Page 84 - Engineering Electromagnetics, 8th Edition

66 ENGINEERING ELECTROMAGNETICS

the partial derivatives. Divergence merely tells us how much flux is leaving a small
volume on a per-unit-volume basis; no direction is associated with it.
We can illustrate the concept of divergence by continuing with the example at
the end of Section 3.4.

EXAMPLE 3.4
Find div D at the origin if D = e −x sin y a x − e −x cos y a y + 2za z .
Solution. We use (10) to obtain

∂D x ∂D y ∂D z
div D = + +
∂x ∂y ∂z
=−e −x sin y + e −x sin y + 2 = 2

The value is the constant 2, regardless of location.
3
If the units of D are C/m , then the units of div D are C/m . This is a volume charge
2
density, a concept discussed in the next section.
D3.7. In each of the following parts, find a numerical value for div D at the
2
2
2
2
point specified: (a) D = (2xyz − y )a x + (x z − 2xy)a y + x ya z C/m at
2
2
2
2
2
2
P A (2, 3, −1); (b) D = 2ρz sin φ a ρ + ρz sin 2φ a φ + 2ρ z sin φ a z C/m at
◦
P B (ρ = 2,φ = 110 , z =−1); (c) D = 2r sin θ cos φ a r + r cos θ cos φ a θ −
2
r sin φ a φ C/m at P C (r = 1.5, θ = 30 , φ = 50 ).
◦
◦
Ans. −10.00; 9.06; 1.29
Finally, we can combine Eqs. (9) and (12) and form the relation between electric
flux density and charge density:
(15)
div D = ρ ν
This is the first of Maxwell’s four equations as they apply to electrostatics and
steady magnetic fields, and it states that the electric flux per unit volume leaving a
vanishingly small volume unit is exactly equal to the volume charge density there.
This equation is aptly called the point form of Gauss’s law. Gauss’s law relates the flux
leaving any closed surface to the charge enclosed, and Maxwell’s first equation makes
an identical statement on a per-unit-volume basis for a vanishingly small volume, or
at a point. Because the divergence may be expressed as the sum of three partial
derivatives, Maxwell’s first equation is also described as the differential-equation
form of Gauss’s law, and conversely, Gauss’s law is recognized as the integral form
of Maxwell’s first equation.
As a specific illustration, let us consider the divergence of D in the region about
a point charge Q located at the origin. We have the field
Q
D = a r
4πr 2