JEE Main 2015PhysicsElectrostaticsHardMCQ

JEE Main 2015Electrostatics Question with Solution

JEE Main 2015 (10 Apr Online)

Question

A thin disc of radius b=2a has a concentric hole of radius a in it (see figure). It carries uniform surface charge σ  on it. If the electric field on its axis at a height hh<<a from its centre is given as Ch then the value of C is

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Correct answer
Aσ4 aϵ0

Step-by-step explanation

 At the axial point of a uniformly charged disc electric field is given by,

E=σ2ϵ01-cosθ



By superposition principle, when inner disc is removed , then, electric field due to remaining disc is,



E=σ2ϵ0 1-cosθ2-1-cosθ1

=σ2ϵ0cosθ1-cosθ2

=σ2ϵ0hh2+a2 -hh2+b2

=σ2ϵ0ha1+h2a2-hb1+h2b2 

h a and b.

E=σ2ϵ0ha-hb

=σ2ϵ0ha-h2a=σh4ϵ0a

C=σ4aϵ0.

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About this question

This is a previous-year question from JEE Main 2015, 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.