Question

Assume each oil drop consists of a capacitance of $C$. If combine $n$ drops to form a biggerdrop, then the capacitance of bigger drop $C$' would be

• A. $C'=\frac{2 n^{\frac{1}{3}}}{3} C$
• B. $C'=\frac{5 n^{\frac{1}{3}}}{4} C$
• C. $C'=\frac{n^{\frac{1}{3}}}{5} C$
• D. $C'=C \cdot n^{\frac{1}{3}}$

Answer 1

Answer: (D)

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
$C^{\prime}=4 \pi \varepsilon R=4 \pi \varepsilon \cdot n^{\frac{1}{3}} \cdot r=n^{\frac{1}{3}} C$