[전자기학 jackson 솔루션] 전자기학 jackson 솔루션 65 4000.pdf

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Physics 742: Jackson, Classical

Electrodynamics

Russell Bloomer1

University of Virginia

Note: There is no guarantee that these are correct, and they should not be copied

1email: [email protected]


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Contents

1 Problem Set 1 11.1 Problem 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Problem 2: Jackson 1.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 Problem 3: Jackson 1.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.4 Problem 4: Jackson 1.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2 Problem Set 2 7

2.1 Problem 1: Jackson 1.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.2 Problem 2: Jackson 1.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.3 Problem 3: Jackson 2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.4 Problem 4: Jackson 2.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

3 Problem Set 3 133.1 Problem 1: Jackson 2.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133.2 Problem 2: Jackson 2.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143.3 Problem 3: Jackson 2.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163.4 Problem 4: Jackson 2.23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

4 Problem Set 4 194.1 Problem 1: Jackson 3.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194.2 Problem 2: Jackson 3.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204.3 Problem 3: Jackson 3.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

5 Problem Set 5 235.1 Problem 1: Jackson 3.17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235.2 Problem 2: Jackson 4.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

6 Problem Set 6 276.1 Problem 1: Jackson 4.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276.2 Problem 2: Jackson 4.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 286.3 Problem 3: Distressed Simple Cubic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

7 Problem Set 7 317.1 Problem 1: Jackson 4.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

8 Problem Set 8 338.1 Problem 1: Jackson 5.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 338.2 Problem 2: Semi-Infinite Cylinder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 348.3 Problem 3: Jackson 5.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 358.4 Problem 4: Jackson 5.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

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10 Problem Set 10 45

10.1 Problem 1: Jackson 5.14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4510.2 Problem 2: Jackson 5.27 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4610.3 Problem 3: Jackson 5.29 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47


12 Problem Set 12 5512.1 Problem 1: Jackson 6.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5512.2 Problem 2: Jackson 6.14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

A Special Functions 61

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Chapter 1Problem Set 1

1.1 Problem 1

Consider a vacuum diode consisting of two parallel plates of area A which are large compared to their separationd. Neglecting edge effects, all quantities will be assumed to depend only on x a coordinate perpendicular to the plates.

Electrons are “boiled” off the cathode plate which is heated and held at potential Φ = 0. They are attracted

to the anode plate which is held at a positive potential V o. The cloud of electrons within the gap (called the spacecharge) quickly builds up to a point where it reduces field at the cathode to zero. From then on a steady currentflows between the plates.

Show that the current in the diode is given by

I = K V 3/2o where K = 4oA

9d2

2e

m (1.1)

The charge density is

ρ = −o∇2V → ρ = −o ∂ 2V

∂x2 (1.2)

because the charge density only depends on the position x. Next the velocity of the electron at any position in thepotential.

1

2mv2 = eV → v =

2eV

m (1.3)

Because this is a steady state, the current is independent of the position x.

dq = ρAdx → dq dt

= ρAdx

dt → I = ρAv (1.4)

Combining equations 3 and 41

I = ρA

2eV

m → ρ = I

A

m

2eV

Now adding in equation 2

ρ = I A

m2eV

→ ∂ 2V ∂x2

= − I oA

m2eV

All that remains is to solve the differential. Start by defining W = ∂V ∂x

. The differential becomes

W ∂W

∂x = − I

oA

2

meV −1/2

∂V

∂x ⇒ W ∂W = − I

oA

2

meV −1/2∂V

1If the ‘e’ on equation is lower case, it is from the homework set, but if it is upper case, it is from ‘Jackson’ 3 rd ed.

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Integrating once to find

1

2W 2 = − 2I

oA

2

meV 1/2

Solving for W

1

2W 2 = − 2I

oA

2

meV 1/2 → W =

8m

e

I

oA

21/4

V 1/4

Now reintroduce W = ∂V ∂x

∂V

∂x =

8m

e

I

oA

21/4V 1/4 ⇒ V −1/4∂V =

8m

e

I

oA

21/4∂x

Integrating

V −1/4∂V =

8m

e

I

oA

21/4∂x ⇒ 4

3V 3/4 =

8m

e

I

oA

21/4x + C

The first boundary condition is at V (0) = 0, which determines the constant to be C = 0. The other boundarycondition is at V (d) = V o. The equation becomes

4

3V 3/4o =

8m

e

I

oA

21/4d

Solving for the current

4

3

4V 3o =

8m

e

I

oA

2d4 ⇒

I

oA

2=

e

8md4

4

3

4V 3o ⇒ I oA =

e

8m

4

3d

2V 3/2o

I = 4oA

9d2

2e

mV 3/2o → I = K V 3/2o

1.2 Problem 2: Jackson 1.5

The time-averaged potential of a neutral hydrogen atom is given by

Φ = q

4πo

e−αr

r

1 +

αr

2

(1.5)

where q is the magnitude of the electric charge, and α−1 = ao/2, ao being the Bohr radius. Find the distribution of charge (both continuous and discrete) that will give this potential and interpret your result physically.

The charge density is related to the potential by

ρ =−

o∇

2Φ→

ρ =−

o∇

2 q 4πo

e−αr

r1 + αr

2

= − q 4π

∇2

e−αr

r

1 +

αr

2

The Laplace in spherical coordinates for a spherically symmetrical object is

∇2 = 1r2

∂

∂r

r2

∂

∂r

2


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Then the charge density becomes

ρ = − q 4π

1

r2∂

∂r

r2

∂

∂r

e−αr

r

1 +

αr

2

= − q 4π

1

r2∂

∂r

r2

∂

∂r

e−αr

r +

αe−αr

2

= − q

4π 1

r2∂

∂r−αr22 e−αr + r2e−αr ∂ ∂r 1r − αr

2

2 e−αr

= − q 4π

1

r2

− α2re−αr + α

3r2

2 e−αr − αe−αr + α2r2e−αr − αr2e−αr ∂

∂r

1

r

+ e−αr

∂

∂r

r2

∂

∂r

1

r

= − q 4π

α3

2 e−αr − α

r2e−αr +

αr2

r2e−αr

r2 + e−αr

1

r2∂

∂r

r2

∂

∂r

1

r

With ∇2 1r

= −4πδ (r), the factor in front of the delta function is 1 at the origin and 0 everywhere else. The above

equation reduces to

ρ = −q

4π

α3

2 e−αr − 4πδ (r)

= qδ (r) − qα

3

8π e−αr (1.6)

The first term is a positive point charge at the origin, which would be the proton. The second term is a negativespherically symmetry object that is decaying off rapidly after the Bohr radius, which is a good description of the

electric cloud.


A simple capacitor is a device formed by two insulated conductors adjacent to each other. If equal and opposite chargesare placed on the conductors, there will be a certain difference potential b etween them. The ratio of magnitude of the charge on one conductor to magnitude of the potential difference is called the capacitance. Using Gauss’s law,calculate the capacitance of

(a) two large, flat, conducting sheets of area A, separated by a small distance d

The electric field outside the plate two plates is zero, because if pillbox containing both plates has no charge thereforeno electric field. Forming a pillbox around one plate by Gauss’s law has |E | = σ/2o, where σ = Q/A. There are twoplates adding constructively in between, so |E | = σ/o. The potential is then

E = V

d

= σ

o

V = σd

o→ V = Qd

Ao

The capacitance is then

C = Q

V =

QQd

Ao

→ C = Aod

(b) two concentric conduction spheres with radii a, b (b > a)

Gauss’s law for a sphere is

E = Q

4πo

1

r2

The potential is then

V = − ab

Ê · dl̂ = − Q4πo

ab

drr2

= Q

4πo

1

r

ab

= Q

4πo

1

a − 1

b

Now the capacitance

C = Q

V =

QQ

4πo

1a − 1

b

= 4πo abb − a

3


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(c) two concentric conducting cylinders of length L, large compared to their radii a, b (b > a)

Using Gauss’s law for cylindrical surface is

E = Q

2πo

1

r

The potential becomes

V = − b

a

Ê · dl̂ = − Q2πo

ba

dr

r

= − Q2πo

ln rb

a =

Q

2πo(ln b − ln a) = Q

2πoln (b/a)

The capacitance is

C = Q

V =

QQ

2πoln (b/a)

= 2πoln (b/a)

(1.7)

(d) What is the inner diameter of the outer conductor in an air-filled coaxial cable whose center conductor is acylindrical wire of diameter 1 mm and whose capacitance is 3 × 10−11F/m? 3 × 10−12F/m?

Using equation 7, and rearranging it

b = ae 2πoC

For 3 × 10−11F/m

b = 1 × e2π·8.85×10−12/3×10−11 = e1.854 = 6.38 mm

For 3 × 10−12F/m

b = 1 × e2π·8.85×10−12/3×10−12 = e18.54 = 1.12 × 108 mm


Prove the mean value theorem: For charge-free space the value of the electrostatic potential at any p oint is equal tothe average of the potential over the surface of any sphere centered on that point.

From Equation 1.36

Φ(x) = 1

4πo

V

ρ(x)R

d3x + 1

4π

S

1

R

dΦ

dn − Φ d

dn

1

R

da (1.8)

For a charge-free space ρ = 0, then

1

4πo

V

ρ(x)R

d3x = 1

4πo

V

0

Rd3x = 0 (1.9)

Using Equation 1.33 for the second part of equation 8

1

R

dΦ

dn =

1

R∇Φ · n̂

From equation 1.16

1

R∇Φ · n̂ = − 1

RE · n̂

The second part of equation 8 becomes

1

4π

S

1

R

dΦ

dnda = − 1

4π

S

1

RE · n̂da

4


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Because this is over the surface it is independent of R, then

− 14πR

S

E · n̂da

By the divergence theorem

− 1

4πR

S E · n̂da = 1

4πoR

V ρ(x)d3

x

Again in a charge-free space

1

4πoR

V

ρ(x)d3x = 0 (1.10)

For the last part of equation 8, ddn

1R

= − 1

R2, because the differential is before the surface. At the surface the

radius is constant so the last part of equation 8 becomes

− 14π

S

Φ d

dn

1

R

da =

1

4πR2

S

Φda (1.11)

So combining equations 9, 10, and 11 to find

Φ(x) = 1

4πR2

SΦda

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6


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Two long, cylindrical conductors of radii a1 and a2 are parallel and separated by a distance d, which is large comparedwith either radius. Show that the capacitance per unit length is given approximately by

C πoln d

a−1

(2.1)

where a is the geometrical mean of the two radii. Approximate what gauge wire (state diameter in millimeters) wouldbe necessary to make a two-wire transmission line with capacitance of 1.2 × 10−11 F/m in the separation of the wireswas 0.5 cm? 1.5 cm? 5.0 cm?

The electric field for a wire is given by

E = Q

2πo

r̂

r

The potential between the two wire is

Φ =

E · d l = Q

2πo

d−a2a1

dr

r +

d−a1a2

dr

r

= Q

2πoln

d − a2a1

+ lnd − a1

a2

= Q

2πo

ln

(d − a2) (d − a1)

a1a2

The capacitance is given by C = QΦ

. Then

C = Q

Φ =

QQ

2πo

ln(d−a2)(d−a1)

a1a2

=

2πo

ln(d−a2)(d−a1)

a1a2

Since the distance between the wires is much large than their radii, d − a1 ≈ d − a2 ≈ d. Then capacitance reduces to

C = 2πo

ln (d−a2)(d−a1)a1a2

≈ 2πoln

d2

a1a2

= πoln

d√ a1a2

Defining the average radius as a =

√ a1a2. The capacitance is then

C = πo

ln

d√ a1a2

= πoln

da

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11/65

For C = 1.2 × 10−11 F/m and d = 0.5cm the radius is

1.2 × 10−11 = πoln

.005a

→ a = .49mmThe diameter is .98 mm. Now for d = 1.5cm

1.2

×10−11 =

πo

ln .015

a → a = 1.5mm

The diameter is 3 mm. Lastly for d = 5.0cm

1.2 × 10−11 = πoln

.050a

→ a = 4.9mmThe diameter is 9.8 mm.


Calculate the attractive force between conductors in the parallel plate capacitor (Problem 1.6a) and parallel cylindercapacitor (Problem 1.7) for

(a) fixed charges on each conductor

For parallel plate capacitor, the work is

W = 1

2CV 2 (2.2)

From the last problem set, C = Aod

, and V = QdAo

. Therefore

W = 1

2

Aod

Qd

Ao

2=

1

2

Q2d

Ao

Force is given by

F = −

∂W

∂l

Q

+

∂W

∂l

V

(2.3)

Because the charge is constant ∂Q∂l

= 0, then the force is

F = −∂W ∂l

= − ∂ ∂l

1

2

Q2d

Ao= −1

2

Q2

Ao

For parallel cylinder capacitor, the work is given by equation 2, with C = πo

ln da

−1and V = Q

πoln

da

. The work

is then

W = 1

2πo

ln

d

a

−1 Q

πoln

d

a

2=

1

2

Q2

πoln

d

a

Because the charge is constant, the force becomes from equation 3

F = − ∂ ∂l

1

2

Q2

πoln

d

a

= −1

2

Q2

πod

(b) fixed potential difference between conductors.

For parallel capacitors, the charge in terms of potential is

V = Qd

Ao→ Q = AoV

d

The work becomes

W = 1

2

Q2d

Ao→ W = 1

2

oAV 2

d

8


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From equation 3, the force is

F =

∂W

∂l

V

= −12

oAV 2

d2

For parallel cylinder capacitor, the charge in terms of potential is

V = Qπo ln

da→ Q = πoV ln d

a

The work becomes

W = 1

2

Q2

πoln

d

a

→ W = 1

2

πoV 2

ln

da

From equation 3, the force is

F =

∂W

∂l

V

= ∂

∂l

1

2

πoV 2

ln

da

= −12

πoV 2

d

ln

da

2


A point charge q is brought to a position a distance d away from infinite plane conductor held at zero potential.Using the method of images find:

(a) the surface-charge density induced on the plane, and plot it.

From the method of images

Φ( x) = q/4πo|x − y| +

q /4πo|x − y| (2.4)

Because it is an infinite plane conductor, by symmetry of problem y = d and y = − d when defining the capacitor onthe xy plane. The charge is then also q = −q . The potential becomes

Φ(x) = q/4πo

|x

− d

|

− q /4πo

|x + d

|Defining the potential in cylindrical coordinates

Φ(ρ,θ,z ) = q/4πo

(z − d)2 − ρ21/2 − q/4πo

(z + d)2 + ρ21/2

The surface-charge density induced on the plane is when z = 0

σ = −o ∂ ∂z

Φ

z=0

= ∂

∂z

−q4π

(z − d)2 − ρ21/2 −−q4π

(z + d)2 + ρ21/2

z=0

=

−

q

4π 2(z − d)−1

2(z − d)

2

+ ρ23/2 −

2(z + d)−12

(z + d)2

+ ρ23/2

z=0

= − q 4π

2d

(d2 + ρ2)3/2

= − q

2πd21

1 +

ρd

23/2See figure 1.

(b) the force between the plane and the charge by using Coulomb’s law for the force between the charge and itsimage.

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Figure 2.1: s(r) = −1/(1 + r2)3/2

The force between the charge and the image is

F = q 1q 24πo

x1 − x2|x1 − x2|3

The separation x1 − x2 = −2dẑ . The charges are q 1 = −q 2 = q . The force becomes

F = q (−q )

4πo

−2dẑ | − 2d|

3 =

q 2

4πo

ẑ

(2d)2

= q 2ẑ

16πod2

(c) the total force acting on the plane by integrating σ2/2o over the whole plane.

The force is

dF = σ2

2oda → dF = 1

2o

q

2πo

11 +

ρd

23/2

2

da

= q 2

8π2od4

1 +

ρd

23 daIntegrating over the entire plane

F = q 2

8π2od4 2π0

∞0

11 +

ρd

23 ρdρdθ= 2π

q 2

8π2od4

∞0

11 +

ρd

23 ρdρ

= q 2

4πod4−d6

4(ρ2 + d2)2

∞

0

= q 2

16πod2

(d) the work necessary to remove the charge q from its position to infinity.

From part b, the force is

F = q 2ẑ

16πoz 2

So the work is the amount of force to move a particle a distance, then the work is

W =

∞d

F · dẑ = ∞

d

q 2dz

16πoz 2 =

q 2

16πo

−1z

∞

d

= q 2

16πod

(e) the potential energy between the charge q and its image [compare the answer to part d and discuss]

10


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The work is given by

W = 1

2

1

4πo

i,j,i=j

q iq j|xi − xj |

For this system,

W = 1

2

1

4πo −q

2

| − 2d| + q 2

|2d|

= − q 2

8πod

The difference between part d and e is that in d the image charge is being moved, and in e the image charge isstationary.

(f ) find the answer to part d in electron volts for an electron originally on angstrom from the surface.

In part d,

W = q 2

16πod

Then for this case

e · 1.60 × 10−1916π (8.85 × 10−12) 10−10 = 3.60eV


Using the method of images, discuss the problem of a point charge q inside a hollow, grounded, conducting sphere of inner radius a. Find

(a) the potential inside the sphere;

The potential is

Φ( x) = 1

4πo

q

|x − y| + q

|x − y|

Now the boundary condition

Φ(x = a) = 0 = q/4πo

a| −

n̂ + y/an̂

| +

q /4πoy

| −n̂ + a/yn̂

|This is identical to the charge on the outside of the sphere, because the magnitude is the concern. Therefore q = −q a

y

and y = a2

y . Then the potential is

Φ(x) = 1

4πo

q

|x − y| + q

|x − y|

(b) the induced surface-charge density;

The surface-charge density is given by

σ = o∂ Φ

∂x

x=a

because it is opposite the normal to the inside. So

σ = ∂

∂x

oq/4πo

(x2 + y2 − 2xy cos γ )1/2 +

oq /4πo

(x2 + y2 − 2xy cos γ )1/2

= 1

4π

q (−1/2)(2x − 2y cos γ )(x2 + y2 − 2xy cos γ )3/2

+ q (−1/2)(2x − 2y cos γ )(x2 + y2 − 2xy cos γ )3/2

= q

4πa

y2 − a2(y2 + a2 − 2ay cos γ )3/2

11


15/65

(c) the magnitude and direction of the force acting on q .

The separation between the charge and its image y − y, so the magnitude is

| F | = 14πo

qq

(y − y)2 = 1

4πo

q 2 ay

(y − a2y

)2

=

1

4πo

q 2ay

(a2 − y2)The direction is radially towards the image charge, because it is attractive.

(d) Is there any change in the solution if the sphere is kept at a fixed potential V ? If the sphere has a total chargeQ on its inner and outer surfaces?

Because this is inside of a sphere, there is no change if there is a fixed p otential or a charge on the sphere. This isdue to spherical symmetry. The only change will be a constant, V , add to the potential

Φ( x) = 1

4πo

q

|x − y| + q

|x − y|

+ V

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Chapter 3

Problem Set 3


A two-dimensional potential is defined by two straight parallel line charges separated by a distance R with equal andopposite linear charge densities λ and −λ,(a) Show by direct construction that the surface of constant potential V is a circular (circle in the transverse

dimensions) and find the coordinates of the axis of the cylinder and its radius in terms of R, λ, and V .

The potential for a line charge is

Φ(r) = λ

2πoln

r

r = V (3.1)

Solving for r ,

r

r

2= e4πoV/λ . But from the geometry r 2 = (r − R)2. Now

r2 = r2e4πoV/λ → (r − R)2 = r2e4πoV/λ

From this

rmin = R

1 + e2πoV/λ

rmax = R

1 − e2πoV/λ

This yields a radius of radius = R2 sinh(2πoV/λ)

. Now the offset will be

rcenter = − R

e2πoV/λ − 1The circle becomes with the axis of the cylinder in the ẑ direction

r +

R

e2πoV/λ − 1

2=

R

2sinh(2πoV /λ)

2

(b) Use the results of part a to show that the capacitance per unit length C of two right-circular conductors, withradii a and b, separated by a distance d > a + b, is

C = 2πo

cosh−1

d2−a2−b22ab

(3.2)

In this case define d = R + d1 + d2, where

d1 = R

eV 1/λ; d2 =

R

e−V 2/λ

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where V 1 = 4πoV a and V 2 = 4πoV b. From the previous part the radii can be defined as

a = ReV 1/λ

eV 1/λ − 1 ; b = Re−V 1/λ

e−V 1/λ − 1Then

d2

− a2

− b2

=

R2 e(V 1−V 2)/λ + 1

(eV 1/λ − 1) (e−V 2/λ − 1)Then

d2 − a2 − b22ab

=

eV 1/λ − 1

e−V 2/λ − 1

2R2eV 1/λe−V 2/λ

×R2

e(V 1−V 2)/λ + 1

(eV 1/λ − 1) (e−V 2/λ − 1)

= e(V 1−V 2)/2λ + e−(V 1−V 2)/2λ

2 = cosh

V 1 − V 2

2λ

Then the potential per length becomes

d2 − a2 − b22ab

= cosh

V 1 − V 2

2λ

⇒ 2cosh−1

d2 − a2 − b2

2ab

=

V 1 − V 2λ

The capacitance per unit length becomes

C = λV 1 − V 2 =

1

2cosh−1

d2−a2−b22ab

=

2πo

2cosh−1

d2−a2−b22ab


A large parallel plate capacitor is made up of two plane conducting sheets with separation D, one of which has a smallhemispherical boss of radius a on its inner surface (D a). The conductor with the boss is kept at zero potential,and the other is at a potential such that far from the boss the electric field between the plates is E o.

(a) Calculate the surface-charge densities at an arbitrary on the plane and on the boss, sketch their behavior as afunction of distance (or angle).

This is similar to a conducting sphere in an electric field, which from class is

Φ(r, θ) = −E o

r − a3

r2

cos θ

For 0 ≤ θ ≤ π/2, the potential can be written in Cartesian coordinates

Φ(x,y,z ) = −E oz + E oa3z

(x2 + y2 + z 2)3/2

The plate is

σ(x, y) = oE z = −o ∂ Φ∂z

z=0

= oE o

1 − a3

(x2 + y2)3/2

For the case of the boss,

σ(r, θ) = oE r = −o ∂ Φ∂r

r=a

= 3oE o cos θ

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Figure 3.1: σ/oE o = 1 − 1/x3

(b) Show that the total charge on the boss has the magnitude 3πoE oa2.

For the total charge on the boss is found from the integral over the area.

Q = 2π0 π/20

3oE o cos θa2 sin θdθdφ

=

2π0

1

2oE oa

2dφ = 3πoE oa2

(c) If, instead of the other conducting sheet at a different potential, a point charge q is placed directly above thehemispherical boss at a distance d from its center , show that the charge induced on the boss is

q = −q

1 − d2 − a2

d√

d2 + a2

(3.3)

The potential is

Φ(r) = 1

4πo q

|r − d| − qa/d

|r − a2/d d| − q

|r + d| + qa/d

|r + a2/d d|

Now in terms of r and θ

Φ(r, θ) = 1

4πo

q √

r2 + d2 − 2rd cos θ− qa/d

r2 + (a2/d)2 − 2r(a2/d)cos θ − q √

r2 + d2 + 2rd cos θ

+ qa/d

r2 + (a2/d)2 − 2r(a2/d)cos θ

Now the surface-charge density needs to be found

σ = −o ∂ Φ∂r

r=a

= q

4π

d2 − a2a2

1

(a2 + d2 − 2ad cos θ)3/2

− 1(a2 + d2 + 2ad cos θ)3/2

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So the induced charge on the boss is

q = π/20

q

4π

d2 − a2a2

1

(a2 + d2 − 2ad cos θ)3/2 − 1

(a2 + d2 + 2ad cos θ)3/2

2πa2 sin θdθ

= −qa2 d2 − a2

2a

1

ad

1

d − a − 1√ a2 + d2

+ 1

a + d − 1√

a2 + d2

= −q

1 −

d2 − a2d√

a2 + d2


(a) Two halves of a long hollow cylinder of inner radius b are separated by a small lengthwise gap on each side, andare kept at different potentials V 1 and V 2. Show that the potential inside is given by

Φ(ρ, φ) = V 1 + V 2

2 +

V 1 − V 2π

tan−1

2bρ

b2 − ρ2 cos φ

(3.4)

The starting point of this is Equation 2.71. In this case, the origin is located inside the cylinder, so bn = 0 to prevent

divergence at the origin. Equation 2.71 b ecomes

Φ(ρ, φ) = a0 +∞

n=1

anρn sin(nφ + αn)

Here Φ(ρ, θ) = Φ(ρ, −θ), so αn = 0. For this case, a0 3π/2−π/2

Φ(b, φ) = 2πa0 = πV 1 + πV 2 → a0 = V 1 + V 22

Now for the remaining terms in the sum

3π/2−π/2

cos nθ cos mθdθ = δ mnπ

Then for Φ 3π/2−π/2

V (θ)cos mθdθ =

n

anbn

3π/2−π/2

cos nθ cos mθdθ = anbnδ mnπ

The coefficients become

an = V 1bnπ

π/2−π/2

cos(nθ)dθ − V 2bnπ

3π/2π/2

cos(nθ)dθ

= 2(V 1 − V 2)

bn

π

sin(nπ/2) = 2(V 1 − V 2)

nπbm

(

−1)

n+32

Because of Φ’s symmetry only odd values for n are allowed. The potential becomes

Φ(ρ, θ) = −i 2(V 1 − V 2)π

n

ρb

n inn

cos(nθ)

= −i 2(V 1 − V 2)π

iIm

n odd

Z n

n

where Z = i(ρ/b)eiθ

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n odd

Zn

n = 1

2 ln 1+Z

1−Z1. From I m ln(A + iB) = tan−1(B/A), then

Im

ln

1 + Z

1 − Z

= tan−1

2ρb cos θ

b2 − ρ2

− i 2(V 1 − V 2)π

iIm

n oddZ n

n

=

2(V 1 − V 2)π

tan−1

2ρb cos θ

b2

−ρ2

Combining the two parts of the potential to find

Φ(ρ, θ) = V 1 + V 2

2 +

2(V 1 − V 2)π

tan−1

2ρb cos θ

b2 − ρ2

(b) Calculate the surface-charge density on each half of the cylinder

The surface-charge density

σ = −o ∂ Φ∂ρ

ρ=b

= −2o V 1 − V 2π

b cos θ b2 + ρ2

b4

−2b2ρ2 + ρ4 + 4ρ2b2 cos2 θ

= −o V 1 − V 2bπ cos θ


A hollow cube has conducting walls defined by six planes x = 0, y = 0, z = 0, and x = a, y = a, z = a. The wallsz = 0 and z = a are held at a constant potential V . The other four sides are at zero potential.

(a) Find the potential Φ(x,y,z ) at any point inside the cube.

From class

Φ(x,y,z )top =∞

m=1∞

n=1Anm sin

nπx

a sin

mπy

a sinh(γ mnz )

where γ =

nπa

2+

mπa

2. Following Equation 2.58

Anm = 4

a2 sinh(γ nma)

a0

dx

a0

dyV sinnπx

a

sinmπy

a

=

16V

π2nm sinh(γ nma), where m,n are both odd

By symmetry the bottom has to be the same except for z → a − z . Therefore the bottom is

Φ(x,y,z )bottom

oddn,m=1

Anm sinnπx

a

sinmπy

a

sinh(γ mn(a − z ))

The total potential is the sum of both top and bottom potential

Φtop + Φbottom =odd

n,m=1

Anm sin

nπxa

sin

mπya

[sinh (γ mnz ) + sinh (γ mn(a − z ))]

(b) Evaluate the potential at the center of the cube numerically, accurate to three significant figures. How manyterms in the series is it necessary to keep in order to attain this accuracy? Compare your numerical resultswith the average value of the potential on the walls. See Problem 2.28.

1Jackson p74-5

17


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To have three digits of accuracy, the first four terms need to be kept, which is

Φa

2, a

2, a

2

=

16V

π2 sinh(√

2π)(1)(1)(2sinh(π

√ 2/2))

+ 16V

3π2 sinh(√

10π)(1)(−1)(2 sinh(π

√ 10/2))

+ 16V

3π2 sinh(√ 10π) (−1)(1)(2 sinh(π√ 10/2))

+ 16V

9π2 sinh(3√

2π)(1)(1)(2 sinh(3π

√ 2/2))

≈ .3329V = .333V

From Problem 2.28 the potential should be 26

V = .333̄V

(c) Find the surface-charge density on the surface z = a.

The surface-charge density in the cube is given by

σ = o∂ Φ

∂z

z=a

=

16oV

π2

oddn,m=1

Anm sinnπx

a

sinmπy

a

[cosh (γ mna) − 1]

= 16oV

π2

oddn,m=1

1

nm sinnπx

a

sinmπy

a

cosh(γ mna) − 1sinh(γ mna)

= 16oV

π2

oddn,m=1

1

nm sinnπx

a

sinmπy

a

tanh

γ mna2

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A thin, flat, conducting, circular disc of radius R is located is located in the x − y plane with its center at the origin,and is maintained at a fixed potential V . With the information that the change density on a disc at fixed potentialis proportional to (R2 − ρ2)−1/2, where ρ is the distance out from the center of the disc,(a) show that for r > R the potential is

Φ(r,θ,φ) = 2V

π

R

r

∞l=0

(−1)l2l + 1

R

r

2l p2l(cos θ) (4.1)

The surface charge is given by

σ(ρ) = λ R2 − ρ2

The volume charge is found by consider a shell of radius r

dq = σ(r)2πrdr =

2π0

dφ

1−1

d(cos θ)r2drf (r)δ (cos θ)

σ(r)2πrdr = 2πr2f (r)dr

The volume charge density is ρ(r) = σ(r)

r δ (cos θ). The potential at the origin is

V = Φ(0) = 1

4πo

R0

λ2πrdr

r√

R2 − r2 = λ

2osin−1

rR

R

0

= λπ

4o

The potential at any point becomes

Φ(x) = 1

4πo

4oV

π

∞l=0

lm=−l

Y lm(θ, φ) 1

rl+1

v

rlY ∗lm(θ, φ)

δ (cos θ)r2drdΩ

r√

R2 − r2

Solving the integral

Φ(x) = V

π2

∞

l=0

P l(cos θ) 2π

rl+1

R

0 1

−1rl+1P l(cos θ

)δ (cos θ)drd(cos θ)

√ R2 − r2

= 2V

π

∞l=0

P l(cos θ)P (0)

rl+1

R0

rl+1dr√ R2 − r2

= 2V

π

∞n=0

P 2n(cos θ)(−1)n(2n + 1)!!

2nn!

R2n+1n!2n

(2n + 1)!!r2n+1

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Then

Φ( x) = 2V

π

∞n=0

(−1)n2n + 1

R

r

2n+1P 2n(cos θ)

(b) find the potential for r < R.

To find the potential for inside the circular radius, use the boundary condition when r = R. The potential inside andoutside has to be equal, therefore

AlRl =

2V

π

(−1)n2n + 1

= BlR−l−1

where l = 2n. Then the equation for inside becomes

Φ = 2V

π

∞n=0

(−1)n2n + 1

rR

2nP 2n(cos θ)

(c) What is the capacitance of the disc?

The capacitance is given by C = Q/V . From part (a) V = λπ4o

. The charge is

Q =

R0

λ√ R2 − r2 2πrdr = −2πλ

R2 − r2

R0

= 2πλR

The capacitance becomes

C = Q

V =

2πλRλπ4o

→ C = 8oR


Two point charges q and −q are located on the z axis at z = +a and z = −a, respectively.(a) Find the electrostatic potential as an expansion in spherical harmonics and powers of r for both r > a and r < a.

From Equation 3.38

Φ(x) = q

4πo

1

| x − a| − 1

|x + a|

= q

4πo

1

r>

∞l=0

r<r>

lP l(cos θ) −

r<r>

lP l(cos(π − θ))

From equation Equation 3.57

Φ = q

4πo

1

r>

∞l=0

r<r>

l 4π

2l + 1 (Y l0(θ, φ) − Y l0(π − θ, φ))

From parity of the spherical harmonics

Φ(x) = q

4πo

1

r>

∞l=0

r<r>

l 4π

2l + 1

1 + (−1)l+1

Y l0(θ, φ)

where r< is the smaller of mod[x, a] and r> is the larger of mod[x, a].

(b) Keeping the product qa ≡ p/2 constant, take the limit of a → 0 and find the potential for r = 0. This is bydefinition a dipole along the z axis and its potential.

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So for a < r

Φ = q

4πo

1

r

∞l=0

ar

l 4π2l + 1

1 + (−1)l+1

Y l0(θ, φ)

In the case a → 0 only the first power of a are examined. The potential is then

Φ = q 4πo

1r

ar

4π2 + 1

(1 + 1) Y 10 (θ, φ)

= q

4πo

ar2

4π3

(2)

3

4π

cos θ

= p cos θ

4πor2

(c) Suppose now that the dipole of part b is surrounded by a grounded spherical shell of radius b concentric withthe origin. By linear superposition find the potential everywhere inside the shell.

The potential is then Φ = Φd + Φs. Here only the Al terms because we are inside a sphere. Then

Φ = p cos θ

4πor2 +

∞l=0

AlrlP l(cos θ)

At the boundary condition Φ(b) = 0, then

0 = p cos θ

4πob2 +

∞l=0

AlblP l(cos θ)

The only term needed is l = 1 because the dipole, so

A1b1 cos θ = − p cos θ

4πob2

A1 = − p4πob3

The potential then becomes

Φ = p cos θ

4πo

1

r2 − fracrb3


For the cylinder in Problem 3.9 the cylindrical surface is made of two equal half-cylinder, one at potential V and theother at potential −V , so that

V (φ, z ) =

V for − π/2 < φ < π/2−V for π/2 < φ < 3π/2 (4.2)

(a) Find the potential inside the cylinder

From class

Φ =∞n

∞ν=0

(Anν sin νφ + Bnν cos νφ) I νnπρ

L

sinnπz

L

Fourier analyst can yield the coefficients. For Anν

Anν = 2

πL

1

I ν

L0

2π0

V (φ, z )sinnπz

L

sin νφdφdz

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But this has two intervals from −π/2 to π/2 and π/2 to 3π/2. In these intervals cosine is symmetric so Anν = 0.Now for Bnν

Bnν = 2

πL

1

I ν

L0

2π0

V (φ, z )sinnπz

L

sin νφdφdz

over the two ranges, only the odd terms survive for ν . For the z component L

0 sin(nπz/L) dz = 2L

(2l+1)π, because

only odd terms survive from n, also. So Bnν becomes

Bnν =

2L

(2l + 1)π

4(−1)k2k + 1

2V

πLI 2k+1

= 16V (−1)k

(2k + 1)(2l + 1)π2I 2k+1

(b) Assuming L b, consider the potential at z = L/2 as a function of ρ and φ and compare it with two-dimensionalProblem 2.13.

From Jackson

I 2k+1

(2l + 1)πρ

L

≈ 1

Γ(2k + 2)

(2l + 1)πρ

2L

k+1

and for small angles sin(2l+1)πL

≈ (−1)l. ThenΦ =

∞l,k=0

16(−1)k+lV π2(2l + 1)(2k + 1)

ρb

2k+1cos((2k + 1)φ)

Using the identity

π

4 = tan−1(1) =

∞l=0

(−1)l2l + 1

(1)2l+1


Φ = 4V

π

∞

k=0

(−1)k2k + 1

ρ

b2k+1

cos((2k + 1)φ)

From problem 2.13

∞k=0

(−1)k2k + 1

ρb

2k+1cos((2k + 1)φ) =

1

2 tan−1

2bρ cos φ

b2 − ρ2


Φ = 2V

π tan−1

2bρ cos φ

b2 − ρ2

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The Dirichlet Green function for the unbounded space between the planes as z = 0 and z = L allows discussion of apoint charge or a distribution of charge between parallel conducting planes held at zero potential.

(a) Using cylindrical coordinates show that one form of the Green function is

G(x, x) = 4L

∞n=1

∞m=−∞

eim(φ−φ) sin

nπz L

sin

nπz L

LI m

nπL

ρ<

K m

nπL

ρ>

(5.1)

We know that

∇2G(x, x) = −4πδ (ρ − ρ)

ρ δ (φ − φ)δ (z − z ) (5.2)

Then expanding the δ -functions

∇2G(x, x) = − 4L

∞m=−∞

∞n=1

e−imφ

eimφ sin

nπz

L

sinnπz

L

δ (ρ − ρ)ρ

=∞

m=−∞

∞n=1

− 4

Le−imφ

sin

nπz

L

δ (ρ − ρ)

ρ

eimφ sin

nπz L

(5.3)

Now expanding the Green function

∇2G(x, x) = ∇2∞

m=−∞

∞n=1

eimφ sinnπz

L

Ψ

where Ψ is the eigenfunction. Expanding the Laplacian

∇2G(x, x) =∞

m=−∞

∞n=1

1

ρ

∂

∂ρρ

∂

∂ρ − m

2

ρ2 − n

2π2

L2

Ψeimφ sin

nπz L

(5.4)

Then equating equation 3 and equation 41

ρ

∂

∂ρρ

∂

∂ρ − m

2

ρ2 − n

2π2

L2

Ψ

1

e−imφsin

nπz

L

= − 4L

δ (ρ − ρ)ρ

Then define gmn = Ψ

e−imφ

sin nπz

L . From Equation 3.98 and 3.99 gmn = C I m(kρ) where k = nπ/L. Now

to find C

∂gmn∂ρ

ρ+

− ∂gmn∂ρ

ρ−

= −4

ρL ⇒Eq3.147 −4

ρL = −Ck 1

kρ → C = 4

L

Now combining the above constant with gmn and equation 3 to find

G(x, x) = 4

L

∞m=−∞

∞n=1

eimφe−imφ

sin

nπz

L

sinnπz

L

I mnπ

L ρ<

K mnπ

L ρ>

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(b) Show that an alternative form of the Green function is

G(x, x) = 2∞

m=−∞

∞0

dk eim(φ−φ)J m(kρ)J m(kρ

)sinh(kz )]

sinh(kL) (5.5)

Expanding equation 2 in φ and ρ

∇2G(x, x) = −2 ∞

m=−∞

∞0

e−imφ

eimφJ m(kρ)J m(kρ)δ (z − z )kdk

=∞

m=−∞

∞0

−2ke−imφJ m(kρ)δ (z − z )

eimφJ m(kρ)dk (5.6)

Now expanding the Green function

∇2G(x, x) = ∇2∞

m=−∞

∞0

eimφJ m(kρ)Ψdk

where Ψ is the eigenfunction. Expanding the Laplacian

∇2G(x, x) =

∞

m=−∞

∞

0 ∂

2

∂z 2

− m2

ρ2

+ 1

ρ

∂

∂ρ

ρ ∂

∂ρ eimφJ m(kρ)Ψdk

=∞

m=−∞

∞0

∂ 2

∂z 2 − k2

ΨeimφJ m(kρ)dk (5.7)

Equating equation 6 and 7, to find ∂ 2

∂z 2 − k2

Ψ

1

e−imφJ m(kρ) = −2kδ (z − z )

Then gm,k = Ψ/e−imφJ m(kρ), but the expansion of the δ -function in z is gm,k = C sinh(kz )). Now

finding C

∂gm,k∂z

z+

− ∂gm,k∂z

z−

= −2k

−kC [sinh(kz )cosh(k(L − z )) + cosh(kz ) sinh(k(L − z ))] = −2k → −kC sinh(kL) = −2k → C = 2sinh(kL)

Now combining the above equation with equation 6 and 7

G(x, x) = 2∞

m=−∞

∞0

eimφe−imφ

J m(kρ)J m(kρ)

sinh(kz ))sinh(kL)


A localized distribution of charge has a charge density

ρ(r) = 1

64πr2e−r sin2 θ (5.8)

(a) Make a multipole expansion of the potential due to this charge density and determine all the nonvanishingmultipole moments. Write down the potential at large distance as a finite expansion in Legendre polynomials.

For Equation 4.1 to be solved first Equations 4.4-4.6 have to be solved. First notice that the charge density isindependent of φ, therefore m = 0. Now notice sin2 θ = 1−cos2 θ, because this is power 2, then l ≤ 2. Now beginningwith Y 20

Y 20 =

5

4π

3

2 cos2 θ − 1

2

→ Y 20 =

5

4π

−3

2 cos2 θ + 1

→ −2

3

4π

5 Y 20 = sin

2 θ − 23

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28/65

Now the next term has no power of θ , so the next term would be from Y 00. Therefore

sin2 θ = 2

3

√ 4πY 00 − 2

3

4π

5 Y 20

Now using Equation 4.3

1√ 4π

V

23√ 4π 164π r4e−r sin θdrdθdφ = 14π1

2

5

4π

V

−23

4π

5

1

64πr6e−r sin3 θdrdθdφ =

−304π

Back to Equation 4.1

Φ(x) = 1

4πo

4π

1

1

4π

P 0r − 4π

5

30

4π

P 2(cos θ)

r3

→ Φ(x) = 1

4πo

P 0r − 6P 2(cos θ)

r3

(b) Determine the potential explicitly at any point in space, and show that near the origin, correct to r2 inclusive,

Φ(r) 14πo

1

4 − r

2

120P 2(cos θ) (5.9)

We know from Equation 4.2

Φ(x) = 1

o

l,m

1

2l + 1

Y ∗lm(θ

, φ)rlρ(x)d3x

Y lm(θ, φ)

rl+1

Looking ahead to the deserved solution and combined with the previous part only l = 0, 2 and m = 0 will be considers.Then

Φ(x) = 1

o

2

3

√ 4πY 00

∞0

1

64πr3e−rdr − 2

3

4π

5 Y 20

r2

5

∞0

1

64πre−rdr

= 1

o

2

3

√ 4π

P 0√ 4π

1

64π6 − 2

3

4π

5

5

4π

r2P 2(cos θ)

5

1

64π1

= 1

o

P 016π

− r2480π

P 2(cos θ)

= 1

4πo

1

4 − r

2

120P 2(cos θ)

(c) If there exists at the origin a nucleus with a quadrupole moment Q = 10−28 m2, determine the magnitude of the interaction energy, assuming that the unit of charge in ρ(r) above is the electronic charge and the unit of length is the hydrogen Bohr radius ao = 4πo/me

2 = 0.529 × 10−10 m. Express your answer as a frequency bydividing by Planck’s constant h.

The quadrupole interaction energy is given in Equation 4.24

W = −16i j

Qij∂E j∂xi

(0) (5.10)

But as it has been shown earlier that m = 0 so the only terms that remain are i = j , but the trace has to be 0, whichis

W = eQ

6

1

2

∂E x∂x

+ 1

2

∂E y∂y

− ∂E z∂z

= eQ

6

1

2∇ · E − 3

2

∂E z∂z

=

eQ

6

1

2

ρ

o− 3

2

∂E z∂z

25


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At the origin ρ(0) =, so the first term drops out, so the energy is then W = eQ4

∂ 2Φ∂z2

. Now the potential has to betransferred into Cartesian coordinates. So for the quadrupole term

Φ(x) = 1

480πor2P 2 (cos θ) → Φ(x) = 1

960πo

2z 2 − x2 − y2

from Merzbacher Quantum Mechanics. So ∂ 2Φ

∂z2 = 1

240πo, then the quadrupole term becomes

W = e2Q

960πoa3o

Then the frequency is

W

h =

e2Q

960πoha3o= 0.98 MHz

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Chapter 6

Problem Set 6


A point charge q is located in free space a distance d from the center of a dielectric sphere of radius a (a < d) anddielectric constant /o.

(a) Find the potential at all points in space as an expansion in spherical harmonics.

Inside the sphere ∇ · D = 0, because there is no charge inside

Φ(r, θ) =∞

l=0

AlrlP l(cos θ) r. Then the outside region will be

Φ(r, θ) =∞

l=0

Blr

−(l+1)P l(cos θ) + 1

4πo

q

|x − x|

Using the typical spherical expansion for 1/|x − x, the expression for all space is

Φ(r, θ) =∞l=0 AlrLP l(cos θ) r


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The other boundary condition is for tangential component

∂ Φ

∂θ

r=a−

= ∂ Φ

∂θ

r=a+

Alal = Bla

−(l+1) + q

4πo

al

dl+1 (6.4)

Then the coefficients can easily be found from equation 3 and 4

Al = 1

/o + (l + 1)/l

2l + 1

l

q

4πodl+1 (6.5)

Bl = 1

/o + (l + 1)/l

1 −

o

qa2l+1

4πodl+1 (6.6)

Combining equations 5 and 6 into equation 1 yields the potential at every point in space.

(b) Calculate the rectangular components of the electric field near the center of the sphere.

Expanding the potential inside the sphere

Φ(r, θ) = A1rP 1(cos θ) + A2r2P 2(cos θ) + . . .

From Quantum Mechanics, rP 1(cos θ) = z and r2P 2(cos θ) = z

2 − x2 + y2, then the potential becomes

Φ(x,y,z ) = q

4πo 3z

(/o + 2)d2 +

5 z 2 − x2 − y2

2(2/o + 3)d3 + . . .

So then E i = −

Φdxi. Then the component in Cartesian space

E x = q

4πod2

5

2/o + 3

x

d + . . .

E y = q

4πod2

5

2/o + 3

y

d + . . .

E z = − q 4πod2

3

/o + 2 +

5

2/o + 3

z

d + . . .

(c) Verify that, in the limit /o → ∞, your result is the same as that for the conducting sphere.Examining equation 2, the equation would blow-up if Al = 0, therefore Al = 0. This is the same for the conductor.Then from that and equation 4,

Bl = − q 4πo

a2l+1dl+1

which is the same as a conduction sphere.


Two concentric conducting spheres of inner and outer radii a and b , respectively, carry charges ±Q. The emptyspace between the spheres is half-filled by a hemispherical shell of dielectric (of dielectric constant /o), as shown inthe figure.1

(a) Find the electric field everywhere between the spheres.

Let’s assume that E ∝ rr3

. Using Gauss’s Law for a sphere between the two spheres:

D · da = Q2πr2 (oE (r) − E (r)) = Q

Q

2π(o + )

r

r3 = E (r)

Now checking the conditions on the solution. The curl of the electric field is obviously 0. The divergent of the fielddoes yield the charge density. By the uniqueness theorem, this is the correct solution.

1Jackson. pg 174

28


32/65

(b) Calculate the surface-charge distribution of the inner sphere.

From Equation 4.40, in particular n · ((D2 − D1) = σfree. The half without the dielectric is

σfree = r · D(a) = r̂ · ra2

Qo2π(o + )

= Qo

2πa2(o + )

Now for part with the dielectric is directly

σfree = Q

2πa2(o + )

(c) Calculate the polarization-charge density induced on the surface of the dielectric at r = a

The electric polarization is given by P = ( − o)E. Because in the free space = o, the electric polarization for thefree space is P = 0. Therefore σpol = 0 for the free space half. Now, the polarization density is σpol = −n̂·(P1−P2).For the dielectric part

σpol = −r̂ · P(a) = −r̂ r̂a2

( − o)Q2π(o + )

σpol = − ( − o)Q2πa2(o + )

6.3 Problem 3: Distressed Simple Cubic

Repeat the analysis of section 4.5 for a cubic crystal which is subjected to a stress such that the lattice separation iselongated along one edge of the cube (the x axis) and contracted along the other cube edges (the y and z axes) bythe amounts

ax = a(1 + δ ) (6.7)

ay,z = a

1 − 1

2δ

(6.8)

where a is the unstressed cubic lattice spacing. Find the susceptibility for two cases, considering that the molecularpolarizabilty γ mol is a constant scalar (Hint: Calculate Enear at a particular dipole by considering only the 6 nearestdipoles in the cubic lattice):

(a) An electric field applied parallel to the x axis

The polarization is given on page 161 in Jackson as P = N pmol and also pmol = oγ mol(E + Ei) From theprevious page Equation 4.63, Ei =

P

3 + Enear. All that remains is to find Enear. From Equation 4.64:

Enear = 3n̂(n̂ · p) − p

r3 (6.9)

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Considering the six nearest neighbors, which occur from the shape(from class notes):

E i = 3n̂(n̂̇ pi) − pi

a3i

E x = 4 px

a3x− 2 px

a3y− 2 px

a3z

E y = 4 py

a3y − 2 py

a3x − 2 py

a3z

E z = 4 pz

a3z− 2 pz

a3x− 2 pz

a3y

Now using the values for the expansion of the nearest neighbors and the fact that δ


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Two long, cylindrical conducting surfaces of radii a and b are lowered vertically into a liquid dielectric. If the liquidrises an average height h between the electrodes when a potential difference V is established between them, showthat the susceptibility of the liquid is

χe =b2 − a

2 ρgh ln(b/a)oV 2 (7.1)

where ρ is the density of the liquid, g is the acceleration due to gravity, and the susceptibility of air is neglected.

The first thing to find is the electric field between the cylinders. There is no charge between the cylinders, so ∇2Φ = 0.From the symmetry of the problem, cylindrical coordinates are the proper choice, Further there is radial symmetry,so there is no polar coordinate. So the Laplacian is

1

r

∂

∂r

r

∂ Φ

∂r

= 0 (7.2)

The familiar solution is

Φ(r) = A ln (r/B)

Now the boundary condition that the difference of the two cylinders is V or more exact

Φ(b) − Φ(a) = A ln(b/a) = V ⇒ A = V ln(b/a)

So the electric field is

E = −∂ Φ∂r

r̂ → E = −V ln(b/a)

r̂

r

Now that the electric field is known, the electric displacement is simply D = E for inside the liquid, and for abovethe liquid, the electric displacement is D = oE. The work to raise the liquid is

W = 1

2

E · Dd3x = 1

2

ba

2πrdr

z−d

E2dz +

lz

oE2dz

where d is the length in the liquid and l is the length above. The work is then

W = 1

2

ba

2πrdr

z−d

E2dz +

lz

E2dz

=

πV 2

ln(b/a) ( (z + d) + o (l − z ))

From Equation 4.102, the force needed to left the liquid is

F z =

∂W

∂z

V

= ( − o) πV 2

ln(b/a)

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35/65

Now finding the volume of liquid that has been lifted is

b2 − a2πh. The mass is then m = b2 − a2 ρπh. Atequilibrium, the force of gravity is equal to the force of the electric field.

b2 − a2 ρπh = ( − o) πV 2

ln(b/a)

b2 − a2 ρπh

o=

(/o − 1) πV 2ln(b/a)

The susceptibility is defined as χe = /o − 1. Then

b2 − a2 ρπho

= χeπV

2

ln(b/a) ⇒ χe =

b2 − a2 ρgh ln(b/a)

oV 2

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Chapter 8

Problem Set 8


A long, right cylindrical, ideal solenoid of arbitrary cross section is created by stacking a large number of identicalcurrent-carrying loops one above the other, with N coils per unit length and each loop carrying a current I .

(a) In the approximation that the solenoidal coil is an ideal current sheet and infinitely long, use Problem 5.1 to

establish that any point outside the coil that H = 0, and that any point inside the coil the magnetic field isaxial and equal to

H = N I (8.1)

This can be solved using two ampere loops. Placing a loop outside the cylinder, it yields B = 0. Placing a secondloop through a side of the surface perpendicular to the bottom surface. So

B · dl = µoI enc

From the first loop that was outside the cylinder, the induced magnetic field outside is 0. This means that the onlyfield is inside the cylinder. The integral becomes

B · dl = µoI enc ⇒ BL = µoNIL

Using right hand rule, Bz = µoN I . In this system M = 0, then H = B/µo, therefore Bz = µoN I ⇒ H = N I . (b) For a realistic solenoid of circular cross section of radius a (Na 1), but still infinite in length, show that the

“smoothed” magnetic field just outside the solenoid (averaged axially over several turns) is not zero, but is thesame in magnitude and direction as that of a single wire on the axis carrying a current I , even if Na → ∞.Compare fields inside and out.

In a real solenoid, it is not simple current loops stacked on each other. There is a little bind in the loop to connectthe loops. This means there is a current in the vertical direction. So what is the current? This can be found by thefact the total current moves up 1/N each loop. In each loop there is I/N in the vertical direction, so over a unitlength there is N loops, so the amount of vertical current is then NI/N = I . Therefore this current in the verticaldirection

B · dl ⇒ 2πρB = µoI ⇒ Bθ = µoI

2πρ

This is the same as a wire along the z -axis. As the number of loops increase per unit length, the more idealit will become. The size of inside is many orders of magnitude large than the field outside. To be more exact,µoNI/(µoI/2πρ) → 2πρN . So the field inside is 2πρN times larger. The directions of the two are perpendicular toeach other.

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8.2 Problem 2: Semi-Infinite Cylinder

Consider a semi-infinite solenoid, which we idealize as a cylindrical current sheet of radius a and azimuthal currentK per unit length, with one end at the origin and the other end at - ∞ along the z -axis.(a) Use the Biot-Savart Law to show that the magnetic induction on the axis is

Bz = µoK 2

1 − z

(a2 + z 2)1/2

(8.2)

Let’s begin by thinking of a single current loop at the origin. The magnetic-flux density is

Bz = µoI

4π

dl × x

|x|3

Because this has to end the ẑ direction, x = a cos φρ̂ + a sin φφ̂ + z ̂z , |x|3 = a2 + z 23/2, and dl = −a sin φρ̂ +a cos φφ̂ + 0ẑ . Then

Bz = µoI

4π

1

(a2 + z 2)3/2

2π0

−az cos φ + az sin φ + a2 dφ

=

µoI

2

a2

(a2 + z 2)3/2 ẑ

Now this has to be integrated from start to finish of the cylinder remembering that I → K

Bz = µoa

2K

2 ẑ

0−∞

dz

(a2 + z 2)3/2

= µoa

2K

2 ẑ

z

(a2 + z 2)1/2

0

−∞=

µoK

2 ẑ cos θ

π

θ1

where θ is the angle that the origin makes with the edge at the end of the cylinder and cos θ = z(a2+z2)3/2

, then

Bz = µoK

2 1 − z

(a2 + z 2)3/2 (b) Use a Taylor series expansion of the field near the axis together with ∇ · B = 0 and ∇× B = 0 to show that the

magnetic induction near the axis is

Bρ ≈ µoK 4

a2ρ

(a2 + z 2)3/2 (8.3)

Bz ≈ µoK 2

1 − z

(a2 + z 2)1/2 − 3

4

a2zρ2

(a2 + z 2)5/2

(8.4)

The two conditions, ∇· B and ∇× B, lead to Laplace’s Equation ∇2Φmag = 0 and B = −∇Φmag. From electrostatics,Φmag = R(ρ)Z (z )Q(θ). The radial solutions are the Bessel and Neumann functions. Then

Φmag

= (AJ ν

(kρ) + BN ν

(kρ)) Ce−kzDeiνθ

From symmetry, ν = 0 and as ρ → 0 has to be well define B = 0. Then Φmag = C J 0(kρ)e−kz. Then the inducedfield is

B = −∇Φmag = −

ρ̂ ∂

∂ρ + ẑ

∂

∂z

Φmag

= −Ck ∂J o(x)∂x

e−kxρ̂ + CkJ o(kρ)e−kzẑ

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From Arfken’s book, ∂J o(x)∂x

= −J 1(x)1. Expanding the two terms2,

Bρ = C kJ 1(kρ)e−kz ≈ Cke−kz

kρ

2 − k

3ρ3

16

Bz = C kJ 0(kρ)e−kz ≈ Cke−kz

1 − k

2ρ2

4

Now, Bz(0, z ) = Cke−kz, this has to be the case, because it has to be continuous. It follows that ∂ nBz(0,z)∂zn =(−1)n Ckn+1e−kz. With this substitution,

Bρ ≈ Cke−kz

kρ

2

≈ ρ

2

∂Bz(0, z )

∂z

Bz ≈ Cke−kz

1 − k2ρ2

4

≈ Bz(0, z ) − ρ

2

4

∂ 2Bz(0, z )

∂z

Using Bz from part one finally

Bρ ≈ρ

2

∂

∂z

µoK

2

1 −

z

(a2 + z2)3/2

⇒ Bρ ≈

µoK

4

a2ρ

(a2 + z2)3/2

Bz ≈µoK

21 −

z

(a2

+ z2

)

3/2 −ρ2

4

∂ 2

∂z

µoK

21 −

z

(a2

+ z2

)

3/2 Bz ≈

µoK

2

1 −

z

(a2 + z2)1/2 −

3

4

a2zρ2

(a2 + z2)5/2


A cylindrical conductor of radius a has a hole of radius b bored parallel to, and centered a distance d from, thecylinder axis (d + b < a). The current density is uniform throughout the remaining metal of the cylinder and isparallel to the axis. Use Ampère’s law and principle of linear superposition to find the magnitude and the directionof the magnetic-flux density in the hole.

From the hint to use superposition, we will first find the magnetic-flux density for the cylindrical conductors. For theone with radius a applying Ampère’s law in the form of Equation 5.24

C

B · dl = µo

S

J · nda

B2πr2 = µoJπr2 → B = µo

2 Jrẑ × r̂

Now for the hole, it has to be the same except that it has a different coordinate system, that is related by a lineartransformation, directly B = µo

2 J rẑ × r̂, because this is a hole the current density is in the opposite direction,

J = −J . Next is relating r r̂ to r r̂. Defining the center of the hole on the −y-axis of the cylinder, the relationshipis r r̂ = rr̂ + dŷ. Now linear superposition can be used to find

Bfinal = B + B =

µo2

J ̂z × (rr̂ − rr̂ − dŷ)

= µo

2 Jdẑ × (−ŷ) ⇒ Bfinal = µoJd

2 x̂


A sphere of radius a carries a uniform surface-charge distribution σ. The sphere is rotated about a diameter withconstant angular velocity ω . Find the vector potential and magnetic-flux density both inside and outside the sphere.

1Arfken. MMP 5thed. pg.6732Arfken. MMP 5thed. pg.670

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The surface current is J = σv = σ( ω × r) = σaω sin θ(r − a)φ̂. The surface current is then related to l = 1 andm = 1 only, because of the sine term. The

36


40/65

vector potential is then

Aφ(r, θ) = µo4π

l

P 1l (cos θ)

l(l + 1)

σωaδ (r − a) r

l<

rl+1>P 1l (cos θ

)r2 sin2 θdrdθdφ

= −µo2 l

P 1l (cos θ)

l(l + 1)

σωa3rl<

rl+1>

π0

P 1l (cos θ)P 11 (cos θ)sin θ

dθ

= −µoσωa3

4

r<r2>

P 11 (cos θ)4

3

= µoσωa

3r<3r2>

sin θ

Now for the magnetic field, B = ∇×A. This needs to be done for the interior (a > r) and exterior (a < r). The vectorpotential interior is Aint =

1

3µoσωar sin θφ̂ . The vector potential for exterior is Aext(r, θ) =

1

3µoσωa

4 sin θ

r2φ̂ So

for the interior

Bint(r, θ) = r̂ 1

r sin θ

∂

∂θ

µoσωa3

sin2 θ

− θ̂ 1r

∂

∂r

µoσωar

2 sin θ

3

Bint(r, θ) = 2

3µoσωa cos θr̂ − sin θθ̂

Now for the exterior

Bext(r, θ) = r̂ 1

r sin θ

∂

∂θ

µoσωa

4

3r2 sin2 θ

− θ̂ 1

r

∂

∂r

µoσωa

4 sin θ

3

1

r

Bext(r, θ) = 1

3

µoσωa4

r3

2cos θr̂ + sin θθ̂

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A current distribution J(x) exists in a medium of unit relative permeability adjacent to a semi-infinite slab of materialhaving relative permeability µr and filling the half space, z 0 the magnetic induction ca be calculated by replacing the medium of permeability µr by animage current distribution, J∗, with components,

µr − 1

µr + 1

J x(x,y,−z),

µr − 1

µr + 1

J y(x,y,−z), −

µr − 1

µr + 1

J z(x,y,−z) (9.1)

The field z > 0 is

B = µo4π

(J + J∗) × (x − x

)|x − x|3 d

3x (9.2)

The field z


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From H + = H −

µrJ z(−z ) + µrJ z+ (z ) = J z−(z ) (9.6)µrJ y(−z ) − µrJ y+ (z ) = J y−(z ) (9.7)µrJ x(−z ) − µrJ x+ (z ) = J x−(z ) (9.8)

From eq 4 and 7, and eq 5 and 8

J x+ (z) = µr − 1

µr + 1J x(−z) J x− (z) =

2µr

µr + 1J x(z) (9.9)

J y+ (z) = µr − 1

µr + 1J y(−z) J y−(z) =

2µr

µr + 1J y (z) (9.10)

Now for the z components.

∇ · J∗− = 0 = ∂J ∗x−

∂x +

∂J ∗y−∂y

+ ∂J ∗z−

∂z =

2µrµr + 1

∂J x∂x

+ ∂J y

∂y

+ µr

∂

∂z (J z + J z+ ) = −

2µrµr + 1

∂J z∂z

+ µr∂

∂z (J z + J z+ )

Then solving and integrating

−µr − 1

µr + 1 ∂J z = ∂J z+ ⇒ −µr − 1

µr + 1J z(−z) = J z+ (z) (9.11)

From equations 9, 10, 11 for J∗+ matches equation 1.

(b) Show that for z < 0 the magnetic induction appears to be due to a current distribution [2µr/(µr + 1)]J in amedium of unit relative permeability.

Combining equation 6 with equation 11 to find

J ∗z−(z ) = 2µrµr + 1

J z (z ) (9.12)

Adding equation 12 with the J − parts of equations 9, 10 to find that J∗− = [2µr/(µr + 1)]J


A magnetically “hard” material is in the shape of a right circular cylinder of length L and radius a. The cylinder hasa permanent magnetization M o, uniform throughout its volume and parallel to its axis.

(a) Determine the magnetic field H and magnetic induction B at all points on the axis of cylinder, both inside andoutside.

Because there is no free current, this is a magnetostatic potential problem. This problem then can be treated as anelectrostatic problem. Because the magnetization is uniform, ρM = −∇ · M = 0. The surface ‘charge’ density is thenσM = n̂ · M = ±M o at the caps. The potential is then

ΦM = 1

4π

σM |x − x| d

3x

This is evaluated at the caps only. Then

Φ1 =

M o

4π

ρ

0

2πρdρ ρ2 + (z −L/2)2

=

M o

2

ρ2

+ (z −L/2)2a

0

= M o

2

a2 + (z −L/2)2 −

(z −L/2)2

This can easily be show for the bottom with symmetry that

Φ2 = −M o2

a2 + (z + L/2)2 −

(z + L/2)2

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The potential over all space is Φ = Φ1 + Φ2.

Φ =M o

2

a2 +

z −

L

2

2−

a2 +z +

L

2

2−

z − L2

+z + L

2

The magnetic field has two cases, one inside

H z = ∂ Φ

∂z

H z = M o

2

z + L/2

a2 + (z + L/2)2−

z − L/2 a2 + (z − L/2)2

− 2

And outside

H z = M o

2

z + L/2

a2 + (z + L/2)2−

z −L/2 a2 + (z −L/2)2

Now the induced field is Bz = µo(H z + M ). In this case, M = M o inside and M = 0 outside. Then the induced fieldboth inside and outside is

Bz =

µoM o

2 z + L/2

a2 + (z + L/2)2−

z −L/2 a2 + (z −L/2)2

(b) Plot the ratios B/µoM o and H/M o on the axis as functions of z for L/a = 5.

See last page


Show that in general a long, straight bar of uniform cross-sectional area A with uniform lengthwise magnetization M ,when placed with its flat end against as infinitely permeable flat surface, adheres with a force given approximatelyby

F µo2

AM 2 (9.13)

Relate your discussion to the electrostatic considerations in Section 1.11.

So consider this as two long straight bars with uniform cross-sectional areas A with uniform magnetization, M . Onerod runs from z = 0 to z = −L and the other from z = 0 to z = L. These two dipoles are “head” to “tail”, sothe components of the B-field cancel everywhere except at z = 0. Now, let’s introduce a small separation z o. Thepotential energy on the real rod from the image rod is

W (z o) = −

M · Bid3x = −M

V

Bz,id3x

Because the image rod is not moved and the rod is much longer than it is any other direction

W (z o) = −M

V

Bz,id3x ≈ −MA

zo+Lzo

Bz,i(z )dz

The force from the change in the vertical separation

F z = −dW (z o)dz o

zo=0

= M A(Bz,i(L) − Bz,i(0))

From the previous problem, if the shape is circular, which is not that bad of an approximation, the induced field is

Bz,i(z ) = µoM

2

z + L a2 + (z + L)2

− z √ a2 + z 2

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Figure 9.1: H/M o = 12

z/L+1/2√ 1/25+(z/L+1/2)2

− z/L−1/2√ 1/25+(z/L−1/2)2

-0 |z/L| > 1/2 or1 |z/L| ≤ 1/2

The force becomes

F z = µoM

2A

2

2L√ a2 + 4L2

− L√ a2 + L2

− L√ a2 + L2

+ 0

Now the condition that a L, F z = − µoM 2A

2

Now comparing the problem with 1.11. From the previous problem inside the rod, σM = M on the end above theplain, and σM = −M below the plain. So

Bz =

µoM inside0 outside

Because of the constant induced field, this is similar to a constant electric field. In fact, the magnetic potential issimilar to the electric potential of plane capacitors with σE/o → µoσM . To create a capacitor, lets separate the realrod, which is the one above the plane, and the image rod. Now on page 43, F/A = σ2/2o for a capacitor. Then

F/A = σ2M µ2o/2µo → F/A = µoσ2M /2. Finally the value of σM , F /A = µoM 2/2 → F = µo2 AM

2

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Figure 9.2: B/µoM o = 12

z/L+1/2√ 1/25+(z/L+1/2)2

− z/L−1/2√ 1/25+(z/L−1/2)2

43


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44


48/65

Chapter 10

Problem Set 10


A long, hollow, right circular cylinder of inner (outer) radius a (b), and of relative permeability µr, is placed in aregion of initially uniform magnetic-flux density Bo at right angles to the field. Find the flux density at all pointsin space, and sketch the logarithm of the ratio of the magnitudes of B on the cylinder axis to Bo as a function of log10 µr for a

2/b2 = 0.5, 0.1. Neglect end effects.

Because there are no end effects, this problem can be reduced to a two dimensional problem. With the z -axis thelength of the cylinder and the Bo in the x-direction. With those being set, this is much like the problem donein class. So ∇ · B = ∇ · (µH) = 0 and for inside and outside ∇ × H = 0. The solution is H = −∇Φm whereΦ = ao + bo ln ρ +

∞n=1

anρn (cos nϕ + sin nϕ) +

∞n=1

bnρ−n (cos nϕ + sin nϕ). All that remains is to solve for the

coefficients. By symmetry the sine terms can be dropped. For ρ > b, an = 0, and for ρ < a, bn = 0. So the potentialis

Φm(ρ, ϕ) =

∞n=1 αnρ

n cos nϕ ρ < a∞n=1

βnρ

n cos nϕ + γ nρ−n cos nϕ

a < ρ < b∞

n=1 δnρ−n cos nϕ ρ > b

H has to be continuous at the boundaries. At ρ = b, the differential with respect to the radial is

Bo

µocos ϕ +

∞

n=1

nb−(n+1)δ n cos nϕ = µr∞

n=1−

nβ nbn−1 − γ nb−(n+1) cos nϕ

From this n = 1, so

Boµo

+ δ 1b−2 = −µrβ 1 + µrγ 1b−2 (10.1)

The tangential continuation at ρ = b

−Boµo

+ δ 1b−2 = β 1 + γ 1b

−2 (10.2)

Likewise at ρ = a

α1 = µrβ 1 − µrγ 1a−2 (10.3)α1 = β 1 + γ 1a−2 (10.4)

Reducing the number of coefficients in equations 1 and 2

β 1 = (µr − 1)

2µrδ 1b

−2 − (µr + 1)2µr

Boµo

(10.5)

γ 1 = (µr + 1)

2µrδ 1 +

(1 − µr)2µr

Boµo

b2 (10.6)

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Doing the same with equations 3 and 4

β 1 = (µr + 1)

2µrα1 (10.7)

γ 1 = (µr − 1)

2µrα1a

2 (10.8)

Now combining equations 5 and 7

α1 = (µr − 1)(µr + 1)

δ 1b−2 − Bo

µo(10.9)

Now for equations 6 and 8

α1 = (µr + 1)

(µr − 1) δ 1a−2 − b

2

a2Boµo

(10.10)

Equating 9 and 10

δ 1 =

1 − a

2

b2

µ2r − 1

b2

(µr + 1)2 b2 − (µr − 1)2 a2

Boµo

b2 (10.11)

With equation 9 and 11

α1 = −4µrb2

(µr + 1)2 b2 − (µr − 1)2 a2

Boµo

(10.12)

Placing equation 12 into equation 7 and also equation 8

β 1 = −2 (µr + 1) b2

(µr + 1)2 b2 − (µr − 1)2 a2

Boµo

(10.13)

γ 1 = −2 (µr − 1) b2

(µr + 1)2 b2 − (µr − 1)2 a2

Boµo

a2 (10.14)

Now the field is

H =

−α1 cos ϕρ̂ + α1 sin ϕϕ̂ ρ < a−β 1 − γ 1ρ2 cos ϕρ̂ + β 1 + γ 1ρ2 sin ϕϕ̂ a < ρ < bBoµo

+ δ1ρ2

cos ϕρ̂ −

Boµo

+ δ1ρ2

sin ϕϕ̂ ρ > b

See last page for graph


A circuit consists of a long thin conducting shell of radius a and a parallel return wire of radius b on axis inside. Of the current is assumed distributed uniformly throughout the cross section of the wire, calculate the self-inductanceper unit length. What is the self-inductance if the inner wire is a thin tube?

First let’s apply Ampere’s law B · dl = µoI enc (10.15)

So quickly, B is

B =

µoI 2π

ρb2

ρ < bµoI 2π

1ρ

b < ρ < a

0 ρ > a

46


50/65

The energy of the magnetic field per length of the system is given by Eq. 5.148

W = 1

2

H · Bd3x = 1

2µo

B2d3x

= 2π

2µo

µoI

2π

2 b0

ρ3dρ

b4 +

ab

dρ

ρ

= πµo

µoI 2π2

14 + ln

ab

= µoI

2

16π

4 ln

ab

+ 1

From Eq. 5.152, W = 12

LI 2. The self-inductance per unit length is

1

2LI 2 =

µoI 2

16π

4 lna

b

+ 1

⇒ L = µoI 2

8π

4 lna

b

+ 1

Now if the inner wire is a tube B = 0 for ρ < b, so there is only the middle term which yields a self-inductance perunit length of

L = µo2π

lna

b


The figure (See Jackson pg. 232) represents a transmission line consisting of two, parallel perfect conductors of arbitrary, but constant, cross section. Current flows down one conductor and returns via the other.Show that the product of the inductance per unit length L and the capacitance per unit length C is

LC = µ (10.16)

where µ and are the permeability and the permittivity of the medium surrounding the conductors.

First using Ampere’s Law for outside the wires B · dl = µI ⇒ B = µI

2πρ

But inside the wire B = 0. The vector potential is

Az =

Bφdρ =

µI 2πρ

dρ ⇒ Az = −µI 2π

ln

ρρo

where ρo = a, b. The work per unit length can be found from Eq. 5.149

W = 1

2

J · Ad3x = 1

2

(J aAdaa + J bAdab)

= 1

2

µI

2π

a0

2πI

πa2 ln

d

b

ρadρa +

b0

2πI

πb2 ln

d

a

ρbdρb

= 1

2

µI

2π

a2πI

πa2 ln

d

b

+

b2πI

πb2 ln

d

a

= µI 2

4π ln

d2

ab

Relating this to the self-inductance per unit length1

2LI 2 =

µI 2

4π ln

d2

ab

⇒ L

[전자기학 jackson 솔루션] 전자기학 jackson 솔루션 65 4000.pdf

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