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

Answer

D

Explanation

The fields at O due to AC and BD cancle each other. The field due to CD is acting in the direction OK and equal in magnitude to E due to AKB.