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(A) Let $r$ be the radius of each small drop and $q$ be the charge on each small drop.The potential of each small drop is $v = \frac{kq}{r} = 10\, V$.

Vedclass · Accepted Answer

$160$

Vedclass · Answer

(A) Let $r$ be the radius of each small drop and $q$ be the charge on each small drop.The potential of each small drop is $v = \frac{kq}{r} = 10\, V$.When $n = 64$ drops combine to form a large drop of radius $R$ and charge $Q$,the volume remains constant:$\frac{4}{3}\pi R^3 = n 	imes \frac{4}{3}\pi r^3 \implies R = n^{1/3}r = (64)^{1/3}r = 4r$.The total charge on the large drop is $Q = nq = 64q$.The potential of the large drop is $V = \frac{kQ}{R} = \frac{k(nq)}{n^{1/3}r} = n^{2/3} 	imes \frac{kq}{r}$.Substituting the values: $V = (64)^{2/3} 	imes 10 = (4^3)^{2/3} 	imes 10 = 4^2 	imes 10 = 16 	imes 10 = 160\, V$.

$64$ identical mercury drops,each charged to a potential of $10\, V$,are combined to form a single larger drop. The potential of the new larger drop is........$V$.

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