Page 75 - Engineering Electromagnetics, 8th Edition

CHAPTER 3 Electric Flux Density, Gauss’s Law, and Divergence 57

which agrees with the results of Chapter 2. The example is a trivial one, and the
objection could be raised that we had to know that the field was symmetrical and
directed radially outward before we could obtain an answer. This is true, and that
leaves the inverse-square-law relationship as the only check obtained from Gauss’s
law. The example does, however, serve to illustrate a method which we may apply
to other problems, including several to which Coulomb’s law is almost incapable of
supplying an answer.
Are there any other surfaces which would have satisfied our two conditions? The
student should determine that such simple surfaces as a cube or a cylinder do not meet
the requirements.
As a second example, let us reconsider the uniform line charge distribution ρ L
lying along the z axis and extending from −∞ to +∞.We must first know the
symmetry of the field, and we may consider this knowledge complete when the
answers to these two questions are known:
1. With which coodinates does the field vary (or of what variables is D a function)?
2. Which components of D are present?
In using Gauss’s law, it is not a question of using symmetry to simplify the
solution, for the application of Gauss’s law depends on symmetry, and if we cannot
show that symmetry exists then we cannot use Gauss’s law to obtain a solution. The
preceding two questions now become “musts.”
From our previous discussion of the uniform line charge, it is evident that only
the radial component of D is present, or

D = D ρ a ρ
and this component is a function of ρ only.
D ρ = f (ρ)
The choice of a closed surface is now simple, for a cylindrical surface is the only
surface to which D ρ is everywhere normal, and it may be closed by plane surfaces
normal to the z axis. A closed right circular cylinder of radius ρ extending from z = 0
to z = L is shown in Figure 3.4.
We apply Gauss’s law,

dS + 0 dS + 0 dS
Q = D S · dS = D S
cyl sides top bottom
L
2π
= D S ρ dφ dz = D S 2πρL
z=0 φ=0
and obtain
Q
D S = D ρ =
2πρL
In terms of the charge density ρ L , the total charge enclosed is

Q = ρ L L