E0⇒EAKB+EACDBEACDB=0=0=(−)EAKB=−E( along KO)=E( along OK)
NEET UG 2008 — Physics Electromagnetism
2008mcqmedium
A thin conducting ring of radius R is given a charge +Q. The electric field at the centre O of the ring due 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 
Official previous-year question
Held on 30 Apr 2008 · Verified 9 Jul 2026.
Options
- A
3 E along OK
- B
3 E along KO
- C
E along OK
- D
E along KO
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