27 identical drops of mercury are charged sim | 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 given by $V = \frac{kq}{r} = 10 \ V$.When

Vedclass · Accepted Answer

$90$

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(A) Let the radius of each small drop be $r$ and the charge on each be $q$. The potential of a small drop is given by $V = \frac{kq}{r} = 10 \ V$.When $n = 27$ drops combine to form a large drop of radius $R$,the volume remains conserved: $\frac{4}{3}\pi R^3 = n \cdot \frac{4}{3}\pi r^3$,which implies $R = n^{1/3}r$.For $n = 27$,$R = (27)^{1/3}r = 3r$.The total charge on the large drop is $Q = nq = 27q$.The potential of the large drop is $V^{\prime} = \frac{kQ}{R} = \frac{k(nq)}{n^{1/3}r} = n^{2/3} \left(\frac{kq}{r}ight) = n^{2/3}V$.Substituting the values: $V^{\prime} = (27)^{2/3} 	imes 10 = (3^3)^{2/3} 	imes 10 = 3^2 	imes 10 = 9 	imes 10 = 90 \ V$.

$27$ identical drops of mercury are charged simultaneously with the same potential of $10 \ V$. Assuming the drops to be spherical,if all the charged drops are made to combine to form one large drop,then its potential will be . . . . . . volt.

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