FIGURE 1 11
Solution The electric field vector E1A at A due to the positive charge
q1 points towards the right and has a magnitude
9
2
-2
8
1A
2
(9
10 Nm C )
(10
C)
(0 05m)
E
−
×
×
=
 =  3 6 × 104  N C–1
The electric field vector E2A at A due to the negative charge q2 points
towards the right and has the same magnitude