Homework Assignment 4 Solution Set
PHYCS 4420
17 February, 2004
Problem 1 (Griffiths 2.37)
If the plates are sufficiently large the field near them does not depend on
d. The field between the plates is zero (the fields from each plate are equal in
strength but oppositely directed), and the field just outside either plate is the
sum of the contributions from each plate. Therefore, the pressure on each plate
is
1 ~
~
P~ = σ(E
outside + Ebetween )
2
1
σ
= σ(2
+ 0)dˆ
2
20
σ2 ˆ
d
=
20
Q2 ˆ
=
d.
20 A2
Of course, this is the same as considering the force on one plate due to an
~ = σ dˆ due to the other plate.
electric field E
20
Problem 2 (Grifffiths 2.38 - optional)
Let’s consider the sphere to be centered at the origin with its ”equator” in
the x − y plane. The electrostatic pressure may be thought of as being due to
the average field surrounding the charge on any given element of surface. The
~ = 1 Q2 r̂. So, for any
field is zero inside the sphere, and outside the field is E
4π0 R
given element of surface the pressure will be
1 ~
~
dP~ = σ (E
outside + Einside )
2
1 1 Q
r̂ + 0)
=σ (
2 4π0 R2
1
Q2
=
r̂
4π0 8πR4
Q2
=
r̂.
32π 2 0 R4
1
Since the pressure is in the radial direction we know that only the components in the z direction will contribute to the overall force. Integrating the z
components of the pressure over the upper hemisphere gives
Z
~
~z
F =
P~ · dA
hemisphere
π
Z 2π
2
Z
dφ(
dθ
=
0
0
Q2
r̂) · (R2 sin θr̂) cos θ
32π 2 0 R4
2
Q
π
32π 2 0 R2
Q2
.
=
32π0 R2
=
Problem 3 (Griffiths 2.34(a))
The field due to this configuration is zero everywhere except for between the
two shells. Thus,
Z
0
~ 2 dτ
U =
E
2 allspace
2
Z 0 b
1 q
=
4πr2 dr
2 a 4π0 r2
q2 1
1 1
=
−
.
2 4π0 a b
Problem 4 (Griffiths 3.26)
To find the approximate potential (in the context of this chapter, anyway)
we should use the multipole expansion and keep only the first non-vanishing
term. The monopole gives
Z
1 1
V0 (~r) =
ρ(~r0 )dτ 0
4π0 r V
Z
Z
Z
1 1 R 0 π 0 2π 0 R
=
dr
dθ
dφ k 02 (R − 2r0 ) sin θ0 r02 sin θ
4π0 r 0
r
0
0
Z R
Z π
1 2π
=
dr0
dθ0 kR(R − 2r0 ) sin2 θ0
4π0 r 0
0
Z
1 π2 R 0
=
dr kR(R − 2r0 )
4π0 r 0
= 0.
The total charge is zero, so the monopole term in the potential is zero. Thus,
we should next check the dipole term. We have
Z
1 1
V1 (~r) =
r0 cos Θρ(~r0 )dτ 0
4π0 r2 V
2
where Θ is the angle between ~r and ~r0 . In this case, we have been asked to find
the potential at a point on the z axis, so Θ = θ0 . Therefore
Z 2π
Z π
Z R
R
1 1
0
0
dφ0 r0 cos θ0 k 02 (R − 2r0 ) sin θ0 r02 sin θ0
dθ
dr
V1 (z) =
4π0 z 2 0
r
0
0
Z
Z
1 2π R 0 π 0 0
=
dθ kr R(R − 2r0 ) sin2 θ0 cos θ0
dr
4π0 z 2 0
0
= 0.
This, too, is zero (this time because of the θ integral)! So, we continue with
Z π
Z 2π
Z R
1 1
3
R
1
0
0
0 02
2 0
dθ
dφ
V2 (z) =
dr
r
cos
k 02 (R − 2r0 ) sin θ0 r02 sin θ0
θ
−
4π0 z 3 0
2
2
r
0
0
Z R
Z π
Z 2π
1 1
1
3
0
2 0
0
0
=
dr
cos θ −
kR(Rr02 − 2r03 ) sin2 θ0
dθ
dφ
4π0 z 3 0
2
2
0
0
Z
1
1 2πkR5 π 0 3
2 0
cos θ −
sin2 θ0
=−
dθ
4π0 6z 3
2
2
0
πkR5
=
.
1920 z 3
This, then, is the approximate potential for points far from our charge distribution.
Problem 5 (Griffiths 3.31)
For a pure dipole the potential and field are given in the book, and they
are exact (they are due to the dipole moment only, since all other terms in the
multipole expansion vanish). We can easily calculate
a
~
F~q (a, 0, 0) = q E(a,
0, 0)
qp
π
π
=
(2 cos r̂ + sin θ̂)
4π0 a3
2
2
qp
θ̂.
=
4π0 a3
Note: at the point (a, 0, 0), θ̂ = −ẑ.
b
~ 0, a)
F~q (0, 0, a) = q E(0,
qp
=
(2 cos 0r̂ + sin 0θ̂)
4π0 a3
qp
=
r̂.
2π0 a3
Note: at the point (0, 0, a), r̂ = ẑ.
3
c The path one takes to move the charge from one point to another should
not change the final answer. The fact that we are using an approximate potential
does not change the fact that we still must satisfy Laplace’s equation, and the
field is still a conservative one. Therefore,
W = ∆U
= U(0,0,a) − U(a,0,0)
qp cos 0 qp cos π2
−
=
4π0 a2
4π0 a2
qp
=
.
4π0 a2
Problem 6 (Griffiths 3.32 - optional)
For such a distribution of point charges the integrals in the multipole expansion become sums over the point charges because ρ(~r) = q1 δ(|~r − ~r1 |) + q2 δ(|~r −
~r2 |) + q3 δ(|~r − ~r3 |). Therefore, the first term in the multipole expansion is
1 1
(q1 + q2 + q3 )
4π0 r
1 q
=−
.
4π0 r
V0 (~r) =
The second term is
V1 (~r) =
1 1
(q1 r1 cos Θ1 + q2 r2 cos Θ2 + q3 r3 cos Θ3 ) .
4π0 r2
Here the only tricky thing is figuring out how to write Θi . However, we can
exploit the fact that ~r · ~ri = rri cos Θi to write
1 1
r̂ · (q1~r1 + q2~r2 + q3~r3 )
4π0 r2
1 1
(sin θ cos φ, sin θ sin φ, cos θ) · (q(0, 0, a) − q(0, a, 0) − q(0, −a, 0)) .
=
4π0 r2
V1 (~r) =
Therefore, including only the two lowest terms in the multipole expansion,
the potential from this three charge configuration is approximately
V (~r) = V1 + V2
i
1 h q
q
=
− + 2 (a cos θ − a sin θ sin φ + a sin θ sin φ)
4π0
r r
1 q
q
=
− + 2 a cos θ .
4π0
r r
Apparently, the dipole contribution from the two charges on the y axis exactly
cancel and we are left with only the dipole term from the charge on the z axis.
4
To find the field we simply take the gradient. Thus,
~ r) = −∇V
~ (~r)
E(~
∂V
1 ∂V
1 ∂V
r̂ +
θ̂ +
φ̂
=−
∂r
r ∂θ
r sin θ ∂φ
a cos θ
a sin θ
q
1
r̂
+
θ̂
=
− 2+
4π0
r
r3
r2
Problem 7 (Griffiths 3.6)
We have seen that the potential (and, therefore, the field) of such a configuration may be modeled by the placement of image charges that are the
mirror-images (across the x − y plane) of the existing charges. Thus, the charge
+q at z = 3d experiences a force due to the field of three other point charges.
It may be written as
2q
q
2q
1
+
−
q −
ẑ
F~ =
4π0
(2d)2
(4d)2
(6d)2
29 1 q 2
ẑ
=−
72 4π0 d
Problem 8 (Griffiths 3.10)
As seen in class, the image charges should be located as though they were
the images in mirrors in the x and y planes. Thus, there are three images in
all: an image −q at the points (a, −b) and (−a, b), and an image q at the point
(−a, −b). The potential, then, is the potential of four point charges:
q 1
1
p
V (x, y) =
+p
2
2
2
4π0
(x − a) + (y − b)
(x + a) + (y + b)2
1
1
−p
−p
(x + a)2 + (y − b)2
(x − a)2 + (y + b)2
The force on the charge q at the point (a, b) will be the same as if it were
due to the three image charges only. So,
1 2
1
2a
1
2b
F~q (a, b) =
q
−
+
x̂
+
−
+
ŷ .
3
3
4π0
(2a)2
(2b)2
((2a)2 + (2b)2 ) 2
((2a)2 + (2b)2 ) 2
The potential due to the three images charges after q is brought to the point
(a, b) is
"
#
1
1
1
1
V (a, b) =
q − −
+p
.
4π0
2a 2b
(2a)2 + (2b)2
So, the work required to bring the charge in from infinity is half the energy due
to this potential (because the image charges accumulate for ”free”). Therefore,
1 q2
1 1
1
√
W =
− − +
.
4π0 4
a b
a2 + b2
5
This method does not work for any arbitrary angle of intersection of two
conducting planes. In order for the reflections to terminate at some finite number of image charges, the set of the real charge and all of its reflections in the
conducting planes must form an abelian group. This is fancy math-talk for
saying that you should be able to get from the charge to any of it’s images by
successive application of a discreet set of operators. In our case, the operator
that takes you from the real charge to one of its images is more than spacial.
It should consist of a rotation about the point of intersection of the planes as
well as a reversal of charge (the first image charge has the same sign as the real
charge, but the next has opposite sign, etc.).
This is a very formal proof, but your reasoning may be more straight forward
and less rigorous. The short story is, only those situations in which the angle
of intersection of the planes is an even divisor of 360◦ will be solvable via the
image charge method. So, for the image method to work we require
θintersection =
360◦
; n ∈ {1, 2, 3, · · ·}.
2n
6