Solution

Volume is conserved: $\frac{4}{3}\pi R^3 = 27 \times \frac{4}{3}\pi r^3 \Rightarrow R = 3r$ $V = \frac{kq}{r} \Rightarrow \frac{V_1}{V_2} = \frac{q_1}{q_2}\cdot\frac{r_2}{r_1}$ $\frac{220}{V_2} = \frac{q}{27q} \times \frac{3r}{r} = \frac{1}{9}$ $V_2 = 220 \times 9 = 1980$ V

Concept potential of a merged drop

Electrostatic Potential and Capacitance in past papers

76 questions from this chapter have appeared across 27 exam years.