Two electrons each are fixed at… - JEE Main 2021 Physics PYQ

JEE Main 2021 — Physics Electromagnetism

2021mcqhard

Two electrons each are fixed at a distance $2d.$ A third charge proton placed at the midpoint is displaced slightly by a distance $x(x\ll d)$ perpendicular to the line joining the two fixed charges. Proton will execute simple harmonic motion having angular frequency:

( $m=$ mass of charged particle)

Official previous-year question

Held on 24 Feb 2021 · Verified 6 Jul 2026.

Options

A
${(\frac{\pi {\epsilon }_{0}m{d}^{3}}{2{q}^{2}})}^{\frac{1}{2}}$
B
${(\frac{2\pi {\epsilon }_{0}m{d}^{3}}{{q}^{2}})}^{\frac{1}{2}}$
C
${(\frac{{q}^{2}}{2\pi {\epsilon }_{0}m{d}^{3}})}^{\frac{1}{2}}$
D
${(\frac{2{q}^{2}}{\pi {\epsilon }_{0}m{d}^{3}})}^{\frac{1}{2}}$

Did you get this right?

Solution

From the given condition, we have

JEE Main 2021 Physics, Electrostatics — solution figure

${F}_{\mathrm{net}q}=-[2{F}_{q/q}\mathrm{cos}\theta ]$

${F}_{\mathrm{net}q}=-2\cdot \frac{1}{4\pi {\epsilon }_{0}}\cdot \frac{{q}^{2}}{{(\sqrt{{d}^{2}+{x}^{2}})}^{2}}\cdot \frac{x}{\sqrt{{d}^{2}+{x}^{2}}}$

$=-\frac{{q}^{2}}{2\pi {\epsilon }_{0}}\frac{x}{{({d}^{2}+{x}^{2})}^{3/2}}$

For $x<<d,$

${F}_{\mathrm{net}q}=-\frac{{q}^{2}}{2\pi {\epsilon }_{0}{d}^{3}}x$

$\therefore a=-\frac{{q}^{2}}{2\pi {\epsilon }_{0}\cdot m{d}^{3}}x$

Comparing with equation of SHM $(a=-{\omega }^{2}x)$

$\therefore \omega =\sqrt{\frac{{q}^{2}}{2\pi {\epsilon }_{0}m{d}^{3}}}$

Your note

JEE Main Electromagnetism in other years

Electromagnetism 2026 Electromagnetism 2025 Electromagnetism 2024 Electromagnetism 2023 Electromagnetism 2022 Electromagnetism 2020 Electromagnetism 2019 Electromagnetism 2018 Electromagnetism 2017 Electromagnetism 2016 Electromagnetism 2015 Electromagnetism 2014

Keep practicing Electromagnetism

Work through every JEE Main Electromagnetism PYQ, year by year.

Browse all Physics topics

Options

${(\frac{\pi {\epsilon }_{0}m{d}^{3}}{2{q}^{2}})}^{\frac{1}{2}}$

${(\frac{2\pi {\epsilon }_{0}m{d}^{3}}{{q}^{2}})}^{\frac{1}{2}}$

${(\frac{{q}^{2}}{2\pi {\epsilon }_{0}m{d}^{3}})}^{\frac{1}{2}}$

${(\frac{2{q}^{2}}{\pi {\epsilon }_{0}m{d}^{3}})}^{\frac{1}{2}}$

Solution

From the given condition, we have

JEE Main 2021 Physics, Electrostatics — solution figure

${F}_{\mathrm{net}q}=-[2{F}_{q/q}\mathrm{cos}\theta ]$

${F}_{\mathrm{net}q}=-2\cdot \frac{1}{4\pi {\epsilon }_{0}}\cdot \frac{{q}^{2}}{{(\sqrt{{d}^{2}+{x}^{2}})}^{2}}\cdot \frac{x}{\sqrt{{d}^{2}+{x}^{2}}}$

$=-\frac{{q}^{2}}{2\pi {\epsilon }_{0}}\frac{x}{{({d}^{2}+{x}^{2})}^{3/2}}$

For $x<<d,$

${F}_{\mathrm{net}q}=-\frac{{q}^{2}}{2\pi {\epsilon }_{0}{d}^{3}}x$

$\therefore a=-\frac{{q}^{2}}{2\pi {\epsilon }_{0}\cdot m{d}^{3}}x$

Comparing with equation of SHM $(a=-{\omega }^{2}x)$

$\therefore \omega =\sqrt{\frac{{q}^{2}}{2\pi {\epsilon }_{0}m{d}^{3}}}$

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

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