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 an
infinite plane conductor held at zero potential. Using the method
of images, find:
(a) the surface-charge density induced on the plane, and plot it;
(b) the force between the plane and the charge by using Coulomb’s
law for the force between the charge and its image;
(c) the total force acting on the plane by integrating σ 2 /20 over
the whole plane;
(d) the work necessary to remove the charge q from its position to
infinity;
(e) the potential energy between the charge q and its image (compare the answer to part d and discuss).
(f ) Find the answer to part d in electron volts for an electron
originally 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
1
1
q
−
Φ(r) =
4π0 |r − dk| |r + dk|
The surface charge induced on the plane is found by differentiating this:
1
2
Homer Reid’s Solutions to Jackson Problems: Chapter 2
σ
dΦ
z=0
dz
(z + d)
q −(z − d)
+
= −
4π |r + dk|3
|r + dk|3
qd
= −
2π(x2 + y 2 + d2 )3/2
= −0
z=0
(1)
We can check this by integrating this over the entire xy plane and verifying
that the total charge is just the value −q of the image charge:
Z
∞
−∞
Z
∞
σ(x, y)dxdy
−∞
Z Z
qd ∞ 2π
rdψdr
2 + d2 )3/2
2π 0
(r
0
Z ∞
rdr
−qd
2 + d2 )3/2
(r
Z0
qd ∞ −3/2
−
u
du
2 d2
∞
qd
−2u−1/2 2
−
2
d
√
−q
= −
=
=
=
=
(b) The point of this problem is that, for points above the z axis, it doesn’t
matter whether there is a charge −q at (0, 0, d) or an infinite grounded sheet
at z = 0. Physics above the z axis is exactly the same whether we have the
charge or the sheet. In particular, the force on the original charge is the same
whether we have the charge or the sheet. That means that, if we assume the
sheet is present instead of the charge, it will feel a reaction force equal to what
the image charge would feel if it were present instead of the sheet. The force
on the image charge would be just F = q 2 /16π0d2 , so this must be what the
sheet feels.
(c) Total force on sheet
=
=
=
=
=
Z ∞ Z 2π
1
σ 2 dA
20 0
0
Z
rdr
q 2 d2 ∞
2
4π0 0 (r + d2 )3
Z
q 2 d2 ∞ −3
u du
8π0 d2
∞
q 2 d2
1
− u−2
8π0
2
2
d
q 2 d2 1 −4
d
8π0 2
Homer Reid’s Solutions to Jackson Problems: Chapter 2
=
q2
16π0 d2
in accordance with the discussion and result of part b.
(d) Work required to remove charge to infinity
=
=
=
=
Z ∞
q2
dz
4π0 d (z + d)2
Z ∞
q2
u−2 du
4π0 2d
q2 1
4π0 2d
q2
8π0 d
(e) Potential energy between charge and its image
=
q2
8π0 d
equal to the result in part d.
(f )
q2
8π0 d
=
(1.6 · 10−19 coulombs )2
8π(8.85 · 10−12 coulombs V−1 m−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 charge
q 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 a
fixed potential V ? If the sphere has a total charge Q on its
inner and outer surfaces?
3
4
Homer Reid’s Solutions to Jackson Problems: Chapter 2
Problem 2.3
A straight-line charge with constant linear charge density λ is located perpendicular
to 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/4π0 ) 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 of
the intersecting planes. Verify explicitly that the potential and the tangential
electric field vanish on the boundary surface.
(b) Determine the surface charge density σ on the plane y = 0, x ≥ 0. Plot σ/λ
versus 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
2
x0
Qx = − λ tan−1
π
y0
What is the total charge on the plane x = 0?
(d) Show
p that far from the origin [ρ ρ0 , where ρ =
x20 + y02 ] the leading term in the potential is
Φ → Φasym =
p
x2 + y 2 and ρ0 =
4λ (x0 )(y0 )(xy)
.
π0
ρ4
Interpret.
(a) The potential can be made to vanish on the specified boundary surfaces
by pretending that we have three image line charges. Two image charges have
charge density −λ and exist at the locations obtained by reflecting the original
image charge across the x and y axes, respectively. The third image charge has
charge density +λ and exists at the location obtained by reflecting the original
charge through the origin. The resulting potential in the first quadrant is
Φ(x, y)
=
=
λ
R2
R2
R2
R2
ln 2 − ln 2 − ln 2 + ln 2
4π0
r1
r2
r3
r4
r2 r3
λ
ln
2π0 r1 r4
where
r12 = [(x − x0 )2 + (y − y0 )2 ]
r22 = [(x + x0 )2 + (y − y0 )2 ]
(2)
5
Homer Reid’s Solutions to Jackson Problems: Chapter 2
r32 = [(x − x0 )2 + (y + y0 )2 ]
r42 = [(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
d
Φ
dy
1 dr2
1 dr3
1 dr1
1 dr4
λ
+
−
−
= −
2π r2 dy
r3 dy
r1 dy
r4 dy
= −0
y=0
We have dr1 /dy = (y − y0 )/r1 and similarly for the other derivatives, so
λ y − y0
y + y0
y − y0
y + y0
σ = −
+
−
−
y=0
2π
r2
r32
r12
r42
2
1
1
y0 λ
−
= −
π
(x − x0 )2 + y02
(x + x0 )2 + y02 )
(c) Total charge per unit length in z
Z ∞
σdx
Qx =
0
Z ∞
Z ∞
y0 λ
dx
dx
= −
−
π
(x − x0 )2 + y02
(x + x0 )2 + y02
0
0
For the first integral the appropriate substitution is (x − x0 ) = y0 tan u, dx =
y0 sec2 udu. A similar substitution works in the second integral.
#
"Z
Z π/2
π/2
λ
du −
du
= −
x
π tan−1 − xy0
tan−1 y 0
0
0
λ π
π
−1 −x0
−1 x0
= −
− tan
− + tan
π 2
y0
2
y0
2λ
x
0
= − tan−1 .
(3)
π
y0
The calculations are obviously symmetric with respect to x0 and y0 . The
total charge on the plane x = 0 is (3) with x0 and y0 interchanged:
Qy = −
2λ
y0
tan−1
π
x0
Since tan−1 x − tan−1 (1/x) = π/2 the total charge induced is
Q = −λ
Homer Reid’s Solutions to Jackson Problems: Chapter 2
6
which is, of course, also the sum of the charge per unit length of the three image
charges.
(d) We have
Φ=
λ
r2 r2
ln 22 32
4π0 r1 r4
Far from the origin,
r12
= [(x − x0 )2 + (y − y0 )2 ]
x0 2
y0 2
2
2
=
x (1 − ) + y (1 − )
x
y
y0
x
0
2
2
≈
x (1 − 2 ) + y (1 − 2
x
y
2
= x − 2x0 x + y 2 − 2y0 y)
xx0 + yy0
2
2
= (x + y ) 1 − 2 2
x + y2
Similarly,
r22
r32
r42
−xx0 + yy0
= (x + y ) 1 − 2
x2 + y 2
xx0 − yy0
2
2
= (x + y ) 1 − 2 2
x + y2
−xx0 − yy0
2
2
= (x + y ) 1 − 2
x2 + y 2
2
2
Next,
r12 r42
r22 r32
so
(xx0 + yy0 )2
= (x + y ) 1 − 4
(x2 + y 2 )2
(xx0 − yy0 )2
2
2 2
= (x + y ) 1 − 4
(x2 + y 2 )2
2
2 2


2
0 −yy0 )
1 − 4 (xx
2
2
2
λ
(x +y )
.
Φ=
ln 
2
0 +yy0 )
4π0
1 − 4 (xx
(x2 +y 2 )2
The (x2 + y 2 ) term in the denominator grows much more quickly than the
(xx0 + yy0 ) term, so in the asymptotic limit we can use ln(1 + ) ≈ to find
(xx0 − yy0 )2
λ
(xx0 + yy0 )2
−4
Φ =
+
4
4π0
(x2 + y 2 )2
(x2 + y 2 )2
λ
−4(x2 x20 + y 2 y02 − 2xyx0 y0 ) + 4(x2 x20 + y 2 y02 + 2xyx0 y0 )
=
4π0
(x2 + y 2 )2
Homer Reid’s Solutions to Jackson Problems: Chapter 2
=
=
7
λ
16xyx0 y0
4π0 (x2 + y 2 )2
√
4λ (xy)(x0 y0 )
.
π0 (x2 + y 2 )2
Problem 2.4
A point charge is placed a distance d > R from the center of an
equally charged, isolated, conducting sphere of radius R.
(a) Inside of what distance from the surface of the sphere is the
point charge attracted rather than repelled by the charged
sphere?
(b) What is the limiting value of the force of attraction when the
point charge is located a distance a(= d−R) from the surface
of the sphere, if a R?
(c) What are the results for parts a and b if the charge on the
sphere is twice (half) as large as the point charge, but still
the same sign?
Let’s call the point charge q. The charged, isolated sphere may be replaced
by two image charges. One image charge, of charge q1 = −(R/d)q at radius
r1 = 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 subtract
the surface charge density induced by the point charge q).
The force on the point charge is the sum of the forces from the two image
charges:
F
"
#
=
1
4π0
=
q2
−dR
d2 + dR
+
4π0 [d2 − R2 ]2
d4
qq1
qq2
+ d2
R2 2
d− d
(4)
(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 terms
looks like d4 , so the dR terms in the numerator cancel and the overall force is
repulsive.
(a) The crossover distance is found by equating the two bracketed terms in (5):
8
Homer Reid’s Solutions to Jackson Problems: Chapter 2
[d2
dR
− R 2 ]2
=
d2 + dR
d4
d4 R = (d + R)[d2 − R2 ]2
0 = d5 − 2d3 R2 − 2d2 R3 + 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) as
a → 0.
F
=
≈
"
a 2
R2 (1 + R
) + (1 +
+
a 4
2
4
a
)
R (1 + R
R2 (1 + R )2 − R2
2
2
−R − aR (2R + 3a)(R − 4a)
q
+
4π0
4a2 R2
R4
q2
4π0
−R2 (1 +
a
R)
a
R)
#
The second term in brackets approaches the constant 2/R 2 as a → 0. The first
term becomes −1/4a2. So we have
F →−
q2
.
16π0 a2
Note that only the first image charge (the one required to make the sphere an
equipotential) contributes to the force as d → a. The second image charge,
the one which represents the difference between the actual charge on the sphere
and 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 of
the 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
dR
2d2 + dR
q2
− 2
+
F =
4π0
[d − R2 ]2
d4
and the relevant equation becomes
0 = 2d5 − 4d3 R2 − 2d2 R3 + 2dR4 + R5 .
Again I solved graphically to find d/R = 1.43. If the charge on the sphere is
half the point charge, then
dR
d2 + 2dR
q2
− 2
+
F =
4π0
[d − R2 ]2
2d4
and the equation is
0 = d5 − 2d3 R2 − 4d2 R3 + dR4 + 2R5 .
The root of this one is d/R=1.88.
Homer Reid’s Solutions to Jackson Problems: Chapter 2
9
Problem 2.5
(a) Show that the work done to remove the charge q from a distance r > a to infinity against the force, Eq. (2.6), of a
grounded conducting sphere is
W =
q2 a
.
8π0 (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 q
against the force, Eq. (2.9), of an isolated charged conducting
sphere. Show that the work done is
q2 a
q 2 a qQ
1
.
−
−
W =
4π0 2(r2 − a2 ) 2r2
r
Relate the work to the electrostatic potential, Eq. (2.8), and
the energy discussion of Section 1.11.
(a) The force is
|F | =
q2 a
1
3
4π0 y (1 − a2 /y 2 )2
directed radially inward. The work is
Z ∞
W = −
F dy
r
Z ∞
dy
q2 a
=
4π0 r y 3 (1 − a2 /y 2 )2
Z ∞
ydy
q2 a
=
4π0 r (y 2 − a2 )2
Z ∞
q2 a
du
=
4π0 r2 −a2 2u2
(6)
∞
=
1
q2 a
−
4π0
2u
=
q2 a
8π0 (r2 − a2 )
r 2 −a2
(7)
To relate this to earlier results, note that the image charge q 0 = −(a/r)q is
located at radius r 0 = a2 /r. The potential energy between the point charge and
10
Homer Reid’s Solutions to Jackson Problems: Chapter 2
its image is
PE
=
=
=
1
qq 0
4π0 |r − r0 |
−q 2 a
1
4π0 r(r − a2 /r)
−q 2 a
1
4π0 r2 − a2
(8)
Result (7) is only half of (8). This would seem to violate energy conservation. It
would seem that we could start with the point charge at infinity and allow it to
fall in to a distance r from the sphere, liberating a quantity of energy (8), which
we could store in a battery or something. Then we could expend an energy
equal to (7) to remove the charge back to infinity, at which point we would
be back where we started, but we would still have half of the energy saved in
the 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 q1 q2 /4π0 r
for the potential energy of two charges comes from calculating the work needed
to bring one charge from infinity to a distance r from the other charge, and it
is assumed that the other charge does not move and keeps a constant charge
during the process. But in this case one of the charges is a fictitious image
charge, and as the point charge q is brought in from infinity the image charge
moves out from the center of the sphere, and its charge increases. So the simple
expression 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 location
as in part a, and another of charge Q − q 0 at the origin. The work needed to
remove the point charge q to infinity is the work needed to remove the point
charge from its image charge, plus the work needed to remove the point charge
from the extra charge at the origin. We calculated the first contribution above.
The second contribution is
Z ∞
Z ∞
q(Q − q 0 )dy
1
qQ q 2 a
−
dy
=
−
+
4π0 y 2
4π0 r
y2
y3
r
1
qQ q 2 a
= −
−
− 2
4π0
y
2y
2
1
qQ q a
= −
+ 2
4π0 r
2r
so the total work done is
W =
1
q2 a
q 2 a qQ
.
−
−
4π0 2(r2 − a2 ) 2r2
r
∞
r
11
Homer Reid’s Solutions to Jackson Problems: Chapter 2
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 since
I define the Green’s function a little differently than Jackson.
The whole technique is based on the divergence theorem. Suppose A(x) is
a vector valued function defined at each point x within a volume V . Then
Z
I
0
0
(∇ · A(x )) dV =
A(x0 ) · dA0
(9)
V
S
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
Z
V
(∇φ(x0 )) · (∇ψ(x0 )) + φ(x0 )∇2 ψ(x0 ) dV 0 =
I
φ(x0 )
S
∂ψ
∂n
dA0
x0
~ with the outward normal to the surface
where ∂ψ/∂n is the dot product of ∇ψ
area element. If we write down this equation with φ and ψ switched and subtract
the two, we come up with
I Z
2
∂ψ
∂φ
dA0 .
(10)
φ
φ∇ ψ − ψ∇2 φ dV 0 =
−ψ
∂n
∂n
S
V
This statement doesn’t appear to be very useful, since it seems to require that
we 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 could
choose ψ(x) in a clever way such that ∇2 ψ = 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
Z
I ∂φ
∂ψx0
dA0 .
− ψx0 (x0 )
φ(x0 ) =
φ(x0 )
ψx0 (x0 )∇2 φ(x0 ) dV 0 +
∂n x0
∂n x0
V
S
If φ is the scalar potential of electrostatics, we know that ∇2 ψ(x0 ) = −ρ(x0 )/0 ,
so we have
I Z
1
0
0
0
0 ∂ψx0
0 ∂φ
φ(x0 ) = −
φ(x )
ψx0 (x )ρ(x )dV +
− ψx0 (x )
dA0 .
0 V
∂n x0
∂n x0
S
(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 φ
or ∂φ/∂n on the boundary. This lack of knowledge can be accommodated by
choosing ψ such that either its value or its normal derivative vanishes on the
boundary surface, so that the term which we can’t evaluate drops out of the
surface integral. More specifically,
Homer Reid’s Solutions to Jackson Problems: Chapter 2
12
• if we know φ but not ∂φ/∂n on the boundary (“Dirichlet” boundary conditions), we choose ψ such that ψ = 0 on the boundary. Then
Z
I
1
∂ψx0
0
0
0
φ(x0 ) = −
dA0 .
(12)
ψx0 (x )ρ(x )dV +
φ(x0 )
0 V
∂n x0
S
• if we know ∂φ/∂n but not φ on the boundary (“Neumann” boundary
conditions), we choose ψ such that ∂ψ/∂n = 0 on the boundary. Then
I
Z
1
∂φ
0
0
0
φ(x0 ) = −
dA0 .
(13)
φx0 (x0 )
ψx0 (x )ρ(x )dV +
0 V
∂n
S
x0
Again, in both cases the function ψx0 (x) has the property that
∇2 ψx0 (x) = δ(x − x0 ).
Solutions to Problems in Jackson,
Classical Electrodynamics, Third Edition
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 a
distance R away from, the axis of a conducting cylinder of radius b held
at 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 the
origin at the axis of the cylinder and the direction from the origin
to the line charge as the x axis), including the asymptotic form far
from the cylinder;
(c) the induced surface-charge density, and plot it as a function of angle
for 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 outside
the conducting sphere, we might expect that the potential on the cylinder can
be made constant by placing an image charge within the cylinder on the line
conducting the line charge with the center of the cylinder, i.e. on the x axis.
Suppose we put the image charge a distance R0 < b from the center of the
cylinder and give it a charge density −τ . Using the expression quoted in Problem
2.3 for the potential of a line charge, the potential at a point x due to the line
charge and its image is
Φ(x)
=
τ
R2
R2
τ
ln
ln
−
4π0 |x − Rî|2
4π0 |x − R0 î|2
1
2
Homer Reid’s Solutions to Jackson Problems: Chapter 2
τ
|x − R0 î|2
ln
.
4π0
|x − Rî|2
=
We want to choose R0 such that the potential is constant when x is on the
cylinder surface. This requires that the argument of the logarithm be equal to
some constant γ at those points:
|x − R0 î|2
|x − Rî|2
or
=γ
b2 + R02 − 2R0 b cos φ = γb2 + γR2 − 2γRb cos φ.
For this to be true everywhere on the cylinder, the φ term must drop out, which
requires R0 = γR. We can then rearrange the remaining terms to find
R0 =
b2
.
R
This is also analogous to the point-charge-and-sphere problem, but there are
differences: in this case the image charge has the same magnitude as the original
line charge, and the potential on the cylinder is constant but not zero.
(b) At a point (ρ, φ), we have
Φ=
τ
ρ2 + R02 − 2ρR0 cos φ
ln 2
.
4π0
ρ + R2 − 2ρR cos φ
For large ρ, this becomes
0
Φ→
1 − 2 Rρ cos φ
τ
ln
.
4π0
1 − 2R
ρ cos φ
Using ln(1 − x) = −(x + x2 /2 + · · ·), we have
Φ →
=
τ 2(R − R0 ) cos φ
4π0
ρ
τ R(1 − b2 /R2 ) cos φ
2π0
ρ
(c)
σ
= −0
∂Φ
∂ρ
r=b
2b − 2R cos φ
τ
2b − 2R0 cos φ
−
= −
4π b2 + R02 − 2bR0 cos φ b2 + R2 − 2bR cos φ
"
#
2
b − bR cos φ
b − R cos φ
τ
= −
−
2π b2 + Rb42 − 2 bR3 cos φ b2 + R2 − 2bR cos φ
Homer Reid’s Solutions to Jackson Problems: Chapter 2
3
Multiplying the first term by R2 /b2 on top and bottom yields
"
#
R2
τ
b −b
σ = −
2π R2 + b2 − 2bR cos φ
R 2 − b2
τ
= −
2πb R2 + b2 − 2bR cos φ
(d) To find the force on the charge, we note that the potential of the image
charge is
τ
C2
.
Φ(x) = −
ln
4π0 |x − R0 î|2
with C some constant. We can differentiate this to find the electric field due to
the image charge:
E(x)
τ
∇ ln |x − R0 î|2
4π0
τ 2(x − R0 î)
= −
.
4π0 |x − R0 î|2
= −∇Φ(x) = −
The original line charge is at x = R, y = 0, and the field there is
E=−
1
τ
R
τ
î = −
î.
2π0 R − R0
2π0 R2 − b2
The force per unit width on the line charge is
F = τE = −
τ2
R
2π0 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 potential
problem with the potential specified on the surface of a cylinder of radius
b, evaluate the coefficients formally, substitute them into the series, and
sum it to obtain the potential inside the cylinder in the form of Poisson’s
integral:
Z 2π
1
b2 − ρ 2
Φ(ρ, φ) =
Φ(b, φ0 ) 2
dφ0
2π 0
b + ρ2 − 2bρ cos(φ0 − φ)
What modification is necessary if the potential is desired in the region of
space bounded by the cylinder and infinity?
Homer Reid’s Solutions to Jackson Problems: Chapter 2
4
Referring to equation (2.71), we know the bn are all zero, because the ln
term and the negative powers of ρ are singular at the origin. We are left with
Φ(ρ, φ) = a0 +
∞
X
n=1
ρn {an sin(nφ) + bn cos(nφ)} .
(1)
Multiplying both sides successively by 1, sin n0 φ, and cos n0 φ and integrating
at ρ = b gives
Z 2π
1
Φ(b, φ)dφ
(2)
a0 =
2π 0
Z 2π
1
an =
Φ(b, φ) sin(nφ)dφ
(3)
πbn 0
Z 2π
1
bn =
Φ(b, φ) cos(nφ)dφ.
(4)
πbn 0
Plugging back into (1), we find
)
(
Z
∞
1 2π
1 X ρ n
0
0
0
Φ(ρ, φ) =
[sin(nφ) sin(nφ ) + cos(nφ) cos(nφ )] dφ0
+
Φ(b, φ )
π 0
2 n=1 b
)
(
Z
∞
1 2π
1 X ρ n
0
0
=
cos n(φ − φ ) .
(5)
+
Φ(b, φ )
π 0
2 n=1 b
The bracketed term can be expressed in closed form. For simplicity define
x = (ρ/b) and α = (φ − φ0 ). Then
∞
1 X n
+
x cos(nα)
2 n=1
=
=
=
=
=
=
∞
1 1 X n inα
+
x e
+ xn e−inα
2 2 n=1
1 1
1
1
+
+
−
2
2 2 1 − xeiα
1 − xe−iα
1 1 1 − xe−iα − xeiα + 1
+
−
2
2 2 1 − xeiα − xe−iα + x2
1
1 − x cos α
+
−
1
2
1 + x2 − 2x cos α
1
x cos α − x2
+
2 1 + x2 − 2x cos α
1
1 − x2
.
2 1 + x2 − 2x cos α
Plugging this back into (5) gives the advertised result.
Homer Reid’s Solutions to Jackson Problems: Chapter 2
5
Problem 2.13
(a) Two halves of a long hollow conducting cylinder of inner radius b are
separated by small lengthwise gaps on each side, and are kept at different
potentials V1 and V2 . Show that the potential inside is given by
V1 − V 2
2bρ
V1 + V 2
−1
+
tan
cos φ
Φ(ρ, φ) =
2
π
b2 − ρ 2
where φ is measured from a plane perpendicular to the plane through the
gap.
(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 an
expression for the potential within the cylinder, the correct expansion is (1)
with expansion coefficients given by (2), (3) and (4):
a0
=
=
=
an
=
=
=
=
bn
=
=
=
1
2π
Z
2π
Φ(b, φ)dφ
0
Z 2π π
1
V1
dφ + V2
dφ
2π
0
π
V1 + V 2
2
Z π
Z 2π
1
sin(nφ)dφ
sin(nφ)dφ
+
V
V
2
1
πbn
π
0
h
i
1
π
2π
−
V
|cos
nφ|
+
V
|cos
nφ|
1
2
0
π
nπbn
1
−
[V1 (cos nπ − 1) + V2 (1 − cos nπ)]
n
nπb
0
, n even
2(V1 − V2 )/(nπbn ) , n odd
Z π
Z 2π
1
cos(nφ)dφ
cos(nφ)dφ
+
V
V
2
1
πbn
π
0
h
i
1
π
2π
V
|sin
nφ|
+
V
|sin
nφ|
1
2
0
π
nπbn
0.
Z
With these coefficients, the potential expansion becomes
Φ(ρ, φ) =
V1 + V 2
2(V1 − V2 ) X 1 ρ n
sin nφ.
+
2
π
n b
n odd
(6)
6
Homer Reid’s Solutions to Jackson Problems: Chapter 2
Here we need an auxiliary result:
X 1
xn sin nφ =
n
n odd
=
=
1 X 1
(x = iy)
(iy)n [einπ − e−inφ ]
2i
n
n odd
∞
1 X (−1)n iφ 2n+1
(ye )
− (ye−iφ )2n+1
2 n=0 2n + 1
1 −1 iφ
tan (ye ) − tan−1 (ye−iφ )
2
(7)
where in the last line we just identified the Taylor series for the inverse tangent
function. Next we need an identity:
γ1 − γ 2
−1
−1
−1
tan γ1 − tan γ2 = tan
.
1 + γ 1 γ2
(I derived this one by drawing some triangles and doing some algebra.) With
this, (7) becomes
X 1
1
2iy sin φ
xn sin nφ =
tan−1
n
2
1 + y2
n odd
2x sin φ
1
−1
tan
.
=
2
1 − x2
Using this in (6) with x = ρ/b gives
V1 − V 2
V1 + V 2
+
tan−1
Φ(ρ, b) =
π
π
2ρb sin φ
b2 − ρ 2
(Evidently, Jackson and I defined the angle φ differently).
.
Homer Reid’s Solutions to Jackson Problems: Chapter 2
7
Problem 2.15
(a) Show that the Green function G(x, y; x0 , y 0 ) appropriate for Dirichlet
boundary conditions for a square two-dimensional region, 0 ≤ x ≤ 1, 0 ≤
y ≤ 1, has an expansion
G(x, y; x0 , y 0 ) = 2
∞
X
gn (y, y 0 ) sin(nπx) sin(nπx0 )
n=1
where gn (y, y 0 ) satisfies
2
∂
2 2
−
n
π
gn (y, y 0 ) = δ(y 0 − y)
∂y 2
and gn (y, 0) = gn (y, 1) = 0.
(b) Taking for gn (y, y 0 ) appropriate linear combinations of sinh(nπy 0 )
and cosh(nπy 0 ) in the two regions y 0 < y and y 0 > y, in accord with the
boundary conditions and the discontinuity in slope required by the source
delta function, show that the explicit form of G is
G(x, y; x0 , y 0 ) =
∞
X
1
−2
sin(nπx) sin(nπx0 ) sinh(nπy< ) sinh[nπ(1 − y> )]
nπ
sinh(nπ)
n=1
where y< (y> ) is the smaller (larger) of y and y 0 .
(I have taken out a factor −4π from the expressions for gn and G, in accordance
with my convention for Green’s functions; see the Green’s functions review
above.)
(a) To use as a Green’s function in a Dirichlet boundary value problem G must
satisfy two conditions. The first is that G vanish on the boundary of the region
of interest. The suggested expansion of G clearly satisfies this. First, sin(nπx0 )
is 0 when x0 is 0 or 1. Second, g(y, y 0 ) vanishes when y 0 is 0 or 1. So G(x, y; x0 , y 0 )
vanishes for points (x0 , y 0 ) on the boundary.
The second condition on G is
2
∂
∂2
∇2 G =
G = δ(x − x0 ) δ(y − y 0 ).
(8)
+
∂x02
∂y 02
With the suggested expansion, we have
∞
X
∂2
G = 2
gn (y, y 0 ) sin(nπx) −n2 π 2 sin(nπx0 )
02
∂x
n=1
∞
X ∂ 02
∂2
G = 2
g (y, y 0 ) sin(nπx) sin(nπx0 )
02
2 n
∂y
∂y
n=1
8
Homer Reid’s Solutions to Jackson Problems: Chapter 2
We can add these together and use the differential equation satisfied by gn to
find
∇2 G = δ(y − y 0 ) · 2
0
∞
X
sin(nπx) sin(nπx0 )
n=1
= δ(y − y ) · δ(x − x0 )
since the infinite sum is just a well-known representation of the δ function.
(b) The suggestion is to take
An1 sinh(nπy 0 ) + Bn1 cosh(nπy 0 ),
gn (y, y 0 ) =
An2 sinh(nπy 0 ) + Bn2 cosh(nπy 0 ),
y 0 < y;
y 0 > 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 g n
(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 except
y = y 0 ) for any choice of the As and Bs. This leaves us free to choose these
coefficients as required to satisfy the boundary conditions and the differential
equation at y = y 0 .
First let’s consider the boundary conditions. Since y is somewhere between
0 and 1, the condition that gn vanish for y 0 = 0 is only relevant to the top line
of (9), where it requires taking Bn1 = 0 but leaves An1 undetermined for now.
The condition that gn vanish for y 0 = 1 only affects the lower line of (9), where
it 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
Bn2 = enπ + An2
→
− An2 + Bn2 = enπ .
Then
so An2 = − cosh(nπ)
and
2An2 = −enπ − e−nπ
Bn2 = sinh(nπ).
With this choice of coefficients, the lower line in (9) becomes
gn (y, y 0 ) = − cosh(nπ) sinh(nπy 0 )+sinh(nπ) cosh(nπy 0 ) = sinh[nπ(1−y 0 )] (11)
for (y 0 > y). Actually, we haven’t completely determined An2 and Bn2 ; we could
multiply (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 0 = y:
An1 sinh(nπy) = γn sinh[nπ(1 − y)].
(12)
9
Homer Reid’s Solutions to Jackson Problems: Chapter 2
70000
60000
g(yprime)
50000
40000
30000
20000
10000
0
0
0.2
0.4
0.6
0.8
1
yprime
Figure 1: gn (y, y 0 ) 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
βn sinh[nπ(1 − y)] sinh(nπy 0 ),
gn (y, y 0 ) =
βn sinh[nπ(1 − y 0 )] sinh(nπy),
= βn sinh[nπ(1 − y> )] sinh(nπy< )
y 0 < y;
y 0 > y.
(13)
with y< and y> defined as in the problem. Figure 1 shows a graph of this
function n = 5, y = .41.
The final step is to choose the normalization constant βn such that gn satisfies its differential equation:
2
∂
2 2
gn (y, y 0 ) = δ(y − y 0 ).
(14)
−
n
π
∂ 2 y 02
To say that the left-hand side “equals” the delta function requires two things:
• that the left-hand side vanish at all points y 0 6= y, and
• that its integral over any interval (y1 , y2 ) equal 1 if the interval contains
the point y 0 = y, and vanish otherwise.
The first condition is clearly satisfied regardless of the choice of βn . The second
condition may be satisfied by making gn continuous, which we have already
done, but giving its first derivative a finite jump of unit magnitude at y 0 = y:
Homer Reid’s Solutions to Jackson Problems: Chapter 2
∂
gn (y, y 0 )
∂y 0
10
y 0 =y +
= 1.
y 0 =y −
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 0 ) = −
sinh[nπ(1 − y> )] sinh(nπy< )
nπ sinh(nπ)
and the composite Green’s function is
G(x, y; x0 , y 0 )
= 2
∞
X
gn (y, y 0 ) sin(nπx) sin(nπx0 )
n=1
∞
X
= −2
sinh[nπ(1 − y> )] sinh(nπy< ) sin(nπx) sin(nπx0 )
(15)
.
nπ sinh(nπ)
n=1
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 square
there 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 be
written as
∞
4 X sin[(2m + 1)πx]
cosh[(2m + 1)π(y − (1/2))]
Φ(x, y) = 3
.
1−
π 0 m=0
(2m + 1)3
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
Z
I 1
∂Φ
∂G
dA0 .
Φ(x0 ) = −
− G(x0 ; x0 )
G(x0 ; x0 )ρ(x0 )dV 0 +
Φ(x0 )
0 V
∂n
∂n
0
0
S
x
x
(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
11
Homer Reid’s Solutions to Jackson Problems: Chapter 2
ρ(x0 )dV 0 = dx0 dy 0 throughout the entire volume. Then we can plug in (15) to
find
Z 1Z 1
∞
2 X
1
Φ(x0 ) =
sinh[nπ(1−y> )] sinh(nπy< ) sin(nπx0 ) sin(nπx0 )dx0 dy 0 .
π0 n=1 n sinh(nπ) 0 0
(17)
The integrals can be done separately. The x integral is
sin(nπx0 )
Z
1
sin(nπx0 )dx0
0
sin(nπx0 )
= −
[cos(nπ) − 1]
nπ
(2 sin(nπx0 ))/nπ , n odd
=
0
, n even
(18)
The y integral is
sinh[nπ(1 − y0 )]
=
=
=
Z
y0
0
0
sinh(nπy )dy + sinh(nπy0 )
0
Z
1
y0
sinh[nπ(1 − y 0 )]dy 0
o
1 n
y0
1
sinh[nπ(1 − y0 )] · cosh(nπy 0 ) 0 − sinh[nπy0 ] · cosh[nπ(1 − y 0 )] y0
nπ
1
{sinh[nπ(1 − y0 )] cosh(nπy0 ) + sinh(nπy0 ) cosh[nπ(1 − y0 )] − sinh(nπy0 ) − sinh[nπ(1 − y0 )]}
nπ
1
{sinh[nπ] − sinh[nπ(1 − y0 )] − sinh(nπy0 )}.
(19)
nπ
Inserting (18) and (19) in (17), we have
X sin(nπx0 ) 4
sinh[nπ(1 − y0 )] + sinh(nπy0 )
Φ(x0 ) = 3
1−
.
π 0
n3
sinh(nπ)
n odd
The thing in brackets is equal to what Jackson has, but this is tedious to show
so I’ll skip the proof.
12
Homer Reid’s Solutions to Jackson Problems: Chapter 2
Problem 2.17
(a) Construct the free-space Green function G(x, y; x0 , y 0 ) for twodimensional electrostatics by integrating 1/R with respect to z 0 − z
between the limits ±Z, where Z is taken to be very large. Show that
apart from an inessential constant, the Green function can be written
alternately as
G(x, y; x0 , y 0 ) = − ln[(x − x0 )2 + (y − y 0 )2 ]
= − ln[ρ2 + ρ02 − 2ρρ0 cos(φ − φ0 )].
(b) Show explicitly by separation of variables in polar coordinates that the
Green function can be expressed as a Fourier series in the azimuthal
coordinate,
∞
1 X im(φ−φ0 )
e
gm (ρ, ρ0 )
G=
2π −∞
where the radial Green functions satisfy
m2
δ(ρ − ρ0 )
1 ∂
0 ∂gm
ρ
− 02 gm =
.
0
0
0
ρ ∂ρ
∂ρ
ρ
ρ
Note that gm (ρ, ρ0 ) for fixed ρ is a different linear combination of the
solutions of the homogeneous radial equation (2.68) for ρ0 < ρ and
for ρ0 > ρ, with a discontinuity of slope at ρ0 = ρ determined by the
source delta function.
(c) Complete the solution and show that the free-space Green function has
the expansion
G(ρ, φ; ρ0 , φ0 ) =
m
∞
1 X 1 ρ<
1
ln(ρ2> ) −
· cos[m(φ − φ0 )]
4π
2π m=1 m ρ>
where ρ< (ρ> ) is the smaller (larger) of ρ and ρ0 .
(As in Problem 2.15, I modified the text of the problem to match with my
convention for Green’s functions.)
(a)
R
= [(x − x0 )2 + (y − y 0 )2 + (z − z 0 )2 ]1/2
≡ [a2 + u2 ]1/2 , a = [(x − x0 )2 + (y − y 0 )2 ]1/2
, u = (z − z 0 ).
Integrating,
Z
Z
−Z
du
2
[a + u2 ]1/2
=
h
i
ln (a2 + u2 )1/2 + u
+Z
−Z
Homer Reid’s Solutions to Jackson Problems: Chapter 2
= ln
= ln
≈ ln
13
(Z 2 + a2 )1/2 + Z
(Z 2 + a2 )1/2 − Z
(1 + (a2 /Z 2 ))1/2 + 1
(1 + (a2 /Z 2 ))1/2 − 1
2+
a2
2Z 2
a2
2Z 2
2
4Z + a2
a2
2
= ln[4Z + a2 ] − ln a2 .
= ln
Since Z is much bigger than a, the first term is essentially independent of a and
is the ’nonessential constant’ Jackson is talking about. The remaining term is
the 2D Green’s function:
G = − ln a2
= − ln[(x − x0 )2 + (y − y 0 )2 ] in rectangular coordinates
= − ln[ρ2 + ρ02 − 2ρρ0 cos(φ − φ0 )] in cylindrical coordinates.
(b) The 2d Green’s function is defined by
Z
∇2 G(ρ, φ; ρ0 , φ0 )ρ0 dρ0 dφ0 = 1
but ∇2 G = 0 at points other than (ρ, φ). These conditions are met if
∇2 G(ρ, φ; ρ0 , φ0 ) =
1
δ(ρ − ρ0 )δ(φ − φ0 ).
ρ0
(20)
You need the ρ0 on the bottom there to cancel out the ρ0 in the area element in
the integral. The Laplacian in two-dimensional cylindrical coordinates is
1 ∂
1 ∂
2
0 ∂
∇ = 0 0 ρ
− 02 02 .
ρ ∂ρ
∂ρ0
ρ ∂φ
Applying this to the suggested expansion for G gives
∞ 0
m2
1 X 1 ∂
0 ∂gm
ρ
− 02 gm eim(φ−φ ) .
∇ G(ρ, φ; ρ , φ ) =
0
0
0
2π −∞ ρ ∂ρ
∂ρ
ρ
2
0
0
If gm satisfies its differential equation as specified in the problem, the term in
brackets equals δ(ρ − ρ0 )/ρ0 for all m and may be removed from the sum, leaving
∇2 G(ρ, φ; ρ0 , φ0 )
∞
1 X im(φ−φ0 )
e
2π −∞
=
δ(ρ − ρ0 )
ρ0
·
=
δ(ρ − ρ0 )
ρ0
δ(φ − φ0 ).
Homer Reid’s Solutions to Jackson Problems: Chapter 2
14
(c) As in Problem 2.15, we’ll construct the functions gm by finding solutions of
the homogenous radial differential equation in the two regions and piecing them
together at ρ = ρ0 such that the function is continuous but its derivative has a
finite jump of magnitude 1/ρ.
For m ≥ 1, the solution to the homogenous equation
m2
1 ∂
0 ∂
ρ
− 02 f (ρ0 ) = 0
ρ0 ∂ρ0
∂ρ0
ρ
is
f (ρ0 ) = Am ρ0m + Bm ρ0−m .
Thus we take
gm =
A1m ρ0m + B1m ρ0−m
A2m ρ0m + B2m ρ0−m
, ρ0 < ρ
, ρ0 > ρ.
In order that the first solution be finite at the origin, and the second solution
be finite at infinity, we have to take B1m = A2m = 0. Then the condition that
the two solutions match at ρ = ρ0 is
A1m ρm = B2m ρ−m
which requires
B2m = ρm γm
A1m = γm ρ−m
for some constant γm . Now we have

m
 γ m ρ0
ρ m
gm =
 γm ρ0
ρ
,
ρ0 < ρ
,
ρ0 > ρ
The finite-derivative step condition is
dgm
dρ0
ρ0 =ρ+
or
−mγm
so
−
dgm
dρ0
1 1
+
ρ ρ
γm = −
=
ρ0 =ρ−
=
1
ρ
1
ρ
1
.
2m
Then
gm

0 m
 − 1 ρ
, ρ0 < ρ
2m
ρ m
=
 − 1 ρ0
, ρ0 > ρ
2m ρ
m
1
ρ<
.
= −
2m ρ>
Homer Reid’s Solutions to Jackson Problems: Chapter 2
15
Plugging this back into the expansion gives
m
∞
0
1 X 1 ρ<
eim(φ−φ )
G = −
4π −∞ m ρ>
m
∞
1 X 1 ρ<
= −
cos[m(φ − φ0 )].
2π 1 m ρ>
Jackson seems to be adding a ln term to this, which comes from the m = 0
solution of the radial equation, but I have left it out because it doesn’t vanish
as ρ0 → ∞.
Problem 2.18
(a) By finding appropriate solutions of the radial equation in part b of
Problem 2.17, find the Green function for the interior Dirichlet problem of a cylinder of radius b [gm (ρ, ρ0 = b) = 0. See (1.40)]. First
find the series expansion akin to the free-space Green function of
Problem 2.17. Then show that it can be written in closed form as
2 02
ρ ρ + b4 − 2ρρ0 b2 cos(φ − φ0 )
G = ln 2 2
b (ρ + ρ02 − 2ρρ0 cos(φ − φ0 ))
or
(b2 − ρ2 )(b2 − ρ02 ) + b2 |ρ − ρ0 |2
.
G = ln
b2 |ρ − ρ0 |2
(b) Show that the solution of the Laplace equation with the potential
given as Φ(b, φ) on the cylinder can be expressed as Poisson’s integral 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 the
closed form? [Note that the exterior Green function is not rigorously
correct because it does not vanish for ρ or ρ0 → ∞. For situations
in which the potential falls of fast enough as ρ → ∞, no mistake is
made in its use.]
(a) As before, we write the general solution of the radial equation for gm in the
two distinct regions:
A1m ρ0m + B1m ρ0−m , ρ0 < ρ
0
gm (ρ, ρ ) =
(21)
A2m ρ0m + B2m ρ0−m , ρ0 > ρ.
The first boundary conditions are that gm remain finite at the origin and
vanish on the cylinder boundary. This requires that
B1m = 0
16
Homer Reid’s Solutions to Jackson Problems: Chapter 2
and
A2m bm + B2m b−m = 0
so
B2m = −γm bm
A2m = γm b−m
for some constant γm .
Next, gm must be continuous at ρ = ρ0 :
m ρ m
b
m
A1m ρ
= γm
−
b
ρ
m b
γ m ρ m
.
−
A1m =
ρm
b
ρ
With this we have
m 0 m
ρ
b
gm (ρ, ρ ) = γm
−
b
ρ
ρ
0 m m b
ρ
−
= γm
,
b
ρ0
ρ m
0
,
ρ0 < ρ
ρ0 > ρ.
Finally, dgm /dρ0 must have a finite jump of magnitude 1/ρ at ρ0 = ρ.
1
ρ
dgm
dρ0
=
= mγm
−
ρ0 =ρ+
m−1
ρ
dgm
dρ0
b
ρ0 =ρ−
m
+ m+1 − mγm
bm
ρ
m
1
b
= 2mγm
ρ
ρ
so
γm =
and
0
gm (ρ, ρ ) =
=
ρ m
b
1 ρ m
2m b
0 m 0 m 1
ρρ
ρ
−
2
2m
b
ρ
0 m m ρ
ρρ
1
−
2
2m
b
ρ0
or
gm (ρ, ρ0 ) =
1
2m
m 1
b
−
ρ
ρ
ρρ0
b2
m
−
ρ<
ρ>
,
ρ0 < ρ
,
ρ0 > ρ.
m .
Plugging into the expansion for G gives
G(ρ, φ, ρ0 , φ0 ) =
0 m m ∞
1 X 1
ρρ
ρ<
cos m(φ − φ0 ).
−
2π n=1 m
b2
ρ>
(22)
Homer Reid’s Solutions to Jackson Problems: Chapter 2
17
Here we need to work out an auxiliary result:
∞
∞ Z x
X
X
1 n
n−1
0
u
du cos m(φ − φ0 )
x cos n(φ − φ ) =
n
0
n=1
n=1
)
Z x( X
∞
1
n
0
=
u cos n(φ − φ ) du
u n=1
0
Z x
cos(φ − φ0 ) − u
du
=
1 + u2 − 2u cos(φ − φ0 )
0
1
x
= − ln(1 − 2u cos(φ − φ0 ) + u2 ) 0
2
1
= − ln[1 − 2x cos(φ − φ0 ) + x2 ].
2
(I summed the infinite series here back in Problem 2.12. The integral in the
second-to-last step can be done by partial fraction decomposition, although I
cheated and looked it up on www.integrals.com). We can apply this result
individually to the two terms in (22):
1
1 + (ρρ0 /b2 )2 − 2(ρρ0 /b2 ) cos(φ − φ0 )
0
0
G(ρ, φ; ρ , φ ) = −
ln
4π
1 + (ρ< /ρ> )2 − 2(ρ< /ρ> ) cos(φ − φ0 )
2 4
ρ> b + ρ2 ρ02 − 2ρρ0 b2 cos(φ − φ0 )
1
= −
ln
4π
b4
ρ2> + ρ2< − 2ρ< ρ> cos(φ − φ0 )
2 4
ρ> b + ρ2 ρ02 − 2ρρ0 b2 cos(φ − φ0 )
1
= −
ln
4π
b4
ρ2> + ρ2< − 2ρ< ρ> cos(φ − φ0 )
2
ρ>
1
= −
ln
4π
b2
4
b + ρ2 ρ02 − 2ρρ0 b2 cos(φ − φ0 )
1
(23)
ln 2 2
−
4π
b (ρ + ρ02 − 2ρρ0 cos(φ − φ0 ))
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 did
include 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 compute
the potential within the cylinder. If there is no charge inside the cylinder, the
volume integral vanishes, and we are left with the surface integral:
Z
∂G
dA0 .
(24)
Φ(ρ, φ) = Φ(b, φ0 )
∂ρ0 ρ0 =b
where the integral is over the surface of the cylinder.
For this we need the normal derivative of (23) on the cylinder:
∂G
1
=−
0
∂ρ
4π
2ρ0 − 2ρ cos(φ − φ0 )
2ρ2 ρ0 − 2ρb2 cos(φ − φ0 )
− 2
4
2
02
0
2
0
b + ρ ρ − 2ρρ b cos(φ − φ ) ρ + ρ02 − 2ρρ0 cos(φ − φ0 )
.
18
Homer Reid’s Solutions to Jackson Problems: Chapter 2
Evaluated at ρ0 = b this is
∂G
∂ρ0
ρ0 =b
1
=−
2π
ρ2 − b 2
b(ρ2 + b2 − 2ρb cos(φ − φ0 ))
.
In the surface integral, the extra factor of b on the bottom is cancelled by the
factor of b in the area element dA0 , and (24) becomes just the result of Problem
2.12.
(c) For the exterior problem we again start with the solution (21). Now the
boundary conditions are different; the condition at ∞ gives A2m = 0, while the
condition at b gives
B1m = −γm bm .
A1m = γm b−m
From the continuity condition at ρ0 = ρ we find
m b
ρ m
−
A2m = γm ρm
.
b
ρ
The finite derivative jump condition gives
m m 1
b
b
1
ρ m
1
ρ m
− mγm
=
−
+
−mγm
b
ρ
ρ
b
ρ
ρ
ρ
or
γm = −
1
2m
m
b
.
ρ
Putting it all together we have for the exterior problem
2 m m 1
b
ρ<
gm =
.
−
2m
ρρ0
ρ>
This is the same gm we came up with before, but with b2 and ρρ0 terms flipped
in first term. But the closed-form expression was symmetrical in those two
expressions (except for the mysterious ln term) so the closed-form expression
for the exterior Green’s function should be the same as the interior Green’s
function.
Solutions to Problems in Jackson,
Classical Electrodynamics, Third Edition
Homer Reid
June 15, 2000
Chapter 3: Problems 1-10
Problem 3.1
Two concentric spheres have radii a, b(b > a) and each is divided into two hemispheres by the same horizontal plane. The upper hemisphere of the inner sphere
and the lower hemisphere of the outer sphere are maintained at potential V . The
other 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 in
the limiting cases b → ∞ and a → 0.
The expansion of the electrostatic potential in spherical coordinates for problems with azimuthal symmetry is
Φ(r, θ) =
∞ h
X
l=0
i
Al rl + Bl r−(l+1) Pl (cos θ).
(1)
We find the coefficients Al and Bl by applying the boundary conditions. Multiplying both sides by Pl0 (cos θ) and integrating from -1 to 1 gives
Z 1
i
2 h
Al rl + Bl r−(l+1) .
Φ(r, θ)Pl (cos θ)d(cos θ) =
2l + 1
−1
At r = a this yields
V
Z
1
Pl (x)dx =
0
i
2 h
Al al + Bl a−(l+1) ,
2l + 1
1
2
Homer Reid’s Solutions to Jackson Problems: Chapter 3
and at r = b,
V
Z
0
Pl (x)dx =
−1
i
2 h l
Al b + Bl b−(l+1) .
2l + 1
The integral from 0 to 1 vanishes for l even, and is given in the text for l odd:
Z 1
(l − 2)!!
1
Pl (x)dx = (− )(l−1)/2 l+1 .
2
2 2 !
0
The integral from -1 to 0 also vanishes for l even, and is just the above result
inverted for l odd. This gives
i
2 h
1
(l − 2)!!
Al al + Bl a−(l+1)
V (− )(l−1)/2 l+1 =
2
2l + 1
2 2 !
i
(l − 2)!!
2 h l
1
−V (− )(l−1)/2 l+1 =
Al b + Bl b−(l+1) .
2
2l + 1
2 2 !
or
αl
−αl
with
1
(2l + 1)(l − 2)!!
αl = V (− )a(l−1)/2
.
2
4 l+1
2 !
The solution is
Al = α l
= Al al + Bl a−(l+1)
= Al bl + Bl b−(l+1)
bl+1 + al+1
a2l+1 − b2l+1
Bl = −αl
al+1 bl+1 (bl + al )
a2l+1 − b2l+1
The first few terms of (1) are
2
(a + b2 )r a2 b2 (a + b)
a4 b4 (a3 + b3 )
3
7 (a4 + b4 )r3
− 2 3
− 4 7
Φ(r, θ) = V
P1 (cos θ)−
P3 (cos θ)+· · ·
4
a3 − b 3
r (a − b3 )
16
a7 − b 7
r (a − b7 )
In the limit as b → ∞, the problem reduces to the exterior problem treated
in Section 2.7 of the text. In that limit, the above expression becomes
7 a 4
3 a 2
P1 (cos θ) − V
P3 (cos θ) + · · ·
Φ(r, θ) → V
4
r
16
r
in agreement with (2.27) with half the potential spacing. When a → 0, the
problem goes over to the interior version of the same problem, as treated in
section 3.3 of the text. In that limit the above expression goes to
3 r 7 r 3
Φ(r, θ) → − V
P1 (cos θ) + V
P3 (cos θ) + · · ·
4
b
16
b
This agrees with equation (3.36) in the text, with the sign of V flipped, because
here the more positive potential is on the lower hemisphere.
Homer Reid’s Solutions to Jackson Problems: Chapter 3
3
Problem 3.2
A spherical surface of radius R has charge uniformly distributed over its surface
with a density Q/4πR2 , except for a spherical cap at the north pole, defined by the
cone θ = α.
(a) Show that the potential inside the spherical surface can be expressed as
Φ=
∞
rl
Q X 1
[Pl+1 (cos α) − Pl−1 (cos α)] l+1 Pl (cos θ)
8π0
2l + 1
R
l=0
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) as
the spherical cap becomes (1)very small, and (2) so large that the area with
charge 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 infinitesimally close to the surface of the sphere, the electric field is
F = −∇Φ = −
so
∂Φ
∂r
=
r=R
σ
r̂
0
σ
.
0
(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
∞
X
∂
Φ(r, θ) =
lAl rl−1 Pl (cos θ)
∂r
l=1
Multiplying by Pl0 and integrating at r = R gives
Z
1 1
2l
σ(θ)Pl (cos θ)d(cos θ) =
Al Rl−1
0 −1
2l + 1
so
2l + 1
Al =
·
2lRl−1
Q
4πR2 0
Z
cos α
Pl (x)dx.
−1
To evaluate the integral we use the identity (eq. 3.28 in the text)
Pl (x) =
d
1
[Pl+1 (x) − Pl−1 (x)]
(2l + 1) dx
Homer Reid’s Solutions to Jackson Problems: Chapter 3
so
Z
cos α
Pl (x)dx =
−1
4
1
[Pl+1 (cos α) − Pl−1 (cos α)] .
2l + 1
(We used the fact that Pl+1 (−1) = Pl−1 (−1) for all l.) With this we have
Al =
Q
[Pl+1 (cos α) − Pl−1 (cos α)]
8π0 lRl+1
so the potential expansion is
Φ(r, θ) =
∞
Q X1
rl
[Pl+1 (cos α) − Pl−1 (cos α)] l+1 Pl (cos θ).
8π0
l
R
l=1
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 in
the potential. I don’t understand how he can determine that constant from the
information contained in the problem; the information about the charge density
only tells you the derivative of the potential. There’s nothing in this problem
that 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
1 ∂Φ
r̂ +
θ̂
∂r
r ∂θ r=0
d
Q
[P
(cos
α)
−
1]
P
(cos
θ)r̂
+
P
(cos
θ)
θ̂
= −
2
1
1
8π0 R2
dθ
h
i
Q
3
3
= −
cos θr̂ − sin θ θ̂
cos2 α −
2
8π0 R 2
2
= −
=
3Q sin2 α
k̂.
16π0 R2
The field points in the positive z direction. That makes sense, since a positive
test charge at the origin would sooner fly up out through the uncharged cap
than through any of the charged surface.
5
Homer Reid’s Solutions to Jackson Problems: Chapter 3
Problem 3.3
A thin, flat, conducting, circular disk of radius R is located in the x − y plane
with its center at the origin, and is maintained at a fixed potential V . With the
information that the 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
∞
2V R X (−1)l
Φ(r, θ, φ) =
π r
2l + 1
l=0
R 2l
r
!
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
1 r 2
K
3 · 1 r 4
5 · 3 · 1 r 6
1+
=
+
+
+···
R
2 R
(2!)(2 · 2) R
(3!)(2 · 2 · 2) R
∞
K X (2n − 1)!! r 2n
(3)
=
R n=0 n! · 2n
R
for some constant K. From the way the problem is worded, I take it we’re
not supposed to try to figure out what K is explicitly, but rather to work the
problem 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 must
be proportional to the surface charge. At the surface of the disk, the normal
direction is the negative θ̂ direction. Hence
1 ∂
Φ(r, θ)
r ∂θ
θ=(π/2)
=±
σ
.
0
(4)
with the plus (minus) sign valid for Φ above (below) the disc.
For r < R the potential expansion is
Φ(r, θ) =
∞
X
Al rl Pl (cos θ).
(5)
l=0
Combining (3), (4), and (5) we have
∞
X
l=0
Al rl−1
d
Pl (cos θ)
dθ
cos θ=0
=±
∞
K X (2n − 1)!! r 2n
.
R0 n=0 n! · 2n
R
(6)
Homer Reid’s Solutions to Jackson Problems: Chapter 3
6
For l even, dPl /dx vanishes at x = 0. For l odd, I used some of the Legendre
polynomial identities to derive the formula
d
P2l+1 (x)
dx
= (−1)l (2l + 1)
x=0
(2l − 1)!!
.
l! · 2l
This formula reminds one strongly of expansion (3). Plugging into (6) and
equating coefficents of powers of r, we find
A2l+1 = ±
so
Φ(r, θ) = A0 ±
(−1)l K
(2l + 1)R2l+1 0
∞
K X (−1)l r 2l+1
P2l+1 (cos θ).
0
2l + 1 R
l=1
I wrote A0 explicitly because we haven’t evaluated it yet–the derivative condition
we used earlier gave no information about it. To find A0 , observe that, on the
surface of the disk (cos θ = 0), all the terms in the above sum vanish ( because
Pl (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 X (−1)l r 2l+1
P2l+1 (cos θ)
0
2l + 1 R
(7)
l=1
where the plus (minus) sign is good for θ less than (greater than)π/2. Note that
the presence of that ± sign preserves symmetry under reflection through the z
axis, 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 be
continuous everywhere–we can’t have anything like the derivative discontinuity
that exists at θ = π/2 for r < R. Since the physical problem is symmetric under
a sign flip in cos θ, the potential expansion can only contain Pl terms for l even.
The expansion is
Φ(r, θ) =
∞
X
B2l r−(2l+1) P2l (cos θ).
l=0
At r = R, this must match up with (7):
V ±
∞
∞
X
K X (−1)l
P2l+1 (cos θ) =
B2l R−(2l+1) P2l (cos θ).
0
2l + 1
l=1
l=0
Multiplying both sides by P2l (cos θ) sin(θ) and integrating gives
Z 0
Z 1
Z 1
∞
K X (−1)l
2R−(2l+1)
P2l+1 (x)Pl (x)dx
Pl (x)dx +
P2l+1 (x)P2l (x)dx +
B2l
= V
−
4l + 1
0
2l + 1
0
−1
−1
l=1
Z
∞
2K X (−1)l 1
= 2V δl,0 +
P2l+1 (x)P2l (x)dx.
0
2l + 1 0
l=1
Homer Reid’s Solutions to Jackson Problems: Chapter 3
7
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 is
the z axis and they are distributed uniformly in the angle φ. (The segments are like
the skin on wedges of an apple, or the earth’s surface between successive meridians
of longitude.) The segments are kept at fixed potentials ±V , alternately.
(a) Set up a series representation for the potential inside the sphere for the general
case of 2n segments, and carry the calculation of the coefficients in the series
far enough to determine exactly which coefficients are different from zero. For
the nonvanishing terms exhibit the coefficients as an integral over cos θ.
(b) For the special case of n = 1 (two hemispheres) determine explicitly the potential up to and including all terms with l = 3. By a coordinate transformation
verify that this reduces to result (3.36) of Section 3.3.
(a) The general potential expansion is
Φ(r, θ, φ) =
∞ X
l
h
X
l=0 m=−l
i
Alm rl + Blm r−(l+1) Ylm (θ, φ).
(8)
For the solution within the sphere, finiteness at the origin requires Blm = 0.
∗
Multiplying by Yl0m0
and integrating over the surface of the sphere we find
Z
1
∗
Alm =
Φ(a, θ, φ) Ylm
(θ, φ) dΩ
al
Z
Z
n
π
2kπ/n
V X
∗
k
Ylm
(θ, φ) sin θ dφ dθ
=
(−1)
al
0
2(k−1)π/n
k=1
(Z
)
1/2 Z 1
X
n
2kπ/n
V 2l + 1 (l − m)!
k
−imφ
m
=
Pl (x) dx
(−1)
e
dφ .
al
4π (l + m)!
−1
2(k−1)π/n
k=1
(9)
The φ integral is easy:
Z 2kπ/n
i
1 h −2imkπ/n
e
− e−2im(k−1)π/n .
e−imφ dφ = −
im
2(k−1)π/n
This is to be summed from k = 1 to n with a factor of (−1)k thrown in:
i
X
1 h −2mπi(1/n)
= −
(e
− 1) − (e−2mπi(2/n) − e−2mπi(1/n) ) + · · · − (1 − e−2mπi((n−1)/n) )
im
o
2 n
=
1 − e−2mπi/n + e2(−2mπi/n) − e3(−2mπi/n) + · · · + e(n−1)(−2mπi/n) .
(10)
im
Homer Reid’s Solutions to Jackson Problems: Chapter 3
8
Putting x = − exp(−2mπi/n), the thing in braces is
1 + x + x2 + x3 + · · · + xn−1 =
1 − xn
1 − e−2mπi
=
,
1−x
1 + e−2imπ/n
Note that the numerator vanishes. Thus the only way this thing can be nonzero
is if the denominator also vanishes, which only happens if the exponent in the
denominator equates to -1. This only happens if m/n = 1/2, 3/2, 5/2, · · · . In
that 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 terms
with a minus sign come out to -1, so all n terms add constructively, and (10)
equates to
(
2n
X
,
m = n/2, 3n/2, 5n/2, · · ·
= im
0,
otherwise.
Then the expression (9) for the coefficients becomes
Alm
1/2 Z 1
2nV 2l + 1 (l − m)!
=
Plm (x)dx,
imal
4π (l + m)!
−1
m=
n 3n
,
, · · · = 0, otherwise.
2 2
(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 with
l = 1, m = ±1, and l = 3, m = ±1 or ±3. We need to evaluate the θ integral
for these terms. We have
Z 1
Z 1
1
P1 (x) dx = −
(1 − x2 )1/2 dx = −π
Z
Z
−1
1
P31 (x) dx
−1
1
−1
P33 (x) dx
= −
Z
−1
1
−1
= −15
Z
(1 − x2 )1/2
1
−1
3π
15 2 3
dx = −
x −
2
2
8
(1 − x2 )3/2 dx = −
15π
.
4
Using these results in (??), we have
A1±1
A3±1
A3±3
1/2
3
4πV i
a
4π · 2
1/2
3πV i 7 · 2
= ±
2a3 4π · 4!
1/2
5πV i
7
= ± 3
a
4π · 6!
= ±
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
9
Homer Reid’s Solutions to Jackson Problems: Chapter 3
coefficient, which I have skipped here.
h r
7 r 3
sin θ(5 cos2 θ − 1) sin φ
sin θ sin φ +
Φ(r, θ, φ) = V 3
a
16 a
7 r 3 3
+
sin θ sin 3φ + · · ·
144 a
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 and
powers 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 the
potential for r 6= 0. This is by definition a dipole along the z azis and its
potential.
(c) Suppose now that the dipole of part b is surrounded by a grounded spherical
shell of radius b concentric with the origin. By linear superposition find the
potential everwhere inside the shell.
(a) First of all, for a point on the z axis the potential is
1
q
1
Φ(z) =
−
4π0 |z − a| z + a
a a 2
a a 2
q
+··· − 1−
···
=
1+
+
+
4π0 z
z
z
z
z
q
a
a 3
=
+
+···
2π0 z
z
z
P
for z > a. Comparing this with the general expansion Φ =
Bl r−(l+1) Pl (cos θ)
at θ = 0 we can identify the Bl s and write
a 3
a
q
P3 (cos θ) + · · ·
P1 (cos θ) +
Φ(r, θ) =
2π0 r
r
r
for r > a. For r < a we can just swap a and r in this equation.
(b)
Φ(r, θ)
=
=
→
qa
2π0 r2
p
4π0 r2
p
4π0 r2
P1 (cos θ) +
P1 (cos θ) +
cos θ
a 2
P3 (cos θ) + · · ·
r
P3 (cos θ) + · · ·
r
a 2
as a → 0.
Homer Reid’s Solutions to Jackson Problems: Chapter 3
10
(c) When we put the grounded sphere around the two charges, a surface charge
distribution forms on the sphere. Let’s denote by Φs the potential due to this
charge 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 charges
within the sphere, in which case we have the general expansion (1), with Bl = 0
to keep us finite at the origin. The total potential is just the sum Φs + Φd :
∞
Φ(r, θ) =
X
p
cos θ +
Al rl Pl (cos θ).
2
4π0 r
l=0
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
p
.
A1 = −
4π0 b3
The total potential inside the sphere is then
r
p P1 (cos θ).
1
−
Φ(r, θ) =
4π0 b2
b
Problem 3.7
Three point charges (q, −2q, q) are located in a straight line with separation a and
with the middle charge (−2q) at the origin of a grounded conducting spherical shell
of radius b, as indicated in the figure.
(a) Write down the potential of the three charges in the absence of the grounded
sphere. Find the limiting form of the potential as a → 0, but the product
qa2 = 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 density
induced on the inner surface at r = b or by image charges located at r > b. Use
linear superposition to satisfy the boundary conditions and find the potential
everywhere inside the sphere for r < a and r > a. Show that in the limit
a → 0,
Q
r5
Φ(r, θ, φ) →
1 − 5 P2 (cos θ).
2π0 r3
b
(a) On the z axis, the potential is
q
2
1
1
Φ(z) =
− +
+
4π0
z |z − a| z + a
a a 2
a a 2
q
··· + 1−
+···
=
−2 + 1 +
+
+
4π0 r
z
z
z
z
q
a 2 a 4
=
+
+··· .
2π0 z
z
z
Homer Reid’s Solutions to Jackson Problems: Chapter 3
11
As before, from this result we can immediately infer the expression for the
potential at all points:
a 4
a 2
q
P2 (cos θ) +
P4 (cos θ) + · · ·
Φ(r, θ) =
2π0 r
r
r
2
2
qa
a
=
P
(cos
θ)
+
·
·
·
P
(cos
θ)
+
4
2
2π0 r3
r
Q
→
P2 (cos θ) as a → 0
(11)
2π0 r3
(b) As in the previous problem, the surface charges on the sphere produce an
extra contribution Φs to the potential within the sphere. Again we can express
Φs with the expansion (1) (with Bl = 0), and we add Φs to (11) to get the full
potential within the sphere:
∞
Φ(r, θ) =
X
Q
P2 (cos θ) +
Al rl Pl (cos θ)
3
2π0 r
l=0
From the condition that Φ vanish at r = b, we determine that only the l = 2
term in the sum contributes, and that
A2 = −
Q
.
2π0 b5
Then the potential within the sphere is
r 5 Q
Φ(r, θ) =
1
−
P2 (cos θ).
2π0 r3
b
Problem 3.9
A hollow right circular cylinder of radius b has its axis coincident with the z axis
and its ends at z = 0 and z = L. The potential on the end faces is zero, while the
potential on the cylindrical surface is given as V (φ, z). Using the appropriate separation of variables in cylindrical coordinates, find a series solution for the potential
anywhere inside the cylinder.
The general solution of the Laplace equation for problems in cylindrical coordinates 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
Homer Reid’s Solutions to Jackson Problems: Chapter 3
12
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 take
k imaginary, i.e.
πn
, n = 1, 2, 3, · · ·
Z(z) = C sin(kn z) with kn =
L
With this form for Z, R must be taken to be of the form
R(ρ) = EIν (kn ρ) + F Kν (kn ρ).
Since we’re looking for the potential on the inside of the cylinder and there is
no charge at the origin, the solution must be finite as ρ → 0, which requires
F = 0. Then the potential expansion becomes
Φ(ρ, φ, z) =
∞ X
∞
X
[Anν sin νφ + Bnν cos νφ] sin(kn z)Iν (kn ρ).
(12)
n=1 ν=0
Multiplying by sin ν 0 φ sin kn0 z and integrating at r = b, we find
Z L Z 2π
πL
Iν (kn b)Anν
V (φ, z) sin νφ sin(kn z) dφ dz =
2
0
0
so
Z L Z 2π
2
Anν =
V (φ, z) sin(νφ) sin(kn z) dφ dz.
πLIν (kn b) 0 0
Similarly,
Z L Z 2π
2
Bnν =
V (φ, z) cos(νφ) sin(kn z) dφ dz.
πLIν (kn b) 0 0
(13)
(14)
Problem 3.10
For the cylinder in Problem 3.9 the cylindrical surface is made of two equal halfcylinders, one at potential V and the other at potential −V , so that
V for −π/2 < φ < π/2
V (φ, z) =
−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 φ
and compare it with two-dimensional Problem 2.13.
The potential expansion is (12) with coefficients given by (13) and (14). The
relevant integrals are
Z L Z 2π
V (φ, z) sin(νφ) sin(kn z) dφ dz
0
0
13
Homer Reid’s Solutions to Jackson Problems: Chapter 3
= V
= 0
(Z
L
sin(kn z) dz
0
Z
= V
=
=
(Z
L
0
Z
) (Z
π/2
−π/2
sin(νφ) dφ −
Z
3π/2
sin(νφ) dφ
π/2
)
2π
V (φ, z) cos(νφ) sin(kn z) dφ dz
0
L
sin(kn z) dz
0
) (Z
π/2
−π/2
cos(νφ) dφ −
Z
3π/2
cos(νφ) dφ
π/2
)
o
2V n
π/2
3π/2
(n odd)
|sin νφ|−π/2 − |sin νφ|π/2
νkn

0 , n or ν even

8V /kn ν , n odd, ν = 1, 5, 9, · · ·

−8V /kn ν , n odd, ν = 3, 7, 11, · · ·
Hence, from (13) and (14),
Anν
Bnν
= 0
= 0,
= (−1)(ν−1)/2 · 16V /(nνπ 2 Iν (kn b)),
n or ν even
n and ν odd
The potential expansion is
Φ(ρ, θ, z) =
16V X (−1)(ν−1)/2
cos(νφ) sin(kn z)Iν (kn ρ)
π 2 n,ν nνIv (kn b)
(15)
where the sum contains only terms with n and ν odd.
(b) At z = L/2 we have
Φ(ρ, θ, L/2) =
Iν (kn ρ)
16V X (−1)(n+ν−2)/2
cos(νφ)
.
2
π n,ν
nν
Iν (kn b)
As L → ∞, the arguments to the I functions become small. Using the limiting
form for Iν quoted in the text as equation (3.102), we have
Φ(ρ, θ) =
ρ ν
16V X (−1)(n+ν−2)/2
cos(νφ)
.
π 2 n,ν
nν
b
The sums over n and ν are now decoupled:
(∞
)( ∞
)
ρ ν
X (−1)ν
16V X (−1)n
Φ(ρ, θ) =
cos(νφ)
π2
2n + 1
2ν + 1
b
n=0
ν=0
(∞
)
ρ ν
16V n π o X (−1)ν
cos(νφ)
=
2
π
4
2ν + 1
b
ν=0
4V
2ρb cos φ
=
tan−1
π
b2 − ρ 2
Homer Reid’s Solutions to Jackson Problems: Chapter 3
14
This agrees with the result of Problem 2.13, with V1 = −V2 = V . The first
series is just the Taylor series for tan−1 (x) at x = 1, so it sums to π/4. The
second 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.
Solutions to Problems in Jackson,
Classical Electrodynamics, Third Edition
Homer Reid
June 15, 2000
Chapter 3: Problems 11-18
Problem 3.11
A modified Bessel-Fourier series on the interval 0 ≤ ρ ≤ a for an arbitrary function
f (ρ) can be based on the ”homogenous” boundary conditions:
At
ρ = 0,
At ρ = a,
d
Jν (k 0 ρ) = 0
dρ
λ
d
ln[Jν (kρ)] = −
dρ
a
ρJν (kρ)
(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 the
usual way.
(b) Find the normalization integral and show that an arbitrary function f (ρ) can
be expanded on the interval in the modified Bessel-Fourier series
f (ρ) =
∞
X
n=1
A n Jν
y
νn
a
with the coefficients An given by
2
An = 2
a
"
ν2
1− 2
yνn
Jν2 (yνn )
+
1
dJν (yνn )
dyνn
2 #−1 Z
a
f (ρ)ρJν
0
y ρ
νn
dρ.
a
2
Homer Reid’s Solutions to Jackson Problems: Chapter 3
(a) The function Jν (kρ) satisfies the equation
d
ν2
1 d
2
ρ Jν (kρ) + k − 2 Jν (kρ) = 0.
ρ dρ dρ
ρ
(1)
Multiplying both sides by ρJν (k 0 ρ) and integrating from 0 to a gives
Z a
d
ν2
d
0
2
0
ρ Jν (kρ) + k ρ −
Jν (k ρ)Jν (kρ) dρ = 0.
Jν (k ρ)
dρ dρ
ρ
0
The first term on the left can be integrated by parts:
Z a
d
d
0
Jν (k ρ)
ρ Jν (kρ) dρ
dρ dρ
0
Z a a
d
d
d
ρ
Jν (k 0 ρ)
Jν (kρ) dρ.
= ρJν (k 0 ρ) Jν (kρ) −
dρ
dρ
dρ
0
0
(2)
(3)
One of the conditions we’re given is that the thing in braces in the first term
here vanishes at ρ = 0. At ρ = a we can invoke the other condition:
d
ln[Jν (kρ)]
dρ
=
ρ=a
1
d
Jν (kρ)
Jν (kρ) dρ
ρ=a
=−
λ
a
d
→ a Jν (ka) = −λJν (ka).
dρ
Plugging this into (3), we have
Z a
d
d
ρ Jν (kρ) dρ
Jν (k 0 ρ)
dρ dρ
0
Z a d
d
0
0
Jν (k ρ)
Jν (kρ) .
ρ
= −λJν (k ρ)Jν (kρ) −
dρ
dρ
0
(4)
This is clearly symmetric in k and k 0 , so when we write down (2) with k and
k 0 switched and subtract from (2), the first integral (along with the ν 2 /ρ term)
vanishes, and we are left with
Z a
02
2
(k − k )
ρJν (k 0 ρ)Jν (kρ) dρ = 0
0
proving orthogonality.
(b) If we multiply (1) by ρ2 J 0 (kρ) and integrate, we find
Z a
Z a
Z a
d
0
2
0
2
0
2
ρJν (kρ) [ρJν (kρ)]dρ+k
Jν (kρ)Jν0 (kρ)dρ = 0.
ρ Jν (kρ)Jν (kρ)dρ−ν
dρ
0
0
0
(5)
3
Homer Reid’s Solutions to Jackson Problems: Chapter 3
R
The first and third integrals are of the form f (x)f 0 (x)dx and can be done
immediately. In the second integral we put f (ρ) = ρ2 Jν (kρ), g 0 (ρ) = Jν0 (kρ)
and integrate by parts:
Z a
Z a
Z a
a
2
2
0
2 2
ρJν (kρ)dρ −
ρ2 Jν (kρ)Jν0 (kρ)dρ
ρ Jν (kρ)Jν (kρ)dρ = ρ Jν (kρ) 0 − 2
0
0
→
Z
a
0
ρ2 Jν (kρ)Jν0 (kρ)dρ =
0
1 2 2
a Jν (ka) −
2
Using this in (5),
a2 02
(ak)2 2
Jν (ka) +
aJν (ka) − k 2
2
2
so
Z
a
0
ρJν2 (kρ)dρ
=
=
ν2
a2
− 2
2
2k
(
a2
2
Z
Z
a
0
ρJν2 (kρ)dρ.
a
0
ρJν2 (kρ)dρ −
ν2 2
J (ka) = 0
2 ν
a2 02
J (ka)
2k 2 ν
2 )
ν2
d
2
1−
Jν (ka) +
Jν (ka)
(ka)2
d(ka)
Jν2 (ka) +
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 a
cut in it. A thin, flat, disc of the same material and slightly smaller radius lies in
the plane, filling the hole, but separated from the sheet by a very narrow insulating
ring. The disc is maintained at a fixed potential V , whilc the infinite sheet is kept
at zero potential.
(a) Using appropriate cylindrical coordinates, find an integral expression involving
Bessel functions for the potential at any point above the plane.
(b) Show that the potential a perpendicular distance z above the center of the disc
is
z
Φ0 (z) = V 1 − √
a2 + z 2
(c) Show that the potential a perpendicular distance z above the edge of the disc
is
kz
V
K(k)
1−
Φa (z) =
2
πa
where k = 2a/(z 2 + 4a2 )1/2 , and K(k) is the complete elliptic integral of the
first kind.
4
Homer Reid’s Solutions to Jackson Problems: Chapter 3
(a) As before, we can write the potential as a sum of terms R(ρ)Q(φ)Z(z). In
this problem there is no φ dependence, so Q = 1. Also, the boundary conditions
on Z are that it vanish at ∞ and be finite at 0, whence Z(z) ∝ exp(−kz) for
any k. Then the potential expansion becomes
Z ∞
Φ(ρ, z) =
A(k)e−kz J0 (kρ) dk.
(6)
0
To evaluate the coefficients A(k), we multiply both sides by ρJ0 (k 0 ρ) and integrate over ρ at z = 0:
Z ∞
Z ∞
Z ∞
ρΦ(ρ, 0)J0 (k 0 ρ) dρ =
A(k)
ρJ0 (kρ)J0 (k 0 ρ) dρ dk
0
0
0
A(k 0 )
k0
=
so
A(k)
= k
Z
= kV
Plugging this back into (6),
Z
Φ(ρ, z) = V
∞
0
Z
∞
ρΦ(ρ, 0)J0 (kρ) dρ
0
Z
a
ρJ0 (kρ)dρ.
0
a
kρ0 e−kz J0 (kρ)J0 (kρ0 ) dρ0 dk.
(7)
0
The ρ0 integral can be done right away. To do it, I appealed to the differential
equation for J0 :
1
J000 (u) + J00 (u) + J0 (u) = 0
u
so
Z x
Z x
Z x
uJ0 (u) du = −
uJ000 du −
J00 (u) du
0
0
0
Z x
Z x
x
0
0
= − |uJ0 (u)|0 +
J0 (u) du −
J00 (u) du
x
0
0
= − |uJ00 (u)|0 = −xJ00 (x) = xJ1 (x).
(In going from the first to second line, I integrated by parts.) Then (7) becomes
Z ∞
Φ(ρ, z) = aV
J1 (ka)J0 (kρ)e−kz dk.
(8)
0
5
Homer Reid’s Solutions to Jackson Problems: Chapter 3
(b) At ρ = 0, (7) becomes
Z
a
Z
∞
−kz
0
ke
J0 (kρ )dk dρ0
Φ(0, z) = V J0 (0)
ρ
0
0
Z a Z ∞
∂
0
−kz
0
= V
ρ −
e
J0 (kρ )dk dρ0
∂z 0
0
!)
Z a (
∂
1
p
dρ0
= V
ρ0 −
∂z
ρ02 + z 2
0
Z a
zρ0
= V
dρ0
02
2 )3/2
(ρ
+
z
0
0
Here we substitute u = ρ02 + z 2 , du = 2ρ0 dρ:
Z 2 2
V zJ0 (0) a +z −3/2
Φ(0, z) =
u
du
2
z2
a2 +z 2
1
= −V z 1/2
u
z2
1
1
−√
= Vz
z
z2 + z2
z
= V 1− √
a2 + z 2
(b) At ρ = a, (8) becomes
Φ(a, z) = aV
Z
∞
J1 (ka)J0 (ka)e−kz dk
0
Problem 3.13
Solve for the potential in Problem 3.1, using the appropriate Green
function obtained in the text, and verify that the answer obtained
in this way agrees with the direct solution from the differential
equation.
For Dirichlet boundary value problems, the basic equation is
Z
I
∂G(x; x0 )
1
0
0
0
G(x; x )ρ(x ) dV +
Φ(x0 )
dA0 .
Φ(x) = −
0 V
∂n
S
x0
(9)
Here there is no charge in the region of interest, so only the surface integral
contributes. The Green’s function for the two-sphere problem is
G(x; x0 ) = −
∞ X
l
∗
X
Ylm
(θ0 , φ0 ) Ylm (θ, φ)
Rl (r; r0 )
2l + 1
l=0 m=−l
(10)
6
Homer Reid’s Solutions to Jackson Problems: Chapter 3
with
1
0
Rl (r; r ) = 1−
a
b
2l+1
l
r<
a2l+1
− l+1
r<
1
l+1
r>
−
l
r>
b2l+1
.
(11)
Actually in this case the potential cannot have any Φ dependence, so all terms
with m 6= 0 in (10) vanish, and we have
G(x; x0 ) = −
∞
1 X
Pl (cos θ)Pl (cos θ0 )Rl (r; r0 ).
4π
l=0
In this case the boundary surfaces are spherical, which means the normal to a
surface element is always in the radial direction:
∞
∂
1 X
∂
0
Pl (cos θ)Pl (cos θ0 ) Rl (r; r0 ).
G(x; x ) = −
∂n
4π
∂n
l=0
The surface integral in (9) has two parts: one integral S1 over the surface of
the inner sphere, and a second integral S2 over the surface of the outer sphere:
Z π Z 2π
∞
∂Rl
1 X
0
0 2
0
0
Pl (cos θ)
S1 = −
Φ(a, θ )Pl (cos θ )a sin θ dφ dθ
4π
∂n r0 =a
0
0
l=0
Z 1
∞
∂Rl
V X 2
a Pl (cos θ)
Pl (x) dx
= −
2
∂n r0 =a
0
l=0
∞
∂Rl
V X 2
a γl Pl (cos θ) ·
= −
2
∂n
l=0
r 0 =a
where
γl
=
Z
1
Pl (x) dx
0
(l − 2)!!
1
,
= (− )(l−1)/2
2
2[(l + 1)/2]!
= 0,
l odd
l even.
A similar calculation gives
S2
= −
=
V
2
∞
∂Rl
V X 2
b Pl (cos θ)
2
∂n
l=0
∞
X
2
b γl Pl (cos θ)
l=0
∂Rl
∂n
r 0 =b
Z
0
Pl (x) dx
−1
r 0 =b
because Pl is odd for l odd, so its integral from -1 to 0 is just the negative of
the integral from 0 to 1. The final potential is the sum of S1 and S2 :
∞
V X
∂Rl
Φ(r, θ) =
γl Pl (cos θ) r02
2
∂n
l=0
r 0 =b
(12)
r 0 =a
Homer Reid’s Solutions to Jackson Problems: Chapter 3
7
Since the point of interest is always between the two spheres, to find the
normal derivative at r = a we differentiate with respect to r< , and at r = b
with respect to r> . Also, at r = a the normal is in the +r direction, while at
r = b the normal is in the negative r direction.
1
al−1
rl
2 ∂
0
2
a
Rl (r; r )
= (2l + 1)a − 2l+1
2l+1
∂n
rl+1
b
1 − ab
r 0 =a
b−(l+2)
a2l+1
0
2
l
2 ∂
Rl (r; r )
= (2l + 1)b b
2l+1 r − l+1
∂n
r
1− a
r 0 =b
b
Combining these with some algebra gives
Φ(r, θ) =
∞
(ab)l+1 (bl + al )r−(l+1) − (al+1 + bl+1 )rl
V X
(2l + 1)γl Pl (cos θ)
2
b2l+1 − a2l+1
l=0
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 charge
density varying as (d2 − z 2 ), where z is the distance from the midpoint. A grounded, conducting spherical shell of inner radius b > d
is centered at the midpoint of the line charge.
(a) Find the potential everywhere inside the spherical shell as an
expansion 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 − z 2 ), and that
the total charge is Q, whence
Q = 2λ
Z
d
0
(d2 − z 2 )dz =
4 3
d λ
3
3Q
.
4d3
In this case we have azimuthal symmetry, so the Green’s function is
→
G(x; x0 ) = −
λ=
∞
1 X
Pl (cos θ0 )Pl (cos θ)Rl (r; r0 )
4π
l=0
(13)
8
Homer Reid’s Solutions to Jackson Problems: Chapter 3
with
l
r<
0
Rl (r; r ) =
1
l+1
r>
−
l
r>
b2l+1
.
Since the potential vanishes on the boundary surface, the potential inside the
sphere is given by
Z
1
G(r, θ; r0 , θ0 )ρ(r0 , θ0 )dV.
Φ(r, θ) = −
0 V
In this case ρ is only nonzero on the z axis, where r = z. Also, Pl (cos θ)=1 for
z > 0, and (−1)l for z < 0. This means that the contributions to the integral
from the portions of the line charge for z > 0 and z < 0 cancel out for odd l,
and add constructively for even l:
#
" Z
∞
d
X
1
Rl (r; z)ρ(z) dz
Pl (cos θ) 2
Φ(r, θ) =
4π0
0
l=0,2,4,...
We have
Z
d
Rl (r; z)ρ(z) dz = λ
0
Z
d
0
l
r<
1
−
l+1
r>
l
r>
b2l+1
(d2 − z 2 ) dz
This is best split up into two separate integrals:
=λ
Z
d
0
l
r<
λ
(d2 − z 2 ) dz − 2l+1
l+1
b
r>
Z
d
0
l l
r<
r> (d2 − z 2 ) dz
The second integral is symmetric between r and r 0 , so we may integrate it
directly:
−
λ
b2l+1
Z
d
0
l l
r<
r> (d2
2
− z ) dz
Z d
λrl
= − 2l+1
z l (d2 − z 2 ) dz
b
0
dl+3
λrl dl+3
−
= − 2l+1
b
l+1 l+3
= −
λrl dl+3
(l + 1)(l + 3)b2l+1
The first integral must be further split into two:
λ
Z
d
0
l
r<
(d2 − z 2 ) dz
l+1
r>
(14)
9
Homer Reid’s Solutions to Jackson Problems: Chapter 3
= λ
(
(
1
rl+1
Z
r
0
z l (d2 − z 2 ) dz + rl
2 l+1
l+3
Z
d
r
d2 − z 2
dz
z l+1
)
d r
r
1
d2
l
−
+
−
+
r
= λ
l+1
l
r
l+1
l+3
lz
(l − 2)z l−2
2
r l
r2
d2
r2
d
2
2
−
+
−
+
= λ
d
l+1 l+3
d
l(l + 2)
l
l+2
r l
r2
2
d2
2
d
= λ
−
+
(l + 2)(l + 3) l(l + 1)
d
l(l + 2)
1
d
r
)
Combining this with (14), we have
d
r l
r2
2
d2
rl dl+3
d2
−
+
−
(l + 2)(l + 3) l(l + 1)
d
l(l + 2) (l + 1)(l + 3)b2l+1
0
(15)
But something is wrong here, because with this result the final potential will
contain terms like r 0 Pl (cos θ) and r2 Pl (cos θ), which do not satisfy the Laplace
equation.
Z
Rl (r; z)ρ(z) dz = λ
Homer Reid’s Solutions to Jackson Problems: Chapter 3
10
Problem 3.15
Consider the following “spherical cow” model of a battery connected to an external circuit. A sphere of radius a and conductivity σ is embedded in a uniform medium of conductivity σ 0 . Inside
the sphere there is a uniform (chemical) force in the z direction
acting on the charge carriers; its strength as an effective electric
field entering Ohm’s law is F . In the steady state, electric fields
exist inside and outside the sphere and surface charge resides on
its surface.
(a) Find the electric field (in addition to F ) and current density
everywhere in space. Determine the surface-charge density
and show that the electric dipole moment of the sphere is
p = 4π0 σa3 F/(σ + 2σ 0 ).
(b) Show that the total current flowing out through the upper
hemisphere of the sphere is
I=
2σσ 0
· πa2 F
σ + 2σ 0
Calculate the total power dissipation outside the sphere. Using the lumped circuit relations, P = I 2 Re = IVe , find the
effective external resistance Re and voltage Ve .
(c) Find the power dissipated within the sphere and deduce the
effective internal resistance Ri and voltage Vi .
(d) Define the total voltage through the relation Vt = (Re + Ri )I
and show that Vt = 4aF/3, as well as Ve + Vi = Vt . Show
that IVt is the power supplied by the “chemical” force.
(a) What’s going on in this problem is that the conductivity has a discontinuity going across the boundary of the sphere, but the current density must be
constant there, which means there must an electric field discontinuity in inverse
proportion to the conductivity discontinuity. To create this electric field discontinuity, there has to be some surface charge on the sphere, and this charge gives
rise to extra fields both inside and outside the sphere.
Since there is no charge inside or outside the sphere, the potential in those
two regions satisfied the Laplace equation, and may be expanded in Legendre
polynomials:
11
Homer Reid’s Solutions to Jackson Problems: Chapter 3
for r < a,
Φ(r, θ) = Φin (r, θ) =
∞
X
Al rl Pl (cos θ)
l=0
∞
X
for r > a,
Φ(r, θ) = Φout (r, θ) =
Bl r−(l+1) Pl (cos θ)
l=0
Continuity at r = a requires that
Al al = Bl a−l+1
Bl = a2l+1 Al
→
so
Φ(r, θ) =
(
Φin (r, θ) =
P∞
Φout (r, θ) =
l=0
P∞
Al rl Pl (cos θ),
l=0
Al a
r<a
2l+1 −(l+1)
r
Pl (cos θ),
r > a.
(16)
Now, in the steady state there can be no discontinuities in the current density, because if there were than there would be more current flowing into some
region 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’t
in steady state. So the current density is continuous everywhere. In particular,
the radial component of the current density is continuous across the boundary
of the sphere, i.e.
Jr (r = a− , θ) = Jr (r = a+ , θ).
(17)
Outside of the sphere, Ohm’s law says that
J = σ 0 E = −σ 0 ∇Φout .
Inside the sphere, there is an extra term coming from the chemical force:
J = σ(E + F k̂) = σ(−∇Φin + F k̂).
Applying (17) to these expressions, we have
∂
∂
+ F cos θ = −σ 0
Φout
σ − Φin
∂r
∂r
r=a
r=a
Using (16), this is
F P1 (cos θ) −
∞
X
l=0
lAl al−1 Pl (cos θ) =
σ0
σ
X
∞
(l + 1)Al al−1 Pl (cos θ).
l=0
Multiplying both sides by Pl0 (cos θ) and integrating from −π to π, we find
0
σ
2A1
(18)
F − A1 =
σ
Homer Reid’s Solutions to Jackson Problems: Chapter 3
12
for l=1, and
−lAl =
σ0
σ
(l + 1)Al
(19)
(20)
for l 6= 1. Since the conductivity ratio is positive, the second relation is impossible to satisfy unless Al = 0 for l 6= 1. The first relation becomes
σ
A1 =
F.
σ + 2σ 0
Then the potential is
Φ(r, θ) =
(
σ
σ+2σ 0 F r cos θ,
r<a
σ
3 −2
σ+2σ 0 F a r
r>a
cos θ,
(21)
The dipole moment p is defined by
Φ(r, θ) →
1 p·r
4π0 r3
as r → ∞.
(22)
The external portion of (21) can be written as
Φ(r, θ) =
F a3 z
σ
σ + 2σ 0 r3
and comparing this with (22) we can read off
σ
F a3 k̂.
p = 4π0
σ + 2σ 0
The electric field is found by taking the gradient of (21):
(
σ
− σ+2σ
r<a
0 F k̂,
E(r, θ) =
3
σ
a
(2 cos θr̂ + sin θ θ̂), r > a
σ+2σ 0 F r
The surface charge σs (θ) on the sphere is proportional to the discontinuity
in the electric field:
σs (θ) = 0 [Er (r = a+ ) − Er (r = a− )]
30 σ
=
F cos θ.
σ + 2σ 0
(b) The current flowing out of the upper hemisphere is just
Z
Z
J · dA = σ (Ein + F k̂) · dA
Z π/2 Z 2π
σ
F
cos θ sin θ a2 dφ dθ
=σ 1−
σ + 2σ 0
0
0
σσ 0
2
· πa F
=2
σ + 2σ 0
(23)
13
Homer Reid’s Solutions to Jackson Problems: Chapter 3
The Ohmic power dissipation in a volume dV is
dP = σE 2 dV
(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 density
there is σEx and the cross-sectional area is dydz, so I = σEx dydz. Also, the
voltage drop in the direction of current flow is V = Ex dx. Hence the power
dissipation due to current in the x direction is IV = σEx2 dV . Adding in the
contributions from the other two directions gives (24).
For the power dissipated outside the sphere we use the expression for the
electric field we found earlier:
Z ∞ Z π Z 2π
Pout = σ 0
E 2 (r, θ, φ)r2 sin θ dφ dθ dr
a
0
0
2
Z ∞Z π
1
σ
2 6
0
F a
(4 cos2 θ + sin2 θ) sin θ dθ dr
= 2πσ
4
σ + 2σ 0
r
0
a
2
8π 0
σ
=
σ
F 2 a3
3
σ + 2σ 0
Dividing by (23), we find the effective external voltage Ve :
Ve = Pout /I =
4
σ
aF ·
3
σ + 2σ 0
and the effective external resistance:
2
.
Re = Pout /I 2 =
3πaσ 0
(c) The power dissipated inside the sphere is
Pin = σ
Z
0
(E + F k̂)2 dV =
4σσ 2
F2
(σ + 2σ 0 )2
0
Z
dV
16σσ 2
=
πa3 F 2
3(σ + 2σ 0 )2
Since we’re in steady state, the current flowing out through the upper hemisphere of the sphere must be replenished by an equal current flowing in through
the lower half of the sphere, so to find the internal voltage and resistance we
can just divide by (23):
8 σ0
Vi = Pin /I =
aF
3 σ + 2σ 0
4
Ri = Pin /I 2 =
.
3πaσ
14
Homer Reid’s Solutions to Jackson Problems: Chapter 3
(c)
(Re + Ri )I =
2
3πa
(Vi + Ve ) =
1
2
+
σ0
σ
·
2σσ 0
4
πa2 F = aF
σ + 2σ 0
3
4aF
4
σ + 2σ 0 = aF
3(σ + 2σ 0 )
3
Problem 3.17
The Dirichlet Green function for the unbounded space between the planes at z = 0
and z = L allows discussion of a point charge or a distribution of charge between
parallel conducting planes held at zero potential.
(a) Using cylindrical coordinates show that one form of the Green function is
G(x, x0 ) = −
∞
∞
X
X
e
im(φ−φ0 )
n=1 m=−∞
sin
nπz L
sin
1
πL
nπz 0
L
×
Im
nπρ0 <
L
Km
nπρ >
L
(b) Show that an alternative form of the Green function is
G(x, x0 ) = −
Z
∞
X
m=−∞
1
×
2π
∞
0
dk eim(φ−φ ) Jm (kρ)Jm (kρ0 )
0
sinh(kz< ) sinh[k(L − z> )]
.
sinh(kL)
In cylindrical coordinates, the solutions of the Laplace equation look like
linear combinations of terms of the form
Tmk (ρ, φ, z) = eimφ Z(kz)Rm (kρ).
(25)
There are two possibilities for the combination Z(kz)Rm (kρ), both of which
solve the Laplace equation:
Z(kz)Rm (kρ) = (Aekz + Be−kz )[CJm (kρ) + DNm (kρ)]
(26)
Z(kz)Rm (kρ) = (Aeikz + Be−ikz )[CIm (kρ) + DKm (kρ)].
(27)
or
The Green’s function G(x; x0 ) must be a solution of the Laplace equation,
and must thus take one of the above forms, at all points x0 6= x. At x0 = x,
G must be continuous, but have a finite discontinuity in its first derivative.
.
Homer Reid’s Solutions to Jackson Problems: Chapter 3
15
Furthermore, G must vanish on the boundary surfaces. These conditions may
be met by dividing space into two regions, one on either side of the source point
x, and taking G to be different linear combinations of terms T (as in (25)) in
the two regions. The question is, in which dimension (i.e., ρ, z, or φ) do we
define the two “sides” of the source point?
(a) The first option is to imagine a cylindrical boundary at ρ0 = ρ, i.e. at the
radius of the source point, and take the inside and outside of the cylinder (i.e.,
ρ0 < ρ and ρ0 > ρ) as the two distinct regions of space. Then, within each
region, the entire range of z must be handled by one function, which means this
one function must vanish at z = 0 and z = L. This cannot happen with terms of
the form (26), so we are forced to take Z and R as in (27), with B = −A and k
restricted to the discrete values kn = nπ/L. Next considering the singularities
of the ρ functions in (27), we see that, to keep G finite everywhere, for the
inner region (ρ0 < ρ) we can only keep the Im (kρ) term, while for the outer
region we can only keep the Km (kρ) term. Then G(x; x0 ) will consist of linear
combinations of terms T as in (25) subject to the restrictions discussed above:
(P
imφ0
sin(kn z 0 )Im (kn ρ0 ),
ρ0 < ρ
0
mn Amn (x)e
G(x; x ) = P
imφ0
0
0
sin(kn z )Km (kn ρ ),
ρ0 > ρ.
mn Bmn (x)e
Clearly, to establish continuity at ρ0 = 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. Then
we can write G as
X
0
γmk (z, φ)eimφ sin(kz 0 )Im (kρ< )Km (kρ> ).
G(x; x0 ) =
mk
The obvious choice of γmk needed to make this a delta function in z and φ is
γmk = (4/L)e−imφ sin(kz). Then we have
G(x; x0 ) =
4 X im(φ0 −φ)
e
sin(kz) sin(kz 0 )Im (kρ< )Km (kρ> ).
L
mk
What I don’t quite understand is that this expression already has the correct
delta function behavior in ρ, even though I never explicitly required this. To
obtain this expression I first demanded that it satisfy the Laplace equation for
all points x0 6= x, that it satisfy the boundary conditions of the geometry, and
that it have the right delta function behavior in z 0 and φ0 . But I never demanded
that it have the correct delta function behavior in ρ0 , and yet it does. I guess
the combination of the requirements that I did impose on this thing is already
enough to ensure that it meets the final requirement.
(b) The second option is to imagine a plane boundary at z 0 = z, and take the
two distinct regions to be the regions above and below the plane. In other words,
the first region is that for which 0 ≤ z 0 ≤ z, and the second region that for which
z ≤ z 0 ≤ L. In this case, within each region the entire range of ρ0 (from 0 to ∞)
must be handled by one function. This requirement excludes terms of the form
Homer Reid’s Solutions to Jackson Problems: Chapter 3
16
(27), because Km is singular at the origin, while Im is singular at infinity, and
there is no linear combination of these functions that will be finite over the whole
range of ρ0 . Hence we must use terms of the form (26). To ensure finiteness at
the origin we must exlude the Nm term, so D = 0. To ensure vanishing at z 0 = 0
we must take A = −B, so the z 0 function in the region 0 ≤ z 0 ≤ z is proportional
to sinh(kz 0 ). To ensure vanishing at z 0 = L we must take A = −Be−2kL , so
the z 0 function in the region z ≤ z 0 ≤ L is proportional to sinh[k(z 0 − L)].
With these restrictions, the differential equation and the boundary conditions
are satisfied for all terms of the form (25) with no limitation on k. Hence the
Green’s function will be an integral, not a sum, over these terms:
P∞ R ∞
imφ0
sinh(kz 0 )Jm (kρ0 ) dk,
0 ≤ z0 ≤ z
0
m=0 R0 Am (k, ρ, φ, z)e
P∞
G(x ; x) =
0
∞
imφ
sinh[k(z 0 − L)]Jm (kρ0 ) dk,
z ≤ z0 ≤ L
m=0 0 Bm (k, ρ, φ, z)e
Problem 3.18
The configuration of Problem 3.12 is modified by placing a conducting plane held
at zero potential parallel to and a distance L away from the plane with the disc
insert in it. For definiteness put the grounded plane at z = 0 and the other plane
with 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 coordinates (z, ρ, φ) as
Z ∞
sinh(λz/a)
Φ(z, ρ) = V
dλJ1 (λ)J0 (λρ/a)
.
sinh(λL/a)
0
(b) Show that in the limit a → ∞ with z, ρ, L fixed the solution of part a reduces
to the expected result. Viewing your result as the lowest order answer in an
expansion in powers of a−1 , consider the question of corrections to the lowest
order expression if a is large compared to ρ and L, but not infinite. Are there
difficulties? 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 the
results of Problem 3.12 are recovered. What about corrections for L a, but
not L → ∞?
(a) The general solution of the Laplace equation in cylindrical coordinates with
angular symmetry that vanishes at z = 0 is
Z ∞
Φ(ρ, z) =
A(k)J0 (kρ) sinh(kz) dk.
(28)
0
17
Homer Reid’s Solutions to Jackson Problems: Chapter 3
Multiplying both sides by ρJ0 (k 0 ρ) and integrating at z = L yields
Z ∞
Z ∞
Z ∞
0
0
ρJ0 (k ρ)Φ(ρ, L) dρ =
A(k) sinh(kL)
ρJ0 (k ρ)J0 (kρ) dρ dk
0
0
0
Z ∞
1
0
δ(k − k ) dk
A(k) sinh(kL)
=
k
0
1
= 0 A(k 0 ) sinh(k 0 L)
k
so
Z ∞
k
ρJ0 (kρ)Φ(ρ, L) dρ
sinh(kL) 0
Z a
Vk
=
ρJ0 (kρ) dρ
sinh(kL) 0
Z ka
V
uJ0 (u) du.
=
k sinh(kL) 0
A(k) =
(29)
I worked out this integral earlier, in Problem 3.12:
Z x
uJ0 (u) du = xJ1 (x).
0
Then (29) becomes
A(k) =
V
· (ka)J1 (ka)
k sinh(kL)
and (28) is
∞
sinh(kz)
dk
sinh(kL)
Z0 ∞
sinh(λz/a)
dλ.
=V
J1 (λ)J0 (λρ/a)
sinh(λL/a)
0
Φ(ρ, z) = V
(b) For x 1,
Z
aJ1 (ka)J0 (kρ)
(30)
1
J0 (x) → 1 − x2 + · · ·
4
and for x 1 and y 1,
x + 16 x3 + · · ·
x
sinh(x)
1 2
2
=
=
(x
−
y
)
+ O(x4 )
1
+
sinh(y)
y
6
y + 16 y 3 + · · ·
With these approximations we may expand the terms containing a in (30):
!
2 ! 2
sinh(λz/a)
1 λρ
z
1 λ
J0 (λρ/a)
≈ 1−
(x2 − y 2 ) (31)
1+
sinh(λL/a)
4 a
L
6 a
"
#
2 1 2
z
λ
1 2
2
=
(L − z ) + ρ + · · ·
1−
(32)
L
a
6
4
Homer Reid’s Solutions to Jackson Problems: Chapter 3
Then the potential expansion (30)
Z ∞
1
Vz
J1 (λ) dλ − 2
Φ(ρ, z) =
L
a
0
18
becomes
Z ∞
1 2
1 2
2
2
λ J1 (λ) dλ + · · ·
(L − z ) + ρ
6
4
0
The first integral evaluates to 1, so for a infinite the potential becomes simply
Φ(z) = V z/L. This is just what we expect to get for the potential between two
infinite sheets, one grounded and the other at potential V.
The second integral, unfortunately, has a bit of an infinity problem. It’s not
hard to see where the problem comes: I derived the expansion above based on
the premise that λ/a is small, but the integral goes over all λ up to ∞, so for
any finite a the expansions eventually become invalid in the integral.
I’m still trying to work out a better procedure for estimating corrections for
finite a.
(c) In this part we’re interested in taking L → ∞ and looking at the potential
a fixed distance away from the plane with the circular insert. Calling the fixed
distance z 0 , the z coordinate of the point we’re interested in is L − z 0 . We have
sinh k(L − z 0 )
sinh(kL) cosh(−kz 0 ) + cosh(kL) sinh(−kz 0 )
=
sinh kL
sinh kL
= cosh(kz 0 ) − coth(kL) sinh(kz 0 )
(33)
Now, coth(kL) differs significantly from 1 only for kLa . 1, in which region
kz 0 . z/L 1, so cosh(kz 0 ) ≈ 1 and sinh(kz 0 ) ≈ 0. By the time k gets big
enough that kz 0 is starting to get significant, coth(kL) has long since started to
look like 1, so the two terms in (33) add directly. The result is that, for all k,
(33) can be approximated as exp(−kz 0 ). Then (30) becomes
Z ∞
0
Φ(ρ, z) = aV
J1 (ka)J0 (kρ)e−kz dk
0
as we found in Problem 3.12.
Solutions to Problems in Jackson,
Classical Electrodynamics, Third Edition
Homer Reid
August 6, 2000
Chapter 3: Problems 19-27
Problem 3.19
Consider a point charge q between two infinite parallel conducting planes held at
zero potential. Let the planes be located at z = 0 and z = L in a cylindrical
coordinate system, with the charge on the z axis at z = z0 , 0 < z0 < L. Use
Green’s reciprocation theorem of Problem 1.12 with Problem 3.18 as the comparison
problem.
(a) Show that the amount of induced charge on the plate at z = L inside a circle
of radius a whose center is on the z axis is given by
QL (a) = −
q
Φ(z0 , 0)
V
(b) Show that the induced charge density on the upper plate can be written as
Z ∞
sinh(kz0 )
q
dk
σ(ρ) = −
kJ0 (kρ)
2π 0
sinh(kL)
(c) Show that the charge density at ρ = 0 is
σ(0) =
πz −πq
0
sec2
2
8L
2L
(a) Green’s reciprocation theorem says that
Z
Z
Z
Z
σΦ0 dA.
ρΦ0 dV +
σ 0 Φ dA =
ρ0 Φ dV +
V
V
S
1
S
(1)
2
Homer Reid’s Solutions to Jackson Problems: Chapter 3
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,
= V,
z = L and r > a
z = L and r < a
=V
Z
∞
dk aJ1 (ak)J0 (rk)
0
sinh(kz)
sinh(kL)
0<z<L
ρ0 (r, z) = qδ(r)δ(z − z0 )
σ 0 (r, z) =?
Φ0 (r, z) = 0,
=?,
z = 0 or z = L
0≤z≤L
Plugging into (1),
Z ∞
Z
sinh(kz0 )
dk aJ1 (ak)
qV
σ 0 (r, z) dA = 0
+V
sinh(kL)
0
z=L,r<a
so
Z
z=L,r<a
σ 0 (r, z) dA = −q
Z
∞
dk aJ1 (ak)
0
sinh(kz0 )
q
= − Φ(z0 , 0)
sinh(kL)
V
(2)
The integral on the left is just the total surface charge contained within a circle
of 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 the
angular part of the integral right away to give
Z ∞
Z a
sinh(kz0 )
dk aJ1 (ak)
2π
σ 0 (r, L)r dr = −q
sinh(kL)
0
0
Differentiating both sides with respect to a, we have
Z ∞
sinh(kz0 )
∂
2πaσ 0 (a, L) = −q
[aJ1 (ak)]
dk
∂a
sinh(kL)
0
(3)
where I’ve blithely assumed that the partial derivative can be passed through
the integral sign. The partial derivative is
∂
∂
[aJ1 (ak)] =
[xJ1 (x)]
∂a
∂x
x=ak
0
xJ1 (x)|x=ak
= |J1 (x) +
= |xJ0 (x)|x=ak = akJ0 (ak)
3
Homer Reid’s Solutions to Jackson Problems: Chapter 3
so (3) becomes
q
σ (a, L) = −
2π
0
Z
∞
dk kJ0 (ak)
0
sinh(kz0 )
sinh(kL)
(4)
(c) At a = 0, (4) becomes
σ 0 (0, L) =
−q
2π
Z
∞
k
0
sinh(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 coordinates
by 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 can
be written as
!
∞
mπ/β
X
1
mπφ0
1 mπ/β
ρ>
mπφ
0
0
ρ<
sin
G(ρ, φ; ρ , φ ) =
−
−
sin
mπ/β
mπ
β
β
a2mπ/β
ρ>
m=1
Problem 2.25 may be of use.
As before, the procedure for determining the Green’s function is to split the
region of interest into two parts (one on each ’side’ of the observation point), find
separate solutions of the Laplace equation that satisfy the boundary conditions
in each region, and then join the two solutions at the source point such that
their 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 two
subregions, defined by 0 ≤ ρ0 ≤ ρ and ρ ≤ ρ0 ≤ a. The general solution of the
Laplace equation in two-dimensional polar coordinates is
Φ(ρ0 , φ0 ) =A0 + B0 ln ρ0
X
+
ρ0n [An sin nφ0 + Bn cos nφ0 ] + ρ0−n [Cn sin nφ0 + Dn cos nφ0 ].
n
The solution in the first region must be admissible down to ρ0 = 0, which
excludes the ln term and the negative powers of ρ. However, these terms may
be included in the solution for the second region. In both regions, the solution
must vanish at φ = 0, which excludes the cos terms (i.e. Bn = Dn = 0). The
solution 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 two
regions:
4
Homer Reid’s Solutions to Jackson Problems: Chapter 3
G(ρ, φ; ρ0 , φ0 )
=
=
∞
X
Am ρ
m=1
∞ h
X
0mπ/β
Bm ρ
sin
0mπ/β
mπφ0
β
+ Cm ρ
,
0−mπ/β
m=1
i
sin
mπφ0
β
,
0 ≤ ρ0 ≤ ρ
(5)
ρ ≤ ρ0 ≤ a
(6)
The solution in the second region must vanish at ρ0 = a for all φ0 , i.e.
Bm amπ/β + Cm a−mπ/β = 0
so
Bm = γm a−mπ/β
and Cm = −γm amπ/β
where γm can be anything. Then (6) becomes
" mπ/β #
∞
0 mπ/β
X
ρ
mπφ0
a
0
0
G(ρ, φ; ρ , φ ) =
γm
sin
−
a
ρ0
β
m=1
ρ ≤ ρ0 ≤ a.
The solutions in the two regions must agree on the boundary between the two
regions, i.e. at ρ0 = ρ. This determines Am and γm :
#
"
ρ mπ/β a mπ/β
γm = λm ρmπ/β
−
Am = λ m
a
ρ
where λm can be anything. Using these expressions for Am , Bm , and Cm we
can write
G(ρ, φ; ρ0 , φ0 )
"
mπ/β #
mπφ0
a
=
λm
ρ0mπ/β sin
−
a
ρ
β
m=1
#
"
∞
mπ/β
mπ/β
X
a
mπφ0
ρ0
mπ/β
−
=
λm
ρ
sin
a
ρ0
β
m=1
∞
X
ρ mπ/β
0 ≤ ρ0 ≤ ρ
ρ ≤ ρ0 ≤ a.
This may be more succintly written as
G(ρ, φ; ρ0 , φ0 ) =
X
λm fm (ρ; ρ0 ) sin
m
where
0
fm (ρ; ρ ) =
"
ρ mπ/β
>
a
−
a
ρ>
mπφ0
β
mπ/β #
mπ/β
ρ<
(7)
.
Homer Reid’s Solutions to Jackson Problems: Chapter 3
5
The final step is to choose the constant λm in (7) such as to make
∇2 G(ρ, φ; ρ0 , φ0 ) =
1
δ(ρ0 − ρ)δ(φ0 − φ).
ρ
(8)
The Laplacian of (7) is
"
#
2
2
X
d2
1 ∂2
∂
mπ
mπφ0
2
0
0
λm
∇ G=
+ 02 02 G =
fm (ρ; ρ ) −
fm (ρ; ρ ) sin
∂ρ02
ρ ∂φ
dρ02
ρ0 β
β
m
This is equal to (8) if
λm = κ m
and
κm
"
d2
fm (ρ; ρ0 ) −
dρ02
1
sin
β
mπ
ρ0 β
2
mπφ
β
0
(9)
#
fm (ρ; ρ ) =
1
δ(ρ0 − ρ).
ρ
At all points ρ0 6= ρ, the latter condition is already satisfied by f as we constructed it earlier. At ρ0 = ρ, the condition is achieved by choosing κm to
satisfy
ρ0 =ρ+
1
d
= .
κm 0 fm (ρ; ρ0 )
(10)
dρ
ρ
0
ρ =ρ−
Referring to (7), we have
d
fm
dρ0
ρ0 +ρ+
d
fm
dρ0
ρ0 +ρ−
"
mπ/β #
mπ ρ mπ/β
a
=
+
ρmπ/β−1
β
a
ρ
"
mπ/β #
mπ ρ mπ/β
a
=
ρmπ/β−1 .
−
β
a
ρ
(11)
(12)
Subtracting (12) from (11) we obtain
dfm
dρ0
ρ0 =ρ+
=
ρ0 =ρ−
2mπ mπ/β 1
a
· .
β
ρ
Then from (10) we read off
κm =
β −mπ/β
a
2mπ
and plugging this into (9) gives
λm =
1 −mπ/β
a
sin
2mπ
mπ
β
φ.
Plugging this into (7) we obtain finally
"
#
ρ ρ mπ/β ρ mπ/β
X 1
mπφ0
mπφ
< >
<
0
0
G(ρ, φ; ρ , φ ) =
sin
−
sin
2mπ
a2
ρ>
β
β
m
I seem to be off by a factor of 2 here, but I can’t find where.
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Solutions to Problems in Jackson,
Classical Electrodynamics, Third Edition
Homer Reid
October 8, 2000
Chapter 4: Problems 8-13
Problem 4.8
A very long, right circular, cylindrical shell of dielectric constant /0 and inner and
outer 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 cylinder
has a dielectric constant of unity.
(a) Determine the potential and electric fields in the three regions, neglecting end
effects.
(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 dielectric
cylinder 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 to
be aligned with the x axis: E0 = E0 î. Since the cylinder is very long and we’re
told to neglect end effects, we can ignore the z direction altogether and treat
this as a two-dimensional problem.
(a) The general solution of the Laplace equation in two dimensional polar coordinates is
X
Φ(r, ϕ) =
[An rn + Bn r−n ][Cn sin(nϕ) + Dn cos(nϕ)]
For the region inside the shell (r < a), the B coefficients must vanish to keep
the potential from blowing up at the origin. Also, in the region outside the shell
1
Homer Reid’s Solutions to Jackson Problems: Chapter 4
2
(r > b), the only positive power of r in the sum must be that which gives rise
to the external electric field, i.e. −E0 r cos ϕ with An = 0 for n > 1. With these
observations we may write expressions for the potential in the three regions:
Φ(r, ϕ) =









X
X
rn [An sin nϕ + Bn cos nϕ],
rn [Cn sin nϕ + Dn cos nϕ] + r−n [En sin nϕ + Fn cos nϕ],
X
−E0 r cos ϕ +
r−n [Gn sin nϕ + Hn cos ϕ],
r<a
a<r<b
r>b
The normal boundary condition at r = a is
0
∂Φ
∂r
=
x=a−
∂Φ
∂r
x=a+
or
0 X n−1
na
[An sin nϕ + Bn cos nϕ] =
X
nan−1 [Cn sin nϕ + Dn cos nϕ] − na−(n+1) [En sin nϕ + Fn cos nϕ]
From this we obtain two equations:
0
An = Cn − En a−2n
0
Bn = Dn − Fn a−2n
(1)
(2)
Next, the tangential boundary condition at r = a is
∂Φ
∂ϕ
or
X
=
x=a+
nan [An cos nϕ − Bn sin nϕ] =
∂Φ
∂ϕ
X
x=a−
nan [Cn cos nϕ − Dn sin nϕ] + na−n [En cos nϕ − Fn sin nϕ]
from which we obtain two more equations:
An = Cn + En a−2n
(3)
−2n
(4)
Bn = Dn + F n a
Similarly, from the normal boundary condition at r = b we obtain
0 X −(n+1)
0
− E0 cos ϕ −
nb
[Gn sin nϕ + Hn cos ϕ] =
X
nbn−1 [Cn sin nϕ + Dn cos nϕ] − nb−(n+1) [En sin nϕ + Fn cos ϕ]
3
Homer Reid’s Solutions to Jackson Problems: Chapter 4
which leads to
0
Gn = Cn b2n − En
0
0
− b2 E0 δn1 − Hn = Dn b2n − Fn
−
(5)
(6)
Finally, we have the tangential boundary condition at r = b:
X
bE0 sin ϕ +
nb−n [Gn cos nϕ − Hn sin nϕ] =
X
nbn [Cn cos nϕ − Dn sin nϕ] + nb−n [En cos nϕ − Fn sin nϕ]
giving
Gn = Cn b2n + En
2
−b E0 δn1 + Hn = Dn b
2n
+ Fn .
(7)
(8)
The four equations (1), (3), (5), and (7) specify a degenerate system of linear
equations, which can only be satisfied by taking An = Cn = En = Gn = 0 for
all n. Next, for n 6= 1, the system of equations (2), (4), (6), and (8) specify the
same degenerate system of equations, so Bn = Dn = Fn = Gn = 0 for n 6= 0.
However, for n = 1, we have
0
B1 = D1 − F1 a−2
D1 =
1
0 1+
B1
2
F1 =
0 1 2
B1 .
a 1−
2
⇒
B1 = D1 + F1 a−2
and
D1 b 2 − F 1
0
0
H1 = b 2 E 0 + D 1 b 2 + F 1
D1 + 1 −
F1
0 = 2b2 E0 + b2 1 +
0
0
−H1 = b2 E0 +
→
Substituting from above,
−4b2 E0 =
or
B1 =
1 2
b ( + 0 )2 − a2 ( − 0 )2 B1
0
b2 (
−40b2
E0 .
+ 0 )2 − a2 ( − 0 )2
Homer Reid’s Solutions to Jackson Problems: Chapter 4
4
Then
−20 ( + 0 )b2
E0
+ 0 )2 − a2 ( − 0 )2
−20 ( − 0 )a2 b2
E0
F1 = 2
b ( + 0 )2 − a2 ( − 0 )2
−b2 (b2 − a2 )(20 − 2 )
H1 = 2
E0 .
b ( + 0 )2 − a2 ( − 0 )2
D1 =
b2 (
The potential is

−40 b2


· E0 rcos ϕ,


b2 ( + 0 )2 − a2 ( − 0 )2




a2
−20 b2
Φ(r, ϕ) =
E0 cos ϕ,
(
+
)r
+
(
−
)
0
0

b2 ( + 0 )2 − a2 ( − 0 )2
r





−(b2 − a2 )(20 − 2 )
b2


· E0 cos ϕ − E0 rcos ϕ,
2
2
2
2
b ( + 0 ) − a ( − 0 )
r
r<a
a<r<b
As → 0 , Φ → −E0 r cos ϕ in all three regions, which is reassuring.
The electric field is

40 b2


E0 [cos ϕr̂ − sin ϕϕ̂] ,
r<a


2

b ( + 0 )2 − a2 ( − 0 )2




a2
20 b2



(
+
)
−
(
−
)
0
0 2 E0 cos ϕr̂

 b2 ( + 0 )2 − a2 ( − 0 )2
r


2
a
E(r, ϕ) =
−
(
+
)
+
(
−
)
E
sin
ϕ
ϕ̂
,
a<r<b

0
0 2
0

r



2



(b2 − a2 )(20 − 2 )
b


E0 [cos ϕr̂ + sin ϕϕ̂]
− 2
·

2
2
2

b ( + 0 ) − a ( − 0 )
r




+E0 [cos ϕr̂ − sin ϕϕ̂] ,
b < r.
(b) In Figure 4.1 I’ve plotted the field lines for b = 2a, = 50 . Also, as an
appendix to this document I’ve included the C program I wrote to generate this
plot.
(c) For a solid dielectric cylinder in a uniform field, we would have a → 0. In
that case the field would look like
 2
0

r<b

 + 0 E0 î,
2
E(r, ϕ) =

(2 − 2 ) b

 E0 î − 0
E0 [cos ϕr̂ + sin ϕϕ̂],
r>b
( + 0 )2 r
On the other hand, a cylindrical cavity in a uniform dielectric corresponds to
b < r.
Homer Reid’s Solutions to Jackson Problems: Chapter 4
5
Figure 1: Field lines in Problem 4.8 for b = 2a, = 50 .
b → ∞, in which case the field becomes

40


E0 î,

( + 0 )2
E(r, ϕ) =

20
20 ( − 0 ) a 2


E0 î −
E0 [cos ϕr̂ + sin ϕϕ̂],
( + 0 )
( + 0 )2
r
r<a
r > a.
Homer Reid’s Solutions to Jackson Problems: Chapter 4
6
Problem 4.9
A point charge q is located in free space a distance d away from the center of a
dielectric sphere of radius a (a < d) and dielectric constant /0 .
(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 of
the sphere.
(c) Verify that, in the limit /0 → ∞, your result is the same as that for the
conducting sphere.
We will take the origin of coordinates at the center of the sphere, and put
the point charge on the z axis at z = +h. Then the problem has azimuthal
symmetry.
(a) Since there is no free charge within the sphere, ∇·D = 0 there. But since the
permittivity is uniform within the sphere, we may also write ∇·(D/) = ∇·E = 0
there. This means that polarization charge only exists on the surface of the
sphere, so within the sphere the potential satisfies the normal Laplace equation,
whence
X
Al rl Pl (cos θ)
(r < a).
Φ(r, θ) =
l
Now, in the region r > a, the potential may be written as the sum of two
components Φ1 and Φ2 , where Φ1 comes from the polarization charge on the
surface of the sphere, while Φ2 comes from the external point charge. Since
Φ1 satisfies the Laplace equation for r > a, we may expand it in Legendre
polynomials:
X
Φ1 (r, θ) =
Bl r−(l+1) Pl (cos θ)
(r > a).
l
On the other hand, Φ2 is just the potential due to a point charge at z = d:

q X rl


Pl (cos θ),
r<d

4π0
dl+1
Φ2 (r, θ) =
(9)
X dl


 q
Pl (cos θ),
r > d.
4π0
rl+1
Putting this all together we may write the potential in the three regions as
 X

Al rl Pl (cos θ),
r<a





l
 X
r
q
Bl r−(l+1) +
Pl (cos θ),
a<r<d
Φ(r, θ) =
l+1
4π0 d



X

qdl



Bl +
r−(l+1) Pl (cos θ),
r > d.
4π0
Homer Reid’s Solutions to Jackson Problems: Chapter 4
7
The normal boundary condition at r = a is
→
∂Φ
∂r
= 0
r=a−
∂Φ
∂r
r=a+
lqal−1
lAl al−1 = −(l + 1)Bl a−(l+2) +
0
4π0 dl+1
q
0 −(l + 1)
−(2l+1)
Bl a
+
→
Al =
l
4π0 dl+1
(10)
The tangential boundary condition at r = a is
∂Φ
∂θ
→
→
=
r=a−
∂Φ
∂θ
r=a+
q
al
4π0 d(l+1)
q a2l+1
Bl = Al a2l+1 −
4π0 dl+1
Al al = Bl a−(l+1) +
(11)
Combining (10) and (11), we obtain
Al =
Bl =
0
1
+ l+1
l
2l + 1
l
q
4π0 dl+1
0
1
+ l+1
l
1−
0
qa2l+1
4π0 dl+1
In particular, as /0 → ∞ we have
Al → 0
as must happen, since the field within a conducting sphere vanishes; and
Bl → −
qa2l+1
.
4π0 dl+1
(12)
With the coefficients (12), the potential outside the sphere due to the polarization charge at the sphere boundary is
l
1
1 qa X a2
Φ1 (r, θ) =
−
Pl (cos θ).
4π0
d
d
rl+1
Comparing with (9) we see that this is just the potential of a charge −qa/d on
the z axis at z = a2 /d. This is just the size and position of the image charge
we found in Chapter 2 for a point charge outside a conducting sphere.
Homer Reid’s Solutions to Jackson Problems: Chapter 4
8
(b) Near the origin, we have
Φ(r, θ) = A1 rP1 (cos θ) + A2 r2 P2 (cos θ) + · · ·
30
50
q
1
2
2
2
(z − x − y ) + · · ·
=
z+
4π0 d2 ( + 20 )
2 d3 (2 + 30 )
so the field components are
q
50 x +···
·
4π0 d2 2 + 30 d
q
50 y Ey =
+···
·
2
4π0 d 2 + 30 d
q
30
50 z +
·
·
·
Ez = −
+
4π0 d2 + 20
2 + 30 d
Ex =
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 hemispherical shell of dielectric (of dielectric constant /0 ), 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 dielectric
at r = a.
We’ll orient this problem such that the boundary between the dielectricfilled space and the empty space is the xy plane. Then the region occupied
by the dielectric is the region a < r < b, 0 < θ < π/2, and the problem has
azimuthal symmetry.
(a) Since the dielectric has uniform permittivity, all the polarization charge
exists on the boundary of the dielectric, so within its body we may take the
potential to be a solution of the normal Laplace equation. The potential in the
region between the spheres may then be written
 X
π

[Al rl + Bl r−(l+1) ]Pl (cos θ),
0<θ<
2
Φ(r, θ) =
X
π

l
−(l+1)
[Cl r + Dl r
]Pl (cos θ),
<θ<π
2
First let’s apply the boundary conditions at the interface between the dielectric and free space. That region is described by θ = π/2, a < r < b, and we
9
Homer Reid’s Solutions to Jackson Problems: Chapter 4
must have
∂Φ
∂θ
∂Φ
∂r
= 0
θ=π/2+
θ=π/2+
∂Φ
∂θ
∂Φ
=
∂r
θ=π/2−
θ=π/2−
which leads to
X Al − Cl Pl0 (0)rl +
Bl − Dl Pl0 (0)r−l+1 = 0
0
0
X
l−1
l [Al − Cl ] P (0)r
− (l + 1) [Bl − Dl ] Pl (0)r−l+2 = 0.
(13)
(14)
Since these equations must be satisfied for all r in the region a < r < b, the
coefficients of each power of r must vanish identically. In (13), this requirement
is automatically satisfied for l even, since Pl0 (0) vanishes for even l. Similarly,
(14) is automatically satisfied for l odd. For other cases the vanishing of the
coefficients must be brought about by taking
Bl = Dl ,
0
Bl = Dl ,
Al = C l
0
Al = C l
l odd
(15)
l even.
(16)
Next let’s consider the charge at the surface of the inner sphere. There are
actually two components of this charge; one component comes from the surface
distribution of the free charge +Q that exists on the sphere, and the other
component comes from the bound polarization charge on the inner surface of
the dielectric
Problem 4.13
Two long, coaxial, cylindrical conducting surfaces of radii a and b are lowered
vertically into a liquid dielectric. If the liquid rises an average height h between the
electrodes when a potential difference V is established between them, show that the
susceptibility of the liquid is
χe =
(b2 − a2 )ρgh ln(b/a)
0 V 2
where ρ is the density of the liquid, g is the acceleration due to gravity, and the
susceptibility of air is neglected.
First let’s work out what happens when a battery of fixed voltage V is connected between two coaxial conducting cylinders with simple vacuum between
them. To begin, we can use Gauss’ law to determine the E field between the
Homer Reid’s Solutions to Jackson Problems: Chapter 4
10
cylinders. For our Gaussian pillbox we take a disk of thickness dz and radius
r, a < r < b centered on the axis of the cylinders. By symmetry there is no
component of E normal to the top or bottom boundary surfaces, and the component normal to the side surfaces (the radial component) is uniform around
the disc. Hence
I
q
1
E · dA = 2π r dzEρ =
= (2π a dz)σ
0
0
→ Eρ (ρ) =
aσ
0 r
where σ is the surface charge on the inner conductor. This must integrate to
give the correct potential difference between the conductors:
Z b
aσ b
ln
V =−
Eρ (ρ)dρ = −
0
a
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=
0 V
a ln(b/a)
(17)
on the cylinder faces (the surface charges are of opposite sign on the two cylinders).
It is useful to figure out the energy per unit length stored in the electric field
between the cylinder plates here. This is just
Z Z
1 b 2π
Wv =
E · D ρ dρ dφ
2 a 0
Z b
= π0
E 2 (ρ)ρ dρ
a
a2 σ 2
=π
ln(b/a)
0
π0 V 2
=
ln(b/a)
(18)
where the v subscript stands for ’vacuum’, since (18) is the energy per unit length
stored in the field between the cylinders with just vacuum between them.
Now suppose we introduce a dielectric material between the cylinders. If the
voltage between the cylinders is kept at V , then the E field must be just the
same as it was in the no-dielectric case, because this field integrated from a to
b must still give the same potential difference. However, in order to establish
this same E field in the presence of the retarding effects of the dielectric, the
battery now has to establish a surface charge that is greater that it was before
by a factor (/0 ). With this greater charge on the electrodes, the D field will
now be bigger by a factor (/0 ) than it was in our above calculation. So the
Homer Reid’s Solutions to Jackson Problems: Chapter 4
11
energy per unit length stored in the field between the cylinders increases by a
factor (/0 − 1) over the result (18):
∆Wd = ( − 0 )
πV 2
.
ln(b/a)
On the other hand, to get to this point the battery has had to flow enough
charge to increase the surface charges to be of magnitude (/0 ) times greater
than (17). In doing this the internal energy of the battery decreases by an
amount equal to the work it had to do to flow the excess charge, namely
∆Wb = −V dQ = V (2π a dσ) = ( − 0 )
2πV 2
ln(b/a)
(per unit length). The energy lost by the battery is twice that gained by the
dielectric, so the system with dielectric between the cylinders has lower overall
energy than the system with vacuum between the cylinders by a factor
∆W = ( − 0 )
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 xy
plane. We’ll take the boundary between the liquid and the air above it to be at
z = 0. With no potential between the cylinder plates, the liquid between the
cylinders is at the same height as the liquid outside.
Now suppose a battery of fixed potential V is connected between the two
cylinder plates. As we showed earlier, the combined system of battery and dielectric can lower its energy by having more of the dielectric rise up between the
cylinders. However, at some point the energy win we get from this is balanced
by the energy hit we take from the gravitational potential energy of having
the excess liquid rise higher between the cylinders. The height at which we no
longer gain by having more liquid between the cylinders is the height to which
the 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 the
liquid outside the electrodes. The decrease in electrostatic energy this affords
over the case with just vacuum filling that space is just (19) times the height,
i.e.
πV 2
Ee = −h( − 0 )
(20)
ln(b/a)
This must be balanced by the gravitational potential energy Eg of the excess
liquid. Eg is easily calculated by noting that the area between the cylinders is
π(b2 − a2 ), so the mass of liquid contained in a height dh between the cylinders
is dm = ρπ(b2 − a2 )dh, and if this mass is at a height h above the liquid surface
its excess gravitational energy is
dEg = (dm)gh = πgρ(b2 − a2 )hdh.
Homer Reid’s Solutions to Jackson Problems: Chapter 4
12
Integrating over the excess height of liquid between the cylinders,
Eg = πgρ(b2 − a2 )
Z
h
h0 dh0 =
0
1
πgρ(b2 − a2 )h2 .
2
(21)
Comparing (20) to (21), we find that the gravitational penalty of the excess
liquid just counterbalances the electrostatic energy reduction when
2( − 0 )V 2
ρg(b2 − a2 ) ln(b/a)
2χe 0 V 2
=
ρg(b2 − a2 ) ln(b/a)
h=
Solving for χe ,
χe =
ρgh(b2 − a2 ) ln(b/a)
.
20 V 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 must
fall, since the total volume of the liquid is conserved. Hence there are really
two other contributions to the energy shift, namely, the change in gravitational
and electrostatic energies of the thin layer of liquid outside the cylinders that
falls away when the liquid rises between the cylinders. But if the surface area of
the vessel containing the liquid is sufficiently larger than the area between the
cylinders, the difference layer will be thin and its energy shifts negligible.
13
Homer Reid’s Solutions to Jackson Problems: Chapter 4
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
#define DELTAY (4.0 * B) / NUMLINES
/* horiz spacing of pts */
/* 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);
}
Homer Reid’s Solutions to Jackson Problems: Chapter 4
14
/*
* 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));
Homer Reid’s Solutions to Jackson Problems: Chapter 4
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);
15
Homer Reid’s Solutions to Jackson Problems: Chapter 4
};
fprintf(g,"e\n");
};
printf("Thank you for your support.\n");
}
16
Solutions to Problems in Jackson,
Classical Electrodynamics, Third Edition
Homer Reid
November 8, 2000
Chapter 5: Problems 1-10
Problem 5.1
Starting with the differential expression
dB =
x − x0
µ0 I 0
dl ×
4π
|x − x0 |3
for the magnetic induction at the point P with coordinate x produced by an increment of current I dl0 at x0 , show explicitly that for a closed loop carrying a current
I the magnetic induction at P is
B=
µ0 I
∇Ω
4π
where Ω is the solid angle subtended by the loop at the point P . This corresponds
to a magnetic scalar potential, ΦM = −µ0 IΩ/4π. The sign convention for the
solid angle is that Ω is positive if the point P views the “inner” side of the surface
spanning the loop, that is, if a unit normal n to the surface is defined by the
direction of current flow via the right-hand rule, Ω is positive if n points away from
the point P , and negative otherwise. This is the same convention as in Section 1.6
for the electric dipole layer.
I like to change the notation slightly: the observation point is r1 , the coordinate of a point on the current loop is r2 , and the displacement vector (pointing
to the observation point) is r12 = r1 − r2 .
The solid angle subtended by the current loop at r1 is given by a surface
integral over the loop:
Z
cos γ dA
Ω=
2
r12
S
1
Solutions to Problems in Jackson,
Classical Electrodynamics, Third Edition
Homer Reid
February 11, 2001
Chapter 5: Problems 10-18
Problem 5.10
A circular current loop of radius a carrying a current I lies in the x − y plane with
its center at the origin.
(a) Show that the only nonvanishing component of the vector potential is
Z
µ0 Ia ∞
Aφ (ρ, z) =
dk cos kz I1 (kρ< )K1 (kρ> )
π
0
where ρ< (ρ> ) is the smaller (larger) of a and ρ.
(b) Show that an alternative expression for Aφ is
Z
µ0 Ia ∞
dke−k|z| J1 (ka)J1 (kρ).
Aφ (ρ, z) =
2
0
(c) Write down integral expressions for the components of magnetic induction,
using the expressions of parts a and b. Evaluate explicitly the components of
B 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 coordinates are (ρ, φ = 0, z). Since there is no current in the z direction, and since the
1
Homer Reid’s Solutions to Jackson Problems: Chapter 5
2
current density is cylindrically symmetric, there is no vector potential in the ρ
or z directions. In the φ direction we have
Aφ = −Ax sin φ + Ay cos φ = Ay
Z
µ0
Jy (x0 )
=
dx0
4π
|x − x0 |
Z
µ0
Jφ (x0 ) cos φ0 0
=
dx
4π
|x − x0 |
Z
0
Jφ (x0 )eiφ
µ0
Re
dx0
=
4π
|x − x0 |
#
"
Z
Z ∞
∞
X
0
0
µ0
2
eim(φ−φ ) cos[k(z − z 0 )]Im (kρ< )Km (kρ> ) dk dx0
Re
Jφ (x0 )eiφ
=
4π
π m=−∞ 0
where we substituted in Jackson’s equation (3.148). Rearranging the order of
integration and remembering that φ = 0, we have
Z ∞ Z
∞
X
µ0
0 i(1−m)φ0
0
0
Aφ =
Re
Jφ (x )e
cos[k(z − z )]Im (kρ< )Km (kρ> )dx dk
2π 2
m=−∞ 0
If m = 1, the φ integral yields 2π; otherwise it vanishes. Thus
Z Z ∞ Z ∞
µ0 ∞
Aφ =
Jφ (r0 , z 0 ) cos[k(z − z 0 )]I1 (kρ< )K1 (kρ> )ρ0 dz 0 dr0 dk
π 0
0
−∞
Substituting (1), we have
Aφ =
Iaµ0
π
Z
∞
cos kz I1 (kρ< )K1 (kρ> ) dk.
0
(b) The procedure for obtaining this expression is identical to the one I just
went through, but with the expression from Problem 3.16(b) used for the Green’s
function instead of equation (3.148).
(c) Let’s suppose that the observation point is in the interior region of the
current loop, so ρ< = ρ, ρ> = a. Then
∂Aφ
Bρ = [∇ × A]ρ = −
∂z
Z
Iaµ0 ∞
=−
k sin kz I1 (kρ)K1 (ka) dk
π
0
1
∂Aφ
Bz = [∇ × A]z = Aφ +
ρ
∂ρ
Z
Iaµ0 ∞
I1 (kρ)
0
+ kI1 (kρ) K1 (ka) dk
=
cos kz
π
ρ
0
3
Homer Reid’s Solutions to Jackson Problems: Chapter 5
As ρ = 0, I1 (ρ) → 0, I1 (ρ)/ρ → 1/2, and I10 (ρ) → 1/2, so
Bρ (ρ = 0) = 0
Z
Iaµ0 ∞
k cos kzK1 (ka) dk
Bz (ρ = 0) =
π
0
Z ∞
Iaµ0 ∂
=
sin kzK1 (ka)dk
π ∂z 0
The integral may be done by parts:
Z ∞
1
sin kzK1 (kz) dk = − sin kzK0 (ka)
a
0
∞
+
0
z
a
Z
∞
cos kzK0 (ka) dk
0
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 is
Jackson’s equation (3.150). Plugging it in to the above,
z
Iµ0 ∂
2
2 ∂z (z + a2 )1/2
a2
Iµ0
=
.
2 (z 2 + a2 )3/2
Bz (ρ = 0) =
Problem 5.11
A circular loop of wire carrying a current I is located with its center at the origin
of coordinates and the normal to its plane having spherical angles θ0 , φ0 . There is
an 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 higher
order contributions? Do they vanish for the circular loop? What about for
other 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 the
current loop has angles θ0 , φ0 . We define the “rotated” frame R0 by aligning
the z 0 axis with the normal to the current loop, so that in R0 the current loop
exists only in the x0 y 0 plane, but the magnetic field now has a z 0 component.
The force on the current loop is
Z
F = (J × B)dV.
(2)
4
Homer Reid’s Solutions to Jackson Problems: Chapter 5
PSfrag replacements
z1 = z
z1
z0
θ0
y 0 = y1
y1
x
φ0
x1
y
x1
x0
R1 → R 0
R → R1
Figure 1: Successive coordinate transformations in Problem 5.11.
The components of J are easy to express in R0 , but more complicated in R; the
opposite is true for B. There are two ways to do the problem: we can work out
the components of J in R and do the integral in R, or we can work out the
components of B in R0 and do the integral in R0 , in which case we would have
to 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 R
and R0 , I imagined that the transformation arose from two separate transformations, as depicted in figure (??). The first transformation is a rotation through
φ0 around the z axis, which takes us from R to an intermediate frame R1 . Then
we rotate through θ0 around the y1 axis, which takes us to R0 . Evidently, the
coordinates of a point in the various frames are related by

 


x1
cos φ0
sin φ0 0
x
 y1  =  − sin φ0 cos φ0 0   y 
(3)
z1
0
0
1
z


 0  
x1
cos θ0 0 − sin θ0
x
  y1 
 y0  =  0
1
0
(4)
z1
sin θ0 0 cos θ0
z0
Multiplying matrices,
 0  
cos θ0 cos φ0
x
 y 0  =  − sin φ0
sin θ0 cos φ0
z0
cos θ0 sin φ0
cos φ0
sin θ0 sin φ0


x
− sin θ0
 y .
0
z
cos θ0
(5)
This matrix also gives us the transformation between unit vectors in the two
5
Homer Reid’s Solutions to Jackson Problems: Chapter 5
frames:
 
î0
cos θ0 cos φ0
 ĵ0  =  − sin φ0
sin θ0 cos φ0
k̂0

cos θ0 sin φ0
cos φ0
sin θ0 sin φ0


î
− sin θ0
  ĵ  .
0
cos θ0
k̂
(6)
We will also the inverse transformation, i.e. the expressions for coordinates in
R in terms of coordinates in R0 :

 
 0 
x
cos θ0 cos φ0 − sin φ0 sin θ0 cos φ0
x
 y  =  cos θ0 sin φ0
cos φ0
sin θ0 sin φ0   y 0  .
(7)
z
− sin θ0
0
cos θ0
z0
To do the integral in (2) it’s convenient to parameterize a point on the
current loop by an angle φ0 reckoned from the x0 axis in R0 . If the loop radius
is a, then the coordinates of a point on the loop are x0 = a cos φ0 , y 0 = a sin φ0 ,
and the current density/volume element product is
J dV = Id l = (Ia dφ0 )φ̂0
= Ia dφ0 [− sin φ0 î0 + cos φ0 ĵ0 ]
h
= Ia dφ0 (− sin φ0 cos θ0 cos φ0 − cos φ0 sin φ0 )î
+ (sin φ0 sin φ0 + cos φ0 cos φ0 )ĵ + (sin φ0 sin θ0 )k̂
i
We also need the components of the B field at a point on the current loop:
B(φ0 ) = B0 [1 + βy(φ0 )]î + B0 [1 + βx(φ0 )]
= B0 [1 + aβ(cos φ0 cos θ0 sin φ0 + sin φ0 cos φ0 )]î + B0 [1 + aβ(cos φ0 cos θ0 cos φ0 − sin φ0 sin φ0 )]ĵ
The components of the cross product are
[J × B]x dV = −Jz By dV
= (· · · )βIa2 B0 dφ0 sin2 φ0 sin θ0 sin φ0
[J × B]y dV = Jz Bx dV
= (· · · ) + βIa2 B0 dφ0 sin2 φ0 sin θ0 cos φ0
[J × B]z dV = (Jx By − Jy Bx ) dV
= (· · · ) + 0
where we only wrote out terms containing a factor of cos2 φ0 or sin2 φ0 , since only
these terms survive after the integral around the current loop (we grouped all
the remaining terms into (· · · )). In the surviving terms, cos2 φ0 and sin2 φ0 turn
into factors of π after the integral around the loop. Then the force components
are
Fx = πβIa2 B0 sin θ0 sin φ0
Fy = πβIa2 B0 sin θ0 cos φ0
Fz = 0.
Homer Reid’s Solutions to Jackson Problems: Chapter 5
6
To compare this with the first-order approximate result, note that the magnetic
moment has magnitude πa2 I and is oriented along the z 0 axis:
m = πa2 I k̂0 = πa2 I sin θ0 cos φ0 î + sin θ0 sin φ0 ĵ + cos θ0 k̂
so
∇ B · m = ∇ B0 (1 + βy)mx + B0 (1 + βx)my
= B0 β my î + mx ĵ
= πβIa2 B0 sin θ0 sin φ0 î + sin θ0 cos φ0 ĵ)
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 0 , respectively (b < a),
have an angle α between their planes. Show that the torque on one of the loops
is about the line of intersection of the two planes containing the loops and has the
magnitude
2 2n
∞
µ0 πII 0 b2 X (n + 1)
Γ(n + 3/2)
b
1
N=
P2n+1
(cos α).
2a
(2n
+
1)
Γ(n
+
2)Γ(3/2)
a
n=0
The torque on the smaller loop is
Z
N = r × Jb (r) × Ba (r) dr
Z n
o
=
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 field
of the larger loop. But r · Jb vanishes, because the current flows in a circle
around the origin—there is no current flowing toward or away from the origin.
Thus
Z
N = rBr (r)Jb (r)dr
(8)
where Br is the radial component of the magnetic field of the larger current
loop.
As in the last problem, it’s convenient to define two reference frames for this
situation. Let R be the frame in which the smaller loop (radius b, current I)
lies in the xy plane, and R0 the frame in which the larger loop lies in the x0 y 0
plane. We might as well take the line of intersection of the two planes to be the
y axis, so y = y 0 . Then the z 0 axis has spherical coordinates (θ = α, φ = 0) in
Homer Reid’s Solutions to Jackson Problems: Chapter 5
7
R, and for transforming back and forth between the two frames we may use the
transformation matrices we derived in the last problem, with θ0 = α, φ0 = 0. If
we choose to evaluate the integral (8) in frame R, the current density is
Jb (r) = Iδ(r − b)δ(θ − π/2) − sin φî + cos φĵ
so the components of the torque are
Z 2π
2
Br (r = b, θ = π/2, φ) sin φ dφ
Nx = −Ib
Ny = Ib2
Z
(9)
0
2π
Br (r = b, θ = π/2, φ) cos φ dφ
(10)
0
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 the
field in R0 : in that frame the field is just that of a circular current loop in the
x0 y 0 plane, Jackson’s equation (5.48):
∞
Br0 (r0 , θ0 ) =
2l+1
µ0 I 0 a X (−1)l (2l + 1)!! r<
P
(cos θ0 ).
2l+2 2l+1
2r0
2l l!
r>
l=0
We are interested in evaluating this field at points along the smaller current
loop, and for all such points r = b; then r< = b, r> = a and we have
Br0 (r0 = b, θ0 ) =
2l
∞
µ0 I 0 X (−1)l (2l + 1)!! b
P2l+1 (cos θ0 ).
2a
2l l!
a
(11)
l=0
To transform this to frame R, we first note that, since the origins of R and R0
coincide, the unit vectors r̂ and r̂0 coincide, so Br = Br0 . Next, (11) expresses
the field in terms of cos θ 0 , the polar angle in frame R0 . How do we write this
in terms of the angles θ and φ in frame R? Well, note that
z0
r
x sin α + z cos α
=
r
r sin θ cos φ sin α + r cos θ cos α
=
r
= sin θ sin α cos φ + cos θ cos α
cos θ0 =
(12)
where in the second line we used the transformation matrix from Problem 5.11
to write down z 0 in terms of x and z. Equation (12) is telling us what our
coordinates in R0 are in terms of our coordinates in R; if a point has angular
coordinates θ, φ in R, then (12) tells us what angle θ 0 it has in R0 . (We could
also work out what the azimuthal angle φ0 would be, but we don’t need to,
because (11) doesn’t depend on φ0 .)
Homer Reid’s Solutions to Jackson Problems: Chapter 5
8
To express the Legendre function in (11) with the argument (12), we may
make use of the addition theorem for associated Legendre polynomials:
Pl (cos θ0 ) = Pl (cos θ cos α + sin θ sin α cos φ)
= Pl (cos θ)Pl (cos α) + 2
l
X
Plm (cos θ)Plm (cos α) cos mφ.
m=1
Of course, the smaller loop exists in the xy plane, so for all points on that loop
we have θ = π/2, whence
Pl (cos θ0 ) = Pl (0)Pl (cos α) + 2
l
X
Plm (0)Plm (cos θ) cos mφ.
m=1
We may now write down an expression for the radial component of the magnetic
field of the larger loop, evaluated at points on the smaller loop, in terms of the
angle φ that goes from 0 to 2π around that loop:
2l (
∞
µ0 I 0 X (−1)l (2l + 1)!! b
P2l+1 (0)P2l+1 (cos α)
Br (φ) =
2a
2l l!
a
l=0
)
2l+1
X
m
m
+2
P2l+1
(0)P2l+1
(cos α) cos mφ .
m=1
This looks ugly, but in fact when we plug it into the integrals (9) and (10)
the sin φ and cos φ terms beat against the cos mφ term, integrating to 0 in the
former case and πδm1 in the latter. The torque is
Nx = 0
2l
∞
πµ0 II 0 b2 X (−1)l (2l + 1)!! b
1
1
Ny =
P2l+1
(0)P2l+1
(cos α).
a
2l l!
a
l=0
To finish we just need to rewrite the numerical factor under the sum:
(−1)l (2l + 1)!! 1
(2l + 1)!!
Γ(l + 3/2)
P2l+1 (0) =
2l l!
2l l!
Γ(l + 1)Γ(3/2)
(2l + 3 − 2)(2l + 3 − 4)(2l + 3 − 6) · · · (5)(3)
Γ(l + 3/2)
=
2l Γ(l + 1)
Γ(l + 1)Γ(3/2)
Γ(l + 3/2)
(l + 3/2 − 1)(l + 3/2 − 2) · · · (5/2)(3/2)
=
Γ(l + 1)
Γ(l + 1)Γ(3/2)
2
Γ(l + 3/2)
=
Γ(l + 1)Γ(3/2)
2
Γ(l + 3/2)
2
= (l + 1)
Γ(l + 2)Γ(3/2)
Homer Reid’s Solutions to Jackson Problems: Chapter 5
9
So my answer is
2 2l
∞
b
Γ(l + 3/2)
πµ0 II 0 b2 X
1
2
P2l+1
(cos α).
(l + 1)
Ny =
a
Γ(l + 2)Γ(3/2)
a
l=0
Evidently I’m off by a factor of 1/(l + 1)(2l + 1) under the sum, but I can’t find
where. Can anybody help?
Problem 5.13
A sphere of radius a carries a uniform surface-charge distribution σ. The sphere
is rotated about a diameter with constant angular velocity ω. Find the vector
potential 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 relative
permeability 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 sketch
the logarithm of the ratio of the magnitudes of B on the cylinder axis to B0 as a
function 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’ll
take B0 along the x axis: B0 = B0 î. To the extent that we ignore end effects,
we may imagine the fields to have no z dependence, so we effectively have a two
dimensional problem.
There are two distinct current distributions in this problem. The first is
a current distribution Jfree giving rise to the uniform field B0 far away from
the cylinder; this current distribution is only nonvanishing at points outside the
cylinder. The second is a current distribution Jbound = ∇ × M existing only
within the cylinder. Since there is no free current within the cylinder or in its
inner 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 H
from 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. To
proceed we may separate the H field in the external region into two components: one that arises from the free current, and one that arises from the bound
currents within the cylinder. The former is just (1/µ0 )B0 and the second is
again derivable from a scalar potential satisfying the Laplace equation. So, in
the external region, H = (1/µ0 )B0 − ∇Φm .
10
Homer Reid’s Solutions to Jackson Problems: Chapter 5
So our task is to find expressions for Φm in the three regions such that the
boundary 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
P
∞
n


n cos nφ + Bn sin nφ
Pn=1 ρ
n A
o
∞
Φm (ρ, φ) =
ρn Cn cos nφ + Dn sin nφ + ρ−n En cos nφ + Fn sin nφ
n=1


P∞ ρ−n G cos nφ + H sin nφ
n
n
n=1
r<a
a<r<b
r>b
Actually, we may argue on symmetry grounds that the sin terms must all
vanish: otherwise, the fields would take different values on the positive and
negative y axes, but there is nothing in the problem distinguishing these axes
from each other. With this simplification we may write down expressions for
the components of the H field in the three regions:
Hr =












−
−
∞
X
∂
Φm =
−nAn ρn−1 cos nφ,
∂r
n=1
∞
X
∂
Φm =
−n Cn ρn−1 − En ρ−(n+1) cos nφ,
∂r
n=1
r<a
a<r<b





∞
i
h

X

∂

−(n+1)

nG
ρ
cos
nφ
,
Φ
=
(1/µ
)B
cos
φ
+
(1/µ
)B
−

n
m
0
0
0
0r

∂r
n=1
Hφ =












r < b.
∞
X
∂
− Φm =
nAn ρn−1 sin nφ,
∂φ
n=1
−
∞
X
∂
Φm =
n Cn ρn−1 + En ρ−(n+1) sin nφ,
∂φ
n=1
r<a
a<r<b





∞

h
i
X

∂

−(n+1)

(1/µ
)B
−
Φ
=
−
(1/µ
)B
sin
φ
+
nG
ρ
sin
nφ
,

0
0φ
m
0
0
n

∂φ
n=1
The boundary conditions at r = b are that µHρ and Hφ be continuous,
where µ = µ0 outside the cylinder and µr µ0 inside. With the above expressions
for the components of H, we have
∞
∞
X
X
1
B0 cos φ +
nGn b−(n+1) cos nφ = µr
−n Cn bn−1 − En b−(n+1) cos nφ
µ0
n=1
n=1
−
∞
∞
X
X
1
B0 sin φ +
nGn b−(n+1) sin nφ =
n Cn bn−1 + En b−(n+1) sin nφ.
µ0
n=1
n=1
We may multiply both sides of these by cos nφ and sin nφ and integrate from
r < b.
11
Homer Reid’s Solutions to Jackson Problems: Chapter 5
0 to 2π to find
−
1
B0 + G1 b−2 = −µr C1 + µr E1 b−2
µ0
Gn b−(n+1) = −µr Cn bn−1 − En b−(n−1) ,
1
B0 + G1 b−2 = C1 + E1 b−2
µ0
Gn b−(n+1) = Cn bn−1 + En b−(n+1) ,
(13)
n 6= 1
(14)
(15)
n 6= 1
(16)
Similarly, at r = a we obtain
A1 = µr C1 − µr E1 a−2
An an−1 = µr Cn an−1 − En a−(n+1) ,
A1 = C1 + E1 a−2
An an−1 = Cn an−1 + En a−(n+1) ,
(17)
n 6= 1
(18)
n 6= 1.
(19)
For n 6= 1, the only solution turns out to be An = Cn = En = Gn = 0. For
n = 1, multiplying (15) by µr and adding and subtracting with (13) yields
2µr C1 = −(µr + 1)
2µr E1 = (1 − µr )
B0
+ (µr − 1)G1 b−2
µ0
B0 2
b + (µr + 1)G1 .
µ0
(20)
(21)
On the other hand, multiplying (18) by µr and adding and subtracting with
(17) yields
2µr C1 = (µr + 1)A1
2
2µr E1 = (µr − 1)a A1 .
Equating (20) with (22), we find
A1 = −
B0
(µr − 1)
+
G1 b−2
µ0
(µr + 1)
while equating (21) with (23) yields
(µr + 1)
B 0 b2
+
A1 = −
G1 a−2
µ0 a 2
(µr − 1)
and now equating these two equations gives
a 2 (µ2r − 1)b2
B0
G1 = 1 −
b2 .
2
2
2
2
b
(µr + 1) b − (µr − 1) a
µ0
(22)
(23)
frag replacements
12
Homer Reid’s Solutions to Jackson Problems: Chapter 5
0
(a/b) = 0.5
(a/b) = 0.1
-0.5
-1
log10 r
-1.5
-2
-2.5
-3
-3.5
-4
-4.5
0
1
2
3
log10 µr
4
5
Figure 2: Damping of field inside cylindrical cylinder of permeability µr .
The other coefficients may be worked out from this one:
−4µr b2
B0
2
2
2
2
(µr + 1) b − (µr − 1) a µ0
B0
−2(µr + 1)b2
C1 =
2
2
2
2
(µr + 1) b − (µr − 1) a µ0
−2(µr − 1)b2
B0 2
E1 =
a .
2
2
2
2
(µr + 1) b − (µr − 1) a µ0
A1 =
The H field is
4µr b2
B0
î,
r<a
(µr +
− (µr − 1)2 a2 µ0
a 2 o
a 2
2b2
B 0 n
(µr + 1) + (µr − 1)
=
cos φr̂ , a < r < b
î − 2(µr − 1)
2
2
2
2
(µr + 1) b − (µr − 1) a µ0
r
r
2
2
2
2
(b − a )(µr − 1)
b
B0
B0
î +
î + 2 sin φ φ̂ ,
r > b.
=
µ
(µr + 1)2 b2 − (µr − 1)2 a2 µ0
r2
H=
1)2 b2
The ratio r of the field within the cylinder to the external field is
r=
(µr +
1)2
This relationship is graphed in Figure
4µr
2 .
− (µr − 1)2 ab2
Homer Reid’s Solutions to Jackson Problems: Chapter 5
13
Problem 5.16
A circular loop of wire of radius a and negligible thickness carries a current I. The
loop 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 of
the medium in the cavity, unity.
(a) In the approximation of b a, show that the magnetic field at the center of
the 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 the
iron. These give rise to two fields B1 and B2 , which must be summed at each
point in space to get the observed field.
B1 is just the field of a planar current loop, which Jackson has already
worked out for us in his section 5.5:

∞
µ0 I X (−1)n (2n + 1)!! r 2n



P2n+1 (cos θ), r < a


2a n=0
2n n!
a
(24)
B1r =
∞

µ0 Ia2 X (−1)n (2n + 1)!! a 2n



P2n+1 (cos θ), r > a.
 2r3
2n n!
r
n=0
B1θ =






∞
µ0 I X (−1)n (2n − 1)!! r 2n 1
P2n+1 (cos θ), r < a
4a n=0
2n−1 n!
a
∞

µ0 Ia2 X (−1)n (2n + 1)!! a 2n 1



P2n+1 (cos θ), r > a.
 − 4r3
2n (n + 1)!
r
n=0
(25)
On the other hand, since J2 vanishes for r < b, the field B2 to which it gives
rise has no divergence or curl in that region, which means that throughout the
region it may be derived from a scalar potential satisfying the Laplace equation:
#
"∞
X
An rn Pn (cos θ)
B2 = −∇Φm = −∇
n=0
→ B2r =
B2θ =
∞
X
n=1
∞
X
n=1
nAn rn−1 Pn (cos θ)
(26)
An rn−1 Pn1 (cos θ)
(27)
Homer Reid’s Solutions to Jackson Problems: Chapter 5
14
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 strictly
radial at the boundary r = b. The An coefficients are thus determined by the
requirement that (27) and (25) sum to zero at r = b:
∞
∞
X
µ0 Ia2 X (−1)n (2n + 1)!! a 2n 1
P2n+1 (cos θ).
An bn−1 Pn1 (cos θ) =
4b3 n=0 2n (n + 1)!
b
n=1
The orthogonality of the associated Legendre polynomials requires that each
term in the sum cancel individually, whence
A2n = 0
A2n+1 =
µ0 Ia2 (−1)n (2n + 1)!! a 2n
.
4b3
2n (n + 1)!
b2
Then the field of the bound current in the iron is determined everywhere in the
region r < b:
∞
µ0 Ia2 X (−1)n (2n + 1)(2n + 1)!! ar 2n
P2n+1 (cos θ)
(28)
B2r =
4b3 n=0
2n (n + 1)!
b2
B2θ
∞
µ0 Ia2 X (−1)n (2n + 1)!! ar 2n 1
=
P2n+1 (cos θ).
4b3 n=0 2n (n + 1)!
b2
(29)
As r → 0, B2θ → 0 and B2r → µ0 Ia2 /4b3 , while B1r → µ0 I/2a, so the total
field at r = 0 is
µ0 Ia2
µ0 I
a3
µ0 I
+
=
1+ 3 .
Br (r = 0) = B1r (r = 0) + B2r (r = 0) =
2a
4b3
2a
2b
(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 made
to 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 in
vacuum with its center a distance d away from a semi-infinite slab of permeability
µ. 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 to
occupy the space z < 0, so that the boundary surface is z = 0.
15
Homer Reid’s Solutions to Jackson Problems: Chapter 5
In the region z < 0, there is no free current, so ∇ × H = 0 everywhere; thus
H may be obtained from a scalar potential, H = −∇Φm , and since ∇ · H = 0
as well we have ∇2 Φm = 0. The azimuthally symmetric solution of the Laplace
equation in cylindrical coordinates that remains finite as z → −∞ is
Z ∞
dk A(k)ekz J0 (kρ),
(30)
Φm (z < 0) =
0
and from this we obtain
Z ∞
∂
Hρ (z < 0) = − Φm = −
dk kA(k)ekz J00 (kρ)
∂ρ
0
Z ∞
=
dk kA(k)ekz J1 (kρ)
0
Z ∞
∂
dk kA(k)ekz J0 (kρ).
Hz (z < 0) = − Φm = −
∂z
0
(31)
(32)
On the other hand, for z > 0 we may decompose the H field into two
components: one component H1 arising from the current loop, and a second
component H2 arising from the bound currents running in the slab. H1 is just
given by the curl of the vector potential we worked out in Problem 5.10:
Z

µ0 Ia ∞


dk e−k(z−d) J1 (ka)J1 (kρ),
 2
1
0
∇×A,
A = Aφ φ̂,
Aφ =
H1 =
Z

µ0
µ Ia ∞

 0
dk e−k(d−z) J1 (ka)J1 (kρ),
2
0
z>d
z < d.
so
H1ρ = −
1 ∂
Aφ
µ0 ∂z
Z
Ia ∞
dk ke−k(z−d) J1 (ka)J1 (kρ),
2 0
=
Z

Ia ∞

−
dk ke−k(d−z) J1 (ka)J1 (kρ),
2 0




H1z
1 1 ∂
=
(ρAφ )
µ0 ρ ∂ρ
z>d
z < d.
(33)
Z

1
Ia ∞
−k(z−d)


dk ke
J1 (ka)
J1 (kρ) − J0 (kρ)
z>d
 2
kρ
0
=
Z ∞

1
Ia


J1 (kρ) − J0 (kρ) ,
z < d.
dk ke−k(d−z) J1 (ka)
2 0
kρ
(34)
In the last two equations we may use Jackson’s identity (3.87),
1
1
J1 (kρ) = [J0 (kρ) + J2 (kρ)]
kρ
2
16
Homer Reid’s Solutions to Jackson Problems: Chapter 5
to rewrite H1z as
Z

Ia ∞


dk e−k(z−d) J1 (ka) [J2 (kρ) − J0 (kρ)] , z > d
 4
H1z =
Z0

Ia ∞


dk e−k(d−z) J1 (ka) [J2 (kρ) − J0 (kρ)] , z < d.
4 0
(35)
Since the H2 field arises entirely from bound currents, it may also be derived
from a scalar potential Φm satisfying the Laplace equation. The azimuthally
symmetric solution of the Laplace equation in cylindrical coordinates that remains finite for all ρ and as z → +∞ is
Z ∞
Φm (z > 0) =
dk B(k)e−kz J0 (kρ)
0
and the components of H2 are
Z ∞
H2r (z > 0) = −
dk kB(k)e−kz J1 (kρ)
0
Z ∞
dk kB(k)e−kz J0 (kρ).
H2z (z > 0) =
(36)
(37)
0
The required forms of the functions A(k) and B(k) are determined by the
boundary conditions on H at the medium boundary, z = 0:
Hρ (z = 0− ) = Hρ (z = 0+ )
µHρ (z = 0− ) = µ0 Hρ (z = 0+ ).
Equating (32) with the sum of (??) and (??), we have
−
Z
0
∞
dk kA(k)J0 (kρ) =
µ0 Ia
2
Z
∞
0
dk ke−kd J1 (ka) (J2 (kρ) − J0 (kρ)) +
Z
∞
dk kB(k)J0 (kρ)
0
Solutions to Problems in Jackson,
Classical Electrodynamics, Third Edition
Homer Reid
April 20, 2001
Chapter 5: Problems 19-27
Problem 5.19
A magnetically “hard” material is in the shape of a right circular cylinder of length L
and radius a. The cylinder has a permanent magnetization M0 , uniform throughout
its volume and parallel to its axis.
(a) Determing the magnetic field H and magnetic induction B at all points on the
axis of the cylinder, both inside and outside.
(b) Plot the ratios B/µ0 M0 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 a
scalar potential Φm (ρ, z) satisfying the Laplace equation. Dividing space into
three regions
Φm =








Z








∞
0
Z
∞
dk A(k)e−kz J0 (kρ),
z > L/2
0
dk B(k)ekz + C(k)e−kz J0 (kρ),
Z ∞
dk D(k)ekz J0 (kρ),
0
1
− L/2 < z < L/2
z < −L/2.
2
Homer Reid’s Solutions to Jackson Problems: Chapter 5
The tangential boundary condition at z = +L/2 is
∂Φm
∂ρ z= L −
2
Z ∞
Z ∞
h
i
−kL/2
⇒
dk kA(k)e
J1 (kρ) =
dk k B(k)ekL/2 + C(k)e−kL/2 J1 (kρ)
∂Φm
∂ρ
=
z= L
2+
0
0
(1)
This must hold for all ρ. Multiplying both sides by ρJ1 (k 0 ρ), integrating from
ρ = 0 to ρ = ∞, and using the identity
Z ∞
1
(2)
dρ ρJn (kρ)Jn (k 0 ρ) = δ(k − k 0 )
k
0
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
µ0 Hz (z = L/2+) = µ0 Hz (z = L/2−) + Mz (z = L/2−)
∂Φm
∂z
∂Φm
+ M (ρ)
∂z z= L −
z= L
2+
2
Z ∞
Z ∞
i
h
−kL/2
⇒
dk kA(k)e
J0 (kρ) =
dk k −B(k)ekL/2 + C(k)e−kL/2 J0 (kρ) + M (ρ)
=
0
0
(4)
where
M (ρ) =
(
M1 ,
0,
ρ<a
ρ > a.
Now we multiply both sides of (4) by ρJ0 (k 0 ρ) and integrate from ρ = 0 to
ρ = ∞ to obtain
A(k) = −B(k)ekL + C(k) + M1 ekL/2
= −B(k)ekL + C(k) + γ(k)
Z
a
ρJ0 (kρ)dρ
0
where we defined
γ(k) = M1 ekL/2
Z
a
ρJ0 (kρ)dρ =
0
aM1 kL/2
e
J1 (ka).
k
(5)
3
Homer Reid’s Solutions to Jackson Problems: Chapter 5
The solution of eqs. (3) and (5) is
B(k) =
1 −kL
e
γ(k)
2
1
C(k) = A(k) − γ(k).
2
(6)
From the boundary conditions at z = −L/2 we may similarly obtain the relations
B(k) + C(k)ekL = D(k)
B(k) − C(k)ekL = D(k) − γ(k)
which may be solved to yield
1
B(k) = D(k) − γ(k)
2
C(k) =
1 −kL
e
γ(k).
2
(7)
Comparing (6) and (7) we find
kL
M1 a
cosh
J1 (ka)
k
2
M1 a −kL/2
B(k) = C(k) =
e
J1 (ka).
2k
A(k) = D(k) =
Then the components of the H field are

Z ∞
kL −kz


e
J1 (ka)J1 (kρ),
M
a
dk cosh

1


2
0


Z ∞

dk e−kL/2 cosh(kz)J1 (ka)J1 (kρ),
Hρ = M 1 a

0


Z ∞


kL kz


dk cosh
e J1 (ka)J1 (kρ),
M1 a

2
0
Z ∞
kL −kz
M1 a
dk cosh
e
J1 (ka)J0 (kρ),
2
0
Z ∞
Hz = −M1 a
dk e−kL/2 sinh(kz)J1 (ka)J0 (kρ),

0


Z ∞


kL kz


−M1 a
dk cosh
e J1 (ka)J0 (kρ),

2
0









z > L/2
− L/2 < z < L/2
z < −L/2
z > L/2
− L/2 < z < L/2
z < −L/2.
4
Homer Reid’s Solutions to Jackson Problems: Chapter 5
Problem 5.23
A right circular cylinder of length L and radius a has a uniform lengthwise magnetization M .
(a) Show that, when it is placed with its flat end against an infinitely permeable
plane surface, it adheres with a force
K(k) − E(k) K(k1 ) − E(k1 )
F = 2µ0 aLM 2
−
k
k1
where
k=√
2a
,
4a2 + L2
k1 = √
a2
a
.
+ 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 find
the magnetic field H0 that exists in all space when the cylinder is pressed up
flat against the infinitely permeable medium. Then we’ll calculate the shift dE
in the energy of the magnetic field incurred by moving the cylinder up a small
distance dz off the surface of the medium. The force on the cylinder is then
readily calculated as F = −dE/dz.
To calculate the energy shift incurred by moving the cylinder a distance dz
away from the permeable medium, we won’t have to go through and completely
recalculate the fields and their energy in the new configuration. Instead, we
can 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 the
face 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 a
fixed magnetization M where previously there was none. Moving the cylinder of
fixed M up a distance dz is thus formally equivalent to keeping the cylinder put
and instead introducing a cylinder of the opposite magnetization −M between 0
and dz, while also introducing a cylinder of magnetization +M between L and
L + dz. The increase in field energy in this latter case is fairly easily calculated
by taking the integral of µ0 Mc˙
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, we
may derive H from a scalar potential satisfying the Laplace equation. To begin we write down the general solutions of the Laplace equation in cylindrical
coordinates, observing first that by symmetry we can only keep terms with no
5
Homer Reid’s Solutions to Jackson Problems: Chapter 5
azimuthal angle dependence:

Z ∞


dk A(k)e−kz J0 (kρ),



0


Z ∞
Φ(m) =
dk [B(k)ekz + C(k)e−kz ]J0 (kρ),

0


Z ∞




dk D(k)e+kz J0 (kρ),

z>L
0<z<L
(8)
z < 0.
0
The boundary conditions at z = 0 are that Hρ and Bz be continuous. Assuming first of all that the medium existing in the region below z = 0 has finite
permeability µ, the tangential boundary condition is
∂Φm
∂ρ z=0+
z=0−
Z ∞
Z ∞
dk k D(k)J1 (kρ) =
dk k [B(k) + C(k)]J1 (kρ).
∂Φm
∂ρ
=
0
(9)
0
Multiplying (9) by ρJ1 (k 0 ρ), integrating from 0 to ∞, and using the identity
(2), we find
D(k) = B(k) + C(k).
(10)
The normal boundary condition at z = 0 is of a mixed type. Below the line
we 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<a
M (ρ) =
(11)
0,
ρ > a.
The normal boundary condition at z = 0 is then
−µ
−
µ
µ0
Z
∂
Φm
∂z
z=0−
= −µ0
∞
0
dk k D(k)J0 (kρ) = −
Z
0
∂
Φm
∂z
+ µ0 M (ρ)
z=0+
∞
dk k [B(k) − C(k)]J0 (kρ) + M (ρ)
Now multiplying by ρJ0 (k 0 ρ), integrating from ρ = 0 to ∞, and using (2) yields
Z ∞
µ
D(k) = −B(k) + C(k) −
ρ M (ρ)J0 (kρ) dρ.
(12)
µ0
0
Using (11), the integral on the RHS is
Z a
Ma
J1 (ka) ≡ γ(k)
M
ρJ0 (kρ) dρ =
k
0
where we defined a convenient shorthand. Then (12) is
µ
D(k) = −B(k) + C(k) − γ(k).
µ0
6
Homer Reid’s Solutions to Jackson Problems: Chapter 5
Now taking µ → ∞, we see that, to keep the B and C coefficients from blowing
up, we must have D → 0. Then equation (??) tells us that B(k) = −C(k), so
the middle entry in (8) may be rewritten:
Z ∞
Φm (z, ρ) =
dk β(k) sinh(kz)J0 (kρ),
(0 < z < L).
0
The boundary conditions at z = L are
∂Φm
∂ρ
−
∂Φm
∂z
=
z=L+
z=L+
∂Φm
∂ρ
=−
z=L−
∂Φm
∂z
+ M (ρ)
z=L−
with M (ρ) defined as above. Working through the same procedure as above
yields 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 of
the H field
Z ∞



M
a
dk e−kz cosh(kL)J0 (kρ)J1 (ka),
z>L

0
Hz (ρ, z) =
Z ∞
(13)


 −M a
dk e−kL cosh(kz)J0 (kρ)J1 (ka),
0 < z < L.
0
Now that we know the field, we want to find the change in energy density
incurred by putting into this field a short cylinder (radius a, height dz) of
magnetization −M k̂ between z = 0 and z = dz, and another cylinder of the
same size but with magnetization +M k̂ between z = L and z = L + dz. The
change in field energy is just the integral of µ0 M · H over the volume in which
the magnetization density has changed:
dU = −2πµ0 M
Z
dz
Z
a
Z
L+dz
Hz (z, ρ)ρ dρ dz + 2πµ0 M
0
0
L
Z
Z a
a
= 2πµ0 M dz
Hz (L, ρ)ρ dρ −
Hz (0, ρ)ρ dρ
0
Z
a
Hz (z, ρ)ρ dρ dz
0
(14)
0
where in the last step we assumed that Hz remains essentially constant over
a distance dz in the z direction, and may thus be taken out of the integral.
Homer Reid’s Solutions to Jackson Problems: Chapter 5
7
Inserting (13) into (), and exchanging the order of integration, we first do the ρ
integral:
Z a
a
J0 (kρ)ρdρ = J1 (ka).
k
0
Then () becomes
Solutions to Problems in Jackson,
Classical Electrodynamics, Third Edition
Homer Reid
March 28, 2002
Chapter 6: Problems 1-8
1
2
Homer Reid’s Solutions to Jackson Problems: Chapter 6
Problem 6.2
The charge and current densities for a single point charge q can be written formally
as
ρ(x0 , t0 ) = qδ[x0 − r(t0 )];
J(x0 , t0 ) = qv(t0 )δ[x0 − r(t0 )]
where r(t0 ) is the charge’s position at time t0 and v(t0 ) is its velocity. In evaluating
expressions involving the retarded time, one must put t0 = tret = t − R(t0 )/c, where
R = x − r(t0 ).
(a) As a preliminary to deriving the Heaviside-Feynman expressions for the electric
and magnetic fields of a point charge, show that
Z
1
d3 x0 δ[x0 − r(tret )] =
κ
where κ = 1 − v · R̂/c. Note that κ is evaluated at the retarded time.
(b) Starting with the Jefimenko generalizations of the Coulomb and Biot-Savart
laws, use the expressions for the charge and current densities for a point charge
and the result of part a to obtain the Heaviside-Feynman expressions for the
electric and magnetic fields of a point charge,
#
("
"
#
)
1 ∂ R̂
q
1 ∂ h v i
R̂
+
E=
− 2
4π0
κR2
c ∂t κR
c ∂t κR ret
ret
ret
and
µ0 q
B=
4π
("
v × R̂
κR2
#
ret
"
#
1 ∂ v × R̂
+
c ∂t
κR
ret
)
(c) In our notation Feynman’s expression for the electric field is
" #
(" #
)
q
R̂
[R]ret ∂ R̂
1 ∂2
E=
+
+ 2 2 [R̂]ret
4π0
R2
c ∂t R2
c ∂t
ret
ret
while Heaviside’s expression for the magnetic field is
("
#
"
# )
v × R̂
1
∂ v × R̂
µ0 q
+
.
B=
4π
κ2 R 2
c[R]ret ∂t
κ
ret
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 its
position is given by
r(t) = (z0 + vz t)k̂.
Homer Reid’s Solutions to Jackson Problems: Chapter 6
3
The retarded time tret (t, z) at a given point z on the z axis is
tret (t, z) = t −
z
c
so
r[tret (t, z)] = (z0 + vz tret (t, z))k̂.
Hence
δ(x − r[tret (x, t)]) = δ(x)δ(y)δ {z − [z0 + vz tret (t, z)]}
n
z o
= δ(x)δ(y)δ z − [z0 + vz (t − )]
c
n
vz o
= δ(x)δ(y)δ z − z0 − vz t + z)]
c
o
n
vz z − (z0 + vz t)
= δ(x)δ(y)δ 1 +
c
By the properties of the δ function we may write this as
z0 + v z t
1
δ(x)δ(y)δ z −
.
=
1 + vz /c
1 + vz /c
The δ function is singling out the point in space from which originates the
electromagnetic disturbance we feel at the origin at time t. Let’s think about
what’s going on here in two limiting cases. First, as vz → 0, the z delta function
becomes δ(z−(z0 +vz t)). This means that the source point for the field we feel at
the origin at time t is just z = z0 − vz t, which is of course just the instantaneous
location of the source particle at time t. In other words, the electromagnetic
disturbance left behind by the particle at time t reaches the origin so quickly
that the particle hasn’t had time to move on. The electromagnetic disturbance
seems 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 −vz t)/2).
This says that the point from which we feel an electromagnetic disturbance at
time t is half as far from the origin as the particle itself is at time t. This
again makes sense. At each point in the particle’s motion, the electromagnetic
disturbance it causes begins propagating toward the origin, while meanwhile
the particle continues propagating away from the origin at the same speed.
Hence when the electromagnetic disturbance has reached the origin, the particle
has 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 disturbance
we are just now feeling was generated.
Homer Reid’s Solutions to Jackson Problems: Chapter 6
4
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 transformed to
Z
1
ΦJ d3 x
Pfield = 2
c
provided the product ΦH falls of rapidly enough at large distances. How
rapidly is “rapidly enough”?
(b) Assuming that the current distribution is localized to a region small compared
to the scale of variation of the electric field, expand the electrostatic potential
in a Taylor series and show that
Pfield =
1
E(0) × m,
c2
where E(0) is the electric field at the current distribution and m is the magnetic moment, (5.54), caused by the current.
(c) Suupose the current distribution is placed instead in a uniform electric field
E0 (filling all space). Show that, no matter how complicated is the localized
J, the result in part a is augmented by a surface integral contribution from
infinity equal to minus one-third of the result of part b, yielding
Pfield =
2
E0 × m.
3c2
Compare this result with that obtained by working directly with (6.117) and
the considerations at the end of Section 5.6.
(a) From the definition of electromagnetic field momentum we have
Z
c2 Pfield = E × H dV
Z
= − (∇Φ) × H dV.
Focusing for now on the z component, we have
c 2 Pz = −
Z
(
∂Φ
∂Φ
Hy −
Hx ) dx dy dz
∂x
∂y
(1)
5
Homer Reid’s Solutions to Jackson Problems: Chapter 6
Let’s take our volume of integration to be a cube of side L, which we will
eventually take to infinity. Integrating the first term by parts with respect to
x, we have
)
Z L Z L (Z L
Z LZ L
Z
x=L
∂Hy
∂Φ
Hy dx dy dz =
dx dy dz.
− Φ
ΦHy
∂x
x=−L
−L −L
−L −L
−L ∂x
Similarly integrating the second term in (1) by parts with respect to y, we may
write (1) as
Z
Z LZ L
Z LZ L
y=L
x=L
ΦHx
dx dz + Φ(∇ × H)z dV
ΦHy
dy dz +
c 2 Pz = −
=−
Z
−L −L
L Z L
−L
−L
x=−L
ΦHy
x=L
x=−L
dy dz +
Z
−L −L
L Z L
−L
−L
y=−L
ΦHx
y=L
y=−L
dx dz +
Z
ΦJz dV
where in going to the last line we used ∇ × H = J since there is no timedependent E field. This equation is just the z component of
Z
Z
2
c P = 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 a
higher power of x) than x2 . Then we are left with just the second term:
Z
2
c P = ΦJ dV.
(3)
(b) We have
Φ(x) = Φ(0) + x · ∇Φ(0) +
∂2Φ
1X
xi xj
+···
2
∂xi ∂xj
We may arbitrarily choose Φ(0) = 0. Also, we are told that the electric field
doesn’t vary much in the region of nonvanishing J, in which case we may ignore
the second derivatives of Φ, to obtain
Φ(x) ≈ x · ∇Φ(0) = −x · E(0).
Plugging into (3),
c2 P = −
We have
Z
x · E(0) J dV.
− x · E(0) J = E(0) × J x − x · E(0) J − E(0) × J x
= E(0) × x × J − E(0) × J x
(4)
Homer Reid’s Solutions to Jackson Problems: Chapter 6
6
where in the first line we added and subtracted a term, and in the second used
the BAC-CAB identity of vector analysis. With this, (4) becomes
Z
Z
c2 P = E(0) ×
x + J dV −
E(0) × J] x dV
Z
E(0) × J] x dV
= 2E(0) × m −
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 that
second term is exactly half the first, but I can’t see how to do this for arbitrary
J. Can anybody help?
(c) From (2) we have
c2 P =
Z
ΦH × dA +
Z
ΦJ dV.
The second term is just equal to (E × m)/c2 , as computed in part b. For the
first term,
Problem 6.13
A parallel plate capacitor is formed of two flat rectangular perfectly conducting
sheets 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. With
the input current and voltage defined at this end of the capacitor, calculate the
input impedance or admittance using the field concepts of Section 6.9.
(a) Calculate the electric and magnetic fields in the capacitor correct to second
order in powers of the frequency, but neglecting fringing fields.
(b) Show that the expansion of the reactance (6.140) in powers of the frequency
to an appropriate order is the same as that obtained for a lumped circuit
consisting of a capacitance C = 0 ab/d in series with an inductance L =
µ0 ad/3b.
(a) We’ll suppose the plates are oriented parallel to the xy plane, with the lower
plate at z = 0 and the upper plate at z = d. We’ll take the edges of side a
parallel to the x axis, and the edges of side b parallel to the y axis. Then the
boundary condition on the current density is
J(0, y, 0) = −J(0, y, d) = J0 ĵ
for 0 < y < b.
7
Homer Reid’s Solutions to Jackson Problems: Chapter 6
With neglect of fringing fields, the electric field between the plates exists
only 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−iωt k̂
B(y) = B(y)e−iωt x̂;
(5)
then time differentiation becomes multiplication by −iω. The Maxwell equations are then
∇·E=0
⇒
∇·B=0
⇒
∂B
∂t
1 ∂E
∇×B= 2
c ∂t
∇×E=−
⇒
⇒
∂E
∂z
∂B
∂x
∂E
∂y
∂B
∂y
=0
=0
(6)
= +iωB
=+
iω
E.
c2
We postulate an expansion in powers of ω for E and B:
E(y) = E0 (y) + ωE1 (y) + ω 2 E2 (y) + · · ·
B(y) = B0 (y) + ωB1 (y) + ω 2 B2 (y) + · · ·
Then the curl equations in (6) become
∂ E0 + ωE1 + ω 2 E2 + · · · = iω B0 + ωB1 + ω 2 B2 + · · ·
∂y
iω ∂ B0 + ωB1 + ω 2 B2 + · · · = 2 E0 + ωE1 + ω 2 E2 + · · ·
∂y
c
(7)
Homer Reid’s Solutions to Jackson Problems: Chapter 6
8
Equating equal powers of ω in these equations, we obtain
∂E0
∂y
∂B0
∂y
∂E1
∂y
∂B1
∂y
∂E2
∂y
∂B2
∂y
∂E3
∂y
∂B3
∂y
∂E4
∂y
∂B4
∂y
=0
⇒
E0 = α
=0
⇒
B0 = β
= iB0 = iβ
⇒
E1 = iβy
⇒
B1 =
⇒
E2
⇒
B2
⇒
E3
⇒
B3
⇒
E4
⇒
B4
i
iα
E0 = 2
2
c
c
α
= iB1 = − 2 y
c
i
β
= 2 E1 = − 2 y
c
c
iβ
= iB2 = − 2 y 2
2c
i
iα
= 2 E2 = − 4 y 2
c
2c
α 3
= iB3 =
y
24c4
i
β
= 2 E3 = + 4 y 3
c
6c
=
iα
y
c2
α
= − 2 y2
2c
β
= − 2 y2
2c
iβ
= − 2 y3
6c
iα
= − 4 y3
6c
iβ 4
=
y
24c2
β 4
=
y
24c4
and so on. Plugging into (7), we obtain
(ky)4
(ky)3
(ky)2
+
+ · · · ) + iβc(ky −
+ ···)
2
24
6
= α cos ky + iβc sin ky
iα
B(y) = β cos ky +
sin ky
c
E(y) = α(1 −
(8)
(9)
(10)
where k = ω/c, and where we simply wrote down what we guessed to be the
sums of the full infinite series from their first few terms.
To complete the problem we need to determine the constants α and β, for
which purpose we appeal to the boundary conditions on the plates. We know
that the discontinuities in the E and B field are proportional to the surface
charge and current densities on the plates. Since these conditions only give
information 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 are
inside. But for the purposes of this problem we’ll just assume there are no
fields outside, so the charge and current densities on the plates determine the
fields inside. I know this is correct in the low-frequency limit, and in the highfrequency limit I’m not yet sure how to compute the radiation fields in the
region outside the plates, so I will ignore them.
9
Homer Reid’s Solutions to Jackson Problems: Chapter 6
The boundary conditions are
σ
0
Bx = −µ0 Ky
Ez = −
where σ and Ky are the charge density and y component of the surface current
density on the top plate (assumed to be identical but with opposite sign on the
bottom plate). Plugging in the solutions (9) and (??), we have
σ = −0 (α cos ky + iβc sin ky)
iα
1
Ky = − (β cos ky +
sin ky)
µ0
c
(11)
As a sanity check, we can verify the continuity relation between charge and
current on the plates:
∂Ky
∂σ
=−
= +iωσ
∂y
∂t
Plugging in (11), the left and right sides of this are
1
ikα
(−kβ sin ky +
cos ky)
µ0
c
RHS = −iω0(α cos ky + iβc sin ky)
LHS = −
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 top
plate is I(t) = I0 cos ωt (with an opposite current taken out of the y = 0 edge
of the bottom plate) then the surface current boundary conditions are
I0
cos ωt
b
Ky (y = a) = 0
Ky (y = 0) =
Comparing with (11), we see that these boundary conditions we have to take
µ0 I 0
cos ωt
b
iµ0 I0 c
cos ωt cot ka
α=−
b
β=−
Homer Reid’s Solutions to Jackson Problems: Chapter 6
10
Plugging into (9) and (10),
iµ0 I0 c
cos ωt [cot ka cos ky + sin ky]
b
iµ0 I0 c
1
=−
[cos ka cos ky + sin ka sin ky]
cos ωt
b
sin ka
iµ0 I0 c
cos[k(y − a)]
=−
cos ωt
b
sin ka
µ0 I 0
cos ωt [− cos ky + cot ka sin ky]
Bz = −
b
µ0 I 0
1
=−
cos ωt
[− sin ka cos ky + cos ka sin ky]
b
sin ka
sin[k(y − a)]
µ0 I 0
cos ωt
=−
b
sin ka
Ez = −
Problem 6.14
An ideal circular parallel plate capacitor of radius a and plate separation d a is
connected to a current source by axial leads, as shown in the sketch. The current
in the wire is I(t) = I0 cos ωt.
(a) Calculate the electric and magnetic fields between the plates to second order in
powers of the frequency (or wave number), neglecting the effects of fringing
fields.
(b) Calculate the volume integrals of we and wm that enter the definition of the
reactance X, (6.140), to second order in ω. Show that in terms of the input
current Ii , defined by Ii = −iωQ, where Q is the total charge on one plate,
these energies are
Z
Z
µ0 |Ii |2 d
1 |Ii |2 d
ω 2 a2
3
3
,
wm d x =
we d x =
1+
4π0 ω 2 a2
4π 8
12c2
(c) Show that the equivalent series circuit has C ≈ π0 a2 /d, L ≈ µ0 d/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 multiplication by −iω. If we neglect the effects of fringing fields, everything is symmetric
in θ, and the electric field between the plates is entirely in the z direction, while
11
Homer Reid’s Solutions to Jackson Problems: Chapter 6
the magnetic field is entirely in the θ direction:
E(x, t) = E(r, z)e−iωt ẑ
B = B(r, z)e−iωt θ̂.
(12)
The Maxwell equations for the fields between the plates are
∂
E=0
∂z
∂
∇·B=0
⇒
B=0
∂θ
(13)
∂E
∂B
⇒
= −iωB
∇×E=−
∂t
∂r
1 ∂
iω
1 ∂E
⇒
(rB) = − 2 E.
∇×B= 2
c ∂t
r ∂r
c
To proceed, let’s propose an expansion of the fields in powers of the frequency:
∇·E=0
⇒
E(r) = E0 (r, z) + ωE1 (r, z) + ω 2 E2 (r, z) + · · ·
2
B(r) = B0 (r, z) + ωB1 (r, z) + ω B2 (r, z) + · · ·
(14)
(15)
Then the curl equations in (13) become
∂ E0 + ωE1 + ω 2 E2 = −iω B0 + ωB1 + ω 2 B2
∂r
iω 1 ∂ rB0 + ωrB1 + ω 2 rB2 = − 2 E0 + ωE1 + ω 2 E2
r ∂r
c
Now we just have to go through and equate like powers of ω in these equations.
For n = 0, we have
∂E0
=0
∂r
for some constant α1 , and
⇒
E 0 = α1
(16)
1 ∂
β
(rB0 ) = 0
⇒
B0 = .
(17)
r ∂r
r
But for nonzero β this blows up at the origin. Hence we must take β = 0, so
B0 = 0. 2 For n = 1, we have
∂E1
= −iB0 = 0
∂r
for some constant α2 , and
i
iα1
1 ∂
(rB1 ) = − 2 E0 = − 2
r ∂r
c
c
Continuing,
∂E2
α1
= −iB1 = − 2 r
∂r
2c
1 ∂
i
iα2
(rB2 ) = − 2 E1 = − 2
r ∂r
c
c
⇒
E 1 = α2
⇒
B1 = −
⇒
E2 = −
⇒
(18)
iα1
r.
2c2
α1 2
r
4c2
iα2
B2 = − 2 r
2c
(19)
(20)
(21)
Homer Reid’s Solutions to Jackson Problems: Chapter 6
α2
∂E3
= −iB2 = − 2 r
∂r
2c
1 ∂
i
iα1
(rB3 ) = − 2 E2 = 4 r2
r ∂r
c
4c
⇒
α2 2
r
4c2
iα1 3
r
B3 =
16c4
E3 = −
⇒
12
(22)
(23)
Evidently E2n and E2n+1 have the same form but differ by the presence of
α1 or α2 , as is true for B2n−1 and B2n . Plugging in equations (16) through (23)
into (14) and (15), we obtain
ω4 4 ω2 2
r +
r +···
2
4c
64c4
(kr)2
(kr)4
= (α1 + ωα2 ) 1 −
+
+···
4
64
i
(kr)2
kr
+···
B(r) = − (α1 + ωα2 )
1−
c
2
8
E(r) = (α1 + ωα2 ) 1 −
These look the first few terms in certain Bessel functions:
E(r) = (α1 + ωα2 )J0 (kr) ≡ βJ0 (kr)
i
B(r) = − βJ1 (kr)
c
where we can define the constant β = (α1 + ωα2 ) since we’re dealing with a
fixed frequency. Inserting into (12) we obtain
E(r, t) = βJ0 (kr)e−iωt k̂
i
B(r, t) = − βJ1 (kr)e−iωt θ̂.
c
(24)
To work out the value of β, we need to apply the boundary conditions at the
capacitor 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 plates
is just Ez (t) = −2σ(t)/0 , where σ(t) is the instantaneous value of the surface
charge 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 flowing
onto that plate:
Z
I0
sin ωt
q = I(t) dt =
ω
and the surface charge is this divided by the plate area (since we are assuming a
low frequency, any charge that flows onto the plate instantaneously equilibrates
with the rest of the charge on the plate, yielding a constant surface charge
density):
I0
σ(t) =
sin ωt.
πa2 ω
Hence the electric field in the low frequency limit is
Ez (ω → 0) = −
2I0
sin ω.
πa2 ω0
Homer Reid’s Solutions to Jackson Problems: Chapter 6
13
Comparing this with(24) in the limit k → 0, we obtain
β=−
2I0 i
.
πa2 ω0
Hence
E(r, t) = −
2I0
J0 (kr) sin ωt k̂
πa2 ω0
B(r, t) =
2µ0 I0 c
J1 (kr) cos ωt θ̂.
πa2 ω
(25)
(b) The average energy densities are
2
0 2
kr
I02
1−
E =
+···
2
2
4
(πa ω) 0
4
2 2
2 2
1 2
µ0 I 0 c
(kr)2
kr
=
B
=
+
·
·
·
1
−
4µ20
(πa2 ω)2 2
8
we =
wm
We only have to keep the first terms in the parentheses to get the energy right
to second order in ω:
Z a
I02
(2πd)(r dr)
Ue ≈
(πa2 ω)2 0 0
I02 d
=
πa2 ω 2 0
Z a
Z a
kr (kr)3
kr (kr)3
µ0 I02 c2
µ0 I02 c2
Um =
−
+
·
·
·
U
=
−
+
·
·
·
(2πd)(rdr)
(2πd)(rdr)
m
(πa2 ω)2 0
2
8
(πa2 ω)2 0
2
8
Solutions to Problems in Jackson,
Classical Electrodynamics, Third Edition
Homer Reid
May 24, 2002
Chapter 8: Waveguide Derivations
Before starting the problems, I thought it would be useful to run through
my 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 out
one spatial coordinate and announce from the start that the fields will have
sinusoidal 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
n
o
E(x) = Ex (x, y)i + Ey (x, y)j + Ez (x, y)k ei(kz−ωt)
n
o
(1)
B(x) = Bx (x, y)i + By (x, y)j) + Bz (x, y)k ei(kz−ωt)
We have here a total of six functions f (x, y) that we must find to satisfy
Maxwell’s equations with the relevant boundary conditions. At first this would
appear tough since the six fields are all coupled by Maxwell’s equations, but
after a little algebra we obtain the following simplified situation: The z− direction fields Ez (x, y) and Bz (x, y) turn out to satisfy (separately) simple onedimensional differential equations, which may be readily solved upon specifying
the boundary conditions for a particular situation. Meanwhile, the remaining
fields (Ex , Ey , Bx , By ) can be expressed simply as linear combinations of Ez and
Bz and their derivatives, so once we obtain the z fields we have everything. In
what follows we’ll derive the differential equations satisfied by Ez and Bz and
the equations giving the remaining fields in terms of them.
1
Homer Reid’s Solutions to Jackson Problems: Chapter 8
2
The differential equations for Ez and Bz
The Maxwell curl equations are
∇×E=−
∂B
,
∂t
∇×B=
1 ∂E
c2m ∂t
where cm is the speed of light in the medium. We begin by applying the first
curl equation to our ansatz (1), obtaining
∂y Ez − ikEy = iωBx
−∂x Ez + ikEx = iωBy
∂x Ey − ∂y Ex = iωBz ,
(2)
(3)
(4)
and we pause to solve the first two of these for Ex and Ey :
Ex =
i
ω
By − ∂ x Ez ,
k
k
ω
i
Ey = − Bx − ∂ y Ez .
k
k
(5)
Next we apply the second curl equation to our ansatz, obtaining
iω
Ex
c2m
ω
−∂x Bz + ikBx = −i 2 Ey
cm
ω
∂x By − ∂y Bx = −i 2 Ez .
cm
∂y Bz − ikBy = −
(6)
(7)
(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 :
ikc2
ω
Bx = 2 m2 2 ∂x Bz + 2 ∂y Ez
(9)
ω − k cm
cm k
ikc2
ω
By = 2 m2 2 ∂y Bz − 2 ∂x Ez .
(10)
ω − k cm
cm k
Finally, with the ansatz (1) the equation ∇ · B = 0 reads
∂Bx
∂By
+
= −ikBz .
∂x
∂y
When we plug (9) and (10) into this, the terms involving Ez fields cancel, and
we obtain an equation involving Bz alone:
2
2
∂2
ω
∂
2
+ 2 Bz +
− k Bz = 0
∂x2
∂y
c2m
or
∂2
∂2
+
∂x2
∂y 2
Bz + γ 2 Bz = 0
(11)
Homer Reid’s Solutions to Jackson Problems: Chapter 8
where
3
r
ω2
− k2.
c2
If we had carried out this derivation in the reverse order we would have obtained
the same equation for Ez :
2
∂2
∂
Ez + γEz = 0.
(12)
+
∂x2
∂y 2
γ=
We can think of equations (11) and (12) as eigenvalue equations that have solutions only for certain values of the parameter γ, which depend on the boundary
conditions.
Armed with equations (11) and (12) and the boundary conditions appropriate 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 of
the E field are given by analogous equations:
ikc2m
ω 2 − k 2 c2m
ikc2
Ey = 2 m2 2
ω − k cm
Ex =
ω
∂y Bz
k
ω
∂ y Ez − ∂ x Bz .
k
∂ x Ez +
(13)
(14)
Boundary Conditions; TE, TM, TEM Modes
The boundary conditions on the fields at the surfaces of the waveguide or cavity
are that Ek and B⊥ be continuous, where ⊥ denotes the component of the
vector normal to the boundary surface and k includes all other components of
the vector. This means that the two eigenvalue equations (11) and (12) must be
solved subject to different boundary conditions, which means in general their
eigenvalues 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 nonzero
solution of (11) for that value of γ, and hence we must have Bz = 0 identically
for the mode at that frequency and wavevector. Since in this case the magnetic
field has nonzero components only transverse to the direction of propagation,
this is called a transverse magnetic mode. Similarly, if (11) can be solved with
nonzero Bz at some γ, then Ez = 0 and we have a transverse electric mode. A
mode for which both Ez and Bz are zero is called a transverse electromagnetic
mode, and can only exist in the region between two conducting surfaces, not
within 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 above
and collect results appropriate to the two cases.
4
Homer Reid’s Solutions to Jackson Problems: Chapter 8
TM Modes
TE Modes
Bz ≡ 0
Ez ≡ 0
∇2t Ez + γ 2 Ez = 0,
Ex =
Ey =
En
∂S
ikc2m
∂ x Ez
ω 2 − k 2 c2m
ikc2m
∂ y Ez
ω 2 − k 2 c2m
iω
∂ y Ez
ω 2 − k 2 c2m
iω
∂ x Ez
By = − 2
ω − k 2 c2m
Bx =
=0
∇2t Bz + γ 2 Bz = 0,
∂Bn
∂n
=0
∂S
iωc2m
∂y Bz
ω 2 − k 2 c2m
iωc2m
Ey = − 2
∂x Bz
ω − k 2 c2m
ikc2
Bx = 2 m2 2 ∂x Bz
ω − k cm
ikc2
By = 2 m2 2 ∂y Bz
ω − k cm
Ex =
(A factor of ei(kz−ωt) is understood in all of these expressions.)
For TM modes, the boundary condition is Ek = 0, and Ez is always perpendicular to the boundary surfaces, so the boundary condition for the eigenvalue
equation 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 the
x direction, so Bx must vanish at this surface; but we just saw that in the TE
case Bx ∝ ∂x Bz , i.e. the derivative of Bz normal to the boundary surface must
vanish. This is general: the boundary condition for the eigenvalue equation in
the TM case is ∂Bz /∂n = 0.
Power flow; Energy Loss
The flow of power down a waveguide is described by the z component of the
Poynting vector S = E × H = µ1 E × B. Using the boxed expressions above, for
the two types of modes we obtain
1
(Ex By − Ey Bx )
µ
kωc2m
=
(∂x Ez )2 + (∂y Ez )2 e2i(kz−ωt)
2
2
2
2
µ(ω − k cm )
SzTM =
or, in the time average,
kωc2m
(∂x Ez )2 + (∂y Ez )2
2
2
2
− k cm )
kω =
(∂x Ez )2 + (∂y Ez )2 .
4
2γ
=
2µ(ω 2
(15)
Homer Reid’s Solutions to Jackson Problems: Chapter 8
5
Similarly, for TE modes we obtain
SzTE =
kω (∂x Bz )2 + (∂y Bz )2 .
4
2µγ
(16)
To address the issue of dissipation in the boundaries, we solve Maxwell’s
equations within the boundary surfaces. The two curl equations are
∇ × E = iωB
∇ × B = µJ − iµωE
= µ (σ − iω) E
≈ µσE
(17)
(18)
since σ ω in most cases. (For example, for a copper waveguide with air
or vacuum interior we have we have σ ≈ 6 · 107 Ω−1 m−1 , while ω ≈ 9 ·
10−12 Ω−1 m−1 · (ω in rad/sec), so the approximation is good up to frequencies
ω ∼ 1019 rad/sec.)
Now we assume that the fields are only changing significantly in the direction normal to the boundary surface (i.e., as we go deeper and deeper into the
boundary surface the fields die out rapidly, whereas as we move along parallel
to the boundary surface the fields don’t change much) and keep only the normal
derivative in the curl equations. If ρ measures the depth of penetration into the
surface, the curl equations become
∂E
= iωB
∂ρ
∂B
ρ̂ ×
= µσE
∂ρ
ρ̂ ×
Differentiating the first of these, taking the cross product with ρ̂ of both sides,
and substituting in the second equation yields
ρ̂ × ρ̂ ×
∂ 2E
= iωµσE
∂ρ2
or, using the bac-cab rule,
∂ 2 Eρ
∂ 2E
ρ̂
−
= iωµσE.
∂ρ2
∂ρ2
Evidently the ρ component of the LHS vanishes here, so Eρ = 0; the electric
field within the conducting boundary has no component normal to the surface.
For the remaining components we obtain
∂ 2 Ek
+ iωµσEk = 0
∂ρ2
Homer Reid’s Solutions to Jackson Problems: Chapter 8
6
with solution
Ek = e ±
=e
√
iωµσρ
±(1+i) ρδ
E0
E0
p
where δ = 2/ωµσ is the skin depth and E0 is the field just at the surface of
the boundary. To keep the solution from blowing up as we penetrate into the
conductor we take the negative sign in the exponent. From (17) we then obtain
B=
ρ
i−1
(ρ̂ × E0 )e−(1+i) δ .
δω
Evaluating this at the surface yields the modified boundary condition on the
fields in the cavity or waveguide:
B0 =
i−1
(ρ̂ × E0 ).
δω
(19)
From this equation we can work out the power loss per unitR length in the
cavity
or waveguide. The power dissipated in a volume dV is (J · E) dV =
R
σ E 2 dV. We integrate over the volume occupied by the boundary surfaces in
a length dz :
I Z ∞
2 −2(1+i) ρδ
dP = dz
σE0 e
dρ dl e2i(kz−ωt)
0
I
σδ
2
= dz
E0 dl e2i(kz−ωt)
2(1 + i)
or, taking the time average,
σδ
dP
= √
dz
4 2
I
E02 dl
(20)
where the line integral is over the cross section of the surface boundary at a
fixed value of z.
Solutions to Problems in Jackson,
Classical Electrodynamics, Third Edition
Homer Reid
May 24, 2002
Chapter 8: Problems 1-8
1
Homer Reid’s Solutions to Jackson Problems: Chapter 8
2
Problem 8.2
A transmission line consisting of two concentric cylinders of metal with conductivity
σ and skin depth δ, as shown, is filled with a uniform lossless dielectric (µ, ). A
TEM mode is propagated along this line.
(a) Show that the time-averaged power flow along the line is
r
b
µ 2
P =
πa |H0 |2 ln
a
where H0 is the peak value of the azimuthal magnetic field at the surface of
the inner conductor.
(b) Show that the transmitted power is attenuated along the line as
P (z) = P0 e−2γz
where
1
γ=
2σδ
r
a1 + 1b
.
µ ln ab
(c) The characteristic impedance Z0 of the line is defined as the ratio of the voltage
between the cylinders to the axial current flowing in one of them at any
position z. Show that for this line
r
1
µ
b
Z0 =
ln
.
2π a
(d) Show that the series resistance and inductance per unit length of the line are
1
1 1
R=
+
2πσδ a b
µ
µc δ 1 1
b
L=
+
.
ln
+
2π
a
4π a b
(a) For the TEM mode, the electric field in the waveguide may be written
E(x, y, z, t) = Et (x, y)e−ikz e−iωt
where Et has only x and y components and may be derived from a scalar
potential, i.e. Et = −∇t Φ. Since Φ satisfies the Laplace equation, we may
write its general form immediately (neglecting an arbitrary constant):
Φ(ρ, θ) = β0 ln ρ +
∞
X
l=1
(αl ρl + βl ρ−l ) sin(lθ + αl ).
Homer Reid’s Solutions to Jackson Problems: Chapter 8
3
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 walls
vanish, i.e.
∂Φ
∂Φ
=
= 0.
∂θ r=b
∂θ r=a
This yields no condition on β0 , since the θ derivative of that term vanishes
anyway, but on the terms in the summation we obtain the conditions
αl al + βl a−l = αl bl + βl b−l = 0
which can only be satisfied if αl = βl = 0 for l 6= 0. Hence we have
Φ(ρ) = β0 ln ρ
−→
1
E = −β0 ρ̂.
ρ
The magnetic field is found from Jackson’s (8.28):
r
1
H=− B=
(z × E)
µ
µ
r
β0
=
θ̂.
µ ρ
(1)
(2)
The time-averaged Poynting vector is
1
1
S = (E × H∗ ) =
2
2
r
|β0 |2
µ
2
1
ẑ
ρ
Integrating over the cross section of the waveguide, we obtain the power transfer:
P =
Z
b
a
Z
2π
0
S · dA
r
Z b
2πρ dρ
1 |β0 |2
2 µ
ρ2
a
r
b
=
|β0 |2 · π ln
µ
a
r
µ
|β0 |2
b
=
·
(πa2 ) ln
a
µ a2
=
(3)
Referring back to (2) to rewrite the term in brackets, we obtain
P =
r
µ
(πa2 ) ln
b
|H(a)|2
a
(4)
(b) Without going back and completely re-solving for the fields in the waveguide
for the case of finite conductivity, we can calculate the power loss per unit length
Homer Reid’s Solutions to Jackson Problems: Chapter 8
4
approximately using Jackson’s equation (8.58):
I
dP
1
|ρ̂ × H|2 dl
−
=
dz
2σδ c
2 !
2
1
β0
β0
=
+ 2πa ·
2πb ·
2σδ µ
b
a
2
1 1
πβ0 =
.
+
σδ µ
b a
Dividing by (3), we obtain
1 dP
1
γ=−
=
2P dz
2σδ
r
(c) The fields inside the waveguide are
a1 + 1b
.
µ ln ab
β0
E(ρ, z, t) = − ei(kz−ωt) ρ̂
ρ
r
β0 i(kz−ωt)
H(ρ, z, t) =
e
θ̂
µ ρ
From the E field we can compute the voltage difference between the cylinders:
V (z, t) = −β0 ei(kz−ωt)
Z
b
a
b
dρ
= −β0 ei(kz−ωt) ln
ρ
a
(5)
while from the H field we can compute the axial current flowing in, say, the
outer cylinder:
r
I = 2πb|Kb | = 2πb|ρ̂ × H(ρ = b)| = 2π
β0 ei(kz−ωt) .
(6)
µ
Dividing (5) by (6), we have
Z=
V
1
=
I
2π
r
µ b
ln .
a
Homer Reid’s Solutions to Jackson Problems: Chapter 8
5
Problem 8.4
Transverse electric and magnetic waves are propagated along a hollow, right circular
cylinder of brass with inner radius R.
(a) Find the cutoff frequencies of the various TE and TM modes. Determine numerically the lowest cutoff frequency (the dominant mode) in terms of the
tube radius and the ratio of cutoff frequencies of the next four higher modes
to that of the dominant mode. For this part assume that hte conductivity of
brass is infinite.
(b) Calculate the attenuation constant of the waveguide as a function of frequency
for the lowest two modes and plot it as a function of frequency.
(a) The equation we have to solve is
(∇2t + γ 2 )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) =
∞
X
JL (γρ)(AL eiLθ + BL e−iLθ ) + NL (γρ)(CL eiLθ + DL eiLθ ).
L=0
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 think
there are any constraints on AL and BL . I guess these guys are determined by
the field configuration one forces into the waveguide at one of its ends.
The allowed values of γ are determined by the boundary conditions. These
are
TM case :
TE case :
Ψ|ρ=R = 0
∂Ψ
∂ρ
=0
=⇒
JL (γR) = 0
(7)
=⇒
JL0 (γR) = 0
(8)
ρ=R
(9)
Hence the allowable eigenvalues are given by
γi =
xi
R
Homer Reid’s Solutions to Jackson Problems: Chapter 8
6
where the xi are the roots of JL (x) = 0 and JL0 (x) = 0. Referring to Jackson’s
tables on pages 114 and 370, we can write down the five lowest-lying eigenvalues:
γ1 =
γ2 =
γ3 =
γ4a =
γ4b =
1.841
,
R
2.405
,
R
3.054
,
R
3.832
,
R
3.832
,
R
TE, L = 1
TM, L = 0
TE, L = 2
TE, L = 1
TM, L = 0.
The last two eigenvalues are degenerate.
The lowest cutoff frequency is
γ1
1.841
ωc = √ = √ .
µ
R µ
(b) The lowest-lying mode is the TE mode with L = 1. For this mode we have
Hz (ρ, θ, z, t) = H0 J1 (γ1 ρ)eiθ ei(kz−ωt)
(10)
with k 2 = µω 2 − γ12 . The tangential component of the field, from Jackson
(8.33), is
k
(11)
Hθ (ρ, θ, z, t) = − 2 Hz
ργ
Using (10) and (11), we can find the current induced in the wall of the conductor
at ρ = R:
k
Keff = ρ̂ × H(ρ = R) = −H0 J1 (γ1 R)eiθ ei(kz−ωt ) θ̂ +
ẑ
.
Rγ12
Then Jackson (8.58) is
"
2 #
dP
1
k
2 2
−
=
H J (γ1 R) · 2πR 1 +
dz
2σδ 0 1
Rγ12
On the other hand, the transmitted power is given by Jackson (8.51):
2 ω 2 1/2 Z
µ ω
λ
1−
Hz∗ Hz dA
ωλ
ω
A
r 2 ω 2 1/2 Z R
µ
ω
λ
ρJ12 (γ1 ρ) dρ
1−
= πH02
ωλ
ω
0
P =
1
2
r
(12)
7
Homer Reid’s Solutions to Jackson Problems: Chapter 8
The integral can be evaluated with Jackson (3.95):
= πH02
r
µ
ω
ωλ
2 1−
ω 2 1/2 R2
λ
ω
2
[J2 (γ1 R)]2
(13)
Dividing (12) by (13), we calculate the attenuation coefficient:
2 "
1/2
2 # r J1 (1.841)
k
ω2
2
1 dP
ωλ 2
1+
=
β=
2P dz
σδR µ J2 (1.841)
Rγ12
ω
ω 2 − ωλ2
2 "
1/2
2 # r 1 dP
J1 (1.841)
Rk
ω2
2
ωλ 2
=
1+
=
2P dz
σδR µ J2 (1.841)
(1.841)2
ω
ω 2 − ωλ2
2 1/2
r 1 dP
2
ω2
J1 (1.841)
µR2 ω 2 ωλ 2
=
=
.
2P dz
σδR µ J2 (1.841)
(1.841)2
ω
ω 2 − ωλ2
Problem 8.5
A waveguide is constructed so that
√ the cross section of the guide forms a right
triangle with sides of length a, a, 2a, as shown. The medium inside has µr = r =
1.
(a) Assuming infinite conductivity for the walls, determine the possible modes of
propagation and their cutoff frequencies.
(b) For the lowest mode of each type calculate the attenuation constant, assuming
that the walls have large, but finite, conductivity. Compare the result with
that 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. The
task is to solve (∇2t + γ 2 )Ψ = 0 subject to the vanishing of Ψ, or its normal
derivative, at the walls. In the text, Jackson finds the form of the solutions
for a rectangular waveguide. A little bit of staring at the triangular waveguide
reveals that appropriate solutions for this geometry can be assembled from linear
combinations of the solutions for the rectangular case. For example, a term like
sin kx x sin ky y, for suitable choices of kx and ky , already vanishes on the two
legs of the triangle. To get it to vanish on the third boundary surface, i.e. the
hypotenuse (x = y), we can simply subtract the same term with kx and ky
swapped. In other words, we take
h mπx nπy nπx mπy i
sin
− sin
sin
Amn sin
a a a a i
h
mπx
nπy
nπx
mπy
X
Hz (x, y) =
Bmn cos
cos
+ cos
cos
a
a
a
a
Ez (x, y) =
X
(TM)
(TE)
Homer Reid’s Solutions to Jackson Problems: Chapter 8
8
The TE case involves the plus sign because in the normal derivative on the
diagonal boundary surface the x derivative comes in with the opposite sign as
the y derivative.
2
These satisfy (∇2t + γmn
)Ψ = 0, where
π 2
2
(n2x + n2y ).
=
γmn
a
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 be
had for the TE1 , 0 mode, in which case
π
γ10 =
a
√
and the cutoff frequency is ωc(1,0) = π/(a µ). 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
πy i
h
πx + cos
Hz = H0 cos
a
a
h
ikπ
πx
πy i
Ht =
î + sin
ĵ .
H0 sin
2
aγ10
a
a
The power loss is
I
1
dP
=
|n × H|2 dl
(14)
−
dz
2σδ
On the lower surface (y = 0) we have
h
i2
πx k2 π2
2 πx
2
2
2
2
+ 1 + 2 4 sin
|n × H| = |Hy + Hz | = H0
cos
a
a γ10
a
The contribution of the lower surface to the integral in (14) is thus
Z
k2 π2
3
+ 2 4
= aH02
2 a γ10
lower
(15)
The contribution of the right (vertical) boundary surface is the same. On the
diagonal boundary surface, we have
1
n = √ (−î + ĵ)
2
=⇒
1
n × H = √ [Hz î + Hz ĵ − (Hy + Hx )k̂]
2
with magnitude
1 2
[Hx + Hy2 + Hz2 + 2Hx Hy ]
2 (
)
2
kπ
H02
2 πγ
2 πγ
4 cos
+4
sin
=
2
a
aγ10
a
|n × H|2 =
9
Homer Reid’s Solutions to Jackson Problems: Chapter 8
where γ = x = y is the common coordinate
as we move from (0, 0) to (a, a). In
√
the integral in (14) we can put dl = 2dγ and integrate over γ from 0 to a to
obtain
Z
√
k2 π2
2
= 2aH0 1 + 2 4 .
a γ10
diagonal
Adding this to two times (15) and inserting into (14), we have
√
√ k2 π2
aH02
dP
=
−
3 + 2 + (2 + 2) 2 4 .
dz
2σδ
a γ10
(16)
On the other hand, from Jackson (8.51) we have
H2
P = 0
2
r
1/2
2 ω
ωλ2
1− 2
ωλ
ω
Z aZ yh
πx πy πx πy i
cos2
×
+ cos2
+ 2 cos
cos
dx dy
a
a
a
a
0
0
µ
By symmetry, the integral is just half of what we would get from integrating
the integrand over a square of side a, which, by inspection, is a2 . Hence
a2 H02
P =
4
r
µ
ω
ωλ
2 ω2
1 − λ2
ω
.1/2
We could at this point proceed to write out the explicit form of the attenuation
constant, but what’s the point?
Problem 8.6
A resonant cavity of copper consists of a hollow, right circular cylinder of inner
radius R and length L, with flat end faces.
(a) Determine the resonant frequencies of the cavity for all types of waves. With
√
(1/ µR) 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 have
the lowest frequency for all R/L ?
(b) If R=2 cm, L=3 cm, and the cavity is made of pure copper, what is the
numerical value of Q for the lowest resonant mode?
(a) Taking the origin at the center of the cavity, the ρ and φ components of the
fields must vanish at z = ±L/2. Since the z dependence of all field components
is e±ikz , the allowed values of k are k = nπ/L, with E ∝ sin kz for k even and
E ∝ cos kz for k odd.
Homer Reid’s Solutions to Jackson Problems: Chapter 8
10
The equation characterizing T M modes is
(∇2t + γ 2 )Ez = 0,
Ez |∂S = 0.
Expanding this in cylindrical coordinates, we obtain
∂ 2 Ez
1 ∂Ez
1 ∂ 2 Ez
+
+
+ γ 2 Ez = 0
∂ρ2
ρ ∂ρ
ρ2 ∂φ2
We put Ez (ρ, φ) = R(ρ)P (φ) to obtain
∂2P
+ νP = 0
∂φ2
ν2
∂ 2 R 1 ∂R
2
+
+ γ − 2 = 0.
∂ρ2
ρ ∂ρ
ρ
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
(
keven
ρ ±iνφ −iωt sin kz,
(TM modes).
e
e
Ez = AJν xνm
R
cos kz,
kodd
For TE modes the relevant equation is
(∇2t + γ 2 )Bz = 0,
∂Bz
∂n
= 0.
∂S
The general solution to the differential equation is the same as above, but now
the boundary condition requires Jν0 (γR) = 0, so γ = yνm /R where yνm is the
mth root of Jν0 (y) = 0. Then the solutions look like
(
ρ ±iνφ −iωt sin kz,
keven
Bz = AJν yνm
(TE modes).
e
e
R
cos kz,
kodd
As we saw above, the allowed wavevectors are k = nπ/L. The frequency is
related to the wavenumber according to
p
2 + k2
ωνmn = cm γνm
n
s

2

cm
R

2 + π2

x
n2 ,
(TM)

νm
 R
L
=
s
2



c
R
m

2
2

yνm + π
n2 ,
(TE)
R
L
11
Homer Reid’s Solutions to Jackson Problems: Chapter 8
PSfrag replacements
9
TE modes
TM modes
ω (units of c/R)
8
7
6
5
4
3
2
1
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
R/L
Figure 1: TM and TE mode frequencies for the resonant cavity of Problem 8.6.
2
Homer Reid’s Solutions to Jackson Problems: Chapter 8
(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
12
Solutions to Problems in Jackson,
Classical Electrodynamics, Third Edition
Homer Reid
January 13, 2003
Chapter 11: Problems 1-8
Problem 11.4
A possible clock is shown in the figure. It consists of a flashtube F and a photocell
P shielded so that each views only the mirror M, located a distance d away, and
mounted rigidly with respect to the flashtube-photocell assembly. The electronic
innards of the box are such that when the photocell responds to a light flash from
the mirror, the flashtube is triggered with a negligible delay and emits a short flash
toward 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 the
line from P F to M, relative to an observer. Using the second postulate
of relativity, show by explicit geometrical or algebraic construction that the
observer 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 P F
to M. Verify that here, too, the clock is observed to tick more slowly, by the
same time dilatation factor.
(a) Suppose that, relative to an observer, the clock is moving with speed v
perpendicular to the direction in which the light travels back and forth. Let dt
be the time measured by the observer for the light to travel from the flashtube
to the mirror. The vertical distance traveled by the light is d, but—as far as
the observer is concerned—during the time dt the mirror has also translated
a horizontal distance vdt. Hence the total distance the observer sees the light
1
Homer Reid’s Solutions to Jackson Problems: Chapter 11
2
p
travel (one-way) is d2 + (vdt)2 . But this distance must equal cdt, since the
light must have the universal speed c in any reference frame. Hence we have
p
cdt = d2 + (vdt)2
or
dt =
1
2
1 + vc2
!1/2
d
.
c
This is the time (as measured by the observer) in which the light makes the trip
from flashtube to mirror. The light takes the same amount of time to travel
back to the photocell, so the total period of the clock is just twice this, or
!1/2
2d
1
period =
v2
c
1 − c2
which is greater than the period of the clock at rest by the normal Lorentz
dilatation factor.
(b) Now we suppose that the clock is moving relative to an observer with speed
v parallel to the direction of motion of the light. We align the z axis with the
direction of motion. Then we may write down the space-time coordinates (in
the clock’s rest frame) of the three relevant events in the operation of the clock
(taking x4 = ct):
xa = (0, 0, 0, 0),
xb = (0, 0, d, d),
xc = (0, 0, 0, 2d),
(light leaves flashtube)
(light reaches mirror)
(light reaches photocell).
The transformation matrix from the clock’s reference frame to the observer’s
reference frame is


1 0 0
0
 0 1 0
0 

Γ=
 0 0 γ γβ 
0 0 γβ γβ
and using this we may write down the spacetime coordinates of the three events
above in the rest frame of the observer:
x0a = (0, 0, 0, 0)
x0b = 0, 0, γ(1 + β)d, γ(1 + β)d
x0c = 0, 0, 2γβd, 2γd
Evidently the difference in the time coordinates of events a and c (which is just
the observed period of the clock) is c∆t = 2γd, so again the observer observes
Homer Reid’s Solutions to Jackson Problems: Chapter 11
3
the clock to have a period of 2γd/c, longer by the factor γ than the period of
the clock in its rest frame.
Problem 11.5
A coordinate system K 0 moves with a velocity v relative to another system K. In K 0
a particle has a velocity u0 and an acceleration a0 . Find the Lorentz transformation
law for acceleration, and show that in the system K the components of acceleration
parallel and perpendicular to v are
1−
ak =
a⊥ =
1−
1+
v2
c2
v·u0
c2
1+
v2
c2
3/2
v·u0 3
c2
a0k
v
0
0
0
3 a⊥ + 2 × (a × u ) .
c
The initial components of the velocity in the moving frame (frame K 0 ) are
and u0⊥ . Using Jackson’s equations 11.31 to transform these to frame K, we
obtain
u0k
u0k + v
uk (0) =
1+
u⊥ (0) =
u0k v
c2
u0⊥
γv 1 +
(1)
uk v .
c2
(2)
After a time dt has passed in frame K, a time dt0 = dt/γ has passed in frame
K0 , and the velocity has increased by the amount a0 dt0 = a0 dt/γ. Then we can
write down the new components of the velocity in K 0 and again transform to
K:
uk (dt) =
u0k + a0k dt0 + v
(u0k +a0k dt0 )v
c2
u0⊥
(u0k +a0k dt0 )v
γv 1 +
2
c
(3)
1+
u⊥ (dt) =
Subtracting (1) from (3), we have
(4)
4
Homer Reid’s Solutions to Jackson Problems: Chapter 11
∆uk =
u0k + a0k dt0 + v
1+
(u0k +a0k dt0 )v
c2
=
1+
−
k
1+
2
u0k v
2
c
v
1 − c2
u0 v
+ 1 + ck2 a0k dt0
a0k dt0
u0 v 2
c2
u0k + v
Using the relation dt0 = dt/γ we can rewrite this as
3/2
2
a0k dt 1 − vc2
=
u0 v
u0 v 2
+ 1 + ck2
1+
1 + ck2
Dividing by dt and taking the limit as dt → 0, we obtain
v 2 1/2
c2
3/2
2
a0k 1 − vc2
∆uk
ak = lim
= .
u0 v 2
dt→0 dt
1 + ck2
a0k dt
So evidently I’m off by 1 in the exponent of the denominator. What am I doing
wrong?
Next, subtracting (2) from (4), we have


1  u0⊥ + a0⊥ dt0
u0⊥ 
∆u⊥ =
−
0
0
u0k v
γv 1 + uk v + ak dt0 v
1
+
2
2
c
c
c2

a0 dt0 v 
0 u
v
k
k
0
0
0
c2
1  a⊥ dt 1 + c2 − u⊥

=


u0k v 2
u0k v a0k dt0 v
γv
1 + c2
+ 1 + c2
2
c


v
0
0
0 0
0 0
0
a
dt
+
u
a
−
u
a
dt
2
⊥ k
k ⊥
c
1  ⊥

=


u0k v 2
u0k v a0k dt0 v
γv + 1 + c2
1 + c2
2
c
Again putting in dt = γdt0 , we have

v
0 0
0 0
0
u
a
−
u
a
a
dt
+
2
⊥
⊥
⊥
k
k dt 
c
1 
= 2 

2
0
0
u v a0k dtv
u v
γv
+ 1 + ck2
1 + ck2
γv c2

Homer Reid’s Solutions to Jackson Problems: Chapter 11
5
As before, when we divide by dt and take the limit as dt → 0 the second term
in the denominator becomes irrelevant and we obtain


h
i
1
v 0 0
∆u⊥

 0
0 0
a
+
a⊥ = lim
= u
a
−
u
a

⊥
⊥ k
k ⊥
u0 v 2
dt→0 dt
c2
γv2 1 + ck2


v2
h
i
v 0 0
 1 − c2  0
0 0
u
a
−
u
a
a
+
= 

⊥
⊥
⊥
k
k
u0 v 2
c2
1 + ck2
Jackson writes the second term in the brackets a little differently. To show that
his expression is equivalent to mine, we use the BAC-CAB rule:
v × (a0 × u0 ) = (v · u0 )a0 − (v · a0 )u0
= vu0k a0k + a0⊥ − va0k u0k + u0⊥
The parallel components cancel, and we are left with
= v(u0k a0⊥ − a0k u0⊥ )
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 twins
born in 2080 remains on earth; the other rides in the rocket. The rocket ship is
so constructed that it has an acceleration g in its own rest frame (this makes the
occupants feel at home). It accelerates in a straight-line path for 5 years (by its own
clocks), decelerates at the same rate for 5 more years, turns around, accelerates for
5 years, decelerates for 5 years, and lands on earth. The twin in the rocket is 40
years 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 as
measured on earth. Using the first result of problem 11.5, we see that the
acceleration of the rocket as measured from earth is
3/2
dv
v(t)2
a=
g,
=± 1− 2
dt
c
Homer Reid’s Solutions to Jackson Problems: Chapter 11
6
or
3/2
dβ
= ± 1 − β2
α
dt
where α = g/c and the plus or minus sign depends on whether we are accelerating or decelerating. Manipulating this a little, we obtain
dβ
= ±αdt.
(1 − β 2 )3/2
Integrating, we obtain
β0
(1 − β 02 )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 the
plus sign in (5). Then we find
β
= αt
(1 − β 2 )1/2
or
β(t) =
αt
[1 + (αt)2 ]1/2
(6)
and
γ(t) = p
1
1 − β 2 (t)
=
p
1 + (αt)2
(7)
Next let’s work out the relation between time as measured on the rocket and
time as measured on earth. With primed (unprimed) quantities referring to the
rocket (to earth), the infinitesimal relation is
dt = γ(t)dt0
or
t02
−
t01
=
=
Z
Z
t2
t1
t2
t1
dt
γ(t)
dt
p
1 + (αt)2
du
1 αt2
√
α αt1
1 + u2
1
=
sinh−1 (αt2 ) − sinh−1 (αt1 )
α
=
Z
7
Homer Reid’s Solutions to Jackson Problems: Chapter 11
For the first leg of the journey this becomes
t2 =
1
sinh(αt02 ).
α
(8)
Now we know that the first leg of the journey lasts until t02 = 5 years, and we
have
9.8 m s−2
g
= 3.27 · 10−8 s−1 .
α= =
c
3 · 108 m s−1
Then the time on earth at the end of the first leg of the journey is, from (8),
1
3.153 · 107 s
−8 −1
t2 =
s
· sinh 3.27 · 10 s (5 yr)
3.27 · 10−8
1 yr
= 2.65 · 109 s
≈ 84 yr.
Finally, the distance the rocket travels during the first leg of its journey is
Z t
d=c
β(t)dt
0
Z t
αt
p
=c
dt
1 + (αt)2
0
Z
c αt
u
√
=
du
α 0
1 + u2
o
1/2
c n
1 + (αt)2
−1
=
α
= 3.0 · 1015 meters.
The behavior of the rocket on the subsequent three legs of the journey is
similar to that in the first leg. In particular, the total distance traveled away
from earth is twice that covered in the first leg, or 6.0 ·1015 meters, and the
total time elapsed on earth during the rocket’s journey is four times that elapsed
during the first leg, or 4·84=336 years. So it should be the year 2436 on earth
by the time the rocket returns home.
Homer Reid’s Solutions to Jackson Problems: Chapter 11
8
Problem 11.13
An infinitely long straight wire of negligible cross-sectional area is at rest and has
a uniform linear charge density q0 in the inertial frame K 0 . The frame K 0 (and the
wire) move with a velocity v parallel to the direction of the wire with respect to
the laboratory frame.
(a) Write down the electric and magnetic fields in cylindrical coordinates in the rest
frame 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 rest
frame? In the laboratory?
(c) From the laboratory charge and current densities, calculate directly the electric
and 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 the
wrong units. I’ll use λ instead. We’ll take our z axis to coincide with the wire
and 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 of
a 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 field
normal to the radial bounding surface is uniform across the circumference. On
the other hand, the charge enclosed in the cylinder is λ dz. Then Gauss’ law is
I
E · dA = 4πQ
=⇒ 2πrdzEr = 4πλdz
2λ
=⇒ Er =
.
r
In terms of cartesian components we have
y0
x0
,
E
=
2λ
.
Ex = 2λ
y
x02 + y 02
x02 + y 02
(9)
9
Homer Reid’s Solutions to Jackson Problems: Chapter 11
The field-strength tensor in the laboratory frame is
F = ΛF 0 Λ̃


γ 0 0 βγ
0 −x0
 0 1 0 0   x0
1
0


= 2λ 02
0
x + y 02  0 0 1 0   y 0
βγ 0 0 γ
0
0


0
0
0 −x
−y
0
 x0
2γλ
0
0
βx0 

.
= 02
0
0
βy 0 
(x + y 02 )  y 0
0 −βx0 −βy 0
0
−y 0
0
0
0

0
γ

0 
 0
0  0
0
βγ
0
1
0
0

0 βγ
0 0 

1 0 
0 γ
Reading off from this the transformed fields, and dropping the primes on x and
y since the z boost leaves these coordinates unchanged, we have
2λ
(xi + yj)
x2 + y 2
2λ
= γ r̂
r
λ
B = βγ
(−yi + xj)
2
2(x + y 2 )
2λ
= βγ φ̂.
r
E=γ
(10)
(11)
(12)
(13)
(b) In the rest frame there is no current and the charge density is ρ0 = λδ(x)δ(y),
so
J 0µ = cλ(δ(x)δ(y), 0, 0, 0).
The transformed current density is
J µ = Λµν J 0ν

γ 0 0
 0 1 0

= cλ 
0 0 1
βγ 0 0

δ(x)δ(y)

0

= cγλ 
0
βδ(x)δ(y)


γβ
δ(x)δ(y)


0 
0





0
0
γ
0


.

(c) Computing the electric field in the laboratory frame is easy, since the charge
density is the same as we had before but with a factor of γ thrown in. Then
the E field is just (9) but with that factor of γ thrown in, i.e.
E=
2γλ
r̂
r
Homer Reid’s Solutions to Jackson Problems: Chapter 11
10
which agrees with (10).
For the magnetic field, we note that the current density in the lab frame is
J = cβγλδ(x)δ(y)ẑ. Then the current piercing a disc of radius r is I = cβγλ.
On the other hand, by symmetry the magnetic field in the azimuthal direction
around the circumference of this disc is constant, so we may use Ampere’s law,
∇ × B = (4π/c)J to write
2πrBφ =
4π
(cβγλ)
c
or
B=
2βγλ
φ̂
r
which agrees with (13).
Problem 11.15
In a certain reference frame a static, uniform, electric field E0 is parallel to the x
axis, 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 reference
frame in which the electric and magnetic fields are parallel. What are the fields in
that frame for θ 1 and θ → (π/2)?
The untransformed fields are
E = E0 i,
B = 2E0 (cos θi + sin θj).
Let’s suppose we boost along the z axis. Then the fields transform according
to Jackson equation (11.149):
E0 = γE0 (1 − 2β sin θ)i + 2γβE0 cos θj
B0 = 2γE0 cos θi + γE0 (2 sin θ + β)j.
The angle between the fields is given by
cos θ =
=
E 0 · B0
|E0 ||B0 |
γ 2 E02 (1
2γ 2 E02 cos θ(1 + β 2 )
− 4β sin θ + 4β 2 )1/2 (4 + 4β sin θ + β 2 )1/2
Setting this equal to unity, we obtain
Homer Reid’s Solutions to Jackson Problems: Chapter 11
11
Problem 11.18
The electric and magnetic fields of a particle of charge q moving in a straight line
with speed v = βc, given by (11.152), become more and more concentrated as
β → 1, as indicated in Fig. 11.9. Choose axes so that the charge moves along
the z axis in the positive direction, passing the origin at t = 0. Let the spatial
coordinates 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 = 2q
r⊥
2 δ(ct − z);
r⊥
B = 2q
v̂ × r⊥
δ(ct − z).
2
r⊥
(b) Show by substitution into the Maxwell equations that these fields are consistent
with a 4-vector source density,
J α = qcv α δ (2) (r⊥ )δ(ct − z)
where the 4-vector v α = (1, v̂).
(c) Show that the fields of part a are derivable from either of the following 4-vector
potentials,
A0 = Az = −2qδ(ct − z) ln(λr⊥ );
A⊥ = 0
or
A0 = Az = 0;
A⊥ = −2qΘ(ct − z)∇⊥ ln(λr⊥ )
where λ is an irrelevant parameter setting the scale of the logarithm. Show
that the two potentials differ by a gauge transformation and find the gauge
function, χ.
(a) In the reference frame in which the particle is at rest at the origin, the fields
are
q
E0 = 02
(x0 i + y 0 j + z 0 k),
B0 = 0.
(x + y 02 + z 02 )3/2
Transforming back to the laboratory frame according to Jackson 11.148, the
electric field is
q
E = 02
(γx0 i + γy 0 j + z 0 k)
02
(x + y + z 02 )3/2
where in this expression the coordinates are still those of the observation point
in the moving frame. The transformation of these to the lab frame is x0 = x,
y = y 0 = y, z 0 = γ(z − βct). Then the correct expression for the transformed
field is
q
E= 2
(γxi + γyj + (z − ct)k).
2
[x + y + γ 2 (z − ct)2 ]3/2
Homer Reid’s Solutions to Jackson Problems: Chapter 11
12
In the limit β → 1, we have γ → ∞. For z 6= ct the γ 2 factor in the denominator
then ensures that all field components are zero. For z = ct, however, although
the z component of the field clearly vanishes, the behavior of the other components is not as immediately clear. To elucidate the behavior of, say, the x
component of the field at z = ct we integrate it from z = ct − to z = ct + :
Z ct+
Z ct+
dz
Ex dz = qγx
2
2
2 3/2
ct−
ct− [r⊥ + γ (z − ct) ]
Z γ/r⊥
qx
du
= 2
r⊥ −γ/r⊥ [1 + u2 ]3/2
γ
2qx
= 2 p 2
.
r⊥
r⊥ + γ 2 2
Taking the limit γ → ∞ for any finite , we find
Z ct+
2qx 2
.
lim
Ex dz =
γ→∞ ct−
r⊥
(14)
On the other hand, integrating between two points on the same side of z = ct,
say from z = ct + to z = ct + 2, we find
"
#
Z ct+2
2γ
γ
2qx 2
p
−p 2
Ex dz =
2 + 4γ 2 2
r⊥
r⊥
r⊥ + γ 2 2
ct+
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
2 δ(z − ct).
r⊥
Ey =
2qy
2 δ(z − ct).
r⊥
and, similarly,
Combining these, we can write
E = 2q
r⊥
2 δ(ct − z).
r⊥
The B field is given by Jackson (11.150) with, in the ultrarelativistic limit,
β = k̂ :
v̂ × r⊥
δ(ct − z).
B = 2q
2
r⊥