Two equal positive point charges are separated by a distance 2 a . The distance of a point from the centre…

Let us assume at distance $y$, force is maximum.

Now, force at a distance $y$ is given by, $F_{net}=2F\cos \theta$

So, $F_{net}=\frac{2Kq{q}_{0}y}{{\left(y^2+a^2\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$

For $F_{net}$ to be maximum, $\frac{d{F}_{net}}{dy}=0$

Or $Kq{q}_0\frac{{\left(y^2+a^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-y{\displaystyle \frac{3}{2}}\times 2y{\left(y^2+a^2\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{{\left(y^2+a^2\right)}^3}=0$

Simplifying, we get $y=\frac{a}{\sqrt{2}}$.

Hence, the value of $x=2$.