JEE Main 2020PhysicsCapacitanceMediumNumerical

JEE Main 2020 — Capacitance Question with Solution

JEE Main 2020 (07 Jan Shift 2)

Question

A 60pF capacitor is fully charged by a 20V supply. It is then disconnected from the supply and is connected to another uncharged 60pF capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in nJ ) ________

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Show full solutionCorrect answer: 6
Correct answer
6

Step-by-step explanation


V0=20V
Heat loss =Ui-Ut
=12Cvo2-212CV022
=CVo24
=60×10-122024J
=6×10-9J=6 nJ

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About this question

This is a previous-year question from JEE Main 2020, 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.