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Solutions to Problem Set 5, Physics 370, Spring 2014
1
TOTAL POINTS POSSIBLE: 70 points.
1. Griffiths Problem 2.43: Find the capacitance per unit length of two
coaxial metal cylindrical tubes, of radii a and b (see Figure 2.53).
HINT: Start by simply considering a length L of the cylinders with
total charge of magnitude Q (on each cylinder).
(10 points possible) The key to figuring out the capacitance per unit
length is to realize capacitance is just (equation 2.53)
Q
C=
(1)
V
Now, we could just consider the total charge Q in a length L of the
cylinders. But to determine the potential difference V between the two
cylinders. we will either need to compute it from the charge distribution, which is not clearly given, or from the electric field.
We can obtain the electric field from Gauss’ law. Start by noting
the electric field must be radial by symmetry. Therefore a cylindrical Gaussian surface of length L (containing charge Q) and radius s
with a < s < b will have zero flux through the end caps and a uniform
electric field on the cylindrical surface. With this, we see the Gauss’
Law (equation 2.13) tells us that
I
~ · d~ℓ = qenc
E
(2a)
ǫ0
S
Q
(2b)
E(2πsL) =
ǫ0
Q 1
E=
(2c)
2πǫ0 L s
Solutions to Problem Set 5, Physics 370, Spring 2014
2
And since I know the force is radial (in the sˆ direction), then
~ =
E
Q 1
sˆ
2πǫ0 L s
(3)
Given equation ?? we can solve for the potential between the cylinders
(via equation 2.22) by integrating in a straight-line radial path from b
to a
Z ~b
~
~ · d~ℓ
V b − V (~a) =
E
(4a)
~a
Z b
Q 1
V (b) − V (a) =
sˆ · dsˆ
s
(4b)
s=a 2πǫ0 L s
Z b
1
Q
ds
(4c)
=
2πǫ0 L s=a s
b
Q
(4d)
ln
V =
2πǫ0 L
a
So the capacitance of a length L of these two cylinders is
C=
Q
V
Q
ln
2πǫ0 L
=
ln ab
=
Q
2πǫ0 L
(5a)
b
a
(5b)
(5c)
And therefore the capacitance per unit length is
2πǫ0
C
.
=
L
ln ab
(6)
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Solutions to Problem Set 5, Physics 370, Spring 2014
2. Griffiths Problem 2.47: Find the net force that the southern hemisphere of a uniformly charged sphere exerts on the northern hemisphere.
Express your answer in terms of the radius R and the total charge Q.
1 3Q2
[Answer : 4πǫ
2 ] HINT: You will need the electric field inside a
0 16R
uniformly charged sphere from Problem 2.12, which was on problem
set #3. Compute a force per unit volume on the northern hemisphere
based on the that electric field, then work out the solution.
(10 points possible) From Problem 2.12, we know the electric field
inside a uniformly charged sphere is
~ =
E
1 Q
rˆ
r.
4πǫ0 R3
(7)
Given total electric force is related to electric field (by equation 2.3):
~
F~ = Qtotal E
(8)
then the force per unit volume can be written:
Qtotal ~
F~
~
E = ρE.
=
f~ =
V
V
(9)
We know this is a uniformly charged sphere of radius R and charge Q,
Q
so ρ = 4 πR
3 and so equation 10 becomes
3
f~ =
Q
4
πR3
3
3
1 Q
rˆ
r=
3
4πǫ0 R
ǫ0
Q
4πR3
2
rˆ
r.
(10)
~
We can consider the force per unit volume could be written as f~ = ddτF ,
therefore we can write the differential force as dF~ = f~dτ . Furhermore,
by symmetry, the total force must be straight “up” in the z-direction, so
we only need to consider the z-component of the force per unit volume:
2
Q
3
~
r cos θ
(11)
fz = f · zˆ =
ǫ0 4πR3
And so the total vertical force on the northern hemisphere due to the
electric field acting on the charges in the northern hemisphere can be
Solutions to Problem Set 5, Physics 370, Spring 2014
computed by integrating the force per volume over the volume
spherical coordinates, so dτ = r 2 sin θdθdφdr) :
Z
Fz = fz dτ
2
Z R Z π/2 Z 2π Q
3
r cos θ(r 2 sin θdθdφdr)
=
3
ǫ
4πR
r=0 θ=0
φ=0 0
2 Z 2π
Z R
Z π/2
3
Q
3
=
dφ
r dr
cos θ sin θdθ
ǫ0 4πR3
φ=0
r=0
θ=0
π/2
2
Q
3
1 4 sin2 θ
=
(2π)( R )
ǫ0 4πR3
4
2
θ=0
2
Q
1
3
−0
=
2
ǫ0 32πR
2
3Q2
Fz =
64πǫ0 R2
4
(using
(12a)
(12b)
(12c)
(12d)
(12e)
(12f)
And so we are done...
And now a fairly subtle point: It is fair to ask how can we use F~ =
~ when the total electric field E
~ is created by the charge distribution
ρE
and a charge can’t push on itself. I will start by noting that the electric
field produced by just those charges in the upper hemisphere can NOT
produce a net force on it, but the electric field we computed is the sum
of the electric field due to charges in both hemispheres
~ =E
~ upper + E
~ lower
E
(13)
and so when we integrate charge density times the total electric field
over the upper hemisphere, we are computing the force on the upper
~ upper , which will work out to zero, and the force
hemisphere due to E
~ lower which is due to the lower hemisphere’s charges repeling the
due E
upper hemisphere. As such, we are safe doing this integral using the
total electric field produced by both hemispheres.
Solutions to Problem Set 5, Physics 370, Spring 2014
5
3. Explain in English (without just jumping to equations) your answers
to the following questions:
(a) Why does the Method of Images work? Specifically, why is it that
we can construct a completely different problem containing not
only our charges, but some addition “image charges” outside our
region of interest and yet use that to solve for the potential within
our region of interest? It would help to say what the critical requirement is of the “completely different” problem you are solving
is in order for it to help you solve your original problem.
(b) When we are placing “image charges” in order to solve an electrostatics problem, why is it that any “image charges” must lie
outside the region we are computing our potential in?
(a) (5 points possible) In a nutshell, the “completely different”
method of images problem must match the boundary conditions
for any potentials in order to be useful for solving our original
problem. This is because via the first uniqueness theorem, the solution to Laplace’s equation (that is, the solution for the potential
in our region of interest) is completely determined if the potential V is specified on the boundary surface S. So by setting up
a “completely different” problem with the same boundary values
for the electric potential, we know its solution will be the solution
in the region of interest.
(b) (5 points possible) If we add an “image charge” in our region
of interest, then we are effectively changing the electric potential
in the region of interest, which means we are no longer solving the
same problem!
Solutions to Problem Set 5, Physics 370, Spring 2014
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4. Griffiths Problem 3.2: In one sentence, justify Earnshaw’s Theorem: A charged particle cannot be held in a stable equilibrium by
electrostatic forces alone. As an example, consider the cubical arrangement of fixed charges in Figure 3.4 (below). It looks, off hand, as though
a positive charge at the center would be suspended in midair, since it
is repelled away from each corner. Where is the leak in this “electrostatic bottle”? [To harness nuclear fusion as a practical energy source
it is necessary to heat a plasma (soup of charged particles) to fantastic temperatures −− so hot that contact would vaporize any ordinary
pot. Earnshaw’s Theorem says that electrostatic containment is also
out of the question. Fortunately it is possible to confine a hot plasma
magnetically.] BIG HINT: What does the potential energy function
look like at a stable equilibrium?
(5 points possible) A particle can be in equilibrium only if it is at the
minimum of the potential energy, however in regions with no charge, we
know from our discussion of the solutions to Laplace’s equation that
there can be no local minima (or maxima) to the electric potential,
therefore a particle cannot be in stable equilibrium in an electrostatic
field. (Yeah, it is somewhat of a run on sentence...)
Another approach is to note that if U is the potential energy, that
for a stable equilibrium, we require that ∇2 U > 0 (that is that the
U is at a local minima), but the potential energy U = qV and we
know in a region with no charge (like the interior of the cube) that
∇2 V = 0 which implies ∇2 U = 0 and that therefore there is no stable
equilibrium. (Yeah, this was even more of a run on sentence...)
You can explain why the equilibrium is unstable in terms of forces,
but it doesn’t really prove it. Consider a positive charge placed at the
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Solutions to Problem Set 5, Physics 370, Spring 2014
center would feel no force since it is equidistant from all the charges
on the corner of the cube and they are symmetrically placed around
it. However, if it moves from the center towards one of the corners the
force on the particle will no longer be symmetric, pushing it farther
from the center. That can help you see why it is unstable, but doesn’t
really provide a proof that no stable configuration can be built with
electrostatic charges.
5. Griffiths Problem 3.03: Find the general solution to Laplace’s equation in spherical coordinates, for the case where V depends only on r.
Do the same for cylindrical coordinates, assuming V depends only on s.
(10 points possible) If V only has a radial dependence, then in spherical coordinates, we can write Laplacian as (see the inside cover of
Griffiths):
1
∂V
1
∂
∂2V
1 ∂
2 ∂V
2
~
r
+ 2
sin θ
+ 2 2
∇V = 2
r ∂r
∂r
r sin θ ∂θ
∂θ
r sin θ ∂φ2
(14a)
1 d
dV
= 2
r2
(14b)
r dr
dr
In the case of the Laplace’s equation, the Laplacian has a value of zero,
therefore:
1 d
2 dV
r
=0
(15)
r 2 dr
dr
A general solution to this can be found by integrating equation 16
r2
dV
= C1
dr
C1
dV = 2 dr
r
C1
V =−
+ C2
r
(16a)
(16b)
(16c)
where C1 and C2 are constants of integration in the final general solution for the potential. We can find the cylindrical coordinates counterpart to this solution by noting (again, checking the inside cover of
Solutions to Problem Set 5, Physics 370, Spring 2014
8
Griffiths) that in cylindrical coordinates, with a potential V that depends only on s, the Laplacian takes the form:
∂V
1 ∂2V
∂2V
1 ∂
2
~
s
+ 2 2 +
(17a)
∇V =
s ∂s
∂s
s ∂φ
∂z 2
1 d
dV
=
s
(17b)
s ds
ds
and again Laplace’s equation can be solved by integration:
dV
1 d
s
=0
s ds
ds
dV
= C1
s
ds
C1
dV =
ds
s
V = C1 ln(s) + C2
(18a)
(18b)
(18c)
(18d)
where again C1 and C2 are constants of integration. We could determine the values of these constants by knowing the boundary conditions
on the potential.
6. Griffiths Problem 3.8: This problem asks you to essentially finish
image problem laid out in Example 3.2. That example asked you to
determine the potential outside a grounded conducting sphere with
radius R with a point charge q located distance a > R from the center
of the sphere as shown in Figure 3.12 (reproduced below):
It argued that the corresponding method of images solution has a potential given by equation 3.17 (in part [a] below), but left it for this
problem for you to verify that. Read that example carefully before
starting this problem.
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Solutions to Problem Set 5, Physics 370, Spring 2014
(a) Using the law of cosines, show that
q
1 q
V (~r) =
+ ′
4πǫ0
(equation 3.17 from the textbook, describing the potential outside
the sphere) can be written as follows:


q
1 
q

√
V (r, θ) =
−q
2
2
4πǫ0
r + a − 2ra cos θ
ra 2
2
R +
− 2ra cos θ
R
(b) Find the induced surface charge on the sphere, as a function of
θ. Integrate this to get the total induced charge. (What should it
be?)
(c) Calculate the energy of this configuration. HINT: Use the force
given by equation 3.18 and the definition of work.
(a) (5 points possible) The potential of a grounded sphere near a
charge q is the same, for points outside the sphere, as the charge
distribution shown in Figure 3.13 (reproduced below), in which
2
q ′ = − Ra q and b = Ra .
The potential of this charge distribution is
q′
q
1
V =
+ ′ .
4πǫ0
According to the law of cosines
√
= r 2 + a2 − 2ar cos θ
(19)
(20)
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Solutions to Problem Set 5, Physics 370, Spring 2014
and
′
=
√
r 2 + b2 − 2rb cos θ
r
R2
R4
cos θ.
= r 2 + 2 − 2r
a
a
(21a)
(21b)
Putting the distances and ′ and the expression for q ′ into the
potential (20) gives


′
1 
q
q

√
V =
+q
4
2
4πǫ0
R
R
r 2 + a2 − 2ra cos θ
r2 +
− 2r cos θ
a2
=
=
=

q
1 
√
−q
4πǫ0
r 2 + a2 − 2ra cos θ
r2 +

q
1 
√
−
4πǫ0
r 2 + a2 − 2ra cos θ

a
R
q
a
R4
a2
−
2
2r Ra
cos θ
q
q
a
r2 +
R
q
1 
√
−q
2
2
4πǫ0
r + a − 2ra cos θ
R2 +
R4
a2
−
q
ra 2
R
2
2r Ra
(22a)


(22b)

cos θ

(22c)


− 2ra cos θ
(22d)
the result we wanted.
(b) (5 points possible) To find the induced surface charge on the
conductor we use equation 2.49 from the textbook which provides
the relationship between potential V and surface charge on a conductor:
∂V
σ = −ǫ0
,
(23)
∂n
where the derivative is in the direction normal to the surface.
For a sphere this means the radial direction, therefore we need to
11
Solutions to Problem Set 5, Physics 370, Spring 2014
evaluate the derivative
!
2
2( ra
−
a
cos
θ)
1
1
2(r − a cos θ)
R2
−
− −
2 (r 2 + a2 − 2ra cos θ)3/2
2 (R2 + ra 2 − 2ra cos θ)3/2
R
(24a)
!
2
ra
r − a cos θ
q
2
−
=−
;
R
4πǫ0 (r 2 + a2 − 2ra cos θ)3/2 (R2 + ra 2 − 2ra cos θ)3/2
R
(24b)
q
∂V
=
∂r
4πǫ0
and we need to evaluate this as the surface of the sphere (r = R):
q
∂V =−
∂r R
4πǫ0
(R2
R − a cos θ
−
+ a2 − 2Ra cos θ)3/2 (R2 +
a2
R
a2
− a cos θ
− 2Ra cos θ)3/2
(25a)
2
R − aR
q
=−
,
4πǫ0 (R2 + a2 − 2Ra cos θ)3/2
(25b)
so that
2
R − aR
q
∂V
=
.
σ = −ǫ0
∂r
4π (R2 + a2 − 2Ra cos θ)3/2
(26)
The total charge on the sphere is the integral of this over the
surface of the sphere; we expect the total charge to be q ′ , the
image charge with which we can replace the sphere. The integral
of the charge density is
′
q =
Z
a2
)
R
π
2π
R2 sin θ
dφdθ
2
2
3/2
θ=0 φ=0 (R + a − 2Ra cos θ)
(27a)
Z π
sin θ
qR 2
(R − a2 )
dθ
=
2
2
3/2
2
θ=0 (R + a − 2Ra cos θ)
(27b)
q(R −
σda =
4π
Z
Z
Making a u-substitution of u = R2 + a2 − 2Ra cos θ and du =
!
Solutions to Problem Set 5, Physics 370, Spring 2014
12
2Ra sin θdθ, we have
Z π
qR 2
1
2
q =
u−3/2 du
(R − a )
2
2Ra θ=0
π
1 qR 2
−2u−1/2 θ=0
(R − a2 )
=
2
2Ra π
qR 2
−2
1
2
√
=
(R − a )
2
2Ra
R2 + a2 − 2Ra cos θ θ=0
1
1
q 2
2
−√
= − (R − a ) √
a
R2 + a2 + 2Ra
R2 + a2 − 2Ra
q
1
1
= − (R2 − a2 )
−
a
R + a |R − a|
2R
q 2
2
= − (R − a )
a
R2 − a2
R
= − q,
a
′
(28a)
(28b)
(28c)
(28d)
(28e)
(28f)
(28g)
as expected from equation 3.15 describing the image charge in this
problem.
(c) (5 points possible) Finally, we calculate the energy of this configuration, which is the work required to bring the charge q in
from infinity. The force on the charge when it is a distance r from
the origin is, from equation 3.18 in the textbook,
q 2 Rr
1
.
F =
4πǫ0 (r 2 − R2 )2
(29)
Noting that this force is in the −ˆ
r direction,
R I can express the work
done on the charge using its definition F~ · d~l where d~l = −drˆ
r
such that
Z a
W =
F dr
(30a)
∞
Z
q2R a
r
=
dr
(30b)
2
4πǫ0 ∞ (r − R2 )2
Again, u-substitution saves the day, with u = r 2 − R2 and du =
Solutions to Problem Set 5, Physics 370, Spring 2014
13
2rdr,
a
−1
q2R
W =
4πǫ0 2(r 2 − R2 ) ∞
q2R
.
=−
8πǫ0 (a2 − R2 )
(31a)
(31b)
(Thanks to Dr. Craig for providing the initial draft of this solution in
LATEX form.)
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Solutions to Problem Set 5, Physics 370, Spring 2014
7. Griffiths Problem 3.11: Two semi-infinite grounded conducting planes
meet at right angles. In the region between them, there is a point charge
q, situated as shown in Figure 3.15 (below).
(a) Set up the image configuration, and calculate the potential in this
region. What charges do you need, and where should they be
located?
(b) What is the force on q?
(c) How much work did it take to bring q in from infinity?
(d) (Extra Credit) Suppose the planes met at some angle other than
90o ; would you still be able to solve the problem by the method of
images? If not, for what particular angles does the method work?
Clearly explain your reasoning. Diagrams may help!
(a) (4 points possible) We can’t use just a single image charge here
because if you try to place one (say at −a, −b) you will not be able
to get V = 0 on the xz and yz planes as required by the boundary
conditions. Instead, the image charge configuration that achieves
the necessary boundary conditions is shown below:
y
–q
+q
2
-b
b
1
a
x
-a
3
+q
4
–q
Solutions to Problem Set 5, Physics 370, Spring 2014
15
The potential in the original region is the sum of the potentials of
the original charge and the three image charges,
V = V1 + V2 + V3 + V4
(32a)
q
q
q
q
1
(32b)
+
−
−
=
4πǫ0 1
4
2
3
1
1
q
p
−p
=
2
2
2
2
4πǫ0
(x − a) + (y − b) + z
(x + a) + (y − b)2 + z 2
1
1
+p
.
−p
(x + a)2 + (y + b)2 + z 2
(x − a)2 + (y + b)2 + z 2
(32c)
(yes, for the proper potential description, you must use the z components to the , since potentials are three-dimensional even when
the charge distribution isn’t)).
(b) (3 points possible) The force on the real charge q is the sum of
the forces due to the other three charges,
2
ˆ 12 ˆ 13 ˆ 14
q
+ 2 + 2 .
F~ =
(33)
2
4πǫ0
12
13
14
y
y
√x+2bˆ
= √aˆax2+bˆ
, and the other two unit
The unit vector ˆ 13 = 22aˆ
a2 +b2
+b2
vectors point along the axes, so
1
a
b
1
q2
~
− 2 xˆ +
xˆ +
yˆ − 2 yˆ
F =
4πǫ0
4a
4(a2 + b2 )3/2
4(a2 + b2 )3/2
4b
(34a)
q2
a
1
b
1
=
− 2 xˆ +
− 2 yˆ
2
2
3/2
2
2
3/2
16πǫ0
(a + b )
a
(a + b )
b
(34b)
(c) (3 points possible) The work required to bring q in from infinity
is just going to be the work done against the potential of the image
charges. By equation 2.39,
W = QV (~r)
(35)
Solutions to Problem Set 5, Physics 370, Spring 2014
16
but remember, this assumes the energy is stored over all space, so
in this case, we divide this by 4 (only 1/4 of all space is in this
volume):
1
W = q[V2 + V3 + V4 ]
4
2
q
q2
q2
1 1
− + √
−
=
4 4πǫ0
2b 2 a2 + b2 2a
q2
1
1 1
√
=
− −
32πǫ0
a2 + b2 a b
(36a)
(36b)
(36c)
(d) (5 points extra credit possible) For this method of images
to work, the angle θ must be either 180o or some integer divisor
of 180o . So for example, 180o , 90o , 60o, or 45o are all OK since
they are 180o divided by 1,2,3, and 4 respectively, but 120o is not
acceptable. To quote Griffith’s solution manual:
Note the strategy: to make the x axis an equipotential (V = 0),
you place the image charge (1) in the reflection point. To make the
45o line an equipotential, you place charge (2) at the image point.
But that screws up the x axis, so you must now insert image (3)
to balance (2). Moreover, to make the 45o line V = 0 you also
Solutions to Problem Set 5, Physics 370, Spring 2014
17
need (4), to balance (1). But now, to restore the x axis to V = 0
you need (5) to balance (4), and so on.
The reason this doesn’t work for arbitrary angles is that you are
eventually forced to place an image charge within the original
region of interest, and that’s not allowed — all image charges
must go outside the region, or you’re changing the potential in
the region of interest and no longer dealing with the same problem
at all.