Question 25 of 311

Two identical positive charges $Q$ each are fixed at a distance of $2a$ apart from each other. Another point charge ${q}_{0}$ with mass $m$ is placed at midpoint between two fixed charges. For a small displacement along the line joining the fixed charges, the charge ${q}_{0}$ executes SHM. The time period of oscillation of charge ${q}_{0}$ will be

A$\sqrt{\dfrac{4{\pi }^{3}{\epsilon }_{0}m{a}^{3}}{{q}_{0}Q}}$B$\sqrt{\dfrac{{q}_{0}Q}{4{\pi }^{3}{\epsilon }_{0}m{a}^{3}}}$C$\sqrt{\dfrac{2{\pi }^{2}{\epsilon }_{0}m{a}^{3}}{{q}_{0}Q}}$D$\sqrt{\dfrac{8{\pi }^{3}{\epsilon }_{0}m{a}^{3}}{{q}_{0}Q}}$

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Answer: (A) $\sqrt{\dfrac{4{\pi }^{3}{\epsilon }_{0}m{a}^{3}}{{q}_{0}Q}}$

Let the charged particle be displaced slightly towards right by distance $x\left(x≪d\right)$.
$\left(QatA,q\right):{F}_{1}=\dfrac{1}{4\pi {\epsilon }_{0}}\dfrac{Qq}{{\left(d+x\right)}^{2}}$, towards right
$\left(Q\text{at}B,q\right):{F}_{2}=\dfrac{1}{4\pi {\epsilon }_{0}}\dfrac{Qq}{{\left(d-x\right)}^{2}}$, towards left
${\vec{F}}_{net}={\vec{F}}_{1}+{\vec{F}}_{2}$, will bring it towards $O$.${\vec{F}}_{net}=\dfrac{1}{4\pi {ϵ}_{0}}\left[\dfrac{1}{{\left(d+x\right)}^{2}}-\dfrac{1}{{\left(d-x\right)}^{2}}\right]\hat{i}$
$=\dfrac{-Qq}{4\pi {ϵ}_{0}}\left[\dfrac{4xd}{{\left({d}^{2}-{x}^{2}\right)}^{2}}\right]\hat{i}$
Since $x≪d,{x}^{2}$ is negligible.
${\vec{F}}_{net}=-\left(\dfrac{Qq}{\pi {ϵ}_{0}{d}^{3}}\right)\times x\hat{i}$
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,
$a=\dfrac{F}{m}=-\dfrac{Qq}{\pi {ϵ}_{0}m{d}^{3}}x....\left(1\right)$
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, $T=\sqrt{\dfrac{4{\pi }^{3}{\epsilon }_{0}m{a}^{3}}{{q}_{0}Q}}$