Page 98 - Engineering Electromagnetics, 8th Edition

80 ENGINEERING ELECTROMAGNETICS

EXAMPLE 4.2
Again find the work required to carry 2C from B to A in the same field, but this time
use the straight-line path from B to A.
Solution. We start by determining the equations of the straight line. Any two of the
following three equations for planes passing through the line are sufficient to define
the line:
y A − y B
y − y B = (x − x B )
x A − x B
z A − z B
z − z B = (y − y B )
y A − y B
x A − x B
x − x B = (z − z B )
z A − z B
From the first equation we have
y =−3(x − 1)
and from the second we obtain
z = 1
Thus,
0.8 0.6 1
W =−2 ydx − 2 xdy − 4 dz
1 0 1
0.8 0.6 y
= 6 (x − 1) dx − 2 1 − dy
1 0 3
=−0.96 J
This is the same answer we found using the circular path between the same
two points, and it again demonstrates the statement (unproved) that the work done is
independent of the path taken in any electrostatic field.
It should be noted that the equations of the straight line show that dy =−3 dx and
1
dx =− dy. These substitutions may be made in the first two integrals, along with
3
a change in limits, and the answer may be obtained by evaluating the new integrals.
This method is often simpler if the integrand is a function of only one variable.
Note that the expressions for dL in our three coordinate systems use the dif-
ferential lengths obtained in Chapter 1 (rectangular in Section 1.3, cylindrical in
Section 1.8, and spherical in Section 1.9):

dL = dx a x + dy a y + dz a z (rectangular) (6)
dL = dρ a ρ + ρ dφa φ + dz a z (cylindrical) (7)
(spherical) (8)
dL = dr a r + rdθ a θ + r sin θ dφ a φ
The interrelationships among the several variables in each expression are determined
from the specific equations for the path.