Q. An oil drop with charge $q$ is held stationary between two plates with an external potential difference of 400 V. If the size of the drop is doubled without any change of charge, the potential difference required to keep the drop stationary will be
A) 400 V
B) 1600 V
C) 3200 V
D) 4000 V
Explanation
(c) $mg = q\ E$ or $\frac{4}{3}\pi r^{3}\ \rho g = \frac{qV}{d}$ or $V \propto r^{3}$
$\therefore\ \ V_{2} = V_{1}\left( \frac{r_{2}}{r_{1}} \right)^{3} = 400 \times \left( \frac{2}{1} \right)^{3} = 3200V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.