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Eight small drops, each of radius $r$ and having the same charge $q$, are combined to form a big drop. The ratio between the potentials of the bigger drop and the smaller drop is

1. 8 : 1
2. 4 : 1
3. 2 : 1
4. 1 : 8

Correct answer: 2 : 1

Solution

When 8 identical drops combine, total volume becomes 8 times, so the new radius becomes $R=2r$. The total charge becomes $8q$. Since potential of a charged sphere is $V=kQ/R$, the new potential is $V' = k(8q)/(2r)=4kq/r$, while the old potential is $V=kq/r$. Hence the ratio is $4:1$.

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