Page 120 - Engineering Electromagnetics, 8th Edition

102 ENGINEERING ELECTROMAGNETICS

V 1 is the potential at the location of Q 1 due to the presence of Q 2 , Q 3 ,... .We
therefore have
m=N
1 1
W E = (Q 1 V 1 + Q 2 V 2 + Q 3 V 3 +· · ·) = 2 Q m V m (41)
2
m=1
In order to obtain an expression for the energy stored in a region of continuous
charge distribution, each charge is replaced by ρ ν dv, and the summation becomes an
integral,

1 ρ ν Vdv (42)
W E =
2
vol
Equations (41) and (42) allow us to find the total potential energy present in a
system of point charges or distributed volume charge density. Similar expressions
may be easily written in terms of line or surface charge density. Usually we prefer
to use Eq. (42) and let it represent all the various types of charge which may have to
be considered. This may always be done by considering point charges, line charge
density, or surface charge density to be continuous distributions of volume charge
density over very small regions. We will illustrate such a procedure with an example
shortly.
Before we undertake any interpretation of this result, we should consider a few
lines of more difficult vector analysis and obtain an expression equivalent to Eq. (42)
but written in terms of E and D.
We begin by making the expression a little bit longer. Using Maxwell’s first
equation, replace ρ ν by its equal ∇ · D and make use of a vector identity which is true
for any scalar function V and any vector function D,
∇ · (V D) ≡ V (∇ · D) + D · (∇V ) (43)
This may be proved readily by expansion in rectangular coordinates. We then have,
successively,

1 1 (∇ · D)Vdv
W E = ρ ν Vdv =
2 2
vol vol

1
= 2 [∇ · (V D) − D · (∇V )] dv
vol
Using the divergence theorem from Chapter 3, the first volume integral of the last
equation is changed into a closed surface integral, where the closed surface surrounds
the volume considered. This volume, first appearing in Eq. (42), must contain every
charge, and there can then be no charges outside of the volume. We may therefore
consider the volume as infinite in extent if we wish. We have

1 1
W E = 2 (V D) · dS − 2 D · (∇V ) dv
S vol
The surface integral is equal to zero, for over this closed surface surrounding the
universe we see that V is approaching zero at least as rapidly as 1/r (the charges
look like point charges from there), and D is approaching zero at least as rapidly as
3
2
1/r . The integrand therefore approaches zero at least as rapidly as 1/r , while the