Q. In Millikan’s oil drop experiment a drop of charge $Q$ and radius $r$ is kept constant between two plates of potential difference of 800 V. Then charge on other drop of radius 2 $r$ which is kept constant with a potential difference of 3200 V is
A) $Q/2$
B) $2\ Q$
C) $4\ Q$
D) $Q/4$
Explanation
(b) $\frac{QV}{l} - \frac{4}{3}\pi r^{3}\ \rho g$
and
$\frac{Q^{'}V'}{l} = \frac{4}{3}\pi(2r)^{3}\ \rho g$
So, $\frac{Q^{'}V^{'}}{QV} = 8$
or $Q^{'} = \frac{8QV}{V^{'}} = 8Q \times \frac{800}{3200} = 2Q$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.