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

Vedclass · Accepted Answer

$160$

Vedclass · Answer

(A) Let the radius of each small drop be $r$ and its charge be $q$. The potential of a small drop is $v = \frac{kq}{r} = 10\,V$.Let the radius of the big drop be $R$ and its charge be $Q$. Since the volume is conserved,the volume of $64$ small drops equals the volume of the big drop: $\frac{4}{3}\pi R^3 = 64 	imes \frac{4}{3}\pi r^3$,which gives $R^3 = 64r^3$,so $R = 4r$.The total charge of the big drop is $Q = 64q$.The potential of the big drop is $V = \frac{kQ}{R} = \frac{k(64q)}{4r} = 16 	imes \frac{kq}{r}$.Substituting $v = \frac{kq}{r} = 10\,V$,we get $V = 16 	imes 10 = 160\,V$.

$64$ drops of mercury,each charged to a potential of $10\,V$,are combined to form one bigger drop. The potential of this bigger drop will be.......$V$ (Assume all the drops to be spherical).

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