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

Let the charges $A$ and $B$ be located at $x = -\dfrac{d}{2}$ and $x = \dfrac{d}{2}$ respectively.

The particle $C$ is displaced to a position $x$ from the origin (midpoint of $AB$).

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

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

The net force on $C$ is $F = F_B - F_A$ (towards $A$):

F = \dfrac{Q q}{4\pi\varepsilon_0} \left[\dfrac{1}{(\dfrac{d}{2} - x)^2} - \dfrac{1}{(\dfrac{d}{2} + x)^2} \right]

$F =$\dfrac{Q q}{4\pi\varepsilon_0} \left[ \dfrac{(\dfrac{d}{2} + x)^2 - (\dfrac{d}{2} - x)^2}{(\dfrac{d^2}{4} - x^2)^2} \right]$<br><br>$F = \dfrac{Q q}{4\pi\varepsilon_0} \left[ \dfrac{2 d x}{(\dfrac{d^2}{4} - x^2)^2} \right]<br><br>Sincex \ll d$, we can neglect$x^2$in the denominator, so (\dfrac{d^2}{4} -$x^2)^2 \approx $\dfrac{d^4}{16}.<br><br>$F \approx $\dfrac{Q q}{4\pi\varepsilon_0} \left( \dfrac{2 d x}{\dfrac{d^4}{16}} \right)$

F \approx \dfrac{Q q}{4\pi\varepsilon_0} \left(\dfrac{32 x}{d^3} \right)

$F \approx$\dfrac{8 Q q x}{\pi\varepsilon_0 d^3}