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
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In balance condition
⇒ ⇒ $QE = mg$ ⇒ ⇒ $Q \frac{v}{d} = \left( \frac{4}{3} \pi r^{3} \rho \right) g$
⇒ ⇒ $Q \propto \frac{r^{3}}{V}$ ⇒ ⇒ $\frac{Q_1}{Q_2} = \left(\frac{r_1}{r_2}\right)^3 \times \frac{V_2}{V_1}$
⇒ ⇒ $\frac{Q}{Q_2} = \left(\frac{r}{r/2}\right)^3 \times \frac{600}{2400} = 2$ ⇒ ⇒ Q 2 = Q / 2
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