Physics - Electrostatics Question with Solution | TestHub

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