Answer to Question #195357 in Engineering for Saim

Question

A negative charge -q is fixed to one corner of a rectangle, as shown in figure 3. What positive charge must be fixed to corner A and what positive charge must be fixed to corner B, so the total electric field at the remaining corner is zero? Express your answer in terms of q.

Answer 1

$$k\frac{q}{2a^2}-k\frac{q'}{2a^2}-k\frac{q'}{2a^2}=0\\q'=0.5q$$