JEE Main 2022 — Physics Electromagnetism

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

$(QatA,q):{F}_{1}=\frac{1}{4\pi {\epsilon }_{0}}\frac{Qq}{{(d+x)}^{2}}$(QatA,q):F1​=4πϵ0​1​(d+x)2Qq​, towards right

$(Q\text{at}B,q):{F}_{2}=\frac{1}{4\pi {\epsilon }_{0}}\frac{Qq}{{(d-x)}^{2}}$(QatB,q):F2​=4πϵ0​1​(d−x)2Qq​, towards left

${\vec{F}}_{net}={\vec{F}}_{1}+{\vec{F}}_{2}$Fnet​=F1​+F2​, will bring it towards $O$O.${\vec{F}}_{net}=\frac{1}{4\pi {\epsilon }_{0}}[\frac{1}{{(d+x)}^{2}}-\frac{1}{{(d-x)}^{2}}]\hat{i}$Fnet​=4πϵ0​1​[(d+x)21​−(d−x)21​]i^

$$=\frac{-Qq}{4\pi {\epsilon }_{0}}[\frac{4xd}{{({d}^{2}-{x}^{2})}^{2}}]\hat{i}$$=4πϵ0​−Qq​[(d2−x2)24xd​]i^

Since $x\ll d,{x}^{2}$x≪d,x2 is negligible.

$${\vec{F}}_{net}=-(\frac{Qq}{\pi {\epsilon }_{0}{d}^{3}})\times x\hat{i}$$Fnet​=−(πϵ0​d3Qq​)×xi^

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

Acceleration of charged particle is given as,

$$a=\frac{F}{m}=-\frac{Qq}{\pi {\epsilon }_{0}m{d}^{3}}x....(1)$$a=mF​=−πϵ0​md3Qq​x....(1)

And we know that for SHM,

$$a=-{\omega }^{2}x\ldots .(2)$$a=−ω2x….(2)

Comparing $(1)$(1) and $(2)$(2), we get

$$\omega =\sqrt{\frac{Qq}{\pi {\epsilon }_{0}m{d}^{3}}}$$ω=πϵ0​md3Qq​​

Hence, the time period, $T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{\pi {\epsilon }_{0}m{d}^{3}}{Qq}}$T=ω2π​=2πQqπϵ0​md3​​

Therefore, $T=\sqrt{\frac{4{\pi }^{3}{\epsilon }_{0}m{a}^{3}}{{q}_{0}Q}}$T=q0​Q4π3ϵ0​ma3​​