Two equal charges are separated by a distance d. A third charge placed on a perpendicular bisector at x distance will experience maximum coulomb force when

Suppose third charge is similar to Q and it is q So net force on it 
 ${\text{F}}_{\text{net}}\text{= 2F cos θ}$                                     
                                                                 
Where $\text{F = }\frac{\text{1}}{{\text{4πε}}_{\text{0}}}\text{.}\frac{\text{Qq}}{\left({\text{x}}^{\text{2}}\text{+}\frac{{\text{d}}^{\text{2}}}{\text{4}}\right)}$ and  $\text{cos θ=}\frac{x}{\sqrt{x^2+\frac{d^2}{4}}}$
$\begin{array}{l}\therefore {\text{F}}_{\text{net}}\text{=}\frac{\text{1}}{{\text{4πε}}_{\text{0}}}.\frac{\text{Qq}}{(x^2+\frac{{\text{d}}^{\text{2}}}{\text{4}})}\times \frac{x}{{(x^2+\frac{{\text{d}}^{\text{2}}}{\text{4}})}^{1/2}}\\ \text{=}\frac{2\text{Qq}x}{{\text{2πε}}_{\text{0}}{\left({\text{x}}^{\text{2}}\text{+}\frac{{\text{d}}^{\text{2}}}{\text{4}}\right)}^{\text{3/2}}}\end{array}$

For ${\text{F}}_{\text{net}}$ to be maximum  $\frac{{\text{dF}}_{\text{net}}}{dx}=0$
$\Rightarrow x\text{\hspace{0.17em}=}\pm \frac{\text{d}}{\text{2}\sqrt{\text{2}}}.$