Question

-3q and 2q charges are placed at some distance. Where you would place third charges q so that the new force on this charge is zero.

Answer 1

$|F_1|=F_2,$

$F_1=\frac{-3kq^2}{(a+x)^2},$

$F_2=\frac{2kq^2}{x^2},$

$\frac{\sqrt3}{a+x}=\frac{\sqrt 2}{x},$

$x=\frac{a\sqrt2}{\sqrt3-\sqrt2}=a(2+\sqrt6)$ at right from the charge $2q.$