Two identical positive charges Q each are fixed at a distance of 2a...

Let the charged particle be displaced slightly towards right by distance $x\left(x ≪ d\right)$.

\left(QatA, q\right): {F}_{1} = \dfrac{1}{4\pi{ε}_{0}}\dfrac{Qq}{{\left(d + x\right)}^{2}}$, towards right</p><p>\left(Q\text{at}B, q\right): {F}_{2} = \dfrac{1}{4\pi{ε}_{0}}\dfrac{Qq}{{\left(d - x\right)}^{2}}$, towards left

$\overset{\to}{F}net = \overset{\to}{F}1 + \overset{\to}{F}2$, will bring it towards $O$.\overset{\to}{F}net = \dfrac{1}{4\pi{ϵ}_{0}}\left[\dfrac{1}{{\left(d + x\right)}^{2}} - \dfrac{1}{{\left(d - x\right)}^{2}}\right]\overset{^}{i}

= \dfrac{- Qq}{4\pi{ϵ}_{0}}\left[\dfrac{4xd}{{\left({d}^{2} - {x}^{2}\right)}^{2}}\right]\overset{^}{i}

Since $x ≪ d, {x}^{2}$ is negligible.

\overset{\to}{F}net = - \left(\dfrac{Qq}{\pi{ϵ}_{0}{d}^{3}}\right) \times x\overset{^}{i}

Clearly, the direction of the force is opposite to the displacement & $F ∝ x$, hence it is an SHM.

Acceleration of charged particle is given as,

$a = \dfrac{F}{m} = - \dfrac{Qq}{\pi{ϵ}_{0}m{d}^{3}}x . . . . \left(1\right)$

And we know that for SHM,

$a = - {\omega}^{2}x \ldots . \left(2\right)$

Comparing \left(1\right)$and \left(2\right)$, we get

$\omega = \sqrt{\dfrac{Qq}{\pi{ϵ}_{0}m{d}^{3}}}$

Hence, the time period, $T = \dfrac{2\pi}{\omega} = 2\pi\sqrt{\dfrac{\pi{ϵ}_{0}m{d}^{3}}{Qq}}$

Therefore, $T = \sqrt{\dfrac{4{\pi}^{3}{ε}_{0}m{a}^{3}}{{q}_{0}Q}}$