JEE Main 2015PhysicsElectrostaticsHardMCQ

JEE Main 2015 — Electrostatics Question with Solution

JEE Main 2015 (11 Apr Online)

Question

A wire of length L=20 cm is bent into a semi-circular arc and the two equal halves of the arc are uniformly charged with charges +Q and -Q as shown in the figure. The magnitude of the charge on each half is Q=103ε0, where ε0 is the permittivity of free the space. The net electric field at the centre O is

Choose an option

Show full solutionCorrect option: A
Correct answer
A25×103i^ N C-1

Step-by-step explanation



L=πR

⇒R= L π = 20 100π   m= 1 5π   m

⇒ due to a charge arc, electric field at centre is given by

E= 2Kλ R sin θ 2



E1=E2=2k⋋Rsin⁡902         {⋋=QπR/2}

undefined

Component along j^  gets cancelled and 

E net = 2 E 1

= 4KQ π R 2

=4×1103ϵ04πϵ0 πR2=4×103ϵ04π2ϵ0R2=102R2=10015π2=25×103

E→net=25×103 N C-1 i^

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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.