Data : n = 125, q = 0.5 × 10-6 C, d = 0.1 m
The radius of each small drop, r = d/2 = 0.05 m
The volume of the larger drop being equal to the volume of the n smaller drops, the radius of the larger drop is
R = = (0.05) = 5 × 0.05 = 0.25 m
The charge on the larger drop,
Q = nq = 125 × (0.5 × 10-6) C
∴ The electric potential of the surface of the larger drop,
V = = (9 × 109) ×
= 9 × 125 × 2 × 103 = 2.25 × 106 V