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(D) The capacitance of a spherical conductor is given by $C = 4 \pi \epsilon_0 R$. When two identical droplets of radius $R$ combine to form a larger

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$2^{1/3} C$

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(D) The capacitance of a spherical conductor is given by $C = 4 \pi \epsilon_0 R$. When two identical droplets of radius $R$ combine to form a larger drop of radius $R'$,the total volume remains constant. Volume of two small drops = Volume of the large drop $2 	imes (\frac{4}{3} \pi R^3) = \frac{4}{3} \pi R'^3$ $R'^3 = 2 R^3$ $R' = 2^{1/3} R$ The capacitance of the new drop is $C' = 4 \pi \epsilon_0 R'$. Substituting $R' = 2^{1/3} R$,we get $C' = 4 \pi \epsilon_0 (2^{1/3} R) = 2^{1/3} (4 \pi \epsilon_0 R) = 2^{1/3} C$.

Consider a spherical drop of mercury of radius $R$ with capacitance $C = 4 \pi \epsilon_0 R$. If two such droplets combine to form a larger one,what would be its capacitance in terms of $C$?

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