Q. $N$ 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) $V \times N^{2/3}$
Explanation
(d) If the drops are conducting, then
$\frac{4}{3}\pi R^{3} = N\left( \frac{4}{3}\pi r^{3} \right) \Rightarrow R = N^{1/3}r.$ Final charge $Q = Nq$
So final potential $V^{'} = \frac{Q}{R} = \frac{Nq}{N^{1/3}r} = V \times N^{2/3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.