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
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
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
a Z a d
d
d
ρ
Jν (k 0 ρ)
Jν (kρ) dρ.
= ρJν (k 0 ρ) Jν (kρ) −
dρ
dρ
dρ
0
0
2
(1)
(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
1
λ
d
ln[Jν (kρ)]
=
Jν (kρ)
=−
dρ
Jν (kρ) dρ
a
ρ=a
ρ=a
→ a
d
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
2 2
a
2
2
0
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
2 2
1 a +z
= −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
0 1
0
0
0
0 ∂G(x; x ) 0
G(x; x )ρ(x ) dV +
Φ(x )
Φ(x) = −
0 dA .
0 V
∂n
S
x
(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)
P
(x)
dx
= −
l
2
∂n r0 =a
0
l=0
∞
∂Rl V X 2
a γl Pl (cos θ) ·
= −
2
∂n r0 =a
l=0
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
Z 0
∞
∂Rl V X 2
b Pl (cos θ)
P
(x)
dx
l
2
∂n r0 =b
−1
l=0
∞
∂Rl V X 2
b γl Pl (cos θ)
2
∂n r0 =b
= −
=
l=0
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 :
0
∞
02 ∂Rl r =b
V X
Φ(r, θ) =
(12)
γl Pl (cos θ) r
2
∂n r0 =a
l=0
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 − ab
r 0 =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
Z
d
r
d2 − z 2
dz
z l+1
)
d )
2
d r
r
1
1
l d
−
+ r − l +
= λ
l+1
l−2
r
l+1
l+3
lz
(l − 2)z r
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)
(
2 l+1
l+3
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
∂
0 ∂
+ F cos θ = −σ
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.