Question

Two charges $+2q$ and $-3q$ are fixed at $(4m,0,0)$ and $(9m,0,0)$ respectively. The distance between two points on the x-axis (in $m$), where the potential is zero is

Answer 1

Let $p$ be a point between both charges, having a position towards right from $+2q$ and towards left from $-3q$.


So, the total electric potential at point $p$
$\frac{k\times 2q}{x}+\frac{-3kq}{5-x}=0$
$10-2x=3x$
$x=2$

Now, let the new position of point $p$ is towards left from $+2q$ charge at distance $y$.

So the new potentail will be
$\frac{k\times 2q}{y}-\frac{3kq}{5+y}=0$
10+ 2y = 3y
$y=10$

$\text{Total distance}=x+y=2+10=12\text{m}$

Or
$x=\frac{d}{(\frac{q_2}{q_1}+1)}(q_2>q_1)$
x is measured from 2q charge.

$x_1=\frac{5}{\frac{3}{2}+1}=2\text{m}$
$x_2=\frac{5}{\frac{3}{2}-1}=10\text{m}$
$\therefore \text{Required distance}=x_1+x_2$
$=12\text{m}$