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(B) Let the radius of each small drop be $r$ and the charge on each be $q$. The potential of each small drop is $V = \frac{1}{4 \pi \varepsilon_0} \fr

Vedclass · Accepted Answer

$Vn^{2/3}$

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(B) Let the radius of each small drop be $r$ and the charge on each be $q$. The potential of each small drop is $V = \frac{1}{4 \pi \varepsilon_0} \frac{q}{r}$.When $n$ drops coalesce 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 gives $R = n^{1/3} r$.The total charge on the large drop is $Q = nq$.The potential of the large drop is $V' = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{R}$.Substituting the values of $Q$ and $R$: $V' = \frac{1}{4 \pi \varepsilon_0} \frac{nq}{n^{1/3} r}$.Simplifying this,we get $V' = n^{1 - 1/3} \left( \frac{1}{4 \pi \varepsilon_0} \frac{q}{r} \right) = n^{2/3} V$.

$n$ small drops of the same size are charged to $V$ volt each. If they coalesce to form a single large drop,then its potential will be:

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