A thin conducting ring of radius `R` is given a charge `+Q`, Fig. The electric field at the center `O` of the ring due to the charge on the part `AKB` of the ring is `E`. The electric field at the center due to the charge on part `ACDB` of the ring is

Options

A. E along KO
B. 3E along OK
C. 3 E along KO
D. E along OK

Answer

D

Explanation

Electric field due to the given charged ring is zero at centre 'O'. So electric field due to AKB is equal and opposite to electric field due to ACDB, from the principle of superposition.

Since `vecE` is field strength of O along `vec(KO)` So electric field strength due to ACDB along `vec(OK)` and it is equal to E.