Q. $n$ Small drops of same size are charged to $V$ volt each. If they coalesce to form a single large drop, then its potential will be
A) $Vn$
B) $Vn^{- 1}$
C) $Vn^{1/3}$
D) $Vn^{2/3}$
Explanation
(d) As $\frac{4}{3}\pi R^{3} = n \times \frac{4}{3}\pi r^{3}\ \therefore R = n^{1/3}r$
New potential $V^{'} = \frac{nq}{4\pi\varepsilon_{0}r} = n^{2/3}V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.