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Assume each oil drop consists of a capacitance of $C$. If combine $n$ drops to form a bigger drop, then the capacitance of bigger $\operatorname{drop} C^{\prime}$ would be
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Verified Answer
The correct answer is:
$C^{\prime}=C \cdot n^{1 / 3}$
Capacitance of an oil drop of radius $r$ is
$C=4 \pi \varepsilon r$
when $n$ drops combine to form a large drop of radius $R$, then
$n \times$ volume of 1 small drop $=$ volume of 1 large drop
$n \times \frac{4}{3} \pi r^3=\frac{4}{3} \pi R^3 \Rightarrow R=n^{1 / 3} \cdot r$
So, capacitance of large drop is
$C^{\prime}=4 \pi \varepsilon R=4 \pi \varepsilon \cdot n^{\frac{1}{3}} \cdot r=n^{\frac{1}{3}} C$
$C=4 \pi \varepsilon r$
when $n$ drops combine to form a large drop of radius $R$, then
$n \times$ volume of 1 small drop $=$ volume of 1 large drop
$n \times \frac{4}{3} \pi r^3=\frac{4}{3} \pi R^3 \Rightarrow R=n^{1 / 3} \cdot r$
So, capacitance of large drop is
$C^{\prime}=4 \pi \varepsilon R=4 \pi \varepsilon \cdot n^{\frac{1}{3}} \cdot r=n^{\frac{1}{3}} C$
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