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C:\User\Teaching\505-506\F10\HW03sol.doc
PHY 505
Classical Electrodynamics I
Fall 2010
Homework 3 with solutions
Problem 3.1
(to be graded of 15 points). Calculate the mutual
capacitance (per unit length) between two cylindrical conductors,
forming a system with the cross-section shown in Fig. on the right, in
the limit
t
<<
w << R
.
Hint
: You may like to use the “
elliptical
” (not “ellipsoidal”!)
coordinates defined by the following equation:
),
cosh(
i
c
iy
x
(*)
with the appropriate choice of constant
c
. In these orthogonal 2D
coordinates, the Laplace operator is very simple:
2
2
2
2
2
2
2
2
2
)
cos
(cosh
1
c
.
(**)
(This is not quite surprising, because Eq. (*) may be also considered as a conformal map
z
=
c
cosh
w
,
where
z
=
x + iy
, and
w
=
+ i
.)
Solution:
On the [
x, y
] plane, the lines of constant
are ellipses with horizontal and vertical
semi-axes
c
cosh
and
c
sinh
,
respectively.
For
= 0, the ellipse degenerates into a straight horizontal
segment –
c
<
x
< +
c
, while for
>> 1, the ellipse is virtually a circle of radius
= (
c
/2)exp
.
As a
result, if we select the axes
x
and
y
as shown in figure, and take
c
=
w
/2, the boundary conditions on the
conductor surfaces (see Fig.) may be satisfied, at
t
<<
w
<<
R
, by a potential distribution
(
), with no
dependence on
:
,
)
4
(ln
,
0
)
0
(
V
W
R
(***)
where
V
is the voltage between the conductors. Hence the boundary problem may be satisfied by a
function
(
) if it satisfies the 1D Laplace equation following from Eq. (**):
.
0
2
2
d
d
This equation shows that
(
) is just a linear function
c
1
+
c
2
. Selecting two constants
c
1,2
to satisfy
boundary conditions (***), we get
)
/
4
ln(
w
R
V
.
To calculate the surface charge of the central conductor, we need to calculate the (normal)
electric field on its surface:
,
sin
)
2
/
)(
/
4
ln(
)
(sinh
sin
)
2
/
)(
/
4
ln(
0
2
/
2
/
0
w
w
R
V
w
w
R
V
y
E
w
x
w
y
n
so that the two-surface charge density
R
w
t