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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 drop then.

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

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

$a & b$

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

answer is D.

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Detailed Solution

$Q = n q$

$R = n^{\frac{1}{3}} r$

$R^{2} = n^{\frac{2}{3}} r^{2}$

$E_{s m a l l} = \frac{k q}{r^{2}}$

$E_{b i g} = \frac{k q}{R^{2}}$

$E_{b i g} = \frac{k n q}{n^{2 / 3} r^{2}}$

$\begin{array}{l} E_{b i g} = n^{\frac{1}{3}} \frac{k q}{r^{2}} \\ E_{b i g} = n^{\frac{1}{3}} E_{s m a l l} \end{array}$

$V_{s m a l l} = \frac{k q}{r}$

$V_{b i g} = \frac{k Q}{R}$

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

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