Q. In Millikan’s oil drop experiment an oil drop carrying a charge $Q$ is held stationary by a potential difference $2400\ V$ between the plates. To keep a drop of half the radius stationary the potential difference had to be made $600\ V.$ What is the charge on the second drop
A) $\frac{Q}{4}$
B) $\frac{Q}{2}$
C) $Q$
D) $\frac{3Q}{2}$
Explanation
(b) In balance condition
$\Rightarrow QE = mg \Rightarrow Q\frac{V}{d} = \left( \frac{4}{3}\pi r^{3}\rho \right)g$
$\Rightarrow Q \propto \frac{r^{3}}{V} \Rightarrow \frac{Q_{1}}{Q_{2}} = \left( \frac{r_{1}}{r_{2}} \right)^{3} \times \frac{V_{2}}{V_{1}}$
$\Rightarrow \frac{Q}{Q_{2}} = \left( \frac{r}{r/2} \right)^{3} \times \frac{600}{2400} = 2 \Rightarrow Q_{2} = Q/2$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.