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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_S = \frac{kq}{r} = 220 \,

Vedclass · Accepted Answer

$1980$

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 $V_S = \frac{kq}{r} = 220 \, V$.When $N = 27$ drops combine to form a bigger drop of radius $R$ and charge $Q$,the volume remains constant.$\frac{4}{3} \pi R^3 = N 	imes \frac{4}{3} \pi r^3 \implies R = N^{1/3} r = (27)^{1/3} r = 3r$.The total charge on the bigger drop is $Q = Nq = 27q$.The potential of the bigger drop is $V_B = \frac{kQ}{R} = \frac{k(Nq)}{N^{1/3}r} = N^{2/3} \frac{kq}{r} = N^{2/3} V_S$.Substituting the values: $V_B = (27)^{2/3} 	imes 220 = (3^3)^{2/3} 	imes 220 = 3^2 	imes 220 = 9 	imes 220 = 1980 \, V$.

Twenty-seven drops of the same size are charged at $220 \, V$ each. They combine to form a bigger drop. Calculate the potential of the bigger drop (in $V$).

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