Question

n identical droplets are charged to V volt each. If they coalesce to form a single drop, then its potential will be

• A. $n^{\frac{2}{3}}V$
• B. $n^{\frac{1}{3}}V$
• C. $nV$
• D. $\frac{V}{n}$

Answer 1

The correct option is A $n^{\frac{2}{3}}V$

Let the radius of each droplet be r units. So, the volume of each droplet is equal to $\frac{4}{3}\pi r^3$.
Thus, n droplets have the total volume equal to $n\left(\begin{array}{c}\frac{4}{3}\pi r^3\end{array}\right)$.
Since, the number of the drop would be equal to the total volume of the droplets hence,
$\Rightarrow R^3=n{r}^3$
$\Rightarrow R=n^{\frac{1}{3}}r\dots \left(i\right)$
The capacitance of each droplet is equal to. $C_d=4\pi {\epsilon }_{0}r$
and thus the charge of each droplet would equal
$q_d=C_d{V}_d=4\pi {\epsilon }_{0}r{V}_d\dots \left(ii\right)$
The capacitance of the bigger drop would be equal to $C=4\pi {\epsilon }_{0}R$.
The potential of the bigger drop would be equal to $V$ (say).
Hence, $V=\frac{n{q}_d}{C}=\frac{n\left(4\pi {\epsilon }_{0}r{V}_d\right)}{4\pi {\epsilon }_0{n}^{\frac{1}{3}r}}=n^{\frac{2}{3}{V}_d}$