Question

TWO point charges $+2nc$ and $-4nc$ are $1m$ apart in air. Find the positions along the line joining the two charges at which resultant potential is zero.

Answer 1

Let the potential at point P is zero which is situated at a distance $x$ to the left of $1nC$ charge as shown in the figure.

Potential due a charge $q$ is given by $V=\frac{kq}{r}$ where $k=\frac{1}{4\pi {ϵ}_o}$

Thus potential at point P $V_p=\frac{k\left(1nC\right)}{x}-\frac{k\left(4nC\right)}{1+x}$

But $V_p=0$

$\therefore$ $\frac{k\left(1nC\right)}{x}-\frac{k\left(4nC\right)}{1+x}=0$

Or $\frac{1}{x}-\frac{4}{1+x}=0$

Or $\frac{1}{x}=\frac{4}{1+x}$

Or $3x=1$

$⟹$ $x=0.33m$