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

Let the two charges $Q$ be placed at $\left(-\dfrac{d}{2}, 0\right)$ and \left(\dfrac{d}{2}, 0\right)$.<br><br>The charge$q$is placed at$(0, y)$on the perpendicular bisector.<br><br>The net force on$q$is directed along the y-axis and its magnitude is given by:<br><br>$F = 2 \times \dfrac{1}{4\pi\varepsilon_0} \dfrac{Qq}{\left(\dfrac{d}{2}\right)^2 + y^2} \cos\theta$<br><br>where$\cos\theta = \dfrac{y}{\sqrt{\left(\dfrac{d}{2}\right)^2 + y^2}}$.<br><br>Thus,$F = \dfrac{2Qq}{4\pi\varepsilon_0} \dfrac{y}{\left( \dfrac{d^2}{4} + y^2 \right)^{3/2}}$.<br><br>For maximum force,$\dfrac{dF}{dy} = 0$.<br><br>$\dfrac{d}{dy} \left[ y \left( \dfrac{d^2}{4} + y^2 \right)^{-3/2} \right] = 0$<br><br>\left( \dfrac{d^2}{4} + y^2 \right)^{-3/2} - \dfrac{3}{2} y \left( \dfrac{d^2}{4} + y^2 \right)^{-5/2} (2y) = 0$

$\dfrac{d^2}{4} + y^2 - 3y^2 = 0 \Rightarrow 2y^2 = \dfrac{d^2}{4} \Rightarrow y = \dfrac{d}{2\sqrt{2}}$.

Substituting $y = \dfrac{d}{2\sqrt{2}}$ into the force equation:

F_\max = \dfrac{2Qq}{4\pi\varepsilon_0} \dfrac{\dfrac{d}{2\sqrt{2}}}{\left( \dfrac{d^2}{4} + \dfrac{d^2}{8} \right)^{3/2}}

F_\max = \dfrac{2Qq}{4\pi\varepsilon_0} \dfrac{\dfrac{d}{2\sqrt{2}}}{\left( \dfrac{3d^2}{8} \right)^{3/2}} = \dfrac{2Qq}{4\pi\varepsilon_0} \dfrac{\dfrac{d}{2\sqrt{2}}}{\dfrac{3\sqrt{3}d^3}{16\sqrt{2}}}

F_\max = \dfrac{16Qq}{3\sqrt{3} (4\pi\varepsilon_0 d^2)}

Since $\dfrac{16}{3\sqrt{3}} \approx \dfrac{16}{5.196} \approx 3.08$, we get:

F_\max \approx \dfrac{3.08 Qq}{4\pi\varepsilon_0 d^2}.