question_answer

A thin conducting ring of radius R is given a charge +Q. The electric field at the centre O of the ring due the 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                

B)  E along OK

C)  E along KO                  

D)  3E along OK

Correct Answer: B

Solution :

 According to symmetry net electric field due to part AKB and the part ACDB at its centre is zero, i.e., \[{{\vec{E}}_{total}}=0\] or         \[{{\vec{E}}_{AKB}}+{{\vec{E}}_{ACDB}}=0\Rightarrow {{\vec{E}}_{1}}+{{\vec{E}}_{2}}=0\] or         \[{{\vec{E}}_{2}}=-{{\vec{E}}_{1}}\] or         \[{{\vec{E}}_{ACDB}}=-\vec{E}(along\,KO)\] Therefore, the electric field at the centre due to the charge on the part ACDB of the ring is E along OK.