Problem Set 2
:
2
.
1
The Coulomb potential for a point charge located at
r
0
is given by
(
r
,
r
0
) =
q

r
−
r
0

a) Find
∇
2
(
r
,
r
0
)
and write down the partial differential equation satisfied by
(
r
,
r
0
)
.
b) Using steps similar to those found on page 12 of Chapter 1 for the wave equation,
find the
fourier transform,
(
k
,
r
0
), of the Coulomb potential.
c) Find
(
k
,
r
0
)
directly from
(
r
,
r
0
) =
1
(
2
)
3
∫∫∫
(
k
,
r
0
)
exp
(
i
k
r
)
d
3
k
d) Evaluate the following integral
∫
0
∞
sin
(
ku
)
du
Solution:
a)
∇
2
(
r
,
r
0
) = ∇
2
q

r
−
r
0

= −
4
q
(
r
−
r
0
)
so
(
r
,
r
0
)
satisfies
∇
2
(
r
,
r
0
) = −
4
q
(
r
−
r
0
)
b)
One can take the Fourier transform of this equation:
∇
2
1
(
2
)
3
∫∫∫
(
k
,
r
0
)
exp
(
i
k
r
)
d
3
k
= −
4
q
(
2
)
3
∫∫∫
exp
(
i
k
(
r
−
r
0
))
d
3
k
∫∫∫
[
(
k
,
r
0
)(−
k
k
) +
4
q
exp
(−
i
k
r
0
)]
exp
i
k
r
)
d
3
k
=
0
Since the exp
i
k
r
are linearly independent functions for each
k
,
(
k
,
r
0
)(−
k
k
) +
4
q
exp
(−
i
k
r
0
) =
0
and
(
k
,
r
0
) =
4
q
exp
(−
i
k
r
0
)
k
k
c)
The direct method for finding
(
k
,
r
0
)
is as follows.:
(
r
,
r
0
) =
q

r
−
r
0

=
1
(
2
)
3
∫∫∫
(
k
,
r
0
)
exp
(
i
k
r
)
d
3
k
and
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