Page 170 - Engineering Electromagnetics, 8th Edition

152 ENGINEERING ELECTROMAGNETICS

which is centered at x = h, y = 0, where
K 1 + 1
h = a
K 1 − 1
Now let us attack a physical problem by considering a zero-potential conducting
plane located at x = 0, and a conducting cylinder of radius b and potential V 0 with
its axis located a distance h from the plane. We solve the last two equations for a and
K 1 in terms of the dimensions b and h,

2
a = h − b 2 (13)
and
√
2

h + h − b 2
K 1 = (14)
b
But the potential of the cylinder is V 0 ,so Eq. (12) leads to
K 1 = e 2π V 0 /ρ L

Therefore,
4π V 0
ρ L = (15)
lnK 1
Thus, given h, b, and V 0 ,we may determine a, ρ L , and the parameter K 1 . The
capacitance between the cylinder and plane is now available. For a length L in the z
direction, we have
ρ L L 4π L 2π L
C = = = √
V 0 lnK 1 ln K 1
or
2π L 2π L
C = √ = −1 (16)
2
2
ln[(h + h − b )/b] cosh (h/b)
The solid line in Figure 6.5 shows the cross section of a cylinder of 5 m radius
at a potential of 100 V in free space, with its axis 13 m distant from a plane at zero
potential. Thus, b = 5, h = 13, V 0 = 100, and we rapidly find the location of the
equivalent line charge from Eq. (13),

2
2
2 2 13 − 5 = 12 m
a = h − b =
the value of the potential parameter K 1 from Eq. (14),
√
2

h + h − b 2 13 + 12
K 1 = = = 5 K 1 = 25
b 5
the strength of the equivalent line charge from Eq. (15),
4π V 0 4π × 8.854 × 10 −12 × 100
ρ L = = = 3.46 nC/m
lnK 1 ln 25
and the capacitance between cylinder and plane from Eq. (16),
2π 2π × 8.854 × 10 −12
C = = = 34.6 pF/m
−1
−1
cosh (h/b) cosh (13/5)