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n charged drops, each of radius r and charge q, coalesce to form a big drop of radius R and charge Q.If V is the electric potential and E is the electric field at the surface of a drop, then

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$E_{s m a l l} = n^{2 / 3} E_{b i g}$

$V_{b i g} = n^{1 / 3} V_{s m a l l}$

$E_{b i g} = n^{2 / 3} E_{s m a l l}$

$V_{b i g} = n^{2 / 3} V_{s m a l l}$

answer is D.

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Detailed Solution

$Q = n q, R = n^{1 / 3} r$

$E_{s m a l l} = \frac{k q}{r^{2}}, E_{b i g} = \frac{k Q}{R^{2}}$

$E_{b i g} = \frac{k n q}{n^{2 / 3} r^{2}} = (n^{1 / 3}) \frac{k q}{r^{2}} = n^{1 / 3} E_{s m a l l}$

$V_{s m a l l} = \frac{k q}{r}, V_{b i g} = \frac{k Q}{R}$

$V_{b i g} = \frac{k n q}{n^{1 / 3} r} = n^{2 / 3} \frac{k q}{r} = n^{2 / 3} V_{s m a l l}$

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