Question

Two charges $3\times {10}^{-8}C$ and $-2\times {10}^{-8}C$ are $15cm$ apart. At what point on the line joining the charges, theelectric potential is zero? Let the electric potential at infinity is zero.

Answer 1

Potential at point P$V_1+V_2=0$
or
$\begin{array}{cc}\frac{1}{4\pi {\epsilon }_0}\frac{q_1}{x}+\frac{1}{4\pi {\epsilon }_0}\frac{q_2}{0.15-x}& =0\\ \frac{1}{4\pi {\epsilon }_0}[\frac{3\times {10}^{-8}}{x}-\frac{2\times {10}^{-8}}{0.15-x}]& =0\\ 9\times {10}^9\times {10}^{-8}[\frac{3}{x}-\frac{2}{0.15-x}]& =0\\ \frac{3}{x}-\frac{2}{0.15-x}& =0\\ \frac{3}{x}& =\frac{2}{0.15-x}\\ or0.45-3x& =2x\end{array}$
or $5x=0.45$
$x=0.09m=9cm$
Total potential at $P^{\prime }$
$\begin{array}{cc}\frac{1}{4\pi {\epsilon }_0}[\frac{3\times {10}^{-8}}{x}-\frac{2\times {10}^{-8}}{x-0.15}]& =0\\ \frac{{10}^{-8}}{4\pi {\epsilon }_0}[\frac{3}{x}-\frac{2}{x-0.15}]& =0\end{array}$
$\frac{3}{x}=\frac{2}{x-0.15}$
$3x-0.45=2x$
$x=0.45m=45cm$