Physics - Electrostatics Question with Solution | TestHub

Volume of 8 small drops $=$ Volume of big drop $\therefore \left(\frac{4}{3}\pi r^3\right)\times 8=\frac{4}{3}\pi R^3$ $\Rightarrow 2r=R$ $\dots$ (i) According to charge conservation (.) (ii) $$8\mathrm{q}=\mathrm{Q}$$ Potential of one small drop $\left({\mathrm{V}}^{\prime }\right)=\frac{\mathrm{q}}{4\pi {\epsilon }_0\mathrm{r}}$ Similarly, potential of big drop $(\mathrm{V})=\frac{Q}{4\pi {\epsilon }_{0}R}$ Now, $\frac{{\mathrm{V}}^{\prime }}{\mathrm{V}}=\frac{\mathrm{q}}{\mathrm{Q}}\times \frac{\mathrm{R}}{\mathrm{r}}\Rightarrow \frac{{\mathrm{V}}^{\prime }}{20}=\frac{9}{8\mathrm{q}}\times \frac{2\mathrm{r}}{\mathrm{r}}$ $\therefore {\mathrm{V}}^{\prime }=5\mathrm{V}$