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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

1. E along KO
2. E along OK
3. E along CO
4. 3E along OK

Correct answer: 3E along OK

Solution

The charge on the part ACDB is three times the charge on the part AKB, as ACDB is three-fourths of the ring while AKB is one-fourth. Since the electric field is directly proportional to the charge, the field due to ACDB will be three times that due to AKB, and it will act along the same direction as OK.

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