Page 162 - Electromagnetics

The integral takes the form (3.100), hence by (3.102) the solution is

a 2 r <
(r) = P 0 cos θ 2 . (3.103)
3 0 r
>
If we are interested only in the potential for r > a, we can use the multipole expansion
(3.95)to obtain

∞
1
1
(r) = n+1 q n P n (cos θ), r > a
4π 0 r
n=0
where
π
n
2

q n = 2π ρ Ps (θ )a P n (cos θ )a sin θ dθ .
0
Substituting for ρ Ps and remembering that cos θ = P 1 (cos θ),wehave
π

q n = 2πa n+2 P 0 P 1 (cos θ )P n (cos θ ) sin θ dθ .
0
Using the orthogonality relation (E.123)we find
2
n+2
q n = 2πa P 0 δ 1n .
2n + 1
Therefore the only nonzero coefficient is
3
4πa P 0
q 1 =
3
and
3
1 1 4πa P 0 P 0 a 3
(r) = P 1 (cos θ) = cos θ, r > a.
4π 0 r 2 3 3 0 r 2
This is a dipole field, and matches (3.103)as expected.
3.2.8 Potential of a dipole layer
Surface charge layers sometimes occur in bipolar form, such as in the membrane sur-
rounding an animal cell. These can be modeled as a dipole layer consisting of parallel
surface charges of opposite sign.
Consider a surface S located in free space. Parallel to this surface, and a distance /2
below, is located a surface charge layer of density ρ s (r) = P s (r). Also parallel to S, but
a distance /2 above, is a surface charge layer of density ρ s (r) =−P s (r). We define the
surface dipole moment density D s as

D s (r) = P s (r). (3.104)

Letting the position vector r point to the surface S we can write the potential (3.61)
0
produced by the two charge layers as
1 1 1 1

(r) = P s (r ) dS − P s (r ) dS .

4π 0 S + |r − r − ˆ n 2 | 4π 0 S − |r − r + ˆ n 2 |
0
0
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