JEE Main 2022PhysicsCapacitanceEasyMCQ

JEE Main 2022 — Capacitance Question with Solution

JEE Main 2022 (29 Jun Shift 1)

Question

A parallel plate capacitor filled with a medium of dielectric constant 10, is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant 15. Then the energy of capacitor will

Choose an option

Show full solutionCorrect option: A
Correct answer
Aincrease by 50%

Step-by-step explanation

When a dielectric of dielectric constant K is inserted between the plates of a capacitor C0, the capacitance becomes KC0.

Energy stored in a capacitor is given by, U=12CV2. Now,

Ui=12K1C0V2 and Uf=12K2C0V2

⇒ΔU=Uf-Ui=12K2-K1C0V2

Percentage change in the energy will be,

ΔUUi×100=12×5×C0212×10×C02×100=50%

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Capacitance chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

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.