Page 172 - Engineering Electromagnetics, 8th Edition

154 ENGINEERING ELECTROMAGNETICS

If we evaluate D x at x = h − b, y = 0, we may obtain ρ S,max

ρ L h − b + a h − b − a
ρ S,max =−D x,x=h−b,y=0 = −
2π (h − b + a) 2 (h − b − a) 2
For our example,
3.46 × 10 −9 13 − 5 + 12 13 − 5 − 12
ρ S,max = − = 0.165 nC/m 2
2π (13 − 5 + 12) 2 (13 − 5 − 12) 2
Similarly, ρ S,min = D x,x=h+b,y=0, and
3.46 × 10 −9 13 + 5 + 12 13 + 5 − 12
ρ S,min = − = 0.073 nC/m 2
2π 30 2 6 2
Thus,

ρ S,max = 2.25ρ S,min
If we apply Eq. (16) to the case of a conductor for which b h, then

2
2

ln h + h − b /b ˙= ln[(h + h)/b]˙= ln(2h/b)
and
2π L
C = (b h) (17)
ln(2h/b)
The capacitance between two circular conductors separated by a distance 2h
is one-half the capacitance given by Eqs. (16) or (17). This last answer is of inter-
est because it gives us an expression for the capacitance of a section of two-wire
transmission line, one of the types of transmission lines studied later in Chapter 13.

D6.3. A conducting cylinder with a radius of 1 cm and at a potential of 20 V is
parallel to a conducting plane which is at zero potential. The plane is 5 cm distant
from the cylinder axis. If the conductors are embedded in a perfect dielectric
for which r = 4.5, find: (a) the capacitance per unit length between cylinder
and plane; (b) ρ S,max on the cylinder.

Ans. 109.2 pF/m; 42.6 nC/m 2

6.5 USING FIELD SKETCHES TO ESTIMATE
CAPACITANCE IN TWO-DIMENSIONAL
PROBLEMS
In capacitance problems in which the conductor configurations cannot be easily de-
scribedusingasinglecoordinatesystem,otheranalysistechniquesareusuallyapplied.
Such methods typically involve a numerical determination of field or potential values
overa grid within the region of interest. In this section, another approach is described
that involves making sketches of field lines and equipotential surfaces in a manner
that follows a few simple rules. This approach, although lacking the accuracy of more