Given q1 = 5 × 10-8 C, r = 16 cm = 0.16 m q2 = -3 × 10-8 C Let potential be zero at a distance × metre from positive charge q1 .
∴ r1 = x meter
r2 = (0.16 – x) metre
S0 V = \(\frac{1}{4\pi\varepsilon_0}\) \(\Big[\frac{q_1}{r_1}+\frac{q_2}{r_2}\Big]\)
or = 9 x 109 \(\Big[\frac{5\times10^8}{x}-\frac{3\times10^{-8}}{0.16-x}\Big]\)
or \(\frac{5\times10^{-8}}{x}=\frac{3\times10^{-8}}{0.16-x}\)
or 0.8 – 5x = 3x
or x = 0.1 m = 10 cm.