JEE Main 2023PhysicsElectrostaticsMediumNumerical

JEE Main 2023 — Electrostatics Question with Solution

JEE Main 2023 (01 Feb Shift 1)

Question

Two equal positive point charges are separated by a distance 2a. The distance of a point from the centre of the line joining two charges on the equatorial line (perpendicular bisector) at which force experienced by a test charge q0 becomes maximum is ax. The value of x is ______.

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Show full solutionCorrect answer: 2
Correct answer
2

Step-by-step explanation

Let us assume at distance y, force is maximum.

Now, force at a distance y is given by, Fnet=2Fcosθ

So, Fnet=2Kqq0yy2+a232

For Fnet to be maximum, dFnetdy=0

Or Kqq0y2+a232-y32×2yy2+a212y2+a23=0

Simplifying, we get y=a2.

Hence, the value of x=2.

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About this question

This is a previous-year question from JEE Main 2023, covering the Electrostatics 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.