Home » Blog » Question of The Day » Problem no.6 , Topic: Electrostatics
Tags: AIEEE, Electrostatics, IIT, IIT JEE, IITJEE, Physics, problems, question, Question of The Day
Shubhanshu dubey December 14, 2010 at 10:04 am
khushali December 14, 2010 at 10:06 am
E1=kqa/pow(a*a+a*a,3/2) E2=kqa/pow(b*b+b*b,3/2) for equilibrium ,F=0,qE1=qE2 so E1=E2 by solving we get, 4*a*a=b*b so a/b=1/2 is answer
khushali December 14, 2010 at 10:07 am
vikash chandola December 15, 2010 at 8:41 pm
electric field through through ring with radius a should equal(in magnitude and opposite in direction) to electric field through radius b so using this find electric field through both the ring and put their sum ZERO it will give answer sqrt(5/3) if i am wrong then please tell me its correct answer.
Bharat December 15, 2010 at 10:02 pm
ankur bhatnagar January 21, 2011 at 10:03 pm
ans is “B”
aniket September 12, 2011 at 11:24 am
Is this the axial position? i dont think it is…
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