Question

$512$ identical drops of mercury are charged to a potential of $2\text{V}$ each. The drops are joined to form a single large drop. The potential of this drop is _____$\text{V}$.

Answer 1

Let:
Charge on each small drop $=q$
Radius of each small drop $=r$

We know that,
$V$ = $\frac{kq}{r}$

$\Rightarrow 2=\frac{kq}{r}...\left(1\right)$

Suppose the radius of large drop is $R$, then,

$\frac{4}{3}\pi R^3=512\times \frac{4}{3}\pi r^3$

$\Rightarrow R=8r$

Now, potential of large drop,

$V^{{}^{\prime }}=\frac{k\left(512q\right)}{R}=\frac{k\left(512q\right)}{8r}=\frac{64kq}{r}$

$\Rightarrow V^{{}^{\prime }}=64\times 2=128\text{V}$ $[$From $\left(1\right)]$