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)

$n^{2/3}v$

Answer 2

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, $C_{d}=4 \pi \varepsilon_{0} r$
and thus the charge of each droplet would equal
$q_{d}=C_{d} V_{d}=4 \pi \varepsilon_{0} r V_{d}\,\,\,\,...(ii)$
The capacitance of the bigger drop would be equal to $C=4 \pi \varepsilon_{0} R$.
The potential of the bigger drop would be equal to $V$ (say).
Hence, $V=\frac{n q_{d}}{C}$
$=\frac{n\left(4 \pi \varepsilon_{0} r V_{d}\right)}{4 \pi \varepsilon_{0} n^{1 / 3} r}$
$=n^{2 / 3} \,V_{d}$