JEE Main 2019PhysicsCapacitanceMediumMCQ

JEE Main 2019 — Capacitance Question with Solution

JEE Main 2019 (10 Jan Shift 2)

Question

A parallel plate capacitor having capacitance 12 pF is charged by a battery to a potential difference of 10 V between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant 6.5 is slipped between the plates. The work done by the capacitor on the slab is

Choose an option

Show full solutionCorrect option: C
Correct answer
C508 pJ

Step-by-step explanation

Initial energy of capacitor

Ui=12q2c
=12×120×12012=600 pJ

Since battery is disconnected so charge remain same. Final energy of capacitor

 Uf=12q2c k

=12×120×12012×6.5=92 pJ

W+Uf=Ui 
W=508 pJ

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About this question

This is a previous-year question from JEE Main 2019, 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.