Page 121 - Engineering Electromagnetics, 8th Edition

CHAPTER 4 Energy and Potential 103

differential area of the surface, looking more and more like a portion of a sphere,
2
is increasing only as r . Consequently, in the limit as r →∞, the integrand and
the integral both approach zero. Substituting E =−∇V in the remaining volume
integral, we have our answer,

1 1 2
W E = 2 D · E dv = 2 0 E dv (44)
vol vol
We may now use this last expression to calculate the energy stored in the elec-
trostatic field of a section of a coaxial cable or capacitor of length L.We found in
Section 3.3 that
aρ S
D ρ =
ρ
Hence,
aρ S
E = a ρ
0 ρ
where ρ S is the surface charge density on the inner conductor, whose radius is a.
Thus,

L 2π b 2 2 2 2 b
1 a ρ S π La ρ S
W E = 0 ρ dρ dφ dz = ln
2 2 2 a
0 0 a ρ 0
0
This same result may be obtained from Eq. (42). We choose the outer conductor
as our zero-potential reference, and the potential of the inner cylinder is then
a a b
aρ S aρ S
V a =− E ρ dρ =− dρ = ln
b b 0 ρ 0 a
The surface charge density ρ S at ρ = a can be interpreted as a volume charge density
1
1
ρ ν = ρ S /t,extending from ρ = a − t to ρ = a + t, where t a. The integrand
2
2
in Eq. (42) is therefore zero everywhere between the cylinders (where the volume
charge density is zero), as well as at the outer cylinder (where the potential is zero).
The integration is therefore performed only within the thin cylindrical shell at ρ = a,
L 2π a+t/2 b
1 1 ρ S ρ S
W E = 2 ρ ν VdV = 2 a ln ρ dρ dφ dz
vol 0 0 a−t/2 t 0 a
from which
2 2
a ρ ln(b/a)
W E = S πL
0
once again.
This expression takes on a more familiar form if we recognize the total charge
on the inner conductor as Q = 2πaLρ S . Combining this with the potential difference
between the cylinders, V a ,we see that
1
W E = QV a
2
which should be familiar as the energy stored in a capacitor.