Question

In identical mercury droplets charged to the same potential $V$ coalesce to form a single bigger drop. The potential of the new drop will be:

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

Answer 1

Answer: (D)

$n^{\frac{2}{3}}V$

We know when $n$ drops are coalesce final volume must be same as initial
$\Rightarrow n\frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3$
$\Rightarrow R=n^{\frac{1}{3}}r$
We know potential $V=\frac{kq}{r}$
Potential of bigger drop $V_b=\frac{k\left(nq\right)}{\left(n^{\frac{1}{3}}r\right)}$
$\Rightarrow V_b=n^{{}^2{/}_3}\times \frac{kq}{r}$= $n^{\frac{2}{3}}V$