Homework 4 Answers, 95.657 Electromagnetic Theory I
Dr. Christopher S. Baird, UMass Lowell
Jackson 2.13
(a) Two halves of a long hollow conducting cylinder of inner radius
b
are separated by small
lengthwise gaps on each side, and are kept at different potentials
V
1
and
V
2
. Show that the potential
inside is given by
,
=
V
1
V
2
2
V
1
−
V
2
tan
−
1
2
b
b
2
−
2
cos
where
is measured from a plane perpendicular to the plane through the gap.
SOLUTION:
Due to the symmetry of the problem, it is apparent that the solution will be best expressed in cylindrical
coordinates. Additionally, because the solution will be independent of the
z
coordinate, the problem
reduces to the two dimensions of polar coordinates
,
. Because the problem contains no charge,
the problem simplifies down to solving the Laplace equation
∇
2
=
0
in polar coordinates and
applying the boundary condition
=
b ,
=
V
where:
V
=
{
V
1
if
/
2
3
/
2
V
2
if
/
2
3
/
2
}
The Laplace equation in polar coordinates is:
1
∂
∂
∂
∂
1
2
∂
2
∂
2
=
0
Separation of variables leads to the general solution:
,
=
a
0
b
0
ln
A
0
B
0
∑
,
≠
0
a
b
−
A
e
i
B
e
−
i
We desire a valid solution at the origin, which is only possible if
b
0
=0 and
b
υ
=0 so that the solution
becomes:
,
=
A
0
B
0
∑
,
≠
0
A
e
i
B
e
−
i
We desire a single, valid solution over the full angular range, so the singlevalue requirement means
,
=
,
2
. When we apply this, we get:
A
0
B
0
∑
,
≠
0
A
e
i
B
e
−
i
=
A
0
B
0
2
∑
,
≠
0
A
e
i
2
B
e
−
i
2
Which leads to
B
0
=0 and
υ
=
n
where
n
=1, 2, ... We now have:
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,
=
A
0
∑
n
=
1
∞
n
A
n
e
in
B
n
e
−
in
Now apply the last boundary condition
=
b ,
=
V
V
=
A
0
∑
n
=
1
∞
b
n
A
n
e
in
B
n
e
−
in
(
Eq.
1)
Let us first find the
A
0
term. Integrate both sides over the full angular sweep.
∫
0
2
V
d
=
∫
0
2
A
0
d
∑
n
=
1
∞
b
n
A
n
∫
0
2
e
in
d
B
n
∫
0
2
e
−
in
d
∫
−/
2
/
2
V
1
d
∫
/
2
3
/
2
V
2
d
=
A
0
2
V
1
V
2
=
A
0
2
A
0
=
V
1
V
2
2
Let us now find the
A
n
coefficients. Multiply (
Eq.
1) on both sides by
e
−
in
'
and integrate over all
angles
:
∫
0
2
V
e
−
in
'
d
=
A
0
∫
0
2
e
−
in
'
d
∑
n
=
1
∞
b
n
A
n
∫
0
2
e
i
n
−
n
'
d
B
n
∫
0
2
e
−
i
n
n
'
d
Use the orthonormality condition
∫
0
2
e
i
k
−
k
'
x
dx
=
2
k , k
'
∫
0
2
d
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 Spring '11
 Avery
 Electrostatics, Magnetism, Mass, Work, Sin, Electric charge

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