JEE Main 2018PhysicsElectrostaticsMediumMCQ

JEE Main 2018 — Electrostatics Question with Solution

JEE Main 2018 (16 Apr Online)

Question

Two identical conducting spheres A and B carry an equal charges. They are separated by a distance much larger than their diameters, and the force between them is F. A third identical conducting sphere, C, is uncharged. Sphere C is first touched to A, then to B, and then removed. As a result, the force between A and B would be equal to:

Choose an option

Show full solutionCorrect option: C
Correct answer
C3F8

Step-by-step explanation



F=kq2r2 when A and C are touched charge on both will be q2. Then when B and C are touched, qB= q2+q2=3q4
F′=kqAqBr2= k ×q2 ×3q4r2=38kq2r2=38F.

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About this question

This is a previous-year question from JEE Main 2018, 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.