Page 168 - Engineering Electromagnetics, 8th Edition

150 ENGINEERING ELECTROMAGNETICS

interface, and they must be equal. Then we may find in succession D 1 , D 2 ,ρ S1 ,ρ S2 ,
and Q, obtaining a capacitance

1 S 1 + 2 S 2
C = = C 1 + C 2 (10)
d
as we should expect.
At this time we can do very little with a capacitor in which two dielectrics
are used in such a way that the interface is not everywhere normal or parallel to
the fields. Certainly we know the boundary conditions at each conductor and at the
dielectric interface; however, we do not know the fields to which to apply the boundary
conditions. Such a problem must be put aside until our knowledge of field theory has
increased and we are willing and able to use more advanced mathematical techniques.

D6.2. Determine the capacitance of: (a)a 1-ft length of 35B/U coaxial cable,
which has an inner conductor 0.1045 in. in diameter, a polyethylene dielectric
( r = 2.26 from Table C.1), and an outer conductor that has an inner diameter of
0.680 in.; (b)a conducting sphere of radius 2.5 mm, covered with a polyethylene
layer 2 mm thick, surrounded by a conducting sphere of radius 4.5 mm; (c)two
rectangular conducting plates, 1 cm by 4 cm, with negligible thickness, between
which are three sheets of dielectric, each 1 cm by 4 cm, and 0.1 mm thick, having
dielectric constants of 1.5, 2.5, and 6.

Ans. 20.5 pF; 1.41 pF; 28.7 pF

6.4 CAPACITANCE OF A TWO-WIRE LINE
We next consider the problem of the two-wire line. The configuration consists of two
parallel conducting cylinders, each of circular cross section, and we will find complete
information about the electric field intensity, the potential field, the surface-charge-
density distribution, and the capacitance. This arrangement is an important type of
transmission line, as is the coaxial cable.
Webeginbyinvestigatingthepotentialfieldoftwoinfinitelinecharges.Figure6.4
shows a positive line charge in the xz plane at x = a and a negative line charge at
x =−a. The potential of a single line charge with zero reference at a radius of R 0 is

ρ L R 0
V = ln
2π R
We now write the expression for the combined potential field in terms of the radial
distances from the positive and negative lines, R 1 and R 2 , respectively,

ρ L R 10 R 20 ρ L R 10 R 2
V = ln − ln = ln
2π R 1 R 2 2π R 20 R 1