Page 183 - Engineering Electromagnetics, 8th Edition

CHAPTER 6 Capacitance 165

potential difference of V 0 by letting V = V 0 at ρ = a, V = 0at ρ = b, b > a, and
obtain

ln(b/ρ)
V = V 0 (35)
ln(b/a)
from which

V 0 1
E = a ρ
ρ ln(b/a)
V 0
D N(ρ=a) =
a ln(b/a)
V 0 2πaL
Q =
a ln(b/a)

2π L
C = (36)
ln(b/a)
which agrees with our result in Section 6.3 (Eq. (5)).

EXAMPLE 6.4
Now assume that V is a function only of φ in cylindrical coordinates. We might look
at the physical problem first for a change and see that equipotential surfaces are given
by φ = constant. These are radial planes. Boundary conditions might be V = 0at
φ = 0 and V = V 0 at φ = α, leading to the physical problem detailed in Figure 6.10.

Figure 6.10 Two infinite radial planes with an
interior angle α.An infinitesimal insulating gap exists
at ρ = 0. The potential field may be found by applying
Laplace’s equation in cylindrical coordinates.