Page 154 - Electromagnetics

ways. We could, for instance, bring small portions of charge (or point charges)together
to form the distribution ρ. Or, we could slowly build up ρ by adding infinitesimal, but
spatially identical, distributions. That is, we can create the distribution ρ from a zero
initial state by repeatedly adding a charge distribution
δρ(r) = ρ(r)/N,

where N is a large number. Whenever we add δρ we must perform the work given by
(3.84), but we also increase the potential proportionately (remembering that all materials
are assumed linear). At each step, more work is required. The total work is

N N

ρ(r) (r)
W = δρ(r)[(n − 1)δ (r)] dV = (n − 1) dV. (3.85)
N N
n=1 V ∞ n=1 V ∞
We must use an infinite number of steps so that no energy is lost to radiation at any step
(since the charge we add each time is infinitesimally small). Using
N

(n − 1) = N(N − 1)/2,
n=1
(3.85)becomes

1
W = ρ(r) (r) dV (3.86)
2
V ∞
as N →∞. Finally, since some assembled charge will be in the form of a volume density
and some in the form of the surface density on conductors, we can generalize (3.86)to
I

1 1
W = ρ(r) (r) dV + Q i V i . (3.87)
2 V 2 i=1
Here V is the region outside the conductors, Q i is the total charge on the ith conductor

(i = 1,..., I), and V i is the absolute potential (referred to infinity)of the ith conductor.
An intriguing property of electrostatic energy is that the charges on the conductors
will arrange themselves, while seeking static equilibrium, into a minimum-energy config-
uration (Thomson’s theorem).
In keeping with our field-centered view of electromagnetics, we now wish to write the
energy (3.86)entirely in terms of the field vectors E and D. Since ρ =∇ · D we have

1
W = [∇· D(r)] (r) dV.
2
V ∞
Then, by (B.42),

1 1
W = ∇· [ (r)D(r)] dV − D(r) · [∇ (r)] dV.
2 2
V ∞ V ∞
Use of the divergence theorem and (3.30)leads to

1 1
W = (r)D(r) · dS + D(r) · E(r) dV
2 2
S ∞ V ∞

© 2001 by CRC Press LLC