Page 46 - Engineering Electromagnetics, 8th Edition

28 ENGINEERING ELECTROMAGNETICS

Figure 2.1 If Q 1 and Q 2 have like
signs, the vector force F 2 on Q 2 is in the
same direction as the vector R 12 .

and is repulsive if the charges are alike in sign or attractive if they are of opposite sign.
Let the vector r 1 locate Q 1 , whereas r 2 locates Q 2 . Then the vector R 12 = r 2 − r 1
represents the directed line segment from Q 1 to Q 2 ,as shown in Figure 2.1. The vector
F 2 is the force on Q 2 and is shown for the case where Q 1 and Q 2 have the same sign.
The vector form of Coulomb’s law is
Q 1 Q 2
F 2 = a 12 (3)
2
4π 0 R 12
where a 12 = a unit vector in the direction of R 12 ,or
R 12 R 12 r 2 − r 1
a 12 = = = (4)
|R 12 | R 12 |r 2 − r 1 |

EXAMPLE 2.1
We illustrate the use of the vector form of Coulomb’s law by locating a charge of
Q 1 = 3 × 10 −4 Cat M(1, 2, 3) and a charge of Q 2 =−10 −4 Cat N(2, 0, 5) in a
vacuum. We desire the force exerted on Q 2 by Q 1 .
Solution. We use (3) and (4) to obtain the vector force. The vector R 12 is

R 12 = r 2 − r 1 = (2 − 1)a x + (0 − 2)a y + (5 − 3)a z = a x − 2a y + 2a z
1
leading to |R 12 |= 3, and the unit vector, a 12 = (a x − 2a y + 2a z ). Thus,
3
−4
−4
3 × 10 (−10 ) a x − 2a y + 2a z
F 2 =
4π(1/36π)10 −9 × 3 2 3

a x − 2a y + 2a z
=−30 N
3
The magnitude of the force is 30 N, and the direction is specified by the unit
vector, which has been left in parentheses to display the magnitude of the force. The
force on Q 2 may also be considered as three component forces,
F 2 =−10a x + 20a y − 20a z