Question

Charges of $5\times {10}^{-6}$ and $-3\times {10}^{-6}C$ are kept 16 cm apart. Find the position of points on the line joining the charges where the electric potential is zero.

Answer 1

Potential at P iss zero

$\frac{k(5\times {10}^{-6})}{x}=\frac{k(3\times {10}^{-6})}{0.16-x}$

$5(0.16-x)=3x$

$0.8=8x$

$x=0.1m$

$x=10cm$