Page 182 - Engineering Electromagnetics, 8th Edition

164 ENGINEERING ELECTROMAGNETICS

Here we have
x
V = V 0
d
V 0
E =− a x
d
V 0
D =− a x
d
D S = D =− V 0 a x
x=0 d
a N = a x
V 0
D N =− = ρ S
d
V 0 S
− V 0
Q = dS =−
S d d
and the capacitance is
|Q| S
C = = (33)
V 0 d
We will use this procedure several times in the examples to follow.

EXAMPLE 6.3
Because no new problems are solved by choosing fields which vary only with y or
with z in rectangular coordinates, we pass on to cylindrical coordinates for our next
example. Variations with respect to z are again nothing new, and we next assume
variation with respect to ρ only. Laplace’s equation becomes

1 ∂ ∂V
ρ = 0
ρ ∂ρ ∂ρ
Noting the ρ in the denominator, we exclude ρ = 0 from our solution and then
multiply by ρ and integrate,
dV
ρ = A
dρ
where a total derivative replaces the partial derivative because V varies only with ρ.
Next, rearrange, and integrate again,
V = A ln ρ + B (34)

The equipotential surfaces are given by ρ = constant and are cylinders, and the
problem is that of the coaxial capacitor or coaxial transmission line. We choose a