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Question

A thin conducting ring of radius R is given a charge +Q uniformly distributed over its circumference. 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
3E along KO
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B
E along OK
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C
E along KO
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D
3E along OK
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Solution

The correct option is B E along OK
The electric field due to part AKB at centre will be along KO with magnitude E.


We know that net electric field at the centre of ring having uniformly distributed over its circumference will be zero.

Enet=0

EACDB+EAKB=0

EACDB=EAKB

Thus, the electric field due to part ACDB will be along OK and of same magnitude as of part AKB.

EACDB=E along OK



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