Question

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

Answer 1

Total number of drops , $N=512$

Potential of the each small drop , $V_s=\frac{1}{4\pi {ϵ}_0}\frac{q}{r}......\left(1\right)$

Potential of the big drop , $V_b=\frac{1}{4\pi {ϵ}_0}\frac{Q}{R}=\frac{1}{4\pi {ϵ}_0}\frac{Nq}{R}.....\left(2\right)$

We know that, Volume of bigger drop is $N$ times the volume of each small drop.

$\frac{4}{3}\pi R^3=N\left(\frac{4}{3}\pi r^3\right)$

$\Rightarrow R=N^{\frac{1}{3}}r$

Equation (2) becomes

$V_b=\frac{1}{4\pi {ϵ}_0}\frac{Nq}{\left(N^{\frac{1}{3}}r\right)}=N^{\frac{2}{3}}\left[\frac{1}{4\pi {ϵ}_0}\frac{q}{r}\right]$

$=N^{\frac{2}{3}}{V}_s$

Potential of the big drop $V_b=\left(512{)}^{\frac{2}{3}}\right(2)$

$=(8^3{)}^{\frac{2}{3}}2=64\times 2=128\text{V}$