Question

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

• A. $V$
• B. $V/N$
• C. $V \times N$
• D. $V \times N^{2/3}$

Answer 1

Answer: (D)

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

Answer 2

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 $V = \frac{Q}{R}$

$= \frac{Nq}{N^{1/3}r} = V \times N^{2/3}$