Physics - Electrostatics Question with Solution | TestHub

The radius of the bigger drop will be$\frac{4}{3}\pi R^3=27\times \frac{4}{3}\pi r^3\Rightarrow R=3r$,  the potential of the small drop,$V=\frac{Kq}{r}$. The potential of the big drop$Q=nq, V\text{'}=\frac{KQ}{R}\Rightarrow V\text{'}=\frac{Knq}{R}$

As we know,$V{\text{'}}_{ }=n^{2/3}V$$=(27{)}^{2/3}\times 220$$=1980 \mathrm{V}$