Page 81 - Engineering Electromagnetics, 8th Edition

CHAPTER 3 Electric Flux Density, Gauss’s Law, and Divergence 63

By exactly the same process we find that

∂D y
+ ˙ = x y z
right left ∂y
and

∂D z
+ ˙ = x y z
top bottom ∂z
and these results may be collected to yield

∂D x ∂D y ∂D z
D · dS ˙= + + x y z
S ∂x ∂y ∂z
or

∂D x ∂D y ∂D z
D · dS = Q ˙= + + ν (7)
S ∂x ∂y ∂z
The expression is an approximation which becomes better as ν becomes
smaller, and in the following section we shall let the volume ν approach zero.
For the moment, we have applied Gauss’s law to the closed surface surrounding the
volume element ν and have as a result the approximation (7) stating that

∂D x ∂D y ∂D z
Charge enclosed in volume ν ˙= + + × volume ν (8)
∂x ∂y ∂z

EXAMPLE 3.3
Find an approximate value for the total charge enclosed in an incremental volume of
3
2
10 −9 m located at the origin, if D = e −x sin y a x − e −x cos y a y + 2za z C/m .
Solution. We first evaluate the three partial derivatives in (8):

∂D x =−e −x sin y
∂x
∂D y = e −x sin y
∂y

∂D z = 2
∂z

At the origin, the first two expressions are zero, and the last is 2. Thus, we find that
the charge enclosed in a small volume element there must be approximately 2 ν.If
3
ν is 10 −9 m , then we have enclosed about 2 nC.