Page 169 - Engineering Electromagnetics, 8th Edition

CHAPTER 6 Capacitance 151

y

R 2 P(x, y, 0)
R
(−a, 0, 0) 1
(a, 0, 0) x
2a

z +r
−r L L
Figure 6.4 Two parallel infinite line charges carrying opposite
charge. The positive line is at x = a, y = 0, and the negative line
is at x =−a, y = 0. A general point P(x, y, 0) in the xy plane is
radially distant R 1 and R 2 from the positive and negative lines,
respectively. The equipotential surfaces are circular cylinders.

We choose R 10 = R 20 , thus placing the zero reference at equal distances from each
line. This surface is the x = 0 plane. Expressing R 1 and R 2 in terms of x and y,

2
2
ρ L (x + a) + y 2 ρ L (x + a) + y 2
V = ln = ln (11)
2
2
2π (x − a) + y 2 4π (x − a) + y 2
In order to recognize the equipotential surfaces and adequately understand the
problem we are going to solve, some algebraic manipulations are necessary. Choosing
an equipotential surface V = V 1 ,we define K 1 as a dimensionless parameter that is
a function of the potential V 1 ,
K 1 = e 4π V 1 /ρ L (12)
so that
2
(x + a) + y 2
K 1 =
2
(x − a) + y 2
After multiplying and collecting like powers, we obtain
K 1 + 1
2
2
2
x − 2ax + y + a = 0
K 1 − 1
We next work through a couple of lines of algebra and complete the square,
K 1 + 1 2a K 1
2 √ 2
2
x − a + y =
K 1 − 1 K 1 − 1
This shows that the V = V 1 equipotential surface is independent of z (or is a cylinder)
and intersects the xy plane in a circle of radius b,
√
2a K 1
b =
K 1 − 1