Physics 2BElectricity and

MagnetismDylan BrennanDylan Brennan

brennan@physics.ucsd.edu

Winter 2006Winter 2006

Dylan P. BrennanDylan P. Brennanbrennan@physics.ucsd.edu455-4144 @455-4144 @ General AtomicsGeneral AtomicsOffice Hours: Office Hours: Tu Tu 9:00 - 11:00 AM9:00 - 11:00 AMLocation TBA Location TBA

TATAJui Jui Yu ChiuYu Chiujchiu@physics.ucsd.eduOffice Hours:Office Hours: W 1:30-2:30 PMW 1:30-2:30 PMF 11:F 11:0000 AM - 12:00 PM AM - 12:00 PMMH2101MH2101

TENTATIVE COURSE OUTLINE

Ch. 34: 1, 2, 3, 4, 5No QuizMaxwell’s Equations, Ch. 3428-30

(Mar. 13)

Ch. 33: 5, 11, 18, 23, 30, 39, 45, 60Chapters 32, 33

(Mar. 10)

Alternating Currents, Ch. 3325-27

(Mar. 6)

Ch. 31: 13, 14, 17, 21, 25, 27, 29,32, 34, 35

Ch. 32: 5, 9, 22, 36, 55, 66, 71

Chapters 30, 31

(Mar. 3)

Induction, Ch. 31

Inductors, Ch. 32

22-24

(Feb. 27)

Ch. 30: 10, 15, 17, 23, 24, 35, 36,40, 45, 57

Chapter 30

(Feb. 24)

Sources of Magnetic Field,Ch. 30

19-21

(Feb. 20)

Ch. 29: 13, 22, 27, 36, 38, 53Chapter 29

(Feb. 17)

Magnetic Field,

Ch. 29

16-18

(Feb. 13)

Ch. 28: 22, 25, 26, 29, 31, 43, 55Chapter 28

(Feb. 10)

Curcuits, Ch. 2813-15

(Feb. 6)

Ch. 26: 15, 36, 37, 46, 54

Ch. 27: 10, 36, 39, 50, 71

Chapters 26, 27

(Feb. 3)

Capacitance, Ch 26

Current, Ch. 27

10-12

(Jan. 30)

Ch. 25: 8, 11, 28, 35, 39, 50Chapter 25

(Jan. 27)

Electric Potential, Ch. 257-9

(Jan. 23)

Ch. 24: 9, 10, 19, 28, 31, 49, 54, 63Chapter 24

(Jan. 20)

Gauss’s Law, Ch. 244-6

(Jan. 16)

Ch. 23: 1, 11, 19, 22, 24, 30, 39,46, 48, 59, 68, 78

Chapter 23

(Jan. 13)

Electric Charge and Field,Coulomb’s Law, Ch. 23

1-3

(Jan. 9)

ASSIGNED PROBLEMSQUIZ SUBJECT

(Date)

TOPICLECTURE NUMBER

(Start Date)

Charge, Charge, CoulombCoulomb’’s Law, s Law, and Electric Fieldand Electric Field

Lectures 1, 2, 3Lectures 1, 2, 3

The Forces of NatureThe Forces of Nature

The Strong Force: (which your book calls the color force The Strong Force: (which your book calls the color force because there are three charges: because there are three charges: red, green, red, green, andand blue) blue).. This is the force between quarks and gluons that holds atomic This is the force between quarks and gluons that holds atomic atomic nuclei together. atomic nuclei together.

TheThe ElectromagneticElectromagnetic ForceForce: about 100 times weaker than the strong force;: about 100 times weaker than the strong force; two charges: two charges: plusplus and and minusminus..

The Weak Force: about 20 orders of magnitude weaker than electricity;The Weak Force: about 20 orders of magnitude weaker than electricity; responsible for nuclear beta decay, changes neutrons responsible for nuclear beta decay, changes neutrons to protons and to protons and vice versa. vice versa. This is how neutrinos interact.This is how neutrinos interact.

Gravitation: the weakest force by far! About 39 orders of magnitude weaker Gravitation: the weakest force by far! About 39 orders of magnitude weaker than electricity! than electricity!

Overview of ElectromagnetismOverview of Electromagnetism

Greeks note that rubbing materials can cause them to attract/repeland they note that certain stones from “Magnesia” attract iron.

18th century “parlor tricks”

Franklin figures out that there are two kinds of electric charge.

19th century: Faraday performs revealing experiments Maxwell discovers equations which unify electricity and magnetism and lead to the prediction of electromagnetic waves and the development of Einstein’s relativity.

20th century: Relativity, the photoelectric effect, quantum mechanics, all lead to the development of quantum electrodynamics. Unification of the electromagnetic and weak interactions.

21st century: ?

Technological RevolutionTechnological Revolution

Coulomb’s LawLike charges repel, opposite charges attract;Like charges repel, opposite charges attract; netnet electric charge is conserved.electric charge is conserved.

We will label the two kinds of charge as eitherWe will label the two kinds of charge as either positivepositive or or negativenegative..

Then the force between twoThen the force between two pointpoint charges is:charges is:

the force exertedthe force exerted onon charge qcharge q22byby the charge qthe charge q11

!

ˆ r = unit vector =r

r- +

!

F12

unit vector pointsunit vector points fromfrom qq11towardstowards qq22

!

F12

CoulombCoulomb’’s Constants Constant

!

k "1

4#$0

% 9.0 &109 N ' m

2

C2

Where the unit of charge that we will use is the Coulomb:Where the unit of charge that we will use is the Coulomb:

!

1 C " 6.241#1018 elementary charges

““permittivitypermittivity”” constant constant

!

"0# 8.85 $10

%12

C2

N & m2

The Shocking Truth About ElectricityThe Shocking Truth About ElectricityThe electric force is phenomenally strong!

Consider a comparison between theConsider a comparison between the electricelectric andand gravitational gravitational forcesforcesacting between the electron and the proton in a hydrogen atom.acting between the electron and the proton in a hydrogen atom.

!

FG =GMpme

r2

~ 10"47

Joule m-1

!

FE =1

4"#0

Qpqe

r2

~ 10$8

Joule m-1

proton

electron

Gravitational Force

Electric Force

!

Mp " 938.26 MeV/c2

me " 0.511 MeV/c2

!

Qp = +1 elementary charge " +1.6022 #10-19 Coulomb

qe = $1 elementary charge " -1.6022#10-19 Coulomb

!

r ~ 10"10

m

The Electric FieldThe Electric FieldIt is convenient to define the It is convenient to define the ELECTRIC FIELDELECTRIC FIELD at a point as the force per unit at a point as the force per unit(positive) charge, so we can determine the force on any magnitude of charge at(positive) charge, so we can determine the force on any magnitude of charge atthat point as that point as qEqE::

!

E =F

q

What are the dimensions of Electric Field?What are the dimensions of Electric Field?

!

E[ ] =N

C=

J

m C=

Volts

m

!

Volt = Joule

Coulomb=

J

C

x x - axis- axis

y y - a

xis

- axi

s

!

ˆ j

!

ˆ i unit vectors point along positive unit vectors point along positive xx and and yy axes axes

x

y

!

r

!

r = xˆ i + yˆ j

r = r ˆ r

r " r = r # r = x2

+ y2

ˆ r = r /r =x

x2

+ y2

$

% & &

'

( ) ) ˆ i +

y

x2

+ y2

$

% & &

'

( ) ) ˆ j

!

ˆ i " ˆ i = ˆ j " ˆ j =1 and ˆ i " ˆ j = 0

!

ˆ r is a unit vector along r, so ˆ r " ˆ r =1

Brief Review: Vectors in 2 DimensionsBrief Review: Vectors in 2 Dimensions

inner (scalar) product of two vectorsinner (scalar) product of two vectors

!

A = Ax( )ˆ i + Ay( )ˆ j

x - component Ax = projection along x - axis = A " ˆ i

y - component Ay = projection along y - axis = A " ˆ j

!

B = Bx( )ˆ i + By( )ˆ j

x - component Bx = projection along x - axis = B " ˆ i

y - component By = projection along y - axis = B " ˆ j

!

inner product

A "B = AxBx + AyBy

= A B cos#

!

where " is the angle

between the two vectors

!

ˆ i

!

ˆ j

Ax

Ay

Bx

By

Point Charges and the Superposition PrinciplePoint Charges and the Superposition PrincipleElectric forces simply add Electric forces simply add vectoriallyvectorially.. The TOTAL FORCE on the object is justthe VECTOR SUM of the individualforces.

!

Ftotal force on Q = Fq>0 + Fq<0 =kQq(2a)

a2 + y

2( )3 / 2

ˆ i

Example: Example: What is the force on qwhen both q1 and q2 arepresent??

!

Fq>0=

kQq

a2

+ y2

a

a2

+ y2

ˆ i +y

a2

+ y2

ˆ j "

# $ $

%

& ' '

!

Fq<0=

kQq

a2

+ y2

a

a2

+ y2

ˆ i "y

a2

+ y2

ˆ j #

$ % %

&

' ( (

Example:Example: Problem 11, Chapter 23:Problem 11, Chapter 23:A proton is on the x-axis at x = 1.6 nm. An electron is on the y-axis at y = 0.85 nm.Find the net force the two exert on a helium(He) nucleus (charge +2e) at the origin.

Solution: Solution: draw a picture; set up the algebra; plug in the numbers with theirdraw a picture; set up the algebra; plug in the numbers with theirunits/dimensions;units/dimensions; cancel out and re-work units to the final units and make surecancel out and re-work units to the final units and make surethey make sense!they make sense!

x - axis

y - a

xis

+2e

-e

+e

Vector force exertedVector force exerted byby electron electron on on He nucleus:He nucleus:

!

Fe,He

=k "e( ) +2e( )

re,He

2"ˆ j ( ) =

k 2e2( )

re,He

2

ˆ j

!

"9 #10

9 N $m

2/C

2( )2 1.602 #10%19

C( )2

0.85 #10%9

m( )2

ˆ j

" 6.38 #10-10

N( )ˆ j = 0.638 nN( )ˆ j !

1 e "1.602 #10$19

C

!

re,He

= 0.85 nm

= 0.85 "10#9

m

Vector force exertedVector force exerted byby proton proton on on He nucleus:He nucleus:

!

Fp,He =k +e( ) +2e( )

rp,He

2"ˆ i ( ) =

k 2e2( )

rp,He

2"ˆ i ( )

!

"9 #10

9 N $m

2/C

2( )2 1.602 #10%19

C( )2

1.6 #10%9

m( )2

%ˆ i ( )

" %0.180 nN( )ˆ i !

Ftot = Fe,He + Fp,He " #0.18 nN( )ˆ i + 0.638 nN( )ˆ j

Vectors in 3 DimensionsVectors in 3 Dimensions

A simple generalization of what we just didA simple generalization of what we just didin the examples in two dimensions!in the examples in two dimensions!

!

F = qE(x,y,z)

At a particular point in space x, y, z,At a particular point in space x, y, z,the (vector) force on a point particlethe (vector) force on a point particlewith charge with charge q q isis

So, you can think of a vector residing at each point in space.So, you can think of a vector residing at each point in space.Of course, the vector will have three components:Of course, the vector will have three components:a scalar function a scalar function EExx(x, y, z),(x, y, z), the component along the the component along the xx-direction-direction;;a scalar functiona scalar function EEyy(x, y, z),(x, y, z), the component along the the component along the yy-direction-direction;;a scalar functiona scalar function EEzz(x, y, z),(x, y, z), the component along the the component along the zz-direction-direction..

!

E(x,y,z) = Ex (x,y,z)ˆ i + Ey (x,y,z)

ˆ j + Ez (x,y,z)ˆ k

!

E(x,y,z) =kq

x2

+ y2

+ z2

x

x2

+ y2

+ z2

ˆ i +y

x2

+ y2

+ z2

ˆ j +z

x2

+ y2

+ z2

ˆ k "

# $ $

%

& ' '

=kq

r2

ˆ r

yy-axis-axis

x-x-axisaxis

zz-axis-axis

!

ˆ i

!

ˆ j

!

ˆ k

yy

xx

zz

q!

r = r ˆ r

EE

!

r = r ˆ r = xˆ i + yˆ j + z ˆ k

ˆ r =x

x2

+ y2

+ z2

ˆ i +y

x2

+ y2

+ z2

ˆ j +z

x2

+ y2

+ z2

ˆ k

r = r = x2

+ y2

+ z2

The radius vector from the origin to point (x, y, z) isThe radius vector from the origin to point (x, y, z) is

Remember your vector algebra . . .Remember your vector algebra . . .

!

r " r = r2ˆ r " ˆ r = r2

= x2

+ y2

+ z2

ˆ r " ˆ r =1

ˆ i " ˆ i = ˆ j " ˆ j = ˆ k " ˆ k =1

ˆ i " ˆ j = ˆ i " ˆ k = ˆ j " ˆ k = 0

inner (scalar) product of two vectorsinner (scalar) product of two vectors

!

A = Ax( )ˆ i + Ay( )ˆ j + Az( ) ˆ k

x - component Ax = projection along x - axis = A " ˆ i

y - component Ay = projection along y - axis = A " ˆ j

z - component Az = projection along z - axis = A " ˆ k

!

B = Bx( )ˆ i + By( )ˆ j + Bz( ) ˆ k

x - component Bx = projection along x - axis = B " ˆ i

y - component By = projection along y - axis = B " ˆ j

z - component Bz = projection along z - axis = B " ˆ k

!

inner product

A "B = AxBx + AyBy + AzBz

= A B cos#

!

where " is the angle

between the two vectors

The Electric Field of a Point ChargeThe Electric Field of a Point Charge

!

q < 0

!

E =kq

r2

ˆ r

A positive chargeA positive chargecauses E field incauses E field inopposite direction,opposite direction,ieie. outward.. outward.

The Electric Field from a Distribution of Point ChargesThe Electric Field from a Distribution of Point Charges

Again, simply use the superposition principle.Again, simply use the superposition principle.

To find the Electric Field at a given point in space,To find the Electric Field at a given point in space,simply add up the contributions from each individual charge.simply add up the contributions from each individual charge.

Calculate the Electric Field contribution from eachCalculate the Electric Field contribution from eachone of these charges as if the other ones didnone of these charges as if the other ones didn’’t exist.t exist.

!

E = E1

+ E2

+ E3

+ " " " = Ei

i

# =kqi

ri2

i

# ˆ r i

Example: Electric DipoleExample: Electric DipoleTwo (point) charges of equal magnitude but opposite signTwo (point) charges of equal magnitude but opposite signat a fixed separation.at a fixed separation.

!

Etotal = Eq>0 + Eq<0 =kq(2a)

a2 + y

2( )3 / 2

ˆ i

"kq(2a)

y3

ˆ i for y >> a!

Eq>0=

kq

a2

+ y2

a

a2

+ y2

ˆ i +y

a2

+ y2

ˆ j "

# $ $

%

& ' '

!

Eq<0=

kq

a2

+ y2

a

a2

+ y2

ˆ i "y

a2

+ y2

ˆ j #

$ % %

&

' ( (

TheThe Dipole MomentDipole Moment is defined to be theis defined to be thevector with magnitude vector with magnitude p = q dp = q dthat points from the negative toward the positive chargethat points from the negative toward the positive charge..

!

p = qd ˆ k

z-ax

isz-

axis

Example: Example: Molecules can be overall charge neutralMolecules can be overall charge neutral but may nevertheless possess an electric but may nevertheless possess an electric dipole moment. dipole moment.

The classic example is the water molecule. The classic example is the water molecule.

!

dipole moment p

Continuous Distributions of ChargeContinuous Distributions of ChargeOf course, real materials are composed of point-like electrons, protons, and neutrons.Of course, real materials are composed of point-like electrons, protons, and neutrons.The protons and electrons can be regarded, for our purposes, as positiveThe protons and electrons can be regarded, for our purposes, as positiveand negative point charges, respectively.and negative point charges, respectively.

Any macroscopic object will contain ~ 10Any macroscopic object will contain ~ 102424 particles, however. In this case particles, however. In this casewe can approximate the distribution of charge as continuous across the volume of thewe can approximate the distribution of charge as continuous across the volume of theobject, given locally by aobject, given locally by a volume charge densityvolume charge density rr (with dimensions (with dimensions Coulombs per unit volumeCoulombs per unit volume, or C/m, or C/m33 ). ).

Likewise, for charge spread out over a surface, we can define a Likewise, for charge spread out over a surface, we can define a surface charge densitysurface charge density ss ((with dimensions Coulombs per unit area, or C/mwith dimensions Coulombs per unit area, or C/m22 ). ).For charge spread out along a line (e.g., an extremely thin wire)For charge spread out along a line (e.g., an extremely thin wire)we can define awe can define a line charge densityline charge density ll ((with dimensions Coulombs perwith dimensions Coulombs perunit length, or C/m).unit length, or C/m).

Then the increment in charge associated with an increment in volume, area, or length is:Then the increment in charge associated with an increment in volume, area, or length is:

!

dq = "dV , dq =#dA , dq = $dx

With the charge in an object broken down into particles or small incrementsWith the charge in an object broken down into particles or small incrementsdqdq, , we can employ the superposition principle to find the Electric Fieldwe can employ the superposition principle to find the Electric Fieldat any position.at any position.

!

dE =k dq

r2

ˆ r

E = dE =k dq

r2

ˆ r ""

ExampleExample: a uniformly charged rod of length: a uniformly charged rod of length ll and charge and charge Q.Q. Find the electric field on the rod axis at distance b.Find the electric field on the rod axis at distance b.

!

dq = " dx =Q

l dx

!

r E = ˆ i

k dq

x2

"2b

"b

#

!

dE =k dq

x2

ˆ i

!

r E = ˆ i

kQ

lx2

"(b + l )

"b

# dx = ˆ i kQ

l(1

b"

1

(b + l))

= ˆ i kQ

b(b + l)

x-x-axisaxis

!

ˆ i

ll

E(E(x=0x=0)?)?

If b>>lIf b>>l

!

r E "

k Q

b2

ˆ i

Point chargePoint charge

Check ifCheck if makes sense:makes sense:

ExampleExample: a uniformly charged ring of radius : a uniformly charged ring of radius aa and charge and charge Q.Q. Find the electric field on the axis through the ring.Find the electric field on the axis through the ring.Assume that the ring is very thin so that we can regard the charge distributionAssume that the ring is very thin so that we can regard the charge distributionon it as being a on it as being a ““line charge.line charge.”” Then the Then the ““charge densitycharge density”” (charge per unit length) (charge per unit length) is is l l = = Q/(2Q/(2pp a). a).

!

dq = " ds =Q

2#a ds !

dEx =k dq

r2

cos"

=k dq

x2

+ a2

x

x2

+ a2

!

dE =k dq

r2

ˆ r

!

E = E = dExring" =

kx dq

x2 + a2( )

3 / 2

ring

" =kx

x2 + a2( )

3 / 2dq

ring

"

=kx

x2 + a2( )

3 / 2

Q

2#a

0

2#a

" ds =kx Q

x2 + a2( )

3 / 2

Example: continued . . .Example: continued . . .Suppose the ring of radius 1 m carries a uniformly distributed charge of 0.01 C.Suppose the ring of radius 1 m carries a uniformly distributed charge of 0.01 C.What is the electric field a distance 8 m from the center of the ring and alongWhat is the electric field a distance 8 m from the center of the ring and alongthe center line?the center line?Well this meets the conditions of what we had just derived. We saw that theWell this meets the conditions of what we had just derived. We saw that theelectric field points along the center line and has magnitude:electric field points along the center line and has magnitude:

!

E = E = dExring" =

kx dq

x2 + a2( )

3 / 2

ring

" =kx

x2 + a2( )

3 / 2dq

ring

"

=kx

x2 + a2( )

3 / 2

Q

2#a

0

2#a

" ds =kx Q

x2 + a2( )

3 / 2

!

E "9 #10

9 N m

2/C

2( ) 8 m( ) 0.01 C( )

8 m( )2

+ 1 m( )2[ ]

3 / 2"1.37 #10

6 N m

2/C

2( ) m C

m3

"1.37 #106 N

C note :

N

C=

Joule

m C=

Volt

m !

C

Kinematics and Dynamics of Matter in Electric FieldsKinematics and Dynamics of Matter in Electric Fields

For point particles of mass For point particles of mass mm and charge and charge qq in an Electric Field in an Electric Field EE we have we have

!

force F = q E

and Newton's law relating force and acceleration F = ma ,

implying that a =q

mE

For example, a uniform(constant in magnitude and directionFor example, a uniform(constant in magnitude and directionin some region of space) electric field produces a constant accelerationin some region of space) electric field produces a constant accelerationof a particle.of a particle.

The Interaction of a Dipole with a uniform Electric Field

!

" + = 1

2d q E sin#

!

"# = 1

2d q E sin$

!

" = p#E

The torque acts to try to line up the dipole along the field.

Calculate the work required to rotate a dipole from aposition orthogonal to the Electric Field directionto a new angle q with respect to the field orientation.

!

W = " d#$

2

#

% = pE sin# d#$

2

#

% = &pE cos#

Associate this work with a change in a potential energywhich we can define as

!

U = "p #E = "p E cos$

Conductors, Insulators, and Dielectrics

Materials in which (some) electrons are free to move in response to anapplied electric field we term conductors.

Insulators are materials where charges are not free to flow as large scaleelectric currents.

However, the molecules in insulators may be ableto respond to an applied electric field, for example, lining up the intrinsic dipole moments of these molecules. Moleculeswithout intrinsic dipole moments may acquire induced dipole momentsin response to the electric field. In either case, we call theseSubstances dielectrics.