Question

$n$ identical spherical drops each of radius $r$ are charged to same potential $V$. They combine to form a bigger drop. The potential of the big drop will be :

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

Answer 1

Answer: (B)

${{n}^{2/3}}V$

Answer 2

After coalescing, volume remains conserved.
Volume of $n$ identical drops = volume of one big drop
i.e, $n \times \frac{4}{3} \pi r^{3}=\frac{4}{3} \pi R^{3}$
or $R=n^{1 / 3} r$ ...(i)
where $R$ is radius of bigger drop and $n$ is radius of each smaller drop.
Potential of each smaller drop,
$V=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r}$ ...(ii)
Potential of bigger drop
$V'=\frac{1}{4 \pi \varepsilon_{0}} \frac{n q}{R}$
$=\frac{1}{4 \pi \varepsilon_{0}} \frac{n q}{n^{1 / 3} r}$ [from E (i)]
$V'=n^{2 / 3} V$ [from E (ii)]