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

Answer

A

Explanation

Positive charge Q is uniformly distributed over the ring and hence electric field intensity at the centre due to it is zero due to cancellation of fields from diametrically opposite ring. Line OK divides portion AB in two equal parts so from symmetry we know that electric field at point O due to AKB will be directed away from K towards O. To make the net electric field intensity zero at the centre, electric field due to ACDB will be equal and opposite to that due to AKB. Hence, the field due to ACDB will also be E but will be directed from O to K.