Page 79 - Engineering Electromagnetics, 8th Edition

CHAPTER 3 Electric Flux Density, Gauss’s Law, and Divergence 61

3.4 APPLICATION OF GAUSS’S LAW:
DIFFERENTIAL VOLUME ELEMENT
We are now going to apply the methods of Gauss’s law to a slightly different type
of problem—one that does not possess any symmetry at all. At first glance, it might
seem that our case is hopeless, for without symmetry, a simple gaussian surface cannot
be chosen such that the normal component of D is constant or zero everywhere on
the surface. Without such a surface, the integral cannot be evaluated. There is only
one way to circumvent these difficulties and that is to choose such a very small
closed surface that D is almost constant over the surface, and the small change in
D may be adequately represented by using the first two terms of the Taylor’s-series
expansion for D. The result will become more nearly correct as the volume enclosed
by the gaussian surface decreases, and we intend eventually to allow this volume to
approach zero.
This example also differs from the preceding ones in that we will not obtain the
value of D as our answer but will instead receive some extremely valuable information
about the way D varies in the region of our small surface. This leads directly to one
of Maxwell’s four equations, which are basic to all electromagnetic theory.
Let us consider any point P, shown in Figure 3.6, located by a rectangular
coordinate system. The value of D at the point P may be expressed in rectangular
components, D 0 = D x0 a x + D y0 a y + D z0 a z .We choose as our closed surface the
small rectangular box, centered at P,having sides of lengths x, y, and z, and
apply Gauss’s law,

D · dS = Q
S

In order to evaluate the integral over the closed surface, the integral must be
broken up into six integrals, one over each face,

D · dS = + + + + +
S front back left right top bottom
Consider the first of these in detail. Because the surface element is very small, D
is essentially constant (over this portion of the entire closed surface) and

˙ = D front · S front
front
˙ = D front · y z a x
˙ = D x,front y z
where we have only to approximate the value of D x at this front face. The front face
is at a distance of x/2 from P, and hence
x
D x,front ˙= D x0 + × rate of change of D x with x
2
x ∂D x
˙ = D x0 +
2 ∂x