JEE Main 2021PhysicsCapacitanceMediumNumerical

JEE Main 2021 — Capacitance Question with Solution

JEE Main 2021 (31 Aug Shift 2)

Question

A parallel plate capacitor of capacitance 200μF is connected to a battery of 200 V. A dielectric slab of dielectric constant 2 is now inserted into the space between plates of capacitor while the battery remain connected. The change in the electrostatic energy in the capacitor will be __________ J.

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Show full solutionCorrect answer: 4
Correct answer
4

Step-by-step explanation

Initially
C=200μF
Ei=12CV2=12×200×10-6×2002
Finally
C'=KC=400μF
Ef=12C'V2=12×400×10-6×(200)2
ΔE=12×400-200×10-6×4×104=4 J

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About this question

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