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

Let the positions of particles $A$ and $B$ be $(-\dfrac{d}{2}, 0)$ and $(\dfrac{d}{2}, 0)$ respectively. The initial position of particle $C$ is $(0, 0)$.

When $C$ is displaced by a distance $x$ perpendicular to $AB$, its new position is $(0, x)$.

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

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 $AB$ cancel each other out. The net force is directed along the perpendicular bisector.

$F_{net} = 2F \cos\theta$

From the geometry of the figure, $\cos\theta = \dfrac{x}{r}$.

Substituting the values of $F$ and $\cos\theta$:

$F_{net} = 2 \left( \dfrac{1}{4\pi\varepsilon_0} \dfrac{Qq}{r^2} \right) \left( \dfrac{x}{r} \right) = \dfrac{2Qqx}{4\pi\varepsilon_0 r^3} = \dfrac{Qqx}{2\pi\varepsilon_0 \left( \left(\dfrac{d}{2}\right)^2 + x^2 \right)^{3/2}}$

Since it is given that $x \ll d$, we can approximate \left(\dfrac{d}{2}\right)^2 + x^2 \approx \left(\dfrac{d}{2}\right)^2$.<br><br>$F_{net} \approx \dfrac{Qqx}{2\pi\varepsilon_0 \left(\dfrac{d}{2}\right)^3} = \dfrac{Qqx}{2\pi\varepsilon_0 \left(\dfrac{d^3}{8}\right)}$<br><br>$F_{net} = \dfrac{4Qqx}{\pi\varepsilon_0 d^3}