Physics - Electrostatics Question with Solution | TestHub

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

$\left(Q at A, q\right):F_1=\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{{\left(d+x\right)}^2}$, towards right

$\left(Q \text{at} B, q\right): F_2=\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{{\left(d-x\right)}^2}$, towards left

${\overrightarrow{F}}_{net}={\overrightarrow{F}}_1+{\overrightarrow{F}}_2$, will bring it towards $O$.${\overrightarrow{F}}_{net}=\frac{1}{4\pi {ϵ}_0}\left[\frac{1}{{{\displaystyle \left(d+x\right)}}^2}-\frac{1}{{\left(d-x\right)}^2}\right]\hat{i}$

$=\frac{-Qq}{4\pi {ϵ}_0}\left[\frac{4xd}{{\left(d^2-x^2\right)}^2}\right]\hat{i}$

Since $x\ll d, x^2$ is negligible.

${\overrightarrow{F}}_{net}=-\left(\frac{Qq}{\pi {ϵ}_0{d}^3}\right)\times x\hat{i}$

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

Acceleration of charged particle is given as,

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

And we know that for SHM,

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

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

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

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

Therefore, $T=\sqrt{\frac{4{\pi }^3{\epsilon }_{0}m{a}^3}{q_{0}Q}}$