Physics---electrostatics3

(a) At D, distance from each of B and C = 0.05m
So VD = (2 × 10-7)/(4π × 8.85 × 10-12 × 0.05) + (-1 × 10-7)/(4π × 8.85 × 10-12 × 0.05)
= +17984 V
(b) VA = (2 × 10-7)/(4π × 8.85 × 10-12 × 0.1) + (-1 × 10-7)/(4π × 8.85 × 10-12 × 0.1)
= +8992 V
(c) Potential difference between D and A is:
VD - VA = 8992 V
Therefore, when bringing a positive charge from D to A, positive work is done which is equal to:
8992 × 3 × 10-7
= 2.7 × 10-3 J