JEE Main 2020 — Physics Electromagnetism

2020mcqmedium

A two point charges $4q$ and $–q$ are fixed on the $x$ -axis at $x=\frac{-d}{2}$ and $x=\frac{d}{2}$ , respectively. If the third point charge ‘ $q$ ’ is taken from the origin to $x=d$ along the semicircle as shown in the figure, the energy of the charge will:

JEE Main 2020 Physics, Electrostatics — question figure

Official previous-year question

Held on 4 Sept 2020 · Verified 6 Jul 2026.

Options

A
Increase by $\frac{3{q}^{2}}{4\pi {\in }_{0}d}$
B
Increase by $\frac{2{q}^{2}}{3\pi {\in }_{0}d}$
C
decrease by $\frac{{q}^{2}}{4\pi {\in }_{0}d}$
D
decrease by $\frac{4{q}^{2}}{3\pi {\in }_{0}d}$

Did you get this right?

Solution

JEE Main 2020 Physics, Electrostatics — solution figure

Potential at origin $O$ , the potential due to two charges is given by

${V}_{0}=\frac{4q}{4\pi {\epsilon }_{0}(\frac{d}{2})}+\frac{(-q)}{4\pi {\epsilon }_{0}(\frac{d}{2})}=\frac{6q}{4\pi {\epsilon }_{0}d}$

Potential at $P$ at a distance, $x=d$ is given by

${V}_{P}=\frac{4q}{4\pi {\epsilon }_{0}(\frac{3d}{2})}+\frac{(-q)}{4\pi {\epsilon }_{0}(\frac{d}{2})}=\frac{q}{6\pi {\epsilon }_{0}d}$

Change in potential energy due the charge motion of charge $q$ is given by

$\Delta U=q\Delta V=q({V}_{f}-{V}_{i})$

$\Rightarrow \Delta U=q({V}_{P}-{V}_{O})$

$\Delta U=q(\frac{q}{6\pi {\epsilon }_{0}d}-\frac{3q}{2\pi {\epsilon }_{0}d})=-\frac{4{q}^{2}}{3\pi {\epsilon }_{0}d}$

Your note

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Solution

JEE Main 2020 Physics, Electrostatics — solution figure

Potential at origin $O$ , the potential due to two charges is given by

${V}_{0}=\frac{4q}{4\pi {\epsilon }_{0}(\frac{d}{2})}+\frac{(-q)}{4\pi {\epsilon }_{0}(\frac{d}{2})}=\frac{6q}{4\pi {\epsilon }_{0}d}$

Potential at $P$ at a distance, $x=d$ is given by

${V}_{P}=\frac{4q}{4\pi {\epsilon }_{0}(\frac{3d}{2})}+\frac{(-q)}{4\pi {\epsilon }_{0}(\frac{d}{2})}=\frac{q}{6\pi {\epsilon }_{0}d}$

Change in potential energy due the charge motion of charge $q$ is given by

$\Delta U=q\Delta V=q({V}_{f}-{V}_{i})$

$\Rightarrow \Delta U=q({V}_{P}-{V}_{O})$

$\Delta U=q(\frac{q}{6\pi {\epsilon }_{0}d}-\frac{3q}{2\pi {\epsilon }_{0}d})=-\frac{4{q}^{2}}{3\pi {\epsilon }_{0}d}$

JEE Main 2020 — Physics Electromagnetism

Options

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Solution

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JEE Main 2020 — Physics 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