First slide

Capacitors and capacitance

Question

If n drops of mercury, each charged to a potential V, coalesce to form a single drop, the potential of the big drop will be

Moderate

A

$\frac{V}{n^{2 / 3}}$
B

$\frac{V}{n^{1 / 3}}$
C

${Vn}^{1 / 3}$
D

${Vn}^{2 / 3}$

Solution

If Q is the charge on each small drop, charge on the big drop is $Q^{'} = nQ . Now Q^{'} = C^{'} V^{'} = Q = CV$ . Therefore
$\frac{V^{'}}{V} = \frac{Q^{'}}{Q} \times \frac{C}{C^{'}} = \frac{n}{n^{1 / 3}} = n^{2 / 3}$
Hence the correct choice is (4).

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