Page 50 - Engineering Electromagnetics, 8th Edition

32 ENGINEERING ELECTROMAGNETICS

If we add more charges at other positions, the field due to n point charges is
n
Q m
E(r) = 2 a m (11)
m=1 4π 0 |r − r m |

EXAMPLE 2.2
In order to illustrate the application of (11), we find E at P(1, 1, 1) caused by four iden-
tical 3-nC (nanocoulomb) charges located at P 1 (1, 1, 0), P 2 (−1, 1, 0), P 3 (−1, −1, 0),
and P 4 (1, −1, 0), as shown in Figure 2.4.

Solution. We find that r = a x + a y + a z , r 1 = a x + a y , and thus r − r 1 = a z .
√ √
The magnitudes are: |r − r 1 |= 1, |r − r 2 |= 5, |r − r 3 |= 3, and |r − r 4 |= 5.
−9
Because Q/4π 0 = 3 × 10 /(4π × 8.854 × 10 −12 ) = 26.96 V · m, we may now
use (11) to obtain

a z 1 2a x + a z 1 2a x + 2a y + a z 1 2a y + a z 1
E = 26.96 + √ + √
2
1 1 2 5 √ + 3 3 2 5 √ 2
5
5
or
E = 6.82a x + 6.82a y + 32.8a z V/m
D2.2. A charge of −0.3 µCis located at A(25, −30, 15) (in cm), and a
second charge of 0.5 µCisat B(−10, 8, 12) cm. Find E at: (a) the origin;
(b) P(15, 20, 50) cm.

Ans. 92.3a x − 77.6a y − 94.2a z kV/m; 11.9a x − 0.519a y + 12.4a z kV/m

Figure 2.4 A symmetrical distribution of four identical 3-nC point
charges produces a field at P, E = 6.82a x + 6.82a y + 32.8a z V/m.