Electric Charges And Fields Ques 15
15. 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
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(a) E along $KO$
(b) E along OK
(c) E along $KO$
(d) 3 E along $OK$
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Answer:
Correct Answer: 15.(b)
Solution:
- (b) By the symmetry of the figure, the electric fields at $O$ due to the portions $AC$ and $BD$ are equal in magnitude and opposite in direction. So, they cancel each other.
Similarly, the field at $O$ due to $CD$ and $AKB$ are equal in magnitude but opposite in direction. Therefore, the electric field at the centre due to the charge on the part $ACDB$ is $E$ along $OK$.
BETA
