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Question: Answered & Verified by Expert
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

PhysicsElectrostaticsJEE MainJEE Main 2015 (11 Apr Online)
Options:
  • A 25×103i^ N C-1
  • B 50×103i^ N C-1
  • C 25×103j^ N C-1
  • D 50×103j^ N C-1
Solution:
2501 Upvotes Verified Answer
The correct answer is: 25×103i^ N C-1


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=2kRsin902         {=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

Enet=25×103 N C-1 i^

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