Q.
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
A) $E\ along\ KO$
B) $3\ E\ along\ OK$
C) $3\ E\ along\ KO$
D) $E\ along\ OK$
Explanation
(d) Electric field at the centre of charged circular ring is zero. Hence electric field at $O$ due to the part $ACDB$ is equal in magnitude and opposite in direction that due to the part $AKB$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.