JEE Main 2014PhysicsElectrostaticsMediumMCQ

JEE Main 2014 — Electrostatics Question with Solution

JEE Main 2014 (19 Apr Online)

Question

The electric field in a region of space is given by, E→=E0i^+2E0j^ where E0=100 N C-1. The flux of this field through a circular surface of radius 0.02 m parallel to the Y‐Z plane is nearly

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Show full solutionCorrect option: C
Correct answer
C0.125 N m2 C-1

Step-by-step explanation

ϕ=E→·A→=E0i^+2E0j^·πr2i^

=E0πr2=0.125 N m2 C-1

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About this question

This is a previous-year question from JEE Main 2014, 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.