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A small drop of mercury of radius ‘r’ lies on a shallow metal disc connected to a steady voltage source V. The drop is then broken into n droplets on the disc. Find the extra charge drawn from the voltage source.

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$4 π ϵ_{0} {rVn}^{2 / 3}$

$4 π ϵ_{0} rV (n^{2 / 3} - 1)$

$4 π ϵ_{0} {rVn}^{1 / 3}$

$4 π ϵ_{0} rV (n^{1 / 3} - 1)$

answer is B.

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Detailed Solution

$r^{'} = n^{- \frac{1}{3}} r$

$V = \frac{kQ}{R}$

$\Rightarrow Q_{0} = \frac{Vr}{k} = 4 {πε}_{0} Vr$

$\Rightarrow \frac{q}{Q_{0}} = \frac{r^{'}}{r} = n^{- \frac{1}{3}}$

$\Rightarrow q = Q_{0} n^{- \frac{1}{3}}$

$\Rightarrow Q = nq = Q_{0} n^{\frac{2}{3}}$

$∆ Q = Q_{0} (n^{\frac{2}{3}} - 1) = 4 {πε}_{0} Vr (n^{\frac{2}{3}} - 1)$

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