Physics - Electrostatics Question with Solution | TestHub

Total volume of the drop will remain unchanged,

So,  $27\times \left(\frac{4}{3}\pi r^3\right)=\frac{4}{3}\pi R^3$

$\Rightarrow R=3r$

Potential of the small drops, $22=\frac{kq}{r}$

For bigger drop, potential will be

$V=\frac{k\left(27q\right)}{\left(3r\right)}=9\left(\frac{kq}{r}\right)=9\times 22=198 \mathrm{V}$