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(B) Let the radius of each small drop be $r$ and the charge on each be $q$. The potential of a small drop is $V = \frac{kq}{r}$.When $n = 8$ drops are

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(B) Let the radius of each small drop be $r$ and the charge on each be $q$. The potential of a small drop is $V = \frac{kq}{r}$.When $n = 8$ drops are joined to form a bigger drop of radius $R$,the volume remains conserved: $\frac{4}{3}\pi R^3 = n \cdot \frac{4}{3}\pi r^3$,which gives $R = n^{1/3}r = 8^{1/3}r = 2r$.The total charge on the bigger drop is $Q = nq = 8q$.The potential of the bigger drop is $V' = \frac{kQ}{R} = \frac{k(nq)}{n^{1/3}r} = n^{2/3} \left( \frac{kq}{r} \right) = n^{2/3}V$.Substituting $n = 8$,we get $V' = 8^{2/3}V = (2^3)^{2/3}V = 2^2 V = 4V$.Therefore,the potential on the bigger drop is $4$ times that of the smaller drop.

If eight identical drops are joined to form a bigger drop,the potential on the bigger drop as compared to that on the smaller drop will be:

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