KALMAN KNIZHNIK - SOLUTIONS TO JACKSON - CHAPTER 1
JACKSON 1.1
Part a
Since a conductor has no E field inside, by Gauss’s law ∇ · E = ρ0 there can also be no charge
density inside the conductor. This means that excess charge must lie entirely on its surface. 2
Part b
You can place a Gaussian surface just inside the conductor. HSince there is no charge density inside,
by Gauss’s law (perhaps more easily seen in integral form) E · da = Qenc
there is also no E field
0
inside, in spite of all of the charges outside the conductor. However, if there are charges inside the
hollow conductor, the same law tells you that Qenc 6= 0, meaning that the field outside is nonzero
too. Hence, it shields its interior from charges outside but does not shield its exterior from charges
inside. 2
Part c
The E field on the surface of the conductor must be normal to its surface, since otherwise charge
would flow on the surface (we are making the assumption that the charge is static. As for the
magnitude, we use Gauss’s law in integral form:
Qenc
(1)
0
= σ n̂ · da, since the only charge that exists is on the surface. Thus, we rewrite Gauss’s
I
Now, dQenc
law as
E · da =
σ
n̂ · da =
0
and since this must hold for any area, we conclude
Z
E · da =
Z
E=
Z
(E −
σ
n̂) · da = 0
0
(2)
σ
n̂ 2
0
(3)
JACKSON 1.2
Starting with Jackson’s definition of the delta function:
D(α; x, y, z) = (2π)3/2 α−3 exp[−
x2 + y 2 + z 2
]
2α2
(4)
we note that D → 0 unless x, y, z → 0 as well (i.e. replace x, y, z with dx, dy, and dz). Now notice
that dx2 + dy 2 + dz 2 ≡ ds2 is the arc length squared. We can write D ∝ exp[−ds2 /2α2 ]. In the
general coordinate system given by Jackson, we can write the square of the arc length as:
ds2 = (
dv
dw
du 2
) + ( )2 + ( )2
U
V
W
(5)
This is not to say that dx = du/U . Rather, I have replaced the entire arc length ds, which is the
same in all coordinate systems.
Let du → u − u0 , dv → v − v 0 , and dw → w − w0 . We obtain:
0 2
D(α; u, v, w) =
e−(u−u ) /2α
√
2πα
2U 2
0 2
e−(v−v ) /2α
√
2πα
2V 2
0 2
e−(w−w ) /2α
√
2πα
2W 2
(6)
This is starting to look like the usual definition of the delta function. Now for some trickery. We
will replace the α’s in each term with αu /U, αv /V, αw /W . We justify this by saying that each of
the α’s go to zero, so the replacement does not change the limit of the function. The result of this
replacement is
0 2
2
0 2
2
0 2
2
e−(u−u ) /2αu e−(v−v ) /2αv e−(w−w ) /2αw
√
√
√
D(α; u, v, w) =
UV W
2παu
2παv
2παw
(7)
This is the result we were looking for. The left hand side (or the right hand side in equation 4) is
δ(x − x0 ). Meanwhile the left hand side is the product of three delta functions time U V W . Thus,
δ(x − x0 ) = δ(u − u0 )δ(v − v 0 )δ(w − w0 )U V W 2
(8)
This result is useful because it can be used to show that for spherical coordinates, for instance,
when u = r, v = θ, w = φ, and the length elements are dr, rdθ, rsinθdφ, (which leads to U = 1, V =
1/r, W = 1/rsinθ) we can find that
δ(x − x0 ) = δ(r − r0 )δ(θ − θ0 )δ(φ − φ0 )
1
r2 sinθ
(9)
JACKSON 1.3
Part a
The charge is uniformly distributed over a spherical shell, so its position is a constant R. Therefore
we can write
ρ(r) ∝ δ(r − R)
(10)
(I Rchoose to switch from x to the more natural spherical r). Of course, I need to normalize it,
3 ρ(r)dV = Q. I therefore write:
Z
Q=
dV Cδ(r − R) = C
Z
r2 sinθδ(r − R)drdθdφ = 4πCR2
(11)
and therefore C = Q/4πR2 , meaning
ρ(r) =
Q
δ(r − R) 2
4πR2
(12)
Part b
By the same reasoning as in part a,
ρ(r) ∝ δ(r − b)
(13)
and normalizing:
Z
λ=
dACδ(r − b) = C
Z
rdθdrδ(r − b) = 2πCb
(14)
meaning that C = λ/2πb, resulting in:
ρ(r) =
λ
δ(r − b) 2
2πb
(15)
There was no integral over z because we were normalizing to the charge density.
Part c
Here we must employ the Heaviside theta function to write that
ρ(r) ∝ Θ(R − r)δ(z)
(16)
where the step function tells us that there is no charge outside of the disk, and the delta function
tells us that the charge is confined to the disk. Normalizing:
Z
Q=
dV CΘ(R − r)δ(z) = C
Z
1
rdzdφdrCΘ(R − r)δ(z) = R2 2πC
2
(17)
we find that C = Q/πR2 (the theta function is akin to an integral from 0 to R, and the integral of
a delta function over all space is unity). Thus,
ρ(r) =
Q
Θ(R − r)δ(z) 2
πR2
(18)
Part d
The same as in part c, except that here instead of being confined to the z-plane, we are confined
to the θ = π/2 plane.
Z
Q=
π
dV CΘ(R − r)δ(θ − ) = C
2
Z
r2 sinθdrdθdφCΘ(R − r)δ(θ −
π
) = 2πR2 C
2
(19)
leading to:
ρ(r) =
Q
π
Θ(R − r)δ(θ − ) 2
2
2πR
2
(20)
JACKSON 1.4
Part a
The first sphere, being a conductor, has no electric field inside (E= 0 for r < a). Outside the
sphere, we can place a Gaussian surface, and since the enclosed charge is Q, it is trivial to show
that outside, E = 4πQ0 r2 r̂ if r > a. 2
Part b
Since the charge density is uniform, we write Qenc = Q(4πr3 /4πa3 ) inside and Qenc = Q outside
and integrate E · da from 0 to either some point r inside the sphere or over its whole area. Thus,
for inside the sphere,
Z r
Qr
1 r3
r̂ 2
(21)
E · da = Q 3 ⇒ Ein =
0 a
4π0 a3
0
and for outside the sphere:
Z a
0
E · da =
Q
1
Q ⇒ Eout =
r̂ 2
0
4π0 a2
(22)
Part c
This sphere has a charge density ρ ∝ rn . Therefore, dQ = Arn dV . Now integrate.
Z r
0
E · da =
1
0
Z
AdrdΩr2 rn =
4πA
0
Z r0
rn+2 dr =
0
4πA 1 n+3
r
0 n + 3
(23)
and recalling that the area of our Gaussian sphere of radius r is 4πr2 , we obtain that the electric
field inside the sphere is:
A rn+1
Ein =
r̂ 2
(24)
0 n + 3
We
are not done, however, because we still want to find the constant A. To do so, recall that
Ra
ρdV
= Q. Performing this integration, with ρ = Arn , it is easy to find that A = Q(n+3)/4πan+3 .
0
Thus, the electric field inside this sphere is:
Ein =
Q rn+1
r̂ 2
4π0 an+3
(25)
Of course, the E field outside is the same in this case as it was in parts a and b. The field outside
does not depend on the form of the charge density inside, only on the total charge.
JACKSON 1.5
Our potential is
q e−αr
αr
Φ=
(1 +
)
(26)
4π0 r
2
Now, one may be tempted to use ρ = −0 ∇2 Φ. However, this is problematic in that the potential
is singular at r = 0. Before dealing with that, let’s take the Laplacian of our potential (multiplied
by −0 ).
∂
∂
qα3 e−αr
−0 ∇2 Φ = −0 r2 Φ = −
(27)
∂r ∂r
8π
Now, for r ≈ 0, we must expand Φ:
Φ≈
q
αr
(1 +
)
4π0 r
2
(28)
where I have used the fact that for x 1, e±x ≈ 1. The second term in the parentheses will
contribute nothing to the Laplacian, since it will have no r dependence when multiplied out. But
the first term looks a lot like the potential of a point charge. Now, from Jackson page 35, we recall
that ∇2 ( 1r )=−4πδ 3 (r). This allows us to write the full charge density (remembering to cancel out
4π’s and 0 ’s):
qα3 e−αr
ρ = qδ 3 (r) −
2
(29)
8π
The first term represents the proton charge, located at the origin, where the potential blows up,
while the second term represents the negatively Rcharged electron cloud in the 1s orbital around
the proton. You can verify that the total charge ρdV = 0, as it should. If you had forgotten to
include the delta function, the total charge would be -q, which is wrong, since the hydrogen atom
consists of the positively charged nucleus and the negatively charged electron cloud, which sum to
0.
JACKSON 1.6
Part a
H
Let’s examine only one side of the capacitor first. From Gauss’s law, E · da = Qenc /0 , and using
a Gaussian pillbox of negligible thickness and two sides of area A, we find that the left hand side
is 2|E|A. Meanwhile, on the right hand side, the charge enclosed is the charge density σ times the
area A of the actual plate. I don’t have a factor of 2 here because the charge is only on a single
plate, whereas the Gaussian pillbox that I am using has two plates, so its area is 2A. Now, setting
the two sides equal, I find that
σ
E=
n̂.
(30)
20
However, the capacitor has two plates, and since the charge is opposite on the second plate, the
electric field will be pointing toward one plate and away from the other, meaning that the electric
field points in the same direction, call it n̂. Then the total electric field between the plates is
E=
σ
n̂.
0
(31)
From the definition of capacitance, C = Q/V , and it is well known that the potential difference of
a capacitor is V = Ed, where d is the spacing between the plates. Thus, V = σd/0 . Dividing the
charge on one plate, Q = σA, by this potential, results in
C=
0 A
2
d
(32)
Part b
The electric field between two concentric conducting spheres is easily found from Gauss’s law to be
E=
Q
r̂ ∀ r ∈ [a, b]
4π0 r2
(33)
since it is just the same as the electric field outside a conducting sphere, but restricted to the region
between a and b. The potential is
V =−
Z a
E · dl =
b
Q 1 1
( − )
4π0 a b
(34)
meaning that the capacitance Q/V is
C = 4π0 (
ab
) 2
b−a
(35)
Part c
We can find the electric field between two concentric cylinders (or look it up) from Gauss’s law.
I
E · da =
Qenc
Q
⇒ E2πrL =
0
0
(36)
where r is the radius of our Gaussian cylinder (remember that the area of a cylinder of length L
and radius r is 2πrL). Thus, E = Q/2πrL0 . Once again, we integrate to find the potential
V =−
Z a
E · dl =
b
Q
2πL0
Z a
dr
r
b
=
b
Q
ln
2πL0 a
(37)
and so the capacitance is
C=
2πL0
2
ln ab
(38)
Part d
For this question, we just plug and chug to solve for b, and remember that we want is the diameter,
i.e. 2b. Also note that we are dealing with the capacitance per unit length.
ln
b
2π8.85 × 10−12
=
⇒ 2b = 6.4 × 10−3 m 2
.5 × 10−3
3 × 10−11
(39)
b
2π8.85 × 10−12
=
⇒ 2b = 1.1 × 105 m 2
(40)
.5 × 10−3
3 × 10−12
Note how much you need to shrink your coaxial capacitor in order to improve your capacitance by
just an order of magnitude.
JACKSON 1.7
The electric field outside a single cylindrical conductor is:
ln
I
E · da =
Qenc
Q
λ
r̂
⇒ E2πrL =
⇒E=
0
0
2πr0
(41)
where λ is the charge per unit length. Thus, the total field along the line joining the two conductors
is
λ
λ
E=(
+
)r̂
(42)
2πr+ 0 2πr− 0
where r+ , r− are the distances from the two conductors. Then the potential difference between
them is
V = Φ+ − Φ− = −
Z +
−
Z d−a2
dr+
λ
E · dl =
{
2π0
a1
r+
+
Z d−a1
dr−
a2
r−
}=
λ
(d − a1 )(d − a2 )
ln
(43)
2π0
a1 a2
Where did I get this from? Well, we need to find the potential from conductor + and subtract
from it the potential due to conductor -. This tells us the potential difference between the two
conductors. Now, the potential due to the first conductor is the integral of the electric field from
its own surface, a1 , to the surface of the second conductor, d − a2 (there is no field inside either
conductor, and d is the line joining the centers of the conductors). The potential from the second
conductor is the integral of the electric field from its own surface, a2 , to the surface of the first
conductor, d − a1 . We can make the approximation
V ≈
λ
d2
ln
2π0 a1 a2
(44)
since d a1 , a2 . Now we are able to find the capacitance per unit length, C = λ/V :
C = 2π0 (ln
d2 −1
)
a1 a2
(45)
Now, for those of you unaware of what geometric mean is, the geometric mean G of a set of numbers
x1 , ..., xn is defined as
√
G = n x1 ...xn
(46)
√
Thus, the geometric mean of the radii is just a1 a2 ≡ a.
C = 2π0 (ln
d2 −1
π0
) = d
2
a
ln a
2
(47)
To find the diameter of the wire, we solve for a:
a = de−π0 /C
(48)
If the capacitance in the wire is 1.2 × 10−11 F/m, then we get a ≈ 0.1d. Plugging in the separation
distances of 0.5cm, 1.5cm, and 5.0cm, we get diameters 2a of 1mm, 3mm, and 10mm. 2
JACKSON 1.8
Part a
In problem 1.6, I found the electric field of these three situations. Electrostatic energy U = 12 CV 2 ,
and energy density is w = 12 0 |E|2 . Using the results from 1.6a,
U=
1 2
1 Q2 d
1 0 A σd 2
( ) =
σ Ad =
2
2 d 0
20
20 A
(49)
while the energy density is
1 σ2
Q2
w = 0 2 =
2
2 0
20 A2
(50)
Note that the energy density of a parallel plate capacitor is constant.
Part b
Using the results from 1.6b
1
ab
Q 1 1 2
Q2 b − a
U = 4π0
(
2
( − )) =
2
b − a 4π0 a b
8π0 ab
(51)
while the energy density is
w=
1 Q2
∀ r ∈ [a, b] 2
32π 2 0 r4
(52)
Part c
Using the results from 1.6c
U=
1 2πL0 Qln ab 2 Q2 ln ab
(
) =
2 ln ab 2πL0
4πL0
2
(53)
while the energy density is
1
Q
Q2
w = 0 (
)2 = 2 2 2 ∀ r ∈ [a, b] 2
2 2πrL0
8π r L 0
(54)
Note that the energy is more strongly concentrated close to the inner conductor for a spherical
capacitor (energy density goes as r−4 ), than for a cylindrical parallel capacitor, where energy
density goes as r−2 .
JACKSON 1.9
Part a
H
The electric field due to a single capacitor plate is calculated by E · da = Q/0 ⇒ E · 2A =
Q/0 ⇒ E = Q/20 A. Thus the force is:
F = QE =
Q2
2
20 A
(55)
Meanwhile, the electric field due to a cylinder was found in problem 1.7, so the force on one cylinder
due to the other is
Q2
Q2
=
F = QE =
2
(56)
2π0 Lr r=d 2π0 Ld
Part b
Recall from problem 1.6a that for a parallel plate capacitor, the potential difference was V =
Qd/0 A. Solving for Q:
0 AV
Q=
(57)
d
and plugging into equation 55, we get:
F =
2
( 0 AV
d )
2
20 A
(58)
For the parallel cylinder capacitor, recall from problem 1.7 that C = π0 /ln ad , meaning that the
charge can be written in terms of the potential difference Q = CV as
π0 V
ln ad
(59)
π0 V 2
2
2Ld(ln ad )2
(60)
Q=
meaning that the force, from equation 56 is:
F =
JACKSON 1.10
In mathematical form, the mean value theorem says
1
Φ(x) =
4πR2
I
Φ(x0 )d3 x0 ,
(61)
S
i.e. that if we take our sphere and integrate the potential over it, then normalize by the area of the
sphere, it will be equal to the potential in the middle. We will use section 1.8 of Jackson as our
guide. We start with Green’s Theorem:
Z
2
2
I
3 0
(φ∇ ψ − ψ∇ φ)d x =
(φ
V
S
∂ψ
∂φ
− ψ )d2 x0
∂n
∂n
(62)
∂ R1
1 ∂Φ 2 0
−
)d x
∂n
R ∂n
(63)
Jackson sets ψ = 1/R and φ = Φ, so we shall copy him:
Z
1
1
(Φ∇
− ∇2 Φ)d3 x0 =
R R
2
V
I
(Φ
S
Let’s examine the left hand side first. We can use the property that ∇2 (1/R) = −4πδ(x − x0 ). This
is convenient because it allows us to easily evaluate the first term. Meanwhile, the second term
on the left hand side is just Poisson’s ∇2 Φ = ρ/0 . Fortunately, the problem asks for charge free
space, so that integral vanishes, since there is no charge inside the region of integration. Thus, the
left hand side becomes:
Z
V
1
1
(Φ∇
− ∇2 Φ)d3 x0 =
R R
2
Z
{Φ(x0 )(−4π)δ(x − x0 ) − (
V
1 ρ
)}d3 x0 = −4πΦ(x)
R 0
(64)
Now the right hand side. The derivative of R with respect to the normal is just −1/R2 , since the
normal is radial to the surface. The right hand side becomes
I
(Φ
S
∂ R1
1 ∂Φ 2 0
−
)d x =
∂n
R ∂n
I
S
−
Φ(x0 )
1 ∂Φ(x0 ) 2 0
−
d x
R2
R ∂n
(65)
To integrate the last term, note that 0 ∂Φ/∂n = σ (see section 2.3.5 of Griffiths). You can see
where this is going. Once again, since we are in charge free space, this term vanshes. In this way,
the right hand side has been reduced to:
I
−
S
1 ∂Φ(x0 ) 2 0
Φ(x0 )
1
−
d x =− 2
R2
R ∂n
R
I
Φ(x0 )d2 x0
(66)
S
Note: R = |x − x0 | is a constant on a sphere, so it can be pulled out of the integral. For a more
complicated surface we could not pull it out. Now, going back to equation 63, and using equations
64 and 66, we get the equality:
1
−4πΦ(x) = − 2
R
I
Φ(x0 )d2 x0
(67)
Φ(x0 )d3 x0 2
(68)
S
Therefore, we have proved the mean value theorem:
1
Φ(x) =
4πR2
I
S
JACKSON 1.11
The figure for this problem is taken from:
http://www-personal.umich.edu/p̃ran/jackson/P505/F07 hw01a.pdf.
Let’s look at the 2-D situation first. We will take a curved Gaussian box aboveH the charged
conductor, such that no charges are enclosed, and use Gauss’s law in integral form: S E · da = 0.
For this box, Gauss’s law says:
I
S
E · da = 0 = Etop atop − Ebottom abottom
(69)
where the a’s corresponds to the top and bottom of the box. There is no contribution from the sides
because they are taken to be normal to the surface. As is clear from the figure, atop = (R + )dθdz,
while abottom = Rdθdz. Plugging this into equation 69 (and canceling out dθ’s and dz’s, we get
Ebottom = Etop (1 +
)
R
(70)
From this we can calculate dE/dn.
Ebottom (1 − R ) − Ebottom
Etop − Ebottom
Ebottom
∂E
= lim
≈ lim
=−
→0
→0
∂n
R
(71)
And therefore:
1 ∂E
1
=−
(72)
E ∂n
R
since Ebottom ≈ Etop as → 0. Now we return to the 3-D situation. This time, the areas of the top
and bottom of the new three-dimensional box are abottom = R1 R2 dΩ and atop = (R1 + )(R2 + )dΩ.
Plugging this into expression 69:
Ebottom = Etop (1 +
)(1 +
)
R1
R2
(73)
This time, taking the limit in equation 71 yields
Ebottom (1 −
Etop − Ebottom
lim
≈ lim
→0
→0
R1 )(1
−
R2 )
− Ebottom
≈ −Ebottom (
1
1
+
)
R1 R2
(74)
and so finally:
1 ∂E
1
1
= −(
+
) 2
E ∂n
R1 R2
(75)
JACKSON 1.12
Starting with Green’s theorem:
Z
2
2
I
3
(φ∇ ψ − ψ∇ φ)d x =
(φ
V
S
∂φ
∂ψ
− ψ )d2 x
∂n
∂n
(76)
We make the substitution ψ → Φ0 , φ → Φ, and using Poisson’s equation ∇2 Φ = −ρ/0 , and
recalling that σ = 0 ∂Φ/∂ n̂ (and similarly for the primed variables), we get:
Z
V
ρ
ρ0
(−Φ + Φ0 )d3 x =
0
0
I
(Φ
S
σ0
σ
− Φ0 )d2 x
0
0
(77)
Rearranging this equation and canceling the 0 ’s yields Green’s reciprocation theorem:
Z
0
3
I
0
Φ ρd x +
Z
2
Φ σd x =
V
I
0 3
Φσ 0 d2 x 2
Φρ d x +
S
V
S
(78)
It is important to emphasize the significance of Green’s reciprocation theorem. Using this theorem,
it is possible to relate a charge configuration with potential Φ to a completely different charge
configuration with potential Φ0 , as will be done in problem 1.13.
JACKSON 1.13
The problem wants us to use Green’s reciprocation theorem
Z
Φ0 ρd3 x +
I
Φ0 σd2 x =
Z
I
Φσ 0 d2 x,
(79)
S
V
S
V
Φρ0 d3 x +
to prove that a charge placed between two parallel conducting planes separated by a distance d
will induce a charge Q = −ql/d on one of the planes, where l is the distance from the other plane.
To do this, we need to find a second, similar, system to compare this with. From the hint, we need
to realize that the primed system in Green’s reciprocation theorem is the electrostatic potential of
a parallel plate capacitor with no charge density (ρ0 = 0):
Φ0 = V
z
d
(80)
where z is the distance from the bottom capacitor and d is the separation distance. You can check
that this formula works for z = 0 and z = d. This system in the problem, the unprimed one in
which there is a charge between the two parallel plates, has the potentials on both plates vanishing
because they are grounded (Φbottom = Φtop = 0), and a charge distribution
ρ(x) =
N
X
qi δ(x − xi ) = qδ(x)δ(y)δ(z − l),
(81)
i=1
if we suppose that charge is located at (0, 0, l). In this way, we have written down Φ, Φ0 , ρ, and ρ0 .
We immediately simplify the reciprocation theorem if we plug in Φ = 0:
Z
0
3
I
Φ ρd x +
V
Φ0 σd2 x = 0.
(82)
S
Now we plug in the charge density from the unprimed coordinate system (equation 81) and the
potential of the primed system (equation 80):
Z
V
z
V qδ(x)δ(y)δ(z − l)d3 x +
d
I
Φ0 σd2 x = 0.
(83)
S
The reason I left the Φ0 in the surface integral is because we need to be careful about it in that the
surface integral needs to be separated into integrals over the top and bottom plates.
Z
V
z
V qδ(x)δ(y)δ(z − l)d3 x +
d
I
Sbottom
Φ0bottom σbottom d2 x
I
+
Stop
Φ0top σtop d2 x = 0.
(84)
The potential of the bottom (top) plate corresponds to plugging z = 0 (z = d) into equation 80,
yielding Φ0 = 0 (Φ0 = V ). Thus, equation 84 reduces to:
Z
V
V
Recalling that
H
Stop
z
qδ(x)δ(y)δ(z − l)d3 x + V
d
I
σd2 x = 0.
(85)
Stop
σtop d2 x = Qtop , this expression is easily evaluated to be:
l
V q + Qtop V = 0
d
(86)
leading immediately to
Qtop = −q
l
2
d
(87)
JACKSON 1.14
This is a straightforward application of Green’s theorem
Z
(φ∇2 ψ − ψ∇2 φ)d3 x =
V
I
(φ
S
∂ψ
∂φ
− ψ )d2 x
∂n
∂n
(88)
where we substitute φ = G(x, y), ψ = G(x’,y).
Z
(G(x, y)∇2 G(x’,y) − G(x’,y)∇2 G(x, y))d3 y =
I
(G(x, y)
S
V
∂G(x’,y)
∂G(x, y) 2
− G(x’,y)
)d y
∂n
∂n
(89)
Then we use the fact that ∇2 G(z, y) = −4πδ(y − z).
−1
(G(x, y)δ(y − x ) − G(x , y)δ(y − x))d y =
4π
∂G(x’,y)
∂G(x, y) 2
− G(x’,y)
)d y
∂n
∂n
V
S
(90)
0
0
This is easily integrated to give an expression for the difference G(x, x ) − G(x , x):
Z
0
0
3
−1
G(x, x ) − G(x , x) =
4π
0
0
I
(G(x, y)
S
I
(G(x, y)
∂G(x’,y)
∂G(x, y) 2
− G(x’,y)
)d y 2
∂n
∂n
(91)
Part a
Dirichlet boundary conditions mean that GD (x, y) = 0 on the surface of S. That means that the
entire surface integral vanishes, and we find that GD (x, x0 ) is symmetric in x and x’:
GD (x, x0 ) = GD (x0 , x) 2
(92)
Note that the physical interpretation of the symmetry of the Green’s function is that the source
and observation points are interchangeable, i.e. exchanging them would give you the same Green’s
function. See page 40 of Jackson.
Part b
The boundary condition that Jackson is referring to is:
∂GN (x, x0 )
4π
=−
0
∂n
S
(93)
which is just the definition of the Neumann boundary condition, where S is the area of the surface.
We plug this into equation 91, to obtain:
1
GN (x, x ) − GN (x , x) =
S
0
0
I
(G(x, y) − G(x0 , y))d2 y ≡ F (x) − F (x0 )
(94)
S
where F (x) = S1 S GN (x, y)d2 y is defined by Jackson. Now this can be rearranged to show that
GN (x, x0 ) − F (x) is symmetric in x and x’:
H
G0N ≡ GN (x, x0 ) − F (x) = GN (x0 , x) − F (x0 ) 2
(95)
Side point: Why is ∂GN (x, x0 )/∂n0 6= 0 for Neumann boundary conditions?
Well, consider the following:
Z
∇2 G(x, x0 )d3 x0 =
V
Z
∇ · {∇G(x, x0 )}d3 x0 =
I
V
But notice:
Z
∇G(x, x0 ) · da0 =
S
0
2
Z
3 0
∇ G(x, x )d x =
V
I
S
∂G(x, x0 ) 0
da
∂n0
−4πδ(x − x0 ) = −4π,
(96)
(97)
V
and therefore:
I
S
∂G(x, x0 ) 0
da = −4π,
∂n0
(98)
meaning that ∂GN /∂n0 cannot be 0! Instead, by convention, ∂GN /∂n0 = −4π/S.
Part c
We will use equation 1.46 in Jackson,
Φ(x) = hΦiS +
1
4π0
Z
ρ(x0 )GN (x, x0 )d3 x0 +
V
1
4π
I
S
∂Φ
GN (x, x0 )da0
∂n0
(99)
which is the Neumann Green’s function solution. Here, hΦiS is the average value of the potential
over the whole surface. Instead of GN , we will plug in G0N ≡ GN − F (x), defined in equation 95:
1
ΦN (x) = hΦiS +
4π0
Z
1
ρ(x )(GN (x, x ) − F (x))d x +
4π
0
V
0
3 0
∂Φ
(GN (x − x0 ) − F (x))da0 . (100)
∂n0
I
S
We can multiply out the GN (x, x0 ) − F (x) terms:
ΦN (x) = Φ(x) −
1
4π0
Z
ρ(x0 )F (x)d3 x0 −
V
1
4π
I
S
∂Φ
F (x)da0
∂n0
(101)
where I have used the fact that the GN terms constitute the original Φ(x) of equation 99. We can
manipulate this further by noting that F (x) doesn’t depend on x0 ,
F (x)
ΦN (x) = Φ(x) −
{
4π0
Z
0
I
3 0
ρ(x )d x + 0
V
S
∂Φ(x0 ) 0
da }
∂n0
(102)
or, using −∂Φ/∂n = −n̂ · ∇Φ = n̂ · E:
F (x)
ΦN (x) = Φ(x) −
{qenc − 0
4π0
I
E(x0 ) · da0 }
(103)
S
and therefore by Gauss’s law, we see that the addition of F (x) does not affect the potential:
ΦN (x) = Φ(x) 2
(104)
JACKSON 1.15
We use the definition of electrostatic energy given by Jackson in equation 1.54:
0
W =
2
Z
|∇Φ|2 d3 x
(105)
V
To minimize the energy, we let Φ → Φ + δΦ. This defines W → W + δW , and plugging this in to
the above formula yields:
W + δW =
0
2
Z
V
|∇Φ + ∇δΦ|2 d3 x =
0
2
Z
V
|∇Φ|2 + |∇δΦ|2 + 2|∇Φ · ∇δΦ|d3 x
(106)
which can be rewritten as
W + δW ≈ W + 0
Z
∇Φ · ∇δΦd3 x
(107)
V
where I have ignored the term of order |∇δΦ|2 . To minimize the energy, I need to set the last term
in equation 107 to 0. To do this, we will integrate by parts:
Z
∇Φ · ∇δΦd3 x = 0 ∇ΦδΦ
δW = 0
V
− 0
Z
∇2 ΦδΦd3 x = 0
(108)
V
S
where S specifies the surfaces in the region (the first term in the integration by parts formula is
always a ‘boundary’ term). Using the fact that there no charges in the volume (∇2 Φ = 0), we find:
0 ∇ΦδΦ
= 0,
(109)
S
i.e. that ∇Φ = 0 on the surfaces, meaning that the surfaces are equipotentials, proving Thomson’s
theorem:
(110)
∇Φ = 0 ⇒ Φ = constant 2
S
JACKSON 1.16
This problem is similar to the last one, but we will use a different approach. We will define the
electrostatic energy of the original configuration to be W , and that of the configuration with a new
conductor to be W 0 . The are:
Z
0
W =
E 2 d3 x
(111)
2 V
Z
0
W0 =
E 02 d3 x
(112)
2 V1
Note that the integration is over different volumes (V of the original configuration and V1 = V − V0
of the new configuration, where V0 is the volume of the new conductor). Similarly the electric fields
are different since the introduction of the new conductor changes the electric field. Now, our goal
is to prove that W 0 < W . To do so, let’s calculate W − W 0 .
0
W −W =
2
Z
0
0
E d x−
2
2 3
V
Z
E 02 d3 x
(113)
V1
we need to rewrite everything in terms of integrals over the volumes V0 and V1 :
0
W −W =
2
0
0
E d x+
2
V0
Z
2 3
Z
(E 2 − E 02 )d3 x
(114)
V1
In this form it is not entirely clear which term is greater, W or W 0 . However, I can rewrite the
second integral using the strange looking reorganization of the term E 2 − E 02 :
E 2 − E 02 = (E − E0 )2 + 2E0 · (E − E0 )
(115)
You should check to make sure this equality works. It seems like an odd thing to do, but let’s keep
going. We’ll plug this equality into equation 116
0
2
Z
0
2
Z
Z
E0 · (E − E0 )d3 x
(116)
E0 · (E − E0 ) = −∇Φ · (E0 − E) = −(∇ · {Φ[E0 − E]} − Φ∇ · {E0 − E})
(117)
W − W0 =
E 2 d3 x +
V0
(E − E0 )2 d3 x + 0
V1
V1
Let’s look at the last term:
where I have used a vector identity to rewrite this. This is useful because we can use the differential
form of Gauss’s law to show that the last term is zero (∇ · E = 0). Knowing this, the expression in
116 is simplified to be
W − W0 =
0
2
Z
V0
E 2 d3 x +
0
2
Z
V1
(E − E0 )2 d3 x − 0
Z
V
∇ · {Φ[E0 − E]}d3 x
(118)
which can be simplified even more if I use the divergence theorem on the last term:
W − W0 =
0
2
Z
E 2 d3 x +
V0
0
2
Z
(E − E0 )2 d3 x − 0
V1
I
Φ[E0 − E] · da
(119)
S
We are almost there. Now note that the potential is constant, since we are are dealing with the
same situation as in problem 1.15, where we found that the energy is minimized if the surfaces
were equipotential. Thus, since the potential is constant, we can pull it out of the integral. Then
all we have left in the last term is Gauss’s law in integral form, which again yields 0! This means
that our expression for W − W 0 reduces to:
W − W0 =
0
2
Z
E 2 d3 x +
V0
0
2
Z
(E − E0 )2 d3 x > 0
(120)
V1
It is greater than 0 because both integrands are positive definite, so the integrals are too. Therefore,
W0 < W 2
(121)
JACKSON 1.17
Part a
We will use equations 1.54 and 1.62 in Jackson:
0
W =
2
Z
|∇Φ|2 d3 x =
V
N
N X
1X
Cij Vi Vj
2 i=1 j=1
(122)
The right hand side is easily evaluated if we recall that all the conductors are at zero potential
except for one. If we suppose that the first conductor is the only one with nonzero potential then
the only nonzero term in the sum on the right hand side is when i = j = 1.
0
2
Z
V
1
|∇Φ|2 d3 x = C11 V12
2
(123)
but V1 is at unit potential (V1 = 1), so this reduces, after canceling out factors of 1/2, to
Z
C = 0
V
|∇Φ|2 d3 x 2
(124)
Part b
The trial function Ψ satisfies the Laplace equation in V, is defined on S, so it satisfies δΨ = 0
on S. Any deviation for Φ(x) can be written as δΦ, and the first order change in the difference
C[Φ] = C[Φ + δΦ] − C[Φ] → 0, since Ψ satisfies the Laplace equation and C[Φ] ≡ C is a stationary
minimum. Therefore
Z
C ≤ C[Ψ] = 0
|∇Φ|2 d3 x 2
(125)
V