JEE Main 2022PhysicsCapacitanceEasyMCQ

JEE Main 2022 — Capacitance Question with Solution

JEE Main 2022 (27 Jun Shift 1)

Question

A force of 10 N acts on a charged particle placed between two plates of a charged capacitor. If one plate of capacitor is removed, then the force acting on that particle will be.

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Show full solutionCorrect option: A
Correct answer
A5 N

Step-by-step explanation

Electric field due to each plate, E1=E2=σ2ε0=Q2Aε0

Net electric field between the plates, Enet=E1+E2=QAε0

Force on charged particle between the plates, F1=qEnet=qQAε0=10 N

Now, in second case, the net electric field, E=σ2ε0 =Q2Aε0=5 N

Force on charged particle, F2=qE=qQ2Aε0=5 N

 

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About this question

This is a previous-year question from JEE Main 2022, covering the Capacitance 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.