Two point charges $Q$ each are placed at a distance $d$ apart. A third point charge $q$ is placed at a distance $x$ from mid-point on the perpendicular bisector. The value of $x$ at which charge $q$ will experience the maximum Coulomb's force is:

The distance between $q$ and $Q$ is $r=\sqrt{x^2+{\left(\frac{d}{2}\right)}^2}=\sqrt{x^2+\frac{d^2}{4}}$.

The force between $q$ and $Q$ will be $F=\frac{kqQ}{r^2}$.

From the given figure we can see that only vertical components of the forces will add, and the horizontal components will get cancelled out. Therefore,

$$\begin{gathered} F_{net}=\frac{2kqQ}{r^2}\cos \theta \\ =\frac{2kqQ}{r^2}\times \frac{x}{r}=\frac{kqQx}{{\left(x^2+{\displaystyle \frac{d^2}{4}}\right)}^{{\displaystyle \frac{3}{2}}}} \end{gathered}$$

For force to be maximum, $\frac{d{F}_{net}}{dx}=0.$

$\Rightarrow kqQ\left[\frac{{\left(x^2+\frac{d^2}{4}\right)}^{\frac{3}{2}}-x\times \frac{3}{2}{\left(x^2+\frac{d^2}{4}\right)}^{\frac{1}{2}}\times 2x}{{\left({\left(x^2+\frac{d^2}{4}\right)}^{\frac{3}{2}}\right)}^2}\right]=0$

$\Rightarrow \left(x^2+\frac{d^2}{4}\right)=3x^2\Rightarrow 2x^2=\frac{d^2}{4}\Rightarrow x=\frac{d}{2\sqrt{2}}$