Electrostatics (Part 1-Student Copy)

8/11/2019 Electrostatics (Part 1-Student Copy)

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THEORIGINOFELECTRICITY

The electrical nature of matter is inherent

in atomic structure.

kg10673.1 27pm

kg10675.1 27

nm

kg1011.9 31em

C1060.1 19e

coulombs


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T

HE

O

RIGIN

OF

E

LECTRICITY

-There are two kinds of electric charge:

positive

and

negative.

The SI unit of electric charge is

the Coulomb C). The magnitude of the charge

on an electron or a proton is

e

= 1.60 x 10

-19

C

-The law of conservation of electric charge

states that the

net electric charge of an isolated

system remains constant during any process.

-

Electric repulsion and attraction:

Like charges

repel, unlike charges attract,

and

charged

bodies attract never repel) neutral bodies.


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18.1 THEORIGINOFELECTRICITY

In nature, atoms are normally

found with equal numbers ofprotons and electrons, so they

are electrically neutral.

By adding or removingelectrons

from matter it will acquire a net

electric charge with magnitude

equal to etimes the number of

electrons added or removed, N.

Neq


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18.2 CHARGEDOBJECTSANDTHEELECTRICFORCE

It is possible to transfer electric charge from oneobject to another.

The body that loses electrons has a deficiency of

negative charge (more protons than electrons),

while the body that gains electrons has an excess

of negative charge (more electrons than protons).


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18.2 CHARGEDOBJECTSANDTHEELECTRICFORCE

Like charges repel and

unl ike

charges attract each o ther.


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SAMPLEPROBLEMS 1. An electron has a charge whose magnitude

is e = 1.60 x 10-19C. How many electrons are

there in one Coulomb of negative charge?

Soln:

N = q/e = -1 C/ -1.60 x 10-19C

= 6 x 1018

electrons 2. In a chemical reaction, an aluminum atom

loses three electrons to become an aluminum

ion. An ion is a charged atom. What is the

charge (in Coulomb) on this ion? Soln:

q = N e = 3 (+1.60 x 10-19C)

= + 4.80 x 10-19 C


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SEATWORK 1. A conducting sphere has a net charge of 2.4 1017C.

What is the approximate number of excess electrons onthe sphere?

2. Each of three objects has a net charge. Objects A and B

attract one another. Objects B and C also attract oneanother, but objects A and C repel one another. What is thesign of each of the net charges on these three objects?


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18.3 CONDUCTORSANDINSULATORS

Not only can electric charge exist on an object, but it can also movethrough an object.

Substances that readily conduct electric charge are called electr ical

conduc tors.

Materials that conduct electric charge poorly are called electr icalinsulators .


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METHODSOFCHARGING-Charging by contact (or conduction):

-the process of giving one object a net

electric charge by placing it in contact with

an object that is already charged.

-Charging by induction

-the process of giving an object a net electric

charge without touching it to a charged

object.

-Charging by friction

-the process of giving an object a net electric

charge by rubbingit with another object with

a different ability to lose (or gain) electron(s).


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18.4 CHARGINGBYCONTACTANDBYINDUCTION

Charging by contact.

18 4 CHARGING BY CONTACT AND BY INDUCTION


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Charging by induction.


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The negatively charged rod induces a slight positive surface charge

on the plastic.


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COULOMBSLAW

-Apoint charge is a charge that occupies so

little space that it can be regarded as amathematical point.

-Coulombs law gives the magnitude Fof the

electric force that two point charges q1and q2

exert on each other:/

q

1

//

q

2

/

F = k -----------

r

2

Where /q1/ and /q2/ are the magnitudes of thecharges and have no algebraic sign. The term

k is a constant equal to 9 x 109Nm2/C2.

18 5 C L


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18.5 COULOMBSLAW

COULOMBS LAW

The magnitude of the electrostatic force exerted by one point charge

on another point charge is directly proportional to the magnitude of the

charges and inversely proportional to the square of the distance between

them.

2

21

r

qqkF

229 CmN1099.841 ok

2212 mNC1085.8


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EXAMPLESOFELECTRICALFORCECALCULATEDI

Example 1 Electric force vs. gravitational force

An particle is the nucleus of a helium atom. It has massm= 6.64 x 10-27kg and charge q= +2e= 3.2 x 10-19C.

Compare the force of the electric repulsion betweentwo particles with the force of gravitational attractionbetween them.

Soln: Fe + Fg + Fe

rFe= kq2/r2 Fg= Gm

2/r2

(9.0 x 109)(3.2 x 10-19C)2

Fe/Fg= kq2/Gm2= -----------------------------------(in SI units)

(6.67 x 10-11)(6.64 x 10-27)2

= 3.1 x 1035

+

+

+

+

+

+

+

+


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18.5 COULOMBSLAW

18 5 COULOMB S LAW


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18.5 COULOMBSLAW

Example 3 A Model of the Hydrogen Atom

In the Bohr model of the hydrogen atom, the electron is in

orbit about the nuclear proton at a radius of 5.29x10-11m.

Determine the speed of the electron, assuming the orbit tobe circular.

2

21

r

qqkF

18 5 COULOMB S LAW


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18.5 COULOMBSLAW

N1022.8m1029.5

C1060.1CmN1099.8 8211

219229

2

21

rqqkF

rmvmaF c2

sm1018.2

kg109.11

m1029.5N1022.8 631-

118

mFrv

ECTOR ADDITION OF ELECTRIC FORCES ON A LINE


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ECTORADDITIONOFELECTRICFORCESONALINE

Example 1: Three point charges are located on thepositive x-axis of a coordinate system, as shown in thediagram. Determine the net force on point charge q1.

18 5 COULOMB S LAW


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18.5 COULOMBSLAW

N7.2m20.0

C100.4C100.3CmN1000.92

66229

2

21

12

r

qqkF

N4.8m15.0

C100.7C100.3CmN1000.9

2

66229

2

31

13

r

qqkF

3qtowardsN5.7or5.7NN4.8N7.21312 FFF


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2.Three point charges are arranged on a line. Charge q3=+5.00 nC and is at the origin. Charge q2= -3.00 nC and isat x = +4.00 cm. Charge q1is at x = +2.00 cm. What is q1

(magnitude and sign) if the net force on q3is zero?

Soln: q3= +5.00 nC q1= ? nC q2= -3.00 nC

0 2 cm 4 cm

vector diagram of forces acting on q3:q3

F31 F32= attractive force by q2repulsive force by q1

Fnet= F32+ F31= 0 F32= 9 x 109

(3 x 10-9

)(5 x 10-9

)/(0.04)2

= 8.4375 x 10-5N8.4375 x 10-5N = F31= 9 x 10

9(5 x 10-9) q1/(0.02)2

q1= +0.750 nC


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1. Three point charges are arranged

along the x-axis. Charge q1= +3.00uC is at the origin, and charge q2=-5.00 uC is at x = 0.200 m. Charge

q3= -8.00 uC. Where is q3locatedif the net force on q1is 7.00 N inthe x-direction? Show the free-

body diagram for q1.


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Soln: q3= -8.00 uC q1= +3.00 uC q2= -5.00 uC

r 0.200 m0 m Assume q3is at a distance r

to the left of q1vector diagram of forces acting on q1:

q1F13 F12= attractive force by q2

F13= attractive force by q3Fnet= F12+ F13= 7.00 N to the x axis-7.00 = 9 x 109(3 x 10-6)(5 x 10-6)/(0.2)2

- (9 x 109)(8 x 10-6)(3 x 10-6)/r2

-7.00 = 3.375 0.216 /r2

-10.375 r2= -0.216r = 0.144 m but we take only the negative value as

assumed. Hence, r = -0.144 m


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2. An average human weighs about 650 N. If two suchgeneric humans each carried 1.0 coulomb of excesscharge, one positive and one negative, how far apart would

they have to be for the electric attraction between themto equal their 650-N weight?

Soln:

Fe= 650 N = 9 x 109(1 C)2/r2

r2= 9 x 109(1)/650

r = 3,721 m

18 5 COULOMB S LAW


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18.5 COULOMBSLAW

18.6 THE ELECTRIC FIELD


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18.6 THEELECTRICFIELD

The positive charge experiences a force which is the vector sum of the

forces exerted by the charges on the rod and the two spheres.

This test chargeshould have a small magnitude so it doesnt affectthe other charge.

18.6 THEELECTRICFIELD


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8 6 C C

Example 6 A Test Charge

The positive test charge has a magnitude of

3.0x10-8C and experiences a force of 6.0x10-8N.

(a) Find the force per coulombthat the test charge

experiences.

(b) Predict the force that a charge of +12x10-8C

would experience if it replaced the test charge.

CN0.2C100.3

N100.6

8

8

oq

F(a)

(b) N1024C100.12CN0.2 88 F

E


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ELECTRICCHARGE

Glass rods, plastic tubes, silk, and fur can be used to demonstratethe movement of electrons and how their presence or absence

make for powerful forces of attraction and repulsion.


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MOVEMENTOFCHARGESCHARGINGBYCONDUCTION

Materials that alloweasy passage ofcharge are calledconductors. Materialsthat resist electronic

flow are calledinsulators. The motionof electrons throughconducts and aboutinsulators allows us to

observe oppositecharges attract andlike charges repel.


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ELECTRONSMOVEFREELYANDCHARGESMAYBEINDUCED

Take a childs toy, a rubber balloon. If you rub the balloon vigorouslyon a fuzzy sweater then bring the balloon slowly toward a painted

concrete or plaster wall, the balloon will stick to the wall and remainfor some time. The electrostatic force between static electrons and theinduced positive charge in the wall attract more strongly than theweight of the balloon.


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STATICELECTRICITYABOUTANINSULATORCANSHIFT

The motion of static charges about a plastic comb and light bits

of paper can cause attractive forces strong enough to overcomethe weight of the paper.

EXAMPLES OF ELECTRICAL FORCE CALCULATED


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EXAMPLESOFELECTRICALFORCECALCULATED

Example 1: An particle (alpha) is the nucleus of a heliumatom. It has mass m = 6.64 x 10-27kg and charge q = +2e =3.2 x 10-19C. Compare the force of the electric repulsion

between two particles with the force of gravitationalattraction between them.

Soln: Fe= 9 x 109(3.2 x 10-19)2/r2

Fg= 6.67 x 10-11(6.64 x 10-27)2/r2

Fe/Fg= 9.2 x 10-28

/2.94 x 10-63

= 3.1 x 1035

E l 2 T i t h l t d th iti


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Example 2: Two point charges are located on the positive x-axis of a coordinate system. Charge q1= 1.0 nC is 2.0 cmfrom the origin, and charge q2= -3.0 nC is 4.0 cm from theorigin. What is the total force exerted by these two charges

on a charge q3= 5.0 nC located at the origin? Gravitationalforces are negligible.

Soln: F31= 9.0 x 109(1.0 x 10-9)(5.0 x 10-9)/(2.0 x 10-2)2

= 1.12 x 10-4N to the left

F32= 9.0 x 109

(3.0 x 10-9

)(5.0 x 10-9

)/(4.0 x 10-2

)2

= 8.4 x 10-5N to the right

= F31+ F32= 28 N, to the left


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Example 1: Two equal positive point chargesq1= q2= 2.0 C are located atx= 0, y =0.30 m andx = 0, y = -0.30 m, respectively.

What are the magnitude and direction ofthe total electric force that these chargesexert on a third point charge Q = 4.0 C at

x = 0.40 m, y = 0 ?

THREE-POINT CHARGES ON A PLANE


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THREE-POINTCHARGESINAPLANE


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Soln:F1= 9 x 10

9(4.0 x 10-6)(2.0 x 10-6)/(0.50)2= 0.29 NF1x= F1cos = (0.29)(0.40/0.50) = 0.23 NF1y= -F1sin = -(0.29)(0.30/0.50) = -0.17 N

The lower charge q2exerts a force with the same magnitudebut at an angle abovethe x-axis. From symmetry, we see thatitsx-component is the same as that due to the upper charge,but its y-component has the opposite sign. So the componentsof the total force Fneton Q are:Fx= 0.23 N + 0.23 N = 0.46 NFy= -0.17 N + 0.17 N = 0The total force on Q is in the +x-direction , with magnitude

0.46 N.


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Example 2: The figure aboveshows three point chargesthat lie in the x,y plane in avacuum. Find the magnitudeand direction of the netelectrostatic force on q1.

THREE-POINT CHARGES ON A PLANE

F12F

q1 F13

F12

F12sin 73o

73o

F12cos 73o

SOLUTION


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SOLUTION

The magnitudes of the forces are:/q1/q2/ (9 x 10

9)(4 x 10-6)(6 x 10-6)

F12= k -------- = ------------------------------------ = 9.6 Nr122 (0.15)2

/q1/q3/ (9 x 109)(4 x 10-6)(5 x 10-6)

F13= k -------- = ------------------------------------ = 18 Nr12

2 (0.10)2

Force x-component y-componentF12 +9.6 (cos 73

o) = +2.8 N +9.6 N (sin 73o) = +9.2 NF13 +18 N 0 NF Fx= +21 N Fy= +9.2 N

The magnitude Fand angle of the net force are:F = Fx

2+ Fy2= (21 N)2+ (9.2 N)2= 23 N

= tan-1(9.2 N/21 N) = 24o


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F32x= cos 60o(16.2 N) = -8.1 N

F32y= sin 60o(16.2 N) = +14 N

Fy= -9.4 N + 14 N = +4.6 N

Fx= -5.4 N 8.1 N = -13.5 N

Fnet1

= (13.5)2+ (4.6)2= 14.3 N, 19oabove the -x-axis

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Electrostatics (Part 1-Student Copy)