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)

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,
$V=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r}$ ...(ii)
Potential of bigger drop
$V^{\prime }=\frac{1}{4\pi {\epsilon }_0}\frac{nq}{R}$
$=\frac{1}{4\pi {\epsilon }_0}\frac{nq}{n^{1/3}r}$ [from Eq. (i)]
$V^{\prime }=n^{2/3}V$ [from Eq. (ii)]