Physics 505
Fall 2007
Homework Assignment #2 — Solutions
Textbook problems: Ch. 2: 2.2, 2.8, 2.10, 2.11
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;
Recall that, if the point charge is outside a grounded conducting sphere, the
method of images gives
Φ(
~x
) =
q
4
π
0
1

~x

~
y


a/y

~x

(
a/y
)
2
~
y

(1)
where
y
=

~
y

, and
~
y
specifies the location of the charge
q
. Although this expres
sion was derived for
y > a
, we note that it is equally valid for
y < a
. After all, in
both cases, the potential satisfies the same boundary condition, Φ(

~x

=
a
) = 0.
Hence, for
q
inside the hollow sphere, the potential is also given by (1). In this
case, the physical charge is inside the sphere, while the image charge lies outside.
The image charge has the opposite sign, and in this case its magnitude is greater
than the physical charge.
b
) the induced surfacecharge density;
The induced surfacecharge density is given by
σ
=

0
∂
Φ
∂n
S
where in this case the unit normal is pointing into the sphere. Although we can
work out this expression with Φ given by (1), it is quicker to note that the result
must the be same (up to a sign change) as that for a point charge outside the
sphere. The difference in sign is due to the inward pointing normal in this case,
as opposed to an outward pointing normal when the point charge is outside. The
result is
σ
=
q
4
πa
2
a
y
1

(
a/y
)
2
(1 + (
a/y
)
2

2(
a/y
) cos
γ
)
3
/
2
where
γ
is the angle between
~x
and
~
y
. Note that
σ
has the opposite sign as
q
.
This is because the numerator in the above expression is actually negative for
y < a
. If desired, this sign can be made explicit by rewriting the above as
σ
=

q
4
πa
2
1

(
y/a
)
2
(1 + (
y/a
)
2

2(
y/a
) cos
γ
)
3
/
2
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By Gauss’ law, the total charge induced on the inside surface of the conducting
sphere must be

q
. This can also be seen by integrating the surfacecharge density
Q
=
Z
σa
2
d
Ω = 2
πa
2
Z
1

1
σ d
cos
γ
=

q
2
(1

(
y/a
)
2
)
Z
1

1
d
cos
γ
(1 + (
y/a
)
2

2(
y/a
) cos
γ
)
3
/
2
=

q
2
a
y
(1

(
y/a
)
2
)
1

1

y/a


1

1 +
y/a

=

q
where we have used the fact that 0
< y/a <
1 when simplifying the absolute
value quantities.
c
) the magnitude and direction of the force acting on
q
.
The force acting on
q
is essentially the force between
q
and its image. Again, the
magnitude of the force is given by the familiar expression
F
=
1
4
π
0
q
2
a
2
a
y
3
1
(1

(
a/y
)
2
)
2
Since the charge
q
is attracted to its image, the direction of the force is given by
ˆ
y
. For
y < a
, it is convenient to rewrite the force as
~
F
=
1
4
π
0
q
2
a
3
~
y
(1

(
y/a
)
2
)
2
This demonstrates that, for
y
a
, the force is linear,
~
F
∼
~
y
.
Because of the
positive sign, however, this is the
opposite
of a restoring force. This demonstrates
that the center of the conducting sphere is a point of unstable equilibrium for the
charge
q
.
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