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One hundred twenty five small liquid drops, each carrying a charge of 0.5 µC and each of diameter 0.1 m form a bigger drop. Calculate the potential at the surface of the bigger drop.

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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 = n3r = 1253 (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 = 14πε0QR = (9 × 109) × 125×(0.5×106)0.25
= 9 × 125 × 2 × 103 = 2.25 × 106 V

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