Page 112 - Engineering Electromagnetics, 8th Edition

94 ENGINEERING ELECTROMAGNETICS

were incorrectly removed from the gradient. Once the physical interpretation of the
gradient, expressed by Eq. (25), is grasped as showing the maximum space rate of
change of a scalar quantity and the direction in which this maximum occurs, the vector
nature of the gradient should be self-evident.
The vector operator
∂ ∂ ∂
∇= a x + a y + a z
∂x ∂y ∂z
may be used formally as an operator on a scalar, T , ∇T , producing
∂T ∂T ∂T
∇T = a x + a y + a z
∂x ∂y ∂z
from which we see that

∇T = grad T

This allows us to use a very compact expression to relate E and V,

E =−∇V (29)

The gradient may be expressed in terms of partial derivatives in other coordinate
systems through the application of its definition Eq. (25). These expressions are
derived in Appendix A and repeated here for convenience when dealing with problems
having cylindrical or spherical symmetry. They also appear inside the back cover.

∂V ∂V ∂V
∇V = a x + a y + a z (rectangular) (30)
∂x ∂y ∂z
∂V 1 ∂V ∂V
∇V = a ρ + a φ + a z (cylindrical) (31)
∂ρ ρ ∂φ ∂z
∂V 1 ∂V 1 ∂V
∇V = a r + a θ + a φ (spherical) (32)
∂r r ∂θ r sin θ ∂φ
Note that the denominator of each term has the form of one of the components of dL in
that coordinate system, except that partial differentials replace ordinary differentials;
for example, r sin θ dφ becomes r sin θ∂φ.
We now illustrate the gradient concept with an example.

EXAMPLE 4.4
2
Given the potential field, V = 2x y − 5z, and a point P(−4, 3, 6), we wish to find
several numerical values at point P: the potential V , the electric field intensity E, the
direction of E, the electric flux density D, and the volume charge density ρ ν .
Solution. The potential at P(−4, 5, 6) is
2
V P = 2(−4) (3) − 5(6) = 66 V