JEE Main 2021PhysicsElectrostaticsHardMCQ

JEE Main 2021 — Electrostatics Question with Solution

JEE Main 2021 (27 Jul Shift 2)

Question

A simple pendulum of mass ' m ', length ' l ' and charge '+q' suspended in the electric field produced by two conducting parallel plates as shown. The value of deflection of pendulum in equilibrium position will be

Choose an option

Show full solutionCorrect option: C
Correct answer
Ctan-1qmg×C2 V1+V2C1+C2(d-t)

Step-by-step explanation

Let E be electric field in air

Tsinθ=qE

Tcosθ=mg

tanθ=qEmg

Q=C1C2C1+C2V1+V2

E=QAϵo=C1C2C1+C2V1+V2Aϵo

C1=ϵ0 A d-t⇒E=C2 V1+V2C1+C2(d-t)

Now θ=tan-1q·Emg

θ=tan-1qmg×C2 V1+V2C1+C2(d-t)

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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.