JEE Main 2023PhysicsElectrostaticsMediumMCQ

JEE Main 2023 — Electrostatics Question with Solution

JEE Main 2023 (13 Apr Shift 2)

Question

A 10 μC  charge is divided into two parts and placed at 1 cm distance so that the repulsive force between them is maximum. The charges of the two parts are:

Choose an option

Show full solutionCorrect option: C
Correct answer
C5μC,5μC

Step-by-step explanation

Let the charges be x μC, q-x μC, where q=10 μC

The force between them is 

F=Kx(q-x)r2.

For the force to be maximum,

dFdx=0⇒dFdx=K(q-2x)r2=0⇒x=q2=5 μC

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Electrostatics 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 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.