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Solutions to Problems in Jackson,

Classical Electrodynamics, Third Edition

Homer Reid

December 8, 1999

Chapter 2

Problem 2.1

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

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

(b) the force between the plane and the charge by using Coulomb’slaw for the force between the charge and its image;

(c) the total force acting on the plane by integrating σ2/2ε0 overthe whole plane;

(d) the work necessary to remove the charge q from its position toinfinity;

(e) the potential energy between the charge q and its image (com-pare the answer to part d and discuss).

(f) Find the answer to part d in electron volts for an electronoriginally one angstrom from the surface.

(a) We’ll take d to be in the z direction, so the charge q is at (x, y, z) = (0, 0, d).The image charge is −q at (0, 0,−d). The potential at a point r is

Φ(r) =q

4πε0

[

1

|r − dk| −1

|r + dk|

]

The surface charge induced on the plane is found by differentiating this:

1

Homer Reid’s Solutions to Jackson Problems: Chapter 2 2

σ = −ε0dΦ

dz

∣

∣

z=0

= − q

4π

[−(z − d)

|r + dk|3 +(z + d)

|r + dk|3]

∣

∣

z=0

= − qd

2π(x2 + y2 + d2)3/2(1)

We can check this by integrating this over the entire xy plane and verifyingthat the total charge is just the value −q of the image charge:

∫

∞

−∞

∫

∞

−∞

σ(x, y)dxdy = − qd

2π

∫

∞

0

∫ 2π

0

rdψdr

(r2 + d2)3/2

= −qd∫

∞

0

rdr

(r2 + d2)3/2

= −qd2

∫

∞

d2

u−3/2du

= −qd2

∣

∣

∣−2u−1/2

∣

∣

∣

∞

d2

= −q √

(b) The point of this problem is that, for points above the z axis, it doesn’tmatter whether there is a charge −q at (0, 0, d) or an infinite grounded sheetat z = 0. Physics above the z axis is exactly the same whether we have thecharge or the sheet. In particular, the force on the original charge is the samewhether we have the charge or the sheet. That means that, if we assume thesheet is present instead of the charge, it will feel a reaction force equal to whatthe image charge would feel if it were present instead of the sheet. The forceon the image charge would be just F = q2/16πε0d

2, so this must be what thesheet feels.

(c) Total force on sheet

=1

2ε0

∫

∞

0

∫ 2π

0

σ2dA

=q2d2

4πε0

∫

∞

0

rdr

(r2 + d2)3

=q2d2

8πε0

∫

∞

d2

u−3du

=q2d2

8πε0

∣

∣

∣

∣

−1

2u−2

∣

∣

∣

∣

∞

d2

=q2d2

8πε0

[

1

2d−4

]


=q2

16πε0d2

in accordance with the discussion and result of part b.

(d) Work required to remove charge to infinity

=q2

4πε0

∫

∞

d

dz

(z + d)2

=q2

4πε0

∫

∞

2d

u−2du

=q2

4πε0

1

2d

=q2

8πε0d

(e) Potential energy between charge and its image

=q2

8πε0d

equal to the result in part d.

(f)

q2

8πε0d=

(1.6 · 10−19 coulombs )2

8π(8.85 · 10−12 coulombsV−1m−1)(10−10 m )

= 7.2 · (1.6 · 10−19 coulombs · 1 V )

= 7.2 eV .

Problem 2.2

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

(a) the potential inside the sphere;

(b) the induced surface-charge density;

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

(d) Is there any change in the solution if the sphere is kept at afixed potential V ? If the sphere has a total charge Q on itsinner and outer surfaces?


Problem 2.3

A straight-line charge with constant linear charge density λ is located perpendicularto the x − y plane in the first quadrant at (x0, y0). The intersecting planes x =0, y ≥ 0 and y = 0, x ≥ 0 are conducting boundary surfaces held at zero potential.Consider the potential, fields, and surface charges in the first quadrant.

(a) The well-known potential for an isolated line charge at (x0, y0) is Φ(x, y) =(λ/4πε0) ln(R2/r2), where r2 = (x − x0)

2 + (y − y0)2 and R is a constant.

Determine the expression for the potential of the line charge in the presence ofthe intersecting planes. Verify explicitly that the potential and the tangentialelectric field vanish on the boundary surface.

(b) Determine the surface charge density σ on the plane y = 0, x ≥ 0. Plot σ/λversus x for (x0 = 2, y0 = 1), (x0 = 1, y0 = 1), and (x0 = 1, y0 = 2).

(c) Show that the total charge (per unit length in z) on the plane y = 0, x ≥ 0 is

Qx = − 2

πλ tan−1

(

x0

y0

)

What is the total charge on the plane x = 0?

(d) Show that far from the origin [ρ ρ0, where ρ =√

x2 + y2 and ρ0 =√

x20 + y2

0 ] the leading term in the potential is

Φ → Φasym =4λ

πε0

(x0)(y0)(xy)

ρ4.

Interpret.

(a) The potential can be made to vanish on the specified boundary surfacesby pretending that we have three image line charges. Two image charges havecharge density −λ and exist at the locations obtained by reflecting the originalimage charge across the x and y axes, respectively. The third image charge hascharge density +λ and exists at the location obtained by reflecting the originalcharge through the origin. The resulting potential in the first quadrant is

Φ(x, y) =λ

4πε0

(

lnR2

r21− ln

R2

r22− ln

R2

r23+ ln

R2

r24

)

=λ

2πε0lnr2r3r1r4

(2)

where

r21 = [(x− x0)2 + (y − y0)

2] r22 = [(x+ x0)2 + (y − y0)

2]


r23 = [(x − x0)2 + (y + y0)

2] r24 = [(x+ x0)2 + (y + y0)

2].

From this you can see that

• when x = 0, r1 = r2 and r3 = r4

• when y = 0, r1 = r3 and r2 = r4

and in both cases the argument of the logarithm in (2) is unity.

(b)

σ = −ε0d

dyΦ

= − λ

2π

(

1

r2

dr2dy

+1

r3

dr3dy

− 1

r1

dr1dy

− 1

r4

dr4dy

)

∣

∣

y=0

We have dr1/dy = (y − y0)/r1 and similarly for the other derivatives, so

σ = − λ

2π

(

y − y0r22

+y + y0r23

− y − y0r21

− y + y0r24

)

∣

∣

y=0

= −y0λπ

(

1

(x− x0)2 + y20

− 1

(x+ x0)2 + y20)

)

(c) Total charge per unit length in z

Qx =

∫

∞

0

σdx

= −y0λπ

[∫

∞

0

dx

(x− x0)2 + y20

−∫

∞

0

dx

(x+ x0)2 + y20

]

For the first integral the appropriate substitution is (x − x0) = y0 tanu, dx =y0 sec2 udu. A similar substitution works in the second integral.

= −λπ

[

∫ π/2

tan−1−

x0

y0

du−∫ π/2

tan−1x0

y0

du

]

= −λπ

[

π

2− tan−1 −x0

y0− π

2+ tan−1 x0

y0

]

= −2λ

πtan−1 x0

y0. (3)

The calculations are obviously symmetric with respect to x0 and y0. Thetotal charge on the plane x = 0 is (3) with x0 and y0 interchanged:

Qy = −2λ

πtan−1 y0

x0

Since tan−1 x− tan−1(1/x) = π/2 the total charge induced is

Q = −λ


which is, of course, also the sum of the charge per unit length of the three imagecharges.

(d) We have

Φ =λ

4πε0lnr22r

23

r21r24

Far from the origin,

r21 = [(x − x0)2 + (y − y0)

2]

=

[

x2(1 − x0

x)2 + y2(1 − y0

y)2

]

≈[

x2(1 − 2x0

x) + y2(1 − 2

y0y

]

=[

x2 − 2x0x+ y2 − 2y0y)]

= (x2 + y2)

[

1 − 2xx0 + yy0x2 + y2

]

Similarly,

r22 = (x2 + y2)

[

1 − 2−xx0 + yy0x2 + y2

]

r23 = (x2 + y2)

[

1 − 2xx0 − yy0x2 + y2

]

r24 = (x2 + y2)

[

1 − 2−xx0 − yy0x2 + y2

]

Next,

r21r24 = (x2 + y2)2

[

1 − 4(xx0 + yy0)

2

(x2 + y2)2

]

r22r23 = (x2 + y2)2

[

1 − 4(xx0 − yy0)

2

(x2 + y2)2

]

so

Φ =λ

4πε0ln

1 − 4 (xx0−yy0)2

(x2+y2)2

1 − 4 (xx0+yy0)2

(x2+y2)2

.

The (x2 + y2) term in the denominator grows much more quickly than the(xx0 + yy0) term, so in the asymptotic limit we can use ln(1 + ε) ≈ ε to find

Φ =λ

4πε0

[

−4(xx0 − yy0)

2

(x2 + y2)2+ 4

(xx0 + yy0)2

(x2 + y2)2

]

=λ

4πε0

[−4(x2x20 + y2y2

0 − 2xyx0y0) + 4(x2x20 + y2y2

0 + 2xyx0y0)

(x2 + y2)2

]


=λ

4πε0

[

16xyx0y0(x2 + y2)2

]

=4λ

πε0

(xy)(x0y0)

(x2 + y2)2.

√

Problem 2.4

A point charge is placed a distance d > R from the center of anequally charged, isolated, conducting sphere of radius R.

(a) Inside of what distance from the surface of the sphere is thepoint charge attracted rather than repelled by the chargedsphere?

(b) What is the limiting value of the force of attraction when thepoint charge is located a distance a(= d−R) from the surfaceof the sphere, if a R?

(c) What are the results for parts a and b if the charge on thesphere is twice (half) as large as the point charge, but stillthe same sign?

Let’s call the point charge q. The charged, isolated sphere may be replacedby two image charges. One image charge, of charge q1 = −(R/d)q at radiusr1 = R2/d, is needed to make the potential equal at all points on the sphere.The second image charge, of charge q2 = q − q1 at the center of the sphere,is necessary to recreate the effect of the additional charge on the sphere (the“additional” charge is the extra charge on the sphere left over after you subtractthe surface charge density induced by the point charge q).

The force on the point charge is the sum of the forces from the two imagecharges:

F =1

4πε0

[

qq1[

d− R2

d

]2 +qq2d2

]

(4)

=q2

4πε0

[ −dR[d2 −R2]2

+d2 + dR

d4

]

(5)

As d → R the denominator of the first term vanishes, so that term wins,and the overall force is attractive. As d → ∞, the denominator of both termslooks like d4, so the dR terms in the numerator cancel and the overall force isrepulsive.

(a) The crossover distance is found by equating the two bracketed terms in (5):


dR

[d2 −R2]2=

d2 + dR

d4

d4R = (d+R)[d2 −R2]2

0 = d5 − 2d3R2 − 2d2R3 + dR4 +R5

I used GnuPlot to solve this one graphically. The root is d/R=1.6178.

(b) The idea here is to set d = R+ a = R(1 + a/R) and find the limit of (4) asa→ 0.

F =q2

4πε0

[

−R2(1 + aR )

[

R2(1 + aR )2 −R2

]2 +R2

[

(1 + aR )2 + (1 + a

R )]

R4(1 + aR )4

]

≈ q2

4πε0

[−R2 − aR

4a2R2+

(2R+ 3a)(R− 4a)

R4

]

The second term in brackets approaches the constant 2/R2 as a→ 0. The firstterm becomes −1/4a2. So we have

F → − q2

16πε0a2.

Note that only the first image charge (the one required to make the sphere anequipotential) contributes to the force as d → a. The second image charge,the one which represents the difference between the actual charge on the sphereand the charge induced by the first image, makes no contribution in this limit.That means that the limiting value of the force will be as above regardless ofthe charge on the sphere.

(c) If the charge on the sphere is twice the point charge, then q2 = 2q − q1 =q(2 +R/d). Then (5) becomes

F =q2

4πε0

[

− dR

[d2 −R2]2+

2d2 + dR

d4

]

and the relevant equation becomes

0 = 2d5 − 4d3R2 − 2d2R3 + 2dR4 +R5.

Again I solved graphically to find d/R = 1.43. If the charge on the sphere ishalf the point charge, then

F =q2

4πε0

[

− dR

[d2 −R2]2+d2 + 2dR

2d4

]

and the equation is

0 = d5 − 2d3R2 − 4d2R3 + dR4 + 2R5.

The root of this one is d/R=1.88.


Problem 2.5

(a) Show that the work done to remove the charge q from a dis-tance r > a to infinity against the force, Eq. (2.6), of agrounded conducting sphere is

W =q2a

8πε0(r2 − a2).

Relate this result to the electrostatic potential, Eq. (2.3),and the energy discussion of Section 1.11.

(b) Repeat the calculation of the work done to remove the charge qagainst the force, Eq. (2.9), of an isolated charged conductingsphere. Show that the work done is

W =1

4πε0

[

q2a

2(r2 − a2)− q2a

2r2− qQ

r

]

.

Relate the work to the electrostatic potential, Eq. (2.8), andthe energy discussion of Section 1.11.

(a) The force is

|F | =q2a

4πε0

1

y3(1 − a2/y2)2

directed radially inward. The work is

W = −∫

∞

r

Fdy (6)

=q2a

4πε0

∫

∞

r

dy

y3(1 − a2/y2)2

=q2a

4πε0

∫

∞

r

ydy

(y2 − a2)2

=q2a

4πε0

∫

∞

r2−a2

du

2u2

=q2a

4πε0

∣

∣

∣

∣

− 1

2u

∣

∣

∣

∣

∞

r2−a2

=q2a

8πε0(r2 − a2)(7)

To relate this to earlier results, note that the image charge q′ = −(a/r)q islocated at radius r′ = a2/r. The potential energy between the point charge and


its image is

PE =1

4πε0

(

qq′

|r − r′|

)

=1

4πε0

( −q2ar(r − a2/r)

)

=1

4πε0

( −q2ar2 − a2

)

(8)

Result (7) is only half of (8). This would seem to violate energy conservation. Itwould seem that we could start with the point charge at infinity and allow it tofall in to a distance r from the sphere, liberating a quantity of energy (8), whichwe could store in a battery or something. Then we could expend an energyequal to (7) to remove the charge back to infinity, at which point we wouldbe back where we started, but we would still have half of the energy saved inthe battery. It would seem that we could keep doing this over and over again,storing up as much energy in the battery as we pleased.

I think the problem is with equation (8). The traditional expression q1q2/4πε0rfor the potential energy of two charges comes from calculating the work neededto bring one charge from infinity to a distance r from the other charge, and itis assumed that the other charge does not move and keeps a constant chargeduring the process. But in this case one of the charges is a fictitious imagecharge, and as the point charge q is brought in from infinity the image chargemoves out from the center of the sphere, and its charge increases. So the simpleexpression doesn’t work to calculate the potential energy of the configuration,and we should take (7) to be the correct result.

(b) In this case there are two image charges: one of the same charge and locationas in part a, and another of charge Q − q′ at the origin. The work needed toremove the point charge q to infinity is the work needed to remove the pointcharge from its image charge, plus the work needed to remove the point chargefrom the extra charge at the origin. We calculated the first contribution above.The second contribution is

−∫

∞

r

q(Q− q′)dy

4πε0y2= − 1

4πε0

∫

∞

r

[

qQ

y2+q2a

y3

]

dy

= − 1

4πε0

∣

∣

∣

∣

−qQy

− q2a

2y2

∣

∣

∣

∣

∞

r

= − 1

4πε0

[

qQ

r+q2a

2r2

]

so the total work done is

W =1

4πε0

[

q2a

2(r2 − a2)− q2a

2r2− qQ

r

]

.


Review of Green’s Functions

Some problems in this and other chapters use the Green’s function technique.It’s useful to review this technique, and also to establish my conventions sinceI define the Green’s function a little differently than Jackson.

The whole technique is based on the divergence theorem. Suppose A(x) isa vector valued function defined at each point x within a volume V . Then

∫

V

(∇ · A(x′)) dV ′ =

∮

S

A(x′) · dA′ (9)

where S is the (closed) surface bounding the volume V . If we take A(x) =φ(x)∇ψ(x) where φ and ψ are scalar functions, (9) becomes

∫

V

[

(∇φ(x′)) · (∇ψ(x′)) + φ(x′)∇2ψ(x′)]

dV ′ =

∮

S

φ(x′)∂ψ

∂n

∣

∣

∣

∣

x′

dA′

where ∂ψ/∂n is the dot product of ~∇ψ with the outward normal to the surfacearea element. If we write down this equation with φ and ψ switched and subtractthe two, we come up with

∫

V

[

φ∇2ψ − ψ∇2φ]

dV ′ =

∮

S

[

φ∂ψ

∂n− ψ

∂φ

∂n

]

dA′. (10)

This statement doesn’t appear to be very useful, since it seems to require thatwe know φ over the whole volume to compute the left side, and both φ and∂φ/∂n on the boundary to compute the right side. However, suppose we couldchoose ψ(x) in a clever way such that ∇2ψ = δ(x − x0) for some point x0

within the volume. (Since this ψ is a function of x which also depends on x0

as a parameter, we might write it as ψx0(x).) Then we could use the sifting

property of the delta function to find

φ(x0) =

∫

V

[

ψx0(x′)∇2φ(x′)

]

dV ′ +

∮

S

[

φ(x′)∂ψx0

∂n

∣

∣

∣

∣

x′

− ψx0(x′)

∂φ

∂n

∣

∣

∣

∣

x′

]

dA′.

If φ is the scalar potential of electrostatics, we know that ∇2ψ(x′) = −ρ(x′)/ε0,so we have

φ(x0) = − 1

ε0

∫

V

ψx0(x′)ρ(x′)dV ′ +

∮

S

[

φ(x′)∂ψx0

∂n

∣

∣

∣

∣

x′

− ψx0(x′)

∂φ

∂n

∣

∣

∣

∣

x′

]

dA′.

(11)Equation (11) allows us to find the potential at an arbitrary point x0 as

long as we know ρ within the volume and both φ and ∂φ/∂n on the boundary.boundary. Usually we do know ρ within the volume, but we only know either φor ∂φ/∂n on the boundary. This lack of knowledge can be accommodated bychoosing ψ such that either its value or its normal derivative vanishes on theboundary surface, so that the term which we can’t evaluate drops out of thesurface integral. More specifically,


• if we know φ but not ∂φ/∂n on the boundary (“Dirichlet” boundary con-ditions), we choose ψ such that ψ = 0 on the boundary. Then

φ(x0) = − 1

ε0

∫

V

ψx0(x′)ρ(x′)dV ′ +

∮

S

φ(x′)∂ψx0

∂n

∣

∣

∣

∣

x′

dA′. (12)

• if we know ∂φ/∂n but not φ on the boundary (“Neumann” boundaryconditions), we choose ψ such that ∂ψ/∂n = 0 on the boundary. Then

φ(x0) = − 1

ε0

∫

V

ψx0(x′)ρ(x′)dV ′ +

∮

S

φx0(x′)

∂φ

∂n

∣

∣

∣

∣

x′

dA′. (13)

Again, in both cases the function ψx0(x) has the property that

∇2ψx0(x) = δ(x − x0).



Homer Reid

December 8, 1999

Chapter 2: Problems 11-20

Problem 2.11

A line charge with linear charge density τ is placed parallel to, and adistance R away from, the axis of a conducting cylinder of radius b heldat fixed voltage such that the potential vanishes at infinity. Find

(a) the magnitude and position of the image charge(s);

(b) the potential at any point (expressed in polar coordinates with theorigin at the axis of the cylinder and the direction from the originto the line charge as the x axis), including the asymptotic form farfrom the cylinder;

(c) the induced surface-charge density, and plot it as a function of anglefor R/b=2,4 in units of τ/2πb;

(d) the force on the charge.

(a) Drawing an analogy to the similar problem of the point charge outsidethe conducting sphere, we might expect that the potential on the cylinder canbe made constant by placing an image charge within the cylinder on the lineconducting the line charge with the center of the cylinder, i.e. on the x axis.Suppose we put the image charge a distance R′ < b from the center of thecylinder and give it a charge density −τ . Using the expression quoted in Problem2.3 for the potential of a line charge, the potential at a point x due to the linecharge and its image is

Φ(x) =τ

4πε0ln

R2

|x − Ri|2− τ

4πε0ln

R2

|x − R′i|2

1


=τ

4πε0ln

|x − R′i|2

|x− Ri|2.

We want to choose R′ such that the potential is constant when x is on thecylinder surface. This requires that the argument of the logarithm be equal tosome constant γ at those points:

|x − R′i|2

|x − Ri|2= γ

orb2 + R′2 − 2R′b cosφ = γb2 + γR2 − 2γRb cosφ.

For this to be true everywhere on the cylinder, the φ term must drop out, whichrequires R′ = γR. We can then rearrange the remaining terms to find

R′ =b2

R.

This is also analogous to the point-charge-and-sphere problem, but there aredifferences: in this case the image charge has the same magnitude as the originalline charge, and the potential on the cylinder is constant but not zero.

(b) At a point (ρ, φ), we have

Φ =τ

4πε0ln

ρ2 + R′2 − 2ρR′ cosφ

ρ2 + R2 − 2ρR cosφ.

For large ρ, this becomes

Φ → τ

4πε0ln

1 − 2R′

ρ cosφ

1 − 2Rρ cosφ

.

Using ln(1 − x) = −(x + x2/2 + · · ·), we have

Φ → τ

4πε0

2(R − R′) cosφ

ρ

=τ

2πε0

R(1 − b2/R2) cosφ

ρ

(c)

σ = −ε0∂Φ

∂ρ

∣

∣

∣

∣

r=b

= − τ

4π

[

2b − 2R′ cosφ

b2 + R′2 − 2bR′ cosφ− 2b − 2R cosφ

b2 + R2 − 2bR cosφ

]

= − τ

2π

[

b − b2

R cosφ

b2 + b4

R2 − 2 b3

R cosφ− b − R cosφ

b2 + R2 − 2bR cosφ

]


Multiplying the first term by R2/b2 on top and bottom yields

σ = − τ

2π

[

R2

b − b

R2 + b2 − 2bR cosφ

]

= − τ

2πb

[

R2 − b2

R2 + b2 − 2bR cosφ

]

(d) To find the force on the charge, we note that the potential of the imagecharge is

Φ(x) = − τ

4πε0ln

C2

|x − R′ i|2.

with C some constant. We can differentiate this to find the electric field due tothe image charge:

E(x) = −∇Φ(x) = − τ

4πε0∇ ln |x − R′ i|2

= − τ

4πε0

2(x− R′ i)

|x − R′i|2.

The original line charge is at x = R, y = 0, and the field there is

E = − τ

2πε0

1

R − R′i = − τ

2πε0

R

R2 − b2i.

The force per unit width on the line charge is

F = τE = − τ2

2πε0

R

R2 − b2

tending to pull the original charge in toward the cylinder.

Problem 2.12

Starting with the series solution (2.71) for the two-dimensional potentialproblem with the potential specified on the surface of a cylinder of radiusb, evaluate the coefficients formally, substitute them into the series, andsum it to obtain the potential inside the cylinder in the form of Poisson’sintegral:

Φ(ρ, φ) =1

2π

∫ 2π

0

Φ(b, φ′)b2 − ρ2

b2 + ρ2 − 2bρ cos(φ′ − φ)dφ′

What modification is necessary if the potential is desired in the region ofspace bounded by the cylinder and infinity?


Referring to equation (2.71), we know the bn are all zero, because the lnterm and the negative powers of ρ are singular at the origin. We are left with

Φ(ρ, φ) = a0 +

∞∑

n=1

ρn an sin(nφ) + bn cos(nφ) . (1)

Multiplying both sides successively by 1, sin n′φ, and cosn′φ and integratingat ρ = b gives

a0 =1

2π

∫ 2π

0

Φ(b, φ)dφ (2)

an =1

πbn

∫ 2π

0

Φ(b, φ) sin(nφ)dφ (3)

bn =1

πbn

∫ 2π

0

Φ(b, φ) cos(nφ)dφ. (4)

Plugging back into (1), we find

Φ(ρ, φ) =1

π

∫ 2π

0

Φ(b, φ′)

1

2+

∞∑

n=1

(ρ

b

)n

[sin(nφ) sin(nφ′) + cos(nφ) cos(nφ′)]

dφ′

=1

π

∫ 2π

0

Φ(b, φ′)

1

2+

∞∑

n=1

(ρ

b

)n

cosn(φ − φ′)

. (5)

The bracketed term can be expressed in closed form. For simplicity definex = (ρ/b) and α = (φ − φ′). Then

1

2+

∞∑

n=1

xn cos(nα) =1

2+

1

2

∞∑

n=1

[

xneinα + xne−inα]

=1

2+

1

2

[

1

1 − xeiα+

1

1 − xe−iα− 2

]

=1

2+

1

2

[

1 − xe−iα − xeiα + 1

1 − xeiα − xe−iα + x2− 2

]

=1

2+

[

1 − x cosα

1 + x2 − 2x cosα− 1

]

=1

2+

x cosα − x2

1 + x2 − 2x cosα

=1

2

[

1 − x2

1 + x2 − 2x cosα

]

.

Plugging this back into (5) gives the advertised result.


Problem 2.13

(a) Two halves of a long hollow conducting cylinder of inner radius b areseparated by small lengthwise gaps on each side, and are kept at differentpotentials V1 and V2. Show that the potential inside is given by

Φ(ρ, φ) =V1 + V2

2+

V1 − V2

πtan−1

(

2bρ

b2 − ρ2cosφ

)

where φ is measured from a plane perpendicular to the plane through thegap.(b) Calculate the surface-charge density on each half of the cylinder.

This problem is just like the previous one. Since we are looking for anexpression for the potential within the cylinder, the correct expansion is (1)with expansion coefficients given by (2), (3) and (4):

a0 =1

2π

∫ 2π

0

Φ(b, φ)dφ

=1

2π

[

V1

∫ π

0

dφ + V2

∫ 2π

π

dφ

]

=V1 + V2

2

an =1

πbn

[

V1

∫ π

0

sin(nφ)dφ + V2

∫ 2π

π

sin(nφ)dφ

]

= − 1

nπbn

[

V1 |cosnφ|π0 + V2 |cosnφ|2ππ

]

= − 1

nπbn[V1(cosnπ − 1) + V2(1 − cosnπ)]

=

0 , n even2(V1 − V2)/(nπbn) , n odd

bn =1

πbn

[

V1

∫ π

0

cos(nφ)dφ + V2

∫ 2π

π

cos(nφ)dφ

]

=1

nπbn

[

V1 |sinnφ|π0 + V2 |sin nφ|2ππ

]

= 0.

With these coefficients, the potential expansion becomes

Φ(ρ, φ) =V1 + V2

2+

2(V1 − V2)

π

∑

n odd

1

n

(ρ

b

)n

sinnφ. (6)


Here we need an auxiliary result:

∑

n odd

1

nxn sin nφ =

1

2i

∑

n odd

1

n(iy)n[einπ − e−inφ] (x = iy)

=1

2

∞∑

n=0

(−1)n

2n + 1

[

(yeiφ)2n+1 − (ye−iφ)2n+1]

=1

2

[

tan−1(yeiφ) − tan−1(ye−iφ)]

(7)

where in the last line we just identified the Taylor series for the inverse tangentfunction. Next we need an identity:

tan−1 γ1 − tan−1 γ2 = tan−1

(

γ1 − γ2

1 + γ1γ2

)

.

(I derived this one by drawing some triangles and doing some algebra.) Withthis, (7) becomes

∑

n odd

1

nxn sin nφ =

1

2tan−1

(

2iy sin φ

1 + y2

)

=1

2tan−1

(

2x sin φ

1 − x2

)

.

Using this in (6) with x = ρ/b gives

Φ(ρ, b) =V1 + V2

π+

V1 − V2

πtan−1

(

2ρb sinφ

b2 − ρ2

)

.

(Evidently, Jackson and I defined the angle φ differently).


Problem 2.15

(a) Show that the Green function G(x, y; x′, y′) appropriate for Dirichletboundary conditions for a square two-dimensional region, 0 ≤ x ≤ 1, 0 ≤y ≤ 1, has an expansion

G(x, y; x′, y′) = 2∞∑

n=1

gn(y, y′) sin(nπx) sin(nπx′)

where gn(y, y′) satisfies

(

∂2

∂y2− n2π2

)

gn(y, y′) = δ(y′ − y) and gn(y, 0) = gn(y, 1) = 0.

(b) Taking for gn(y, y′) appropriate linear combinations of sinh(nπy′)and cosh(nπy′) in the two regions y′ < y and y′ > y, in accord with theboundary conditions and the discontinuity in slope required by the sourcedelta function, show that the explicit form of G is

G(x, y; x′, y′) =

−2

∞∑

n=1

1

nπ sinh(nπ)sin(nπx) sin(nπx′) sinh(nπy<) sinh[nπ(1 − y>)]

where y< (y>) is the smaller (larger) of y and y′.

(I have taken out a factor −4π from the expressions for gn and G, in accordancewith my convention for Green’s functions; see the Green’s functions reviewabove.)

(a) To use as a Green’s function in a Dirichlet boundary value problem G mustsatisfy two conditions. The first is that G vanish on the boundary of the regionof interest. The suggested expansion of G clearly satisfies this. First, sin(nπx′)is 0 when x′ is 0 or 1. Second, g(y, y′) vanishes when y′ is 0 or 1. So G(x, y; x′, y′)vanishes for points (x′, y′) on the boundary.

The second condition on G is

∇2G =

(

∂2

∂x′2+

∂2

∂y′2

)

G = δ(x − x′) δ(y − y′). (8)

With the suggested expansion, we have

∂2

∂x′2G = 2

∞∑

n=1

gn(y, y′) sin(nπx)[

−n2π2 sin(nπx′)]

∂2

∂y′2G = 2

∞∑

n=1

∂′2

∂y2gn(y, y′) sin(nπx) sin(nπx′)


We can add these together and use the differential equation satisfied by gn tofind

∇2G = δ(y − y′) · 2∞∑

n=1

sin(nπx) sin(nπx′)

= δ(y − y′) · δ(x − x′)

since the infinite sum is just a well-known representation of the δ function.

(b) The suggestion is to take

gn(y, y′) =

An1 sinh(nπy′) + Bn1 cosh(nπy′), y′ < y;An2 sinh(nπy′) + Bn2 cosh(nπy′), y′ > y.

(9)

The idea to use hyperbolic sines and cosines comes from the fact that sinh(nπy)and cosh(nπy) satisfy a homogeneous version of the differential equation for gn

(i.e. satisfy that differential equation with the δ function replaced by zero).Thus gn as defined in (9) satisfies its differential equation (at all points excepty = y′) for any choice of the As and Bs. This leaves us free to choose thesecoefficients as required to satisfy the boundary conditions and the differentialequation at y = y′.

First let’s consider the boundary conditions. Since y is somewhere between0 and 1, the condition that gn vanish for y′ = 0 is only relevant to the top lineof (9), where it requires taking Bn1 = 0 but leaves An1 undetermined for now.The condition that gn vanish for y′ = 1 only affects the lower line of (9), whereit requires that

0 = An2 sinh(nπ) + Bn2 cosh(nπ)

= (An2 + Bn2)enπ + (−An2 + Bn2)e

−nπ (10)

One way to make this work is to take

An2 + Bn2 = −e−nπ and − An2 + Bn2 = enπ.

Then

Bn2 = enπ + An2 → 2An2 = −enπ − e−nπ

so An2 = − cosh(nπ) and Bn2 = sinh(nπ).

With this choice of coefficients, the lower line in (9) becomes

gn(y, y′) = − cosh(nπ) sinh(nπy′)+sinh(nπ) cosh(nπy′) = sinh[nπ(1−y′)] (11)

for (y′ > y). Actually, we haven’t completely determined An2 and Bn2; we couldmultiply (11) by an arbitrary constant γn and (10) would still be satisfied.

Next we need to make sure that the two halves of (9) match up at y′ = y:

An1 sinh(nπy) = γn sinh[nπ(1 − y)]. (12)


0

10000

20000

30000

40000

50000

60000

70000

0 0.2 0.4 0.6 0.8 1

g(yp

rime)

yprime

Figure 1: gn(y, y′) from Problem 2.15 with n=5, y=.41

This obviously happens when

An1 = βn sinh[nπ(1 − y)] and γn = βn sinh(nπy)

where βn is any constant. In other words, we have

gn(y, y′) =

βn sinh[nπ(1 − y)] sinh(nπy′), y′ < y;βn sinh[nπ(1 − y′)] sinh(nπy), y′ > y.

= βn sinh[nπ(1 − y>)] sinh(nπy<) (13)

with y< and y> defined as in the problem. Figure 1 shows a graph of thisfunction n = 5, y = .41.

The final step is to choose the normalization constant βn such that gn sat-isfies its differential equation:

(

∂2

∂2y′2− n2π2

)

gn(y, y′) = δ(y − y′). (14)

To say that the left-hand side “equals” the delta function requires two things:

• that the left-hand side vanish at all points y′ 6= y, and

• that its integral over any interval (y1, y2) equal 1 if the interval containsthe point y′ = y, and vanish otherwise.

The first condition is clearly satisfied regardless of the choice of βn. The secondcondition may be satisfied by making gn continuous, which we have alreadydone, but giving its first derivative a finite jump of unit magnitude at y′ = y:


∂

∂y′gn(y, y′)

∣

∣

∣

∣

y′=y+

y′=y−

= 1.

Differentiating (13), we find this condition to require

nπβn [− cosh[nπ(1 − y)] sinh(nπy) − sinh[nπ(1 − y)] cosh(nπy)] = −nπβn sinh(nπ) = 1

so (14) is satisfied if

βn = − 1

nπ sinh(nπ).

Then (13) is

gn(y, y′) = − sinh[nπ(1 − y>)] sinh(nπy<)

nπ sinh(nπ)

and the composite Green’s function is

G(x, y; x′, y′) = 2

∞∑

n=1

gn(y, y′) sin(nπx) sin(nπx′)

= −2

∞∑

n=1

sinh[nπ(1 − y>)] sinh(nπy<) sin(nπx) sin(nπx′)

nπ sinh(nπ).(15)

Problem 2.16

A two-dimensional potential exists on a unit square area (0 ≤ x ≤ 1, 0 ≤y ≤ 1) bounded by “surfaces” held at zero potential. Over the entire squarethere is a uniform charge density of unit strength (per unit length in z).Using the Green function of Problem 2.15, show that the solution can bewritten as

Φ(x, y) =4

π3ε0

∞∑

m=0

sin[(2m + 1)πx]

(2m + 1)3

1 − cosh[(2m + 1)π(y − (1/2))]

cosh[(2m + 1)π/2]

.

Referring to my Green’s functions review above, the potential at a point x0

within the square is given by

Φ(x0) = − 1

ε0

∫

V

G(x0;x′)ρ(x′)dV ′ +

∮

S

[

Φ(x′)∂G

∂n

∣

∣

∣

∣

x′

− G(x0;x′)

∂Φ

∂n

∣

∣

∣

∣

x′

]

dA′.

(16)In this case the surface integral vanishes, because we’re given that Φ vanishes

on the boundary, and G vanishes there by construction. We’re also given that


ρ(x′)dV ′ = dx′dy′ throughout the entire volume. Then we can plug in (15) tofind

Φ(x0) =2

πε0

∞∑

n=1

1

n sinh(nπ)

∫ 1

0

∫ 1

0

sinh[nπ(1−y>)] sinh(nπy<) sin(nπx0) sin(nπx′)dx′dy′.

(17)The integrals can be done separately. The x integral is

sin(nπx0)

∫ 1

0

sin(nπx′)dx′ = − sin(nπx0)

nπ[cos(nπ) − 1]

=

(2 sin(nπx0))/nπ , n odd0 , n even

(18)

The y integral is

sinh[nπ(1 − y0)]

∫ y0

0

sinh(nπy′)dy′ + sinh(nπy0)

∫ 1

y0

sinh[nπ(1 − y′)]dy′

=1

nπ

sinh[nπ(1 − y0)] ·∣

∣ cosh(nπy′)∣

∣

y0

0− sinh[nπy0] ·

∣

∣ cosh[nπ(1 − y′)]∣

∣

1

y0

=1

nπsinh[nπ(1 − y0)] cosh(nπy0) + sinh(nπy0) cosh[nπ(1 − y0)] − sinh(nπy0) − sinh[nπ(1 − y0)]

=1

nπsinh[nπ] − sinh[nπ(1 − y0)] − sinh(nπy0). (19)

Inserting (18) and (19) in (17), we have

Φ(x0) =4

π3ε0

∑

n odd

sin(nπx0)

n3

1 − sinh[nπ(1 − y0)] + sinh(nπy0)

sinh(nπ)

.

The thing in brackets is equal to what Jackson has, but this is tedious to showso I’ll skip the proof.


Problem 2.17

(a) Construct the free-space Green function G(x, y; x′, y′) for two-dimensional electrostatics by integrating 1/R with respect to z ′ − zbetween the limits ±Z, where Z is taken to be very large. Show thatapart from an inessential constant, the Green function can be writtenalternately as

G(x, y; x′, y′) = − ln[(x − x′)2 + (y − y′)2]

= − ln[ρ2 + ρ′2 − 2ρρ′ cos(φ − φ′)].

(b) Show explicitly by separation of variables in polar coordinates that theGreen function can be expressed as a Fourier series in the azimuthalcoordinate,

G =1

2π

∞∑

−∞

eim(φ−φ′)gm(ρ, ρ′)

where the radial Green functions satisfy

1

ρ′∂

∂ρ′

(

ρ′∂gm

∂ρ′

)

− m2

ρ′2gm =

δ(ρ − ρ′)

ρ.

Note that gm(ρ, ρ′) for fixed ρ is a different linear combination of thesolutions of the homogeneous radial equation (2.68) for ρ′ < ρ andfor ρ′ > ρ, with a discontinuity of slope at ρ′ = ρ determined by thesource delta function.

(c) Complete the solution and show that the free-space Green function hasthe expansion

G(ρ, φ; ρ′, φ′) =1

4πln(ρ2

>) − 1

2π

∞∑

m=1

1

m

(

ρ<

ρ>

)m

· cos[m(φ − φ′)]

where ρ<(ρ>) is the smaller (larger) of ρ and ρ′.

(As in Problem 2.15, I modified the text of the problem to match with myconvention for Green’s functions.)(a)

R = [(x − x′)2 + (y − y′)2 + (z − z′)2]1/2

≡ [a2 + u2]1/2 , a = [(x − x′)2 + (y − y′)2]1/2 , u = (z − z′).

Integrating,∫ Z

−Z

du

[a2 + u2]1/2=

∣

∣

∣ln

[

(a2 + u2)1/2 + u] ∣

∣

∣

+Z

−Z


= ln(Z2 + a2)1/2 + Z

(Z2 + a2)1/2 − Z

= ln(1 + (a2/Z2))1/2 + 1

(1 + (a2/Z2))1/2 − 1

≈ ln2 + a2

2Z2

a2

2Z2

= ln4Z2 + a2

a2

= ln[4Z2 + a2] − ln a2.

Since Z is much bigger than a, the first term is essentially independent of a andis the ’nonessential constant’ Jackson is talking about. The remaining term isthe 2D Green’s function:

G = − lna2 = − ln[(x − x′)2 + (y − y′)2] in rectangular coordinates

= − ln[ρ2 + ρ′2 − 2ρρ′ cos(φ − φ′)] in cylindrical coordinates.

(b) The 2d Green’s function is defined by

∫

∇2G(ρ, φ; ρ′, φ′)ρ′dρ′dφ′ = 1

but ∇2G = 0 at points other than (ρ, φ). These conditions are met if

∇2G(ρ, φ; ρ′, φ′) =1

ρ′δ(ρ − ρ′)δ(φ − φ′). (20)

You need the ρ′ on the bottom there to cancel out the ρ′ in the area element inthe integral. The Laplacian in two-dimensional cylindrical coordinates is

∇2 =1

ρ′∂

∂ρ′

(

ρ′∂

∂ρ′

)

− 1

ρ′2∂

∂φ′2.

Applying this to the suggested expansion for G gives

∇2G(ρ, φ; ρ′, φ′) =1

2π

∞∑

−∞

1

ρ′∂

∂ρ′

(

ρ′∂gm

∂ρ′

)

− m2

ρ′2gm

eim(φ−φ′).

If gm satisfies its differential equation as specified in the problem, the term inbrackets equals δ(ρ−ρ′)/ρ′ for all m and may be removed from the sum, leaving

∇2G(ρ, φ; ρ′, φ′) =

(

δ(ρ − ρ′)

ρ′

)

· 1

2π

∞∑

−∞

eim(φ−φ′)

=

(

δ(ρ − ρ′)

ρ′

)

δ(φ − φ′).


(c) As in Problem 2.15, we’ll construct the functions gm by finding solutions ofthe homogenous radial differential equation in the two regions and piecing themtogether at ρ = ρ′ such that the function is continuous but its derivative has afinite jump of magnitude 1/ρ.

For m ≥ 1, the solution to the homogenous equation

1

ρ′∂

∂ρ′

(

ρ′∂

∂ρ′

)

− m2

ρ′2

f(ρ′) = 0

isf(ρ′) = Amρ′m + Bmρ′−m.

Thus we take

gm =

A1mρ′m + B1mρ′−m , ρ′ < ρA2mρ′m + B2mρ′−m , ρ′ > ρ.

In order that the first solution be finite at the origin, and the second solutionbe finite at infinity, we have to take B1m = A2m = 0. Then the condition thatthe two solutions match at ρ = ρ′ is

A1mρm = B2mρ−m

which requiresA1m = γmρ−m B2m = ρmγm

for some constant γm. Now we have

gm =

γm

(

ρ′

ρ

)m

, ρ′ < ρ

γm

(

ρρ′

)m

, ρ′ > ρ

The finite-derivative step condition is

dgm

dρ′

∣

∣

∣

∣

ρ′=ρ+

− dgm

dρ′

∣

∣

∣

∣

ρ′=ρ−

=1

ρ

or

−mγm

(

1

ρ+

1

ρ

)

=1

ρ

so

γm = − 1

2m.

Then

gm =

− 12m

(

ρ′

ρ

)m

, ρ′ < ρ

− 12m

(

ρρ′

)m

, ρ′ > ρ

= − 1

2m

(

ρ<

ρ>

)m

.


Plugging this back into the expansion gives

G = − 1

4π

∞∑

−∞

1

m

(

ρ<

ρ>

)m

eim(φ−φ′)

= − 1

2π

∞∑

1

1

m

(

ρ<

ρ>

)m

cos[m(φ − φ′)].

Jackson seems to be adding a ln term to this, which comes from the m = 0solution of the radial equation, but I have left it out because it doesn’t vanishas ρ′ → ∞.

Problem 2.18

(a) By finding appropriate solutions of the radial equation in part b ofProblem 2.17, find the Green function for the interior Dirichlet prob-lem of a cylinder of radius b [gm(ρ, ρ′ = b) = 0. See (1.40)]. Firstfind the series expansion akin to the free-space Green function ofProblem 2.17. Then show that it can be written in closed form as

G = ln

[

ρ2ρ′2 + b4 − 2ρρ′b2 cos(φ − φ′)

b2(ρ2 + ρ′2 − 2ρρ′ cos(φ − φ′))

]

or

G = ln

[

(b2 − ρ2)(b2 − ρ′2) + b2|ρ − ρ′|2b2|ρ − ρ′|2

]

.

(b) Show that the solution of the Laplace equation with the potentialgiven as Φ(b, φ) on the cylinder can be expressed as Poisson’s inte-gral of Problem 2.12.

(c) What changes are necessary for the Green function for the exterior

problem (b < ρ < ∞), for both the Fourier expansion and theclosed form? [Note that the exterior Green function is not rigorouslycorrect because it does not vanish for ρ or ρ′ → ∞. For situationsin which the potential falls of fast enough as ρ → ∞, no mistake ismade in its use.]

(a) As before, we write the general solution of the radial equation for gm in thetwo distinct regions:

gm(ρ, ρ′) =

A1mρ′m + B1mρ′−m , ρ′ < ρA2mρ′m + B2mρ′−m , ρ′ > ρ.

(21)

The first boundary conditions are that gm remain finite at the origin andvanish on the cylinder boundary. This requires that

B1m = 0


andA2mbm + B2mb−m = 0

soA2m = γmb−m B2m = −γmbm

for some constant γm.Next, gm must be continuous at ρ = ρ′:

A1mρm = γm

[

(ρ

b

)m

−(

b

ρ

)m]

A1m =γm

ρm

[

(ρ

b

)m

−(

b

ρ

)m]

.

With this we have

gm(ρ, ρ′) = γm

[

(ρ

b

)m

−(

b

ρ

)m] (

ρ′

ρ

)m

, ρ′ < ρ

= γm

[(

ρ′

b

)m

−(

b

ρ′

)m]

, ρ′ > ρ.

Finally, dgm/dρ′ must have a finite jump of magnitude 1/ρ at ρ′ = ρ.

1

ρ=

dgm

dρ′

∣

∣

∣

∣

ρ′=ρ+

− dgm

dρ′

∣

∣

∣

∣

ρ′=ρ−

= mγm

[

ρm−1

bm+

bm

ρm+1

]

− mγm

[

(ρ

b

)m

−(

b

ρ

)m]

1

ρ

= 2mγm

(

b

ρ

)m1

ρ

so

γm =1

2m

(ρ

b

)m

and

gm(ρ, ρ′) =1

2m

[(

ρρ′

b2

)m

−(

ρ′

ρ

)m]

, ρ′ < ρ

=1

2m

[(

ρρ′

b2

)m

−(

ρ

ρ′

)m]

, ρ′ > ρ.

or

gm(ρ, ρ′) =1

2m

[(

ρρ′

b2

)m

−(

ρ<

ρ>

)m]

.

Plugging into the expansion for G gives

G(ρ, φ, ρ′, φ′) =1

2π

∞∑

n=1

1

m

[(

ρρ′

b2

)m

−(

ρ<

ρ>

)m]

cosm(φ − φ′). (22)


Here we need to work out an auxiliary result:

∞∑

n=1

1

nxn cosn(φ − φ′) =

∞∑

n=1

[∫ x

0

un−1du

]

cosm(φ − φ′)

=

∫ x

0

1

u

∞∑

n=1

un cosn(φ − φ′)

du

=

∫ x

0

cos(φ − φ′) − u

1 + u2 − 2u cos(φ − φ′)

du

= −1

2

∣

∣ln(1 − 2u cos(φ − φ′) + u2)∣

∣

x

0

= −1

2ln[1 − 2x cos(φ − φ′) + x2].

(I summed the infinite series here back in Problem 2.12. The integral in thesecond-to-last step can be done by partial fraction decomposition, although Icheated and looked it up on www.integrals.com). We can apply this resultindividually to the two terms in (22):

G(ρ, φ; ρ′, φ′) = − 1

4πln

[

1 + (ρρ′/b2)2 − 2(ρρ′/b2) cos(φ − φ′)

1 + (ρ</ρ>)2 − 2(ρ</ρ>) cos(φ − φ′)

]

= − 1

4πln

[(

ρ2>

b4

)

b4 + ρ2ρ′2 − 2ρρ′b2 cos(φ − φ′)

ρ2> + ρ2

< − 2ρ<ρ> cos(φ − φ′)

]

= − 1

4πln

[(

ρ2>

b4

)

b4 + ρ2ρ′2 − 2ρρ′b2 cos(φ − φ′)

ρ2> + ρ2

< − 2ρ<ρ> cos(φ − φ′)

]

= − 1

4πln

(

ρ2>

b2

)

− 1

4πln

[

b4 + ρ2ρ′2 − 2ρρ′b2 cos(φ − φ′)

b2(ρ2 + ρ′2 − 2ρρ′ cos(φ − φ′))

]

(23)

This is Jackson’s result, with an additional ln term thrown in for good measure.I’m not sure why Jackson didn’t quote this term as part of his answer; he didinclude it in his answer to problem 2.17 (c). Did I do something wrong?

(b) Now we want to plug the expression for G above into (16) to computethe potential within the cylinder. If there is no charge inside the cylinder, thevolume integral vanishes, and we are left with the surface integral:

Φ(ρ, φ) =

∫

Φ(b, φ′)∂G

∂ρ′

∣

∣

∣

∣

ρ′=b

dA′. (24)

where the integral is over the surface of the cylinder.For this we need the normal derivative of (23) on the cylinder:

∂G

∂ρ′= − 1

4π

2ρ2ρ′ − 2ρb2 cos(φ − φ′)

b4 + ρ2ρ′2 − 2ρρ′b2 cos(φ − φ′)− 2ρ′ − 2ρ cos(φ − φ′)

ρ2 + ρ′2 − 2ρρ′ cos(φ − φ′)

.


Evaluated at ρ′ = b this is

∂G

∂ρ′

∣

∣

∣

∣

ρ′=b

= − 1

2π

ρ2 − b2

b(ρ2 + b2 − 2ρb cos(φ − φ′))

.

In the surface integral, the extra factor of b on the bottom is cancelled by thefactor of b in the area element dA′, and (24) becomes just the result of Problem2.12.

(c) For the exterior problem we again start with the solution (21). Now theboundary conditions are different; the condition at ∞ gives A2m = 0, while thecondition at b gives

A1m = γmb−m B1m = −γmbm.

From the continuity condition at ρ′ = ρ we find

A2m = γmρm

[

(ρ

b

)m

−(

b

ρ

)m]

.

The finite derivative jump condition gives

−mγm

[

(ρ

b

)m

−(

b

ρ

)m]

1

ρ− mγm

[

(ρ

b

)m

+

(

b

ρ

)m]

1

ρ=

1

ρ

or

γm = − 1

2m

(

b

ρ

)m

.

Putting it all together we have for the exterior problem

gm =1

2m

[(

b2

ρρ′

)m

−(

ρ<

ρ>

)m]

.

This is the same gm we came up with before, but with b2 and ρρ′ terms flippedin first term. But the closed-form expression was symmetrical in those twoexpressions (except for the mysterious ln term) so the closed-form expressionfor the exterior Green’s function should be the same as the interior Green’sfunction.



Homer Reid

June 15, 2000


Problem 3.1

Two concentric spheres have radii a, b(b > a) and each is divided into two hemi-spheres by the same horizontal plane. The upper hemisphere of the inner sphereand the lower hemisphere of the outer sphere are maintained at potential V . Theother hemispheres are at zero potential.Detemine the potential in the region a ≤ r ≤ b as a series in Legendre polynomials.Include terms at least up to l = 4. Check your solution against known results inthe limiting cases b → ∞ and a → 0.

The expansion of the electrostatic potential in spherical coordinates for prob-lems with azimuthal symmetry is

Φ(r, θ) =

∞∑

l=0

[

Alrl + Blr

−(l+1)]

Pl(cos θ). (1)

We find the coefficients Al and Bl by applying the boundary conditions. Mul-tiplying both sides by Pl′(cos θ) and integrating from -1 to 1 gives

∫ 1

−1

Φ(r, θ)Pl(cos θ)d(cos θ) =2

2l + 1

[

Alrl + Blr

−(l+1)]

.

At r = a this yields

V

∫ 1

0

Pl(x)dx =2

2l + 1

[

Alal + Bla

−(l+1)]

,

1


and at r = b,

V

∫ 0

−1

Pl(x)dx =2

2l + 1

[

Albl + Blb

−(l+1)]

.

The integral from 0 to 1 vanishes for l even, and is given in the text for l odd:

∫ 1

0

Pl(x)dx = (−1

2)(l−1)/2 (l − 2)!!

2(

l+12

)

!.

The integral from -1 to 0 also vanishes for l even, and is just the above resultinverted for l odd. This gives

V (−1

2)(l−1)/2 (l − 2)!!

2(

l+12

)

!=

2

2l + 1

[

Alal + Bla

−(l+1)]

−V (−1

2)(l−1)/2 (l − 2)!!

2(

l+12

)

!=

2

2l + 1

[

Albl + Blb

−(l+1)]

.

or

αl = Alal + Bla

−(l+1)

−αl = Albl + Blb

−(l+1)

with

αl = V (−1

2)a(l−1)/2 (2l + 1)(l − 2)!!

4(

l+12

)

!.

The solution is

Al = αl

[

bl+1 + al+1

a2l+1 − b2l+1

]

Bl = −αl

[

al+1bl+1(bl + al)

a2l+1 − b2l+1

]

The first few terms of (1) are

Φ(r, θ) =3

4V

[

(a2 + b2)r

a3 − b3− a2b2(a + b)

r2(a3 − b3)

]

P1(cos θ)− 7

16

[

(a4 + b4)r3

a7 − b7− a4b4(a3 + b3)

r4(a7 − b7)

]

P3(cos θ)+· · ·

In the limit as b → ∞, the problem reduces to the exterior problem treatedin Section 2.7 of the text. In that limit, the above expression becomes

Φ(r, θ) → 3

4V(a

r

)2

P1(cos θ) − 7

16V(a

r

)4

P3(cos θ) + · · ·

in agreement with (2.27) with half the potential spacing. When a → 0, theproblem goes over to the interior version of the same problem, as treated insection 3.3 of the text. In that limit the above expression goes to

Φ(r, θ) → −3

4V(r

b

)

P1(cos θ) +7

16V(r

b

)3

P3(cos θ) + · · ·

This agrees with equation (3.36) in the text, with the sign of V flipped, becausehere the more positive potential is on the lower hemisphere.


Problem 3.2

A spherical surface of radius R has charge uniformly distributed over its surfacewith a density Q/4πR2, except for a spherical cap at the north pole, defined by thecone θ = α.

(a) Show that the potential inside the spherical surface can be expressed as

Φ =Q

8πε0

∞∑

l=0

1

2l + 1[Pl+1(cosα) − Pl−1(cosα)]

rl

Rl+1Pl(cos θ)

where, for l = 0, Pl−1(cos α) = −1. What is the potential outside?

(b) Find the magnitude and direction of the electric field at the origin.

(c) Discuss the limiting forms of the potential (part a) and electric field (part b) asthe spherical cap becomes (1)very small, and (2) so large that the area withcharge on it becomes a very small cap at the south pole.

(a) Let’s denote the charge density on the sphere by σ(θ). At a point infinites-imally close to the surface of the sphere, the electric field is

F = −∇Φ = − σ

ε0r

so∂Φ

∂r

∣

∣

∣

∣

r=R

=σ

ε0. (2)

The expression for the potential within the sphere must be finite at the origin,so the Bl in (1) are zero. Differentiating that expansion, (2) becomes

∂

∂rΦ(r, θ) =

∞∑

l=1

lAlrl−1Pl(cos θ)

Multiplying by Pl′ and integrating at r = R gives

1

ε0

∫ 1

−1

σ(θ)Pl(cos θ)d(cos θ) =2l

2l + 1AlR

l−1

so

Al =2l + 1

2lRl−1·(

Q

4πR2ε0

)∫ cos α

−1

Pl(x)dx.

To evaluate the integral we use the identity (eq. 3.28 in the text)

Pl(x) =1

(2l + 1)

d

dx[Pl+1(x) − Pl−1(x)]


so∫ cos α

−1

Pl(x)dx =1

2l + 1[Pl+1(cosα) − Pl−1(cosα)] .

(We used the fact that Pl+1(−1) = Pl−1(−1) for all l.) With this we have

Al =Q

8πε0lRl+1[Pl+1(cosα) − Pl−1(cosα)]

so the potential expansion is

Φ(r, θ) =Q

8πε0

∞∑

l=1

1

l[Pl+1(cosα) − Pl−1(cosα)]

rl

Rl+1Pl(cos θ).

Within the body of the sum, I have an l where Jackson has a 2l + 1. Also,he includes the l = 0 term in the sum, corresponding to a constant term inthe potential. I don’t understand how he can determine that constant from theinformation contained in the problem; the information about the charge densityonly tells you the derivative of the potential. There’s nothing in this problemthat fixes the value of the potential on the surface beyond an arbitrary constant.

(b) The field at the origin comes from the l = 1 term in the potential:

E(r = 0) = −∇Φ|r=0 = −∣

∣

∣

∣

∂Φ

∂rr +

1

r

∂Φ

∂θθ

∣

∣

∣

∣

r=0

= − Q

8πε0R2[P2(cosα) − 1]

[

P1(cos θ)r +d

dθP1(cos θ)θ

]

= − Q

8πε0R2

[

3

2cos2 α − 3

2

]

[

cos θr − sin θθ]

=3Q sin2 α

16πε0R2k.

The field points in the positive z direction. That makes sense, since a positivetest charge at the origin would sooner fly up out through the uncharged capthan through any of the charged surface.


Problem 3.3

A thin, flat, conducting, circular disk of radius R is located in the x − y planewith its center at the origin, and is maintained at a fixed potential V . With theinformation that the charge density on a disc at fixed potential is 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)l

2l + 1

(

R

r

2l)

P2l(cos θ)

(b) find the potential for r < R.

(c) What is the capacitance of the disk?

We are told that the surface charge density on the disk goes like

σ(r) = K(R2 − r2)−1/2

=K

R

[

1 +1

2

( r

R

)2

+3 · 1

(2!)(2 · 2)

( r

R

)4

+5 · 3 · 1

(3!)(2 · 2 · 2)

( r

R

)6

+ · · ·]

=K

R

∞∑

n=0

(2n − 1)!!

n! · 2n

( r

R

)2n

(3)

for some constant K. From the way the problem is worded, I take it we’renot supposed to try to figure out what K is explicitly, but rather to work theproblem knowing only the form of (3).

At a point infinitesimally close to the surface of the disk (i.e., as θ → π/2),the component of ∇Φ in the direction normal to the surface of the disk mustbe proportional to the surface charge. At the surface of the disk, the normaldirection is the negative θ direction. Hence

1

r

∂

∂θΦ(r, θ)

∣

∣

∣

∣

θ=(π/2)

= ± σ

ε0. (4)

with the plus (minus) sign valid for Φ above (below) the disc.For r < R the potential expansion is

Φ(r, θ) =

∞∑

l=0

AlrlPl(cos θ). (5)

Combining (3), (4), and (5) we have

∞∑

l=0

Alrl−1 d

dθPl(cos θ)

∣

∣

∣

∣

cos θ=0

= ± K

Rε0

∞∑

n=0

(2n − 1)!!

n! · 2n

( r

R

)2n

. (6)


For l even, dPl/dx vanishes at x = 0. For l odd, I used some of the Legendrepolynomial identities to derive the formula

d

dxP2l+1(x)

∣

∣

∣

∣

x=0

= (−1)l(2l + 1)(2l − 1)!!

l! · 2l.

This formula reminds one strongly of expansion (3). Plugging into (6) andequating coefficents of powers of r, we find

A2l+1 = ± (−1)lK

(2l + 1)R2l+1ε0

so

Φ(r, θ) = A0 ±K

ε0

∞∑

l=1

(−1)l

2l + 1

( r

R

)2l+1

P2l+1(cos θ).

I wrote A0 explicitly because we haven’t evaluated it yet–the derivative conditionwe used earlier gave no information about it. To find A0, observe that, on thesurface of the disk (cos θ = 0), all the terms in the above sum vanish ( becausePl(0) is 0 for odd l) so Φ = A0 on the disk. But Φ = V on the disk. Therefore,A0 = V . We have

Φ(r, θ) = V ± K

ε0

∞∑

l=1

(−1)l

2l + 1

( r

R

)2l+1

P2l+1(cos θ) (7)

where the plus (minus) sign is good for θ less than (greater than)π/2. Note thatthe presence of that ± sign preserves symmetry under reflection through the zaxis, a symmetry that is clearly present in the physical problem.

(a) For r > R, there is no charge. Thus the potential and its derivative must becontinuous everywhere–we can’t have anything like the derivative discontinuitythat exists at θ = π/2 for r < R. Since the physical problem is symmetric undera sign flip in cos θ, the potential expansion can only contain Pl terms for l even.The expansion is

Φ(r, θ) =

∞∑

l=0

B2lr−(2l+1)P2l(cos θ).

At r = R, this must match up with (7):

V ± K

ε0

∞∑

l=1

(−1)l

2l + 1P2l+1(cos θ) =

∞∑

l=0

B2lR−(2l+1)P2l(cos θ).

Multiplying both sides by P2l(cos θ) sin(θ) and integrating gives

B2l2R−(2l+1)

4l + 1= V

∫ 1

−1

Pl(x)dx +K

ε0

∞∑

l=1

(−1)l

2l + 1

−∫ 0

−1

P2l+1(x)P2l(x)dx +

∫ 1

0

P2l+1(x)Pl(x)dx

= 2V δl,0 +2K

ε0

∞∑

l=1

(−1)l

2l + 1

∫ 1

0

P2l+1(x)P2l(x)dx.


but I can’t do this last integral.

Problem 3.4

The surface of a hollow conducting sphere of inner radius a is divided into an even

number of equal segments by a set of planes; their common line of intersection isthe z axis and they are distributed uniformly in the angle φ. (The segments are likethe skin on wedges of an apple, or the earth’s surface between successive meridiansof longitude.) The segments are kept at fixed potentials ±V , alternately.

(a) Set up a series representation for the potential inside the sphere for the generalcase of 2n segments, and carry the calculation of the coefficients in the seriesfar enough to determine exactly which coefficients are different from zero. Forthe nonvanishing terms exhibit the coefficients as an integral over cos θ.

(b) For the special case of n = 1 (two hemispheres) determine explicitly the poten-tial up to and including all terms with l = 3. By a coordinate transformationverify that this reduces to result (3.36) of Section 3.3.

(a) The general potential expansion is

Φ(r, θ, φ) =

∞∑

l=0

l∑

m=−l

[

Almrl + Blmr−(l+1)]

Ylm(θ, φ). (8)

For the solution within the sphere, finiteness at the origin requires Blm = 0.Multiplying by Y ∗

l′m′and integrating over the surface of the sphere we find

Alm =1

al

∫

Φ(a, θ, φ) Y ∗

lm(θ, φ) dΩ

=V

al

n∑

k=1

(−1)k

∫ π

0

∫ 2kπ/n

2(k−1)π/n

Y ∗

lm(θ, φ) sin θ dφ dθ

=V

al

[

2l + 1

4π

(l − m)!

(l + m)!

]1/2∫ 1

−1

P ml (x) dx

n∑

k=1

(−1)k

∫ 2kπ/n

2(k−1)π/n

e−imφ dφ

.

(9)

The φ integral is easy:∫ 2kπ/n

2(k−1)π/n

e−imφ dφ = − 1

im

[

e−2imkπ/n − e−2im(k−1)π/n]

.

This is to be summed from k = 1 to n with a factor of (−1)k thrown in:

∑

= − 1

im

[

(e−2mπi(1/n) − 1) − (e−2mπi(2/n) − e−2mπi(1/n)) + · · · − (1 − e−2mπi((n−1)/n))]

=2

im

1 − e−2mπi/n + e2(−2mπi/n) − e3(−2mπi/n) + · · · + e(n−1)(−2mπi/n)

. (10)


Putting x = − exp(−2mπi/n), the thing in braces is

1 + x + x2 + x3 + · · · + xn−1 =1 − xn

1 − x=

1 − e−2mπi

1 + e−2imπ/n,

Note that the numerator vanishes. Thus the only way this thing can be nonzerois if the denominator also vanishes, which only happens if the exponent in thedenominator equates to -1. This only happens if m/n = 1/2, 3/2, 5/2, · · · . Inthat case, the 2mπi/n term in the exponent of the terms in (10) equates to πi,so all the terms with a plus sign in (10) come out to +1, while all the termswith a minus sign come out to -1, so all n terms add constructively, and (10)equates to

∑

=

2nim , m = n/2, 3n/2, 5n/2, · · ·0, otherwise.

Then the expression (9) for the coefficients becomes

Alm =2nV

imal

[

2l + 1

4π

(l − m)!

(l + m)!

]1/2 ∫ 1

−1

P ml (x)dx, m =

n

2,3n

2, · · · = 0, otherwise.

(b) As shown above, the only terms that contribute are those with m = n/2,m = 3n/2, et cetera. Of course there is also the constraint that m < l. Then,with n = 2, up to l = 3 the only nonzero terms in the series (9) are those withl = 1, m = ±1, and l = 3, m = ±1 or ±3. We need to evaluate the θ integralfor these terms. We have

∫ 1

−1

P 11 (x) dx = −

∫ 1

−1

(1 − x2)1/2 dx = −π

∫ 1

−1

P 13 (x) dx = −

∫ 1

−1

(1 − x2)1/2

[

15

2x2 − 3

2

]

dx = −3π

8∫ 1

−1

P 33 (x) dx = −15

∫ 1

−1

(1 − x2)3/2 dx = −15π

4.

Using these results in (??), we have

A1±1 = ±4πV i

a

[

3

4π · 2

]1/2

A3±1 = ±3πV i

2a3

[

7 · 24π · 4!

]1/2

A3±3 = ±5πV i

a3

[

7

4π · 6!

]1/2

Now we can plug these coefficients into (8) to piece together the solution.This involves some arithmetic in combining all the numerical factors in each


coefficient, which I have skipped here.

Φ(r, θ, φ) = V[

3(r

a

)

sin θ sin φ +7

16

( r

a

)3

sin θ(5 cos2 θ − 1) sinφ

+7

144

(r

a

)3

sin3 θ sin 3φ + · · ·]

Problem 3.6

Two point charges q and −q are located on the z azis at z = +a and z = −a,respectively.

(a) Find the electrostatic potential as an expansion in spherical harmonics andpowers of r for both r > a and r < a.

(b) Keeping the product qa = p/2 constant, take the limit of a → 0 and find thepotential for r 6= 0. This is by definition a dipole along the z azis and itspotential.

(c) Suppose now that the dipole of part b is surrounded by a grounded sphericalshell of radius b concentric with the origin. By linear superposition find thepotential everwhere inside the shell.

(a) First of all, for a point on the z axis the potential is

Φ(z) =q

4πε0

[

1

|z − a| −1

z + a

]

=q

4πε0z

[

1 +(a

z

)

+(a

z

)2

+ · · · −(

1 −(a

z

)

+(a

z

)2

· · ·)]

=q

2πε0z

[

(a

z

)

+(a

z

)3

+ · · ·]

for z > a. Comparing this with the general expansion Φ =∑

Blr−(l+1)Pl(cos θ)

at θ = 0 we can identify the Bls and write

Φ(r, θ) =q

2πε0r

[

(a

r

)

P1(cos θ) +(a

r

)3

P3(cos θ) + · · ·]

for r > a. For r < a we can just swap a and r in this equation.

(b)

Φ(r, θ) =qa

2πε0r2

[

P1(cos θ) +(a

r

)2

P3(cos θ) + · · ·]

=p

4πε0r2

[

P1(cos θ) +(a

r

)2

P3(cos θ) + · · ·]

→ p

4πε0r2cos θ as a → 0.


(c) When we put the grounded sphere around the two charges, a surface chargedistribution forms on the sphere. Let’s denote by Φs the potential due to thischarge distribution alone (not including the potential of the dipole) and by Φd

the potential due to the dipole. To calculate Φs, we pretend there are no chargeswithin the sphere, in which case we have the general expansion (1), with Bl = 0to keep us finite at the origin. The total potential is just the sum Φs + Φd:

Φ(r, θ) =p

4πε0r2cos θ +

∞∑

l=0

AlrlPl(cos θ).

The condition that this vanish at r = b ensures, by the orthogonality of the Pl,that only the l = 1 term in the sum contribute, and that

A1 = − p

4πε0b3.

The total potential inside the sphere is then

Φ(r, θ) =p

4πε0b2

(

1 − r

b

)

P1(cos θ).

Problem 3.7

Three point charges (q,−2q, q) are located in a straight line with separation a andwith the middle charge (−2q) at the origin of a grounded conducting spherical shellof radius b, as indicated in the figure.

(a) Write down the potential of the three charges in the absence of the groundedsphere. Find the limiting form of the potential as a → 0, but the productqa2 = Q remains finite. Write this latter answer in spherical coordinates.

(b) The presence of the grounded sphere of radius b alters the potential for r < b.The added potential can be viewed as caused by the surface-charge densityinduced on the inner surface at r = b or by image charges located at r > b. Uselinear superposition to satisfy the boundary conditions and find the potentialeverywhere inside the sphere for r < a and r > a. Show that in the limita → 0,

Φ(r, θ, φ) → Q

2πε0r3

(

1 − r5

b5

)

P2(cos θ).

(a) On the z axis, the potential is

Φ(z) =q

4πε0

[

−2

z+

1

|z − a| +1

z + a

]

=q

4πε0r

[

−2 +

(

1 +(a

z

)

+(a

z

)2

· · ·)

+

(

1 −(a

z

)

+(a

z

)2

+ · · ·)]

=q

2πε0z

[

(a

z

)2

+(a

z

)4

+ · · ·]

.


As before, from this result we can immediately infer the expression for thepotential at all points:

Φ(r, θ) =q

2πε0r

[

(a

r

)2

P2(cos θ) +(a

r

)4

P4(cos θ) + · · ·]

=qa2

2πε0r3

[

P2(cos θ) +(a

r

)2

P4(cos θ) + · · ·]

→ Q

2πε0r3P2(cos θ) as a → 0 (11)

(b) As in the previous problem, the surface charges on the sphere produce anextra contribution Φs to the potential within the sphere. Again we can expressΦs with the expansion (1) (with Bl = 0), and we add Φs to (11) to get the fullpotential within the sphere:

Φ(r, θ) =Q

2πε0r3P2(cos θ) +

∞∑

l=0

AlrlPl(cos θ)

From the condition that Φ vanish at r = b, we determine that only the l = 2term in the sum contributes, and that

A2 = − Q

2πε0b5.

Then the potential within the sphere is

Φ(r, θ) =Q

2πε0r3

[

1 −(r

b

)5]

P2(cos θ).

Problem 3.9

A hollow right circular cylinder of radius b has its axis coincident with the z axisand its ends at z = 0 and z = L. The potential on the end faces is zero, while thepotential on the cylindrical surface is given as V (φ, z). Using the appropriate sepa-ration of variables in cylindrical coordinates, find a series solution for the potentialanywhere inside the cylinder.

The general solution of the Laplace equation for problems in cylindrical coordi-nates consists of a sum of terms of the form

R(ρ)Q(φ)Z(z).

The φ function is of the form

Q(φ) = A sin νφ + B cos νφ


with ν an integer. The z function is of the form

Z(z) = Cekz + De−kz .

In this case, Z must vanish at z = 0 and z = L, which means we have to takek imaginary, i.e.

Z(z) = C sin(knz) with kn =πn

L, n = 1, 2, 3, · · ·

With this form for Z, R must be taken to be of the form

R(ρ) = EIν(knρ) + FKν(knρ).

Since we’re looking for the potential on the inside of the cylinder and there isno charge at the origin, the solution must be finite as ρ → 0, which requiresF = 0. Then the potential expansion becomes

Φ(ρ, φ, z) =

∞∑

n=1

∞∑

ν=0

[Anν sin νφ + Bnν cos νφ] sin(knz)Iν(knρ). (12)

Multiplying by sin ν′φ sin kn′z and integrating at r = b, we find∫ L

0

∫ 2π

0

V (φ, z) sin νφ sin(knz) dφ dz =πL

2Iν(knb)Anν

so

Anν =2

πLIν(knb)

∫ L

0

∫ 2π

0

V (φ, z) sin(νφ) sin(knz) dφ dz. (13)

Similarly,

Bnν =2

πLIν(knb)

∫ L

0

∫ 2π

0

V (φ, z) cos(νφ) sin(knz) dφ dz. (14)

Problem 3.10

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

V (φ, z) =

V for −π/2 < φ < π/2−V for π/2 < φ < 3π/2

(a) Find the potential inside the cylinder.

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

The potential expansion is (12) with coefficients given by (13) and (14). Therelevant integrals are

∫ L

0

∫ 2π

0

V (φ, z) sin(νφ) sin(knz) dφ dz


= V

∫ L

0

sin(knz) dz

∫ π/2

−π/2

sin(νφ) dφ −∫ 3π/2

π/2

sin(νφ) dφ

= 0∫ L

0

∫ 2π

0

V (φ, z) cos(νφ) sin(knz) dφ dz

= V

∫ L

0

sin(knz) dz

∫ π/2

−π/2

cos(νφ) dφ −∫ 3π/2

π/2

cos(νφ) dφ

=2V

νkn

|sin νφ|π/2−π/2 − |sin νφ|3π/2

π/2

(n odd)

=

0 , n or ν even8V/knν , n odd, ν = 1, 5, 9, · · ·

−8V/knν , n odd, ν = 3, 7, 11, · · ·

Hence, from (13) and (14),

Anν = 0Bnν = 0, n or ν even

= (−1)(ν−1)/2 · 16V/(nνπ2Iν(knb)), n and ν odd

The potential expansion is

Φ(ρ, θ, z) =16V

π2

∑

n,ν

(−1)(ν−1)/2

nνIv(knb)cos(νφ) sin(knz)Iν(knρ) (15)

where the sum contains only terms with n and ν odd.

(b) At z = L/2 we have

Φ(ρ, θ, L/2) =16V

π2

∑

n,ν

(−1)(n+ν−2)/2

nνcos(νφ)

Iν (knρ)

Iν (knb).

As L → ∞, the arguments to the I functions become small. Using the limitingform for Iν quoted in the text as equation (3.102), we have

Φ(ρ, θ) =16V

π2

∑

n,ν

(−1)(n+ν−2)/2

nνcos(νφ)

(ρ

b

)ν

.

The sums over n and ν are now decoupled:

Φ(ρ, θ) =16V

π2

∞∑

n=0

(−1)n

2n + 1

∞∑

ν=0

(−1)ν

2ν + 1cos(νφ)

(ρ

b

)ν

=16V

π2

π

4

∞∑

ν=0

(−1)ν

2ν + 1cos(νφ)

(ρ

b

)ν

=4V

πtan−1

(

2ρb cosφ

b2 − ρ2

)


This agrees with the result of Problem 2.13, with V1 = −V2 = V . The firstseries is just the Taylor series for tan−1(x) at x = 1, so it sums to π/4. Thesecond series can also be put into the form of the Taylor series for tan−1(x),using tricks exactly analogous to what I did in my solution for Problem 2.13.



Homer Reid

June 15, 2000


Problem 3.11

A modified Bessel-Fourier series on the interval 0 ≤ ρ ≤ a for an arbitrary functionf(ρ) can be based on the ”homogenous” boundary conditions:

At ρ = 0, ρJν(kρ)d

dρJν(k′ρ) = 0

At ρ = a,d

dρln[Jν(kρ)] = −λ

a(λ real)

The first condition restricts ν. The second condition yields eigenvalues k = yνn/a,where yνn is the nth positive root of x dJν(x)/dx + λJν(x) = 0.

(a) Show that the Bessel functions of different eigenvalues are orthogonal in theusual way.

(b) Find the normalization integral and show that an arbitrary function f(ρ) canbe expanded on the interval in the modified Bessel-Fourier series

f(ρ) =

∞∑

n=1

AnJν

(yνn

a

)

with the coefficients An given by

An =2

a2

[

(

1 − ν2

y2νn

)

J2ν (yνn) +

(

dJν(yνn)

dyνn

)2]

−1∫ a

0

f(ρ)ρJν

(yνnρ

a

)

dρ.

1


(a) The function Jν(kρ) satisfies the equation

1

ρ

d

dρ

[

ρd

dρJν(kρ)

]

+

(

k2 − ν2

ρ2

)

Jν(kρ) = 0. (1)

Multiplying both sides by ρJν(k′ρ) and integrating from 0 to a gives

∫ a

0

Jν(k′ρ)d

dρ

[

ρd

dρJν(kρ)

]

+

(

k2ρ − ν2

ρ

)

Jν(k′ρ)Jν(kρ)

dρ = 0. (2)

The first term on the left can be integrated by parts:

∫ a

0

Jν(k′ρ)d

dρ

[

ρd

dρJν(kρ)

]

dρ

=

∣

∣

∣

∣

ρJν(k′ρ)d

dρJν(kρ)

∣

∣

∣

∣

a

0

−∫ a

0

ρ

[

d

dρJν(k′ρ)

] [

d

dρJν(kρ)

]

dρ. (3)

One of the conditions we’re given is that the thing in braces in the first termhere vanishes at ρ = 0. At ρ = a we can invoke the other condition:

d

dρln[Jν(kρ)]

∣

∣

∣

∣

ρ=a

=1

Jν(kρ)

d

dρJν(kρ)

∣

∣

∣

∣

ρ=a

= −λ

a

→ ad

dρJν(ka) = −λJν(ka).

Plugging this into (3), we have

∫ a

0

Jν(k′ρ)d

dρ

[

ρd

dρJν(kρ)

]

dρ

= −λJν(k′ρ)Jν(kρ) −∫ a

0

ρ

[

d

dρJν(k′ρ)

] [

d

dρJν(kρ)

]

. (4)

This is clearly symmetric in k and k′, so when we write down (2) with k andk′ switched and subtract from (2), the first integral (along with the ν2/ρ term)vanishes, and we are left with

(k′2 − k2)

∫ a

0

ρJν(k′ρ)Jν(kρ) dρ = 0

proving orthogonality.

(b) If we multiply (1) by ρ2J ′(kρ) and integrate, we find

∫ a

0

ρJ ′

ν(kρ)d

dρ[ρJ ′

ν(kρ)]dρ+k2

∫ a

0

ρ2Jν(kρ)J ′

ν(kρ)dρ−ν2

∫ a

0

Jν(kρ)J ′

ν(kρ)dρ = 0.

(5)


The first and third integrals are of the form∫

f(x)f ′(x)dx and can be doneimmediately. In the second integral we put f(ρ) = ρ2Jν(kρ), g′(ρ) = J ′

ν(kρ)and integrate by parts:∫ a

0

ρ2Jν(kρ)J ′

ν(kρ)dρ =∣

∣ρ2J2ν (kρ)

∣

∣

a

0− 2

∫ a

0

ρJ2ν (kρ)dρ −

∫ a

0

ρ2Jν(kρ)J ′

ν(kρ)dρ

→∫ a

0

ρ2Jν(kρ)J ′

ν(kρ)dρ =1

2a2J2

ν (ka) −∫ a

0

ρJ2ν (kρ)dρ.

Using this in (5),

a2

2J ′2

ν (ka) +(ak)2

2aJ2

ν (ka) − k2

∫ a

0

ρJ2ν (kρ)dρ − ν2

2J2

ν (ka) = 0

so∫ a

0

ρJ2ν (kρ)dρ =

(

a2

2− ν2

2k2

)

J2ν (ka) +

a2

2k2J ′2

ν (ka)

=a2

2

(

1 − ν2

(ka)2

)

J2ν (ka) +

[

d

d(ka)Jν(ka)

]2

This agrees with what Jackson has if you note that k is chosen such that ka =ynm.

Problem 3.12

An infinite, thin, plane sheet of conducting material has a circular hole of radius acut in it. A thin, flat, disc of the same material and slightly smaller radius lies inthe plane, filling the hole, but separated from the sheet by a very narrow insulatingring. The disc is maintained at a fixed potential V , whilc the infinite sheet is keptat zero potential.

(a) Using appropriate cylindrical coordinates, find an integral expression involvingBessel functions for the potential at any point above the plane.

(b) Show that the potential a perpendicular distance z above the center of the discis

Φ0(z) = V

(

1 − z√a2 + z2

)

(c) Show that the potential a perpendicular distance z above the edge of the discis

Φa(z) =V

2

[

1 − kz

πaK(k)

]

where k = 2a/(z2 + 4a2)1/2, and K(k) is the complete elliptic integral of thefirst kind.


(a) As before, we can write the potential as a sum of terms R(ρ)Q(φ)Z(z). Inthis problem there is no φ dependence, so Q = 1. Also, the boundary conditionson Z are that it vanish at ∞ and be finite at 0, whence Z(z) ∝ exp(−kz) forany k. Then the potential expansion becomes

Φ(ρ, z) =

∫

∞

0

A(k)e−kzJ0(kρ) dk. (6)

To evaluate the coefficients A(k), we multiply both sides by ρJ0(k′ρ) and inte-

grate over ρ at z = 0:

∫

∞

0

ρΦ(ρ, 0)J0(k′ρ) dρ =

∫

∞

0

A(k)

∫

∞

0

ρJ0(kρ)J0(k′ρ) dρ

dk

=A(k′)

k′

so

A(k) = k

∫

∞

0

ρΦ(ρ, 0)J0(kρ) dρ

= kV

∫ a

0

ρJ0(kρ)dρ.

Plugging this back into (6),

Φ(ρ, z) = V

∫

∞

0

∫ a

0

kρ′e−kzJ0(kρ)J0(kρ′) dρ′ dk. (7)

The ρ′ integral can be done right away. To do it, I appealed to the differentialequation for J0:

J ′′

0 (u) +1

uJ ′

0(u) + J0(u) = 0

so∫ x

0

uJ0(u) du = −∫ x

0

uJ ′′

0 du −∫ x

0

J ′

0(u) du

= − |uJ ′

0(u)|x0 +

∫ x

0

J ′

0(u) du −∫ x

0

J ′

0(u) du

= − |uJ ′

0(u)|x0 = −xJ ′

0(x) = xJ1(x).

(In going from the first to second line, I integrated by parts.) Then (7) becomes

Φ(ρ, z) = aV

∫

∞

0

J1(ka)J0(kρ)e−kz dk. (8)


(b) At ρ = 0, (7) becomes

Φ(0, z) = V J0(0)

∫ a

0

ρ′∫

∞

0

ke−kzJ0(kρ′)dk

dρ′

= V

∫ a

0

ρ′

− ∂

∂z

∫

∞

0

e−kzJ0(kρ′)dk

dρ′

= V

∫ a

0

ρ′

− ∂

∂z

(

1√

ρ′2 + z2

)

dρ′

= V

∫ a

0

zρ′

(ρ′2 + z2)3/2dρ′

Here we substitute u = ρ′2 + z2, du = 2ρ′dρ:

Φ(0, z) =V zJ0(0)

2

∫ a2+z2

z2

u−3/2 du

= −V z

∣

∣

∣

∣

1

u1/2

∣

∣

∣

∣

a2+z2

z2

= V z

[

1

z− 1√

z2 + z2

]

= V

[

1 − z√a2 + z2

]

(b) At ρ = a, (8) becomes

Φ(a, z) = aV

∫

∞

0

J1(ka)J0(ka)e−kz dk

Problem 3.13

Solve for the potential in Problem 3.1, using the appropriate Greenfunction obtained in the text, and verify that the answer obtainedin this way agrees with the direct solution from the differentialequation.

For Dirichlet boundary value problems, the basic equation is

Φ(x) = − 1

ε0

∫

V

G(x;x′)ρ(x′) dV ′ +

∮

S

Φ(x′)∂G(x;x′)

∂n

∣

∣

∣

∣

x′

dA′. (9)

Here there is no charge in the region of interest, so only the surface integralcontributes. The Green’s function for the two-sphere problem is

G(x;x′) = −∞∑

l=0

l∑

m=−l

Y ∗

lm(θ′, φ′) Ylm(θ, φ)

2l + 1Rl(r; r

′) (10)


with

Rl(r; r′) =

1[

1 −(

ab

)]2l+1

(

rl< − a2l+1

rl+1<

)(

1

rl+1>

− rl>

b2l+1

)

. (11)

Actually in this case the potential cannot have any Φ dependence, so all termswith m 6= 0 in (10) vanish, and we have

G(x;x′) = − 1

4π

∞∑

l=0

Pl(cos θ)Pl(cos θ′)Rl(r; r′).

In this case the boundary surfaces are spherical, which means the normal to asurface element is always in the radial direction:

∂

∂nG(x;x′) = − 1

4π

∞∑

l=0

Pl(cos θ)Pl(cos θ′)∂

∂nRl(r; r

′).

The surface integral in (9) has two parts: one integral S1 over the surface ofthe inner sphere, and a second integral S2 over the surface of the outer sphere:

S1 = − 1

4π

∞∑

l=0

Pl(cos θ)∂Rl

∂n

∣

∣

∣

∣

r′=a

∫ π

0

∫ 2π

0

Φ(a, θ′)Pl(cos θ′)a2 sin θ′ dφ dθ′

= −V

2

∞∑

l=0

a2Pl(cos θ)∂Rl

∂n

∣

∣

∣

∣

r′=a

∫ 1

0

Pl(x) dx

= −V

2

∞∑

l=0

a2γlPl(cos θ) · ∂Rl

∂n

∣

∣

∣

∣

r′=a

where

γl =

∫ 1

0

Pl(x) dx

= (−1

2)(l−1)/2 (l − 2)!!

2[(l + 1)/2]!, l odd

= 0, l even.

A similar calculation gives

S2 = −V

2

∞∑

l=0

b2Pl(cos θ)∂Rl

∂n

∣

∣

∣

∣

r′=b

∫ 0

−1

Pl(x) dx

=V

2

∞∑

l=0

b2γlPl(cos θ)∂Rl

∂n

∣

∣

∣

∣

r′=b

because Pl is odd for l odd, so its integral from -1 to 0 is just the negative ofthe integral from 0 to 1. The final potential is the sum of S1 and S2:

Φ(r, θ) =V

2

∞∑

l=0

γlPl(cos θ)

∣

∣

∣

∣

r′2∂Rl

∂n

∣

∣

∣

∣

r′=b

r′=a

(12)


Since the point of interest is always between the two spheres, to find thenormal derivative at r = a we differentiate with respect to r<, and at r = bwith respect to r>. Also, at r = a the normal is in the +r direction, while atr = b the normal is in the negative r direction.

a2 ∂

∂nRl(r; r

′)

∣

∣

∣

∣

r′=a

= (2l + 1)a2 al−1

[

1 −(

ab

)]2l+1

(

1

rl+1− rl

b2l+1

)

b2 ∂

∂nRl(r; r

′)

∣

∣

∣

∣

r′=b

= (2l + 1)b2 b−(l+2)

[

1 −(

ab

)]2l+1

(

rl − a2l+1

rl+1

)

Combining these with some algebra gives

Φ(r, θ) =V

2

∞∑

l=0

(2l + 1)γlPl(cos θ)

[

(ab)l+1(bl + al)r−(l+1) − (al+1 + bl+1)rl

b2l+1 − a2l+1

]

in agreement with what we found in Problem 3.1.

Problem 3.14

A line charge of length 2d with a total charge Q has a linear chargedensity varying as (d2 − z2), where z is the distance from the mid-point. A grounded, conducting spherical shell of inner radius b > dis centered at the midpoint of the line charge.

(a) Find the potential everywhere inside the spherical shell as anexpansion in Legendre polynomials.

(b) Calculate the surface-charge density induced on the shell.

(c) Discuss your answers to parts a and b in the limit that d << b.

First of all, we are told that the charge density ρ(z) = λ(d2 − z2), and thatthe total charge is Q, whence

Q = 2λ

∫ d

0

(d2 − z2)dz =4

3d3λ

→ λ =3Q

4d3.

In this case we have azimuthal symmetry, so the Green’s function is

G(x;x′) = − 1

4π

∞∑

l=0

Pl(cos θ′)Pl(cos θ)Rl(r; r′) (13)


with

Rl(r; r′) = rl

<

(

1

rl+1>

− rl>

b2l+1

)

.

Since the potential vanishes on the boundary surface, the potential inside thesphere is given by

Φ(r, θ) = − 1

ε0

∫

V

G(r, θ; r′, θ′)ρ(r′, θ′)dV.

In this case ρ is only nonzero on the z axis, where r = z. Also, Pl(cos θ)=1 forz > 0, and (−1)l for z < 0. This means that the contributions to the integralfrom the portions of the line charge for z > 0 and z < 0 cancel out for odd l,and add constructively for even l:

Φ(r, θ) =1

4πε0

∞∑

l=0,2,4,...

Pl(cos θ)

[

2

∫ d

0

Rl(r; z)ρ(z) dz

]

We have

∫ d

0

Rl(r; z)ρ(z) dz = λ

∫ d

0

rl<

(

1

rl+1>

− rl>

b2l+1

)

(d2 − z2) dz

This is best split up into two separate integrals:

= λ

∫ d

0

rl<

rl+1>

(d2 − z2) dz − λ

b2l+1

∫ d

0

rl<rl

>(d2 − z2) dz

The second integral is symmetric between r and r′, so we may integrate itdirectly:

− λ

b2l+1

∫ d

0

rl<rl

>(d2 − z2) dz = − λrl

b2l+1

∫ d

0

zl(d2 − z2) dz

= − λrl

b2l+1

[

dl+3

l + 1− dl+3

l + 3

]

= − λrldl+3

(l + 1)(l + 3)b2l+1(14)

The first integral must be further split into two:

λ

∫ d

0

rl<

rl+1>

(d2 − z2) dz


= λ

1

rl+1

∫ r

0

zl(d2 − z2) dz + rl

∫ d

r

d2 − z2

zl+1dz

= λ

1

rl+1

[

d2rl+1

l + 1− rl+3

l + 3

]

+ rl

∣

∣

∣

∣

− d2

lzl+

1

(l − 2)zl−2

∣

∣

∣

∣

d

r

= λ

d2

l + 1− r2

l + 3+(r

d

)l

d2 2

l(l + 2)− d2

l+

r2

l + 2

= λ

r2

(l + 2)(l + 3)− d2

l(l + 1)+(r

d

)l

d2 2

l(l + 2)

Combining this with (14), we have

∫ d

0

Rl(r; z)ρ(z) dz = λ

r2

(l + 2)(l + 3)− d2

l(l + 1)+(r

d

)l

d2 2

l(l + 2)− rldl+3

(l + 1)(l + 3)b2l+1

(15)But something is wrong here, because with this result the final potential willcontain terms like r0Pl(cos θ) and r2Pl(cos θ), which do not satisfy the Laplaceequation.


Problem 3.15

Consider the following “spherical cow” model of a battery con-nected to an external circuit. A sphere of radius a and conductiv-ity σ is embedded in a uniform medium of conductivity σ′. Insidethe sphere there is a uniform (chemical) force in the z directionacting on the charge carriers; its strength as an effective electricfield entering Ohm’s law is F . In the steady state, electric fieldsexist inside and outside the sphere and surface charge resides onits surface.

(a) Find the electric field (in addition to F ) and current densityeverywhere in space. Determine the surface-charge densityand show that the electric dipole moment of the sphere isp = 4πε0σa3F/(σ + 2σ′).

(b) Show that the total current flowing out through the upperhemisphere of the sphere is

I =2σσ′

σ + 2σ′· πa2F

Calculate the total power dissipation outside the sphere. Us-ing the lumped circuit relations, P = I2Re = IVe, find theeffective external resistance Re and voltage Ve.

(c) Find the power dissipated within the sphere and deduce theeffective internal resistance Ri and voltage Vi.

(d) Define the total voltage through the relation Vt = (Re + Ri)Iand show that Vt = 4aF/3, as well as Ve + Vi = Vt. Showthat IVt is the power supplied by the “chemical” force.

(a) What’s going on in this problem is that the conductivity has a discontinu-ity going across the boundary of the sphere, but the current density must beconstant there, which means there must an electric field discontinuity in inverseproportion to the conductivity discontinuity. To create this electric field discon-tinuity, there has to be some surface charge on the sphere, and this charge givesrise to extra fields both inside and outside the sphere.

Since there is no charge inside or outside the sphere, the potential in thosetwo regions satisfied the Laplace equation, and may be expanded in Legendrepolynomials:


for r < a, Φ(r, θ) = Φin(r, θ) =

∞∑

l=0

AlrlPl(cos θ)

for r > a, Φ(r, θ) = Φout(r, θ) =

∞∑

l=0

Blr−(l+1)Pl(cos θ)

Continuity at r = a requires that

Alal = Bla

−l+1 → Bl = a2l+1Al

so

Φ(r, θ) =

Φin(r, θ) =∑

∞

l=0 AlrlPl(cos θ), r < a

Φout(r, θ) =∑

∞

l=0 Ala2l+1r−(l+1)Pl(cos θ), r > a.

(16)

Now, in the steady state there can be no discontinuities in the current den-sity, because if there were than there would be more current flowing into someregion of space than out of it, which means charge would pile up in that region,which would be a growing source of electric field, which would mean we aren’tin steady state. So the current density is continuous everywhere. In particular,the radial component of the current density is continuous across the boundaryof the sphere, i.e.

Jr(r = a−, θ) = Jr(r = a+, θ). (17)

Outside of the sphere, Ohm’s law says that

J = σ′E = −σ′∇Φout.

Inside the sphere, there is an extra term coming from the chemical force:

J = σ(E + F k) = σ(−∇Φin + F k).

Applying (17) to these expressions, we have

σ

(

− ∂

∂rΦin

∣

∣

∣

∣

r=a

+ F cos θ

)

= −σ′∂

∂rΦout

∣

∣

∣

∣

r=a

Using (16), this is

FP1(cos θ) −∞∑

l=0

lAlal−1Pl(cos θ) =

(

σ′

σ

) ∞∑

l=0

(l + 1)Alal−1Pl(cos θ).

Multiplying both sides by Pl′ (cos θ) and integrating from −π to π, we find

F − A1 =

(

σ′

σ

)

2A1 (18)


for l=1, and

−lAl =

(

σ′

σ

)

(l + 1)Al (19)

(20)

for l 6= 1. Since the conductivity ratio is positive, the second relation is impos-sible to satisfy unless Al = 0 for l 6= 1. The first relation becomes

A1 =σ

σ + 2σ′F.

Then the potential is

Φ(r, θ) =

σσ+2σ′

Fr cos θ, r < a

σσ+2σ′

Fa3r−2 cos θ, r > a(21)

The dipole moment p is defined by

Φ(r, θ) → 1

4πε0

p · rr3

as r → ∞. (22)

The external portion of (21) can be written as

Φ(r, θ) =σ

σ + 2σ′

Fa3z

r3

and comparing this with (22) we can read off

p = 4πε0σ

σ + 2σ′Fa3k.

The electric field is found by taking the gradient of (21):

E(r, θ) =

− σσ+2σ′

F k, r < a

σσ+2σ′

F(

ar

)3(2 cos θr + sin θθ), r > a

The surface charge σs(θ) on the sphere is proportional to the discontinuityin the electric field:

σs(θ) = ε0[Er(r = a+) − Er(r = a−)]

=3ε0σ

σ + 2σ′F cos θ.

(b) The current flowing out of the upper hemisphere is just∫

J · dA = σ

∫

(Ein + F k) · dA

= σ

(

1 − σ

σ + 2σ′

)

F

∫ π/2

0

∫ 2π

0

cos θ sin θ a2 dφ dθ

= 2σσ′

σ + 2σ′· πa2F (23)


The Ohmic power dissipation in a volume dV is

dP = σE2dV (24)

To see this, suppose we have a rectangular volume element with sides dx, dy,and dz. Consider first the current flowing in the x direction. The current densitythere is σEx and the cross-sectional area is dydz, so I = σExdydz. Also, thevoltage drop in the direction of current flow is V = Exdx. Hence the powerdissipation due to current in the x direction is IV = σE2

xdV . Adding in thecontributions from the other two directions gives (24).

For the power dissipated outside the sphere we use the expression for theelectric field we found earlier:

Pout = σ′

∫

∞

a

∫ π

0

∫ 2π

0

E2(r, θ, φ)r2 sin θ dφ dθ dr

= 2πσ′

(

σ

σ + 2σ′

)2

F 2a6

∫

∞

a

∫ π

0

1

r4(4 cos2 θ + sin2 θ) sin θ dθ dr

=8π

3σ′

(

σ

σ + 2σ′

)2

F 2a3

Dividing by (23), we find the effective external voltage Ve:

Ve = Pout/I =4

3aF · σ

σ + 2σ′

and the effective external resistance:

Re = Pout/I2 =2

3πaσ′.

(c) The power dissipated inside the sphere is

Pin = σ

∫

(E + F k)2dV =4σσ2′

(σ + 2σ′)2F 2

∫

dV

=16σσ2′

3(σ + 2σ′)2πa3F 2

Since we’re in steady state, the current flowing out through the upper hemi-sphere of the sphere must be replenished by an equal current flowing in throughthe lower half of the sphere, so to find the internal voltage and resistance wecan just divide by (23):

Vi = Pin/I =8

3

σ′

σ + 2σ′aF

Ri = Pin/I2 =4

3πaσ.


(c)

(Re + Ri)I =2

3πa

(

1

σ′+

2

σ

)

· 2σσ′

σ + 2σ′πa2F =

4

3aF

(Vi + Ve) =4aF

3(σ + 2σ′)σ + 2σ′ =

4

3aF

Problem 3.17

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

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

G(x,x′) = − 1

πL×

∞∑

n=1

∞∑

m=−∞

eim(φ−φ′) sin(nπz

L

)

sin

(

nπz′

L

)

Im

(nπρ′<L

)

Km

(nπρ>

L

)

.

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

G(x,x′) = − 1

2π×

∞∑

m=−∞

∫

∞

0

dk eim(φ−φ′)Jm(kρ)Jm(kρ′)sinh(kz<) sinh[k(L − z>)]

sinh(kL).

In cylindrical coordinates, the solutions of the Laplace equation look likelinear combinations of terms of the form

Tmk(ρ, φ, z) = eimφZ(kz)Rm(kρ). (25)

There are two possibilities for the combination Z(kz)Rm(kρ), both of whichsolve the Laplace equation:

Z(kz)Rm(kρ) = (Aekz + Be−kz)[CJm(kρ) + DNm(kρ)] (26)

or

Z(kz)Rm(kρ) = (Aeikz + Be−ikz)[CIm(kρ) + DKm(kρ)]. (27)

The Green’s function G(x;x′) must be a solution of the Laplace equation,and must thus take one of the above forms, at all points x′ 6= x. At x′ = x,G must be continuous, but have a finite discontinuity in its first derivative.


Furthermore, G must vanish on the boundary surfaces. These conditions maybe met by dividing space into two regions, one on either side of the source pointx, and taking G to be different linear combinations of terms T (as in (25)) inthe two regions. The question is, in which dimension (i.e., ρ, z, or φ) do wedefine the two “sides” of the source point?

(a) The first option is to imagine a cylindrical boundary at ρ′ = ρ, i.e. at theradius of the source point, and take the inside and outside of the cylinder (i.e.,ρ′ < ρ and ρ′ > ρ) as the two distinct regions of space. Then, within eachregion, the entire range of z must be handled by one function, which means thisone function must vanish at z = 0 and z = L. This cannot happen with terms ofthe form (26), so we are forced to take Z and R as in (27), with B = −A and krestricted to the discrete values kn = nπ/L. Next considering the singularitiesof the ρ functions in (27), we see that, to keep G finite everywhere, for theinner region (ρ′ < ρ) we can only keep the Im(kρ) term, while for the outerregion we can only keep the Km(kρ) term. Then G(x;x′) will consist of linearcombinations of terms T as in (25) subject to the restrictions discussed above:

G(x;x′) =

∑

mn Amn(x)eimφ′

sin(knz′)Im(knρ′), ρ′ < ρ∑

mn Bmn(x)eimφ′

sin(knz′)Km(knρ′), ρ′ > ρ.

Clearly, to establish continuity at ρ′ = ρ, we need to take Amk(x) = γmk(z, φ)Km(kρ)and Bmk(x) = γmk(z, φ)Im(kρ), where γmk is any function of z and φ. Thenwe can write G as

G(x;x′) =∑

mk

γmk(z, φ)eimφ′

sin(kz′)Im(kρ<)Km(kρ>).

The obvious choice of γmk needed to make this a delta function in z and φ isγmk = (4/L)e−imφ sin(kz). Then we have

G(x;x′) =4

L

∑

mk

eim(φ′−φ) sin(kz) sin(kz′)Im(kρ<)Km(kρ>).

What I don’t quite understand is that this expression already has the correctdelta function behavior in ρ, even though I never explicitly required this. Toobtain this expression I first demanded that it satisfy the Laplace equation forall points x′ 6= x, that it satisfy the boundary conditions of the geometry, andthat it have the right delta function behavior in z′ and φ′. But I never demandedthat it have the correct delta function behavior in ρ′, and yet it does. I guessthe combination of the requirements that I did impose on this thing is alreadyenough to ensure that it meets the final requirement.

(b) The second option is to imagine a plane boundary at z ′ = z, and take thetwo distinct regions to be the regions above and below the plane. In other words,the first region is that for which 0 ≤ z′ ≤ z, and the second region that for whichz ≤ z′ ≤ L. In this case, within each region the entire range of ρ′ (from 0 to ∞)must be handled by one function. This requirement excludes terms of the form


(27), because Km is singular at the origin, while Im is singular at infinity, andthere is no linear combination of these functions that will be finite over the wholerange of ρ′. Hence we must use terms of the form (26). To ensure finiteness atthe origin we must exlude the Nm term, so D = 0. To ensure vanishing at z′ = 0we must take A = −B, so the z′ function in the region 0 ≤ z′ ≤ z is proportionalto sinh(kz′). To ensure vanishing at z′ = L we must take A = −Be−2kL, sothe z′ function in the region z ≤ z′ ≤ L is proportional to sinh[k(z′ − L)].With these restrictions, the differential equation and the boundary conditionsare satisfied for all terms of the form (25) with no limitation on k. Hence theGreen’s function will be an integral, not a sum, over these terms:

G(x′;x) =

∑

∞

m=0

∫

∞

0Am(k, ρ, φ, z)eimφ′

sinh(kz′)Jm(kρ′) dk, 0 ≤ z′ ≤ z∑

∞

m=0

∫

∞

0 Bm(k, ρ, φ, z)eimφ′

sinh[k(z′ − L)]Jm(kρ′) dk, z ≤ z′ ≤ L

Problem 3.18

The configuration of Problem 3.12 is modified by placing a conducting plane heldat zero potential parallel to and a distance L away from the plane with the discinsert in it. For definiteness put the grounded plane at z = 0 and the other planewith the center of the disc on the z axis at z = L.

(a) Show that the potential between the planes can be written in cylindrical coor-dinates (z, ρ, φ) as

Φ(z, ρ) = V

∫

∞

0

dλJ1(λ)J0(λρ/a)sinh(λz/a)

sinh(λL/a).

(b) Show that in the limit a → ∞ with z, ρ, L fixed the solution of part a reducesto the expected result. Viewing your result as the lowest order answer in anexpansion in powers of a−1, consider the question of corrections to the lowestorder expression if a is large compared to ρ and L, but not infinite. Are theredifficulties? Can you obtain an explicit estimate of the corrections?

(c) Consider the limit of L → ∞ with (L − z), a and ρ fixed and show that theresults of Problem 3.12 are recovered. What about corrections for L a, butnot L → ∞?

(a) The general solution of the Laplace equation in cylindrical coordinates withangular symmetry that vanishes at z = 0 is

Φ(ρ, z) =

∫

∞

0

A(k)J0(kρ) sinh(kz) dk. (28)


Multiplying both sides by ρJ0(k′ρ) and integrating at z = L yields

∫

∞

0

ρJ0(k′ρ)Φ(ρ, L) dρ =

∫

∞

0

A(k) sinh(kL)

∫

∞

0

ρJ0(k′ρ)J0(kρ) dρ

dk

=

∫

∞

0

A(k) sinh(kL)

1

kδ(k − k′)

dk

=1

k′A(k′) sinh(k′L)

so

A(k) =k

sinh(kL)

∫

∞

0

ρJ0(kρ)Φ(ρ, L) dρ

=V k

sinh(kL)

∫ a

0

ρJ0(kρ) dρ

=V

k sinh(kL)

∫ ka

0

uJ0(u) du. (29)

I worked out this integral earlier, in Problem 3.12:∫ x

0

uJ0(u) du = xJ1(x).

Then (29) becomes

A(k) =V

k sinh(kL)· (ka)J1(ka)

and (28) is

Φ(ρ, z) = V

∫

∞

0

aJ1(ka)J0(kρ)sinh(kz)

sinh(kL)dk

= V

∫

∞

0

J1(λ)J0(λρ/a)sinh(λz/a)

sinh(λL/a)dλ. (30)

(b) For x 1,

J0(x) → 1 − 1

4x2 + · · ·

and for x 1 and y 1,

sinh(x)

sinh(y)=

x + 16x3 + · · ·

y + 16y3 + · · · =

x

y

[

1 +1

6(x2 − y2)

]

+ O(x4)

With these approximations we may expand the terms containing a in (30):

J0(λρ/a)sinh(λz/a)

sinh(λL/a)≈(

1− 1

4

(

λρ

a

)2)

( z

L

)

(

1 +1

6

(

λ

a

)2

(x2 − y2)

)

(31)

=z

L

[

1 −(

λ

a

)2(1

6(L2 − z2) +

1

4ρ2

)

+ · · ·]

(32)


Then the potential expansion (30) becomes

Φ(ρ, z) =V z

L

[∫

∞

0

J1(λ) dλ − 1

a2

[

1

6(L2 − z2) +

1

4ρ2

]∫

∞

0

λ2J1(λ) dλ + · · ·]

The first integral evaluates to 1, so for a infinite the potential becomes simplyΦ(z) = V z/L. This is just what we expect to get for the potential between twoinfinite sheets, one grounded and the other at potential V.

The second integral, unfortunately, has a bit of an infinity problem. It’s nothard to see where the problem comes: I derived the expansion above based onthe premise that λ/a is small, but the integral goes over all λ up to ∞, so forany finite a the expansions eventually become invalid in the integral.

I’m still trying to work out a better procedure for estimating corrections forfinite a.

(c) In this part we’re interested in taking L → ∞ and looking at the potentiala fixed distance away from the plane with the circular insert. Calling the fixeddistance z′, the z coordinate of the point we’re interested in is L− z′. We have

sinh k(L − z′)

sinh kL=

sinh(kL) cosh(−kz′) + cosh(kL) sinh(−kz′)

sinh kL= cosh(kz′) − coth(kL) sinh(kz′) (33)

Now, coth(kL) differs significantly from 1 only for kLa . 1, in which regionkz′ . z/L 1, so cosh(kz′) ≈ 1 and sinh(kz′) ≈ 0. By the time k gets bigenough that kz′ is starting to get significant, coth(kL) has long since started tolook like 1, so the two terms in (33) add directly. The result is that, for all k,(33) can be approximated as exp(−kz′). Then (30) becomes

Φ(ρ, z) = aV

∫

∞

0

J1(ka)J0(kρ)e−kz′

dk

as we found in Problem 3.12.



Homer Reid

August 6, 2000


Problem 3.19

Consider a point charge q between two infinite parallel conducting planes held atzero potential. Let the planes be located at z = 0 and z = L in a cylindricalcoordinate system, with the charge on the z axis at z = z0, 0 < z0 < L. UseGreen’s reciprocation theorem of Problem 1.12 with Problem 3.18 as the comparisonproblem.

(a) Show that the amount of induced charge on the plate at z = L inside a circleof radius a whose center is on the z axis is given by

QL(a) = − q

VΦ(z0, 0)

(b) Show that the induced charge density on the upper plate can be written as

σ(ρ) = − q

2π

∫

∞

0

dksinh(kz0)

sinh(kL)kJ0(kρ)

(c) Show that the charge density at ρ = 0 is

σ(0) =−πq

8L2sec2

(πz0

2L

)

(a) Green’s reciprocation theorem says that∫

V

ρ′Φ dV +

∫

S

σ′Φ dA =

∫

V

ρΦ′ dV +

∫

S

σΦ′ dA. (1)

1


We’ll use the unprimed symbols to refer to the quantities of Problem 3.18,and the primed symbols to refer to those of Problem 3.19. Then

ρ(r, z) = 0

σ(r, z) =?

Φ(r, z) = 0, z = 0

= 0, z = L and r > a

= V, z = L and r < a

= V

∫

∞

0

dk aJ1(ak)J0(rk)sinh(kz)

sinh(kL)0 < z < L

ρ′(r, z) = qδ(r)δ(z − z0)

σ′(r, z) =?

Φ′(r, z) = 0, z = 0 or z = L

=?, 0 ≤ z ≤ L

Plugging into (1),

qV

∫

∞

0

dk aJ1(ak)sinh(kz0)

sinh(kL)+ V

∫

z=L,r<a

σ′(r, z) dA = 0

so∫

z=L,r<a

σ′(r, z) dA = −q

∫

∞

0

dk aJ1(ak)sinh(kz0)

sinh(kL)= − q

VΦ(z0, 0) (2)

The integral on the left is just the total surface charge contained within a circleof radius a around the origin of the plane z = L.

(b) The integrand on the left of (2) doesn’t depend on φ, so we can do theangular part of the integral right away to give

2π

∫ a

0

σ′(r, L)r dr = −q

∫

∞

0

dk aJ1(ak)sinh(kz0)

sinh(kL)

Differentiating both sides with respect to a, we have

2πaσ′(a, L) = −q

∫

∞

0

dk∂

∂a[aJ1(ak)]

sinh(kz0)

sinh(kL)(3)

where I’ve blithely assumed that the partial derivative can be passed throughthe integral sign. The partial derivative is

∂

∂a[aJ1(ak)] =

∣

∣

∣

∣

∂

∂x[xJ1(x)]

∣

∣

∣

∣

x=ak

= |J1(x) + xJ ′

1(x)|x=ak

= |xJ0(x)|x=ak = akJ0(ak)


so (3) becomes

σ′(a, L) = − q

2π

∫

∞

0

dk kJ0(ak)sinh(kz0)

sinh(kL)(4)

(c) At a = 0, (4) becomes

σ′(0, L) =−q

2π

∫

∞

0

ksinh(kz0)

sinh(kL).

I have no idea how to do this integral.

Problem 3.22

The geometry of a two-dimensional potential problem is defined in polar coordinatesby the surfaces φ = 0, φ = β, and ρ = a, as indicated in the sketch.Using separation of variables in polar coordinates, show the the Green function canbe written as

G(ρ, φ; ρ′, φ′) =

∞∑

m=1

− 1

mπρ

mπ/β<

(

1

ρmπ/β>

− ρmπ/β>

a2mπ/β

)

sin

(

mπφ

β

)

sin

(

mπφ′

β

)

Problem 2.25 may be of use.

As before, the procedure for determining the Green’s function is to split theregion of interest into two parts (one on each ’side’ of the observation point), findseparate solutions of the Laplace equation that satisfy the boundary conditionsin each region, and then join the two solutions at the source point such thattheir values match up but the first derivative (in whichever dimension we chose’sides’) has a finite discontinuity.

Suppose the observation point is (ρ, φ). Let’s break the region into twosubregions, defined by 0 ≤ ρ′ ≤ ρ and ρ ≤ ρ′ ≤ a. The general solution of theLaplace equation in two-dimensional polar coordinates is

Φ(ρ′, φ′) =A0 + B0 ln ρ′

+∑

n

ρ′n[An sin nφ′ + Bn cosnφ′] + ρ′−n[Cn sin nφ′ + Dn cosnφ′].

The solution in the first region must be admissible down to ρ′ = 0, whichexcludes the ln term and the negative powers of ρ. However, these terms maybe included in the solution for the second region. In both regions, the solutionmust vanish at φ = 0, which excludes the cos terms (i.e. Bn = Dn = 0). Thesolution must also vanish at φ = β, which requires that n = mπ/β, m = 1, 2, · · · .With these considerations we may write down the solutions for G in the tworegions:


G(ρ, φ; ρ′, φ′)

=

∞∑

m=1

Amρ′mπ/β sin

(

mπφ′

β

)

, 0 ≤ ρ′ ≤ ρ (5)

=

∞∑

m=1

[

Bmρ′mπ/β + Cmρ′−mπ/β]

sin

(

mπφ′

β

)

, ρ ≤ ρ′ ≤ a (6)

The solution in the second region must vanish at ρ′ = a for all φ′, i.e.

Bmamπ/β + Cma−mπ/β = 0

soBm = γma−mπ/β and Cm = −γmamπ/β

where γm can be anything. Then (6) becomes

G(ρ, φ; ρ′, φ′) =

∞∑

m=1

γm

[

(

ρ′

a

)mπ/β

−(

a

ρ′

)mπ/β]

sin

(

mπφ′

β

)

ρ ≤ ρ′ ≤ a.

The solutions in the two regions must agree on the boundary between the tworegions, i.e. at ρ′ = ρ. This determines Am and γm:

Am = λm

[

(ρ

a

)mπ/β

−(

a

ρ

)mπ/β]

γm = λmρmπ/β

where λm can be anything. Using these expressions for Am, Bm, and Cm wecan write

G(ρ, φ; ρ′, φ′)

=∞∑

m=1

λm

[

(ρ

a

)mπ/β

−(

a

ρ

)mπ/β]

ρ′mπ/β sin

(

mπφ′

β

)

0 ≤ ρ′ ≤ ρ

=

∞∑

m=1

λm

[

(

ρ′

a

)mπ/β

−(

a

ρ′

)mπ/β]

ρmπ/β sin

(

mπφ′

β

)

ρ ≤ ρ′ ≤ a.

This may be more succintly written as

G(ρ, φ; ρ′, φ′) =∑

m

λmfm(ρ; ρ′) sin

(

mπφ′

β

)

(7)

where

fm(ρ; ρ′) =

[

(ρ>

a

)mπ/β

−(

a

ρ>

)mπ/β]

ρmπ/β< .


The final step is to choose the constant λm in (7) such as to make

∇2G(ρ, φ; ρ′, φ′) =1

ρδ(ρ′ − ρ)δ(φ′ − φ). (8)

The Laplacian of (7) is

∇2G =

[

∂2

∂ρ′2+

1

ρ′2∂2

∂φ′2

]

G =∑

m

λm

[

d2

dρ′2fm(ρ; ρ′) −

(

mπ

ρ′β

)2

fm(ρ; ρ′)

]

sin

(

mπφ′

β

)

This is equal to (8) if

λm = κm1

βsin

(

mπφ

β

)

(9)

and

κm

[

d2

dρ′2fm(ρ; ρ′) −

(

mπ

ρ′β

)2

fm(ρ; ρ′)

]

=1

ρδ(ρ′ − ρ).

At all points ρ′ 6= ρ, the latter condition is already satisfied by f as we con-structed it earlier. At ρ′ = ρ, the condition is achieved by choosing κm tosatisfy

κmd

dρ′fm(ρ; ρ′)

∣

∣

∣

∣

ρ′=ρ+

ρ′=ρ−

=1

ρ. (10)

Referring to (7), we have

d

dρ′fm

∣

∣

∣

∣

ρ′+ρ+

=mπ

β

[

(ρ

a

)mπ/β

+

(

a

ρ

)mπ/β]

ρmπ/β−1 (11)

d

dρ′fm

∣

∣

∣

∣

ρ′+ρ−

=mπ

β

[

(ρ

a

)mπ/β

−(

a

ρ

)mπ/β]

ρmπ/β−1. (12)

Subtracting (12) from (11) we obtain

dfm

dρ′

∣

∣

∣

∣

ρ′=ρ+

ρ′=ρ−

=2mπ

βamπ/β · 1

ρ.

Then from (10) we read off

κm =β

2mπa−mπ/β

and plugging this into (9) gives

λm =1

2mπa−mπ/β sin

(

mπ

β

)

φ.

Plugging this into (7) we obtain finally

G(ρ, φ; ρ′, φ′) =∑

m

1

2mπ

[

(ρ<ρ>

a2

)mπ/β

−(

ρ<

ρ>

)mπ/β]

sin

(

mπφ

β

)

sin

(

mπφ′

β

)

I seem to be off by a factor of 2 here, but I can’t find where.

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Homer Reid

October 8, 2000


Problem 4.8

A very long, right circular, cylindrical shell of dielectric constant ε/ε0 and inner andouter radii a and b, respectively, is placed in a previously uniform electric field E0

with its axis perpendicular to the field. The medium inside and outside the cylinderhas a dielectric constant of unity.

(a) Determine the potential and electric fields in the three regions, neglecting endeffects.

(b) Sketch the lines of force for a typical case of b ≈ 2a.

(c) Discuss the limiting forms of your solution appropriate for a solid dielectriccylinder in a uniform field, and a cylindrical cavity in a uniform dielectric.

We will take the axis of the cylinder to be the z axis and the electric field tobe aligned with the x axis: E0 = E0i. Since the cylinder is very long and we’retold to neglect end effects, we can ignore the z direction altogether and treatthis as a two-dimensional problem.

(a) The general solution of the Laplace equation in two dimensional polar co-ordinates is

Φ(r, ϕ) =∑

[Anrn + Bnr−n][Cn sin(nϕ) + Dn cos(nϕ)]

For the region inside the shell (r < a), the B coefficients must vanish to keepthe potential from blowing up at the origin. Also, in the region outside the shell

1


(r > b), the only positive power of r in the sum must be that which gives riseto the external electric field, i.e. −E0r cosϕ with An = 0 for n > 1. With theseobservations we may write expressions for the potential in the three regions:

Φ(r, ϕ) =

∑

rn[An sinnϕ + Bn cosnϕ], r < a∑

rn[Cn sin nϕ + Dn cosnϕ] + r−n[En sin nϕ + Fn cosnϕ], a < r < b

−E0r cosϕ +∑

r−n[Gn sin nϕ + Hn cosϕ], r > b

The normal boundary condition at r = a is

ε0∂Φ

∂r

∣

∣

∣

∣

x=a−

= ε∂Φ

∂r

∣

∣

∣

∣

x=a+

or

ε0ε

∑

nan−1[An sin nϕ + Bn cosnϕ] =∑

nan−1[Cn sin nϕ + Dn cosnϕ] − na−(n+1)[En sin nϕ + Fn cosnϕ]

From this we obtain two equations:

ε0ε

An = Cn − Ena−2n (1)

ε0ε

Bn = Dn − Fna−2n (2)

Next, the tangential boundary condition at r = a is

∂Φ

∂ϕ

∣

∣

∣

∣

x=a+

=∂Φ

∂ϕ

∣

∣

∣

∣

x=a−

or

∑

nan[An cosnϕ − Bn sin nϕ] =∑

nan[Cn cosnϕ − Dn sinnϕ] + na−n[En cosnϕ − Fn sin nϕ]

from which we obtain two more equations:

An = Cn + Ena−2n (3)

Bn = Dn + Fna−2n (4)

Similarly, from the normal boundary condition at r = b we obtain

−ε0ε

E0 cosϕ − ε0ε

∑

nb−(n+1)[Gn sinnϕ + Hn cosϕ] =∑

nbn−1[Cn sin nϕ + Dn cosnϕ] − nb−(n+1)[En sin nϕ + Fn cosϕ]


which leads to

−ε0ε

Gn = Cnb2n − En (5)

−ε0ε

b2E0δn1 −ε0ε

Hn = Dnb2n − Fn (6)

Finally, we have the tangential boundary condition at r = b:

bE0 sin ϕ +∑

nb−n[Gn cosnϕ − Hn sinnϕ] =∑

nbn[Cn cosnϕ − Dn sinnϕ] + nb−n[En cosnϕ − Fn sin nϕ]

giving

Gn = Cnb2n + En (7)

−b2E0δn1 + Hn = Dnb2n + Fn. (8)

The four equations (1), (3), (5), and (7) specify a degenerate system of linearequations, which can only be satisfied by taking An = Cn = En = Gn = 0 forall n. Next, for n 6= 1, the system of equations (2), (4), (6), and (8) specify thesame degenerate system of equations, so Bn = Dn = Fn = Gn = 0 for n 6= 0.However, for n = 1, we have

ε0ε

B1 = D1 − F1a−2 D1 =

1

2

(

1 +ε0ε

)

B1

⇒

B1 = D1 + F1a−2 F1 =

1

2a2

(

1 − ε0ε

)

B1.

and

−H1 = b2E0 +ε

ε0D1b

2 − ε

ε0F1

H1 = b2E0 + D1b2 + F1

→ 0 = 2b2E0 + b2

(

1 +ε

ε0

)

D1 +

(

1 − ε

ε0

)

F1

Substituting from above,

−4b2E0 =1

εε0

[

b2(ε + ε0)2 − a2(ε − ε0)

2]

B1

or

B1 =−4εε0b

2

b2(ε + ε0)2 − a2(ε − ε0)2E0.


Then

D1 =−2ε0(ε + ε0)b

2

b2(ε + ε0)2 − a2(ε − ε0)2E0

F1 =−2ε0(ε − ε0)a

2b2

b2(ε + ε0)2 − a2(ε − ε0)2E0

H1 =−b2(b2 − a2)(ε20 − ε2)

b2(ε + ε0)2 − a2(ε − ε0)2E0.

The potential is

Φ(r, ϕ) =

−4εε0b2

b2(ε + ε0)2 − a2(ε − ε0)2· E0rcos ϕ, r < a

−2ε0b2

b2(ε + ε0)2 − a2(ε − ε0)2

(

(ε + ε0)r + (ε − ε0)a2

r

)

E0cos ϕ, a < r < b

−(b2 − a2)(ε20 − ε2)

b2(ε + ε0)2 − a2(ε − ε0)2· b2

rE0cos ϕ − E0rcos ϕ, b < r.

As ε → ε0, Φ → −E0r cosϕ in all three regions, which is reassuring.The electric field is

E(r, ϕ) =

4εε0b2

b2(ε + ε0)2 − a2(ε − ε0)2E0 [cosϕr − sin ϕϕ] , r < a

2ε0b2

b2(ε + ε0)2 − a2(ε − ε0)2

[

(ε + ε0) − (ε − ε0)a2

r2

]

E0 cosϕr

−[

(ε + ε0) + (ε − ε0)a2

r2

]

E0 sin ϕϕ

, a < r < b

− (b2 − a2)(ε20 − ε2)

b2(ε + ε0)2 − a2(ε − ε0)2·(

b

r

)2

E0 [cosϕr + sin ϕϕ]

+E0 [cosϕr − sin ϕϕ] , b < r.

(b) In Figure 4.1 I’ve plotted the field lines for b = 2a, ε = 5ε0. Also, as anappendix to this document I’ve included the C program I wrote to generate thisplot.

(c) For a solid dielectric cylinder in a uniform field, we would have a → 0. Inthat case the field would look like

E(r, ϕ) =

2ε0ε + ε0

E0i, r < b

E0i−(ε20 − ε2)

(ε + ε0)2

(

b

r

)2

E0[cos ϕr + sin ϕϕ], r > b

On the other hand, a cylindrical cavity in a uniform dielectric corresponds to


Figure 1: Field lines in Problem 4.8 for b = 2a, ε = 5ε0.

b → ∞, in which case the field becomes

E(r, ϕ) =

4εε0(ε + ε0)2

E0i, r < a

2ε0(ε + ε0)

E0i−2ε0(ε − ε0)

(ε + ε0)2

(a

r

)2

E0[cos ϕr + sin ϕϕ], r > a.


Problem 4.9

A point charge q is located in free space a distance d away from the center of adielectric sphere of radius a (a < d) and dielectric constant ε/ε0.

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

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

(c) Verify that, in the limit ε/ε0 → ∞, your result is the same as that for theconducting sphere.

We will take the origin of coordinates at the center of the sphere, and putthe point charge on the z axis at z = +h. Then the problem has azimuthalsymmetry.

(a) Since there is no free charge within the sphere, ∇·D = 0 there. But since thepermittivity is uniform within the sphere, we may also write ∇·(D/ε) = ∇·E = 0there. This means that polarization charge only exists on the surface of thesphere, so within the sphere the potential satisfies the normal Laplace equation,whence

Φ(r, θ) =∑

l

AlrlPl(cos θ) (r < a).

Now, in the region r > a, the potential may be written as the sum of twocomponents Φ1 and Φ2, where Φ1 comes from the polarization charge on thesurface of the sphere, while Φ2 comes from the external point charge. SinceΦ1 satisfies the Laplace equation for r > a, we may expand it in Legendrepolynomials:

Φ1(r, θ) =∑

l

Blr−(l+1)Pl(cos θ) (r > a).

On the other hand, Φ2 is just the potential due to a point charge at z = d:

Φ2(r, θ) =

q

4πε0

∑ rl

dl+1Pl(cos θ), r < d

q

4πε0

∑ dl

rl+1Pl(cos θ), r > d.

(9)

Putting this all together we may write the potential in the three regions as

Φ(r, θ) =

∑

AlrlPl(cos θ), r < a

∑

[

Blr−(l+1) +

q

4πε0

rl

dl+1

]

Pl(cos θ), a < r < d

∑

[

Bl +qdl

4πε0

]

r−(l+1)Pl(cos θ), r > d.


The normal boundary condition at r = a is

ε∂Φ

∂r

∣

∣

∣

∣

r=a−

= ε0∂Φ

∂r

∣

∣

∣

∣

r=a+

→ ε

ε0lAla

l−1 = −(l + 1)Bla−(l+2) +

lqal−1

4πε0dl+1

→ Al =ε0ε

[−(l + 1)

lBla

−(2l+1) +q

4πε0dl+1

]

(10)

The tangential boundary condition at r = a is

∂Φ

∂θ

∣

∣

∣

∣

r=a−

=∂Φ

∂θ

∣

∣

∣

∣

r=a+

→ Alal = Bla

−(l+1) +q

4πε0

al

d(l+1)

→ Bl = Ala2l+1 − q

4πε0

a2l+1

dl+1(11)

Combining (10) and (11), we obtain

Al =1

εε0

+ l+1l

(

2l + 1

l

)

q

4πε0dl+1

Bl =1

εε0

+ l+1l

(

1 − ε

ε0

)

qa2l+1

4πε0dl+1

In particular, as ε/ε0 → ∞ we have

Al → 0

as must happen, since the field within a conducting sphere vanishes; and

Bl → − qa2l+1

4πε0dl+1. (12)

With the coefficients (12), the potential outside the sphere due to the polariza-tion charge at the sphere boundary is

Φ1(r, θ) =1

4πε0

(

−qa

d

)

∑

(

a2

d

)l1

rl+1Pl(cos θ).

Comparing with (9) we see that this is just the potential of a charge −qa/d onthe z axis at z = a2/d. This is just the size and position of the image chargewe found in Chapter 2 for a point charge outside a conducting sphere.


(b) Near the origin, we have

Φ(r, θ) = A1rP1(cos θ) + A2r2P2(cos θ) + · · ·

=q

4πε0

[

3ε0d2(ε + 2ε0)

z +1

2

(

5ε0d3(2ε + 3ε0)

)

(z2 − x2 − y2) + · · ·]

so the field components are

Ex =q

4πε0d2· 5ε02ε + 3ε0

(x

d

)

+ · · ·

Ey =q

4πε0d2· 5ε02ε + 3ε0

(y

d

)

+ · · ·

Ez = − q

4πε0d2

[

3ε0ε + 2ε0

+5ε0

2ε + 3ε0

(z

d

)

+ · · ·]

Problem 4.10

Two concentric conducting spheres of inner and outer radii a and b, respectively,carry charges ±Q. The empty space between the spheres is half-filled by a hemi-spherical shell of dielectric (of dielectric constant ε/ε0), as shown in the figure.

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

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

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

We’ll orient this problem such that the boundary between the dielectric-filled space and the empty space is the xy plane. Then the region occupiedby the dielectric is the region a < r < b, 0 < θ < π/2, and the problem hasazimuthal symmetry.

(a) Since the dielectric has uniform permittivity, all the polarization chargeexists on the boundary of the dielectric, so within its body we may take thepotential to be a solution of the normal Laplace equation. The potential in theregion between the spheres may then be written

Φ(r, θ) =

∑

[Alrl + Blr

−(l+1)]Pl(cos θ), 0 < θ <π

2∑

[Clrl + Dlr

−(l+1)]Pl(cos θ),π

2< θ < π

First let’s apply the boundary conditions at the interface between the di-electric and free space. That region is described by θ = π/2, a < r < b, and we


must have

ε∂Φ

∂θ

∣

∣

∣

∣

θ=π/2+

= ε0∂Φ

∂θ

∣

∣

∣

∣

θ=π/2−

∂Φ

∂r

∣

∣

∣

∣

θ=π/2+

=∂Φ

∂r

∣

∣

∣

∣

θ=π/2−

which leads to

∑

[

ε

ε0Al − Cl

]

P ′

l (0)rl +

[

ε

ε0Bl − Dl

]

P ′

l (0)r−l+1 = 0 (13)

∑

l [Al − Cl] P (0)rl−1 − (l + 1) [Bl − Dl]Pl(0)r−l+2 = 0. (14)

Since these equations must be satisfied for all r in the region a < r < b, thecoefficients of each power of r must vanish identically. In (13), this requirementis automatically satisfied for l even, since P ′

l (0) vanishes for even l. Similarly,(14) is automatically satisfied for l odd. For other cases the vanishing of thecoefficients must be brought about by taking

ε

ε0Al = Cl

ε

ε0Bl = Dl, l odd (15)

Al = Cl Bl = Dl, l even. (16)

Next let’s consider the charge at the surface of the inner sphere. There areactually two components of this charge; one component comes from the surfacedistribution of the free charge +Q that exists on the sphere, and the othercomponent comes from the bound polarization charge on the inner surface ofthe dielectric

Problem 4.13

Two long, coaxial, cylindrical conducting surfaces of radii a and b are loweredvertically into a liquid dielectric. If the liquid rises an average height h between theelectrodes when a potential difference V is established between them, show that thesusceptibility of the liquid is

χe =(b2 − a2)ρgh ln(b/a)

ε0V 2

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

First let’s work out what happens when a battery of fixed voltage V is con-nected between two coaxial conducting cylinders with simple vacuum betweenthem. To begin, we can use Gauss’ law to determine the E field between the


cylinders. For our Gaussian pillbox we take a disk of thickness dz and radiusr, a < r < b centered on the axis of the cylinders. By symmetry there is nocomponent of E normal to the top or bottom boundary surfaces, and the com-ponent normal to the side surfaces (the radial component) is uniform aroundthe disc. Hence

∮

E · dA = 2π r dzEρ =q

ε0=

1

ε0(2π a dz)σ

→ Eρ(ρ) =aσ

ε0r

where σ is the surface charge on the inner conductor. This must integrate togive the correct potential difference between the conductors:

V = −∫ b

a

Eρ(ρ)dρ = −aσ

ε0ln

b

a

which tells us that, to establish a potential difference V between the conductors,the battery has to flow enough charge to establish a surface charge of magnitude

σ =ε0V

a ln(b/a)(17)

on the cylinder faces (the surface charges are of opposite sign on the two cylin-ders).

It is useful to figure out the energy per unit length stored in the electric fieldbetween the cylinder plates here. This is just

Wv =1

2

∫ b

a

∫ 2π

0

E ·D ρ dρ dφ

= πε0

∫ b

a

E2(ρ)ρ dρ

= πa2σ2

ε0ln(b/a)

=πε0V

2

ln(b/a)(18)

where the v subscript stands for ’vacuum’, since (18) is the energy per unit lengthstored in the field between the cylinders with just vacuum between them.

Now suppose we introduce a dielectric material between the cylinders. If thevoltage between the cylinders is kept at V , then the E field must be just thesame as it was in the no-dielectric case, because this field integrated from a tob must still give the same potential difference. However, in order to establishthis same E field in the presence of the retarding effects of the dielectric, thebattery now has to establish a surface charge that is greater that it was beforeby a factor (ε/ε0). With this greater charge on the electrodes, the D field willnow be bigger by a factor (ε/ε0) than it was in our above calculation. So the


energy per unit length stored in the field between the cylinders increases by afactor (ε/ε0 − 1) over the result (18):

∆Wd = (ε − ε0)πV 2

ln(b/a).

On the other hand, to get to this point the battery has had to flow enoughcharge to increase the surface charges to be of magnitude (ε/ε0) times greaterthan (17). In doing this the internal energy of the battery decreases by anamount equal to the work it had to do to flow the excess charge, namely

∆Wb = −V dQ = V (2π a dσ) = (ε − ε0)2πV 2

ln(b/a)

(per unit length). The energy lost by the battery is twice that gained by thedielectric, so the system with dielectric between the cylinders has lower overallenergy than the system with vacuum between the cylinders by a factor

∆W = (ε − ε0)πV 2

ln(b/a)(19)

(per unit length).Turning now to the situation in this problem, we’ll take the axis of the

cylinders as the z axis, so that the surface of the liquid is parallel to the xyplane. We’ll take the boundary between the liquid and the air above it to be atz = 0. With no potential between the cylinder plates, the liquid between thecylinders is at the same height as the liquid outside.

Now suppose a battery of fixed potential V is connected between the twocylinder plates. As we showed earlier, the combined system of battery and di-electric can lower its energy by having more of the dielectric rise up between thecylinders. However, at some point the energy win we get from this is balancedby the energy hit we take from the gravitational potential energy of havingthe excess liquid rise higher between the cylinders. The height at which we nolonger gain by having more liquid between the cylinders is the height to whichthe system will settle.

So suppose that, with a battery keeping a voltage V between the electrodes,the liquid between the electrodes rises to a height h above the surface of theliquid outside the electrodes. The decrease in electrostatic energy this affordsover the case with just vacuum filling that space is just (19) times the height,i.e.

Ee = −h(ε − ε0)πV 2

ln(b/a)(20)

This must be balanced by the gravitational potential energy Eg of the excessliquid. Eg is easily calculated by noting that the area between the cylinders isπ(b2 − a2), so the mass of liquid contained in a height dh between the cylindersis dm = ρπ(b2 −a2)dh, and if this mass is at a height h above the liquid surfaceits excess gravitational energy is

dEg = (dm)gh = πgρ(b2 − a2)hdh.


Integrating over the excess height of liquid between the cylinders,

Eg = πgρ(b2 − a2)

∫ h

0

h′ dh′ =1

2πgρ(b2 − a2)h2. (21)

Comparing (20) to (21), we find that the gravitational penalty of the excessliquid just counterbalances the electrostatic energy reduction when

h =2(ε − ε0)V

2

ρg(b2 − a2) ln(b/a)

=2χeε0V

2

ρg(b2 − a2) ln(b/a)

Solving for χe,

χe =ρgh(b2 − a2) ln(b/a)

2ε0V 2.

So I seem to be off by a factor of 2 somewhere.Actually we should note one detail here. When the surface of the liquid

between the cylinders rises, the surface of the liquid outside the cylinders mustfall, since the total volume of the liquid is conserved. Hence there are reallytwo other contributions to the energy shift, namely, the change in gravitationaland electrostatic energies of the thin layer of liquid outside the cylinders thatfalls away when the liquid rises between the cylinders. But if the surface area ofthe vessel containing the liquid is sufficiently larger than the area between thecylinders, the difference layer will be thin and its energy shifts negligible.


Appendix

Source code for field line plotting program used in Problem 4.8.

/*

* Program to draw field lines for Jackson problem 4.8.

* Homer Reid October 2000

*/

#include <stdio.h>

#include <math.h>

#include "/usr2/homer/include/GnuPlot.c"

#define EZ 1.0 /* permittivity of free space */

#define EPS 5.0 /* permittivity of cylinder */

#define E0 1.0 /* external field (irrelevant here) */

#define A 4.0 /* radius of inner cylinder */

#define B 8.0 /* radius of outer cylinder */

#define NUMLINES 25.0 /* number of field lines to draw */

#define NUMPOINTS 250.0 /* no. of pts to plot for each line */

#define DELTAX (4.0 * B) / NUMPOINTS /* horiz spacing of pts */

#define DELTAY (4.0 * B) / NUMLINES /* vert spacing of initial pts */

#define DENOM (B*B*(EPS+EZ)*(EPS+EZ) - A*A*(EPS-EZ)*(EPS-EZ))

/*

* Return r component of electric field at position (r,phi).

*/

double Er(double r, double phi)

double Coeff;

if ( r < A )

Coeff=(4.0*EPS*EZ*B*B)/DENOM;

else if ( r < B )

Coeff=(2*EPS*B*B/DENOM)*( (EPS+EZ) - (EPS-EZ)*(A*A)/(r*r) );

else

Coeff=1.0 - ((B*B - A*A)*(EZ*EZ-EPS*EPS)*(B*B)/(r*r*DENOM));

return Coeff*E0*cos(phi);


/*

* Return phi component of electric field at (r,phi).

*/

double Ephi(double r, double phi)

double Coeff;

if ( r < A )

Coeff=(4.0*EPS*EZ*B*B)/DENOM;

else if ( r < B )

Coeff=(2*EPS*B*B/DENOM)*( (EPS+EZ) + (EPS-EZ)*(A*A)/(r*r) );

else

Coeff=1.0 + ((B*B - A*A)*(EZ*EZ-EPS*EPS)*(B*B)/(r*r*DENOM));

return -Coeff*E0*sin(phi);

void main()

double i,j,r,phi,x,y,dx,dy;

double RComp,PhiComp;

FILE *g;

g=GnuPlot("Field lines");

/*

* Send basic GnuPlot configuration commands.

*/

fprintf(g,"set terminal postscript portrait color\n");

fprintf(g,"set output ’fig4.1.eps’\n");

fprintf(g,"set multiplot \n");

fprintf(g,"set size square\n");

fprintf(g,"set noxtics\n");

fprintf(g,"set noytics\n");

fprintf(g,"set xrange [%g:%g]\n",-2.0*B,2.0*B);

fprintf(g,"set yrange [%g:%g]\n",-2.0*B,2.0*B);

/*

* Draw circles at r=a and r=b.

*/

fprintf(g,"plot ’-’ t ’’, ’-’ t ’’ with lines, ’-’ t ’’ with lines\n");

fprintf(g,"e\n");

for(phi=0; phi<=2*M_PI; phi+=(2*M_PI/100))

fprintf(g,"%g %g\n",A*cos(phi),A*sin(phi));


fprintf(g,"e\n");

for(phi=0; phi<=2*M_PI; phi+=(2*M_PI/100))

fprintf(g,"%g %g\n",B*cos(phi),B*sin(phi));

fprintf(g,"e\n");

/*

* Draw field lines.

*/

for (i=1.0; i<=NUMLINES; i+=1.0)

/*

* Compute starting x and y coordinates and initiate plot.

*/

x=-2.0*B; y=2.0*B * ((NUMLINES - 2.0*i)/NUMLINES);

fprintf(g,"plot ’-’ t ’’ with lines\n");

/*

* Plot NUMPOINTS points for this field line.

*/

for (j=0.0; j<NUMPOINTS; j+=1.0)

/*

* compute polar coordinates of present location

*/

r=sqrt(x*x + y*y);

if (x==0.0)

phi=(y>0.0) ? M_PI/2.0 : -M_PI/2.0;

else

phi=atan(y/x);

/*

* compute rise and run of electric field

*/

RComp=Er(r,phi);

PhiComp=Ephi(r,phi);

dx=cos(phi)*RComp - sin(phi)*PhiComp;

dy=sin(phi)*RComp + cos(phi)*PhiComp;

/*

* bump x coordinate forward a fixed amount, and y

* coordinate up or down by an amount depending on

* the direction of the electric field at this point

*/

x+=DELTAX;

y+=DELTAX * (dy/dx);

fprintf(g,"%g %g\n",x,y);


;

fprintf(g,"e\n");

;

printf("Thank you for your support.\n");



Homer Reid

November 8, 2000


Problem 5.1

Starting with the differential expression

dB =µ0I

4πdl′ × x − x′

|x − x′|3

for the magnetic induction at the point P with coordinate x produced by an incre-ment of current I dl′ at x′, show explicitly that for a closed loop carrying a currentI the magnetic induction at P is

B =µ0I

4π∇Ω

where Ω is the solid angle subtended by the loop at the point P . This correspondsto a magnetic scalar potential, ΦM = −µ0IΩ/4π. The sign convention for thesolid angle is that Ω is positive if the point P views the “inner” side of the surfacespanning the loop, that is, if a unit normal n to the surface is defined by thedirection of current flow via the right-hand rule, Ω is positive if n points away fromthe point P , and negative otherwise. This is the same convention as in Section 1.6for the electric dipole layer.

I like to change the notation slightly: the observation point is r1, the coordi-nate of a point on the current loop is r2, and the displacement vector (pointingto the observation point) is r12 = r1 − r2.

The solid angle subtended by the current loop at r1 is given by a surfaceintegral over the loop:

Ω =

∫

S

cos γ dA

r212

1



Homer Reid

February 11, 2001


Problem 5.10

A circular current loop of radius a carrying a current I lies in the x− y plane withits center at the origin.

(a) Show that the only nonvanishing component of the vector potential is

Aφ(ρ, z) =µ0Ia

π

∫

∞

0

dk cos kz I1(kρ<)K1(kρ>)

where ρ<(ρ>) is the smaller (larger) of a and ρ.

(b) Show that an alternative expression for Aφ is

Aφ(ρ, z) =µ0Ia

2

∫

∞

0

dke−k|z|J1(ka)J1(kρ).

(c) Write down integral expressions for the components of magnetic induction,using the expressions of parts a and b. Evaluate explicitly the components ofB on the z axis by performing the necessary integrations.

(a) Translating Jackson’s equation (5.33) into cylindrical coordinates, we have

Jφ = Iδ(z)δ(ρ − a) (1)

Following Jackson, we take the observation point x on the x axis, so its coordi-nates are (ρ, φ = 0, z). Since there is no current in the z direction, and since the

1


current density is cylindrically symmetric, there is no vector potential in the ρor z directions. In the φ direction we have

Aφ = −Ax sin φ + Ay cosφ = Ay

=µ0

4π

∫

Jy(x′)

|x − x′| dx′

=µ0

4π

∫

Jφ(x′) cosφ′

|x − x′| dx′

=µ0

4πRe

∫

Jφ(x′)eiφ′

|x− x′| dx′

=µ0

4πRe

∫

Jφ(x′)eiφ′

[

2

π

∞∑

m=−∞

∫

∞

0

eim(φ−φ′) cos[k(z − z′)]Im(kρ<)Km(kρ>) dk

]

dx′

where we substituted in Jackson’s equation (3.148). Rearranging the order ofintegration and remembering that φ = 0, we have

Aφ =µ0

2π2Re

∞∑

m=−∞

∫

∞

0

[∫

Jφ(x′)ei(1−m)φ′

cos[k(z − z′)]Im(kρ<)Km(kρ>)dx′

]

dk

If m = 1, the φ integral yields 2π; otherwise it vanishes. Thus

Aφ =µ0

π

∫

∞

0

[∫

∞

0

∫

∞

−∞

Jφ(r′, z′) cos[k(z − z′)]I1(kρ<)K1(kρ>)ρ′ dz′ dr′]

dk

Substituting (1), we have

Aφ =Iaµ0

π

∫

∞

0

cos kz I1(kρ<)K1(kρ>) dk.

(b) The procedure for obtaining this expression is identical to the one I justwent through, but with the expression from Problem 3.16(b) used for the Green’sfunction instead of equation (3.148).

(c) Let’s suppose that the observation point is in the interior region of thecurrent loop, so ρ< = ρ, ρ> = a. Then

Bρ = [∇×A]ρ = −∂Aφ

∂z

= −Iaµ0

π

∫

∞

0

k sin kz I1(kρ)K1(ka) dk

Bz = [∇×A]z =1

ρAφ +

∂Aφ

∂ρ

=Iaµ0

π

∫

∞

0

cos kz

[

I1(kρ)

ρ+ kI ′1(kρ)

]

K1(ka) dk


As ρ = 0, I1(ρ) → 0, I1(ρ)/ρ → 1/2, and I ′

1(ρ) → 1/2, so

Bρ(ρ = 0) = 0

Bz(ρ = 0) =Iaµ0

π

∫

∞

0

k cos kzK1(ka) dk

=Iaµ0

π

∂

∂z

∫

∞

0

sin kzK1(ka)dk

The integral may be done by parts:

∫

∞

0

sin kzK1(kz) dk =

∣

∣

∣

∣

−1

asin kzK0(ka)

∣

∣

∣

∣

∞

0

+z

a

∫

∞

0

cos kzK0(ka) dk

K0 is finite at zero but sin vanishes there, and sin is finite at infinity but K0

vanishes there, so the first term vanishes. The integral in the second term isJackson’s equation (3.150). Plugging it in to the above,

Bz(ρ = 0) =Iµ0

2

∂

∂z

z

(z2 + a2)1/2

=Iµ0

2

a2

(z2 + a2)3/2.

Problem 5.11

A circular loop of wire carrying a current I is located with its center at the originof coordinates and the normal to its plane having spherical angles θ0, φ0. There isan applied magnetic field, Bx = B0(1 + βy) and By = B0(1 + βx).

(a) Calculate the force acting on the loop without making any approximations.Compare your result with the approximate result (5.69). Comment.

(b) Calculate the torque in lowest order. Can you deduce anything about the higherorder contributions? Do they vanish for the circular loop? What about forother shapes?

(a) Basically we’re dealing with two different reference frames here. In the “lab”frame, R, the magnetic field exists only in the xy plane, and the normal to thecurrent loop has angles θ0, φ0. We define the “rotated” frame R′ by aligningthe z′ axis with the normal to the current loop, so that in R′ the current loopexists only in the x′y′ plane, but the magnetic field now has a z′ component.

The force on the current loop is

F =

∫

(J ×B)dV. (2)


PSfrag replacements

x y

z1 = z

x1x1

y1

z1

x′

y′ = y1

z′

φ0

θ0

R → R1 R1 → R′

Figure 1: Successive coordinate transformations in Problem 5.11.

The components of J are easy to express in R′, but more complicated in R; theopposite is true for B. There are two ways to do the problem: we can work outthe components of J in R and do the integral in R, or we can work out thecomponents of B in R′ and do the integral in R′, in which case we would haveto transform the components of the force back to R to get the answer we desire.I think the former approach is easier.

To derive the transformation matrix relating the coordinates of a point in Rand R′, I imagined that the transformation arose from two separate transforma-tions, as depicted in figure (??). The first transformation is a rotation throughφ0 around the z axis, which takes us from R to an intermediate frame R1. Thenwe rotate through θ0 around the y1 axis, which takes us to R′. Evidently, thecoordinates of a point in the various frames are related by

x1

y1

z1

=

cosφ0 sin φ0 0− sinφ0 cosφ0 0

0 0 1

xyz

(3)

x′

y′

z′

=

cos θ0 0 − sin θ0

0 1 0sin θ0 0 cos θ0

x1

y1

z1

(4)

Multiplying matrices,

x′

y′

z′

=

cos θ0 cosφ0 cos θ0 sin φ0 − sin θ0

− sinφ0 cosφ0 0sin θ0 cosφ0 sin θ0 sin φ0 cos θ0

xyz

. (5)

This matrix also gives us the transformation between unit vectors in the two


frames:

i′

j′

k′

=

cos θ0 cosφ0 cos θ0 sin φ0 − sin θ0

− sinφ0 cosφ0 0sin θ0 cosφ0 sin θ0 sin φ0 cos θ0

i

j

k

. (6)

We will also the inverse transformation, i.e. the expressions for coordinates inR in terms of coordinates in R′:

xyz

=

cos θ0 cosφ0 − sin φ0 sin θ0 cosφ0

cos θ0 sin φ0 cosφ0 sin θ0 sinφ0

− sin θ0 0 cos θ0

x′

y′

z′

. (7)

To do the integral in (2) it’s convenient to parameterize a point on thecurrent loop by an angle φ′ reckoned from the x′ axis in R′. If the loop radiusis a, then the coordinates of a point on the loop are x′ = a cosφ′, y′ = a sinφ′,and the current density/volume element product is

J dV = Id l = (Ia dφ′)φ′

= Ia dφ′[− sinφ′i′ + cosφ′j′]

= Ia dφ′

[

(− sinφ′ cos θ0 cosφ0 − cosφ′ sin φ0 )i

+ (sin φ′ sin φ0 + cosφ′ cosφ0 )j + (sin φ′ sin θ0)k]

We also need the components of the B field at a point on the current loop:

B(φ′) = B0[1 + βy(φ′)]i + B0[1 + βx(φ′)]

= B0[1 + aβ(cosφ′ cos θ0 sin φ0 + sin φ′ cosφ0)]i + B0[1 + aβ(cosφ′ cos θ0 cosφ0 − sin φ′ sin φ0)]j

The components of the cross product are

[J×B]x dV = −JzBy dV

= (· · · )βIa2B0dφ′(

sin2 φ′ sin θ0 sin φ0

)

[J ×B]y dV = JzBx dV

= (· · · ) + βIa2B0 dφ′(

sin2 φ′ sin θ0 cosφ0

)

[J ×B]z dV = (JxBy − JyBx) dV

= (· · · ) + 0

where we only wrote out terms containing a factor of cos2 φ′ or sin2 φ′, since onlythese terms survive after the integral around the current loop (we grouped allthe remaining terms into (· · · )). In the surviving terms, cos2 φ′ and sin2 φ′ turninto factors of π after the integral around the loop. Then the force componentsare

Fx = πβIa2B0 sin θ0 sin φ0

Fy = πβIa2B0 sin θ0 cosφ0

Fz = 0.


To compare this with the first-order approximate result, note that the magneticmoment has magnitude πa2I and is oriented along the z′ axis:

m = πa2Ik′ = πa2I(

sin θ0 cosφ0i + sin θ0 sin φ0j + cos θ0k)

so

∇(

B · m)

= ∇(

B0(1 + βy)mx + B0(1 + βx)my

)

= B0β(

my i + mxj)

= πβIa2B0

(

sin θ0 sinφ0i + sin θ0 cosφ0j)

in exact agreement with the result we calculated so laboriously above.

Problem 5.12

Two concentric circular loops of radii a, b and currents I, I ′, respectively (b < a),have an angle α between their planes. Show that the torque on one of the loopsis about the line of intersection of the two planes containing the loops and has themagnitude

N =µ0πII ′b2

2a

∞∑

n=0

(n + 1)

(2n + 1)

[

Γ(n + 3/2)

Γ(n + 2)Γ(3/2)

]2 (

b

a

)2n

P 12n+1(cosα).

The torque on the smaller loop is

N =

∫

r ×(

Jb(r) ×Ba(r))

dr

=

∫

[

r ·Ba(r)]

Jb(r) −[

r · Jb(r)]

Ba(r)

dr.

where Jb is the current density of the smaller loop and Ba is the magnetic fieldof the larger loop. But r · Jb vanishes, because the current flows in a circlearound the origin—there is no current flowing toward or away from the origin.Thus

N =

∫

rBr(r)Jb(r)dr (8)

where Br is the radial component of the magnetic field of the larger currentloop.

As in the last problem, it’s convenient to define two reference frames for thissituation. Let R be the frame in which the smaller loop (radius b, current I)lies in the xy plane, and R′ the frame in which the larger loop lies in the x′y′

plane. We might as well take the line of intersection of the two planes to be they axis, so y = y′. Then the z′ axis has spherical coordinates (θ = α, φ = 0) in


R, and for transforming back and forth between the two frames we may use thetransformation matrices we derived in the last problem, with θ0 = α, φ0 = 0. Ifwe choose to evaluate the integral (8) in frame R, the current density is

Jb(r) = Iδ(r − b)δ(θ − π/2)[

− sin φi + cos φj]

so the components of the torque are

Nx = −Ib2

∫ 2π

0

Br(r = b, θ = π/2, φ) sinφ dφ (9)

Ny = Ib2

∫ 2π

0

Br(r = b, θ = π/2, φ) cosφ dφ (10)

To do the integral in (8), we need an expression for the radial component Br

of the field of the larger loop. Of course, we already have an expression for thefield in R′: in that frame the field is just that of a circular current loop in thex′y′ plane, Jackson’s equation (5.48):

B′

r(r′, θ′) =

µ0I′a

2r′

∞∑

l=0

(−1)l(2l + 1)!!

2ll!

r2l+1<

r2l+2>

P2l+1(cos θ′).

We are interested in evaluating this field at points along the smaller currentloop, and for all such points r = b; then r< = b, r> = a and we have

B′

r(r′ = b, θ′) =

µ0I′

2a

∞∑

l=0

(−1)l(2l + 1)!!

2ll!

(

b

a

)2l

P2l+1(cos θ′). (11)

To transform this to frame R, we first note that, since the origins of R and R′

coincide, the unit vectors r and r′ coincide, so Br = B′

r. Next, (11) expressesthe field in terms of cos θ′, the polar angle in frame R′. How do we write thisin terms of the angles θ and φ in frame R? Well, note that

cos θ′ =z′

r

=x sin α + z cosα

r

=r sin θ cosφ sin α + r cos θ cosα

r= sin θ sin α cosφ + cos θ cosα (12)

where in the second line we used the transformation matrix from Problem 5.11to write down z′ in terms of x and z. Equation (12) is telling us what ourcoordinates in R′ are in terms of our coordinates in R; if a point has angularcoordinates θ, φ in R, then (12) tells us what angle θ′ it has in R′. (We couldalso work out what the azimuthal angle φ′ would be, but we don’t need to,because (11) doesn’t depend on φ′.)


To express the Legendre function in (11) with the argument (12), we maymake use of the addition theorem for associated Legendre polynomials:

Pl(cos θ′) = Pl(cos θ cosα + sin θ sin α cosφ)

= Pl(cos θ)Pl(cosα) + 2

l∑

m=1

P ml (cos θ)P m

l (cosα) cos mφ.

Of course, the smaller loop exists in the xy plane, so for all points on that loopwe have θ = π/2, whence

Pl(cos θ′) = Pl(0)Pl(cosα) + 2

l∑

m=1

P ml (0)P m

l (cos θ) cos mφ.

We may now write down an expression for the radial component of the magneticfield of the larger loop, evaluated at points on the smaller loop, in terms of theangle φ that goes from 0 to 2π around that loop:

Br(φ) =µ0I

′

2a

∞∑

l=0

(−1)l(2l + 1)!!

2ll!

(

b

a

)2l

P2l+1(0)P2l+1(cosα)

+ 2

2l+1∑

m=1

P m2l+1(0)P m

2l+1(cos α) cosmφ

.

This looks ugly, but in fact when we plug it into the integrals (9) and (10)the sin φ and cosφ terms beat against the cosmφ term, integrating to 0 in theformer case and πδm1 in the latter. The torque is

Nx = 0

Ny =πµ0II ′b2

a

∞∑

l=0

(−1)l(2l + 1)!!

2ll!

(

b

a

)2l

P 12l+1(0)P 1

2l+1(cosα).

To finish we just need to rewrite the numerical factor under the sum:

(−1)l(2l + 1)!!

2ll!P 1

2l+1(0) =(2l + 1)!!

2ll!

[

Γ(l + 3/2)

Γ(l + 1)Γ(3/2)

]

=(2l + 3 − 2)(2l + 3 − 4)(2l + 3 − 6) · · · (5)(3)

2lΓ(l + 1)

[

Γ(l + 3/2)

Γ(l + 1)Γ(3/2)

]

=(l + 3/2− 1)(l + 3/2− 2) · · · (5/2)(3/2)

Γ(l + 1)

[

Γ(l + 3/2)

Γ(l + 1)Γ(3/2)

]

=

[

Γ(l + 3/2)

Γ(l + 1)Γ(3/2)

]2

= (l + 1)2[

Γ(l + 3/2)

Γ(l + 2)Γ(3/2)

]2


So my answer is

Ny =πµ0II ′b2

a

∞∑

l=0

(l + 1)2[

Γ(l + 3/2)

Γ(l + 2)Γ(3/2)

]2 (

b

a

)2l

P 12l+1(cosα).

Evidently I’m off by a factor of 1/(l + 1)(2l + 1) under the sum, but I can’t findwhere. Can anybody help?

Problem 5.13

A sphere of radius a carries a uniform surface-charge distribution σ. The sphereis rotated about a diameter with constant angular velocity ω. Find the vectorpotential and magnetic-flux density both inside and outside the sphere.

Problem 5.14

A long, hollow, right circular cylinder of inner (outer) radius a (b), and of relativepermeability µr, is placed in a region of initially uniform magnetic-flux density B0

at right angles to the field. Find the flux density at all points in space, and sketchthe logarithm of the ratio of the magnitudes of B on the cylinder axis to B0 as afunction of log10 µr for a2/b2 = 0.5, 0.1. Neglect end effects.

We’ll take the cylinder axis as the z axis of our coordinate system, and we’lltake B0 along the x axis: B0 = B0i. To the extent that we ignore end effects,we may imagine the fields to have no z dependence, so we effectively have a twodimensional problem.

There are two distinct current distributions in this problem. The first isa current distribution Jfree giving rise to the uniform field B0 far away fromthe cylinder; this current distribution is only nonvanishing at points outside thecylinder. The second is a current distribution Jbound = ∇× M existing onlywithin the cylinder. Since there is no free current within the cylinder or in itsinner region, the equations determining H in those regions are

∇ ·B = ∇ · (µH) = 0, ∇×H = Jfree = 0.

These imply that, within the cylinder and in its inner region, we may derive Hfrom a scalar potential: H = −∇Φm, with Φm satisfying the Laplace equation.

In the external region, there is free current, so things are not so simple. Toproceed we may separate the H field in the external region into two compo-nents: one that arises from the free current, and one that arises from the boundcurrents within the cylinder. The former is just (1/µ0)B0 and the second isagain derivable from a scalar potential satisfying the Laplace equation. So, inthe external region, H = (1/µ0)B0 −∇Φm.


So our task is to find expressions for Φm in the three regions such that theboundary conditions on B and H are satisfied at the borders of the regions.Writing down the solutions of the 2-D Laplace equation in the three regions,and excluding terms which blow up as ρ → 0 or ρ → ∞, we have

Φm(ρ, φ) =

∑

∞

n=1 ρn[

An cosnφ + Bn sinnφ]

r < a∑

∞

n=1

ρn[

Cn cosnφ + Dn sin nφ]

+ ρ−n[

En cosnφ + Fn sin nφ]

a < r < b∑

∞

n=1 ρ−n[

Gn cosnφ + Hn sinnφ]

r > b

Actually, we may argue on symmetry grounds that the sin terms must allvanish: otherwise, the fields would take different values on the positive andnegative y axes, but there is nothing in the problem distinguishing these axesfrom each other. With this simplification we may write down expressions forthe components of the H field in the three regions:

Hr =

− ∂

∂rΦm =

∞∑

n=1

−nAnρn−1 cosnφ, r < a

− ∂

∂rΦm =

∞∑

n=1

−n(

Cnρn−1 − Enρ−(n+1))

cosnφ, a < r < b

(1/µ0)B0r −∂

∂rΦm =

[

(1/µ0)B0 cosφ +∞∑

n=1

nGnρ−(n+1) cosnφ]

, r < b.

Hφ =

− ∂

∂φΦm =

∞∑

n=1

nAnρn−1 sinnφ, r < a

− ∂

∂φΦm =

∞∑

n=1

n(

Cnρn−1 + Enρ−(n+1))

sinnφ, a < r < b

(1/µ0)B0φ − ∂

∂φΦm =

[

− (1/µ0)B0 sin φ +

∞∑

n=1

nGnρ−(n+1) sin nφ]

, r < b.

The boundary conditions at r = b are that µHρ and Hφ be continuous,where µ = µ0 outside the cylinder and µrµ0 inside. With the above expressionsfor the components of H, we have

1

µ0B0 cosφ +

∞∑

n=1

nGnb−(n+1) cosnφ = µr

∞∑

n=1

−n(

Cnbn−1 − Enb−(n+1))

cosnφ

− 1

µ0B0 sinφ +

∞∑

n=1

nGnb−(n+1) sin nφ =∞∑

n=1

n(

Cnbn−1 + Enb−(n+1))

sin nφ.

We may multiply both sides of these by cosnφ and sin nφ and integrate from


0 to 2π to find

1

µ0B0 + G1b

−2 = −µrC1 + µrE1b−2 (13)

Gnb−(n+1) = −µr

(

Cnbn−1 − Enb−(n−1))

, n 6= 1 (14)

− 1

µ0B0 + G1b

−2 = C1 + E1b−2 (15)

Gnb−(n+1) =(

Cnbn−1 + Enb−(n+1))

, n 6= 1 (16)

Similarly, at r = a we obtain

A1 = µrC1 − µrE1a−2 (17)

Anan−1 = µr

(

Cnan−1 − Ena−(n+1))

, n 6= 1

A1 = C1 + E1a−2 (18)

Anan−1 =(

Cnan−1 + Ena−(n+1))

, n 6= 1.

(19)

For n 6= 1, the only solution turns out to be An = Cn = En = Gn = 0. Forn = 1, multiplying (15) by µr and adding and subtracting with (13) yields

2µrC1 = −(µr + 1)B0

µ0+ (µr − 1)G1b

−2 (20)

2µrE1 = (1 − µr)B0

µ0b2 + (µr + 1)G1. (21)

On the other hand, multiplying (18) by µr and adding and subtracting with(17) yields

2µrC1 = (µr + 1)A1 (22)

2µrE1 = (µr − 1)a2A1. (23)

Equating (20) with (22), we find

A1 = −B0

µ0+

(µr − 1)

(µr + 1)G1b

−2

while equating (21) with (23) yields

A1 = −B0

µ0

(

b2

a2

)

+(µr + 1)

(µr − 1)G1a

−2

and now equating these two equations gives

G1 =

[

1 −(a

b

)2]

(µ2r − 1)b2

(µr + 1)2b2 − (µr − 1)2a2

(

B0

µ0

)

b2.


-4.5

-4

-3.5

-3

-2.5

-2

-1.5

-1

-0.5

0

0 1 2 3 4 5

PSfra

grep

lacem

ents

log10 µr

log10r

(a/b) = 0.5

(a/b) = 0.1

Figure 2: Damping of field inside cylindrical cylinder of permeability µr.

The other coefficients may be worked out from this one:

A1 =−4µrb

2

(µr + 1)2b2 − (µr − 1)2a2

B0

µ0

C1 =−2(µr + 1)b2

(µr + 1)2b2 − (µr − 1)2a2

B0

µ0

E1 =−2(µr − 1)b2

(µr + 1)2b2 − (µr − 1)2a2

B0

µ0a2.

The H field is

H =4µrb

2

(µr + 1)2b2 − (µr − 1)2a2

B0

µ0i, r < a

=2b2

(µr + 1)2b2 − (µr − 1)2a2

B0

µ0

[

(µr + 1) + (µr − 1)(a

r

)2]

i− 2(µr − 1)(a

r

)2

cosφr

, a < r < b

=B0

µi +

(b2 − a2)(µ2r − 1)

(µr + 1)2b2 − (µr − 1)2a2

(

B0

µ0

) (

b2

r2

)

[

i + 2 sin φ φ]

, r > b.

The ratio r of the field within the cylinder to the external field is

r =4µr

(µr + 1)2 − (µr − 1)2 a2

b2

.

This relationship is graphed in Figure


Problem 5.16

A circular loop of wire of radius a and negligible thickness carries a current I . Theloop is centered in a spherical cavity of radius b > a in a large block of soft iron.Assume that the relative permeability of the iron is effectively infinite and that ofthe medium in the cavity, unity.

(a) In the approximation of b a, show that the magnetic field at the center ofthe loop is augmented by a factor (1 + a3/2b3) by the presence of the iron.

(b) What is the radius of the ”image” current loop (carrying the same current)that simulates the effect of the iron for r < b?

(a) There are two distinct current distributions in this problem: the free current

density J1 flowing in the loop, and the bound current density J2 flowing in theiron. These give rise to two fields B1 and B2, which must be summed at eachpoint in space to get the observed field.

B1 is just the field of a planar current loop, which Jackson has alreadyworked out for us in his section 5.5:

B1r =

µ0I

2a

∞∑

n=0

(−1)n(2n + 1)!!

2nn!

( r

a

)2n

P2n+1(cos θ), r < a

µ0Ia2

2r3

∞∑

n=0

(−1)n(2n + 1)!!

2nn!

(a

r

)2n

P2n+1(cos θ), r > a.

(24)

B1θ =

µ0I

4a

∞∑

n=0

(−1)n(2n − 1)!!

2n−1n!

(r

a

)2n

P 12n+1(cos θ), r < a

−µ0Ia2

4r3

∞∑

n=0

(−1)n(2n + 1)!!

2n(n + 1)!

(a

r

)2n

P 12n+1(cos θ), r > a.

(25)

On the other hand, since J2 vanishes for r < b, the field B2 to which it givesrise has no divergence or curl in that region, which means that throughout theregion it may be derived from a scalar potential satisfying the Laplace equation:

B2 = −∇Φm = −∇[

∞∑

n=0

AnrnPn(cos θ)

]

→ B2r =∞∑

n=1

nAnrn−1Pn(cos θ) (26)

B2θ =

∞∑

n=1

Anrn−1P 1n(cos θ) (27)


Since the iron filling the space r > b is assumed to have infinite permeability,the H field (and hence the B field, since B = H for r < b) must be strictlyradial at the boundary r = b. The An coefficients are thus determined by therequirement that (27) and (25) sum to zero at r = b:

∞∑

n=1

Anbn−1P 1n(cos θ) =

µ0Ia2

4b3

∞∑

n=0

(−1)n(2n + 1)!!

2n(n + 1)!

(a

b

)2n

P 12n+1(cos θ).

The orthogonality of the associated Legendre polynomials requires that eachterm in the sum cancel individually, whence

A2n = 0

A2n+1 =µ0Ia2

4b3

(−1)n(2n + 1)!!

2n(n + 1)!

( a

b2

)2n

.

Then the field of the bound current in the iron is determined everywhere in theregion r < b:

B2r =µ0Ia2

4b3

∞∑

n=0

(−1)n(2n + 1)(2n + 1)!!

2n(n + 1)!

(ar

b2

)2n

P2n+1(cos θ) (28)

B2θ =µ0Ia2

4b3

∞∑

n=0

(−1)n(2n + 1)!!

2n(n + 1)!

(ar

b2

)2n

P 12n+1(cos θ). (29)

As r → 0, B2θ → 0 and B2r → µ0Ia2/4b3, while B1r → µ0I/2a, so the totalfield at r = 0 is

Br(r = 0) = B1r(r = 0) + B2r(r = 0) =µ0I

2a+

µ0Ia2

4b3=

µ0I

2a

[

1 +a3

2b3

]

.

(b) The B2 field may be attributed to an image current ring outside r = b if,for suitable redefinitions of I and a, the expressions (28) and (29) can be madeto look like the r < a versions of (24) and (25).

Problem 5.18

A circular loop of wire having a radius a and carrying a current I is located invacuum with its center a distance d away from a semi-infinite slab of permeabilityµ. Find the force acting on the loop when

(a) the plane of the loop is parallel to the face of the slab,

(b) the plane of the loop is perpendicular to the face of the slab.

(c) Determine the limiting form of your answer to parts a and b when d a.Can you obtain these limiting values in some simple and direct way?

(a) We’ll take the loop to be at z = +d, and the slab of permeability µ tooccupy the space z < 0, so that the boundary surface is z = 0.


In the region z < 0, there is no free current, so ∇×H = 0 everywhere; thusH may be obtained from a scalar potential, H = −∇Φm, and since ∇ · H = 0as well we have ∇2Φm = 0. The azimuthally symmetric solution of the Laplaceequation in cylindrical coordinates that remains finite as z → −∞ is

Φm(z < 0) =

∫

∞

0

dk A(k)ekzJ0(kρ), (30)

and from this we obtain

Hρ(z < 0) = − ∂

∂ρΦm = −

∫

∞

0

dk kA(k)ekzJ ′

0(kρ)

=

∫

∞

0

dk kA(k)ekzJ1(kρ) (31)

Hz(z < 0) = − ∂

∂zΦm = −

∫

∞

0

dk kA(k)ekzJ0(kρ). (32)

On the other hand, for z > 0 we may decompose the H field into twocomponents: one component H1 arising from the current loop, and a secondcomponent H2 arising from the bound currents running in the slab. H1 is justgiven by the curl of the vector potential we worked out in Problem 5.10:

H1 =1

µ0∇×A, A = Aφφ, Aφ =

µ0Ia

2

∫

∞

0

dk e−k(z−d)J1(ka)J1(kρ), z > d

µ0Ia

2

∫

∞

0

dk e−k(d−z)J1(ka)J1(kρ), z < d.

so

H1ρ = − 1

µ0

∂

∂zAφ

=

Ia

2

∫

∞

0

dk ke−k(z−d)J1(ka)J1(kρ), z > d

−Ia

2

∫

∞

0

dk ke−k(d−z)J1(ka)J1(kρ), z < d.

(33)

H1z =1

µ0

1

ρ

∂

∂ρ(ρAφ)

=

Ia

2

∫

∞

0

dk ke−k(z−d)J1(ka)

[

1

kρJ1(kρ) − J0(kρ)

]

z > d

Ia

2

∫

∞

0

dk ke−k(d−z)J1(ka)

[

1

kρJ1(kρ) − J0(kρ)

]

, z < d.

(34)

In the last two equations we may use Jackson’s identity (3.87),

1

kρJ1(kρ) =

1

2[J0(kρ) + J2(kρ)]


to rewrite H1z as

H1z =

Ia

4

∫

∞

0

dk e−k(z−d)J1(ka) [J2(kρ) − J0(kρ)] , z > d

Ia

4

∫

∞

0

dk e−k(d−z)J1(ka) [J2(kρ) − J0(kρ)] , z < d.

(35)

Since the H2 field arises entirely from bound currents, it may also be derivedfrom a scalar potential Φm satisfying the Laplace equation. The azimuthallysymmetric solution of the Laplace equation in cylindrical coordinates that re-mains finite for all ρ and as z → +∞ is

Φm(z > 0) =

∫

∞

0

dk B(k)e−kzJ0(kρ)

and the components of H2 are

H2r(z > 0) = −∫

∞

0

dk kB(k)e−kzJ1(kρ) (36)

H2z(z > 0) =

∫

∞

0

dk kB(k)e−kzJ0(kρ). (37)

The required forms of the functions A(k) and B(k) are determined by theboundary conditions on H at the medium boundary, z = 0:

Hρ(z = 0−) = Hρ(z = 0+) µHρ(z = 0−) = µ0Hρ(z = 0+).

Equating (32) with the sum of (??) and (??), we have

−∫

∞

0

dk kA(k)J0(kρ) =µ0Ia

2

∫

∞

0

dk ke−kdJ1(ka) (J2(kρ) − J0(kρ)) +

∫

∞

0

dk kB(k)J0(kρ)



Homer Reid

April 20, 2001


Problem 5.19

A magnetically “hard” material is in the shape of a right circular cylinder of length Land radius a. The cylinder has a permanent magnetization M0, uniform throughoutits volume and parallel to its axis.

(a) Determing the magnetic field H and magnetic induction B at all points on theaxis of the cylinder, both inside and outside.

(b) Plot the ratios B/µ0M0 and H/M0 at all points on the axis of the cylinder,both inside and outside.

There is no free current in this problem, so H(ρ, z) may be derived from ascalar potential Φm(ρ, z) satisfying the Laplace equation. Dividing space intothree regions

Φm =

∫

∞

0

dk A(k)e−kzJ0(kρ), z > L/2

∫

∞

0

dk[

B(k)ekz + C(k)e−kz]

J0(kρ), − L/2 < z < L/2

∫

∞

0

dk D(k)ekzJ0(kρ), z < −L/2.

1


The tangential boundary condition at z = +L/2 is

∂Φm

∂ρ

∣

∣

∣

∣

z= L

2+

=∂Φm

∂ρ

∣

∣

∣

∣

z= L

2−

⇒∫

∞

0

dk kA(k)e−kL/2J1(kρ) =

∫

∞

0

dk k[

B(k)ekL/2 + C(k)e−kL/2]

J1(kρ)

(1)

This must hold for all ρ. Multiplying both sides by ρJ1(k′ρ), integrating from

ρ = 0 to ρ = ∞, and using the identity

∫

∞

0

dρ ρJn(kρ)Jn(k′ρ) =1

kδ(k − k′) (2)

we obtain from (1) the relation

A(k) = B(k)ekL + C(k). (3)

The perpendicular boundary condition at z = +L/2 is

Bz(z = L/2+) = Bz(L/2−)

or

µ0Hz(z = L/2+) = µ0

[

Hz(z = L/2−) + Mz(z = L/2−)]

∂Φm

∂z

∣

∣

∣

∣

z= L

2+

=∂Φm

∂z

∣

∣

∣

∣

z= L

2−

+ M(ρ)

⇒∫

∞

0

dk kA(k)e−kL/2J0(kρ) =

∫

∞

0

dk k[

−B(k)ekL/2 + C(k)e−kL/2]

J0(kρ) + M(ρ)

(4)

where

M(ρ) =

M1, ρ < a

0, ρ > a.

Now we multiply both sides of (4) by ρJ0(k′ρ) and integrate from ρ = 0 to

ρ = ∞ to obtain

A(k) = −B(k)ekL + C(k) + M1ekL/2

∫ a

0

ρJ0(kρ)dρ

= −B(k)ekL + C(k) + γ(k) (5)

where we defined

γ(k) = M1ekL/2

∫ a

0

ρJ0(kρ)dρ =aM1

kekL/2J1(ka).


The solution of eqs. (3) and (5) is

B(k) =1

2e−kLγ(k) C(k) = A(k) − 1

2γ(k). (6)

From the boundary conditions at z = −L/2 we may similarly obtain the rela-tions

B(k) + C(k)ekL = D(k)

B(k) − C(k)ekL = D(k) − γ(k)

which may be solved to yield

B(k) = D(k) − 1

2γ(k) C(k) =

1

2e−kLγ(k). (7)

Comparing (6) and (7) we find

A(k) = D(k) =M1a

kcosh

kL

2J1(ka)

B(k) = C(k) =M1a

2ke−kL/2J1(ka).

Then the components of the H field are

Hρ =

M1a

∫

∞

0

dk coshkL

2e−kzJ1(ka)J1(kρ), z > L/2

M1a

∫

∞

0

dk e−kL/2 cosh(kz)J1(ka)J1(kρ), − L/2 < z < L/2

M1a

∫

∞

0

dk coshkL

2ekzJ1(ka)J1(kρ), z < −L/2

Hz =

M1a

∫

∞

0

dk coshkL

2e−kzJ1(ka)J0(kρ), z > L/2

−M1a

∫

∞

0

dk e−kL/2 sinh(kz)J1(ka)J0(kρ), − L/2 < z < L/2

−M1a

∫

∞

0

dk coshkL

2ekzJ1(ka)J0(kρ), z < −L/2.


Problem 5.23

A right circular cylinder of length L and radius a has a uniform lengthwise magne-tization M .

(a) Show that, when it is placed with its flat end against an infinitely permeableplane surface, it adheres with a force

F = 2µ0aLM2

[

K(k) − E(k)

k− K(k1) − E(k1)

k1

]

where

k =2a√

4a2 + L2, k1 =

a√a2 + L2

.

(b) Find the limiting form of the force if L a.

We’ll define our coordinate system so that the z axis is the cylinder axis,and we’ll take the surface of the permeable medium at z = 0.

Our general strategy for this problem will be as follows. First, we’ll findthe magnetic field H0 that exists in all space when the cylinder is pressed upflat against the infinitely permeable medium. Then we’ll calculate the shift dEin the energy of the magnetic field incurred by moving the cylinder up a smalldistance dz off the surface of the medium. The force on the cylinder is thenreadily calculated as F = −dE/dz.

To calculate the energy shift incurred by moving the cylinder a distance dzaway from the permeable medium, we won’t have to go through and completelyrecalculate the fields and their energy in the new configuration. Instead, wecan use the following little trick. When we move the cylinder up a distance dz,two things happen. First a gap of height dz opens between the surface and theface of the cylinder, where previously there had been a fixed magnetization M,but now there is just free space. Second, between L and L + dz there is now afixed magnetization M where previously there was none. Moving the cylinder offixed M up a distance dz is thus formally equivalent to keeping the cylinder putand instead introducing a cylinder of the opposite magnetization −M between 0and dz, while also introducing a cylinder of magnetization +M between L andL + dz. The increase in field energy in this latter case is fairly easily calculatedby taking the integral of µ0Mc

H0 over the regions in which the fixed magnetization changes.So the first task is to find the field that exists when the cylinder is pressed

flat against the surface. Since there are no free currents in the problem, wemay derive H from a scalar potential satisfying the Laplace equation. To be-gin we write down the general solutions of the Laplace equation in cylindricalcoordinates, observing first that by symmetry we can only keep terms with no


azimuthal angle dependence:

Φ(m) =

∫

∞

0

dk A(k)e−kzJ0(kρ), z > L

∫

∞

0

dk [B(k)ekz + C(k)e−kz ]J0(kρ), 0 < z < L

∫

∞

0

dk D(k)e+kzJ0(kρ), z < 0.

(8)

The boundary conditions at z = 0 are that Hρ and Bz be continuous. Assum-ing first of all that the medium existing in the region below z = 0 has finitepermeability µ, the tangential boundary condition is

∂Φm

∂ρ

∣

∣

∣

∣

z=0−

=∂Φm

∂ρ

∣

∣

∣

∣

z=0+∫

∞

0

dk k D(k)J1(kρ) =

∫

∞

0

dk k [B(k) + C(k)]J1(kρ). (9)

Multiplying (9) by ρJ1(k′ρ), integrating from 0 to ∞, and using the identity

(2), we findD(k) = B(k) + C(k). (10)

The normal boundary condition at z = 0 is of a mixed type. Below the linewe have simply Bz = µHz. Above the line we may write Bz = µ0[Hz + M(ρ)],where M(ρ) represents the fixed magnetic polarization of the cylinder:

M(ρ) =

M, ρ < a

0, ρ > a.(11)

The normal boundary condition at z = 0 is then

−µ∂

∂zΦm

∣

∣

∣

∣

z=0−

= −µ0∂

∂zΦm

∣

∣

∣

∣

z=0+

+ µ0M(ρ)

− µ

µ0

∫

∞

0

dk k D(k)J0(kρ) = −∫

∞

0

dk k [B(k) − C(k)]J0(kρ) + M(ρ)

Now multiplying by ρJ0(k′ρ), integrating from ρ = 0 to ∞, and using (2) yields

µ

µ0D(k) = −B(k) + C(k) −

∫

∞

0

ρ M(ρ)J0(kρ) dρ. (12)

Using (11), the integral on the RHS is

M

∫ a

0

ρJ0(kρ) dρ =Ma

kJ1(ka) ≡ γ(k)

where we defined a convenient shorthand. Then (12) is

µ

µ0D(k) = −B(k) + C(k) − γ(k).


Now taking µ → ∞, we see that, to keep the B and C coefficients from blowingup, we must have D → 0. Then equation (??) tells us that B(k) = −C(k), sothe middle entry in (8) may be rewritten:

Φm(z, ρ) =

∫

∞

0

dk β(k) sinh(kz)J0(kρ), (0 < z < L).

The boundary conditions at z = L are

∂Φm

∂ρ

∣

∣

∣

∣

z=L+

=∂Φm

∂ρ

∣

∣

∣

∣

z=L−

− ∂Φm

∂z

∣

∣

∣

∣

z=L+

= −∂Φm

∂z

∣

∣

∣

∣

z=L−

+ M(ρ)

with M(ρ) defined as above. Working through the same procedure as aboveyields the conditions

A(k)e−kL = β(k) sinh(kL)

A(k)e−kL = β(k) cosh(kL) + γ(k)

with γ(k) defined as above. The solution is

β(k) = −γ(k)e+kL

A(k) = γ(k) sinh(kL).

Plugging these back into (8) and differentiating, we find for the z component ofthe H field

Hz(ρ, z) =

Ma

∫

∞

0

dk e−kz cosh(kL)J0(kρ)J1(ka), z > L

−Ma

∫

∞

0

dk e−kL cosh(kz)J0(kρ)J1(ka), 0 < z < L.

(13)

Now that we know the field, we want to find the change in energy densityincurred by putting into this field a short cylinder (radius a, height dz) ofmagnetization −M k between z = 0 and z = dz, and another cylinder of thesame size but with magnetization +M k between z = L and z = L + dz. Thechange in field energy is just the integral of µ0M · H over the volume in whichthe magnetization density has changed:

dU = −2πµ0M

∫ dz

0

∫ a

0

Hz(z, ρ)ρ dρ dz + 2πµ0M

∫ L+dz

L

∫ a

0

Hz(z, ρ)ρ dρ dz

= 2πµ0Mdz(

∫ a

0

Hz(L, ρ)ρ dρ −∫ a

0

Hz(0, ρ)ρ dρ)

(14)

where in the last step we assumed that Hz remains essentially constant overa distance dz in the z direction, and may thus be taken out of the integral.


Inserting (13) into (), and exchanging the order of integration, we first do the ρintegral:

∫ a

0

J0(kρ)ρdρ =a

kJ1(ka).

Then () becomes



Homer Reid

March 28, 2002


1


Problem 6.2

The charge and current densities for a single point charge q can be written formallyas

ρ(x′, t′) = qδ[x′ − r(t′)]; J(x′, t′) = qv(t′)δ[x′ − r(t′)]

where r(t′) is the charge’s position at time t′ and v(t′) is its velocity. In evaluatingexpressions involving the retarded time, one must put t′ = tret = t−R(t′)/c, whereR = x − r(t′).

(a) As a preliminary to deriving the Heaviside-Feynman expressions for the electricand magnetic fields of a point charge, show that

∫

d3x′ δ[x′ − r(tret)] =1

κ

where κ = 1 − v · R/c. Note that κ is evaluated at the retarded time.

(b) Starting with the Jefimenko generalizations of the Coulomb and Biot-Savartlaws, use the expressions for the charge and current densities for a point chargeand the result of part a to obtain the Heaviside-Feynman expressions for theelectric and magnetic fields of a point charge,

E =q

4πε0

[

R

κR2

]

ret

+1

c

∂

∂t

[

R

κR

]

ret

− 1

c2

∂

∂t

[ v

κR

]

ret

and

B =µ0q

4π

[

v × R

κR2

]

ret

+1

c

∂

∂t

[

v × R

κR

]

ret

(c) In our notation Feynman’s expression for the electric field is

E =q

4πε0

[

R

R2

]

ret

+[R]ret

c

∂

∂t

[

R

R2

]

ret

+1

c2

∂2

∂t2[R]ret

while Heaviside’s expression for the magnetic field is

B =µ0q

4π

[

v × R

κ2R2

]

ret

+1

c[R]ret

∂

∂t

[

v × R

κ

]

ret

.

Show the equivalence of the two sets of expressions for the fields.

(a) Let’s first assume that the charge is traveling along the z axis, so that itsposition is given by

r(t) = (z0 + vzt)k.


The retarded time tret(t, z) at a given point z on the z axis is

tret(t, z) = t − z

c

sor[tret(t, z)] = (z0 + vztret(t, z))k.

Hence

δ(x − r[tret(x, t)]) = δ(x)δ(y)δ z − [z0 + vztret(t, z)]

= δ(x)δ(y)δ

z − [z0 + vz(t −z

c)]

= δ(x)δ(y)δ

z − z0 − vzt +vz

cz)]

= δ(x)δ(y)δ(

1 +vz

c

)

z − (z0 + vzt)

By the properties of the δ function we may write this as

=

(

1

1 + vz/c

)

δ(x)δ(y)δ

z − z0 + vzt

1 + vz/c

.

The δ function is singling out the point in space from which originates theelectromagnetic disturbance we feel at the origin at time t. Let’s think aboutwhat’s going on here in two limiting cases. First, as vz → 0, the z delta functionbecomes δ(z−(z0+vzt)). This means that the source point for the field we feel atthe origin at time t is just z = z0−vzt, which is of course just the instantaneouslocation of the source particle at time t. In other words, the electromagneticdisturbance left behind by the particle at time t reaches the origin so quicklythat the particle hasn’t had time to move on. The electromagnetic disturbanceseems to be coming from the instantaneous location of the particle itself.

In the opposite limit vz → c, the z delta function becomes δ(z−(z0−vzt)/2).This says that the point from which we feel an electromagnetic disturbance attime t is half as far from the origin as the particle itself is at time t. Thisagain makes sense. At each point in the particle’s motion, the electromagneticdisturbance it causes begins propagating toward the origin, while meanwhilethe particle continues propagating away from the origin at the same speed.Hence when the electromagnetic disturbance has reached the origin, the particlehas traveled as far as the electromagnetic disturbance did, but in the opposite

direction, so it is now twice as far from the origin as it was when the disturbancewe are just now feeling was generated.


Problem 6.5

A localized electric charge distribution produces an electrostatic field, E = −∇Φ.Into this field is placed a small localized time-independent current density J(x),which generates a magnetic field H.

(a) Show that the momentum of these electromagnetic fields, (6.117), can be trans-formed to

Pfield =1

c2

∫

ΦJ d3x

provided the product ΦH falls of rapidly enough at large distances. Howrapidly is “rapidly enough”?

(b) Assuming that the current distribution is localized to a region small comparedto the scale of variation of the electric field, expand the electrostatic potentialin a Taylor series and show that

Pfield =1

c2E(0) ×m,

where E(0) is the electric field at the current distribution and m is the mag-netic moment, (5.54), caused by the current.

(c) Suupose the current distribution is placed instead in a uniform electric fieldE0 (filling all space). Show that, no matter how complicated is the localizedJ, the result in part a is augmented by a surface integral contribution frominfinity equal to minus one-third of the result of part b, yielding

Pfield =2

3c2E0 ×m.

Compare this result with that obtained by working directly with (6.117) andthe considerations at the end of Section 5.6.

(a) From the definition of electromagnetic field momentum we have

c2Pfield =

∫

E×H dV

= −∫

(∇Φ) ×H dV.

Focusing for now on the z component, we have

c2Pz = −∫

(∂Φ

∂xHy − ∂Φ

∂yHx) dx dy dz (1)


Let’s take our volume of integration to be a cube of side L, which we willeventually take to infinity. Integrating the first term by parts with respect tox, we have

∫ L

−L

∫ L

−L

∫ L

−L

∂Φ

∂xHy dx

dy dz =

∫ L

−L

∫ L

−L

ΦHy

∣

∣

∣

x=L

x=−L−

∫

Φ∂Hy

∂xdx

dy dz.

Similarly integrating the second term in (1) by parts with respect to y, we maywrite (1) as

c2Pz = −∫ L

−L

∫ L

−L

ΦHy

∣

∣

∣

x=L

x=−Ldy dz +

∫ L

−L

∫ L

−L

ΦHx

∣

∣

∣

y=L

y=−Ldx dz +

∫

Φ(∇×H)z dV

= −∫ L

−L

∫ L

−L

ΦHy

∣

∣

∣

x=L

x=−Ldy dz +

∫ L

−L

∫ L

−L

ΦHx

∣

∣

∣

y=L

y=−Ldx dz +

∫

ΦJz dV

where in going to the last line we used ∇ × H = J since there is no time-dependent E field. This equation is just the z component of

c2P =

∫

ΦH× dA +

∫

ΦJ dV. (2)

If we now take L → ∞, the first integral (which describes surface effects)vanishes providing the product Φ(x)H(x) vanishes more quickly (i.e. like ahigher power of x) than x2. Then we are left with just the second term:

c2P =

∫

ΦJ dV. (3)

(b) We have

Φ(x) = Φ(0) + x · ∇Φ(0) +1

2

∑

xixj∂2Φ

∂xi∂xj+ · · ·

We may arbitrarily choose Φ(0) = 0. Also, we are told that the electric fielddoesn’t vary much in the region of nonvanishing J, in which case we may ignorethe second derivatives of Φ, to obtain

Φ(x) ≈ x · ∇Φ(0) = −x ·E(0).

Plugging into (3),

c2P = −∫

[

x · E(0)]

J dV. (4)

We have

−[

x ·E(0)]

J =[

E(0) × J]

x−[

x ·E(0)]

J −[

E(0) × J]

x

= E(0) ×[

x × J]

−[

E(0) × J]

x


where in the first line we added and subtracted a term, and in the second usedthe BAC-CAB identity of vector analysis. With this, (4) becomes

c2P = E(0) ×∫

[

x + J]

dV −∫

[

E(0) × J] x dV

= 2E(0) ×m−∫

[

E(0) × J] x dV

where in the first term we have identified the definition of the dipole moment m.Evidently to get this to match up with what Jackson has we need to argue thatsecond term is exactly half the first, but I can’t see how to do this for arbitraryJ. Can anybody help?

(c) From (2) we have

c2P =

∫

ΦH× dA +

∫

ΦJ dV.

The second term is just equal to (E × m)/c2, as computed in part b. For thefirst term,

Problem 6.13

A parallel plate capacitor is formed of two flat rectangular perfectly conductingsheets of dimensions a and b separated by a distance d small compared to a or b.Current is fed in and taken out uniformly along adjacent edges of length b. Withthe input current and voltage defined at this end of the capacitor, calculate theinput impedance or admittance using the field concepts of Section 6.9.

(a) Calculate the electric and magnetic fields in the capacitor correct to secondorder in powers of the frequency, but neglecting fringing fields.

(b) Show that the expansion of the reactance (6.140) in powers of the frequencyto an appropriate order is the same as that obtained for a lumped circuitconsisting of a capacitance C = ε0ab/d in series with an inductance L =µ0ad/3b.

(a) We’ll suppose the plates are oriented parallel to the xy plane, with the lowerplate at z = 0 and the upper plate at z = d. We’ll take the edges of side aparallel to the x axis, and the edges of side b parallel to the y axis. Then theboundary condition on the current density is

J(0, y, 0) = −J(0, y, d) = J0j

for 0 < y < b.


With neglect of fringing fields, the electric field between the plates existsonly in the z direction, while the magnetic field exists only in the x direction.We assume harmonic time dependence and write

E(y) = E(y)e−iωtk B(y) = B(y)e−iωtx; (5)

then time differentiation becomes multiplication by −iω. The Maxwell equa-tions are then

∇ · E = 0 ⇒ ∂E

∂z= 0

∇ · B = 0 ⇒ ∂B

∂x= 0

∇×E = −∂B

∂t⇒ ∂E

∂y= +iωB

∇×B =1

c2

∂E

∂t⇒ ∂B

∂y= +

iω

c2E.

(6)

We postulate an expansion in powers of ω for E and B:

E(y) = E0(y) + ωE1(y) + ω2E2(y) + · · ·B(y) = B0(y) + ωB1(y) + ω2B2(y) + · · ·

(7)

Then the curl equations in (6) become

∂

∂y

[

E0 + ωE1 + ω2E2 + · · ·]

= iω[

B0 + ωB1 + ω2B2 + · · ·]

∂

∂y

[

B0 + ωB1 + ω2B2 + · · ·]

=iω

c2

[

E0 + ωE1 + ω2E2 + · · ·]


Equating equal powers of ω in these equations, we obtain

∂E0

∂y= 0 ⇒ E0 = α

∂B0

∂y= 0 ⇒ B0 = β

∂E1

∂y= iB0 = iβ ⇒ E1 = iβy

∂B1

∂y=

i

c2E0 =

iα

c2⇒ B1 =

iα

c2y

∂E2

∂y= iB1 = − α

c2y ⇒ E2 = − α

2c2y2

∂B2

∂y=

i

c2E1 = − β

c2y ⇒ B2 = − β

2c2y2

∂E3

∂y= iB2 = − iβ

2c2y2 ⇒ E3 = − iβ

6c2y3

∂B3

∂y=

i

c2E2 = − iα

2c4y2 ⇒ B3 = − iα

6c4y3

∂E4

∂y= iB3 =

α

24c4y3 ⇒ E4 =

iβ

24c2y4

∂B4

∂y=

i

c2E3 = +

β

6c4y3 ⇒ B4 =

β

24c4y4

and so on. Plugging into (7), we obtain

E(y) = α(1 − (ky)2

2+

(ky)4

24+ · · · ) + iβc(ky − (ky)3

6+ · · · ) (8)

= α cos ky + iβc sin ky (9)

B(y) = β cos ky +iα

csin ky (10)

where k = ω/c, and where we simply wrote down what we guessed to be thesums of the full infinite series from their first few terms.

To complete the problem we need to determine the constants α and β, forwhich purpose we appeal to the boundary conditions on the plates. We knowthat the discontinuities in the E and B field are proportional to the surfacecharge and current densities on the plates. Since these conditions only giveinformation on the differences between the fields outside and between the plates,we ostensibly have to know what the fields are outside to get what they areinside. But for the purposes of this problem we’ll just assume there are nofields outside, so the charge and current densities on the plates determine thefields inside. I know this is correct in the low-frequency limit, and in the high-frequency limit I’m not yet sure how to compute the radiation fields in theregion outside the plates, so I will ignore them.


The boundary conditions are

Ez = − σ

ε0

Bx = −µ0Ky

where σ and Ky are the charge density and y component of the surface currentdensity on the top plate (assumed to be identical but with opposite sign on thebottom plate). Plugging in the solutions (9) and (??), we have

σ = −ε0(α cos ky + iβc sin ky)

Ky = − 1

µ0(β cos ky +

iα

csin ky)

(11)

As a sanity check, we can verify the continuity relation between charge andcurrent on the plates:

∂Ky

∂y= −∂σ

∂t= +iωσ

Plugging in (11), the left and right sides of this are

LHS = − 1

µ0(−kβ sin ky +

ikα

ccos ky)

RHS = −iωε0(α cos ky + iβc sin ky)

and the two are evidently equal.The forcing function in this problem is the surface current density specified

on the edges of length b. If the total current fed into the y = 0 edge of the topplate is I(t) = I0 cosωt (with an opposite current taken out of the y = 0 edgeof the bottom plate) then the surface current boundary conditions are

Ky(y = 0) =I0

bcosωt

Ky(y = a) = 0

Comparing with (11), we see that these boundary conditions we have to take

β = −µ0I0

bcosωt

α = − iµ0I0c

bcosωt cot ka


Plugging into (9) and (10),

Ez = − iµ0I0c

bcosωt [cot ka cos ky + sin ky]

= − iµ0I0c

bcosωt

[

1

sin ka

]

[cos ka cos ky + sin ka sin ky]

= − iµ0I0c

bcosωt

cos[k(y − a)]

sin ka

Bz = −µ0I0

bcosωt [− cosky + cot ka sin ky]

= −µ0I0

bcosωt

1

sin ka[− sin ka cosky + cos ka sin ky]

= −µ0I0

bcosωt

sin[k(y − a)]

sin ka

Problem 6.14

An ideal circular parallel plate capacitor of radius a and plate separation d a isconnected to a current source by axial leads, as shown in the sketch. The currentin the wire is I(t) = I0 cosωt.

(a) Calculate the electric and magnetic fields between the plates to second order inpowers of the frequency (or wave number), neglecting the effects of fringingfields.

(b) Calculate the volume integrals of we and wm that enter the definition of thereactance X, (6.140), to second order in ω. Show that in terms of the inputcurrent Ii, defined by Ii = −iωQ, where Q is the total charge on one plate,these energies are

∫

we d3x =1

4πε0

|Ii|2dω2a2

,

∫

wm d3x =µ0

4π

|Ii|2d8

(

1 +ω2a2

12c2

)

(c) Show that the equivalent series circuit has C ≈ πε0a2/d, L ≈ µ0d/8π, and

that an estimate for the resonant frequency of the system is ωres = 2√

2c/a.Compare with the first root of J0(x).

(a) We work in cylindrical coordinates and assume harmonic time dependence(∝ e−iωt) for all quantities; then time differentiation is replaced by multiplica-tion by −iω. If we neglect the effects of fringing fields, everything is symmetricin θ, and the electric field between the plates is entirely in the z direction, while


the magnetic field is entirely in the θ direction:

E(x, t) = E(r, z)e−iωtz B = B(r, z)e−iωtθ. (12)

The Maxwell equations for the fields between the plates are

∇ ·E = 0 ⇒ ∂

∂zE = 0

∇ ·B = 0 ⇒ ∂

∂θB = 0

∇×E = −∂B

∂t⇒ ∂E

∂r= −iωB

∇×B =1

c2

∂E

∂t⇒ 1

r

∂

∂r(rB) = − iω

c2E.

(13)

To proceed, let’s propose an expansion of the fields in powers of the frequency:

E(r) = E0(r, z) + ωE1(r, z) + ω2E2(r, z) + · · · (14)

B(r) = B0(r, z) + ωB1(r, z) + ω2B2(r, z) + · · · (15)

Then the curl equations in (13) become

∂

∂r

[

E0 + ωE1 + ω2E2

]

= −iω[

B0 + ωB1 + ω2B2

]

1

r

∂

∂r

[

rB0 + ωrB1 + ω2rB2

]

= − iω

c2

[

E0 + ωE1 + ω2E2

]

Now we just have to go through and equate like powers of ω in these equations.For n = 0, we have

∂E0

∂r= 0 ⇒ E0 = α1 (16)

for some constant α1, and

1

r

∂

∂r(rB0) = 0 ⇒ B0 =

β

r. (17)

But for nonzero β this blows up at the origin. Hence we must take β = 0, soB0 = 0. 2 For n = 1, we have

∂E1

∂r= −iB0 = 0 ⇒ E1 = α2 (18)

for some constant α2, and

1

r

∂

∂r(rB1) = − i

c2E0 = − iα1

c2⇒ B1 = − iα1

2c2r. (19)

Continuing,

∂E2

∂r= −iB1 = − α1

2c2r ⇒ E2 = − α1

4c2r2 (20)

1

r

∂

∂r(rB2) = − i

c2E1 = − iα2

c2⇒ B2 = − iα2

2c2r (21)


∂E3

∂r= −iB2 = − α2

2c2r ⇒ E3 = − α2

4c2r2 (22)

1

r

∂

∂r(rB3) = − i

c2E2 =

iα1

4c4r2 ⇒ B3 =

iα1

16c4r3 (23)

Evidently E2n and E2n+1 have the same form but differ by the presence ofα1 or α2, as is true for B2n−1 and B2n. Plugging in equations (16) through (23)into (14) and (15), we obtain

E(r) = (α1 + ωα2)(

1 − ω2

4c2r2 +

ω4

64c4r4

)

+ · · ·

= (α1 + ωα2)(

1 − (kr)2

4+

(kr)4

64+ · · ·

)

B(r) = − i

c(α1 + ωα2)

(

kr

2

)

(

1 − (kr)2

8+ · · ·

)

These look the first few terms in certain Bessel functions:

E(r) = (α1 + ωα2)J0(kr) ≡ βJ0(kr)

B(r) = − i

cβJ1(kr)

where we can define the constant β = (α1 + ωα2) since we’re dealing with afixed frequency. Inserting into (12) we obtain

E(r, t) = βJ0(kr)e−iωtk B(r, t) = − i

cβJ1(kr)e−iωt θ. (24)

To work out the value of β, we need to apply the boundary conditions at thecapacitor plates. An easy way to do this is to consider what happens as ω → 0.In that limit there is no magnetic field, and the electric field between the platesis just Ez(t) = −2σ(t)/ε0, where σ(t) is the instantaneous value of the surfacecharge induced on each plate (positive on the top plate, negative on the bottom).Now, the total charge on the top plate is just the integral of the current flowingonto that plate:

q =

∫

I(t) dt =I0

ωsin ωt

and the surface charge is this divided by the plate area (since we are assuming alow frequency, any charge that flows onto the plate instantaneously equilibrateswith the rest of the charge on the plate, yielding a constant surface chargedensity):

σ(t) =I0

πa2ωsin ωt.

Hence the electric field in the low frequency limit is

Ez(ω → 0) = − 2I0

πa2ωε0sin ω.


Comparing this with(24) in the limit k → 0, we obtain

β = − 2I0i

πa2ωε0.

Hence

E(r, t) = − 2I0

πa2ωε0J0(kr) sin ωt k B(r, t) =

2µ0I0c

πa2ωJ1(kr) cos ωt θ. (25)

(b) The average energy densities are

we =ε04

E2 =I20

(πa2ω)2ε0

(

1 − kr

4+ · · ·

)2

wm =1

4µ20

B2 =µ0I

20 c2

(πa2ω)2

(

kr

2

)2(

1 − (kr)2

8+ · · ·

)2

We only have to keep the first terms in the parentheses to get the energy rightto second order in ω:

Ue ≈ I20

(πa2ω)2ε0

∫ a

0

(2πd)(r dr)

=I20d

πa2ω2ε0

Um =µ0I

20 c2

(πa2ω)2

∫ a

0

(2πd)(rdr)(kr

2− (kr)3

8+ · · ·

)

Um =µ0I

20 c2

(πa2ω)2

∫ a

0

(2πd)(rdr)(kr

2− (kr)3

8+ · · ·

)



Homer Reid

May 24, 2002

Chapter 8: Waveguide Derivations

Before starting the problems, I thought it would be useful to run throughmy own derivations of some of the formulas from this chapter.

Waveguides and cavities: basic pedagogy

The unifying feature of waveguide and cavity problems is that we single outone spatial coordinate and announce from the start that the fields will havesinusoidal dependence on that coordinate. Taking the special coordinate to be z,this means that all components of all fields have the functional form f(x, y)eikz

for some wavevector k.Assuming harmonic time dependence, we write explicitly

E(x) =

Ex(x, y)i + Ey(x, y)j + Ez(x, y)k

ei(kz−ωt)

B(x) =

Bx(x, y)i + By(x, y)j) + Bz(x, y)k

ei(kz−ωt)(1)

We have here a total of six functions f(x, y) that we must find to satisfyMaxwell’s equations with the relevant boundary conditions. At first this wouldappear tough since the six fields are all coupled by Maxwell’s equations, butafter a little algebra we obtain the following simplified situation: The z− di-rection fields Ez(x, y) and Bz(x, y) turn out to satisfy (separately) simple one-dimensional differential equations, which may be readily solved upon specifyingthe boundary conditions for a particular situation. Meanwhile, the remainingfields (Ex, Ey, Bx, By) can be expressed simply as linear combinations of Ez andBz and their derivatives, so once we obtain the z fields we have everything. Inwhat follows we’ll derive the differential equations satisfied by Ez and Bz andthe equations giving the remaining fields in terms of them.

1


The differential equations for Ez

and Bz

The Maxwell curl equations are

∇×E = −∂B

∂t, ∇×B =

1

c2m

∂E

∂t

where cm is the speed of light in the medium. We begin by applying the firstcurl equation to our ansatz (1), obtaining

∂yEz − ikEy = iωBx (2)

−∂xEz + ikEx = iωBy (3)

∂xEy − ∂yEx = iωBz, (4)

and we pause to solve the first two of these for Ex and Ey :

Ex =ω

kBy − i

k∂xEz , Ey = −ω

kBx − i

k∂yEz . (5)

Next we apply the second curl equation to our ansatz, obtaining

∂yBz − ikBy = − iω

c2m

Ex (6)

−∂xBz + ikBx = −iω

c2m

Ey (7)

∂xBy − ∂yBx = −iω

c2m

Ez . (8)

But in (5) we solved for Ex and Ey , and if we then plug those solutions into (6)and (7) we can solve for Bx and By in terms of Bz and Ez :

Bx =ikc2

m

ω2 − k2c2m

(

∂xBz +ω

c2mk

∂yEz

)

(9)

By =ikc2

m

ω2 − k2c2m

(

∂yBz − ω

c2mk

∂xEz

)

. (10)

Finally, with the ansatz (1) the equation ∇ · B = 0 reads

∂Bx

∂x+

∂By

∂y= −ikBz.

When we plug (9) and (10) into this, the terms involving Ez fields cancel, andwe obtain an equation involving Bz alone:

(

∂2

∂x2+

∂2

∂y2

)

Bz +

(

ω2

c2m

− k2

)

Bz = 0

or

(

∂2

∂x2+

∂2

∂y2

)

Bz + γ2Bz = 0 (11)


where

γ =

√

ω2

c2− k2.

If we had carried out this derivation in the reverse order we would have obtainedthe same equation for Ez :

(

∂2

∂x2+

∂2

∂y2

)

Ez + γEz = 0. (12)

We can think of equations (11) and (12) as eigenvalue equations that have solu-tions only for certain values of the parameter γ, which depend on the boundaryconditions.

Armed with equations (11) and (12) and the boundary conditions appropri-ate to our problem we can now solve for Bz and Ez and then use (9) and (10)to find the remaining components of the B field. The remaining components ofthe E field are given by analogous equations:

Ex =ikc2

m

ω2 − k2c2m

(

∂xEz +ω

k∂yBz

)

(13)

Ey =ikc2

m

ω2 − k2c2m

(

∂yEz − ω

k∂xBz

)

. (14)

Boundary Conditions; TE, TM, TEM Modes

The boundary conditions on the fields at the surfaces of the waveguide or cavityare that E‖ and B⊥ be continuous, where ⊥ denotes the component of thevector normal to the boundary surface and ‖ includes all other components ofthe vector. This means that the two eigenvalue equations (11) and (12) must besolved subject to different boundary conditions, which means in general theireigenvalues will be different. If we have a solution of (12) for some value of γ(i.e. for some combination of values of ω and k), then there will be no nonzerosolution of (11) for that value of γ, and hence we must have Bz = 0 identicallyfor the mode at that frequency and wavevector. Since in this case the magneticfield has nonzero components only transverse to the direction of propagation,this is called a transverse magnetic mode. Similarly, if (11) can be solved withnonzero Bz at some γ, then Ez = 0 and we have a transverse electric mode. Amode for which both Ez and Bz are zero is called a transverse electromagneticmode, and can only exist in the region between two conducting surfaces, notwithin a single conductor as is possible for TE and TM modes.

Since either Ez or Bz is zero, we can simplify some of the equations aboveand collect results appropriate to the two cases.


TM Modes TE Modes

Bz ≡ 0

∇2t Ez + γ2Ez = 0, En

∣

∣

∣

∂S= 0

Ex =ikc2

m

ω2 − k2c2m

∂xEz

Ey =ikc2

m

ω2 − k2c2m

∂yEz

Bx =iω

ω2 − k2c2m

∂yEz

By = − iω

ω2 − k2c2m

∂xEz

Ez ≡ 0

∇2t Bz + γ2Bz = 0,

∂Bn

∂n

∣

∣

∣

∣

∂S

= 0

Ex =iωc2

m

ω2 − k2c2m

∂yBz

Ey = − iωc2m

ω2 − k2c2m

∂xBz

Bx =ikc2

m

ω2 − k2c2m

∂xBz

By =ikc2

m

ω2 − k2c2m

∂yBz

(A factor of ei(kz−ωt) is understood in all of these expressions.)

For TM modes, the boundary condition is E‖ = 0, and Ez is always perpen-dicular to the boundary surfaces, so the boundary condition for the eigenvalueequation is Ez = 0. For the case of TE modes, the boundary condition is B⊥ = 0.Suppose one boundary surface is the yz plane. The normal to this plane is thex direction, so Bx must vanish at this surface; but we just saw that in the TEcase Bx ∝ ∂xBz, i.e. the derivative of Bz normal to the boundary surface mustvanish. This is general: the boundary condition for the eigenvalue equation inthe TM case is ∂Bz/∂n = 0.

Power flow; Energy Loss

The flow of power down a waveguide is described by the z component of thePoynting vector S = E×H = 1

µE×B. Using the boxed expressions above, forthe two types of modes we obtain

STM

z =1

µ(ExBy − EyBx)

=kωc2

m

µ(ω2 − k2c2m)2

[

(∂xEz)2 + (∂yEz)

2]

e2i(kz−ωt)

or, in the time average,

=kωc2

m

2µ(ω2 − k2c2m)2

[

(∂xEz)2 + (∂yEz)

2]

=εkω

2γ4

[

(∂xEz)2 + (∂yEz)

2]

. (15)


Similarly, for TE modes we obtain

STE

z =kω

2µγ4

[

(∂xBz)2 + (∂yBz)

2]

. (16)

To address the issue of dissipation in the boundaries, we solve Maxwell’sequations within the boundary surfaces. The two curl equations are

∇×E = iωB (17)

∇×B = µJ− iµεωE

= µ (σ − iεω)E

≈ µσE (18)

since σ εω in most cases. (For example, for a copper waveguide with airor vacuum interior we have we have σ ≈ 6 · 107 Ω−1 m−1, while εω ≈ 9 ·10−12 Ω−1 m−1 · (ω in rad/sec), so the approximation is good up to frequenciesω ∼ 1019 rad/sec.)

Now we assume that the fields are only changing significantly in the direc-tion normal to the boundary surface (i.e., as we go deeper and deeper into theboundary surface the fields die out rapidly, whereas as we move along parallelto the boundary surface the fields don’t change much) and keep only the normalderivative in the curl equations. If ρ measures the depth of penetration into thesurface, the curl equations become

ρ × ∂E

∂ρ= iωB

ρ × ∂B

∂ρ= µσE

Differentiating the first of these, taking the cross product with ρ of both sides,and substituting in the second equation yields

ρ × ρ × ∂2E

∂ρ2= iωµσE

or, using the bac-cab rule,

∂2Eρ

∂ρ2ρ − ∂2E

∂ρ2= iωµσE.

Evidently the ρ component of the LHS vanishes here, so Eρ = 0; the electricfield within the conducting boundary has no component normal to the surface.For the remaining components we obtain

∂2E‖

∂ρ2+ iωµσE‖ = 0


with solution

E‖ = e±√

iωµσρE0

= e±(1+i) ρ

δ E0

where δ =√

2/ωµσ is the skin depth and E0 is the field just at the surface ofthe boundary. To keep the solution from blowing up as we penetrate into theconductor we take the negative sign in the exponent. From (17) we then obtain

B =i − 1

δω(ρ × E0)e

−(1+i) ρ

δ .

Evaluating this at the surface yields the modified boundary condition on thefields in the cavity or waveguide:

B0 =i − 1

δω(ρ ×E0). (19)

From this equation we can work out the power loss per unit length in thecavity or waveguide. The power dissipated in a volume dV is

∫

(J · E) dV =σ

∫

E2 dV. We integrate over the volume occupied by the boundary surfaces ina length dz :

dP = dz

∮ ∫

∞

0

σE20e−2(1+i) ρ

δ dρ dl

e2i(kz−ωt)

= dz

σδ

2(1 + i)

∮

E20 dl

e2i(kz−ωt)

or, taking the time average,

dP

dz=

σδ

4√

2

∮

E20 dl (20)

where the line integral is over the cross section of the surface boundary at afixed value of z.



Homer Reid

May 24, 2002


1


Problem 8.2

A transmission line consisting of two concentric cylinders of metal with conductivityσ and skin depth δ, as shown, is filled with a uniform lossless dielectric (µ, ε). ATEM mode is propagated along this line.

(a) Show that the time-averaged power flow along the line is

P =

√

µ

επa2|H0|2 ln

(

b

a

)

where H0 is the peak value of the azimuthal magnetic field at the surface ofthe inner conductor.

(b) Show that the transmitted power is attenuated along the line as

P (z) = P0e−2γz

where

γ =1

2σδ

√

ε

µ

(

1a + 1

b

)

ln(

ba

) .

(c) The characteristic impedance Z0 of the line is defined as the ratio of the voltagebetween the cylinders to the axial current flowing in one of them at anyposition z. Show that for this line

Z0 =1

2π

√

µ

εln

(

b

a

)

.

(d) Show that the series resistance and inductance per unit length of the line are

R =1

2πσδ

(

1

a+

1

b

)

L =

µ

2πln

(

b

a

)

+µcδ

4π

(

1

a+

1

b

)

.

(a) For the TEM mode, the electric field in the waveguide may be written

E(x, y, z, t) = Et(x, y)e−ikze−iωt

where Et has only x and y components and may be derived from a scalarpotential, i.e. Et = −∇tΦ. Since Φ satisfies the Laplace equation, we maywrite its general form immediately (neglecting an arbitrary constant):

Φ(ρ, θ) = β0 ln ρ +

∞∑

l=1

(αlρl + βlρ

−l) sin(lθ + αl).


In this part of the problem we’ll neglect dissipation in the waveguide walls,so the boundary condition on Et is that its components transverse to the wallsvanish, i.e.

∂Φ

∂θ

∣

∣

∣

∣

r=b

=∂Φ

∂θ

∣

∣

∣

∣

r=a

= 0.

This yields no condition on β0, since the θ derivative of that term vanishesanyway, but on the terms in the summation we obtain the conditions

αlal + βla

−l = αlbl + βlb

−l = 0

which can only be satisfied if αl = βl = 0 for l 6= 0. Hence we have

Φ(ρ) = β0 ln ρ −→ E = −β01

ρρ. (1)

The magnetic field is found from Jackson’s (8.28):

H = − 1

µB =

√

ε

µ(z ×E)

=

√

ε

µ

β0

ρθ. (2)

The time-averaged Poynting vector is

S =1

2(E ×H∗) =

1

2

√

ε

µ|β0|2

(

1

ρ

)2

z

Integrating over the cross section of the waveguide, we obtain the power transfer:

P =

∫ b

a

∫ 2π

0

S · dA

=1

2

√

ε

µ|β0|2

∫ b

a

2πρ dρ

ρ2

=

√

ε

µ|β0|2 · π ln

(

b

a

)

(3)

=

√

µ

ε(πa2) ln

(

b

a

)

·[

ε

µ

|β0|2a2

]

Referring back to (2) to rewrite the term in brackets, we obtain

P =

√

µ

ε(πa2) ln

(

b

a

)

|H(a)|2 (4)

(b) Without going back and completely re-solving for the fields in the waveguidefor the case of finite conductivity, we can calculate the power loss per unit length


approximately using Jackson’s equation (8.58):

−dP

dz=

1

2σδ

∮

c

|ρ ×H|2 dl

=1

2σδ

(

ε

µ

)

(

2πb ·(

β0

b

)2

+ 2πa ·(

β0

a

)2)

=πβ2

0

σδ

(

ε

µ

)(

1

b+

1

a

)

.

Dividing by (3), we obtain

γ = − 1

2P

dP

dz=

1

2σδ

√

ε

µ

(

1a + 1

b

)

ln(

ba

) .

(c) The fields inside the waveguide are

E(ρ, z, t) = −β0

ρei(kz−ωt)ρ

H(ρ, z, t) =

√

ε

µ

β0

ρei(kz−ωt)θ

From the E field we can compute the voltage difference between the cylinders:

V (z, t) = −β0ei(kz−ωt)

∫ b

a

dρ

ρ= −β0e

i(kz−ωt) lnb

a(5)

while from the H field we can compute the axial current flowing in, say, theouter cylinder:

I = 2πb|Kb| = 2πb|ρ ×H(ρ = b)| = 2π

√

ε

µβ0e

i(kz−ωt). (6)

Dividing (5) by (6), we have

Z =V

I=

1

2π

√

µ

εln

b

a.


Problem 8.4

Transverse electric and magnetic waves are propagated along a hollow, right circularcylinder of brass with inner radius R.

(a) Find the cutoff frequencies of the various TE and TM modes. Determine nu-merically the lowest cutoff frequency (the dominant mode) in terms of thetube radius and the ratio of cutoff frequencies of the next four higher modesto that of the dominant mode. For this part assume that hte conductivity ofbrass is infinite.

(b) Calculate the attenuation constant of the waveguide as a function of frequencyfor the lowest two modes and plot it as a function of frequency.

(a) The equation we have to solve is

(∇2t + γ2)Ψ(ρ, θ) = 0,

i.e. the Helmholtz equation. Ψ is Ez for the TM case and Hz for the TE case.The boundary conditions are Ψ(ρ = R) = 0 for the TM case, and (∂Ψ/∂ρ)(ρ =R) = 0 for the TE case.

The general solution of Helmholtz in 2D is

Ψ(ρ, θ) =∞∑

L=0

JL(γρ)(ALeiLθ + BLe−iLθ) + NL(γρ)(CLeiLθ + DLeiLθ).

Since this solution must be valid everywhere in the interior of the waveguide,including at ρ = 0, the part of the solution involving NL must vanish. Also,for a physical solution we must have L an integer. But otherwise I don’t thinkthere are any constraints on AL and BL. I guess these guys are determined bythe field configuration one forces into the waveguide at one of its ends.

The allowed values of γ are determined by the boundary conditions. Theseare

TM case : Ψ|ρ=R = 0 =⇒ JL(γR) = 0 (7)

TE case :∂Ψ

∂ρ

∣

∣

∣

∣

ρ=R

= 0 =⇒ J ′

L(γR) = 0 (8)

(9)

Hence the allowable eigenvalues are given by

γi =xi

R


where the xi are the roots of JL(x) = 0 and J ′

L(x) = 0. Referring to Jackson’stables on pages 114 and 370, we can write down the five lowest-lying eigenvalues:

γ1 =1.841

R, TE, L = 1

γ2 =2.405

R, TM, L = 0

γ3 =3.054

R, TE, L = 2

γ4a =3.832

R, TE, L = 1

γ4b =3.832

R, TM, L = 0.

The last two eigenvalues are degenerate.The lowest cutoff frequency is

ωc =γ1√µε

=1.841

R√

µε.

(b) The lowest-lying mode is the TE mode with L = 1. For this mode we have

Hz(ρ, θ, z, t) = H0J1(γ1ρ)eiθei(kz−ωt) (10)

with k2 = µεω2 − γ21 . The tangential component of the field, from Jackson

(8.33), is

Hθ(ρ, θ, z, t) = − k

ργ2Hz (11)

Using (10) and (11), we can find the current induced in the wall of the conductorat ρ = R:

Keff = ρ ×H(ρ = R) = −H0J1(γ1R)eiθei(kz−ωt)

[

θ +k

Rγ21

z

]

.

Then Jackson (8.58) is

−dP

dz=

1

2σδH2

0J21 (γ1R) · 2πR

[

1 +

(

k

Rγ21

)2]

(12)

On the other hand, the transmitted power is given by Jackson (8.51):

P =1

2

√

µ

ε

(

ω

ωλ

)2 [

1 −(ωλ

ω

)2]1/2 ∫

A

H∗

z Hz dA

= πH20

√

µ

ε

(

ω

ωλ

)2 [

1 −(ωλ

ω

)2]1/2 ∫ R

0

ρJ21 (γ1ρ) dρ


The integral can be evaluated with Jackson (3.95):

= πH20

√

µ

ε

(

ω

ωλ

)2 [

1 −(ωλ

ω

)2]1/2

R2

2[J2(γ1R)]2

(13)

Dividing (12) by (13), we calculate the attenuation coefficient:

β =1

2P

dP

dz=

2

σδR

√

ε

µ

[

J1(1.841)

J2(1.841)

]2[

1 +

(

k

Rγ21

)2]

(ωλ

ω

)2[

ω2

ω2 − ω2λ

]1/2

=1

2P

dP

dz=

2

σδR

√

ε

µ

[

J1(1.841)

J2(1.841)

]2[

1 +

(

Rk

(1.841)2

)2]

(ωλ

ω

)2[

ω2

ω2 − ω2λ

]1/2

=1

2P

dP

dz=

2

σδR

√

ε

µ

[

J1(1.841)

J2(1.841)

]2 [µεR2ω2

(1.841)2

]

(ωλ

ω

)2[

ω2

ω2 − ω2λ

]1/2

.

Problem 8.5

A waveguide is constructed so that the cross section of the guide forms a righttriangle with sides of length a, a,

√2a, as shown. The medium inside has µr = εr =

1.

(a) Assuming infinite conductivity for the walls, determine the possible modes ofpropagation and their cutoff frequencies.

(b) For the lowest mode of each type calculate the attenuation constant, assumingthat the walls have large, but finite, conductivity. Compare the result withthat for a square guide of side a made from the same material.

(a) We’ll take the origin of coordinates at the lower left corner of the triangle.Then the boundary surfaces are defined by x = 0, y = a, and x = y. Thetask is to solve (∇2

t + γ2)Ψ = 0 subject to the vanishing of Ψ, or its normalderivative, at the walls. In the text, Jackson finds the form of the solutionsfor a rectangular waveguide. A little bit of staring at the triangular waveguidereveals that appropriate solutions for this geometry can be assembled from linearcombinations of the solutions for the rectangular case. For example, a term likesin kxx sin kyy, for suitable choices of kx and ky, already vanishes on the twolegs of the triangle. To get it to vanish on the third boundary surface, i.e. thehypotenuse (x = y), we can simply subtract the same term with kx and ky

swapped. In other words, we take

Ez(x, y) =∑

Amn

[

sin(mπx

a

)

sin(nπy

a

)

− sin(nπx

a

)

sin(mπy

a

)]

(TM)

Hz(x, y) =∑

Bmn

[

cos(mπx

a

)

cos(nπy

a

)

+ cos(nπx

a

)

cos(mπy

a

)]

(TE)


The TE case involves the plus sign because in the normal derivative on thediagonal boundary surface the x derivative comes in with the opposite sign asthe y derivative.

These satisfy (∇2t + γ2

mn)Ψ = 0, where

γ2mn =

(π

a

)2

(n2x + n2

y).

In contrast to the rectangular case, TM modes with m = n vanish identically.For both TM and TE modes, mode (m, n) is the same mode as (n, m).

As in the case of the rectangular waveguide, the smallest value of γ is to behad for the TE1, 0 mode, in which case

γ10 =(π

a

)

and the cutoff frequency is ωc(1,0) = π/(a√

µε). For the TM case the lowest

propagating mode is (m, n) = (2, 1), for which γ21 =√

5π/a and ωc(2,1) =√5ωc(1,0).

(b) The lowest-frequency TE mode has

Hz = H0

[

cos(πx

a

)

+ cos(πy

a

)]

Ht =ikπ

aγ210

H0

[

sin(πx

a

)

i + sin(πy

a

)

j]

.

The power loss is

−dP

dz=

1

2σδ

∮

|n×H|2 dl (14)

On the lower surface (y = 0) we have

|n ×H|2 = |H2y + H2

z | = H20

[

cos(πx

a

)

+ 1]2

+k2π2

a2γ410

sin2(πx

a

)

The contribution of the lower surface to the integral in (14) is thus∫

lower= aH2

0

[

3

2+

k2π2

a2γ410

]

(15)

The contribution of the right (vertical) boundary surface is the same. On thediagonal boundary surface, we have

n =1√2(−i + j) =⇒ n ×H =

1√2[Hzi + Hz j − (Hy + Hx)k]

with magnitude

|n ×H|2 =1

2[H2

x + H2y + H2

z + 2HxHy]

=H2

0

2

4 cos2(πγ

a

)

+ 4

(

kπ

aγ10

)2

sin2(πγ

a

)


where γ = x = y is the common coordinate as we move from (0, 0) to (a, a). Inthe integral in (14) we can put dl =

√2dγ and integrate over γ from 0 to a to

obtain∫

diagonal=

√2aH2

0

[

1 +k2π2

a2γ410

]

.

Adding this to two times (15) and inserting into (14), we have

−dP

dz=

aH20

2σδ

[

3 +√

2 + (2 +√

2)k2π2

a2γ410

]

. (16)

On the other hand, from Jackson (8.51) we have

P =H2

0

2

√

µ

ε

(

ω

ωλ

)2 (

1 − ω2λ

ω2

)1/2

×∫ a

0

∫ y

0

[

cos2(πx

a

)

+ cos2(πy

a

)

+ 2 cos(πx

a

)

cos(πy

a

)]

dx dy

By symmetry, the integral is just half of what we would get from integratingthe integrand over a square of side a, which, by inspection, is a2. Hence

P =a2H2

0

4

√

µ

ε

(

ω

ωλ

)2(

1 − ω2λ

ω2

)

.1/2

We could at this point proceed to write out the explicit form of the attenuationconstant, but what’s the point?

Problem 8.6

A resonant cavity of copper consists of a hollow, right circular cylinder of innerradius R and length L, with flat end faces.

(a) Determine the resonant frequencies of the cavity for all types of waves. With(1/

√µεR) as a unit of frequency, plot the lowest four resonant frequencies of

each type as a function of R/L for 0 < R/L < 2. Does the same mode havethe lowest frequency for all R/L ?

(b) If R=2 cm, L=3 cm, and the cavity is made of pure copper, what is thenumerical value of Q for the lowest resonant mode?

(a) Taking the origin at the center of the cavity, the ρ and φ components of thefields must vanish at z = ±L/2. Since the z dependence of all field componentsis e±ikz , the allowed values of k are k = nπ/L, with E ∝ sin kz for k even andE ∝ cos kz for k odd.


The equation characterizing TM modes is

(∇2t + γ2)Ez = 0, Ez |∂S = 0.

Expanding this in cylindrical coordinates, we obtain

∂2Ez

∂ρ2+

1

ρ

∂Ez

∂ρ+

1

ρ2

∂2Ez

∂φ2+ γ2Ez = 0

We put Ez(ρ, φ) = R(ρ)P (φ) to obtain

∂2P

∂φ2+ νP = 0

∂2R

∂ρ2+

1

ρ

∂R

∂ρ+

(

γ2 − ν2

ρ2

)

= 0.

The solutions are

P (φ) = e±iνφ

R(ρ) = Jν(γρ).

For single-valuedness we require ν ∈ , and to ensure Ez(ρ = R) = 0 we require

γ = xνm/R where xνm is the mth root of Jν(x) = 0. Hence

Ez = AJν

(

xνmρ

R

)

e±iνφe−iωt

sin kz, keven

cos kz, kodd(TM modes).

For TE modes the relevant equation is

(∇2t + γ2)Bz = 0,

∂Bz

∂n

∣

∣

∣

∣

∂S

= 0.

The general solution to the differential equation is the same as above, but nowthe boundary condition requires J ′

ν(γR) = 0, so γ = yνm/R where yνm is themth root of J ′

ν(y) = 0. Then the solutions look like

Bz = AJν

(

yνmρ

R

)

e±iνφe−iωt

sin kz, keven

cos kz, kodd(TE modes).

As we saw above, the allowed wavevectors are k = nπ/L. The frequency isrelated to the wavenumber according to

ωνmn = cm

√

γ2νm + k2

n

=

cm

R

√

x2νm + π2

(

R

L

)2

n2, (TM)

cm

R

√

y2νm + π2

(

R

L

)2

n2, (TE)


1

2

3

4

5

6

7

8

9

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

TE modesTM modes

PSfra

grep

lacem

ents

R/L

ω(u

nits

ofc/

R)

Figure 1: TM and TE mode frequencies for the resonant cavity of Problem 8.6.


(The m subscript on cm is not related to the m subscripts on ω and ν).The lowest four TM and TE mode frequencies are shown in Figure 1.

(b) The lowest resonant mode is the TE1,1,1 mode.info



Homer Reid

January 13, 2003


Problem 11.4

A possible clock is shown in the figure. It consists of a flashtube F and a photocellP shielded so that each views only the mirror M, located a distance d away, andmounted rigidly with respect to the flashtube-photocell assembly. The electronicinnards of the box are such that when the photocell responds to a light flash fromthe mirror, the flashtube is triggered with a negligible delay and emits a short flashtoward the mirror. The clock thus “ticks” once every (2d/c) seconds when at rest.

(a) Suppose that the clock moves with a uniform velocity v, perpendicular to theline from PF to M, relative to an observer. Using the second postulateof relativity, show by explicit geometrical or algebraic construction that theobserver sees the relativistic time dilatation as the clock moves by.

(b) Suppose that the clock moves with a velocity v parallel to the line from PFto M. Verify that here, too, the clock is observed to tick more slowly, by thesame time dilatation factor.

(a) Suppose that, relative to an observer, the clock is moving with speed vperpendicular to the direction in which the light travels back and forth. Let dtbe the time measured by the observer for the light to travel from the flashtubeto the mirror. The vertical distance traveled by the light is d, but—as far asthe observer is concerned—during the time dt the mirror has also translateda horizontal distance vdt. Hence the total distance the observer sees the light

1


travel (one-way) is√

d2 + (vdt)2. But this distance must equal cdt, since thelight must have the universal speed c in any reference frame. Hence we have

cdt =√

d2 + (vdt)2

or

dt =

(

1

1 + v2

c2

)1/2d

c.

This is the time (as measured by the observer) in which the light makes the tripfrom flashtube to mirror. The light takes the same amount of time to travelback to the photocell, so the total period of the clock is just twice this, or

period =

(

1

1 − v2

c2

)1/22d

c

which is greater than the period of the clock at rest by the normal Lorentzdilatation factor.

(b) Now we suppose that the clock is moving relative to an observer with speedv parallel to the direction of motion of the light. We align the z axis with thedirection of motion. Then we may write down the space-time coordinates (inthe clock’s rest frame) of the three relevant events in the operation of the clock(taking x4 = ct):

xa = (0, 0, 0, 0), (light leaves flashtube)

xb = (0, 0, d, d), (light reaches mirror)

xc = (0, 0, 0, 2d), (light reaches photocell).

The transformation matrix from the clock’s reference frame to the observer’sreference frame is

Γ =

1 0 0 00 1 0 00 0 γ γβ0 0 γβ γβ

and using this we may write down the spacetime coordinates of the three eventsabove in the rest frame of the observer:

x′

a = (0, 0, 0, 0)

x′

b =(

0, 0, γ(1 + β)d, γ(1 + β)d)

x′

c =(

0, 0, 2γβd, 2γd)

Evidently the difference in the time coordinates of events a and c (which is justthe observed period of the clock) is c∆t = 2γd, so again the observer observes


the clock to have a period of 2γd/c, longer by the factor γ than the period ofthe clock in its rest frame.

Problem 11.5

A coordinate system K ′ moves with a velocity v relative to another system K. In K ′

a particle has a velocity u′ and an acceleration a′. Find the Lorentz transformationlaw for acceleration, and show that in the system K the components of accelerationparallel and perpendicular to v are

a‖ =

(

1 − v2

c2

)3/2

(

1 + v·u′

c2

)3 a′

‖

a⊥ =

(

1 − v2

c2

)

(

1 + v·u′

c2

)3

(

a′

⊥+

v

c2× (a′ × u′)

)

.

The initial components of the velocity in the moving frame (frame K ′) areu′

‖and u′

⊥. Using Jackson’s equations 11.31 to transform these to frame K, we

obtain

u‖(0) =u′

‖+ v

1 +u′

‖v

c2

(1)

u⊥(0) =u′

⊥

γv

(

1 +u‖v

c2

) . (2)

After a time dt has passed in frame K, a time dt′ = dt/γ has passed in frameK′, and the velocity has increased by the amount a′dt′ = a′dt/γ. Then we canwrite down the new components of the velocity in K ′ and again transform toK:

u‖(dt) =u′

‖+ a′

‖dt′ + v

1 +(u′

‖+a′

‖dt′)v

c2

(3)

u⊥(dt) =u′

⊥

γv

(

1 +(u′

‖+a′

‖dt′)v

c2

) (4)

Subtracting (1) from (3), we have


∆u‖ =u′

‖+ a′

‖dt′ + v

1 +(u′

‖+a′

‖dt′)v

c2

−u′

‖+ v

1 +u′

‖v

c2

=a′

‖dt′(

1 − v2

c2

)

(

1 +u′

‖v

c2

)2

+(

1 +u′

‖v

c2

)

a′

‖dt′

Using the relation dt′ = dt/γ we can rewrite this as

=a′

‖dt(

1 − v2

c2

)3/2

(

1 +u′

‖v

c2

)2

+(

1 +u′

‖v

c2

)

(

1 + v2

c2

)1/2a′

‖dt

Dividing by dt and taking the limit as dt → 0, we obtain

a‖ = limdt→0

∆u‖

dt=

a′

‖

(

1 − v2

c2

)3/2

(

1 +u′

‖v

c2

)2 .

So evidently I’m off by 1 in the exponent of the denominator. What am I doingwrong?

Next, subtracting (2) from (4), we have

∆u⊥ =1

γv

u′

⊥+ a′

⊥dt′

1 +u′

‖v

c2 +a′

‖dt′v

c2

− u′

⊥

1 +u′

‖v

c2

=1

γv

a′

⊥dt′(

1 +u′

‖v

c2

)

− u′

⊥

(

a′

‖dt′v

c2

)

(

1 +u′

‖v

c2

)2

+(

1 +u′

‖v

c2

)

a′

‖dt′v

c2

=1

γv

a′

⊥dt′ + v

c2

(

u′

‖a′

⊥− u′

⊥a′

‖

)

dt′

(

1 +u′

‖v

c2

)2

+(

1 +u′

‖v

c2

)

a′

‖dt′v

c2

Again putting in dt = γdt′, we have

=1

γ2v

a′

⊥dt + v

c2

(

u′

‖a′

⊥− u′

⊥a′

‖

)

dt(

1 +u′

‖v

c2

)2

+(

1 +u′

‖v

c2

)

a′

‖dtv

γvc2


As before, when we divide by dt and take the limit as dt → 0 the second termin the denominator becomes irrelevant and we obtain

a⊥ = limdt→0

∆u⊥

dt=

1

γ2v

(

1 +u′

‖v

c2

)2

[

a′

⊥+

v

c2

(

u′

‖a′

⊥− u′

⊥a′

‖

)]

=

1 − v2

c2

(

1 +u′

‖v

c2

)2

[

a′

⊥+

v

c2

(

u′

‖a′

⊥− u′

⊥a′

‖

)]

Jackson writes the second term in the brackets a little differently. To show thathis expression is equivalent to mine, we use the BAC-CAB rule:

v × (a′ × u′) = (v · u′)a′ − (v · a′)u′

= vu′

‖

(

a′

‖+ a′

⊥

)

− va′

‖

(

u′

‖+ u′

⊥

)

The parallel components cancel, and we are left with

= v(u′

‖a′

⊥− a′

‖u′

⊥)

which is the way I wrote it above.But I’m still off by 1 in the exponent of the term in the denominator! What

am I missing?

Problem 11.6

Assume that a rocket ship leaves the earth in the year 2010. One of a set of twinsborn in 2080 remains on earth; the other rides in the rocket. The rocket ship isso constructed that it has an acceleration g in its own rest frame (this makes theoccupants feel at home). It accelerates in a straight-line path for 5 years (by its ownclocks), decelerates at the same rate for 5 more years, turns around, accelerates for5 years, decelerates for 5 years, and lands on earth. The twin in the rocket is 40years old.

(a) What year is it on earth?

(b) How far away from the earth did the rocket ship travel?

Let v(t) be the speed of the rocket, as observed on earth, at a time t asmeasured on earth. Using the first result of problem 11.5, we see that theacceleration of the rocket as measured from earth is

a =dv

dt= ±

(

1 − v(t)2

c2

)3/2

g,


ordβ

dt= ±

(

1 − β2)3/2

α

where α = g/c and the plus or minus sign depends on whether we are acceler-ating or decelerating. Manipulating this a little, we obtain

dβ

(1 − β2)3/2= ±αdt.

Integrating, we obtain

∣

∣

∣

∣

β′

(1 − β′2)1/2

∣

∣

∣

∣

β2

β1

= ±α(t2 − t1). (5)

For the first leg of the rocket’s journey, we have t1 = β1 = 0 and we take theplus sign in (5). Then we find

β

(1 − β2)1/2= αt

or

β(t) =αt

[1 + (αt)2]1/2(6)

and

γ(t) =1

√

1 − β2(t)=√

1 + (αt)2

(7)

Next let’s work out the relation between time as measured on the rocket andtime as measured on earth. With primed (unprimed) quantities referring to therocket (to earth), the infinitesimal relation is

dt = γ(t)dt′

or

t′2 − t′1 =

∫ t2

t1

dt

γ(t)

=

∫ t2

t1

dt√

1 + (αt)2

=1

α

∫ αt2

αt1

du√1 + u2

=1

α

[

sinh−1(αt2) − sinh−1(αt1)]


For the first leg of the journey this becomes

t2 =1

αsinh(αt′2). (8)

Now we know that the first leg of the journey lasts until t′2 = 5 years, and wehave

α =g

c=

9.8 m s−2

3 · 108 m s−1= 3.27 · 10−8 s−1.

Then the time on earth at the end of the first leg of the journey is, from (8),

t2 =

(

1

3.27 · 10−8

)

· sinh

[

3.27 · 10−8 s−1 (5 yr)

(

3.153 · 107 s

1 yr

)]

s

= 2.65 · 109 s

≈ 84 yr.

Finally, the distance the rocket travels during the first leg of its journey is

d = c

∫ t

0

β(t)dt

= c

∫ t

0

αt√

1 + (αt)2dt

=c

α

∫ αt

0

u√1 + u2

du

=c

α

[

1 + (αt)2]1/2 − 1

= 3.0 · 1015 meters.

The behavior of the rocket on the subsequent three legs of the journey issimilar to that in the first leg. In particular, the total distance traveled awayfrom earth is twice that covered in the first leg, or 6.0 ·1015 meters, and thetotal time elapsed on earth during the rocket’s journey is four times that elapsedduring the first leg, or 4·84=336 years. So it should be the year 2436 on earthby the time the rocket returns home.


Problem 11.13

An infinitely long straight wire of negligible cross-sectional area is at rest and hasa uniform linear charge density q0 in the inertial frame K ′. The frame K ′ (and thewire) move with a velocity v parallel to the direction of the wire with respect tothe laboratory frame.

(a) Write down the electric and magnetic fields in cylindrical coordinates in the restframe of the wire. Using the Lorentz transformation properties of the fields,find the components of the electric and magnetic fields in the laboratory.

(b) What are the charge and current densities associated with the wire in its restframe? In the laboratory?

(c) From the laboratory charge and current densities, calculate directly the electricand magnetic fields in the laboratory. Compare with the results of part (a).

I don’t like q0 as a symbol for charge density, because it appears to have thewrong units. I’ll use λ instead. We’ll take our z axis to coincide with the wireand take v in the positive z direction.

(a) In the rest frame there is no current and the E field is static; hence B = 0.The electric field is found by considering a Gaussian pillbox in the shape ofa right circular cylinder coaxial with the wire and of radius r and length dz.There is no electric field normal to the upper and lower surfaces, and the fieldnormal to the radial bounding surface is uniform across the circumference. Onthe other hand, the charge enclosed in the cylinder is λ dz. Then Gauss’ law is

∮

E · dA = 4πQ

=⇒ 2πrdzEr = 4πλdz

=⇒ Er =2λ

r. (9)

In terms of cartesian components we have

Ex = 2λ

(

x′

x′2 + y′2

)

, Ey = 2λ

(

y′

x′2 + y′2

)

.


The field-strength tensor in the laboratory frame is

F = ΛF ′Λ

= 2λ1

x′2 + y′2

γ 0 0 βγ0 1 0 00 0 1 0

βγ 0 0 γ

0 −x′ −y′ 0x′ 0 0 0y′ 0 0 00 0 0 0

γ 0 0 βγ0 1 0 00 0 1 0

βγ 0 0 γ

=2γλ

(x′2 + y′2)

0 −x′ −y′ 0x′ 0 0 βx′

y′ 0 0 βy′

0 −βx′ −βy′ 0

.

Reading off from this the transformed fields, and dropping the primes on x andy since the z boost leaves these coordinates unchanged, we have

E = γ2λ

x2 + y2(xi + yj) (10)

= γ2λ

rr (11)

B = βγλ

2(x2 + y2)(−yi + xj) (12)

= βγ2λ

rφ. (13)

(b) In the rest frame there is no current and the charge density is ρ′ = λδ(x)δ(y),so

J ′µ = cλ(δ(x)δ(y), 0, 0, 0).

The transformed current density is

J µ = ΛµνJ ′ν

= cλ

γ 0 0 γβ0 1 0 00 0 1 0

βγ 0 0 γ

δ(x)δ(y)000

= cγλ

δ(x)δ(y)00

βδ(x)δ(y)

.

(c) Computing the electric field in the laboratory frame is easy, since the chargedensity is the same as we had before but with a factor of γ thrown in. Thenthe E field is just (9) but with that factor of γ thrown in, i.e.

E =2γλ

rr


which agrees with (10).For the magnetic field, we note that the current density in the lab frame is

J = cβγλδ(x)δ(y)z. Then the current piercing a disc of radius r is I = cβγλ.On the other hand, by symmetry the magnetic field in the azimuthal directionaround the circumference of this disc is constant, so we may use Ampere’s law,∇×B = (4π/c)J to write

2πrBφ =4π

c(cβγλ)

or

B =2βγλ

rφ

which agrees with (13).

Problem 11.15

In a certain reference frame a static, uniform, electric field E0 is parallel to the xaxis, and a static, uniform, magnetic induction B0 = 2E0 lies in the x − y plane,making an angle θ with the axis. Determine the relative velocity of a referenceframe in which the electric and magnetic fields are parallel. What are the fields inthat frame for θ 1 and θ → (π/2)?

The untransformed fields are

E = E0i, B = 2E0(cos θi + sin θj).

Let’s suppose we boost along the z axis. Then the fields transform accordingto Jackson equation (11.149):

E′ = γE0(1 − 2β sin θ)i + 2γβE0 cos θj

B′ = 2γE0 cos θi + γE0(2 sin θ + β)j.

The angle between the fields is given by

cos θ =E′ · B′

|E′||B′|

=2γ2E2

0 cos θ(1 + β2)

γ2E20(1 − 4β sin θ + 4β2)1/2(4 + 4β sin θ + β2)1/2

Setting this equal to unity, we obtain


Problem 11.18

The electric and magnetic fields of a particle of charge q moving in a straight linewith speed v = βc, given by (11.152), become more and more concentrated asβ → 1, as indicated in Fig. 11.9. Choose axes so that the charge moves alongthe z axis in the positive direction, passing the origin at t = 0. Let the spatialcoordinates of the observation point be (x, y, z) and define the transverse vector r⊥,with components x and y. Consider the fields and the source in the limit of β = 1.

(a) Show that the fields can be written as

E = 2qr⊥r2⊥

δ(ct − z); B = 2qv × r⊥

r2⊥

δ(ct − z).

(b) Show by substitution into the Maxwell equations that these fields are consistentwith a 4-vector source density,

Jα = qcvαδ(2)(r⊥)δ(ct − z)

where the 4-vector vα = (1, v).

(c) Show that the fields of part a are derivable from either of the following 4-vectorpotentials,

A0 = Az = −2qδ(ct − z) ln(λr⊥); A⊥ = 0

orA0 = Az = 0; A⊥ = −2qΘ(ct− z)∇⊥ ln(λr⊥)

where λ is an irrelevant parameter setting the scale of the logarithm. Showthat the two potentials differ by a gauge transformation and find the gaugefunction, χ.

(a) In the reference frame in which the particle is at rest at the origin, the fieldsare

E′ =q

(x′2 + y′2 + z′2)3/2(x′i + y′j + z′k), B′ = 0.

Transforming back to the laboratory frame according to Jackson 11.148, theelectric field is

E =q

(x′2 + y′2 + z′2)3/2(γx′i + γy′j + z′k)

where in this expression the coordinates are still those of the observation pointin the moving frame. The transformation of these to the lab frame is x′ = x,y = y′ = y, z′ = γ(z − βct). Then the correct expression for the transformedfield is

E =q

[x2 + y2 + γ2(z − ct)2]3/2(γxi + γyj + (z − ct)k).


In the limit β → 1, we have γ → ∞. For z 6= ct the γ2 factor in the denominatorthen ensures that all field components are zero. For z = ct, however, althoughthe z component of the field clearly vanishes, the behavior of the other com-ponents is not as immediately clear. To elucidate the behavior of, say, the xcomponent of the field at z = ct we integrate it from z = ct − ε to z = ct + ε :

∫ ct+ε

ct−ε

Exdz = qγx

∫ ct+ε

ct−ε

dz

[r2⊥

+ γ2(z − ct)2]3/2

=qx

r2⊥

∫ γε/r⊥

−γε/r⊥

du

[1 + u2]3/2

=2qx

r2⊥

γε√

r2⊥

+ γ2ε2.

Taking the limit γ → ∞ for any finite ε, we find

limγ→∞

∫ ct+ε

ct−ε

Exdz =2qx

r⊥

2

. (14)

On the other hand, integrating between two points on the same side of z = ct,say from z = ct + ε to z = ct + 2ε, we find

∫ ct+2ε

ct+ε

Exdz =2qx

r⊥

2[

2γε√

r2⊥

+ 4γ2ε2− γε√

r2⊥

+ γ2ε2

]

which vanishes as γ → ∞.Since Ex vanishes at any point z 6= ct but yields something nonzero when

integrated across that point, we conclude that it is just a δ function in (z − ct)with coefficient given by (14):

Ex =2qx

r2⊥

δ(z − ct).

and, similarly,

Ey =2qy

r2⊥

δ(z − ct).

Combining these, we can write

E = 2qr⊥r2⊥

δ(ct − z).

The B field is given by Jackson (11.150) with, in the ultrarelativistic limit,β = k :

B = 2qv × r⊥

r2⊥

δ(ct − z).