Question

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

• A. $n^{2/3}v$
• B. $n^{1/3}v$
• C. $nv$
• D. $v/n$

Answer 1

Answer: (A)

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 $n\left(\frac{4}{3}\pi r^3\right)$
Since, the number of the drop would be equal to the total volume of the droplets hence,
$\Rightarrow R^3=n{r}^3$
$\Rightarrow R=n^{1/3}r...(i)$
The capacitance of each droplet is equal to, $C_d=4\pi {\epsilon }_{0}r$
and thus the charge of each droplet would equal
$q_d=C_d{V}_d=4\pi {\epsilon }_{0}r{V}_d...(ii)$
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, $V=\frac{n{q}_d}{C}$
$=\frac{n\left(4\pi {\epsilon }_{0}r{V}_d\right)}{4\pi {\epsilon }_0{n}^{1/3}r}$
$=n^{2/3}{V}_d$