Page 160 - Electromagnetics

according to (3.35). Separating the volume into regions with bounding surfaces S i across
which the permittivity is discontinuous, we may write

1 ρ(r ) 1 ρ s (r )

(r) = dV + dS +

4π 0 V |r − r | 4π 0 S |r − r |

1 −∇ · P(r ) 1 ˆ n · P(r )

+ dV + dS , (3.99)

4π 0 |r − r | 4π 0 |r − r |
i V i S i
where ˆ n points outward from region i. Using the divergence theorem on the fourth term
and employing (B.42), we obtain

1 ρ(r ) 1 ρ s (r )

(r) = dV + dS +

4π 0 V |r − r | 4π 0 S |r − r |

1 1

+ P(r ) ·∇ dV .

4π 0 |r − r |
i V i
ˆ
2
Since ∇ (1/R) = R/R , the third term is a sum of integrals of the form

ˆ
1 R

P(r ) · dV.
4π R 2
V i
Comparing this to the second term of (3.92), we see that this integral represents a volume
superposition of dipole terms where P is a volume density of dipole moments.
Thus, a dielectric with permittivity is equivalent to a volume distribution of dipoles
in free space. No higher-order moments are required, and no zero-order moments are
needed since any net charge is included in ρ. Note that we have arrived at this conclusion
based only on Maxwell’s equations and the assumption of a linear, isotropic relationship
between D and E. Assuming our macroscopic theory is correct, we are tempted to make
assumptions about the behavior of matter on a microscopic level (e.g., atoms exposed to
fields are polarized and their electron clouds are displaced from their positively charged
nuclei), but this area of science is better studied from the viewpoints of particle physics
and quantum mechanics.
Potential of an azimuthally-symmetric charged spherical surface. In several
of our example problems we shall be interested in evaluating the potential of a charged
spherical surface. When the charge is azimuthally-symmetric, the potential is particularly
simple.
We will need the value of the integral
1 f (θ )

F(r) = dS (3.100)
4π S |r − r |

where r = r ˆ r describes an arbitrary observation point and r = aˆ r identifies the source

point on the surface of the sphere of radius a. The integral is most easily done using the
−1
expansion (E.200)for |r − r | in spherical harmonics. We have
∞ n n π π

Y nm (θ, φ) r <
2

∗

F(r) = a f (θ )Y (θ ,φ ) sin θ dθ dφ
n+1 nm
n=0 m=−n 2n + 1 r > −π 0
© 2001 by CRC Press LLC