Two identical positive charges… - JEE Main 2022 Physics PYQ

JEE Main 2022 — Physics Electromagnetism

2022mcqmedium

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

Official previous-year question

Held on 27 Jul 2022 · Verified 6 Jul 2026.

Options

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

Did you get this right?

Solution

JEE Main 2022 Physics, Electrostatics — solution figure

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

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

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

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

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

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

${\vec{F}}_{net}=-(\frac{Qq}{\pi {\epsilon }_{0}{d}^{3}})\times x\hat{i}$

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,

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

And we know that for SHM,

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

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

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

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

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

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Options

$\sqrt{\frac{4{\pi }^{3}{\epsilon }_{0}m{a}^{3}}{{q}_{0}Q}}$

$\sqrt{\frac{{q}_{0}Q}{4{\pi }^{3}{\epsilon }_{0}m{a}^{3}}}$

$\sqrt{\frac{2{\pi }^{2}{\epsilon }_{0}m{a}^{3}}{{q}_{0}Q}}$

$\sqrt{\frac{8{\pi }^{3}{\epsilon }_{0}m{a}^{3}}{{q}_{0}Q}}$

Solution

JEE Main 2022 Physics, Electrostatics — solution figure

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

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

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

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

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

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

${\vec{F}}_{net}=-(\frac{Qq}{\pi {\epsilon }_{0}{d}^{3}})\times x\hat{i}$

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,

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

And we know that for SHM,

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

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

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

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

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

Options

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Solution

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Options

Did you get this right?

Solution

Your note

JEE Main Electromagnetism in other years

More JEE Main Electromagnetism questions

Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Electromagnetism