Q. In Milikan’s experiment, an oil drop having charge $q$ gets stationary on applying a potential difference $V$ in between two plates separated by a distance $'d^{'}.$ The weight of the drop is
A) $qVd$
B) $q\frac{d}{V}$
C) $\frac{q}{Vd}$
D) $q\frac{V}{d}$
Explanation
(d) $QE = mg \Rightarrow mg = \frac{QV}{d}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.