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(D) Let $r$ be the radius of each small drop and $q$ be the charge on each small drop.The potential of a small drop is given by $V = \frac{kq}{r}$.Whe

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$n^{2/3}V$

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(D) Let $r$ be the radius of each small drop and $q$ be the charge on each small drop.The potential of a small drop is given by $V = \frac{kq}{r}$.When $n$ small drops coalesce to form a big drop of radius $R$ and charge $Q$, the total volume remains conserved.So, $n 	imes (\frac{4}{3}\pi r^3) = \frac{4}{3}\pi R^3$, which implies $R^3 = nr^3$ or $R = n^{1/3}r$.The total charge on the big drop is $Q = nq$.The potential of the big drop $V_{big}$ is given by $V_{big} = \frac{kQ}{R}$.Substituting the values of $Q$ and $R$, we get $V_{big} = \frac{k(nq)}{n^{1/3}r} = n^{1 - 1/3} 	imes \frac{kq}{r} = n^{2/3}V$.

$n$ small spherical drops of the same size, each charged to a potential $V$, coalesce to form a single big drop. The potential of the big drop is:

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