Twenty-seven drops of water of the same size | vedclass

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Question

(A) 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 $n = 27$ drops uni

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(A) 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 $n = 27$ drops unite to form a big drop of radius $R$,the volume remains conserved:$\frac{4}{3} \pi R^3 = n \cdot \frac{4}{3} \pi r^3 \implies R^3 = n r^3 \implies R = n^{1/3} r$.For $n = 27$,$R = (27)^{1/3} r = 3r$.The total charge on the big drop is $Q = nq = 27q$.The potential of the big 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 $n = 27$,we get $V = (27)^{2/3} v = (3^3)^{2/3} v = 3^2 v = 9v$.Thus,the potential increases by a factor of $9$.

Twenty-seven drops of water of the same size are equally and similarly charged. They are then united to form a bigger drop. By what factor will the electrical potential change?

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