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)

$\frac{Q}{2}$

Answer 2

In balance condition

⇒ $QE = mg$ ⇒ $Q\frac{V}{d} = \left( \frac{4}{3}\pi r^{3}\rho \right)g$

⇒ $Q \propto \frac{r^{3}}{V}$ ⇒ $\frac{Q_{1}}{Q_{2}} = \left( \frac{r_{1}}{r_{2}} \right)^{3} \times \frac{V_{2}}{V_{1}}$

⇒ $\frac{Q}{Q_{2}} = \left( \frac{r}{r/2} \right)^{3} \times \frac{600}{2400} = 2$ ⇒ Q₂ = Q / 2