Page 97 - Engineering Electromagnetics, 8th Edition

CHAPTER 4 Energy and Potential 79

For this special case of a uniform electric field intensity, we should note that the
work involved in moving the charge depends only on Q, E, and L BA ,avector drawn
from the initial to the final point of the path chosen. It does not depend on the particular
path we have selected along which to carry the charge. We may proceed from B to A
on a straight line or via the Old Chisholm Trail; the answer is the same. We show in
Section4.5thatanidenticalstatementmaybemadeforanynonuniform(static)Efield.
Let us use several examples to illustrate the mechanics of setting up the line
integral appearing in Eq. (5).

EXAMPLE 4.1
We are given the nonuniform field

E = ya x + xa y + 2a z

and we are asked to determine the work expended in carrying 2C from B(1, 0, 1) to
A(0.8, 0.6, 1) along the shorter arc of the circle

2
2
x + y = 1 z = 1
Solution. We use W =−Q B A E · dL, where E is not necessarily constant. Working
in rectangular coordinates, the differential path dL is dxa x + dya y + dza z , and the
integral becomes

A
W =−Q E · dL
B
A
=−2 (ya x + xa y + 2a z ) · (dx a x + dy a y + dz a z )
B
0.8 0.6 1
=−2 ydx − 2 xdy − 4 dz
1 0 1
where the limits on the integrals have been chosen to agree with the initial and final
values of the appropriate variable of integration. Using the equation of the circular
path (and selecting the sign of the radical which is correct for the quadrant involved),
we have

0.8 0.6
2
2
W =−2 1 − x dx − 2 1 − y dy − 0
1 0
0.8 0.6

=− x 1 − x + sin −1 x − y 1 − y + sin −1 y
2
2
1 0
=−(0.48 + 0.927 − 0 − 1.571) − (0.48 + 0.644 − 0 − 0)
=−0.96 J