(i) Let C be the point on the line joining the two charges, where electric potential is zero, then $V_C$ = 0 or $V_{CA} + V_{CB}$ = 0 or $V_{CA}$ = - $V_{CB}$ or $\frac{1}{4\pi\varepsilon_0} \frac{q_A}{r_{CA}}$ = - $\frac{1}{4\pi\varepsilon_0} \frac{q_B}{r_{CB}}$ or $\frac{5\times10^{-8}\mathrm{C}}{x\times10^{-2}\mathrm{m}}$ = - $\frac{(-3\times10^{-8}\mathrm{C})}{[(6-x)\times10^{-2}\mathrm{m}]}$ or $\frac{5}{x}$ = $\frac{3}{16-x}$ or 80 – 5x = 3x or 80 = 8x or x = $\frac{80}{8}$ or x = 10 cm So, electric potential is zero at distance of 10 cm from charge of 5 × $10^{8}$ C on line joining the two charges between them. If point C is not between the two charges, then $V_{CA}+V_{CB}$ = 0 or $V_{CA}$ = - $V_{CB}$ or $\frac{1}{4\pi\varepsilon_0} \frac{q_Q}{r_{CA}}$ = - $\frac{1}{4\pi\varepsilon_0} \frac{q_B}{r_{CB}}$ or $\frac{5\times10^{-8}\mathrm{C}}{[(16+x)\times10^{-2}\mathrm{m}]}$ = - $\frac{(-3\times10^{-8}\mathrm{C})}{[x\times10^{-2}\mathrm{m}]}$ $\frac{5}{16+x}$ = $\frac{3}{x}$ or 5x = 48 + 3x or 2x = 48 or x = 24 cm So, electric potential is also equal to zero at a distance of 24 cm from charge of $-3\times10^{-8}$ C and at a distance of (24 + 16) = 40 cm from charge of 5 × $10^{-8}$ C, on the side of charge of – 3 × $10^{-8}$ C