Question

$\text{N}$ drops of mercury of equal radii and possessing equal charges combine to form a big spherical drop. If each individual drop has capacitance $C$ ,then the capacitance of the bigger drop compared to each individual drop is

• A. $\text{N}$ times
• B. ${\text{N}}^{\frac{2}{3}}$ times
• C. ${\text{N}}^{\frac{1}{3}}$ times
• D. ${\text{N}}^{\frac{5}{3}}$ times

Answer 1

The correct option is C ${\text{N}}^{\frac{1}{3}}$ times

Let the radius of small drop be $r$ and big drop be $R$.

Capacitance of isolated spherical conductor is given by $$C=4\pi {ϵ}_{0}r$$So, Capacitance of small drop of mercury is $$C_s=4\pi {ϵ}_{0}r$$Capacitance of bigger drop is $$C_b=4\pi {ϵ}_{0}R$$
The combined volume of the $\text{N}$ drops is equal to the volume of bigger drop.

$\therefore N\times \frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3$

$R={\text{N}}^{\frac{1}{3}}r$

Thus, the capacitance of bigger drop is
$C_b={\text{N}}^{\frac{1}{3}}{C}_a$

Hence, option (c) is the correct answer.