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

Let the positions of particles $A$ and $B$ be at $-\dfrac{d}{2}$ and $\dfrac{d}{2}$ respectively, with the origin at the midpoint.

When particle $C$ is displaced by a distance $x$ towards $B$, its new distance from $A$ is $\dfrac{d}{2} + x$ and from $B$ is $\dfrac{d}{2} - x$.

The electric force on $C$ due to $A$ is $F_A = \dfrac{1}{4\pi\varepsilon_0} \dfrac{Q q}{\left(\dfrac{d}{2} + x\right)^2}$ directed towards $B$.

The electric force on $C$ due to $B$ is $F_B = \dfrac{1}{4\pi\varepsilon_0} \dfrac{Q q}{\left(\dfrac{d}{2} - x\right)^2}$ directed towards $A$.

The net electric force experienced by $C$ is the difference between these two forces:

F = F_B - F_A = \dfrac{Q q}{4\pi\varepsilon_0} \left[ \dfrac{1}{ \left(\dfrac{d}{2} -x\right)^2}-\dfrac{1}{\left(\dfrac{d}{2} +x\right)^2} \right]<br><br>F = \dfrac{Q q}{4\pi\varepsilon_0} \left[ \dfrac{ \left(\dfrac{d}{2} +x\right)^2 - \left(\dfrac{d}{2} - x\right)^2}{ \left(\dfrac{d^2}{4} -x^2\right)^2} \right]$<br><br>Using the identity$(a+b)^2 - (a-b)^2 = 4ab$, the numerator becomes$4 $\left(\dfrac{d}{2}\right)x = 2dx$.

F = \dfrac{Q q}{4\pi\varepsilon_0} \left[\dfrac{2dx}{ \left(\dfrac{d^2}{4} -x^2\right)^2} \right]

$F = \dfrac{Q q d x}{2\pi\varepsilon_0 \left(\dfrac{d^2}{4} - x²\right)^2}$