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

Let the charges $Q$ be placed at $(-d/2, 0)$ and (d/2, 0)$.<br><br>The particle$C$is displaced to$(0, x)$along the perpendicular bisector.<br><br>The distance of$C$from each charge$Q$is$r = \sqrt{(d/2)^2 + x^2}$.<br><br>The attractive electrostatic force between$C$and each charge$Q$is$F = \dfrac{1}{4\pi\varepsilon_0} \dfrac{Qq}{r^2}$.<br><br>The components of the forces parallel to$AB$cancel each other, while the components perpendicular to$AB$add up to provide the restoring force.<br><br>The net restoring force is$F_{net} = 2F \cos\theta$, where$\cos\theta = \dfrac{x}{r}$.<br><br>$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}$<br><br>Since$x \ll d$,$r \approx \dfrac{d}{2}$.<br><br>$F_{net} = \dfrac{2Qqx}{4\pi\varepsilon_0 (d/2)^3} = \dfrac{16Qqx}{4\pi\varepsilon_0 d^3} = \dfrac{4Qq}{\pi\varepsilon_0 d^3} x$<br><br>The restoring force is proportional to$x$, so the motion is simple harmonic with force constant$k = \dfrac{4Qq}{\pi\varepsilon_0 d^3}$.<br><br>The time period of oscillation is$T = 2\pi \sqrt{\dfrac{m}{k}} = 2\pi \sqrt{\dfrac{m \pi\varepsilon_0 d^3}{4Qq}} = $\left(\dfrac{m\pi^3 \varepsilon_0 d^3}{Qq}\right)^{1/2}$