Question

In Millikan's oil drop experiment, an oil drop carrying a charge $Q$ is held stationary by a potential difference $2400V$ between the plates. To keep a drop of half the radius stationary, the potential difference had to be made $600V$. 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, $F=QE$
or $F=\frac{QV}{d}(\because V=Ed)$
For drop to be stationary, weight of drop = force due to charge
i.e., $mg=\frac{QV}{d}$
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 $Q_2=\frac{Q}{2}$