Question

Two charges $q$ and $-3q$ are placed fixed on $x$ -axis separated by distance ${}^{\prime }{d}^{\prime }.$ Where should a third charge $2q$ be placed such that it will not experience any force?

Answer 1

Let the charge at point $2q$ be placed at point $P$ as shown. The force due to $q$ is to the left and that due to $-3q$ is to the right.
$\therefore \frac{2q^2}{4\pi {ϵ}_0{x}^2}=\frac{6q^2}{4\pi {ϵ}_0(d+x{)}^2}\Rightarrow (d+x{)}^2=3x^2$
$\therefore 2x^2-2dx-d^2=0\Rightarrow x=\frac{d}{2}\pm \frac{\sqrt{3}d}{2}$
($-ve$ sign would be between $q$ and $-3q$ and hence is unacceptable.)
$\Rightarrow x=\frac{d}{2}+\frac{\sqrt{3}d}{2}=\frac{d}{2}(1+\sqrt{3})$ to the left of $q$.