Two point charges Q each are placed at a distance d apart. A third...

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

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

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,

${F}_{net} = \dfrac{2kqQ}{{r}^{2}}cos\theta = \dfrac{2kqQ}{{r}^{2}} \times \dfrac{x}{r} = \dfrac{kqQx}{{\left({x}^{2} + \dfrac{{d}^{2}}{4}\right)}^{\dfrac{3}{2}}}$

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

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

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