N identical spherical drops charged to the same potential V…

Physics Electrostatics Potential & Capacitance Capacitance Single Correct MCQ

Published on: September 12, 2026

N identical spherical drops charged to the same potential V are combined to form a big drop. The potential of the new drop will be

V / N

$V \times N$

$V \times N^{2/3}$

Text Solution

Verified by Experts

The correct answer is:

If the drops are conducting, then

$\frac{4}{3} \pi R^{3} = N \left( \frac{4}{3} \pi r^{3} \right)$ ⇒ ⇒ $R = N^{1/3} r$ . Final charge Q = Nq

So final potential $V = \frac{Q}{R}$ $= \frac{Nq}{N^{1/3} r} = V \times N^{2/3}$

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