Question

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} $

Answer 1

Answer: (B)

Force on charge, $F=Q E$
or $F=\frac{Q V}{d} (\because V=E d)$
For drop to be stationary, weight of drop = force due to charge
i.e., $m g=\frac{Q V}{d}$
For two drops, $\frac{Q_{1}}{Q_{2}} \cdot \frac{V_{1}}{V_{2}}=\frac{m_{1}}{m_{2}}$
or $\frac{Q_{1}}{Q_{2}} \cdot \frac{V_{1}}{V_{2}}=\frac{\frac{4}{3} \pi r_{1}^{3} \rho}{\frac{4}{3} \pi r_{2}^{3} \rho}$
or $\frac{Q_{2}}{Q_{1}}=\frac{r_{2}^{3}}{r_{1}^{3}} \times \frac{V_{1}}{V_{2}}$
$\therefore \frac{Q_{2}}{Q}=\frac{(r / 2)^{3}}{r^{3}} \times \frac{2400}{600}$
or $Q_{2}=\frac{Q}{2}$