Page 159 - Electromagnetics

where

2
n

q n = 2π ρ(r ,θ )r P n (cos θ )r sin θ dθ dr .

r θ
As a simple example consider a spherical distribution of charge given by
3Q
ρ(r) = cos θ, r ≤ a.
πa 3
This can be viewed as two adjacent hemispheres carrying total charges ±Q. Since cos θ =
P 1 (cos θ), we compute
a π 3Q
n
2

q n = 2π 3 P 1 (cos θ )r P n (cos θ )r sin θ dθ dr
0 0 πa
3Q a n+3 π

= 2π P 1 (cos θ)P n (cos θ ) sin θ dθ .

3
πa n + 3 0
Using the orthogonality relation (E.123)we find
3Q a n+3 2
q n = 2π δ 1n .
3
πa n + 3 2n + 1
Hence the only nonzero coefficient is q 1 = Qa and
1 1 Qa
(r) = QaP 1 (cos θ) = cos θ.
4π r 2 4π r 2
This is the potential of a dipole having moment p = ˆ zQa. Thus we could replace the
sphere with point charges ∓Q at z =∓a/2 without changing the field for r > a.
Physical interpretation of the polarization vector in a dielectric. We have
used the Maxwell–Minkowski equations to determine the electrostatic potential of a
charge distribution in the presence of a dielectric medium. Alternatively, we can use
the Maxwell–Boffi equations
∇× E = 0, (3.96)
1
∇· E = (ρ −∇ · P). (3.97)
0
Equation (3.96)allows us to define a scalar potential through (3.30). Substitution into
(3.97)gives
1
2
∇ (r) =− [ρ(r) + ρ P (r)] (3.98)
0
where ρ P =−∇ · P. This has the form of Poisson’s equation (3.50), but with charge
density term ρ(r) + ρ P (r). Hence the solution is

1 ρ(r ) −∇ · P(r )

(r) = dV .

4π 0 V |r − r |
To this we must add any potential produced by surface sources such as ρ s . If there is a
discontinuity in the dielectric region, there is also a surface polarization source ρ Ps = ˆ n·P
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