Two particles A and B , each carrying a charge Q , are held fixed...

Let the positions of particles $A$ and $B$ be at a distance $\dfrac{d}{2}$ from the origin on the line connecting them. Particle $C$ is displaced by a distance $x$ along the perpendicular bisector.

The distance of particle $C$ from both $A$ and $B$ is $r = \sqrt{x^2 + \dfrac{d^2}{4}}$.

The magnitude of the electric force exerted by each charge $Q$ on $q$ is $F = \dfrac{1}{4\pi\varepsilon_0} \dfrac{Qq}{r^2}$.

By symmetry, the components of the forces parallel to the line $AB$ are equal and opposite, thus cancelling each other out.

The components of the forces perpendicular to $AB$ (along the displacement direction) add up. The net force is $F_{net} = 2F \cos\theta$, where $\theta$ is the angle between the force vector and the perpendicular bisector.

From the geometry, $\cos\theta = \dfrac{x}{r} = \dfrac{x}{\sqrt{x^2 + \dfrac{d^2}{4}}}$.

Substituting $F$ and $\cos\theta$ into the net force equation:

$F_{net} = 2 \left( \dfrac{1}{4\pi\varepsilon_0} \dfrac{Qq}{x^2 + \dfrac{d^2}{4}} \right) \dfrac{x}{\sqrt{x^2 + \dfrac{d^2}{4}}}$

$F_{net} = \dfrac{Qqx}{2\pi\varepsilon_0 \left(x^2 + \dfrac{d^2}{4}\right)^{3/2}}$