Page 184 - Engineering Electromagnetics, 8th Edition

166 ENGINEERING ELECTROMAGNETICS

Laplace’s equation is now
2
1 ∂ V = 0
2
ρ ∂φ 2
We exclude ρ = 0 and have
2
d V = 0
dφ 2
The solution is
V = Aφ + B
The boundary conditions determine A and B, and

φ
V = V 0 (37)
α
Taking the gradient of Eq. (37) produces the electric field intensity,

V 0 a φ
E =− (38)
αρ
and it is interesting to note that E is a function of ρ and not of φ. This does not
contradict our original assumptions, which were restrictions only on the potential
field. Note, however, that the vector field E is in the φ direction.
A problem involving the capacitance of these two radial planes is included at the
end of the chapter.

EXAMPLE 6.5
We now turn to spherical coordinates, dispose immediately of variations with respect
to φ only as having just been solved, and treat first V = V (r).
The details are left for a problem later, but the final potential field is given by
1 1
−
r b
V = V 0
1 1 (39)
−
a b
where the boundary conditions are evidently V = 0at r = b and V = V 0 at r = a,
b > a. The problem is that of concentric spheres. The capacitance was found previ-
ously in Section 6.3 (by a somewhat different method) and is

4π
C =
1 1 (40)
−
a b