Page 77 - Engineering Electromagnetics, 8th Edition

CHAPTER 3 Electric Flux Density, Gauss’s Law, and Divergence 59

Figure 3.5 The two coaxial
cylindrical conductors forming a
coaxial cable provide an electric
flux density within the cylinders,
given by D ρ = aρ S/ρ.

The total charge on a length L of the inner conductor is
L 2π
Q = ρ S adφ dz = 2πaLρ S
z=0 φ=0
from which we have
aρ S aρ S
D S = D = a ρ (a <ρ < b)
ρ ρ
This result might be expressed in terms of charge per unit length because the inner
conductor has 2πaρ S coulombs on a meter length, and hence, letting ρ L = 2πaρ S ,

ρ L
D = a ρ
2πρ
and the solution has a form identical with that of the infinite line charge.
Because every line of electric flux starting from the charge on the inner cylinder
must terminate on a negative charge on the inner surface of the outer cylinder, the
total charge on that surface must be
Q outer cyl =−2πaLρ S,inner cyl
and the surface charge on the outer cylinder is found as

2πbLρ S,outer cyl =−2πaLρ S,inner cyl
or
a
ρ S,outer cyl =− ρ S,inner cyl
b
What would happen if we should use a cylinder of radius ρ, ρ > b, for the
gaussian surface? The total charge enclosed would then be zero, for there are equal
and opposite charges on each conducting cylinder. Hence
0 = D S 2πρL (ρ> b)
D S = 0 (ρ> b)