Question

Eight drops of Hg (equal radii), which have equal charge constitute a bigger drop. The capacitance of bigger drop in comparison to small drop is :

• A. two times
• B. four times
• C. eight times
• D. sixteen times

Answer 1

The correct option is A two times

Let radius of big drop $=R$, radius of small drop $=r$, charge of each small drop $=q$ and charge of big drop $=Q$.
Thus, $Q=8q$ and $\frac{4}{3}\pi R^3=8\frac{4}{3}\pi r^3\Rightarrow R=2r$
$V_{small}=\frac{kq}{r}$
$V_{big}=\frac{kQ}{R}=\frac{k8q}{2r}=4\frac{kq}{r}=4V_{small}$
$C_{big}=\frac{Q}{V_{big}}=\frac{8q}{4V_{small}}=2\frac{q}{V_{small}}=2C_{small}$