A thin conducting ring of radius R is given a charge +Q. The electric field at the centre O of the ring due the to the charge on the part AKB of the ring is E. The electric field at the centre due to the charge on the part ACDB of the ring is

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

E along OK

E along KO

3E along KO

3E along OK

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

According to symmetry net electric field due to part AKB and the part ACDB at its centre is zero, i.e.,

${\vec{E}}_{total} = 0 or {\vec{E}}_{AKB} + {\vec{E}}_{ACDB} = 0 \Rightarrow {\vec{E}}_{1} + {\vec{E}}_{2} = 0 or {\vec{E}}_{2} = - {\vec{E}}_{1} or {\vec{E}}_{ACDB} = - \vec{E} (along KO)$

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE