 q2 _____ 4πε0r

2

Therefore

W = ndash q1 q2 _____ 4πε0r

2 times r = ndash q1 q2 _____ 4πε0r

= ndashq1 times q2 _____ 4πε0r

= ndashq1V

That is V q

4πε0r--------------- =

Orsimply

V kqr------

=

Example

Determinehowmuchworkisdonebytheelectricfieldofpointcharge150μCwhenachargeof200μCismovedfrom infinity to apoint 0400m from thepoint charge(Assumenoaccelerationofthecharges)

Solution

The work done by the electric field is W = -qV = -14πε0 times q times (Q rinfin - Q r0400)

W = (- 200 times 10-6 C times 900 times 109 NmC-2 times 150 times 10-6 C) divide 0400 m = - 0675 J

An external force would have to do +0675 J of work

Electric field strength and electric potential gradient

Letus lookbackatFigure1006Supposeagain that thecharge+qismovedasmalldistancebyaforceFfromAtoBsothattheforcecanbeconsideredconstantTheworkdoneisgivenby

∆W F x∆times=

TheforceFandtheelectricfieldEareoppositelydirectedandweknowthat

F = -qE and ΔW = q ΔV

Thereforetheworkdonecanbegivenas

q ΔV = -q E Δx

Therefore E V∆x∆-------

ndash=

The rate of change of potential ΔV at a point with respect to distance Δx in the direction in which the change is maximum is called the potential gradient We say that the electric field = - the potential gradient and the units are Vm-

1 From the equation we can see that in a graph of electric potential versus distance the gradient of the straight line equals the electric field strength

In reality if a charged particle enters a uniform electric field it will be accelerated uniformly by the field and its kinetic energy will increase This is why we had to assume no acceleration in the last worked example

Ek∆ 12---

mv2 q E xsdot sdot q Vx---

xsdot sdot q Vsdot= = = =

Example

Determine how far apart two parallel plates must besituated so that a potential difference of 150 times 102 V produces an electric field strength of 100 times 103 NC-1

Solution

Using E V∆x------- xhArrndash V∆

E-------15 102 Vtimes

100 103 N C 1ndashtimes------------------------------------------= = =

= 150 times 10-1

The plates are 150 times 10-1 m apart

The electric field and the electric potential at a point due to an evenly distributed charge +q on a sphere can be represented graphically as in Figure 1008

r

r

V

E

r0

r0

r0

Charge of +Q evenly distributed over surface

E 14πε0------------ Q

r2----- r r0gt=

V 14πε0------------Q

r----r r0gt=

Figure 1008 Electric field and potential due to a charged sphere

PHYSICS 2015indb 313 150514 337 PM

Chapter 10

314

AH

L

When the sphere becomes charged we know that thechargedistributes itselfevenlyoverthesurfaceThereforeeverypartofthematerialoftheconductorisatthesamepotential As the electric potential at a point is defined as beingnumericallyequaltotheworkdoneinbringingaunitpositivechargefrominfinitytothatpointithasaconstantvalueineverypartofthematerialoftheconductor

Since the potential is the same at all points on the conducting surfacethen∆V∆xiszeroButE=ndash∆V∆xThereforethe electric field inside the conductor is zero There is no electric field inside the conductor

SomefurtherobservationsofthegraphsinFigure1009are

bull OutsidethespherethegraphsobeytherelationshipsgivenasE α 1 r2 and V α 1 r

bull Atthesurfacer = r0Thereforetheelectricfieldandpotentialhavetheminimumvalueforr at this point and this infers a maximum field and potential

bull Insidethespheretheelectricfieldiszerobull Insidethespherenoworkisdonetomoveacharge

fromapointinsidetothesurfaceThereforethereisno potential difference and the potential is the same asitiswhenr = r0

Similargraphscanbedrawnfortheelectricfieldintensityandthe electric potential as a function of distance from conducting parallelplatesandsurfacesandthesearegiveninFigure1009

Potential plot E eld plotE eld

++++

ndashndashndashndash

xx x

+

x x

ndashx x

+ ndashx

x

+ +

x

x

Figure 1009 Electric field and electric potential at a distance from a charged surface

Potential due to one or more point charges

The potential due to one point charge can be determined byusingtheequationformula

V = kq r

Example 1

Determine the electric potential at a point 20 times 10-1 m fromthecentreofan isolatedconducting spherewithapointchargeof40pCinair

Solution

Using the formula

V = kq r we have

V90 10 9times( ) 40 10 12ndashtimes( )times

20 10 1ndashtimes( )------------------------------------------------------------------ 018 V= =

the potential at the point is 180 times 10-1 V

The potential due to a number of point charges can be determined by adding up the potentials due to individual point charges because the electric potential at any point outside a conducting sphere will be the same as if all the charge was concentrated at its centre

Example 2

Threepointchargesofareplacedattheverticesofaright-angledtriangleasshowninthediagrambelowDeterminetheabsolutepotentialat the+20μCchargeduetothetwoothercharges

- 6μC

+3μC +2μC

4 m

3 m

Solution

The electric potential of the +2 μC charge due to the ndash 6 μC charge is

V = (9 times 109 Nm2C-2 times -6 times 10-6 C) divide (radic 32 + 42) m = - 108 times 104 V

PHYSICS 2015indb 314 150514 338 PM

Fields

315

AH

L

Insummarywecanconcludethat

bull No work is done to move a charge along an equipotential

bull Equipotentials are always perpendicular to the electric lines of force

Figure1011and1012showsomeequipotential lines fortwo oppositely charged and identically positive spheresseparated by a distance

+ve ndashve

equipotential lines

Figure 1011 Equipotential lines between two opposite charges

equipotential lines

Figure 1012 Equipotential lines between two charges which are the same

+

ndash ndash ndash ndash

+ + +

40 V

30 V

20 V

10 V

50 V

Figure 1013 Equipotential lines between charged parallel plates

Figure1013showstheequipotentiallinesbetweenchargedparallel plates Throughout this chapter the similarities and differences between gravitational fields and electric fieldshavebeendiscussedTherelationshipsthatexistsbetweengravitationalandelectricquantitiesandtheeffectsofpointmassesandchargesissummarisedinTable1014

The electric potential of the +2 μC charge due to the +3 μC charge is

V = (9 times 109 Nm2C-2 times 3 times 10-6 C) divide 3m = 9 times 103 V

The net absolute potential is the sum of the 2 potentials- 108 times 104 V + 9 times 103 V =- 18 times 103 V

The absolute potential at the point is - 18 times 103 V

Equipotential surfacesRegions in space where the electric potential of a chargedistributionhasaconstantvaluearecalledequipotentials The placeswherethepotentialisconstantinthreedimensionsarecalled equipotential surfacesandwheretheyareconstantintwodimensionstheyarecalledequipotential lines

TheyareinsomewaysanalogoustothecontourlinesontopographicmapsInthiscasethegravitationalpotentialenergy is constant as amassmoves around the contourlines because the mass remains at the same elevationabovetheEarthrsquossurfaceThegravitationalfieldstrengthacts in a direction perpendicular to a contour line

Similarlybecausetheelectricpotentialonanequipotentiallinehas thesamevalueanelectric forcecandonoworkwhena testchargemovesonanequipotentialThereforethe electric field cannot have a component along anequipotentialandthusitmustbeeverywhereperpendiculartotheequipotentialsurfaceorequipotentiallineThisfactmakesiteasytoplotequipotentialsifthelinesofforceorlinesofelectricfluxofanelectricfieldareknown

Forexamplethereareaseriesofequipotentiallinesbetweentwoparallelplateconductorsthatareperpendiculartotheelectric fieldTherewill be a series of concentric circles(eachcirclefurtherapartthanthepreviousone)thatmapouttheequipotentialsaroundanisolatedpositivesphereThe lines of force and some equipotential lines for anisolatedpositivesphereareshowninFigure1010

Lines ofequipotential

Figure 1010 Equipotentials around an isolated positive sphere

PHYSICS 2015indb 315 150514 338 PM

Chapter 10

316

AH

L

Gravitational quantity Electrical quantityQ

uant

ities

g V∆x∆-------ndash= E V∆

x∆-------ndash=

Poin

t mas

ses a

nd ch

arge

s

V Gmr----

ndash= V 14πε0------------q

r---=

g G m

r2-----ndash= E 14πε0------------ q

r2-----=

F Gm1 m2

r2--------------= F 1

4πε0------------

q1q2

r2-----------=

Figure 1014 Formulas

Example

Deduce the electric potential on the surface of a gold nucleus that has a radius of 62 fm

Solution

Using the formula

V = kq r and knowing the atomic number of gold is 79 We will assume the nucleus is spherical and it behaves as if it were a point charge at its centre (relative to outside points)

V = 90 times 109 Nm2C-2 times 79 times 16 times 10-19 C divide 62 times 10-15 m = 18 times 107 V

The potential at the point is 18 MV

Example

Deduce the ionisation energy in electron-volts of theelectron in the hydrogen atom if the electron is in its ground state and it is in a circular orbit at a distance of 53 times 10-11 m from the proton

Solution

This problem is an energy coulombic circular motion question based on Bohrrsquos model of the atom (not the accepted quantum mechanics model) The ionisation energy is the energy required to remove the electron from the ground state to infinity The electron travels in a circular orbit and therefore has a centripetal acceleration The ionisation energy will counteract the coulombic force and the movement of the electron will be in the opposite direction to the centripetal force

Total energy = Ek electron + Ep due to the proton-electron interaction

ΣF = kqQ r2 = mv2 r and as such mv2 = = kqQ r

Therefore Ek electron = frac12 kqQ r

Ep due to the proton-electron interaction = - kqQ r

Total energy = frac12 kqQ r + - kqQ r = -frac12 kqQ r

= - 90 times 109 Nm2C-2 times (16 times 10-19 C)2 divide 53 times 10-11 m = -217 times 10-18 J

= -217 times 10-18 J divide 16 times 10-19 = -136 eV

The ionisation energy is 136 eV

Exercise 1021 ApointchargePisplacedmidwaybetweentwo

identicalnegativechargesWhichoneofthefollowingiscorrectwithregardstoelectricfieldandelectricpotentialatpointP

Electric field Electric potentialA non-zero zeroB zero non-zeroC non-zero non-zeroD zero zero

2 Twopositivechargedspheresaretiedtogetherinavacuumsomewhereinspacewheretherearenoexternal forces A has a mass of 25 g and a charge of20μCandBhasamassof15gandachargeof30μCThedistancebetweenthemis40cmTheyarethenreleasedasshowninthediagram

BEFORE AFTER

v1 v2

A B

(a) Determinetheirinitialelectricpotentialenergy in the before situation

(b) DeterminethespeedofsphereBafterrelease

3 Thediagrambelowrepresentstwoequipotentiallines in separated by a distance of 5 cm in a uniform electric field

+ + + + + + + +

ndash ndash ndash ndash ndash ndash ndash ndash

40 V

20 V 5 cm

PHYSICS 2015indb 316 150514 338 PM

Fields

317

AH

L

Determine the strength of the electric field

4 Thisquestionisabouttheelectricfieldduetoacharged sphere and the motion of electrons in that fieldThediagrambelowshowsanisolatedmetalsphereinavacuumthatcarriesanegativeelectriccharge of 60 μC

ndash

(a) Onthediagramdrawtheconventionalwayto represent the electric field pattern due to the charged sphere and lines to represent threeequipotentialsurfacesintheregionoutside the sphere

(b) Explainhowthelinesrepresentingtheequipotentialsurfacesthatyouhavesketchedindicatethatthestrengthoftheelectricfieldisdecreasingwithdistancefrom the centre of the sphere

(c) Theelectricfieldstrengthatthesurfaceofthe sphere and at points outside the sphere can be determined by assuming that the sphere acts as a point charge of magnitude 60 μC at its centre The radius of the sphere is 25 times 10ndash2 m Deduce that the magnitude of the field strength at the surface of the sphere is 86 times 107 Vmndash1

An electron is initially at rest on the surface of the sphere

(d) (i) Describethepathfollowedbytheelectronasitleavesthesurfaceofthesphere

(ii) Calculatetheinitialaccelerationofthe electron

5 Determinetheamountofworkthatisdoneinmovingachargeof100nCthroughapotentialdifferenceof150times102 V

6 Three identical 20 μC conducting spheres are placedatthecornersofanequilateraltriangleofsides 25 cm The triangle has one apex C pointing upthepageand2baseanglesAandBDeterminetheabsolutepotentialatB

7 Determinehowfaraparttwoparallelplatesmustbesituatedsothatapotentialdifferenceof250 times 102 V produces an electric field strength of 200 times 103 NC-1

8 Thegapbetweentwoparallelplatesis10times10-3mandthereisapotentialdifferenceof10times104 V betweentheplatesCalculate

i theworkdonebyanelectroninmovingfrom one plate to the other

ii thespeedwithwhichtheelectronreachesthe second plate if released from rest

iii theelectricfieldintensitybetweentheplates

9 Anelectronguninapicturetubeisacceleratedby a potential 25 times 103VDeterminethekineticenergygainedbytheelectroninelectron-volts

10 Determine the electric potential 20 times 10-2 m from achargeof-10times10-5 C

11 Determinetheelectricpotentialatapointmid-waybetweenachargeofndash20pCandanotherof+5pConthelinejoiningtheircentresifthecharges are 10 cm apart

12 During a thunderstorm the electric potential differencebetweenacloudandthegroundis10 times 109 V Determine the magnitude of the change in electric potential energy of an electron thatmovesbetweenthesepointsinelectron-volts

13 A charge of 15 μC is placed in a uniform electric fieldoftwooppositelychargedparallelplateswithamagnitudeof14times103 NC-1

(a) Determinetheworkthatmustbedoneagainstthefieldtomovethepointchargeadistance of 55 cm

(b) Calculatethepotentialdifferencebetweenthe final and initial positions of the charge

(c) Determinethepotentialdifferencebetweenthe plates if their separation distance is 15 cm

14 Duringaflashoflightningthepotentialdifferencebetweenacloudandthegroundwas12 times 109 V and the amount of transferred charge was32C

(a) Determinethechangeinenergyofthetransferred charge

(b) Iftheenergyreleasedwasallusedtoacceleratea1tonnecardeduceitsfinalspeed

(c) Iftheenergyreleasedcouldbeusedtomelticeat0degCdeducetheamountoficethatcould be melted

PHYSICS 2015indb 317 150514 338 PM

Chapter 10

318

AH

L

15 SupposethatwhenanelectronmovedfromAtoBin the diagram along an electric field line that the electric field does 36 times 10-19Jofworkonit

Determinethedifferencesinelectricpotential

(a) VB ndash VA(b) VC ndash VA(c) VC ndash VB

16 DeterminethepotentialatpointPthatislocatedatthecentreofthesquareasshowninthediagrambelow

- 6μC

+3μC +2μC

1m

1m

5μC

P

102 Fields at work

Essential idea Similar approaches can be taken inanalysingelectricalandgravitationalpotentialproblems

Understandingsbull Potentialandpotentialenergybull Potentialgradientbull Potentialdifferencebull Escapespeedbull Orbitalmotionorbitalspeedandorbitalenergybull Forcesandinverse-squarelawbehaviour

Orbital motion orbital speed and orbital energy

Although orbital motion may be circular elliptical orparabolic this sub-topic only dealswith circular orbitsThis sub-topic is not fundamentally new physics butan application that synthesizes ideas from gravitationcircularmotiondynamicsandenergy

TheMoonorbitstheEarthandinthissenseitisoftenreferredtoasasatelliteoftheEarthBefore1957itwastheonlyEarthsatelliteHoweverin1957theRussianslaunchedthefirstmanmadesatelliteSputnik1SincethisdatemanymoresatelliteshavebeenlaunchedandtherearenowliterallythousandsofthemorbitingtheEarthSomeareusedtomonitortheweathersome used to enable people to find accurately their position on thesurfaceoftheEarthmanyareusedincommunicationsandnodoubtsomeareusedtospyonothercountriesFigure1015showshowinprincipleasatellitecanbeputintoorbit

Theperson(whosesizeisgreatlyexaggeratedwithrespectto Earth) standing on the surface on the Earth throwssome stonesThe greater the speed with which a stoneis thrown the further it will land from her The pathsfollowedbythethrownstonesareparabolasByastretchof the imaginationwecanvisualisea situation inwhichastoneis thrownwithsuchaspeedthatbecauseof thecurvatureoftheEarthitwillnotlandonthesurfaceoftheEarthbutgointoldquoorbitrdquo(Path4onFigure1015)

NATURE OF SCIENCECommunication of scientific explanations The ability to apply field theory to the unobservable (charges) and the massively scaled (motion of satellites) required scientists to develop new ways to investigate analyse and report findings to a general public used to scientific discoveries based on tangible and discernible evidence (51)

copy IBO 2014

PHYSICS 2015indb 318 150514 338 PM

Fields

319

AH

L

1

2

34Earth

Figure 1015 Throwing a stone into orbit

TheforcethatcausesthestonestofollowaparabolicpathandtofalltoEarthisgravityandsimilarlytheforcethatkeeps the stone in orbit is gravity For circular motionto occur we have seen that a force must act at rightanglestothevelocityofanobjectthatistheremustbeacentripetalforceHenceinthesituationwedescribeherethe centripetal force for circular orbital motion about the EarthisprovidedbygravitationalattractionoftheEarth

We can calculate the speedwithwhich a stonemustbethrowninordertoputitintoorbitjustabovethesurfaceof the Earth

If the stone has mass m and speed v thenwehave fromNewtonrsquos2ndlaw

mv2

RE---------- G

MEm

RE2------------=

whereRE is the radius of the Earth and ME is the mass of the Earth

Bearinginmindthatg 0 GME

RE2--------= then

v g RE 10 64 106timestimes 8 103times= = =

Thatisthestonemustbethrownat8times103m sndash1

Clearly we are not going to get a satellite into orbit soclosetothesurfaceoftheEarthMovingatthisspeedthefrictionduetoairresistancewouldmeltthesatellitebeforeithadtravelledacoupleofkilometresInrealitythereforeasatelliteisputintoorbitabouttheEarthbysendingitattachedtoarocketbeyondtheEarthrsquosatmosphereandthengivingitacomponentofvelocityperpendiculartoaradialvectorfromtheEarthSeeFigure1016

Earth Satellite carried by rocket to hereSatellite orbit

Tangential component of velocity

Figure 1016 Getting a satellite into orbit

Keplerrsquos third law(This work of Kepler and Newtonrsquos synthesis of the work is an excellent example of the scientific method and makes for a good TOK discussion)

In 1627 Johannes Kepler (1571-1630)publishedhis lawsof planetary motion The laws are empirical in natureandwere deduced from the observations of theDanishastronomer Tycho de Brahe (1546-1601)The third lawgives a relationship between the radius of orbit R of a planet and its period TofrevolutionabouttheSunThelawisexpressedmathematicallyas

2

3 constantTR

=

We shall nowuseNewtonrsquos LawofGravitation to showhowitisthattheplanetsmoveinaccordancewithKeplerrsquosthirdlaw

InessenceNewtonwasabletousehislawofgravitytopredict the motion of the planets since all he had to do was factor theF givenby this law intohis second law F = ma to find their accelerations and hence their future positions

InFigure1017theEarthisshownorbitingtheSunandthedistancebetweentheircentresisR

RSun

Earth

F es

F se

Figure 1017 Planets move according to Keplerrsquos third law

Fes is the force that the Earth exerts on the Sun and Fse is the force that the Sun exerts on the Earth The forces are equalandoppositeandtheSunandtheEarthwillactuallyorbitaboutacommoncentreHoweversince theSun issoverymuchmoremassivethantheEarththiscommoncentrewillbeclosetothecentreoftheSunandsowecanregard the Earth as orbiting about the centre of the Sun Theotherthingthatweshallassumeisthatwecanignorethe forces that the other planets exert on the Earth (This wouldnotbeawisethingtodoifyouwereplanningtosend a space ship to theMoon for example)We shallalso assume that we have followed Newtonrsquos exampleandindeedprovedthataspherewillactasapointmasssituated at the centre of the sphere

Kepler had postulated that the orbits of the planets areellipticalbut since theeccentricityof theEarthrsquosorbit issmallweshallassumeacircularorbit

PHYSICS 2015indb 319 150514 338 PM

Chapter 10

320

AH

L

TheaccelerationoftheEarthtowardstheSunisa = Rω2

whereω ω 2πT------=

Hencea R 2π

T------ times

2 4π2 RT 2------------- = =

But the acceleration is given by Newtonrsquos Second Law F = mawhereFisnowgivenbytheLawofGravitationSo in this situation

F maGMs Me

R2-------------------= = butwealsohavethat

a a R 2πT

------ times

2 4π2 RT 2------------- = = and m = Me so that

G MsMe

R2------------------- Me4π2R

T2-------------times

GMs

4π2-----------rArr R3

T2------= =

Butthequantity

isaconstantthathasthesamevalueforeachoftheplanetssowehaveforalltheplanetsnotjustEarththat

R3

T2------ k =

wherekisaconstantWhichisofcourseKeplerrsquosthirdlaw

ThisisindeedanamazingbreakthroughItisdifficulttorefutethe idea that all particles attract each other in accordance withtheLawofGravitationwhenthelawisabletoaccountfortheobservedmotionoftheplanetsabouttheSun

The gravitational effects of the planets upon each othershould produce perturbations in their orbits Such is the predictivepoweroftheUniversalGravitationalLawthatit enabled physicists to compute these perturbations The telescopehadbeeninventedin1608andbythemiddleofthe 18th Century had reached a degree of perfection in design that enabled astronomers to actually measure the orbital perturbations of the planets Their measurements werealways inagreementwith thepredictionsmadebyNewtonrsquoslawHoweverin1781anewplanetUranuswasdiscovered and the orbit of this planet did not fit withthe orbit predicted by Universal Gravitation Such wasthe physicistrsquos faith in theNewtonianmethod that theysuspectedthatthediscrepancywasduetothepresenceofayetundetectedplanetUsingtheLawofGravitationtheFrenchastronomerJLeverrier and the English astronomer J C Adamswereable tocalculate justhowmassive thisnewplanetmustbeandalsowhereitshouldbeIn1846the planetNeptunewas discovered justwhere they hadpredicted Ina similarwaydiscrepancies in theorbitofNeptune ledto thepredictionandsubsequentdiscoveryin1930oftheplanetPlutoNewtonrsquosLawofGravitation

had passed the ultimate test of any theory it is not only abletoexplainexistingdatabutalsotomakepredictions

Satellite energyWhenasatelliteisinorbitaboutaplanetitwillhavebothkineticenergyandgravitationalpotentialenergySupposewe consider a satellite ofmass m that is in an orbit of radius r about a planet of mass M

Thegravitationalpotentialduetotheplanetatdistancer from its centre is

ThegravitationalpotentialenergyofthesatelliteVsat

is therefore GMe m

r----------------ndash

ThatisVsat GMem

r----------------ndash=

Thegravitationalfieldstrengthatthesurfaceoftheplanetisgivenby

g 0GMe

Re2------------=

Hencewecanwrite

Vsat g0 Re

2

rndash=

m

ThekineticenergyofthesatelliteKsatisequaltofrac12mv2wherev is its orbital speed

Byequatingthegravitationalforceactingonthesatellitetoitscentripetalaccelerationwehave

G Mem

r2---------------- mv2

r---------- mv2hArr

GMemr----------------= =

Fromwhich12---

mv2 12---

GMemr----------------times=

= 0 e

2g R

r

2m

Which is actually quite interesting since it shows thatirrespective of the orbital radius the KE is numericallyequal to half the PE Also the total energy E tot K sat Vsat+ 1

2---G Me m

r----------------times G Me m

r----------------ndash + 1

2---G Me m

r----------------ndash= = =of the satelliteisalwaysnegativesince

E tot K sat Vsat+ 12---

G Me mr----------------times

G Me mr----------------ndash

+ 12---

G Me mr----------------ndash= = =

PHYSICS 2015indb 320 150514 338 PM

Fields

321

AH

L

The energies of an orbiting satellite as a function of radial distancefromthecentreofaplanetareshownplottedinFigure1018

12

100806

0402

0ndash02

ndash04

ndash06ndash08ndash10

ndash12

2 4 6 8 10 12

ener

gya

rbita

ry u

nits

distance timesR

kinetic energytotal energy

potential energy

Figure 1018 Energy of an orbiting satellite as a function of distance from the centre of a planet

WeightlessnessSuppose that you are inanelevator(lift)thatisdescendingat constant speedandyou let goof abook thatyouareholdinginyourhandThebookwillfalltothefloorwithaccelerationequaltotheaccelerationduetogravityIfthecablethatsupportstheelevatorweretosnap(asituationthatItrustwillneverhappentoanyofyou)andyounowletgothebookthatyouareholdinginyourotherhandthis book will not fall to the floor - it will stay exactlyin linewithyourhandThis isbecause thebook isnowfallingwith the sameaccelerationas theelevatorandassuch the book cannot ldquocatchrdquo up with the floor of theelevatorFurthermoreifyouhappenedtobestandingonasetofbathroomscalesthescaleswouldnowreadzero-youwouldbeapparentlyweightlessItisthisideaoffreefallthatexplainstheapparentweightlessnessofastronautsin an orbiting satellite These astronauts are in free fall inthesensethattheyareacceleratingtowardsthecentre of the Earth

It is actually possible to define theweight of a body inseveral different waysWe can define it for example asthegravitationalforceexertedonthebodybyaspecifiedobject such as the EarthThiswe have seen thatwe doinlotsofsituationswherewedefinetheweightasbeingequaltomgIfweusethisdefinitionthenanobjectinfreefallcannotbydefinitionbeweightlesssinceitisstillinagravitationalfieldHoweverifwedefinetheweightofanobjectintermsofaldquoweighingrdquoprocesssuchasthereadingonasetofbathroomscaleswhichineffectmeasuresthecontact force between the object and the scales thenclearlyobjectsinfreefallareweightlessOnenowhasto

askthequestionwhetherornotitispossibleForexampletomeasurethegravitationalforceactingonanastronautin orbit about the Earth

WecanalsodefineweightintermsofthenetgravitationalforceactingonabodyduetoseveraldifferentobjectsForexample for an object out in space its weight could bedefined in terms of the resultant of the forces exerted on it bytheSuntheMoontheEarthandalltheotherplanetsin the Solar System If this resultant is zero at a particular pointthenthebodyisweightlessatthispoint

Inviewofthevariousdefinitionsofweightthatareavailabletousitisimportantthatwhenweusethewordldquoweightrdquoweareawareofthecontextinwhichitisbeingused

Example

Calculate the height above the surface of the Earth atwhichageo-stationarysatelliteorbits

Solution

A geo-stationary satellite is one that orbits the Earth in such a way that it is stationary with respect to a point on the surface of the Earth This means that its orbital period must be the same as the time for the Earth to spin once on its axis ie 24 hours

From Keplerrsquos third law we have G Ms

4π2----------- R3

T2------=

That is

RMe

mh

using the fact that the force of attraction between the satellite and the Earth is given by

FG Mem

R2----------------=

and that F = ma

where m is the mass of the satellite and a 4π2R

T2-------------=

we have

GMem

R2---------------- m 4π2R

T2-------------times

GMe

4π2------------rArr R3

T2------= =

PHYSICS 2015indb 321 150514 338 PM

Chapter 10

322

AH

L

Now the mass of the Earth is 60 times 1024 kg and the period T measured in seconds is given by T = 86400 s

So substitution gives R = 42 times 106 m

The radius of the Earth is 64 times 106 m so that the orbital height h is about 36 times 107 m

Example

Calculatetheminimumenergyrequiredtoputasatelliteofmass500kgintoanorbitthatisasaheightequaltotheEarthrsquosradiusabovethesurfaceoftheEarth

Solution

We have seen that when dealing with gravitational fields and potential it is useful to remember that

g 0G M

Re2---------= or g0 Re

2 GM=

The gravitational potential at the surface of the Earth is

g 0ndash Re GMRe

---------ndash=

The gravitational potential at a distance R

from the centre of the Earth is

=

The difference in potential between the surface of the Earth and a point distance R from the centre is therefore

V∆ g 0Re 1ReR------ndash

=

If R 2Re= then V∆g0 Re

2------------=

This means that the work required to ldquoliftrdquo the satellite into orbit is g0Rm where m is the mass of the satellite This is equal to

10 32 106 500timestimestimes = 16000 MJ

However the satellite must also have kinetic energy in order to orbit Earth This will be equal to

g 0 mRe2

2R-----------------g 0mRe

2

4----------------- 8000 MJ= =

The minimum energy required is therefore

24000 MJ

Exercise 1031 The speed needed to put a satellite in orbit does

not depend on

A the radius of the orbitB theshapeoftheorbitC thevalueofgattheorbitD the mass of the satellite

2 EstimatethespeedofanEarthsatellitewhoseorbitis400kmabovetheEarthrsquossurfaceAlsodetermine the period of the orbit

3 Calculatethespeedofa200kgsatelliteorbitingthe Earth at a height of 70 times 106 m

Assume that g = 82 m sndash2 for this orbit

4 TheradiioftwosatellitesXandYorbitingthe Earth are 2r and 8rwherer is the radius of the Earth Calculate the ratio of the periods of revolutionofXtoY

5 A satellite of mass mkgissentfromEarthrsquossurfaceinto an orbit of radius 5RwhereR is the radius of theEarthWritedownanexpressionfor

(a) thepotentialenergyofthesatelliteinorbit

(b) thekineticenergyofthesatelliteinorbit

(c) theminimumworkrequiredtosendthesatellitefromrestattheEarthrsquossurfaceintoits orbit

6 A satellite in an orbit of 10rfallsbacktoEarth(radius r)afteramalfunctionDeterminethespeedwithwhichitwillhittheEarthrsquossurface

7 The radius of the moon is frac14 that of the Earth AssumingEarthandtheMoontohavethesamedensitycomparetheaccelerationsoffreefallatthe surface of Earth to that at the surface of the Moon

8 UsethefollowingdatatodeterminethegravitationalfieldstrengthatthesurfaceoftheMoon and hence determine the escape speed from the surface of the Moon

Mass of the Moon = 73 times 1022kg

RadiusoftheMoon=17times106 m

PHYSICS 2015indb 322 150514 338 PM

Fields

313

AH

L

Electric potential due to a point charge

Letus takeapoint rmetres froma chargedobjectThepotentialatthispointcanbecalculatedusingthefollowing

W = ndashFr = ndashqV and F = ndash q1 q2 _____ 4πε0r

2

Therefore

W = ndash q1 q2 _____ 4πε0r

2 times r = ndash q1 q2 _____ 4πε0r

= ndashq1 times q2 _____ 4πε0r

= ndashq1V

That is V q

4πε0r--------------- =

Orsimply

V kqr------

=

Example

Determinehowmuchworkisdonebytheelectricfieldofpointcharge150μCwhenachargeof200μCismovedfrom infinity to apoint 0400m from thepoint charge(Assumenoaccelerationofthecharges)

Solution

The work done by the electric field is W = -qV = -14πε0 times q times (Q rinfin - Q r0400)

W = (- 200 times 10-6 C times 900 times 109 NmC-2 times 150 times 10-6 C) divide 0400 m = - 0675 J

An external force would have to do +0675 J of work

Electric field strength and electric potential gradient

Letus lookbackatFigure1006Supposeagain that thecharge+qismovedasmalldistancebyaforceFfromAtoBsothattheforcecanbeconsideredconstantTheworkdoneisgivenby

∆W F x∆times=

TheforceFandtheelectricfieldEareoppositelydirectedandweknowthat

F = -qE and ΔW = q ΔV

Thereforetheworkdonecanbegivenas

q ΔV = -q E Δx

Therefore E V∆x∆-------

ndash=

The rate of change of potential ΔV at a point with respect to distance Δx in the direction in which the change is maximum is called the potential gradient We say that the electric field = - the potential gradient and the units are Vm-

1 From the equation we can see that in a graph of electric potential versus distance the gradient of the straight line equals the electric field strength

In reality if a charged particle enters a uniform electric field it will be accelerated uniformly by the field and its kinetic energy will increase This is why we had to assume no acceleration in the last worked example

Ek∆ 12---

mv2 q E xsdot sdot q Vx---

xsdot sdot q Vsdot= = = =

Example

Determine how far apart two parallel plates must besituated so that a potential difference of 150 times 102 V produces an electric field strength of 100 times 103 NC-1

Solution

Using E V∆x------- xhArrndash V∆

E-------15 102 Vtimes

100 103 N C 1ndashtimes------------------------------------------= = =

= 150 times 10-1

The plates are 150 times 10-1 m apart

The electric field and the electric potential at a point due to an evenly distributed charge +q on a sphere can be represented graphically as in Figure 1008

r

r

V

E

r0

r0

r0

Charge of +Q evenly distributed over surface

E 14πε0------------ Q

r2----- r r0gt=

V 14πε0------------Q

r----r r0gt=

Figure 1008 Electric field and potential due to a charged sphere

PHYSICS 2015indb 313 150514 337 PM

Chapter 10

314

AH

L

When the sphere becomes charged we know that thechargedistributes itselfevenlyoverthesurfaceThereforeeverypartofthematerialoftheconductorisatthesamepotential As the electric potential at a point is defined as beingnumericallyequaltotheworkdoneinbringingaunitpositivechargefrominfinitytothatpointithasaconstantvalueineverypartofthematerialoftheconductor

Since the potential is the same at all points on the conducting surfacethen∆V∆xiszeroButE=ndash∆V∆xThereforethe electric field inside the conductor is zero There is no electric field inside the conductor

SomefurtherobservationsofthegraphsinFigure1009are

bull OutsidethespherethegraphsobeytherelationshipsgivenasE α 1 r2 and V α 1 r

bull Atthesurfacer = r0Thereforetheelectricfieldandpotentialhavetheminimumvalueforr at this point and this infers a maximum field and potential

bull Insidethespheretheelectricfieldiszerobull Insidethespherenoworkisdonetomoveacharge

fromapointinsidetothesurfaceThereforethereisno potential difference and the potential is the same asitiswhenr = r0

Similargraphscanbedrawnfortheelectricfieldintensityandthe electric potential as a function of distance from conducting parallelplatesandsurfacesandthesearegiveninFigure1009

Potential plot E eld plotE eld

++++

ndashndashndashndash

xx x

+

x x

ndashx x

+ ndashx

x

+ +

x

x

Figure 1009 Electric field and electric potential at a distance from a charged surface

Potential due to one or more point charges

The potential due to one point charge can be determined byusingtheequationformula

V = kq r

Example 1

Determine the electric potential at a point 20 times 10-1 m fromthecentreofan isolatedconducting spherewithapointchargeof40pCinair

Solution

Using the formula

V = kq r we have

V90 10 9times( ) 40 10 12ndashtimes( )times

20 10 1ndashtimes( )------------------------------------------------------------------ 018 V= =

the potential at the point is 180 times 10-1 V

The potential due to a number of point charges can be determined by adding up the potentials due to individual point charges because the electric potential at any point outside a conducting sphere will be the same as if all the charge was concentrated at its centre

Example 2

Threepointchargesofareplacedattheverticesofaright-angledtriangleasshowninthediagrambelowDeterminetheabsolutepotentialat the+20μCchargeduetothetwoothercharges

- 6μC

+3μC +2μC

4 m

3 m

Solution

The electric potential of the +2 μC charge due to the ndash 6 μC charge is

V = (9 times 109 Nm2C-2 times -6 times 10-6 C) divide (radic 32 + 42) m = - 108 times 104 V

PHYSICS 2015indb 314 150514 338 PM

Fields

315

AH

L

Insummarywecanconcludethat

bull No work is done to move a charge along an equipotential

bull Equipotentials are always perpendicular to the electric lines of force

Figure1011and1012showsomeequipotential lines fortwo oppositely charged and identically positive spheresseparated by a distance

+ve ndashve

equipotential lines

Figure 1011 Equipotential lines between two opposite charges

equipotential lines

Figure 1012 Equipotential lines between two charges which are the same

+

ndash ndash ndash ndash

+ + +

40 V

30 V

20 V

10 V

50 V

Figure 1013 Equipotential lines between charged parallel plates

Figure1013showstheequipotentiallinesbetweenchargedparallel plates Throughout this chapter the similarities and differences between gravitational fields and electric fieldshavebeendiscussedTherelationshipsthatexistsbetweengravitationalandelectricquantitiesandtheeffectsofpointmassesandchargesissummarisedinTable1014

The electric potential of the +2 μC charge due to the +3 μC charge is

V = (9 times 109 Nm2C-2 times 3 times 10-6 C) divide 3m = 9 times 103 V

The net absolute potential is the sum of the 2 potentials- 108 times 104 V + 9 times 103 V =- 18 times 103 V

The absolute potential at the point is - 18 times 103 V

Equipotential surfacesRegions in space where the electric potential of a chargedistributionhasaconstantvaluearecalledequipotentials The placeswherethepotentialisconstantinthreedimensionsarecalled equipotential surfacesandwheretheyareconstantintwodimensionstheyarecalledequipotential lines

TheyareinsomewaysanalogoustothecontourlinesontopographicmapsInthiscasethegravitationalpotentialenergy is constant as amassmoves around the contourlines because the mass remains at the same elevationabovetheEarthrsquossurfaceThegravitationalfieldstrengthacts in a direction perpendicular to a contour line

Similarlybecausetheelectricpotentialonanequipotentiallinehas thesamevalueanelectric forcecandonoworkwhena testchargemovesonanequipotentialThereforethe electric field cannot have a component along anequipotentialandthusitmustbeeverywhereperpendiculartotheequipotentialsurfaceorequipotentiallineThisfactmakesiteasytoplotequipotentialsifthelinesofforceorlinesofelectricfluxofanelectricfieldareknown

Forexamplethereareaseriesofequipotentiallinesbetweentwoparallelplateconductorsthatareperpendiculartotheelectric fieldTherewill be a series of concentric circles(eachcirclefurtherapartthanthepreviousone)thatmapouttheequipotentialsaroundanisolatedpositivesphereThe lines of force and some equipotential lines for anisolatedpositivesphereareshowninFigure1010

Lines ofequipotential

Figure 1010 Equipotentials around an isolated positive sphere

PHYSICS 2015indb 315 150514 338 PM

Chapter 10

316

AH

L

Gravitational quantity Electrical quantityQ

uant

ities

g V∆x∆-------ndash= E V∆

x∆-------ndash=

Poin

t mas

ses a

nd ch

arge

s

V Gmr----

ndash= V 14πε0------------q

r---=

g G m

r2-----ndash= E 14πε0------------ q

r2-----=

F Gm1 m2

r2--------------= F 1

4πε0------------

q1q2

r2-----------=

Figure 1014 Formulas

Example

Deduce the electric potential on the surface of a gold nucleus that has a radius of 62 fm

Solution

Using the formula

V = kq r and knowing the atomic number of gold is 79 We will assume the nucleus is spherical and it behaves as if it were a point charge at its centre (relative to outside points)

V = 90 times 109 Nm2C-2 times 79 times 16 times 10-19 C divide 62 times 10-15 m = 18 times 107 V

The potential at the point is 18 MV

Example

Deduce the ionisation energy in electron-volts of theelectron in the hydrogen atom if the electron is in its ground state and it is in a circular orbit at a distance of 53 times 10-11 m from the proton

Solution

This problem is an energy coulombic circular motion question based on Bohrrsquos model of the atom (not the accepted quantum mechanics model) The ionisation energy is the energy required to remove the electron from the ground state to infinity The electron travels in a circular orbit and therefore has a centripetal acceleration The ionisation energy will counteract the coulombic force and the movement of the electron will be in the opposite direction to the centripetal force

Total energy = Ek electron + Ep due to the proton-electron interaction

ΣF = kqQ r2 = mv2 r and as such mv2 = = kqQ r

Therefore Ek electron = frac12 kqQ r

Ep due to the proton-electron interaction = - kqQ r

Total energy = frac12 kqQ r + - kqQ r = -frac12 kqQ r

= - 90 times 109 Nm2C-2 times (16 times 10-19 C)2 divide 53 times 10-11 m = -217 times 10-18 J

= -217 times 10-18 J divide 16 times 10-19 = -136 eV

The ionisation energy is 136 eV

Exercise 1021 ApointchargePisplacedmidwaybetweentwo

identicalnegativechargesWhichoneofthefollowingiscorrectwithregardstoelectricfieldandelectricpotentialatpointP

Electric field Electric potentialA non-zero zeroB zero non-zeroC non-zero non-zeroD zero zero

2 Twopositivechargedspheresaretiedtogetherinavacuumsomewhereinspacewheretherearenoexternal forces A has a mass of 25 g and a charge of20μCandBhasamassof15gandachargeof30μCThedistancebetweenthemis40cmTheyarethenreleasedasshowninthediagram

BEFORE AFTER

v1 v2

A B

(a) Determinetheirinitialelectricpotentialenergy in the before situation

(b) DeterminethespeedofsphereBafterrelease

3 Thediagrambelowrepresentstwoequipotentiallines in separated by a distance of 5 cm in a uniform electric field

+ + + + + + + +

ndash ndash ndash ndash ndash ndash ndash ndash

40 V

20 V 5 cm

PHYSICS 2015indb 316 150514 338 PM

Fields

317

AH

L

Determine the strength of the electric field

4 Thisquestionisabouttheelectricfieldduetoacharged sphere and the motion of electrons in that fieldThediagrambelowshowsanisolatedmetalsphereinavacuumthatcarriesanegativeelectriccharge of 60 μC

ndash

(a) Onthediagramdrawtheconventionalwayto represent the electric field pattern due to the charged sphere and lines to represent threeequipotentialsurfacesintheregionoutside the sphere

(b) Explainhowthelinesrepresentingtheequipotentialsurfacesthatyouhavesketchedindicatethatthestrengthoftheelectricfieldisdecreasingwithdistancefrom the centre of the sphere

(c) Theelectricfieldstrengthatthesurfaceofthe sphere and at points outside the sphere can be determined by assuming that the sphere acts as a point charge of magnitude 60 μC at its centre The radius of the sphere is 25 times 10ndash2 m Deduce that the magnitude of the field strength at the surface of the sphere is 86 times 107 Vmndash1

An electron is initially at rest on the surface of the sphere

(d) (i) Describethepathfollowedbytheelectronasitleavesthesurfaceofthesphere

(ii) Calculatetheinitialaccelerationofthe electron

5 Determinetheamountofworkthatisdoneinmovingachargeof100nCthroughapotentialdifferenceof150times102 V

6 Three identical 20 μC conducting spheres are placedatthecornersofanequilateraltriangleofsides 25 cm The triangle has one apex C pointing upthepageand2baseanglesAandBDeterminetheabsolutepotentialatB

7 Determinehowfaraparttwoparallelplatesmustbesituatedsothatapotentialdifferenceof250 times 102 V produces an electric field strength of 200 times 103 NC-1

8 Thegapbetweentwoparallelplatesis10times10-3mandthereisapotentialdifferenceof10times104 V betweentheplatesCalculate

i theworkdonebyanelectroninmovingfrom one plate to the other

ii thespeedwithwhichtheelectronreachesthe second plate if released from rest

iii theelectricfieldintensitybetweentheplates

9 Anelectronguninapicturetubeisacceleratedby a potential 25 times 103VDeterminethekineticenergygainedbytheelectroninelectron-volts

10 Determine the electric potential 20 times 10-2 m from achargeof-10times10-5 C

11 Determinetheelectricpotentialatapointmid-waybetweenachargeofndash20pCandanotherof+5pConthelinejoiningtheircentresifthecharges are 10 cm apart

12 During a thunderstorm the electric potential differencebetweenacloudandthegroundis10 times 109 V Determine the magnitude of the change in electric potential energy of an electron thatmovesbetweenthesepointsinelectron-volts

13 A charge of 15 μC is placed in a uniform electric fieldoftwooppositelychargedparallelplateswithamagnitudeof14times103 NC-1

(a) Determinetheworkthatmustbedoneagainstthefieldtomovethepointchargeadistance of 55 cm

(b) Calculatethepotentialdifferencebetweenthe final and initial positions of the charge

(c) Determinethepotentialdifferencebetweenthe plates if their separation distance is 15 cm

14 Duringaflashoflightningthepotentialdifferencebetweenacloudandthegroundwas12 times 109 V and the amount of transferred charge was32C

(a) Determinethechangeinenergyofthetransferred charge

(b) Iftheenergyreleasedwasallusedtoacceleratea1tonnecardeduceitsfinalspeed

(c) Iftheenergyreleasedcouldbeusedtomelticeat0degCdeducetheamountoficethatcould be melted

PHYSICS 2015indb 317 150514 338 PM

Chapter 10

318

AH

L

15 SupposethatwhenanelectronmovedfromAtoBin the diagram along an electric field line that the electric field does 36 times 10-19Jofworkonit

Determinethedifferencesinelectricpotential

(a) VB ndash VA(b) VC ndash VA(c) VC ndash VB

16 DeterminethepotentialatpointPthatislocatedatthecentreofthesquareasshowninthediagrambelow

- 6μC

+3μC +2μC

1m

1m

5μC

P

102 Fields at work

Essential idea Similar approaches can be taken inanalysingelectricalandgravitationalpotentialproblems

Understandingsbull Potentialandpotentialenergybull Potentialgradientbull Potentialdifferencebull Escapespeedbull Orbitalmotionorbitalspeedandorbitalenergybull Forcesandinverse-squarelawbehaviour

Orbital motion orbital speed and orbital energy

Although orbital motion may be circular elliptical orparabolic this sub-topic only dealswith circular orbitsThis sub-topic is not fundamentally new physics butan application that synthesizes ideas from gravitationcircularmotiondynamicsandenergy

TheMoonorbitstheEarthandinthissenseitisoftenreferredtoasasatelliteoftheEarthBefore1957itwastheonlyEarthsatelliteHoweverin1957theRussianslaunchedthefirstmanmadesatelliteSputnik1SincethisdatemanymoresatelliteshavebeenlaunchedandtherearenowliterallythousandsofthemorbitingtheEarthSomeareusedtomonitortheweathersome used to enable people to find accurately their position on thesurfaceoftheEarthmanyareusedincommunicationsandnodoubtsomeareusedtospyonothercountriesFigure1015showshowinprincipleasatellitecanbeputintoorbit

Theperson(whosesizeisgreatlyexaggeratedwithrespectto Earth) standing on the surface on the Earth throwssome stonesThe greater the speed with which a stoneis thrown the further it will land from her The pathsfollowedbythethrownstonesareparabolasByastretchof the imaginationwecanvisualisea situation inwhichastoneis thrownwithsuchaspeedthatbecauseof thecurvatureoftheEarthitwillnotlandonthesurfaceoftheEarthbutgointoldquoorbitrdquo(Path4onFigure1015)

NATURE OF SCIENCECommunication of scientific explanations The ability to apply field theory to the unobservable (charges) and the massively scaled (motion of satellites) required scientists to develop new ways to investigate analyse and report findings to a general public used to scientific discoveries based on tangible and discernible evidence (51)

copy IBO 2014

PHYSICS 2015indb 318 150514 338 PM

Fields

319

AH

L

1

2

34Earth

Figure 1015 Throwing a stone into orbit

TheforcethatcausesthestonestofollowaparabolicpathandtofalltoEarthisgravityandsimilarlytheforcethatkeeps the stone in orbit is gravity For circular motionto occur we have seen that a force must act at rightanglestothevelocityofanobjectthatistheremustbeacentripetalforceHenceinthesituationwedescribeherethe centripetal force for circular orbital motion about the EarthisprovidedbygravitationalattractionoftheEarth

We can calculate the speedwithwhich a stonemustbethrowninordertoputitintoorbitjustabovethesurfaceof the Earth

If the stone has mass m and speed v thenwehave fromNewtonrsquos2ndlaw

mv2

RE---------- G

MEm

RE2------------=

whereRE is the radius of the Earth and ME is the mass of the Earth

Bearinginmindthatg 0 GME

RE2--------= then

v g RE 10 64 106timestimes 8 103times= = =

Thatisthestonemustbethrownat8times103m sndash1

Clearly we are not going to get a satellite into orbit soclosetothesurfaceoftheEarthMovingatthisspeedthefrictionduetoairresistancewouldmeltthesatellitebeforeithadtravelledacoupleofkilometresInrealitythereforeasatelliteisputintoorbitabouttheEarthbysendingitattachedtoarocketbeyondtheEarthrsquosatmosphereandthengivingitacomponentofvelocityperpendiculartoaradialvectorfromtheEarthSeeFigure1016

Earth Satellite carried by rocket to hereSatellite orbit

Tangential component of velocity

Figure 1016 Getting a satellite into orbit

Keplerrsquos third law(This work of Kepler and Newtonrsquos synthesis of the work is an excellent example of the scientific method and makes for a good TOK discussion)

In 1627 Johannes Kepler (1571-1630)publishedhis lawsof planetary motion The laws are empirical in natureandwere deduced from the observations of theDanishastronomer Tycho de Brahe (1546-1601)The third lawgives a relationship between the radius of orbit R of a planet and its period TofrevolutionabouttheSunThelawisexpressedmathematicallyas

2

3 constantTR

=

We shall nowuseNewtonrsquos LawofGravitation to showhowitisthattheplanetsmoveinaccordancewithKeplerrsquosthirdlaw

InessenceNewtonwasabletousehislawofgravitytopredict the motion of the planets since all he had to do was factor theF givenby this law intohis second law F = ma to find their accelerations and hence their future positions

InFigure1017theEarthisshownorbitingtheSunandthedistancebetweentheircentresisR

RSun

Earth

F es

F se

Figure 1017 Planets move according to Keplerrsquos third law

Fes is the force that the Earth exerts on the Sun and Fse is the force that the Sun exerts on the Earth The forces are equalandoppositeandtheSunandtheEarthwillactuallyorbitaboutacommoncentreHoweversince theSun issoverymuchmoremassivethantheEarththiscommoncentrewillbeclosetothecentreoftheSunandsowecanregard the Earth as orbiting about the centre of the Sun Theotherthingthatweshallassumeisthatwecanignorethe forces that the other planets exert on the Earth (This wouldnotbeawisethingtodoifyouwereplanningtosend a space ship to theMoon for example)We shallalso assume that we have followed Newtonrsquos exampleandindeedprovedthataspherewillactasapointmasssituated at the centre of the sphere

Kepler had postulated that the orbits of the planets areellipticalbut since theeccentricityof theEarthrsquosorbit issmallweshallassumeacircularorbit

PHYSICS 2015indb 319 150514 338 PM

Chapter 10

320

AH

L

TheaccelerationoftheEarthtowardstheSunisa = Rω2

whereω ω 2πT------=

Hencea R 2π

T------ times

2 4π2 RT 2------------- = =

But the acceleration is given by Newtonrsquos Second Law F = mawhereFisnowgivenbytheLawofGravitationSo in this situation

F maGMs Me

R2-------------------= = butwealsohavethat

a a R 2πT

------ times

2 4π2 RT 2------------- = = and m = Me so that

G MsMe

R2------------------- Me4π2R

T2-------------times

GMs

4π2-----------rArr R3

T2------= =

Butthequantity

isaconstantthathasthesamevalueforeachoftheplanetssowehaveforalltheplanetsnotjustEarththat

R3

T2------ k =

wherekisaconstantWhichisofcourseKeplerrsquosthirdlaw

ThisisindeedanamazingbreakthroughItisdifficulttorefutethe idea that all particles attract each other in accordance withtheLawofGravitationwhenthelawisabletoaccountfortheobservedmotionoftheplanetsabouttheSun

The gravitational effects of the planets upon each othershould produce perturbations in their orbits Such is the predictivepoweroftheUniversalGravitationalLawthatit enabled physicists to compute these perturbations The telescopehadbeeninventedin1608andbythemiddleofthe 18th Century had reached a degree of perfection in design that enabled astronomers to actually measure the orbital perturbations of the planets Their measurements werealways inagreementwith thepredictionsmadebyNewtonrsquoslawHoweverin1781anewplanetUranuswasdiscovered and the orbit of this planet did not fit withthe orbit predicted by Universal Gravitation Such wasthe physicistrsquos faith in theNewtonianmethod that theysuspectedthatthediscrepancywasduetothepresenceofayetundetectedplanetUsingtheLawofGravitationtheFrenchastronomerJLeverrier and the English astronomer J C Adamswereable tocalculate justhowmassive thisnewplanetmustbeandalsowhereitshouldbeIn1846the planetNeptunewas discovered justwhere they hadpredicted Ina similarwaydiscrepancies in theorbitofNeptune ledto thepredictionandsubsequentdiscoveryin1930oftheplanetPlutoNewtonrsquosLawofGravitation

had passed the ultimate test of any theory it is not only abletoexplainexistingdatabutalsotomakepredictions

Satellite energyWhenasatelliteisinorbitaboutaplanetitwillhavebothkineticenergyandgravitationalpotentialenergySupposewe consider a satellite ofmass m that is in an orbit of radius r about a planet of mass M

Thegravitationalpotentialduetotheplanetatdistancer from its centre is

ThegravitationalpotentialenergyofthesatelliteVsat

is therefore GMe m

r----------------ndash

ThatisVsat GMem

r----------------ndash=

Thegravitationalfieldstrengthatthesurfaceoftheplanetisgivenby

g 0GMe

Re2------------=

Hencewecanwrite

Vsat g0 Re

2

rndash=

m

ThekineticenergyofthesatelliteKsatisequaltofrac12mv2wherev is its orbital speed

Byequatingthegravitationalforceactingonthesatellitetoitscentripetalaccelerationwehave

G Mem

r2---------------- mv2

r---------- mv2hArr

GMemr----------------= =

Fromwhich12---

mv2 12---

GMemr----------------times=

= 0 e

2g R

r

2m

Which is actually quite interesting since it shows thatirrespective of the orbital radius the KE is numericallyequal to half the PE Also the total energy E tot K sat Vsat+ 1

2---G Me m

r----------------times G Me m

r----------------ndash + 1

2---G Me m

r----------------ndash= = =of the satelliteisalwaysnegativesince

E tot K sat Vsat+ 12---

G Me mr----------------times

G Me mr----------------ndash

+ 12---

G Me mr----------------ndash= = =

PHYSICS 2015indb 320 150514 338 PM

Fields

321

AH

L

The energies of an orbiting satellite as a function of radial distancefromthecentreofaplanetareshownplottedinFigure1018

12

100806

0402

0ndash02

ndash04

ndash06ndash08ndash10

ndash12

2 4 6 8 10 12

ener

gya

rbita

ry u

nits

distance timesR

kinetic energytotal energy

potential energy

Figure 1018 Energy of an orbiting satellite as a function of distance from the centre of a planet

WeightlessnessSuppose that you are inanelevator(lift)thatisdescendingat constant speedandyou let goof abook thatyouareholdinginyourhandThebookwillfalltothefloorwithaccelerationequaltotheaccelerationduetogravityIfthecablethatsupportstheelevatorweretosnap(asituationthatItrustwillneverhappentoanyofyou)andyounowletgothebookthatyouareholdinginyourotherhandthis book will not fall to the floor - it will stay exactlyin linewithyourhandThis isbecause thebook isnowfallingwith the sameaccelerationas theelevatorandassuch the book cannot ldquocatchrdquo up with the floor of theelevatorFurthermoreifyouhappenedtobestandingonasetofbathroomscalesthescaleswouldnowreadzero-youwouldbeapparentlyweightlessItisthisideaoffreefallthatexplainstheapparentweightlessnessofastronautsin an orbiting satellite These astronauts are in free fall inthesensethattheyareacceleratingtowardsthecentre of the Earth

It is actually possible to define theweight of a body inseveral different waysWe can define it for example asthegravitationalforceexertedonthebodybyaspecifiedobject such as the EarthThiswe have seen thatwe doinlotsofsituationswherewedefinetheweightasbeingequaltomgIfweusethisdefinitionthenanobjectinfreefallcannotbydefinitionbeweightlesssinceitisstillinagravitationalfieldHoweverifwedefinetheweightofanobjectintermsofaldquoweighingrdquoprocesssuchasthereadingonasetofbathroomscaleswhichineffectmeasuresthecontact force between the object and the scales thenclearlyobjectsinfreefallareweightlessOnenowhasto

askthequestionwhetherornotitispossibleForexampletomeasurethegravitationalforceactingonanastronautin orbit about the Earth

WecanalsodefineweightintermsofthenetgravitationalforceactingonabodyduetoseveraldifferentobjectsForexample for an object out in space its weight could bedefined in terms of the resultant of the forces exerted on it bytheSuntheMoontheEarthandalltheotherplanetsin the Solar System If this resultant is zero at a particular pointthenthebodyisweightlessatthispoint

Inviewofthevariousdefinitionsofweightthatareavailabletousitisimportantthatwhenweusethewordldquoweightrdquoweareawareofthecontextinwhichitisbeingused

Example

Calculate the height above the surface of the Earth atwhichageo-stationarysatelliteorbits

Solution

A geo-stationary satellite is one that orbits the Earth in such a way that it is stationary with respect to a point on the surface of the Earth This means that its orbital period must be the same as the time for the Earth to spin once on its axis ie 24 hours

From Keplerrsquos third law we have G Ms

4π2----------- R3

T2------=

That is

RMe

mh

using the fact that the force of attraction between the satellite and the Earth is given by

FG Mem

R2----------------=

and that F = ma

where m is the mass of the satellite and a 4π2R

T2-------------=

we have

GMem

R2---------------- m 4π2R

T2-------------times

GMe

4π2------------rArr R3

T2------= =

PHYSICS 2015indb 321 150514 338 PM

Chapter 10

322

AH

L

Now the mass of the Earth is 60 times 1024 kg and the period T measured in seconds is given by T = 86400 s

So substitution gives R = 42 times 106 m

The radius of the Earth is 64 times 106 m so that the orbital height h is about 36 times 107 m

Example

Calculatetheminimumenergyrequiredtoputasatelliteofmass500kgintoanorbitthatisasaheightequaltotheEarthrsquosradiusabovethesurfaceoftheEarth

Solution

We have seen that when dealing with gravitational fields and potential it is useful to remember that

g 0G M

Re2---------= or g0 Re

2 GM=

The gravitational potential at the surface of the Earth is

g 0ndash Re GMRe

---------ndash=

The gravitational potential at a distance R

from the centre of the Earth is

=

The difference in potential between the surface of the Earth and a point distance R from the centre is therefore

V∆ g 0Re 1ReR------ndash

=

If R 2Re= then V∆g0 Re

2------------=

This means that the work required to ldquoliftrdquo the satellite into orbit is g0Rm where m is the mass of the satellite This is equal to

10 32 106 500timestimestimes = 16000 MJ

However the satellite must also have kinetic energy in order to orbit Earth This will be equal to

g 0 mRe2

2R-----------------g 0mRe

2

4----------------- 8000 MJ= =

The minimum energy required is therefore

24000 MJ

Exercise 1031 The speed needed to put a satellite in orbit does

not depend on

A the radius of the orbitB theshapeoftheorbitC thevalueofgattheorbitD the mass of the satellite

2 EstimatethespeedofanEarthsatellitewhoseorbitis400kmabovetheEarthrsquossurfaceAlsodetermine the period of the orbit

3 Calculatethespeedofa200kgsatelliteorbitingthe Earth at a height of 70 times 106 m

Assume that g = 82 m sndash2 for this orbit

4 TheradiioftwosatellitesXandYorbitingthe Earth are 2r and 8rwherer is the radius of the Earth Calculate the ratio of the periods of revolutionofXtoY

5 A satellite of mass mkgissentfromEarthrsquossurfaceinto an orbit of radius 5RwhereR is the radius of theEarthWritedownanexpressionfor

(a) thepotentialenergyofthesatelliteinorbit

(b) thekineticenergyofthesatelliteinorbit

(c) theminimumworkrequiredtosendthesatellitefromrestattheEarthrsquossurfaceintoits orbit

6 A satellite in an orbit of 10rfallsbacktoEarth(radius r)afteramalfunctionDeterminethespeedwithwhichitwillhittheEarthrsquossurface

7 The radius of the moon is frac14 that of the Earth AssumingEarthandtheMoontohavethesamedensitycomparetheaccelerationsoffreefallatthe surface of Earth to that at the surface of the Moon

8 UsethefollowingdatatodeterminethegravitationalfieldstrengthatthesurfaceoftheMoon and hence determine the escape speed from the surface of the Moon

Mass of the Moon = 73 times 1022kg

RadiusoftheMoon=17times106 m

PHYSICS 2015indb 322 150514 338 PM

Chapter 10

314

AH

L

When the sphere becomes charged we know that thechargedistributes itselfevenlyoverthesurfaceThereforeeverypartofthematerialoftheconductorisatthesamepotential As the electric potential at a point is defined as beingnumericallyequaltotheworkdoneinbringingaunitpositivechargefrominfinitytothatpointithasaconstantvalueineverypartofthematerialoftheconductor

Since the potential is the same at all points on the conducting surfacethen∆V∆xiszeroButE=ndash∆V∆xThereforethe electric field inside the conductor is zero There is no electric field inside the conductor

SomefurtherobservationsofthegraphsinFigure1009are

bull OutsidethespherethegraphsobeytherelationshipsgivenasE α 1 r2 and V α 1 r

bull Atthesurfacer = r0Thereforetheelectricfieldandpotentialhavetheminimumvalueforr at this point and this infers a maximum field and potential

bull Insidethespheretheelectricfieldiszerobull Insidethespherenoworkisdonetomoveacharge

fromapointinsidetothesurfaceThereforethereisno potential difference and the potential is the same asitiswhenr = r0

Similargraphscanbedrawnfortheelectricfieldintensityandthe electric potential as a function of distance from conducting parallelplatesandsurfacesandthesearegiveninFigure1009

Potential plot E eld plotE eld

++++

ndashndashndashndash

xx x

+

x x

ndashx x

+ ndashx

x

+ +

x

x

Figure 1009 Electric field and electric potential at a distance from a charged surface

Potential due to one or more point charges

The potential due to one point charge can be determined byusingtheequationformula

V = kq r

Example 1

Determine the electric potential at a point 20 times 10-1 m fromthecentreofan isolatedconducting spherewithapointchargeof40pCinair

Solution

Using the formula

V = kq r we have

V90 10 9times( ) 40 10 12ndashtimes( )times

20 10 1ndashtimes( )------------------------------------------------------------------ 018 V= =

the potential at the point is 180 times 10-1 V

The potential due to a number of point charges can be determined by adding up the potentials due to individual point charges because the electric potential at any point outside a conducting sphere will be the same as if all the charge was concentrated at its centre

Example 2

Threepointchargesofareplacedattheverticesofaright-angledtriangleasshowninthediagrambelowDeterminetheabsolutepotentialat the+20μCchargeduetothetwoothercharges

- 6μC

+3μC +2μC

4 m

3 m

Solution

The electric potential of the +2 μC charge due to the ndash 6 μC charge is

V = (9 times 109 Nm2C-2 times -6 times 10-6 C) divide (radic 32 + 42) m = - 108 times 104 V

PHYSICS 2015indb 314 150514 338 PM

Fields

315

AH

L

Insummarywecanconcludethat

bull No work is done to move a charge along an equipotential

bull Equipotentials are always perpendicular to the electric lines of force

Figure1011and1012showsomeequipotential lines fortwo oppositely charged and identically positive spheresseparated by a distance

+ve ndashve

equipotential lines

Figure 1011 Equipotential lines between two opposite charges

equipotential lines

Figure 1012 Equipotential lines between two charges which are the same

+

ndash ndash ndash ndash

+ + +

40 V

30 V

20 V

10 V

50 V

Figure 1013 Equipotential lines between charged parallel plates

Figure1013showstheequipotentiallinesbetweenchargedparallel plates Throughout this chapter the similarities and differences between gravitational fields and electric fieldshavebeendiscussedTherelationshipsthatexistsbetweengravitationalandelectricquantitiesandtheeffectsofpointmassesandchargesissummarisedinTable1014

The electric potential of the +2 μC charge due to the +3 μC charge is

V = (9 times 109 Nm2C-2 times 3 times 10-6 C) divide 3m = 9 times 103 V

The net absolute potential is the sum of the 2 potentials- 108 times 104 V + 9 times 103 V =- 18 times 103 V

The absolute potential at the point is - 18 times 103 V

Equipotential surfacesRegions in space where the electric potential of a chargedistributionhasaconstantvaluearecalledequipotentials The placeswherethepotentialisconstantinthreedimensionsarecalled equipotential surfacesandwheretheyareconstantintwodimensionstheyarecalledequipotential lines

TheyareinsomewaysanalogoustothecontourlinesontopographicmapsInthiscasethegravitationalpotentialenergy is constant as amassmoves around the contourlines because the mass remains at the same elevationabovetheEarthrsquossurfaceThegravitationalfieldstrengthacts in a direction perpendicular to a contour line

Similarlybecausetheelectricpotentialonanequipotentiallinehas thesamevalueanelectric forcecandonoworkwhena testchargemovesonanequipotentialThereforethe electric field cannot have a component along anequipotentialandthusitmustbeeverywhereperpendiculartotheequipotentialsurfaceorequipotentiallineThisfactmakesiteasytoplotequipotentialsifthelinesofforceorlinesofelectricfluxofanelectricfieldareknown

Forexamplethereareaseriesofequipotentiallinesbetweentwoparallelplateconductorsthatareperpendiculartotheelectric fieldTherewill be a series of concentric circles(eachcirclefurtherapartthanthepreviousone)thatmapouttheequipotentialsaroundanisolatedpositivesphereThe lines of force and some equipotential lines for anisolatedpositivesphereareshowninFigure1010

Lines ofequipotential

Figure 1010 Equipotentials around an isolated positive sphere

PHYSICS 2015indb 315 150514 338 PM

Chapter 10

316

AH

L

Gravitational quantity Electrical quantityQ

uant

ities

g V∆x∆-------ndash= E V∆

x∆-------ndash=

Poin

t mas

ses a

nd ch

arge

s

V Gmr----

ndash= V 14πε0------------q

r---=

g G m

r2-----ndash= E 14πε0------------ q

r2-----=

F Gm1 m2

r2--------------= F 1

4πε0------------

q1q2

r2-----------=

Figure 1014 Formulas

Example

Deduce the electric potential on the surface of a gold nucleus that has a radius of 62 fm

Solution

Using the formula

V = kq r and knowing the atomic number of gold is 79 We will assume the nucleus is spherical and it behaves as if it were a point charge at its centre (relative to outside points)

V = 90 times 109 Nm2C-2 times 79 times 16 times 10-19 C divide 62 times 10-15 m = 18 times 107 V

The potential at the point is 18 MV

Example

Deduce the ionisation energy in electron-volts of theelectron in the hydrogen atom if the electron is in its ground state and it is in a circular orbit at a distance of 53 times 10-11 m from the proton

Solution

This problem is an energy coulombic circular motion question based on Bohrrsquos model of the atom (not the accepted quantum mechanics model) The ionisation energy is the energy required to remove the electron from the ground state to infinity The electron travels in a circular orbit and therefore has a centripetal acceleration The ionisation energy will counteract the coulombic force and the movement of the electron will be in the opposite direction to the centripetal force

Total energy = Ek electron + Ep due to the proton-electron interaction

ΣF = kqQ r2 = mv2 r and as such mv2 = = kqQ r

Therefore Ek electron = frac12 kqQ r

Ep due to the proton-electron interaction = - kqQ r

Total energy = frac12 kqQ r + - kqQ r = -frac12 kqQ r

= - 90 times 109 Nm2C-2 times (16 times 10-19 C)2 divide 53 times 10-11 m = -217 times 10-18 J

= -217 times 10-18 J divide 16 times 10-19 = -136 eV

The ionisation energy is 136 eV

Exercise 1021 ApointchargePisplacedmidwaybetweentwo

identicalnegativechargesWhichoneofthefollowingiscorrectwithregardstoelectricfieldandelectricpotentialatpointP

Electric field Electric potentialA non-zero zeroB zero non-zeroC non-zero non-zeroD zero zero

2 Twopositivechargedspheresaretiedtogetherinavacuumsomewhereinspacewheretherearenoexternal forces A has a mass of 25 g and a charge of20μCandBhasamassof15gandachargeof30μCThedistancebetweenthemis40cmTheyarethenreleasedasshowninthediagram

BEFORE AFTER

v1 v2

A B

(a) Determinetheirinitialelectricpotentialenergy in the before situation

(b) DeterminethespeedofsphereBafterrelease

3 Thediagrambelowrepresentstwoequipotentiallines in separated by a distance of 5 cm in a uniform electric field

+ + + + + + + +

ndash ndash ndash ndash ndash ndash ndash ndash

40 V

20 V 5 cm

PHYSICS 2015indb 316 150514 338 PM

Fields

317

AH

L

Determine the strength of the electric field

4 Thisquestionisabouttheelectricfieldduetoacharged sphere and the motion of electrons in that fieldThediagrambelowshowsanisolatedmetalsphereinavacuumthatcarriesanegativeelectriccharge of 60 μC

ndash

(a) Onthediagramdrawtheconventionalwayto represent the electric field pattern due to the charged sphere and lines to represent threeequipotentialsurfacesintheregionoutside the sphere

(b) Explainhowthelinesrepresentingtheequipotentialsurfacesthatyouhavesketchedindicatethatthestrengthoftheelectricfieldisdecreasingwithdistancefrom the centre of the sphere

(c) Theelectricfieldstrengthatthesurfaceofthe sphere and at points outside the sphere can be determined by assuming that the sphere acts as a point charge of magnitude 60 μC at its centre The radius of the sphere is 25 times 10ndash2 m Deduce that the magnitude of the field strength at the surface of the sphere is 86 times 107 Vmndash1

An electron is initially at rest on the surface of the sphere

(d) (i) Describethepathfollowedbytheelectronasitleavesthesurfaceofthesphere

(ii) Calculatetheinitialaccelerationofthe electron

5 Determinetheamountofworkthatisdoneinmovingachargeof100nCthroughapotentialdifferenceof150times102 V

6 Three identical 20 μC conducting spheres are placedatthecornersofanequilateraltriangleofsides 25 cm The triangle has one apex C pointing upthepageand2baseanglesAandBDeterminetheabsolutepotentialatB

7 Determinehowfaraparttwoparallelplatesmustbesituatedsothatapotentialdifferenceof250 times 102 V produces an electric field strength of 200 times 103 NC-1

8 Thegapbetweentwoparallelplatesis10times10-3mandthereisapotentialdifferenceof10times104 V betweentheplatesCalculate

i theworkdonebyanelectroninmovingfrom one plate to the other

ii thespeedwithwhichtheelectronreachesthe second plate if released from rest

iii theelectricfieldintensitybetweentheplates

9 Anelectronguninapicturetubeisacceleratedby a potential 25 times 103VDeterminethekineticenergygainedbytheelectroninelectron-volts

10 Determine the electric potential 20 times 10-2 m from achargeof-10times10-5 C

11 Determinetheelectricpotentialatapointmid-waybetweenachargeofndash20pCandanotherof+5pConthelinejoiningtheircentresifthecharges are 10 cm apart

12 During a thunderstorm the electric potential differencebetweenacloudandthegroundis10 times 109 V Determine the magnitude of the change in electric potential energy of an electron thatmovesbetweenthesepointsinelectron-volts

13 A charge of 15 μC is placed in a uniform electric fieldoftwooppositelychargedparallelplateswithamagnitudeof14times103 NC-1

(a) Determinetheworkthatmustbedoneagainstthefieldtomovethepointchargeadistance of 55 cm

(b) Calculatethepotentialdifferencebetweenthe final and initial positions of the charge

(c) Determinethepotentialdifferencebetweenthe plates if their separation distance is 15 cm

14 Duringaflashoflightningthepotentialdifferencebetweenacloudandthegroundwas12 times 109 V and the amount of transferred charge was32C

(a) Determinethechangeinenergyofthetransferred charge

(b) Iftheenergyreleasedwasallusedtoacceleratea1tonnecardeduceitsfinalspeed

(c) Iftheenergyreleasedcouldbeusedtomelticeat0degCdeducetheamountoficethatcould be melted

PHYSICS 2015indb 317 150514 338 PM

Chapter 10

318

AH

L

15 SupposethatwhenanelectronmovedfromAtoBin the diagram along an electric field line that the electric field does 36 times 10-19Jofworkonit

Determinethedifferencesinelectricpotential

(a) VB ndash VA(b) VC ndash VA(c) VC ndash VB

16 DeterminethepotentialatpointPthatislocatedatthecentreofthesquareasshowninthediagrambelow

- 6μC

+3μC +2μC

1m

1m

5μC

P

102 Fields at work

Essential idea Similar approaches can be taken inanalysingelectricalandgravitationalpotentialproblems

Understandingsbull Potentialandpotentialenergybull Potentialgradientbull Potentialdifferencebull Escapespeedbull Orbitalmotionorbitalspeedandorbitalenergybull Forcesandinverse-squarelawbehaviour

Orbital motion orbital speed and orbital energy

Although orbital motion may be circular elliptical orparabolic this sub-topic only dealswith circular orbitsThis sub-topic is not fundamentally new physics butan application that synthesizes ideas from gravitationcircularmotiondynamicsandenergy

TheMoonorbitstheEarthandinthissenseitisoftenreferredtoasasatelliteoftheEarthBefore1957itwastheonlyEarthsatelliteHoweverin1957theRussianslaunchedthefirstmanmadesatelliteSputnik1SincethisdatemanymoresatelliteshavebeenlaunchedandtherearenowliterallythousandsofthemorbitingtheEarthSomeareusedtomonitortheweathersome used to enable people to find accurately their position on thesurfaceoftheEarthmanyareusedincommunicationsandnodoubtsomeareusedtospyonothercountriesFigure1015showshowinprincipleasatellitecanbeputintoorbit

Theperson(whosesizeisgreatlyexaggeratedwithrespectto Earth) standing on the surface on the Earth throwssome stonesThe greater the speed with which a stoneis thrown the further it will land from her The pathsfollowedbythethrownstonesareparabolasByastretchof the imaginationwecanvisualisea situation inwhichastoneis thrownwithsuchaspeedthatbecauseof thecurvatureoftheEarthitwillnotlandonthesurfaceoftheEarthbutgointoldquoorbitrdquo(Path4onFigure1015)

NATURE OF SCIENCECommunication of scientific explanations The ability to apply field theory to the unobservable (charges) and the massively scaled (motion of satellites) required scientists to develop new ways to investigate analyse and report findings to a general public used to scientific discoveries based on tangible and discernible evidence (51)

copy IBO 2014

PHYSICS 2015indb 318 150514 338 PM

Fields

319

AH

L

1

2

34Earth

Figure 1015 Throwing a stone into orbit

TheforcethatcausesthestonestofollowaparabolicpathandtofalltoEarthisgravityandsimilarlytheforcethatkeeps the stone in orbit is gravity For circular motionto occur we have seen that a force must act at rightanglestothevelocityofanobjectthatistheremustbeacentripetalforceHenceinthesituationwedescribeherethe centripetal force for circular orbital motion about the EarthisprovidedbygravitationalattractionoftheEarth

We can calculate the speedwithwhich a stonemustbethrowninordertoputitintoorbitjustabovethesurfaceof the Earth

If the stone has mass m and speed v thenwehave fromNewtonrsquos2ndlaw

mv2

RE---------- G

MEm

RE2------------=

whereRE is the radius of the Earth and ME is the mass of the Earth

Bearinginmindthatg 0 GME

RE2--------= then

v g RE 10 64 106timestimes 8 103times= = =

Thatisthestonemustbethrownat8times103m sndash1

Clearly we are not going to get a satellite into orbit soclosetothesurfaceoftheEarthMovingatthisspeedthefrictionduetoairresistancewouldmeltthesatellitebeforeithadtravelledacoupleofkilometresInrealitythereforeasatelliteisputintoorbitabouttheEarthbysendingitattachedtoarocketbeyondtheEarthrsquosatmosphereandthengivingitacomponentofvelocityperpendiculartoaradialvectorfromtheEarthSeeFigure1016

Earth Satellite carried by rocket to hereSatellite orbit

Tangential component of velocity

Figure 1016 Getting a satellite into orbit

Keplerrsquos third law(This work of Kepler and Newtonrsquos synthesis of the work is an excellent example of the scientific method and makes for a good TOK discussion)

In 1627 Johannes Kepler (1571-1630)publishedhis lawsof planetary motion The laws are empirical in natureandwere deduced from the observations of theDanishastronomer Tycho de Brahe (1546-1601)The third lawgives a relationship between the radius of orbit R of a planet and its period TofrevolutionabouttheSunThelawisexpressedmathematicallyas

2

3 constantTR

=

We shall nowuseNewtonrsquos LawofGravitation to showhowitisthattheplanetsmoveinaccordancewithKeplerrsquosthirdlaw

InessenceNewtonwasabletousehislawofgravitytopredict the motion of the planets since all he had to do was factor theF givenby this law intohis second law F = ma to find their accelerations and hence their future positions

InFigure1017theEarthisshownorbitingtheSunandthedistancebetweentheircentresisR

RSun

Earth

F es

F se

Figure 1017 Planets move according to Keplerrsquos third law

Fes is the force that the Earth exerts on the Sun and Fse is the force that the Sun exerts on the Earth The forces are equalandoppositeandtheSunandtheEarthwillactuallyorbitaboutacommoncentreHoweversince theSun issoverymuchmoremassivethantheEarththiscommoncentrewillbeclosetothecentreoftheSunandsowecanregard the Earth as orbiting about the centre of the Sun Theotherthingthatweshallassumeisthatwecanignorethe forces that the other planets exert on the Earth (This wouldnotbeawisethingtodoifyouwereplanningtosend a space ship to theMoon for example)We shallalso assume that we have followed Newtonrsquos exampleandindeedprovedthataspherewillactasapointmasssituated at the centre of the sphere

Kepler had postulated that the orbits of the planets areellipticalbut since theeccentricityof theEarthrsquosorbit issmallweshallassumeacircularorbit

PHYSICS 2015indb 319 150514 338 PM

Chapter 10

320

AH

L

TheaccelerationoftheEarthtowardstheSunisa = Rω2

whereω ω 2πT------=

Hencea R 2π

T------ times

2 4π2 RT 2------------- = =

But the acceleration is given by Newtonrsquos Second Law F = mawhereFisnowgivenbytheLawofGravitationSo in this situation

F maGMs Me

R2-------------------= = butwealsohavethat

a a R 2πT

------ times

2 4π2 RT 2------------- = = and m = Me so that

G MsMe

R2------------------- Me4π2R

T2-------------times

GMs

4π2-----------rArr R3

T2------= =

Butthequantity

isaconstantthathasthesamevalueforeachoftheplanetssowehaveforalltheplanetsnotjustEarththat

R3

T2------ k =

wherekisaconstantWhichisofcourseKeplerrsquosthirdlaw

ThisisindeedanamazingbreakthroughItisdifficulttorefutethe idea that all particles attract each other in accordance withtheLawofGravitationwhenthelawisabletoaccountfortheobservedmotionoftheplanetsabouttheSun

The gravitational effects of the planets upon each othershould produce perturbations in their orbits Such is the predictivepoweroftheUniversalGravitationalLawthatit enabled physicists to compute these perturbations The telescopehadbeeninventedin1608andbythemiddleofthe 18th Century had reached a degree of perfection in design that enabled astronomers to actually measure the orbital perturbations of the planets Their measurements werealways inagreementwith thepredictionsmadebyNewtonrsquoslawHoweverin1781anewplanetUranuswasdiscovered and the orbit of this planet did not fit withthe orbit predicted by Universal Gravitation Such wasthe physicistrsquos faith in theNewtonianmethod that theysuspectedthatthediscrepancywasduetothepresenceofayetundetectedplanetUsingtheLawofGravitationtheFrenchastronomerJLeverrier and the English astronomer J C Adamswereable tocalculate justhowmassive thisnewplanetmustbeandalsowhereitshouldbeIn1846the planetNeptunewas discovered justwhere they hadpredicted Ina similarwaydiscrepancies in theorbitofNeptune ledto thepredictionandsubsequentdiscoveryin1930oftheplanetPlutoNewtonrsquosLawofGravitation

had passed the ultimate test of any theory it is not only abletoexplainexistingdatabutalsotomakepredictions

Satellite energyWhenasatelliteisinorbitaboutaplanetitwillhavebothkineticenergyandgravitationalpotentialenergySupposewe consider a satellite ofmass m that is in an orbit of radius r about a planet of mass M

Thegravitationalpotentialduetotheplanetatdistancer from its centre is

ThegravitationalpotentialenergyofthesatelliteVsat

is therefore GMe m

r----------------ndash

ThatisVsat GMem

r----------------ndash=

Thegravitationalfieldstrengthatthesurfaceoftheplanetisgivenby

g 0GMe

Re2------------=

Hencewecanwrite

Vsat g0 Re

2

rndash=

m

ThekineticenergyofthesatelliteKsatisequaltofrac12mv2wherev is its orbital speed

Byequatingthegravitationalforceactingonthesatellitetoitscentripetalaccelerationwehave

G Mem

r2---------------- mv2

r---------- mv2hArr

GMemr----------------= =

Fromwhich12---

mv2 12---

GMemr----------------times=

= 0 e

2g R

r

2m

Which is actually quite interesting since it shows thatirrespective of the orbital radius the KE is numericallyequal to half the PE Also the total energy E tot K sat Vsat+ 1

2---G Me m

r----------------times G Me m

r----------------ndash + 1

2---G Me m

r----------------ndash= = =of the satelliteisalwaysnegativesince

E tot K sat Vsat+ 12---

G Me mr----------------times

G Me mr----------------ndash

+ 12---

G Me mr----------------ndash= = =

PHYSICS 2015indb 320 150514 338 PM

Fields

321

AH

L

The energies of an orbiting satellite as a function of radial distancefromthecentreofaplanetareshownplottedinFigure1018

12

100806

0402

0ndash02

ndash04

ndash06ndash08ndash10

ndash12

2 4 6 8 10 12

ener

gya

rbita

ry u

nits

distance timesR

kinetic energytotal energy

potential energy

Figure 1018 Energy of an orbiting satellite as a function of distance from the centre of a planet

WeightlessnessSuppose that you are inanelevator(lift)thatisdescendingat constant speedandyou let goof abook thatyouareholdinginyourhandThebookwillfalltothefloorwithaccelerationequaltotheaccelerationduetogravityIfthecablethatsupportstheelevatorweretosnap(asituationthatItrustwillneverhappentoanyofyou)andyounowletgothebookthatyouareholdinginyourotherhandthis book will not fall to the floor - it will stay exactlyin linewithyourhandThis isbecause thebook isnowfallingwith the sameaccelerationas theelevatorandassuch the book cannot ldquocatchrdquo up with the floor of theelevatorFurthermoreifyouhappenedtobestandingonasetofbathroomscalesthescaleswouldnowreadzero-youwouldbeapparentlyweightlessItisthisideaoffreefallthatexplainstheapparentweightlessnessofastronautsin an orbiting satellite These astronauts are in free fall inthesensethattheyareacceleratingtowardsthecentre of the Earth

It is actually possible to define theweight of a body inseveral different waysWe can define it for example asthegravitationalforceexertedonthebodybyaspecifiedobject such as the EarthThiswe have seen thatwe doinlotsofsituationswherewedefinetheweightasbeingequaltomgIfweusethisdefinitionthenanobjectinfreefallcannotbydefinitionbeweightlesssinceitisstillinagravitationalfieldHoweverifwedefinetheweightofanobjectintermsofaldquoweighingrdquoprocesssuchasthereadingonasetofbathroomscaleswhichineffectmeasuresthecontact force between the object and the scales thenclearlyobjectsinfreefallareweightlessOnenowhasto

askthequestionwhetherornotitispossibleForexampletomeasurethegravitationalforceactingonanastronautin orbit about the Earth

WecanalsodefineweightintermsofthenetgravitationalforceactingonabodyduetoseveraldifferentobjectsForexample for an object out in space its weight could bedefined in terms of the resultant of the forces exerted on it bytheSuntheMoontheEarthandalltheotherplanetsin the Solar System If this resultant is zero at a particular pointthenthebodyisweightlessatthispoint

Inviewofthevariousdefinitionsofweightthatareavailabletousitisimportantthatwhenweusethewordldquoweightrdquoweareawareofthecontextinwhichitisbeingused

Example

Calculate the height above the surface of the Earth atwhichageo-stationarysatelliteorbits

Solution

A geo-stationary satellite is one that orbits the Earth in such a way that it is stationary with respect to a point on the surface of the Earth This means that its orbital period must be the same as the time for the Earth to spin once on its axis ie 24 hours

From Keplerrsquos third law we have G Ms

4π2----------- R3

T2------=

That is

RMe

mh

using the fact that the force of attraction between the satellite and the Earth is given by

FG Mem

R2----------------=

and that F = ma

where m is the mass of the satellite and a 4π2R

T2-------------=

we have

GMem

R2---------------- m 4π2R

T2-------------times

GMe

4π2------------rArr R3

T2------= =

PHYSICS 2015indb 321 150514 338 PM

Chapter 10

322

AH

L

Now the mass of the Earth is 60 times 1024 kg and the period T measured in seconds is given by T = 86400 s

So substitution gives R = 42 times 106 m

The radius of the Earth is 64 times 106 m so that the orbital height h is about 36 times 107 m

Example

Calculatetheminimumenergyrequiredtoputasatelliteofmass500kgintoanorbitthatisasaheightequaltotheEarthrsquosradiusabovethesurfaceoftheEarth

Solution

We have seen that when dealing with gravitational fields and potential it is useful to remember that

g 0G M

Re2---------= or g0 Re

2 GM=

The gravitational potential at the surface of the Earth is

g 0ndash Re GMRe

---------ndash=

The gravitational potential at a distance R

from the centre of the Earth is

=

The difference in potential between the surface of the Earth and a point distance R from the centre is therefore

V∆ g 0Re 1ReR------ndash

=

If R 2Re= then V∆g0 Re

2------------=

This means that the work required to ldquoliftrdquo the satellite into orbit is g0Rm where m is the mass of the satellite This is equal to

10 32 106 500timestimestimes = 16000 MJ

However the satellite must also have kinetic energy in order to orbit Earth This will be equal to

g 0 mRe2

2R-----------------g 0mRe

2

4----------------- 8000 MJ= =

The minimum energy required is therefore

24000 MJ

Exercise 1031 The speed needed to put a satellite in orbit does

not depend on

A the radius of the orbitB theshapeoftheorbitC thevalueofgattheorbitD the mass of the satellite

2 EstimatethespeedofanEarthsatellitewhoseorbitis400kmabovetheEarthrsquossurfaceAlsodetermine the period of the orbit

3 Calculatethespeedofa200kgsatelliteorbitingthe Earth at a height of 70 times 106 m

Assume that g = 82 m sndash2 for this orbit

4 TheradiioftwosatellitesXandYorbitingthe Earth are 2r and 8rwherer is the radius of the Earth Calculate the ratio of the periods of revolutionofXtoY

5 A satellite of mass mkgissentfromEarthrsquossurfaceinto an orbit of radius 5RwhereR is the radius of theEarthWritedownanexpressionfor

(a) thepotentialenergyofthesatelliteinorbit

(b) thekineticenergyofthesatelliteinorbit

(c) theminimumworkrequiredtosendthesatellitefromrestattheEarthrsquossurfaceintoits orbit

6 A satellite in an orbit of 10rfallsbacktoEarth(radius r)afteramalfunctionDeterminethespeedwithwhichitwillhittheEarthrsquossurface

7 The radius of the moon is frac14 that of the Earth AssumingEarthandtheMoontohavethesamedensitycomparetheaccelerationsoffreefallatthe surface of Earth to that at the surface of the Moon

8 UsethefollowingdatatodeterminethegravitationalfieldstrengthatthesurfaceoftheMoon and hence determine the escape speed from the surface of the Moon

Mass of the Moon = 73 times 1022kg

RadiusoftheMoon=17times106 m

PHYSICS 2015indb 322 150514 338 PM

Fields

315

AH

L

Insummarywecanconcludethat

bull No work is done to move a charge along an equipotential

bull Equipotentials are always perpendicular to the electric lines of force

Figure1011and1012showsomeequipotential lines fortwo oppositely charged and identically positive spheresseparated by a distance

+ve ndashve

equipotential lines

Figure 1011 Equipotential lines between two opposite charges

equipotential lines

Figure 1012 Equipotential lines between two charges which are the same

+

ndash ndash ndash ndash

+ + +

40 V

30 V

20 V

10 V

50 V

Figure 1013 Equipotential lines between charged parallel plates

Figure1013showstheequipotentiallinesbetweenchargedparallel plates Throughout this chapter the similarities and differences between gravitational fields and electric fieldshavebeendiscussedTherelationshipsthatexistsbetweengravitationalandelectricquantitiesandtheeffectsofpointmassesandchargesissummarisedinTable1014

The electric potential of the +2 μC charge due to the +3 μC charge is

V = (9 times 109 Nm2C-2 times 3 times 10-6 C) divide 3m = 9 times 103 V

The net absolute potential is the sum of the 2 potentials- 108 times 104 V + 9 times 103 V =- 18 times 103 V

The absolute potential at the point is - 18 times 103 V

Equipotential surfacesRegions in space where the electric potential of a chargedistributionhasaconstantvaluearecalledequipotentials The placeswherethepotentialisconstantinthreedimensionsarecalled equipotential surfacesandwheretheyareconstantintwodimensionstheyarecalledequipotential lines

TheyareinsomewaysanalogoustothecontourlinesontopographicmapsInthiscasethegravitationalpotentialenergy is constant as amassmoves around the contourlines because the mass remains at the same elevationabovetheEarthrsquossurfaceThegravitationalfieldstrengthacts in a direction perpendicular to a contour line

Similarlybecausetheelectricpotentialonanequipotentiallinehas thesamevalueanelectric forcecandonoworkwhena testchargemovesonanequipotentialThereforethe electric field cannot have a component along anequipotentialandthusitmustbeeverywhereperpendiculartotheequipotentialsurfaceorequipotentiallineThisfactmakesiteasytoplotequipotentialsifthelinesofforceorlinesofelectricfluxofanelectricfieldareknown

Forexamplethereareaseriesofequipotentiallinesbetweentwoparallelplateconductorsthatareperpendiculartotheelectric fieldTherewill be a series of concentric circles(eachcirclefurtherapartthanthepreviousone)thatmapouttheequipotentialsaroundanisolatedpositivesphereThe lines of force and some equipotential lines for anisolatedpositivesphereareshowninFigure1010

Lines ofequipotential

Figure 1010 Equipotentials around an isolated positive sphere

PHYSICS 2015indb 315 150514 338 PM

Chapter 10

316

AH

L

Gravitational quantity Electrical quantityQ

uant

ities

g V∆x∆-------ndash= E V∆

x∆-------ndash=

Poin

t mas

ses a

nd ch

arge

s

V Gmr----

ndash= V 14πε0------------q

r---=

g G m

r2-----ndash= E 14πε0------------ q

r2-----=

F Gm1 m2

r2--------------= F 1

4πε0------------

q1q2

r2-----------=

Figure 1014 Formulas

Example

Deduce the electric potential on the surface of a gold nucleus that has a radius of 62 fm

Solution

Using the formula

V = kq r and knowing the atomic number of gold is 79 We will assume the nucleus is spherical and it behaves as if it were a point charge at its centre (relative to outside points)

V = 90 times 109 Nm2C-2 times 79 times 16 times 10-19 C divide 62 times 10-15 m = 18 times 107 V

The potential at the point is 18 MV

Example

Deduce the ionisation energy in electron-volts of theelectron in the hydrogen atom if the electron is in its ground state and it is in a circular orbit at a distance of 53 times 10-11 m from the proton

Solution

This problem is an energy coulombic circular motion question based on Bohrrsquos model of the atom (not the accepted quantum mechanics model) The ionisation energy is the energy required to remove the electron from the ground state to infinity The electron travels in a circular orbit and therefore has a centripetal acceleration The ionisation energy will counteract the coulombic force and the movement of the electron will be in the opposite direction to the centripetal force

Total energy = Ek electron + Ep due to the proton-electron interaction

ΣF = kqQ r2 = mv2 r and as such mv2 = = kqQ r

Therefore Ek electron = frac12 kqQ r

Ep due to the proton-electron interaction = - kqQ r

Total energy = frac12 kqQ r + - kqQ r = -frac12 kqQ r

= - 90 times 109 Nm2C-2 times (16 times 10-19 C)2 divide 53 times 10-11 m = -217 times 10-18 J

= -217 times 10-18 J divide 16 times 10-19 = -136 eV

The ionisation energy is 136 eV

Exercise 1021 ApointchargePisplacedmidwaybetweentwo

identicalnegativechargesWhichoneofthefollowingiscorrectwithregardstoelectricfieldandelectricpotentialatpointP

Electric field Electric potentialA non-zero zeroB zero non-zeroC non-zero non-zeroD zero zero

2 Twopositivechargedspheresaretiedtogetherinavacuumsomewhereinspacewheretherearenoexternal forces A has a mass of 25 g and a charge of20μCandBhasamassof15gandachargeof30μCThedistancebetweenthemis40cmTheyarethenreleasedasshowninthediagram

BEFORE AFTER

v1 v2

A B

(a) Determinetheirinitialelectricpotentialenergy in the before situation

(b) DeterminethespeedofsphereBafterrelease

3 Thediagrambelowrepresentstwoequipotentiallines in separated by a distance of 5 cm in a uniform electric field

+ + + + + + + +

ndash ndash ndash ndash ndash ndash ndash ndash

40 V

20 V 5 cm

PHYSICS 2015indb 316 150514 338 PM

Fields

317

AH

L

Determine the strength of the electric field

4 Thisquestionisabouttheelectricfieldduetoacharged sphere and the motion of electrons in that fieldThediagrambelowshowsanisolatedmetalsphereinavacuumthatcarriesanegativeelectriccharge of 60 μC

ndash

(a) Onthediagramdrawtheconventionalwayto represent the electric field pattern due to the charged sphere and lines to represent threeequipotentialsurfacesintheregionoutside the sphere

(b) Explainhowthelinesrepresentingtheequipotentialsurfacesthatyouhavesketchedindicatethatthestrengthoftheelectricfieldisdecreasingwithdistancefrom the centre of the sphere

(c) Theelectricfieldstrengthatthesurfaceofthe sphere and at points outside the sphere can be determined by assuming that the sphere acts as a point charge of magnitude 60 μC at its centre The radius of the sphere is 25 times 10ndash2 m Deduce that the magnitude of the field strength at the surface of the sphere is 86 times 107 Vmndash1

An electron is initially at rest on the surface of the sphere

(d) (i) Describethepathfollowedbytheelectronasitleavesthesurfaceofthesphere

(ii) Calculatetheinitialaccelerationofthe electron

5 Determinetheamountofworkthatisdoneinmovingachargeof100nCthroughapotentialdifferenceof150times102 V

6 Three identical 20 μC conducting spheres are placedatthecornersofanequilateraltriangleofsides 25 cm The triangle has one apex C pointing upthepageand2baseanglesAandBDeterminetheabsolutepotentialatB

7 Determinehowfaraparttwoparallelplatesmustbesituatedsothatapotentialdifferenceof250 times 102 V produces an electric field strength of 200 times 103 NC-1

8 Thegapbetweentwoparallelplatesis10times10-3mandthereisapotentialdifferenceof10times104 V betweentheplatesCalculate

i theworkdonebyanelectroninmovingfrom one plate to the other

ii thespeedwithwhichtheelectronreachesthe second plate if released from rest

iii theelectricfieldintensitybetweentheplates

9 Anelectronguninapicturetubeisacceleratedby a potential 25 times 103VDeterminethekineticenergygainedbytheelectroninelectron-volts

10 Determine the electric potential 20 times 10-2 m from achargeof-10times10-5 C

11 Determinetheelectricpotentialatapointmid-waybetweenachargeofndash20pCandanotherof+5pConthelinejoiningtheircentresifthecharges are 10 cm apart

12 During a thunderstorm the electric potential differencebetweenacloudandthegroundis10 times 109 V Determine the magnitude of the change in electric potential energy of an electron thatmovesbetweenthesepointsinelectron-volts

13 A charge of 15 μC is placed in a uniform electric fieldoftwooppositelychargedparallelplateswithamagnitudeof14times103 NC-1

(a) Determinetheworkthatmustbedoneagainstthefieldtomovethepointchargeadistance of 55 cm

(b) Calculatethepotentialdifferencebetweenthe final and initial positions of the charge

(c) Determinethepotentialdifferencebetweenthe plates if their separation distance is 15 cm

14 Duringaflashoflightningthepotentialdifferencebetweenacloudandthegroundwas12 times 109 V and the amount of transferred charge was32C

(a) Determinethechangeinenergyofthetransferred charge

(b) Iftheenergyreleasedwasallusedtoacceleratea1tonnecardeduceitsfinalspeed

(c) Iftheenergyreleasedcouldbeusedtomelticeat0degCdeducetheamountoficethatcould be melted

PHYSICS 2015indb 317 150514 338 PM

Chapter 10

318

AH

L

15 SupposethatwhenanelectronmovedfromAtoBin the diagram along an electric field line that the electric field does 36 times 10-19Jofworkonit

Determinethedifferencesinelectricpotential

(a) VB ndash VA(b) VC ndash VA(c) VC ndash VB

16 DeterminethepotentialatpointPthatislocatedatthecentreofthesquareasshowninthediagrambelow

- 6μC

+3μC +2μC

1m

1m

5μC

P

102 Fields at work

Essential idea Similar approaches can be taken inanalysingelectricalandgravitationalpotentialproblems

Understandingsbull Potentialandpotentialenergybull Potentialgradientbull Potentialdifferencebull Escapespeedbull Orbitalmotionorbitalspeedandorbitalenergybull Forcesandinverse-squarelawbehaviour

Orbital motion orbital speed and orbital energy

Although orbital motion may be circular elliptical orparabolic this sub-topic only dealswith circular orbitsThis sub-topic is not fundamentally new physics butan application that synthesizes ideas from gravitationcircularmotiondynamicsandenergy

TheMoonorbitstheEarthandinthissenseitisoftenreferredtoasasatelliteoftheEarthBefore1957itwastheonlyEarthsatelliteHoweverin1957theRussianslaunchedthefirstmanmadesatelliteSputnik1SincethisdatemanymoresatelliteshavebeenlaunchedandtherearenowliterallythousandsofthemorbitingtheEarthSomeareusedtomonitortheweathersome used to enable people to find accurately their position on thesurfaceoftheEarthmanyareusedincommunicationsandnodoubtsomeareusedtospyonothercountriesFigure1015showshowinprincipleasatellitecanbeputintoorbit

Theperson(whosesizeisgreatlyexaggeratedwithrespectto Earth) standing on the surface on the Earth throwssome stonesThe greater the speed with which a stoneis thrown the further it will land from her The pathsfollowedbythethrownstonesareparabolasByastretchof the imaginationwecanvisualisea situation inwhichastoneis thrownwithsuchaspeedthatbecauseof thecurvatureoftheEarthitwillnotlandonthesurfaceoftheEarthbutgointoldquoorbitrdquo(Path4onFigure1015)

NATURE OF SCIENCECommunication of scientific explanations The ability to apply field theory to the unobservable (charges) and the massively scaled (motion of satellites) required scientists to develop new ways to investigate analyse and report findings to a general public used to scientific discoveries based on tangible and discernible evidence (51)

copy IBO 2014

PHYSICS 2015indb 318 150514 338 PM

Fields

319

AH

L

1

2

34Earth

Figure 1015 Throwing a stone into orbit

TheforcethatcausesthestonestofollowaparabolicpathandtofalltoEarthisgravityandsimilarlytheforcethatkeeps the stone in orbit is gravity For circular motionto occur we have seen that a force must act at rightanglestothevelocityofanobjectthatistheremustbeacentripetalforceHenceinthesituationwedescribeherethe centripetal force for circular orbital motion about the EarthisprovidedbygravitationalattractionoftheEarth

We can calculate the speedwithwhich a stonemustbethrowninordertoputitintoorbitjustabovethesurfaceof the Earth

If the stone has mass m and speed v thenwehave fromNewtonrsquos2ndlaw

mv2

RE---------- G

MEm

RE2------------=

whereRE is the radius of the Earth and ME is the mass of the Earth

Bearinginmindthatg 0 GME

RE2--------= then

v g RE 10 64 106timestimes 8 103times= = =

Thatisthestonemustbethrownat8times103m sndash1

Clearly we are not going to get a satellite into orbit soclosetothesurfaceoftheEarthMovingatthisspeedthefrictionduetoairresistancewouldmeltthesatellitebeforeithadtravelledacoupleofkilometresInrealitythereforeasatelliteisputintoorbitabouttheEarthbysendingitattachedtoarocketbeyondtheEarthrsquosatmosphereandthengivingitacomponentofvelocityperpendiculartoaradialvectorfromtheEarthSeeFigure1016

Earth Satellite carried by rocket to hereSatellite orbit

Tangential component of velocity

Figure 1016 Getting a satellite into orbit

Keplerrsquos third law(This work of Kepler and Newtonrsquos synthesis of the work is an excellent example of the scientific method and makes for a good TOK discussion)

In 1627 Johannes Kepler (1571-1630)publishedhis lawsof planetary motion The laws are empirical in natureandwere deduced from the observations of theDanishastronomer Tycho de Brahe (1546-1601)The third lawgives a relationship between the radius of orbit R of a planet and its period TofrevolutionabouttheSunThelawisexpressedmathematicallyas

2

3 constantTR

=

We shall nowuseNewtonrsquos LawofGravitation to showhowitisthattheplanetsmoveinaccordancewithKeplerrsquosthirdlaw

InessenceNewtonwasabletousehislawofgravitytopredict the motion of the planets since all he had to do was factor theF givenby this law intohis second law F = ma to find their accelerations and hence their future positions

InFigure1017theEarthisshownorbitingtheSunandthedistancebetweentheircentresisR

RSun

Earth

F es

F se

Figure 1017 Planets move according to Keplerrsquos third law

Fes is the force that the Earth exerts on the Sun and Fse is the force that the Sun exerts on the Earth The forces are equalandoppositeandtheSunandtheEarthwillactuallyorbitaboutacommoncentreHoweversince theSun issoverymuchmoremassivethantheEarththiscommoncentrewillbeclosetothecentreoftheSunandsowecanregard the Earth as orbiting about the centre of the Sun Theotherthingthatweshallassumeisthatwecanignorethe forces that the other planets exert on the Earth (This wouldnotbeawisethingtodoifyouwereplanningtosend a space ship to theMoon for example)We shallalso assume that we have followed Newtonrsquos exampleandindeedprovedthataspherewillactasapointmasssituated at the centre of the sphere

Kepler had postulated that the orbits of the planets areellipticalbut since theeccentricityof theEarthrsquosorbit issmallweshallassumeacircularorbit

PHYSICS 2015indb 319 150514 338 PM

Chapter 10

320

AH

L

TheaccelerationoftheEarthtowardstheSunisa = Rω2

whereω ω 2πT------=

Hencea R 2π

T------ times

2 4π2 RT 2------------- = =

But the acceleration is given by Newtonrsquos Second Law F = mawhereFisnowgivenbytheLawofGravitationSo in this situation

F maGMs Me

R2-------------------= = butwealsohavethat

a a R 2πT

------ times

2 4π2 RT 2------------- = = and m = Me so that

G MsMe

R2------------------- Me4π2R

T2-------------times

GMs

4π2-----------rArr R3

T2------= =

Butthequantity

isaconstantthathasthesamevalueforeachoftheplanetssowehaveforalltheplanetsnotjustEarththat

R3

T2------ k =

wherekisaconstantWhichisofcourseKeplerrsquosthirdlaw

ThisisindeedanamazingbreakthroughItisdifficulttorefutethe idea that all particles attract each other in accordance withtheLawofGravitationwhenthelawisabletoaccountfortheobservedmotionoftheplanetsabouttheSun

The gravitational effects of the planets upon each othershould produce perturbations in their orbits Such is the predictivepoweroftheUniversalGravitationalLawthatit enabled physicists to compute these perturbations The telescopehadbeeninventedin1608andbythemiddleofthe 18th Century had reached a degree of perfection in design that enabled astronomers to actually measure the orbital perturbations of the planets Their measurements werealways inagreementwith thepredictionsmadebyNewtonrsquoslawHoweverin1781anewplanetUranuswasdiscovered and the orbit of this planet did not fit withthe orbit predicted by Universal Gravitation Such wasthe physicistrsquos faith in theNewtonianmethod that theysuspectedthatthediscrepancywasduetothepresenceofayetundetectedplanetUsingtheLawofGravitationtheFrenchastronomerJLeverrier and the English astronomer J C Adamswereable tocalculate justhowmassive thisnewplanetmustbeandalsowhereitshouldbeIn1846the planetNeptunewas discovered justwhere they hadpredicted Ina similarwaydiscrepancies in theorbitofNeptune ledto thepredictionandsubsequentdiscoveryin1930oftheplanetPlutoNewtonrsquosLawofGravitation

had passed the ultimate test of any theory it is not only abletoexplainexistingdatabutalsotomakepredictions

Satellite energyWhenasatelliteisinorbitaboutaplanetitwillhavebothkineticenergyandgravitationalpotentialenergySupposewe consider a satellite ofmass m that is in an orbit of radius r about a planet of mass M

Thegravitationalpotentialduetotheplanetatdistancer from its centre is

ThegravitationalpotentialenergyofthesatelliteVsat

is therefore GMe m

r----------------ndash

ThatisVsat GMem

r----------------ndash=

Thegravitationalfieldstrengthatthesurfaceoftheplanetisgivenby

g 0GMe

Re2------------=

Hencewecanwrite

Vsat g0 Re

2

rndash=

m

ThekineticenergyofthesatelliteKsatisequaltofrac12mv2wherev is its orbital speed

Byequatingthegravitationalforceactingonthesatellitetoitscentripetalaccelerationwehave

G Mem

r2---------------- mv2

r---------- mv2hArr

GMemr----------------= =

Fromwhich12---

mv2 12---

GMemr----------------times=

= 0 e

2g R

r

2m

Which is actually quite interesting since it shows thatirrespective of the orbital radius the KE is numericallyequal to half the PE Also the total energy E tot K sat Vsat+ 1

2---G Me m

r----------------times G Me m

r----------------ndash + 1

2---G Me m

r----------------ndash= = =of the satelliteisalwaysnegativesince

E tot K sat Vsat+ 12---

G Me mr----------------times

G Me mr----------------ndash

+ 12---

G Me mr----------------ndash= = =

PHYSICS 2015indb 320 150514 338 PM

Fields

321

AH

L

The energies of an orbiting satellite as a function of radial distancefromthecentreofaplanetareshownplottedinFigure1018

12

100806

0402

0ndash02

ndash04

ndash06ndash08ndash10

ndash12

2 4 6 8 10 12

ener

gya

rbita

ry u

nits

distance timesR

kinetic energytotal energy

potential energy

Figure 1018 Energy of an orbiting satellite as a function of distance from the centre of a planet

WeightlessnessSuppose that you are inanelevator(lift)thatisdescendingat constant speedandyou let goof abook thatyouareholdinginyourhandThebookwillfalltothefloorwithaccelerationequaltotheaccelerationduetogravityIfthecablethatsupportstheelevatorweretosnap(asituationthatItrustwillneverhappentoanyofyou)andyounowletgothebookthatyouareholdinginyourotherhandthis book will not fall to the floor - it will stay exactlyin linewithyourhandThis isbecause thebook isnowfallingwith the sameaccelerationas theelevatorandassuch the book cannot ldquocatchrdquo up with the floor of theelevatorFurthermoreifyouhappenedtobestandingonasetofbathroomscalesthescaleswouldnowreadzero-youwouldbeapparentlyweightlessItisthisideaoffreefallthatexplainstheapparentweightlessnessofastronautsin an orbiting satellite These astronauts are in free fall inthesensethattheyareacceleratingtowardsthecentre of the Earth

It is actually possible to define theweight of a body inseveral different waysWe can define it for example asthegravitationalforceexertedonthebodybyaspecifiedobject such as the EarthThiswe have seen thatwe doinlotsofsituationswherewedefinetheweightasbeingequaltomgIfweusethisdefinitionthenanobjectinfreefallcannotbydefinitionbeweightlesssinceitisstillinagravitationalfieldHoweverifwedefinetheweightofanobjectintermsofaldquoweighingrdquoprocesssuchasthereadingonasetofbathroomscaleswhichineffectmeasuresthecontact force between the object and the scales thenclearlyobjectsinfreefallareweightlessOnenowhasto

askthequestionwhetherornotitispossibleForexampletomeasurethegravitationalforceactingonanastronautin orbit about the Earth

WecanalsodefineweightintermsofthenetgravitationalforceactingonabodyduetoseveraldifferentobjectsForexample for an object out in space its weight could bedefined in terms of the resultant of the forces exerted on it bytheSuntheMoontheEarthandalltheotherplanetsin the Solar System If this resultant is zero at a particular pointthenthebodyisweightlessatthispoint

Inviewofthevariousdefinitionsofweightthatareavailabletousitisimportantthatwhenweusethewordldquoweightrdquoweareawareofthecontextinwhichitisbeingused

Example

Calculate the height above the surface of the Earth atwhichageo-stationarysatelliteorbits

Solution

A geo-stationary satellite is one that orbits the Earth in such a way that it is stationary with respect to a point on the surface of the Earth This means that its orbital period must be the same as the time for the Earth to spin once on its axis ie 24 hours

From Keplerrsquos third law we have G Ms

4π2----------- R3

T2------=

That is

RMe

mh

using the fact that the force of attraction between the satellite and the Earth is given by

FG Mem

R2----------------=

and that F = ma

where m is the mass of the satellite and a 4π2R

T2-------------=

we have

GMem

R2---------------- m 4π2R

T2-------------times

GMe

4π2------------rArr R3

T2------= =

PHYSICS 2015indb 321 150514 338 PM

Chapter 10

322

AH

L

Now the mass of the Earth is 60 times 1024 kg and the period T measured in seconds is given by T = 86400 s

So substitution gives R = 42 times 106 m

The radius of the Earth is 64 times 106 m so that the orbital height h is about 36 times 107 m

Example

Calculatetheminimumenergyrequiredtoputasatelliteofmass500kgintoanorbitthatisasaheightequaltotheEarthrsquosradiusabovethesurfaceoftheEarth

Solution

We have seen that when dealing with gravitational fields and potential it is useful to remember that

g 0G M

Re2---------= or g0 Re

2 GM=

The gravitational potential at the surface of the Earth is

g 0ndash Re GMRe

---------ndash=

The gravitational potential at a distance R

from the centre of the Earth is

=

The difference in potential between the surface of the Earth and a point distance R from the centre is therefore

V∆ g 0Re 1ReR------ndash

=

If R 2Re= then V∆g0 Re

2------------=

This means that the work required to ldquoliftrdquo the satellite into orbit is g0Rm where m is the mass of the satellite This is equal to

10 32 106 500timestimestimes = 16000 MJ

However the satellite must also have kinetic energy in order to orbit Earth This will be equal to

g 0 mRe2

2R-----------------g 0mRe

2

4----------------- 8000 MJ= =

The minimum energy required is therefore

24000 MJ

Exercise 1031 The speed needed to put a satellite in orbit does

not depend on

A the radius of the orbitB theshapeoftheorbitC thevalueofgattheorbitD the mass of the satellite

2 EstimatethespeedofanEarthsatellitewhoseorbitis400kmabovetheEarthrsquossurfaceAlsodetermine the period of the orbit

3 Calculatethespeedofa200kgsatelliteorbitingthe Earth at a height of 70 times 106 m

Assume that g = 82 m sndash2 for this orbit

4 TheradiioftwosatellitesXandYorbitingthe Earth are 2r and 8rwherer is the radius of the Earth Calculate the ratio of the periods of revolutionofXtoY

5 A satellite of mass mkgissentfromEarthrsquossurfaceinto an orbit of radius 5RwhereR is the radius of theEarthWritedownanexpressionfor

(a) thepotentialenergyofthesatelliteinorbit

(b) thekineticenergyofthesatelliteinorbit

(c) theminimumworkrequiredtosendthesatellitefromrestattheEarthrsquossurfaceintoits orbit

6 A satellite in an orbit of 10rfallsbacktoEarth(radius r)afteramalfunctionDeterminethespeedwithwhichitwillhittheEarthrsquossurface

7 The radius of the moon is frac14 that of the Earth AssumingEarthandtheMoontohavethesamedensitycomparetheaccelerationsoffreefallatthe surface of Earth to that at the surface of the Moon

8 UsethefollowingdatatodeterminethegravitationalfieldstrengthatthesurfaceoftheMoon and hence determine the escape speed from the surface of the Moon

Mass of the Moon = 73 times 1022kg

RadiusoftheMoon=17times106 m

PHYSICS 2015indb 322 150514 338 PM

Chapter 10

316

AH

L

Gravitational quantity Electrical quantityQ

uant

ities

g V∆x∆-------ndash= E V∆

x∆-------ndash=

Poin

t mas

ses a

nd ch

arge

s

V Gmr----

ndash= V 14πε0------------q

r---=

g G m

r2-----ndash= E 14πε0------------ q

r2-----=

F Gm1 m2

r2--------------= F 1

4πε0------------

q1q2

r2-----------=

Figure 1014 Formulas

Example

Deduce the electric potential on the surface of a gold nucleus that has a radius of 62 fm

Solution

Using the formula

V = kq r and knowing the atomic number of gold is 79 We will assume the nucleus is spherical and it behaves as if it were a point charge at its centre (relative to outside points)

V = 90 times 109 Nm2C-2 times 79 times 16 times 10-19 C divide 62 times 10-15 m = 18 times 107 V

The potential at the point is 18 MV

Example

Deduce the ionisation energy in electron-volts of theelectron in the hydrogen atom if the electron is in its ground state and it is in a circular orbit at a distance of 53 times 10-11 m from the proton

Solution

This problem is an energy coulombic circular motion question based on Bohrrsquos model of the atom (not the accepted quantum mechanics model) The ionisation energy is the energy required to remove the electron from the ground state to infinity The electron travels in a circular orbit and therefore has a centripetal acceleration The ionisation energy will counteract the coulombic force and the movement of the electron will be in the opposite direction to the centripetal force

Total energy = Ek electron + Ep due to the proton-electron interaction

ΣF = kqQ r2 = mv2 r and as such mv2 = = kqQ r

Therefore Ek electron = frac12 kqQ r

Ep due to the proton-electron interaction = - kqQ r

Total energy = frac12 kqQ r + - kqQ r = -frac12 kqQ r

= - 90 times 109 Nm2C-2 times (16 times 10-19 C)2 divide 53 times 10-11 m = -217 times 10-18 J

= -217 times 10-18 J divide 16 times 10-19 = -136 eV

The ionisation energy is 136 eV

Exercise 1021 ApointchargePisplacedmidwaybetweentwo

identicalnegativechargesWhichoneofthefollowingiscorrectwithregardstoelectricfieldandelectricpotentialatpointP

Electric field Electric potentialA non-zero zeroB zero non-zeroC non-zero non-zeroD zero zero

2 Twopositivechargedspheresaretiedtogetherinavacuumsomewhereinspacewheretherearenoexternal forces A has a mass of 25 g and a charge of20μCandBhasamassof15gandachargeof30μCThedistancebetweenthemis40cmTheyarethenreleasedasshowninthediagram

BEFORE AFTER

v1 v2

A B

(a) Determinetheirinitialelectricpotentialenergy in the before situation

(b) DeterminethespeedofsphereBafterrelease

3 Thediagrambelowrepresentstwoequipotentiallines in separated by a distance of 5 cm in a uniform electric field

+ + + + + + + +

ndash ndash ndash ndash ndash ndash ndash ndash

40 V

20 V 5 cm

PHYSICS 2015indb 316 150514 338 PM

Fields

317

AH

L

Determine the strength of the electric field

4 Thisquestionisabouttheelectricfieldduetoacharged sphere and the motion of electrons in that fieldThediagrambelowshowsanisolatedmetalsphereinavacuumthatcarriesanegativeelectriccharge of 60 μC

ndash

(a) Onthediagramdrawtheconventionalwayto represent the electric field pattern due to the charged sphere and lines to represent threeequipotentialsurfacesintheregionoutside the sphere

(b) Explainhowthelinesrepresentingtheequipotentialsurfacesthatyouhavesketchedindicatethatthestrengthoftheelectricfieldisdecreasingwithdistancefrom the centre of the sphere

(c) Theelectricfieldstrengthatthesurfaceofthe sphere and at points outside the sphere can be determined by assuming that the sphere acts as a point charge of magnitude 60 μC at its centre The radius of the sphere is 25 times 10ndash2 m Deduce that the magnitude of the field strength at the surface of the sphere is 86 times 107 Vmndash1

An electron is initially at rest on the surface of the sphere

(d) (i) Describethepathfollowedbytheelectronasitleavesthesurfaceofthesphere

(ii) Calculatetheinitialaccelerationofthe electron

5 Determinetheamountofworkthatisdoneinmovingachargeof100nCthroughapotentialdifferenceof150times102 V

6 Three identical 20 μC conducting spheres are placedatthecornersofanequilateraltriangleofsides 25 cm The triangle has one apex C pointing upthepageand2baseanglesAandBDeterminetheabsolutepotentialatB

7 Determinehowfaraparttwoparallelplatesmustbesituatedsothatapotentialdifferenceof250 times 102 V produces an electric field strength of 200 times 103 NC-1

8 Thegapbetweentwoparallelplatesis10times10-3mandthereisapotentialdifferenceof10times104 V betweentheplatesCalculate

i theworkdonebyanelectroninmovingfrom one plate to the other

ii thespeedwithwhichtheelectronreachesthe second plate if released from rest

iii theelectricfieldintensitybetweentheplates

9 Anelectronguninapicturetubeisacceleratedby a potential 25 times 103VDeterminethekineticenergygainedbytheelectroninelectron-volts

10 Determine the electric potential 20 times 10-2 m from achargeof-10times10-5 C

11 Determinetheelectricpotentialatapointmid-waybetweenachargeofndash20pCandanotherof+5pConthelinejoiningtheircentresifthecharges are 10 cm apart

12 During a thunderstorm the electric potential differencebetweenacloudandthegroundis10 times 109 V Determine the magnitude of the change in electric potential energy of an electron thatmovesbetweenthesepointsinelectron-volts

13 A charge of 15 μC is placed in a uniform electric fieldoftwooppositelychargedparallelplateswithamagnitudeof14times103 NC-1

(a) Determinetheworkthatmustbedoneagainstthefieldtomovethepointchargeadistance of 55 cm

(b) Calculatethepotentialdifferencebetweenthe final and initial positions of the charge

(c) Determinethepotentialdifferencebetweenthe plates if their separation distance is 15 cm

14 Duringaflashoflightningthepotentialdifferencebetweenacloudandthegroundwas12 times 109 V and the amount of transferred charge was32C

(a) Determinethechangeinenergyofthetransferred charge

(b) Iftheenergyreleasedwasallusedtoacceleratea1tonnecardeduceitsfinalspeed

(c) Iftheenergyreleasedcouldbeusedtomelticeat0degCdeducetheamountoficethatcould be melted

PHYSICS 2015indb 317 150514 338 PM

Chapter 10

318

AH

L

15 SupposethatwhenanelectronmovedfromAtoBin the diagram along an electric field line that the electric field does 36 times 10-19Jofworkonit

Determinethedifferencesinelectricpotential

(a) VB ndash VA(b) VC ndash VA(c) VC ndash VB

16 DeterminethepotentialatpointPthatislocatedatthecentreofthesquareasshowninthediagrambelow

- 6μC

+3μC +2μC

1m

1m

5μC

P

102 Fields at work

Essential idea Similar approaches can be taken inanalysingelectricalandgravitationalpotentialproblems

Understandingsbull Potentialandpotentialenergybull Potentialgradientbull Potentialdifferencebull Escapespeedbull Orbitalmotionorbitalspeedandorbitalenergybull Forcesandinverse-squarelawbehaviour

Orbital motion orbital speed and orbital energy

Although orbital motion may be circular elliptical orparabolic this sub-topic only dealswith circular orbitsThis sub-topic is not fundamentally new physics butan application that synthesizes ideas from gravitationcircularmotiondynamicsandenergy

TheMoonorbitstheEarthandinthissenseitisoftenreferredtoasasatelliteoftheEarthBefore1957itwastheonlyEarthsatelliteHoweverin1957theRussianslaunchedthefirstmanmadesatelliteSputnik1SincethisdatemanymoresatelliteshavebeenlaunchedandtherearenowliterallythousandsofthemorbitingtheEarthSomeareusedtomonitortheweathersome used to enable people to find accurately their position on thesurfaceoftheEarthmanyareusedincommunicationsandnodoubtsomeareusedtospyonothercountriesFigure1015showshowinprincipleasatellitecanbeputintoorbit

Theperson(whosesizeisgreatlyexaggeratedwithrespectto Earth) standing on the surface on the Earth throwssome stonesThe greater the speed with which a stoneis thrown the further it will land from her The pathsfollowedbythethrownstonesareparabolasByastretchof the imaginationwecanvisualisea situation inwhichastoneis thrownwithsuchaspeedthatbecauseof thecurvatureoftheEarthitwillnotlandonthesurfaceoftheEarthbutgointoldquoorbitrdquo(Path4onFigure1015)

NATURE OF SCIENCECommunication of scientific explanations The ability to apply field theory to the unobservable (charges) and the massively scaled (motion of satellites) required scientists to develop new ways to investigate analyse and report findings to a general public used to scientific discoveries based on tangible and discernible evidence (51)

copy IBO 2014

PHYSICS 2015indb 318 150514 338 PM

Fields

319

AH

L

1

2

34Earth

Figure 1015 Throwing a stone into orbit

TheforcethatcausesthestonestofollowaparabolicpathandtofalltoEarthisgravityandsimilarlytheforcethatkeeps the stone in orbit is gravity For circular motionto occur we have seen that a force must act at rightanglestothevelocityofanobjectthatistheremustbeacentripetalforceHenceinthesituationwedescribeherethe centripetal force for circular orbital motion about the EarthisprovidedbygravitationalattractionoftheEarth

We can calculate the speedwithwhich a stonemustbethrowninordertoputitintoorbitjustabovethesurfaceof the Earth

If the stone has mass m and speed v thenwehave fromNewtonrsquos2ndlaw

mv2

RE---------- G

MEm

RE2------------=

whereRE is the radius of the Earth and ME is the mass of the Earth

Bearinginmindthatg 0 GME

RE2--------= then

v g RE 10 64 106timestimes 8 103times= = =

Thatisthestonemustbethrownat8times103m sndash1

Clearly we are not going to get a satellite into orbit soclosetothesurfaceoftheEarthMovingatthisspeedthefrictionduetoairresistancewouldmeltthesatellitebeforeithadtravelledacoupleofkilometresInrealitythereforeasatelliteisputintoorbitabouttheEarthbysendingitattachedtoarocketbeyondtheEarthrsquosatmosphereandthengivingitacomponentofvelocityperpendiculartoaradialvectorfromtheEarthSeeFigure1016

Earth Satellite carried by rocket to hereSatellite orbit

Tangential component of velocity

Figure 1016 Getting a satellite into orbit

Keplerrsquos third law(This work of Kepler and Newtonrsquos synthesis of the work is an excellent example of the scientific method and makes for a good TOK discussion)

In 1627 Johannes Kepler (1571-1630)publishedhis lawsof planetary motion The laws are empirical in natureandwere deduced from the observations of theDanishastronomer Tycho de Brahe (1546-1601)The third lawgives a relationship between the radius of orbit R of a planet and its period TofrevolutionabouttheSunThelawisexpressedmathematicallyas

2

3 constantTR

=

We shall nowuseNewtonrsquos LawofGravitation to showhowitisthattheplanetsmoveinaccordancewithKeplerrsquosthirdlaw

InessenceNewtonwasabletousehislawofgravitytopredict the motion of the planets since all he had to do was factor theF givenby this law intohis second law F = ma to find their accelerations and hence their future positions

InFigure1017theEarthisshownorbitingtheSunandthedistancebetweentheircentresisR

RSun

Earth

F es

F se

Figure 1017 Planets move according to Keplerrsquos third law

Fes is the force that the Earth exerts on the Sun and Fse is the force that the Sun exerts on the Earth The forces are equalandoppositeandtheSunandtheEarthwillactuallyorbitaboutacommoncentreHoweversince theSun issoverymuchmoremassivethantheEarththiscommoncentrewillbeclosetothecentreoftheSunandsowecanregard the Earth as orbiting about the centre of the Sun Theotherthingthatweshallassumeisthatwecanignorethe forces that the other planets exert on the Earth (This wouldnotbeawisethingtodoifyouwereplanningtosend a space ship to theMoon for example)We shallalso assume that we have followed Newtonrsquos exampleandindeedprovedthataspherewillactasapointmasssituated at the centre of the sphere

Kepler had postulated that the orbits of the planets areellipticalbut since theeccentricityof theEarthrsquosorbit issmallweshallassumeacircularorbit

PHYSICS 2015indb 319 150514 338 PM

Chapter 10

320

AH

L

TheaccelerationoftheEarthtowardstheSunisa = Rω2

whereω ω 2πT------=

Hencea R 2π

T------ times

2 4π2 RT 2------------- = =

But the acceleration is given by Newtonrsquos Second Law F = mawhereFisnowgivenbytheLawofGravitationSo in this situation

F maGMs Me

R2-------------------= = butwealsohavethat

a a R 2πT

------ times

2 4π2 RT 2------------- = = and m = Me so that

G MsMe

R2------------------- Me4π2R

T2-------------times

GMs

4π2-----------rArr R3

T2------= =

Butthequantity

isaconstantthathasthesamevalueforeachoftheplanetssowehaveforalltheplanetsnotjustEarththat

R3

T2------ k =

wherekisaconstantWhichisofcourseKeplerrsquosthirdlaw

ThisisindeedanamazingbreakthroughItisdifficulttorefutethe idea that all particles attract each other in accordance withtheLawofGravitationwhenthelawisabletoaccountfortheobservedmotionoftheplanetsabouttheSun

The gravitational effects of the planets upon each othershould produce perturbations in their orbits Such is the predictivepoweroftheUniversalGravitationalLawthatit enabled physicists to compute these perturbations The telescopehadbeeninventedin1608andbythemiddleofthe 18th Century had reached a degree of perfection in design that enabled astronomers to actually measure the orbital perturbations of the planets Their measurements werealways inagreementwith thepredictionsmadebyNewtonrsquoslawHoweverin1781anewplanetUranuswasdiscovered and the orbit of this planet did not fit withthe orbit predicted by Universal Gravitation Such wasthe physicistrsquos faith in theNewtonianmethod that theysuspectedthatthediscrepancywasduetothepresenceofayetundetectedplanetUsingtheLawofGravitationtheFrenchastronomerJLeverrier and the English astronomer J C Adamswereable tocalculate justhowmassive thisnewplanetmustbeandalsowhereitshouldbeIn1846the planetNeptunewas discovered justwhere they hadpredicted Ina similarwaydiscrepancies in theorbitofNeptune ledto thepredictionandsubsequentdiscoveryin1930oftheplanetPlutoNewtonrsquosLawofGravitation

had passed the ultimate test of any theory it is not only abletoexplainexistingdatabutalsotomakepredictions

Satellite energyWhenasatelliteisinorbitaboutaplanetitwillhavebothkineticenergyandgravitationalpotentialenergySupposewe consider a satellite ofmass m that is in an orbit of radius r about a planet of mass M

Thegravitationalpotentialduetotheplanetatdistancer from its centre is

ThegravitationalpotentialenergyofthesatelliteVsat

is therefore GMe m

r----------------ndash

ThatisVsat GMem

r----------------ndash=

Thegravitationalfieldstrengthatthesurfaceoftheplanetisgivenby

g 0GMe

Re2------------=

Hencewecanwrite

Vsat g0 Re

2

rndash=

m

ThekineticenergyofthesatelliteKsatisequaltofrac12mv2wherev is its orbital speed

Byequatingthegravitationalforceactingonthesatellitetoitscentripetalaccelerationwehave

G Mem

r2---------------- mv2

r---------- mv2hArr

GMemr----------------= =

Fromwhich12---

mv2 12---

GMemr----------------times=

= 0 e

2g R

r

2m

Which is actually quite interesting since it shows thatirrespective of the orbital radius the KE is numericallyequal to half the PE Also the total energy E tot K sat Vsat+ 1

2---G Me m

r----------------times G Me m

r----------------ndash + 1

2---G Me m

r----------------ndash= = =of the satelliteisalwaysnegativesince

E tot K sat Vsat+ 12---

G Me mr----------------times

G Me mr----------------ndash

+ 12---

G Me mr----------------ndash= = =

PHYSICS 2015indb 320 150514 338 PM

Fields

321

AH

L

The energies of an orbiting satellite as a function of radial distancefromthecentreofaplanetareshownplottedinFigure1018

12

100806

0402

0ndash02

ndash04

ndash06ndash08ndash10

ndash12

2 4 6 8 10 12

ener

gya

rbita

ry u

nits

distance timesR

kinetic energytotal energy

potential energy

Figure 1018 Energy of an orbiting satellite as a function of distance from the centre of a planet

WeightlessnessSuppose that you are inanelevator(lift)thatisdescendingat constant speedandyou let goof abook thatyouareholdinginyourhandThebookwillfalltothefloorwithaccelerationequaltotheaccelerationduetogravityIfthecablethatsupportstheelevatorweretosnap(asituationthatItrustwillneverhappentoanyofyou)andyounowletgothebookthatyouareholdinginyourotherhandthis book will not fall to the floor - it will stay exactlyin linewithyourhandThis isbecause thebook isnowfallingwith the sameaccelerationas theelevatorandassuch the book cannot ldquocatchrdquo up with the floor of theelevatorFurthermoreifyouhappenedtobestandingonasetofbathroomscalesthescaleswouldnowreadzero-youwouldbeapparentlyweightlessItisthisideaoffreefallthatexplainstheapparentweightlessnessofastronautsin an orbiting satellite These astronauts are in free fall inthesensethattheyareacceleratingtowardsthecentre of the Earth

It is actually possible to define theweight of a body inseveral different waysWe can define it for example asthegravitationalforceexertedonthebodybyaspecifiedobject such as the EarthThiswe have seen thatwe doinlotsofsituationswherewedefinetheweightasbeingequaltomgIfweusethisdefinitionthenanobjectinfreefallcannotbydefinitionbeweightlesssinceitisstillinagravitationalfieldHoweverifwedefinetheweightofanobjectintermsofaldquoweighingrdquoprocesssuchasthereadingonasetofbathroomscaleswhichineffectmeasuresthecontact force between the object and the scales thenclearlyobjectsinfreefallareweightlessOnenowhasto

askthequestionwhetherornotitispossibleForexampletomeasurethegravitationalforceactingonanastronautin orbit about the Earth

WecanalsodefineweightintermsofthenetgravitationalforceactingonabodyduetoseveraldifferentobjectsForexample for an object out in space its weight could bedefined in terms of the resultant of the forces exerted on it bytheSuntheMoontheEarthandalltheotherplanetsin the Solar System If this resultant is zero at a particular pointthenthebodyisweightlessatthispoint

Inviewofthevariousdefinitionsofweightthatareavailabletousitisimportantthatwhenweusethewordldquoweightrdquoweareawareofthecontextinwhichitisbeingused

Example

Calculate the height above the surface of the Earth atwhichageo-stationarysatelliteorbits

Solution

A geo-stationary satellite is one that orbits the Earth in such a way that it is stationary with respect to a point on the surface of the Earth This means that its orbital period must be the same as the time for the Earth to spin once on its axis ie 24 hours

From Keplerrsquos third law we have G Ms

4π2----------- R3

T2------=

That is

RMe

mh

using the fact that the force of attraction between the satellite and the Earth is given by

FG Mem

R2----------------=

and that F = ma

where m is the mass of the satellite and a 4π2R

T2-------------=

we have

GMem

R2---------------- m 4π2R

T2-------------times

GMe

4π2------------rArr R3

T2------= =

PHYSICS 2015indb 321 150514 338 PM

Chapter 10

322

AH

L

Now the mass of the Earth is 60 times 1024 kg and the period T measured in seconds is given by T = 86400 s

So substitution gives R = 42 times 106 m

The radius of the Earth is 64 times 106 m so that the orbital height h is about 36 times 107 m

Example

Calculatetheminimumenergyrequiredtoputasatelliteofmass500kgintoanorbitthatisasaheightequaltotheEarthrsquosradiusabovethesurfaceoftheEarth

Solution

We have seen that when dealing with gravitational fields and potential it is useful to remember that

g 0G M

Re2---------= or g0 Re

2 GM=

The gravitational potential at the surface of the Earth is

g 0ndash Re GMRe

---------ndash=

The gravitational potential at a distance R

from the centre of the Earth is

=

The difference in potential between the surface of the Earth and a point distance R from the centre is therefore

V∆ g 0Re 1ReR------ndash

=

If R 2Re= then V∆g0 Re

2------------=

This means that the work required to ldquoliftrdquo the satellite into orbit is g0Rm where m is the mass of the satellite This is equal to

10 32 106 500timestimestimes = 16000 MJ

However the satellite must also have kinetic energy in order to orbit Earth This will be equal to

g 0 mRe2

2R-----------------g 0mRe

2

4----------------- 8000 MJ= =

The minimum energy required is therefore

24000 MJ

Exercise 1031 The speed needed to put a satellite in orbit does

not depend on

A the radius of the orbitB theshapeoftheorbitC thevalueofgattheorbitD the mass of the satellite

2 EstimatethespeedofanEarthsatellitewhoseorbitis400kmabovetheEarthrsquossurfaceAlsodetermine the period of the orbit

3 Calculatethespeedofa200kgsatelliteorbitingthe Earth at a height of 70 times 106 m

Assume that g = 82 m sndash2 for this orbit

4 TheradiioftwosatellitesXandYorbitingthe Earth are 2r and 8rwherer is the radius of the Earth Calculate the ratio of the periods of revolutionofXtoY

5 A satellite of mass mkgissentfromEarthrsquossurfaceinto an orbit of radius 5RwhereR is the radius of theEarthWritedownanexpressionfor

(a) thepotentialenergyofthesatelliteinorbit

(b) thekineticenergyofthesatelliteinorbit

(c) theminimumworkrequiredtosendthesatellitefromrestattheEarthrsquossurfaceintoits orbit

6 A satellite in an orbit of 10rfallsbacktoEarth(radius r)afteramalfunctionDeterminethespeedwithwhichitwillhittheEarthrsquossurface

7 The radius of the moon is frac14 that of the Earth AssumingEarthandtheMoontohavethesamedensitycomparetheaccelerationsoffreefallatthe surface of Earth to that at the surface of the Moon

8 UsethefollowingdatatodeterminethegravitationalfieldstrengthatthesurfaceoftheMoon and hence determine the escape speed from the surface of the Moon

Mass of the Moon = 73 times 1022kg

RadiusoftheMoon=17times106 m

PHYSICS 2015indb 322 150514 338 PM

Fields

317

AH

L

Determine the strength of the electric field

4 Thisquestionisabouttheelectricfieldduetoacharged sphere and the motion of electrons in that fieldThediagrambelowshowsanisolatedmetalsphereinavacuumthatcarriesanegativeelectriccharge of 60 μC

ndash

(a) Onthediagramdrawtheconventionalwayto represent the electric field pattern due to the charged sphere and lines to represent threeequipotentialsurfacesintheregionoutside the sphere

(b) Explainhowthelinesrepresentingtheequipotentialsurfacesthatyouhavesketchedindicatethatthestrengthoftheelectricfieldisdecreasingwithdistancefrom the centre of the sphere

(c) Theelectricfieldstrengthatthesurfaceofthe sphere and at points outside the sphere can be determined by assuming that the sphere acts as a point charge of magnitude 60 μC at its centre The radius of the sphere is 25 times 10ndash2 m Deduce that the magnitude of the field strength at the surface of the sphere is 86 times 107 Vmndash1

An electron is initially at rest on the surface of the sphere

(d) (i) Describethepathfollowedbytheelectronasitleavesthesurfaceofthesphere

(ii) Calculatetheinitialaccelerationofthe electron

5 Determinetheamountofworkthatisdoneinmovingachargeof100nCthroughapotentialdifferenceof150times102 V

6 Three identical 20 μC conducting spheres are placedatthecornersofanequilateraltriangleofsides 25 cm The triangle has one apex C pointing upthepageand2baseanglesAandBDeterminetheabsolutepotentialatB

7 Determinehowfaraparttwoparallelplatesmustbesituatedsothatapotentialdifferenceof250 times 102 V produces an electric field strength of 200 times 103 NC-1

8 Thegapbetweentwoparallelplatesis10times10-3mandthereisapotentialdifferenceof10times104 V betweentheplatesCalculate

i theworkdonebyanelectroninmovingfrom one plate to the other

ii thespeedwithwhichtheelectronreachesthe second plate if released from rest

iii theelectricfieldintensitybetweentheplates

9 Anelectronguninapicturetubeisacceleratedby a potential 25 times 103VDeterminethekineticenergygainedbytheelectroninelectron-volts

10 Determine the electric potential 20 times 10-2 m from achargeof-10times10-5 C

11 Determinetheelectricpotentialatapointmid-waybetweenachargeofndash20pCandanotherof+5pConthelinejoiningtheircentresifthecharges are 10 cm apart

12 During a thunderstorm the electric potential differencebetweenacloudandthegroundis10 times 109 V Determine the magnitude of the change in electric potential energy of an electron thatmovesbetweenthesepointsinelectron-volts

13 A charge of 15 μC is placed in a uniform electric fieldoftwooppositelychargedparallelplateswithamagnitudeof14times103 NC-1

(a) Determinetheworkthatmustbedoneagainstthefieldtomovethepointchargeadistance of 55 cm

(b) Calculatethepotentialdifferencebetweenthe final and initial positions of the charge

(c) Determinethepotentialdifferencebetweenthe plates if their separation distance is 15 cm

14 Duringaflashoflightningthepotentialdifferencebetweenacloudandthegroundwas12 times 109 V and the amount of transferred charge was32C

(a) Determinethechangeinenergyofthetransferred charge

(b) Iftheenergyreleasedwasallusedtoacceleratea1tonnecardeduceitsfinalspeed

(c) Iftheenergyreleasedcouldbeusedtomelticeat0degCdeducetheamountoficethatcould be melted

PHYSICS 2015indb 317 150514 338 PM

Chapter 10

318

AH

L

15 SupposethatwhenanelectronmovedfromAtoBin the diagram along an electric field line that the electric field does 36 times 10-19Jofworkonit

Determinethedifferencesinelectricpotential

(a) VB ndash VA(b) VC ndash VA(c) VC ndash VB

16 DeterminethepotentialatpointPthatislocatedatthecentreofthesquareasshowninthediagrambelow

- 6μC

+3μC +2μC

1m

1m

5μC

P

102 Fields at work

Essential idea Similar approaches can be taken inanalysingelectricalandgravitationalpotentialproblems

Understandingsbull Potentialandpotentialenergybull Potentialgradientbull Potentialdifferencebull Escapespeedbull Orbitalmotionorbitalspeedandorbitalenergybull Forcesandinverse-squarelawbehaviour

Orbital motion orbital speed and orbital energy

Although orbital motion may be circular elliptical orparabolic this sub-topic only dealswith circular orbitsThis sub-topic is not fundamentally new physics butan application that synthesizes ideas from gravitationcircularmotiondynamicsandenergy

TheMoonorbitstheEarthandinthissenseitisoftenreferredtoasasatelliteoftheEarthBefore1957itwastheonlyEarthsatelliteHoweverin1957theRussianslaunchedthefirstmanmadesatelliteSputnik1SincethisdatemanymoresatelliteshavebeenlaunchedandtherearenowliterallythousandsofthemorbitingtheEarthSomeareusedtomonitortheweathersome used to enable people to find accurately their position on thesurfaceoftheEarthmanyareusedincommunicationsandnodoubtsomeareusedtospyonothercountriesFigure1015showshowinprincipleasatellitecanbeputintoorbit

Theperson(whosesizeisgreatlyexaggeratedwithrespectto Earth) standing on the surface on the Earth throwssome stonesThe greater the speed with which a stoneis thrown the further it will land from her The pathsfollowedbythethrownstonesareparabolasByastretchof the imaginationwecanvisualisea situation inwhichastoneis thrownwithsuchaspeedthatbecauseof thecurvatureoftheEarthitwillnotlandonthesurfaceoftheEarthbutgointoldquoorbitrdquo(Path4onFigure1015)

NATURE OF SCIENCECommunication of scientific explanations The ability to apply field theory to the unobservable (charges) and the massively scaled (motion of satellites) required scientists to develop new ways to investigate analyse and report findings to a general public used to scientific discoveries based on tangible and discernible evidence (51)

copy IBO 2014

PHYSICS 2015indb 318 150514 338 PM

Fields

319

AH

L

1

2

34Earth

Figure 1015 Throwing a stone into orbit

TheforcethatcausesthestonestofollowaparabolicpathandtofalltoEarthisgravityandsimilarlytheforcethatkeeps the stone in orbit is gravity For circular motionto occur we have seen that a force must act at rightanglestothevelocityofanobjectthatistheremustbeacentripetalforceHenceinthesituationwedescribeherethe centripetal force for circular orbital motion about the EarthisprovidedbygravitationalattractionoftheEarth

We can calculate the speedwithwhich a stonemustbethrowninordertoputitintoorbitjustabovethesurfaceof the Earth

If the stone has mass m and speed v thenwehave fromNewtonrsquos2ndlaw

mv2

RE---------- G

MEm

RE2------------=

whereRE is the radius of the Earth and ME is the mass of the Earth

Bearinginmindthatg 0 GME

RE2--------= then

v g RE 10 64 106timestimes 8 103times= = =

Thatisthestonemustbethrownat8times103m sndash1

Clearly we are not going to get a satellite into orbit soclosetothesurfaceoftheEarthMovingatthisspeedthefrictionduetoairresistancewouldmeltthesatellitebeforeithadtravelledacoupleofkilometresInrealitythereforeasatelliteisputintoorbitabouttheEarthbysendingitattachedtoarocketbeyondtheEarthrsquosatmosphereandthengivingitacomponentofvelocityperpendiculartoaradialvectorfromtheEarthSeeFigure1016

Earth Satellite carried by rocket to hereSatellite orbit

Tangential component of velocity

Figure 1016 Getting a satellite into orbit

Keplerrsquos third law(This work of Kepler and Newtonrsquos synthesis of the work is an excellent example of the scientific method and makes for a good TOK discussion)

In 1627 Johannes Kepler (1571-1630)publishedhis lawsof planetary motion The laws are empirical in natureandwere deduced from the observations of theDanishastronomer Tycho de Brahe (1546-1601)The third lawgives a relationship between the radius of orbit R of a planet and its period TofrevolutionabouttheSunThelawisexpressedmathematicallyas

2

3 constantTR

=

We shall nowuseNewtonrsquos LawofGravitation to showhowitisthattheplanetsmoveinaccordancewithKeplerrsquosthirdlaw

InessenceNewtonwasabletousehislawofgravitytopredict the motion of the planets since all he had to do was factor theF givenby this law intohis second law F = ma to find their accelerations and hence their future positions

InFigure1017theEarthisshownorbitingtheSunandthedistancebetweentheircentresisR

RSun

Earth

F es

F se

Figure 1017 Planets move according to Keplerrsquos third law

Fes is the force that the Earth exerts on the Sun and Fse is the force that the Sun exerts on the Earth The forces are equalandoppositeandtheSunandtheEarthwillactuallyorbitaboutacommoncentreHoweversince theSun issoverymuchmoremassivethantheEarththiscommoncentrewillbeclosetothecentreoftheSunandsowecanregard the Earth as orbiting about the centre of the Sun Theotherthingthatweshallassumeisthatwecanignorethe forces that the other planets exert on the Earth (This wouldnotbeawisethingtodoifyouwereplanningtosend a space ship to theMoon for example)We shallalso assume that we have followed Newtonrsquos exampleandindeedprovedthataspherewillactasapointmasssituated at the centre of the sphere

Kepler had postulated that the orbits of the planets areellipticalbut since theeccentricityof theEarthrsquosorbit issmallweshallassumeacircularorbit

PHYSICS 2015indb 319 150514 338 PM

Chapter 10

320

AH

L

TheaccelerationoftheEarthtowardstheSunisa = Rω2

whereω ω 2πT------=

Hencea R 2π

T------ times

2 4π2 RT 2------------- = =

But the acceleration is given by Newtonrsquos Second Law F = mawhereFisnowgivenbytheLawofGravitationSo in this situation

F maGMs Me

R2-------------------= = butwealsohavethat

a a R 2πT

------ times

2 4π2 RT 2------------- = = and m = Me so that

G MsMe

R2------------------- Me4π2R

T2-------------times

GMs

4π2-----------rArr R3

T2------= =

Butthequantity

isaconstantthathasthesamevalueforeachoftheplanetssowehaveforalltheplanetsnotjustEarththat

R3

T2------ k =

wherekisaconstantWhichisofcourseKeplerrsquosthirdlaw

ThisisindeedanamazingbreakthroughItisdifficulttorefutethe idea that all particles attract each other in accordance withtheLawofGravitationwhenthelawisabletoaccountfortheobservedmotionoftheplanetsabouttheSun

The gravitational effects of the planets upon each othershould produce perturbations in their orbits Such is the predictivepoweroftheUniversalGravitationalLawthatit enabled physicists to compute these perturbations The telescopehadbeeninventedin1608andbythemiddleofthe 18th Century had reached a degree of perfection in design that enabled astronomers to actually measure the orbital perturbations of the planets Their measurements werealways inagreementwith thepredictionsmadebyNewtonrsquoslawHoweverin1781anewplanetUranuswasdiscovered and the orbit of this planet did not fit withthe orbit predicted by Universal Gravitation Such wasthe physicistrsquos faith in theNewtonianmethod that theysuspectedthatthediscrepancywasduetothepresenceofayetundetectedplanetUsingtheLawofGravitationtheFrenchastronomerJLeverrier and the English astronomer J C Adamswereable tocalculate justhowmassive thisnewplanetmustbeandalsowhereitshouldbeIn1846the planetNeptunewas discovered justwhere they hadpredicted Ina similarwaydiscrepancies in theorbitofNeptune ledto thepredictionandsubsequentdiscoveryin1930oftheplanetPlutoNewtonrsquosLawofGravitation

had passed the ultimate test of any theory it is not only abletoexplainexistingdatabutalsotomakepredictions

Satellite energyWhenasatelliteisinorbitaboutaplanetitwillhavebothkineticenergyandgravitationalpotentialenergySupposewe consider a satellite ofmass m that is in an orbit of radius r about a planet of mass M

Thegravitationalpotentialduetotheplanetatdistancer from its centre is

ThegravitationalpotentialenergyofthesatelliteVsat

is therefore GMe m

r----------------ndash

ThatisVsat GMem

r----------------ndash=

Thegravitationalfieldstrengthatthesurfaceoftheplanetisgivenby

g 0GMe

Re2------------=

Hencewecanwrite

Vsat g0 Re

2

rndash=

m

ThekineticenergyofthesatelliteKsatisequaltofrac12mv2wherev is its orbital speed

Byequatingthegravitationalforceactingonthesatellitetoitscentripetalaccelerationwehave

G Mem

r2---------------- mv2

r---------- mv2hArr

GMemr----------------= =

Fromwhich12---

mv2 12---

GMemr----------------times=

= 0 e

2g R

r

2m

Which is actually quite interesting since it shows thatirrespective of the orbital radius the KE is numericallyequal to half the PE Also the total energy E tot K sat Vsat+ 1

2---G Me m

r----------------times G Me m

r----------------ndash + 1

2---G Me m

r----------------ndash= = =of the satelliteisalwaysnegativesince

E tot K sat Vsat+ 12---

G Me mr----------------times

G Me mr----------------ndash

+ 12---

G Me mr----------------ndash= = =

PHYSICS 2015indb 320 150514 338 PM

Fields

321

AH

L

The energies of an orbiting satellite as a function of radial distancefromthecentreofaplanetareshownplottedinFigure1018

12

100806

0402

0ndash02

ndash04

ndash06ndash08ndash10

ndash12

2 4 6 8 10 12

ener

gya

rbita

ry u

nits

distance timesR

kinetic energytotal energy

potential energy

Figure 1018 Energy of an orbiting satellite as a function of distance from the centre of a planet

WeightlessnessSuppose that you are inanelevator(lift)thatisdescendingat constant speedandyou let goof abook thatyouareholdinginyourhandThebookwillfalltothefloorwithaccelerationequaltotheaccelerationduetogravityIfthecablethatsupportstheelevatorweretosnap(asituationthatItrustwillneverhappentoanyofyou)andyounowletgothebookthatyouareholdinginyourotherhandthis book will not fall to the floor - it will stay exactlyin linewithyourhandThis isbecause thebook isnowfallingwith the sameaccelerationas theelevatorandassuch the book cannot ldquocatchrdquo up with the floor of theelevatorFurthermoreifyouhappenedtobestandingonasetofbathroomscalesthescaleswouldnowreadzero-youwouldbeapparentlyweightlessItisthisideaoffreefallthatexplainstheapparentweightlessnessofastronautsin an orbiting satellite These astronauts are in free fall inthesensethattheyareacceleratingtowardsthecentre of the Earth

It is actually possible to define theweight of a body inseveral different waysWe can define it for example asthegravitationalforceexertedonthebodybyaspecifiedobject such as the EarthThiswe have seen thatwe doinlotsofsituationswherewedefinetheweightasbeingequaltomgIfweusethisdefinitionthenanobjectinfreefallcannotbydefinitionbeweightlesssinceitisstillinagravitationalfieldHoweverifwedefinetheweightofanobjectintermsofaldquoweighingrdquoprocesssuchasthereadingonasetofbathroomscaleswhichineffectmeasuresthecontact force between the object and the scales thenclearlyobjectsinfreefallareweightlessOnenowhasto

askthequestionwhetherornotitispossibleForexampletomeasurethegravitationalforceactingonanastronautin orbit about the Earth

WecanalsodefineweightintermsofthenetgravitationalforceactingonabodyduetoseveraldifferentobjectsForexample for an object out in space its weight could bedefined in terms of the resultant of the forces exerted on it bytheSuntheMoontheEarthandalltheotherplanetsin the Solar System If this resultant is zero at a particular pointthenthebodyisweightlessatthispoint

Inviewofthevariousdefinitionsofweightthatareavailabletousitisimportantthatwhenweusethewordldquoweightrdquoweareawareofthecontextinwhichitisbeingused

Example

Calculate the height above the surface of the Earth atwhichageo-stationarysatelliteorbits

Solution

A geo-stationary satellite is one that orbits the Earth in such a way that it is stationary with respect to a point on the surface of the Earth This means that its orbital period must be the same as the time for the Earth to spin once on its axis ie 24 hours

From Keplerrsquos third law we have G Ms

4π2----------- R3

T2------=

That is

RMe

mh

using the fact that the force of attraction between the satellite and the Earth is given by

FG Mem

R2----------------=

and that F = ma

where m is the mass of the satellite and a 4π2R

T2-------------=

we have

GMem

R2---------------- m 4π2R

T2-------------times

GMe

4π2------------rArr R3

T2------= =

PHYSICS 2015indb 321 150514 338 PM

Chapter 10

322

AH

L

Now the mass of the Earth is 60 times 1024 kg and the period T measured in seconds is given by T = 86400 s

So substitution gives R = 42 times 106 m

The radius of the Earth is 64 times 106 m so that the orbital height h is about 36 times 107 m

Example

Calculatetheminimumenergyrequiredtoputasatelliteofmass500kgintoanorbitthatisasaheightequaltotheEarthrsquosradiusabovethesurfaceoftheEarth

Solution

We have seen that when dealing with gravitational fields and potential it is useful to remember that

g 0G M

Re2---------= or g0 Re

2 GM=

The gravitational potential at the surface of the Earth is

g 0ndash Re GMRe

---------ndash=

The gravitational potential at a distance R

from the centre of the Earth is

=

The difference in potential between the surface of the Earth and a point distance R from the centre is therefore

V∆ g 0Re 1ReR------ndash

=

If R 2Re= then V∆g0 Re

2------------=

This means that the work required to ldquoliftrdquo the satellite into orbit is g0Rm where m is the mass of the satellite This is equal to

10 32 106 500timestimestimes = 16000 MJ

However the satellite must also have kinetic energy in order to orbit Earth This will be equal to

g 0 mRe2

2R-----------------g 0mRe

2

4----------------- 8000 MJ= =

The minimum energy required is therefore

24000 MJ

Exercise 1031 The speed needed to put a satellite in orbit does

not depend on

A the radius of the orbitB theshapeoftheorbitC thevalueofgattheorbitD the mass of the satellite

2 EstimatethespeedofanEarthsatellitewhoseorbitis400kmabovetheEarthrsquossurfaceAlsodetermine the period of the orbit

3 Calculatethespeedofa200kgsatelliteorbitingthe Earth at a height of 70 times 106 m

Assume that g = 82 m sndash2 for this orbit

4 TheradiioftwosatellitesXandYorbitingthe Earth are 2r and 8rwherer is the radius of the Earth Calculate the ratio of the periods of revolutionofXtoY

5 A satellite of mass mkgissentfromEarthrsquossurfaceinto an orbit of radius 5RwhereR is the radius of theEarthWritedownanexpressionfor

(a) thepotentialenergyofthesatelliteinorbit

(b) thekineticenergyofthesatelliteinorbit

(c) theminimumworkrequiredtosendthesatellitefromrestattheEarthrsquossurfaceintoits orbit

6 A satellite in an orbit of 10rfallsbacktoEarth(radius r)afteramalfunctionDeterminethespeedwithwhichitwillhittheEarthrsquossurface

7 The radius of the moon is frac14 that of the Earth AssumingEarthandtheMoontohavethesamedensitycomparetheaccelerationsoffreefallatthe surface of Earth to that at the surface of the Moon

8 UsethefollowingdatatodeterminethegravitationalfieldstrengthatthesurfaceoftheMoon and hence determine the escape speed from the surface of the Moon

Mass of the Moon = 73 times 1022kg

RadiusoftheMoon=17times106 m

PHYSICS 2015indb 322 150514 338 PM

Chapter 10

318

AH

L

15 SupposethatwhenanelectronmovedfromAtoBin the diagram along an electric field line that the electric field does 36 times 10-19Jofworkonit

Determinethedifferencesinelectricpotential

(a) VB ndash VA(b) VC ndash VA(c) VC ndash VB

16 DeterminethepotentialatpointPthatislocatedatthecentreofthesquareasshowninthediagrambelow

- 6μC

+3μC +2μC

1m

1m

5μC

P

102 Fields at work

Essential idea Similar approaches can be taken inanalysingelectricalandgravitationalpotentialproblems

Understandingsbull Potentialandpotentialenergybull Potentialgradientbull Potentialdifferencebull Escapespeedbull Orbitalmotionorbitalspeedandorbitalenergybull Forcesandinverse-squarelawbehaviour

Orbital motion orbital speed and orbital energy

Although orbital motion may be circular elliptical orparabolic this sub-topic only dealswith circular orbitsThis sub-topic is not fundamentally new physics butan application that synthesizes ideas from gravitationcircularmotiondynamicsandenergy

TheMoonorbitstheEarthandinthissenseitisoftenreferredtoasasatelliteoftheEarthBefore1957itwastheonlyEarthsatelliteHoweverin1957theRussianslaunchedthefirstmanmadesatelliteSputnik1SincethisdatemanymoresatelliteshavebeenlaunchedandtherearenowliterallythousandsofthemorbitingtheEarthSomeareusedtomonitortheweathersome used to enable people to find accurately their position on thesurfaceoftheEarthmanyareusedincommunicationsandnodoubtsomeareusedtospyonothercountriesFigure1015showshowinprincipleasatellitecanbeputintoorbit

Theperson(whosesizeisgreatlyexaggeratedwithrespectto Earth) standing on the surface on the Earth throwssome stonesThe greater the speed with which a stoneis thrown the further it will land from her The pathsfollowedbythethrownstonesareparabolasByastretchof the imaginationwecanvisualisea situation inwhichastoneis thrownwithsuchaspeedthatbecauseof thecurvatureoftheEarthitwillnotlandonthesurfaceoftheEarthbutgointoldquoorbitrdquo(Path4onFigure1015)

NATURE OF SCIENCECommunication of scientific explanations The ability to apply field theory to the unobservable (charges) and the massively scaled (motion of satellites) required scientists to develop new ways to investigate analyse and report findings to a general public used to scientific discoveries based on tangible and discernible evidence (51)

copy IBO 2014

PHYSICS 2015indb 318 150514 338 PM

Fields

319

AH

L

1

2

34Earth

Figure 1015 Throwing a stone into orbit

TheforcethatcausesthestonestofollowaparabolicpathandtofalltoEarthisgravityandsimilarlytheforcethatkeeps the stone in orbit is gravity For circular motionto occur we have seen that a force must act at rightanglestothevelocityofanobjectthatistheremustbeacentripetalforceHenceinthesituationwedescribeherethe centripetal force for circular orbital motion about the EarthisprovidedbygravitationalattractionoftheEarth

We can calculate the speedwithwhich a stonemustbethrowninordertoputitintoorbitjustabovethesurfaceof the Earth

If the stone has mass m and speed v thenwehave fromNewtonrsquos2ndlaw

mv2

RE---------- G

MEm

RE2------------=

whereRE is the radius of the Earth and ME is the mass of the Earth

Bearinginmindthatg 0 GME

RE2--------= then

v g RE 10 64 106timestimes 8 103times= = =

Thatisthestonemustbethrownat8times103m sndash1

Clearly we are not going to get a satellite into orbit soclosetothesurfaceoftheEarthMovingatthisspeedthefrictionduetoairresistancewouldmeltthesatellitebeforeithadtravelledacoupleofkilometresInrealitythereforeasatelliteisputintoorbitabouttheEarthbysendingitattachedtoarocketbeyondtheEarthrsquosatmosphereandthengivingitacomponentofvelocityperpendiculartoaradialvectorfromtheEarthSeeFigure1016

Earth Satellite carried by rocket to hereSatellite orbit

Tangential component of velocity

Figure 1016 Getting a satellite into orbit

Keplerrsquos third law(This work of Kepler and Newtonrsquos synthesis of the work is an excellent example of the scientific method and makes for a good TOK discussion)

In 1627 Johannes Kepler (1571-1630)publishedhis lawsof planetary motion The laws are empirical in natureandwere deduced from the observations of theDanishastronomer Tycho de Brahe (1546-1601)The third lawgives a relationship between the radius of orbit R of a planet and its period TofrevolutionabouttheSunThelawisexpressedmathematicallyas

2

3 constantTR

=

We shall nowuseNewtonrsquos LawofGravitation to showhowitisthattheplanetsmoveinaccordancewithKeplerrsquosthirdlaw

InessenceNewtonwasabletousehislawofgravitytopredict the motion of the planets since all he had to do was factor theF givenby this law intohis second law F = ma to find their accelerations and hence their future positions

InFigure1017theEarthisshownorbitingtheSunandthedistancebetweentheircentresisR

RSun

Earth

F es

F se

Figure 1017 Planets move according to Keplerrsquos third law

Fes is the force that the Earth exerts on the Sun and Fse is the force that the Sun exerts on the Earth The forces are equalandoppositeandtheSunandtheEarthwillactuallyorbitaboutacommoncentreHoweversince theSun issoverymuchmoremassivethantheEarththiscommoncentrewillbeclosetothecentreoftheSunandsowecanregard the Earth as orbiting about the centre of the Sun Theotherthingthatweshallassumeisthatwecanignorethe forces that the other planets exert on the Earth (This wouldnotbeawisethingtodoifyouwereplanningtosend a space ship to theMoon for example)We shallalso assume that we have followed Newtonrsquos exampleandindeedprovedthataspherewillactasapointmasssituated at the centre of the sphere

Kepler had postulated that the orbits of the planets areellipticalbut since theeccentricityof theEarthrsquosorbit issmallweshallassumeacircularorbit

PHYSICS 2015indb 319 150514 338 PM

Chapter 10

320

AH

L

TheaccelerationoftheEarthtowardstheSunisa = Rω2

whereω ω 2πT------=

Hencea R 2π

T------ times

2 4π2 RT 2------------- = =

But the acceleration is given by Newtonrsquos Second Law F = mawhereFisnowgivenbytheLawofGravitationSo in this situation

F maGMs Me

R2-------------------= = butwealsohavethat

a a R 2πT

------ times

2 4π2 RT 2------------- = = and m = Me so that

G MsMe

R2------------------- Me4π2R

T2-------------times

GMs

4π2-----------rArr R3

T2------= =

Butthequantity

isaconstantthathasthesamevalueforeachoftheplanetssowehaveforalltheplanetsnotjustEarththat

R3

T2------ k =

wherekisaconstantWhichisofcourseKeplerrsquosthirdlaw

ThisisindeedanamazingbreakthroughItisdifficulttorefutethe idea that all particles attract each other in accordance withtheLawofGravitationwhenthelawisabletoaccountfortheobservedmotionoftheplanetsabouttheSun

The gravitational effects of the planets upon each othershould produce perturbations in their orbits Such is the predictivepoweroftheUniversalGravitationalLawthatit enabled physicists to compute these perturbations The telescopehadbeeninventedin1608andbythemiddleofthe 18th Century had reached a degree of perfection in design that enabled astronomers to actually measure the orbital perturbations of the planets Their measurements werealways inagreementwith thepredictionsmadebyNewtonrsquoslawHoweverin1781anewplanetUranuswasdiscovered and the orbit of this planet did not fit withthe orbit predicted by Universal Gravitation Such wasthe physicistrsquos faith in theNewtonianmethod that theysuspectedthatthediscrepancywasduetothepresenceofayetundetectedplanetUsingtheLawofGravitationtheFrenchastronomerJLeverrier and the English astronomer J C Adamswereable tocalculate justhowmassive thisnewplanetmustbeandalsowhereitshouldbeIn1846the planetNeptunewas discovered justwhere they hadpredicted Ina similarwaydiscrepancies in theorbitofNeptune ledto thepredictionandsubsequentdiscoveryin1930oftheplanetPlutoNewtonrsquosLawofGravitation

had passed the ultimate test of any theory it is not only abletoexplainexistingdatabutalsotomakepredictions

Satellite energyWhenasatelliteisinorbitaboutaplanetitwillhavebothkineticenergyandgravitationalpotentialenergySupposewe consider a satellite ofmass m that is in an orbit of radius r about a planet of mass M

Thegravitationalpotentialduetotheplanetatdistancer from its centre is

ThegravitationalpotentialenergyofthesatelliteVsat

is therefore GMe m

r----------------ndash

ThatisVsat GMem

r----------------ndash=

Thegravitationalfieldstrengthatthesurfaceoftheplanetisgivenby

g 0GMe

Re2------------=

Hencewecanwrite

Vsat g0 Re

2

rndash=

m

ThekineticenergyofthesatelliteKsatisequaltofrac12mv2wherev is its orbital speed

Byequatingthegravitationalforceactingonthesatellitetoitscentripetalaccelerationwehave

G Mem

r2---------------- mv2

r---------- mv2hArr

GMemr----------------= =

Fromwhich12---

mv2 12---

GMemr----------------times=

= 0 e

2g R

r

2m

Which is actually quite interesting since it shows thatirrespective of the orbital radius the KE is numericallyequal to half the PE Also the total energy E tot K sat Vsat+ 1

2---G Me m

r----------------times G Me m

r----------------ndash + 1

2---G Me m

r----------------ndash= = =of the satelliteisalwaysnegativesince

E tot K sat Vsat+ 12---

G Me mr----------------times

G Me mr----------------ndash

+ 12---

G Me mr----------------ndash= = =

PHYSICS 2015indb 320 150514 338 PM

Fields

321

AH

L

The energies of an orbiting satellite as a function of radial distancefromthecentreofaplanetareshownplottedinFigure1018

12

100806

0402

0ndash02

ndash04

ndash06ndash08ndash10

ndash12

2 4 6 8 10 12

ener

gya

rbita

ry u

nits

distance timesR

kinetic energytotal energy

potential energy

Figure 1018 Energy of an orbiting satellite as a function of distance from the centre of a planet

WeightlessnessSuppose that you are inanelevator(lift)thatisdescendingat constant speedandyou let goof abook thatyouareholdinginyourhandThebookwillfalltothefloorwithaccelerationequaltotheaccelerationduetogravityIfthecablethatsupportstheelevatorweretosnap(asituationthatItrustwillneverhappentoanyofyou)andyounowletgothebookthatyouareholdinginyourotherhandthis book will not fall to the floor - it will stay exactlyin linewithyourhandThis isbecause thebook isnowfallingwith the sameaccelerationas theelevatorandassuch the book cannot ldquocatchrdquo up with the floor of theelevatorFurthermoreifyouhappenedtobestandingonasetofbathroomscalesthescaleswouldnowreadzero-youwouldbeapparentlyweightlessItisthisideaoffreefallthatexplainstheapparentweightlessnessofastronautsin an orbiting satellite These astronauts are in free fall inthesensethattheyareacceleratingtowardsthecentre of the Earth

It is actually possible to define theweight of a body inseveral different waysWe can define it for example asthegravitationalforceexertedonthebodybyaspecifiedobject such as the EarthThiswe have seen thatwe doinlotsofsituationswherewedefinetheweightasbeingequaltomgIfweusethisdefinitionthenanobjectinfreefallcannotbydefinitionbeweightlesssinceitisstillinagravitationalfieldHoweverifwedefinetheweightofanobjectintermsofaldquoweighingrdquoprocesssuchasthereadingonasetofbathroomscaleswhichineffectmeasuresthecontact force between the object and the scales thenclearlyobjectsinfreefallareweightlessOnenowhasto

askthequestionwhetherornotitispossibleForexampletomeasurethegravitationalforceactingonanastronautin orbit about the Earth

WecanalsodefineweightintermsofthenetgravitationalforceactingonabodyduetoseveraldifferentobjectsForexample for an object out in space its weight could bedefined in terms of the resultant of the forces exerted on it bytheSuntheMoontheEarthandalltheotherplanetsin the Solar System If this resultant is zero at a particular pointthenthebodyisweightlessatthispoint

Inviewofthevariousdefinitionsofweightthatareavailabletousitisimportantthatwhenweusethewordldquoweightrdquoweareawareofthecontextinwhichitisbeingused

Example

Calculate the height above the surface of the Earth atwhichageo-stationarysatelliteorbits

Solution

A geo-stationary satellite is one that orbits the Earth in such a way that it is stationary with respect to a point on the surface of the Earth This means that its orbital period must be the same as the time for the Earth to spin once on its axis ie 24 hours

From Keplerrsquos third law we have G Ms

4π2----------- R3

T2------=

That is

RMe

mh

using the fact that the force of attraction between the satellite and the Earth is given by

FG Mem

R2----------------=

and that F = ma

where m is the mass of the satellite and a 4π2R

T2-------------=

we have

GMem

R2---------------- m 4π2R

T2-------------times

GMe

4π2------------rArr R3

T2------= =

PHYSICS 2015indb 321 150514 338 PM

Chapter 10

322

AH

L

Now the mass of the Earth is 60 times 1024 kg and the period T measured in seconds is given by T = 86400 s

So substitution gives R = 42 times 106 m

The radius of the Earth is 64 times 106 m so that the orbital height h is about 36 times 107 m

Example

Calculatetheminimumenergyrequiredtoputasatelliteofmass500kgintoanorbitthatisasaheightequaltotheEarthrsquosradiusabovethesurfaceoftheEarth

Solution

We have seen that when dealing with gravitational fields and potential it is useful to remember that

g 0G M

Re2---------= or g0 Re

2 GM=

The gravitational potential at the surface of the Earth is

g 0ndash Re GMRe

---------ndash=

The gravitational potential at a distance R

from the centre of the Earth is

=

The difference in potential between the surface of the Earth and a point distance R from the centre is therefore

V∆ g 0Re 1ReR------ndash

=

If R 2Re= then V∆g0 Re

2------------=

This means that the work required to ldquoliftrdquo the satellite into orbit is g0Rm where m is the mass of the satellite This is equal to

10 32 106 500timestimestimes = 16000 MJ

However the satellite must also have kinetic energy in order to orbit Earth This will be equal to

g 0 mRe2

2R-----------------g 0mRe

2

4----------------- 8000 MJ= =

The minimum energy required is therefore

24000 MJ

Exercise 1031 The speed needed to put a satellite in orbit does

not depend on

A the radius of the orbitB theshapeoftheorbitC thevalueofgattheorbitD the mass of the satellite

2 EstimatethespeedofanEarthsatellitewhoseorbitis400kmabovetheEarthrsquossurfaceAlsodetermine the period of the orbit

3 Calculatethespeedofa200kgsatelliteorbitingthe Earth at a height of 70 times 106 m

Assume that g = 82 m sndash2 for this orbit

4 TheradiioftwosatellitesXandYorbitingthe Earth are 2r and 8rwherer is the radius of the Earth Calculate the ratio of the periods of revolutionofXtoY

5 A satellite of mass mkgissentfromEarthrsquossurfaceinto an orbit of radius 5RwhereR is the radius of theEarthWritedownanexpressionfor

(a) thepotentialenergyofthesatelliteinorbit

(b) thekineticenergyofthesatelliteinorbit

(c) theminimumworkrequiredtosendthesatellitefromrestattheEarthrsquossurfaceintoits orbit

6 A satellite in an orbit of 10rfallsbacktoEarth(radius r)afteramalfunctionDeterminethespeedwithwhichitwillhittheEarthrsquossurface

7 The radius of the moon is frac14 that of the Earth AssumingEarthandtheMoontohavethesamedensitycomparetheaccelerationsoffreefallatthe surface of Earth to that at the surface of the Moon

8 UsethefollowingdatatodeterminethegravitationalfieldstrengthatthesurfaceoftheMoon and hence determine the escape speed from the surface of the Moon

Mass of the Moon = 73 times 1022kg

RadiusoftheMoon=17times106 m

PHYSICS 2015indb 322 150514 338 PM

Fields

319

AH

L

1

2

34Earth

Figure 1015 Throwing a stone into orbit

TheforcethatcausesthestonestofollowaparabolicpathandtofalltoEarthisgravityandsimilarlytheforcethatkeeps the stone in orbit is gravity For circular motionto occur we have seen that a force must act at rightanglestothevelocityofanobjectthatistheremustbeacentripetalforceHenceinthesituationwedescribeherethe centripetal force for circular orbital motion about the EarthisprovidedbygravitationalattractionoftheEarth

We can calculate the speedwithwhich a stonemustbethrowninordertoputitintoorbitjustabovethesurfaceof the Earth

If the stone has mass m and speed v thenwehave fromNewtonrsquos2ndlaw

mv2

RE---------- G

MEm

RE2------------=

whereRE is the radius of the Earth and ME is the mass of the Earth

Bearinginmindthatg 0 GME

RE2--------= then

v g RE 10 64 106timestimes 8 103times= = =

Thatisthestonemustbethrownat8times103m sndash1

Clearly we are not going to get a satellite into orbit soclosetothesurfaceoftheEarthMovingatthisspeedthefrictionduetoairresistancewouldmeltthesatellitebeforeithadtravelledacoupleofkilometresInrealitythereforeasatelliteisputintoorbitabouttheEarthbysendingitattachedtoarocketbeyondtheEarthrsquosatmosphereandthengivingitacomponentofvelocityperpendiculartoaradialvectorfromtheEarthSeeFigure1016

Earth Satellite carried by rocket to hereSatellite orbit

Tangential component of velocity

Figure 1016 Getting a satellite into orbit

Keplerrsquos third law(This work of Kepler and Newtonrsquos synthesis of the work is an excellent example of the scientific method and makes for a good TOK discussion)

In 1627 Johannes Kepler (1571-1630)publishedhis lawsof planetary motion The laws are empirical in natureandwere deduced from the observations of theDanishastronomer Tycho de Brahe (1546-1601)The third lawgives a relationship between the radius of orbit R of a planet and its period TofrevolutionabouttheSunThelawisexpressedmathematicallyas

2

3 constantTR

=

We shall nowuseNewtonrsquos LawofGravitation to showhowitisthattheplanetsmoveinaccordancewithKeplerrsquosthirdlaw

InessenceNewtonwasabletousehislawofgravitytopredict the motion of the planets since all he had to do was factor theF givenby this law intohis second law F = ma to find their accelerations and hence their future positions

InFigure1017theEarthisshownorbitingtheSunandthedistancebetweentheircentresisR

RSun

Earth

F es

F se

Figure 1017 Planets move according to Keplerrsquos third law

Fes is the force that the Earth exerts on the Sun and Fse is the force that the Sun exerts on the Earth The forces are equalandoppositeandtheSunandtheEarthwillactuallyorbitaboutacommoncentreHoweversince theSun issoverymuchmoremassivethantheEarththiscommoncentrewillbeclosetothecentreoftheSunandsowecanregard the Earth as orbiting about the centre of the Sun Theotherthingthatweshallassumeisthatwecanignorethe forces that the other planets exert on the Earth (This wouldnotbeawisethingtodoifyouwereplanningtosend a space ship to theMoon for example)We shallalso assume that we have followed Newtonrsquos exampleandindeedprovedthataspherewillactasapointmasssituated at the centre of the sphere

Kepler had postulated that the orbits of the planets areellipticalbut since theeccentricityof theEarthrsquosorbit issmallweshallassumeacircularorbit

PHYSICS 2015indb 319 150514 338 PM

Chapter 10

320

AH

L

TheaccelerationoftheEarthtowardstheSunisa = Rω2

whereω ω 2πT------=

Hencea R 2π

T------ times

2 4π2 RT 2------------- = =

But the acceleration is given by Newtonrsquos Second Law F = mawhereFisnowgivenbytheLawofGravitationSo in this situation

F maGMs Me

R2-------------------= = butwealsohavethat

a a R 2πT

------ times

2 4π2 RT 2------------- = = and m = Me so that

G MsMe

R2------------------- Me4π2R

T2-------------times

GMs

4π2-----------rArr R3

T2------= =

Butthequantity

isaconstantthathasthesamevalueforeachoftheplanetssowehaveforalltheplanetsnotjustEarththat

R3

T2------ k =

wherekisaconstantWhichisofcourseKeplerrsquosthirdlaw

ThisisindeedanamazingbreakthroughItisdifficulttorefutethe idea that all particles attract each other in accordance withtheLawofGravitationwhenthelawisabletoaccountfortheobservedmotionoftheplanetsabouttheSun

The gravitational effects of the planets upon each othershould produce perturbations in their orbits Such is the predictivepoweroftheUniversalGravitationalLawthatit enabled physicists to compute these perturbations The telescopehadbeeninventedin1608andbythemiddleofthe 18th Century had reached a degree of perfection in design that enabled astronomers to actually measure the orbital perturbations of the planets Their measurements werealways inagreementwith thepredictionsmadebyNewtonrsquoslawHoweverin1781anewplanetUranuswasdiscovered and the orbit of this planet did not fit withthe orbit predicted by Universal Gravitation Such wasthe physicistrsquos faith in theNewtonianmethod that theysuspectedthatthediscrepancywasduetothepresenceofayetundetectedplanetUsingtheLawofGravitationtheFrenchastronomerJLeverrier and the English astronomer J C Adamswereable tocalculate justhowmassive thisnewplanetmustbeandalsowhereitshouldbeIn1846the planetNeptunewas discovered justwhere they hadpredicted Ina similarwaydiscrepancies in theorbitofNeptune ledto thepredictionandsubsequentdiscoveryin1930oftheplanetPlutoNewtonrsquosLawofGravitation

had passed the ultimate test of any theory it is not only abletoexplainexistingdatabutalsotomakepredictions

Satellite energyWhenasatelliteisinorbitaboutaplanetitwillhavebothkineticenergyandgravitationalpotentialenergySupposewe consider a satellite ofmass m that is in an orbit of radius r about a planet of mass M

Thegravitationalpotentialduetotheplanetatdistancer from its centre is

ThegravitationalpotentialenergyofthesatelliteVsat

is therefore GMe m

r----------------ndash

ThatisVsat GMem

r----------------ndash=

Thegravitationalfieldstrengthatthesurfaceoftheplanetisgivenby

g 0GMe

Re2------------=

Hencewecanwrite

Vsat g0 Re

2

rndash=

m

ThekineticenergyofthesatelliteKsatisequaltofrac12mv2wherev is its orbital speed

Byequatingthegravitationalforceactingonthesatellitetoitscentripetalaccelerationwehave

G Mem

r2---------------- mv2

r---------- mv2hArr

GMemr----------------= =

Fromwhich12---

mv2 12---

GMemr----------------times=

= 0 e

2g R

r

2m

Which is actually quite interesting since it shows thatirrespective of the orbital radius the KE is numericallyequal to half the PE Also the total energy E tot K sat Vsat+ 1

2---G Me m

r----------------times G Me m

r----------------ndash + 1

2---G Me m

r----------------ndash= = =of the satelliteisalwaysnegativesince

E tot K sat Vsat+ 12---

G Me mr----------------times

G Me mr----------------ndash

+ 12---

G Me mr----------------ndash= = =

PHYSICS 2015indb 320 150514 338 PM

Fields

321

AH

L

The energies of an orbiting satellite as a function of radial distancefromthecentreofaplanetareshownplottedinFigure1018

12

100806

0402

0ndash02

ndash04

ndash06ndash08ndash10

ndash12

2 4 6 8 10 12

ener

gya

rbita

ry u

nits

distance timesR

kinetic energytotal energy

potential energy

Figure 1018 Energy of an orbiting satellite as a function of distance from the centre of a planet

WeightlessnessSuppose that you are inanelevator(lift)thatisdescendingat constant speedandyou let goof abook thatyouareholdinginyourhandThebookwillfalltothefloorwithaccelerationequaltotheaccelerationduetogravityIfthecablethatsupportstheelevatorweretosnap(asituationthatItrustwillneverhappentoanyofyou)andyounowletgothebookthatyouareholdinginyourotherhandthis book will not fall to the floor - it will stay exactlyin linewithyourhandThis isbecause thebook isnowfallingwith the sameaccelerationas theelevatorandassuch the book cannot ldquocatchrdquo up with the floor of theelevatorFurthermoreifyouhappenedtobestandingonasetofbathroomscalesthescaleswouldnowreadzero-youwouldbeapparentlyweightlessItisthisideaoffreefallthatexplainstheapparentweightlessnessofastronautsin an orbiting satellite These astronauts are in free fall inthesensethattheyareacceleratingtowardsthecentre of the Earth

It is actually possible to define theweight of a body inseveral different waysWe can define it for example asthegravitationalforceexertedonthebodybyaspecifiedobject such as the EarthThiswe have seen thatwe doinlotsofsituationswherewedefinetheweightasbeingequaltomgIfweusethisdefinitionthenanobjectinfreefallcannotbydefinitionbeweightlesssinceitisstillinagravitationalfieldHoweverifwedefinetheweightofanobjectintermsofaldquoweighingrdquoprocesssuchasthereadingonasetofbathroomscaleswhichineffectmeasuresthecontact force between the object and the scales thenclearlyobjectsinfreefallareweightlessOnenowhasto

askthequestionwhetherornotitispossibleForexampletomeasurethegravitationalforceactingonanastronautin orbit about the Earth

WecanalsodefineweightintermsofthenetgravitationalforceactingonabodyduetoseveraldifferentobjectsForexample for an object out in space its weight could bedefined in terms of the resultant of the forces exerted on it bytheSuntheMoontheEarthandalltheotherplanetsin the Solar System If this resultant is zero at a particular pointthenthebodyisweightlessatthispoint

Inviewofthevariousdefinitionsofweightthatareavailabletousitisimportantthatwhenweusethewordldquoweightrdquoweareawareofthecontextinwhichitisbeingused

Example

Calculate the height above the surface of the Earth atwhichageo-stationarysatelliteorbits

Solution

A geo-stationary satellite is one that orbits the Earth in such a way that it is stationary with respect to a point on the surface of the Earth This means that its orbital period must be the same as the time for the Earth to spin once on its axis ie 24 hours

From Keplerrsquos third law we have G Ms

4π2----------- R3

T2------=

That is

RMe

mh

using the fact that the force of attraction between the satellite and the Earth is given by

FG Mem

R2----------------=

and that F = ma

where m is the mass of the satellite and a 4π2R

T2-------------=

we have

GMem

R2---------------- m 4π2R

T2-------------times

GMe

4π2------------rArr R3

T2------= =

PHYSICS 2015indb 321 150514 338 PM

Chapter 10

322

AH

L

Now the mass of the Earth is 60 times 1024 kg and the period T measured in seconds is given by T = 86400 s

So substitution gives R = 42 times 106 m

The radius of the Earth is 64 times 106 m so that the orbital height h is about 36 times 107 m

Example

Calculatetheminimumenergyrequiredtoputasatelliteofmass500kgintoanorbitthatisasaheightequaltotheEarthrsquosradiusabovethesurfaceoftheEarth

Solution

We have seen that when dealing with gravitational fields and potential it is useful to remember that

g 0G M

Re2---------= or g0 Re

2 GM=

The gravitational potential at the surface of the Earth is

g 0ndash Re GMRe

---------ndash=

The gravitational potential at a distance R

from the centre of the Earth is

=

The difference in potential between the surface of the Earth and a point distance R from the centre is therefore

V∆ g 0Re 1ReR------ndash

=

If R 2Re= then V∆g0 Re

2------------=

This means that the work required to ldquoliftrdquo the satellite into orbit is g0Rm where m is the mass of the satellite This is equal to

10 32 106 500timestimestimes = 16000 MJ

However the satellite must also have kinetic energy in order to orbit Earth This will be equal to

g 0 mRe2

2R-----------------g 0mRe

2

4----------------- 8000 MJ= =

The minimum energy required is therefore

24000 MJ

Exercise 1031 The speed needed to put a satellite in orbit does

not depend on

A the radius of the orbitB theshapeoftheorbitC thevalueofgattheorbitD the mass of the satellite

2 EstimatethespeedofanEarthsatellitewhoseorbitis400kmabovetheEarthrsquossurfaceAlsodetermine the period of the orbit

3 Calculatethespeedofa200kgsatelliteorbitingthe Earth at a height of 70 times 106 m

Assume that g = 82 m sndash2 for this orbit

4 TheradiioftwosatellitesXandYorbitingthe Earth are 2r and 8rwherer is the radius of the Earth Calculate the ratio of the periods of revolutionofXtoY

5 A satellite of mass mkgissentfromEarthrsquossurfaceinto an orbit of radius 5RwhereR is the radius of theEarthWritedownanexpressionfor

(a) thepotentialenergyofthesatelliteinorbit

(b) thekineticenergyofthesatelliteinorbit

(c) theminimumworkrequiredtosendthesatellitefromrestattheEarthrsquossurfaceintoits orbit

6 A satellite in an orbit of 10rfallsbacktoEarth(radius r)afteramalfunctionDeterminethespeedwithwhichitwillhittheEarthrsquossurface

7 The radius of the moon is frac14 that of the Earth AssumingEarthandtheMoontohavethesamedensitycomparetheaccelerationsoffreefallatthe surface of Earth to that at the surface of the Moon

8 UsethefollowingdatatodeterminethegravitationalfieldstrengthatthesurfaceoftheMoon and hence determine the escape speed from the surface of the Moon

Mass of the Moon = 73 times 1022kg

RadiusoftheMoon=17times106 m

PHYSICS 2015indb 322 150514 338 PM

Chapter 10

320

AH

L

TheaccelerationoftheEarthtowardstheSunisa = Rω2

whereω ω 2πT------=

Hencea R 2π

T------ times

2 4π2 RT 2------------- = =

But the acceleration is given by Newtonrsquos Second Law F = mawhereFisnowgivenbytheLawofGravitationSo in this situation

F maGMs Me

R2-------------------= = butwealsohavethat

a a R 2πT

------ times

2 4π2 RT 2------------- = = and m = Me so that

G MsMe

R2------------------- Me4π2R

T2-------------times

GMs

4π2-----------rArr R3

T2------= =

Butthequantity

isaconstantthathasthesamevalueforeachoftheplanetssowehaveforalltheplanetsnotjustEarththat

R3

T2------ k =

wherekisaconstantWhichisofcourseKeplerrsquosthirdlaw

ThisisindeedanamazingbreakthroughItisdifficulttorefutethe idea that all particles attract each other in accordance withtheLawofGravitationwhenthelawisabletoaccountfortheobservedmotionoftheplanetsabouttheSun

The gravitational effects of the planets upon each othershould produce perturbations in their orbits Such is the predictivepoweroftheUniversalGravitationalLawthatit enabled physicists to compute these perturbations The telescopehadbeeninventedin1608andbythemiddleofthe 18th Century had reached a degree of perfection in design that enabled astronomers to actually measure the orbital perturbations of the planets Their measurements werealways inagreementwith thepredictionsmadebyNewtonrsquoslawHoweverin1781anewplanetUranuswasdiscovered and the orbit of this planet did not fit withthe orbit predicted by Universal Gravitation Such wasthe physicistrsquos faith in theNewtonianmethod that theysuspectedthatthediscrepancywasduetothepresenceofayetundetectedplanetUsingtheLawofGravitationtheFrenchastronomerJLeverrier and the English astronomer J C Adamswereable tocalculate justhowmassive thisnewplanetmustbeandalsowhereitshouldbeIn1846the planetNeptunewas discovered justwhere they hadpredicted Ina similarwaydiscrepancies in theorbitofNeptune ledto thepredictionandsubsequentdiscoveryin1930oftheplanetPlutoNewtonrsquosLawofGravitation

had passed the ultimate test of any theory it is not only abletoexplainexistingdatabutalsotomakepredictions

Satellite energyWhenasatelliteisinorbitaboutaplanetitwillhavebothkineticenergyandgravitationalpotentialenergySupposewe consider a satellite ofmass m that is in an orbit of radius r about a planet of mass M

Thegravitationalpotentialduetotheplanetatdistancer from its centre is

ThegravitationalpotentialenergyofthesatelliteVsat

is therefore GMe m

r----------------ndash

ThatisVsat GMem

r----------------ndash=

Thegravitationalfieldstrengthatthesurfaceoftheplanetisgivenby

g 0GMe

Re2------------=

Hencewecanwrite

Vsat g0 Re

2

rndash=

m

ThekineticenergyofthesatelliteKsatisequaltofrac12mv2wherev is its orbital speed

Byequatingthegravitationalforceactingonthesatellitetoitscentripetalaccelerationwehave

G Mem

r2---------------- mv2

r---------- mv2hArr

GMemr----------------= =

Fromwhich12---

mv2 12---

GMemr----------------times=

= 0 e

2g R

r

2m

Which is actually quite interesting since it shows thatirrespective of the orbital radius the KE is numericallyequal to half the PE Also the total energy E tot K sat Vsat+ 1

2---G Me m

r----------------times G Me m

r----------------ndash + 1

2---G Me m

r----------------ndash= = =of the satelliteisalwaysnegativesince

E tot K sat Vsat+ 12---

G Me mr----------------times

G Me mr----------------ndash

+ 12---

G Me mr----------------ndash= = =

PHYSICS 2015indb 320 150514 338 PM

Fields

321

AH

L

The energies of an orbiting satellite as a function of radial distancefromthecentreofaplanetareshownplottedinFigure1018

12

100806

0402

0ndash02

ndash04

ndash06ndash08ndash10

ndash12

2 4 6 8 10 12

ener

gya

rbita

ry u

nits

distance timesR

kinetic energytotal energy

potential energy

Figure 1018 Energy of an orbiting satellite as a function of distance from the centre of a planet

WeightlessnessSuppose that you are inanelevator(lift)thatisdescendingat constant speedandyou let goof abook thatyouareholdinginyourhandThebookwillfalltothefloorwithaccelerationequaltotheaccelerationduetogravityIfthecablethatsupportstheelevatorweretosnap(asituationthatItrustwillneverhappentoanyofyou)andyounowletgothebookthatyouareholdinginyourotherhandthis book will not fall to the floor - it will stay exactlyin linewithyourhandThis isbecause thebook isnowfallingwith the sameaccelerationas theelevatorandassuch the book cannot ldquocatchrdquo up with the floor of theelevatorFurthermoreifyouhappenedtobestandingonasetofbathroomscalesthescaleswouldnowreadzero-youwouldbeapparentlyweightlessItisthisideaoffreefallthatexplainstheapparentweightlessnessofastronautsin an orbiting satellite These astronauts are in free fall inthesensethattheyareacceleratingtowardsthecentre of the Earth

It is actually possible to define theweight of a body inseveral different waysWe can define it for example asthegravitationalforceexertedonthebodybyaspecifiedobject such as the EarthThiswe have seen thatwe doinlotsofsituationswherewedefinetheweightasbeingequaltomgIfweusethisdefinitionthenanobjectinfreefallcannotbydefinitionbeweightlesssinceitisstillinagravitationalfieldHoweverifwedefinetheweightofanobjectintermsofaldquoweighingrdquoprocesssuchasthereadingonasetofbathroomscaleswhichineffectmeasuresthecontact force between the object and the scales thenclearlyobjectsinfreefallareweightlessOnenowhasto

askthequestionwhetherornotitispossibleForexampletomeasurethegravitationalforceactingonanastronautin orbit about the Earth

WecanalsodefineweightintermsofthenetgravitationalforceactingonabodyduetoseveraldifferentobjectsForexample for an object out in space its weight could bedefined in terms of the resultant of the forces exerted on it bytheSuntheMoontheEarthandalltheotherplanetsin the Solar System If this resultant is zero at a particular pointthenthebodyisweightlessatthispoint

Inviewofthevariousdefinitionsofweightthatareavailabletousitisimportantthatwhenweusethewordldquoweightrdquoweareawareofthecontextinwhichitisbeingused

Example

Calculate the height above the surface of the Earth atwhichageo-stationarysatelliteorbits

Solution

A geo-stationary satellite is one that orbits the Earth in such a way that it is stationary with respect to a point on the surface of the Earth This means that its orbital period must be the same as the time for the Earth to spin once on its axis ie 24 hours

From Keplerrsquos third law we have G Ms

4π2----------- R3

T2------=

That is

RMe

mh

using the fact that the force of attraction between the satellite and the Earth is given by

FG Mem

R2----------------=

and that F = ma

where m is the mass of the satellite and a 4π2R

T2-------------=

we have

GMem

R2---------------- m 4π2R

T2-------------times

GMe

4π2------------rArr R3

T2------= =

PHYSICS 2015indb 321 150514 338 PM

Chapter 10

322

AH

L

Now the mass of the Earth is 60 times 1024 kg and the period T measured in seconds is given by T = 86400 s

So substitution gives R = 42 times 106 m

The radius of the Earth is 64 times 106 m so that the orbital height h is about 36 times 107 m

Example

Calculatetheminimumenergyrequiredtoputasatelliteofmass500kgintoanorbitthatisasaheightequaltotheEarthrsquosradiusabovethesurfaceoftheEarth

Solution

We have seen that when dealing with gravitational fields and potential it is useful to remember that

g 0G M

Re2---------= or g0 Re

2 GM=

The gravitational potential at the surface of the Earth is

g 0ndash Re GMRe

---------ndash=

The gravitational potential at a distance R

from the centre of the Earth is

=

The difference in potential between the surface of the Earth and a point distance R from the centre is therefore

V∆ g 0Re 1ReR------ndash

=

If R 2Re= then V∆g0 Re

2------------=

This means that the work required to ldquoliftrdquo the satellite into orbit is g0Rm where m is the mass of the satellite This is equal to

10 32 106 500timestimestimes = 16000 MJ

However the satellite must also have kinetic energy in order to orbit Earth This will be equal to

g 0 mRe2

2R-----------------g 0mRe

2

4----------------- 8000 MJ= =

The minimum energy required is therefore

24000 MJ

Exercise 1031 The speed needed to put a satellite in orbit does

not depend on

A the radius of the orbitB theshapeoftheorbitC thevalueofgattheorbitD the mass of the satellite

2 EstimatethespeedofanEarthsatellitewhoseorbitis400kmabovetheEarthrsquossurfaceAlsodetermine the period of the orbit

3 Calculatethespeedofa200kgsatelliteorbitingthe Earth at a height of 70 times 106 m

Assume that g = 82 m sndash2 for this orbit

4 TheradiioftwosatellitesXandYorbitingthe Earth are 2r and 8rwherer is the radius of the Earth Calculate the ratio of the periods of revolutionofXtoY

5 A satellite of mass mkgissentfromEarthrsquossurfaceinto an orbit of radius 5RwhereR is the radius of theEarthWritedownanexpressionfor

(a) thepotentialenergyofthesatelliteinorbit

(b) thekineticenergyofthesatelliteinorbit

(c) theminimumworkrequiredtosendthesatellitefromrestattheEarthrsquossurfaceintoits orbit

6 A satellite in an orbit of 10rfallsbacktoEarth(radius r)afteramalfunctionDeterminethespeedwithwhichitwillhittheEarthrsquossurface

7 The radius of the moon is frac14 that of the Earth AssumingEarthandtheMoontohavethesamedensitycomparetheaccelerationsoffreefallatthe surface of Earth to that at the surface of the Moon

8 UsethefollowingdatatodeterminethegravitationalfieldstrengthatthesurfaceoftheMoon and hence determine the escape speed from the surface of the Moon

Mass of the Moon = 73 times 1022kg

RadiusoftheMoon=17times106 m

PHYSICS 2015indb 322 150514 338 PM

Fields

321

AH

L

The energies of an orbiting satellite as a function of radial distancefromthecentreofaplanetareshownplottedinFigure1018

12

100806

0402

0ndash02

ndash04

ndash06ndash08ndash10

ndash12

2 4 6 8 10 12

ener

gya

rbita

ry u

nits

distance timesR

kinetic energytotal energy

potential energy

Figure 1018 Energy of an orbiting satellite as a function of distance from the centre of a planet

WeightlessnessSuppose that you are inanelevator(lift)thatisdescendingat constant speedandyou let goof abook thatyouareholdinginyourhandThebookwillfalltothefloorwithaccelerationequaltotheaccelerationduetogravityIfthecablethatsupportstheelevatorweretosnap(asituationthatItrustwillneverhappentoanyofyou)andyounowletgothebookthatyouareholdinginyourotherhandthis book will not fall to the floor - it will stay exactlyin linewithyourhandThis isbecause thebook isnowfallingwith the sameaccelerationas theelevatorandassuch the book cannot ldquocatchrdquo up with the floor of theelevatorFurthermoreifyouhappenedtobestandingonasetofbathroomscalesthescaleswouldnowreadzero-youwouldbeapparentlyweightlessItisthisideaoffreefallthatexplainstheapparentweightlessnessofastronautsin an orbiting satellite These astronauts are in free fall inthesensethattheyareacceleratingtowardsthecentre of the Earth

It is actually possible to define theweight of a body inseveral different waysWe can define it for example asthegravitationalforceexertedonthebodybyaspecifiedobject such as the EarthThiswe have seen thatwe doinlotsofsituationswherewedefinetheweightasbeingequaltomgIfweusethisdefinitionthenanobjectinfreefallcannotbydefinitionbeweightlesssinceitisstillinagravitationalfieldHoweverifwedefinetheweightofanobjectintermsofaldquoweighingrdquoprocesssuchasthereadingonasetofbathroomscaleswhichineffectmeasuresthecontact force between the object and the scales thenclearlyobjectsinfreefallareweightlessOnenowhasto

askthequestionwhetherornotitispossibleForexampletomeasurethegravitationalforceactingonanastronautin orbit about the Earth

WecanalsodefineweightintermsofthenetgravitationalforceactingonabodyduetoseveraldifferentobjectsForexample for an object out in space its weight could bedefined in terms of the resultant of the forces exerted on it bytheSuntheMoontheEarthandalltheotherplanetsin the Solar System If this resultant is zero at a particular pointthenthebodyisweightlessatthispoint

Inviewofthevariousdefinitionsofweightthatareavailabletousitisimportantthatwhenweusethewordldquoweightrdquoweareawareofthecontextinwhichitisbeingused

Example

Calculate the height above the surface of the Earth atwhichageo-stationarysatelliteorbits

Solution

A geo-stationary satellite is one that orbits the Earth in such a way that it is stationary with respect to a point on the surface of the Earth This means that its orbital period must be the same as the time for the Earth to spin once on its axis ie 24 hours

From Keplerrsquos third law we have G Ms

4π2----------- R3

T2------=

That is

RMe

mh

using the fact that the force of attraction between the satellite and the Earth is given by

FG Mem

R2----------------=

and that F = ma

where m is the mass of the satellite and a 4π2R

T2-------------=

we have

GMem

R2---------------- m 4π2R

T2-------------times

GMe

4π2------------rArr R3

T2------= =

PHYSICS 2015indb 321 150514 338 PM

Chapter 10

322

AH

L

Now the mass of the Earth is 60 times 1024 kg and the period T measured in seconds is given by T = 86400 s

So substitution gives R = 42 times 106 m

The radius of the Earth is 64 times 106 m so that the orbital height h is about 36 times 107 m

Example

Calculatetheminimumenergyrequiredtoputasatelliteofmass500kgintoanorbitthatisasaheightequaltotheEarthrsquosradiusabovethesurfaceoftheEarth

Solution

We have seen that when dealing with gravitational fields and potential it is useful to remember that

g 0G M

Re2---------= or g0 Re

2 GM=

The gravitational potential at the surface of the Earth is

g 0ndash Re GMRe

---------ndash=

The gravitational potential at a distance R

from the centre of the Earth is

=

The difference in potential between the surface of the Earth and a point distance R from the centre is therefore

V∆ g 0Re 1ReR------ndash

=

If R 2Re= then V∆g0 Re

2------------=

This means that the work required to ldquoliftrdquo the satellite into orbit is g0Rm where m is the mass of the satellite This is equal to

10 32 106 500timestimestimes = 16000 MJ

However the satellite must also have kinetic energy in order to orbit Earth This will be equal to

g 0 mRe2

2R-----------------g 0mRe

2

4----------------- 8000 MJ= =

The minimum energy required is therefore

24000 MJ

Exercise 1031 The speed needed to put a satellite in orbit does

not depend on

A the radius of the orbitB theshapeoftheorbitC thevalueofgattheorbitD the mass of the satellite

2 EstimatethespeedofanEarthsatellitewhoseorbitis400kmabovetheEarthrsquossurfaceAlsodetermine the period of the orbit

3 Calculatethespeedofa200kgsatelliteorbitingthe Earth at a height of 70 times 106 m

Assume that g = 82 m sndash2 for this orbit

4 TheradiioftwosatellitesXandYorbitingthe Earth are 2r and 8rwherer is the radius of the Earth Calculate the ratio of the periods of revolutionofXtoY

5 A satellite of mass mkgissentfromEarthrsquossurfaceinto an orbit of radius 5RwhereR is the radius of theEarthWritedownanexpressionfor

(a) thepotentialenergyofthesatelliteinorbit

(b) thekineticenergyofthesatelliteinorbit

(c) theminimumworkrequiredtosendthesatellitefromrestattheEarthrsquossurfaceintoits orbit

6 A satellite in an orbit of 10rfallsbacktoEarth(radius r)afteramalfunctionDeterminethespeedwithwhichitwillhittheEarthrsquossurface

7 The radius of the moon is frac14 that of the Earth AssumingEarthandtheMoontohavethesamedensitycomparetheaccelerationsoffreefallatthe surface of Earth to that at the surface of the Moon

8 UsethefollowingdatatodeterminethegravitationalfieldstrengthatthesurfaceoftheMoon and hence determine the escape speed from the surface of the Moon

Mass of the Moon = 73 times 1022kg

RadiusoftheMoon=17times106 m

PHYSICS 2015indb 322 150514 338 PM

Chapter 10

322

AH

L

Now the mass of the Earth is 60 times 1024 kg and the period T measured in seconds is given by T = 86400 s

So substitution gives R = 42 times 106 m

The radius of the Earth is 64 times 106 m so that the orbital height h is about 36 times 107 m

Example

Calculatetheminimumenergyrequiredtoputasatelliteofmass500kgintoanorbitthatisasaheightequaltotheEarthrsquosradiusabovethesurfaceoftheEarth

Solution

We have seen that when dealing with gravitational fields and potential it is useful to remember that

g 0G M

Re2---------= or g0 Re

2 GM=

The gravitational potential at the surface of the Earth is

g 0ndash Re GMRe

---------ndash=

The gravitational potential at a distance R

from the centre of the Earth is

=

The difference in potential between the surface of the Earth and a point distance R from the centre is therefore

V∆ g 0Re 1ReR------ndash

=

If R 2Re= then V∆g0 Re

2------------=

This means that the work required to ldquoliftrdquo the satellite into orbit is g0Rm where m is the mass of the satellite This is equal to

10 32 106 500timestimestimes = 16000 MJ

However the satellite must also have kinetic energy in order to orbit Earth This will be equal to

g 0 mRe2

2R-----------------g 0mRe

2

4----------------- 8000 MJ= =

The minimum energy required is therefore

24000 MJ

Exercise 1031 The speed needed to put a satellite in orbit does

not depend on

A the radius of the orbitB theshapeoftheorbitC thevalueofgattheorbitD the mass of the satellite

2 EstimatethespeedofanEarthsatellitewhoseorbitis400kmabovetheEarthrsquossurfaceAlsodetermine the period of the orbit

3 Calculatethespeedofa200kgsatelliteorbitingthe Earth at a height of 70 times 106 m

Assume that g = 82 m sndash2 for this orbit

4 TheradiioftwosatellitesXandYorbitingthe Earth are 2r and 8rwherer is the radius of the Earth Calculate the ratio of the periods of revolutionofXtoY

5 A satellite of mass mkgissentfromEarthrsquossurfaceinto an orbit of radius 5RwhereR is the radius of theEarthWritedownanexpressionfor

(a) thepotentialenergyofthesatelliteinorbit

(b) thekineticenergyofthesatelliteinorbit

(c) theminimumworkrequiredtosendthesatellitefromrestattheEarthrsquossurfaceintoits orbit

6 A satellite in an orbit of 10rfallsbacktoEarth(radius r)afteramalfunctionDeterminethespeedwithwhichitwillhittheEarthrsquossurface

7 The radius of the moon is frac14 that of the Earth AssumingEarthandtheMoontohavethesamedensitycomparetheaccelerationsoffreefallatthe surface of Earth to that at the surface of the Moon

8 UsethefollowingdatatodeterminethegravitationalfieldstrengthatthesurfaceoftheMoon and hence determine the escape speed from the surface of the Moon

Mass of the Moon = 73 times 1022kg

RadiusoftheMoon=17times106 m

PHYSICS 2015indb 322 150514 338 PM