JEE Main 2021PhysicsElectrostaticsHardNumerical

JEE Main 2021Electrostatics Question with Solution

JEE Main 2021 (25 Jul Shift 1)

Question

A particle of mass 1 mg and charge q is lying at the mid-point of two stationary particles kept at a distance 2 m when each is carrying same charge q. If the free charged particle is displaced from its equilibrium position through distance x x<<1 m. The particle executes SHM. Its angular frequency of oscillation will be _______ ×105 rad s-1 (if q2=10C2)

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Show full solutionCorrect answer: 6000
Correct answer
6000

Step-by-step explanation

Net force on free charged particle, 

F=kq2d+x2-kq2d-x2

F=-kq24dxd2-x22

a=-4kq2dmxd4

a=-4kq2md3x

So, angular frequency 

ω=4 kq2md3

ω=4×9×109×101×10-6×13

ω=6×108 rad sec-1

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About this question

This is a previous-year question from JEE Main 2021, 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.