JEE Main 2020PhysicsElectrostaticsMediumMCQ

JEE Main 2020 — Electrostatics Question with Solution

JEE Main 2020 (04 Sep Shift 1)

Question

A two point charges 4q and –q are fixed on the x -axis at x=−d2 and x=d2, 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:

 

Choose an option

Show full solutionCorrect option: D
Correct answer
Ddecrease by 4q23π∈0d

Step-by-step explanation

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

V0=4q4πε0d2+−q4πε0d2=6q4πε0d

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

VP=4q4πε03d2+−q4πε0d2=q6πε0d

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

 ∆U=qΔV=qVf−Vi

⇒∆U=qVP−VO

∆U=qq6πε0d−3q2πε0d=−4q23πε0d

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About this question

This is a previous-year question from JEE Main 2020, covering the Electrostatics chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.