Page 185 - Engineering Electromagnetics, 8th Edition

CHAPTER 6 Capacitance 167

EXAMPLE 6.6
In spherical coordinates we now restrict the potential function to V = V (θ), obtaining
1 d dV
sin θ = 0
2
r sin θ dθ dθ
We exclude r = 0 and θ = 0or π and have
dV
sin θ = A
dθ
The second integral is then
Adθ

V = + B
sin θ
which is not as obvious as the previous ones. From integral tables (or a good memory)
we have
θ

V = A ln tan + B (41)
2
The equipotential surfaces of Eq. (41) are cones. Figure 6.11 illustrates the case
where V = 0at θ = π/2 and V = V 0 at θ = α, α< π/2. We obtain

θ
ln tan
2
(42)
V = V 0 α
ln tan
2

Figure 6.11 For the cone θ = α at V 0 and the
plane θ = π/2at V = 0, the potential field is given by
V = V 0 [ln(tan θ/2)]/[ln(tan α/2)].