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 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          [AIPMPT (S) 2008]

A)  3E along KO    

B)              E along OK

C)  E along KO     

D)  3E along OK

Correct Answer: B

Solution :

[b] Key Idea: It is given that the ring is conducting.

As the ring is conducting, so electric field at its centre is zero,

i.e.,       \[{{\overrightarrow{E}}_{total}}=0\]

or         \[{{\overrightarrow{E}}_{AKB}}+{{\overrightarrow{E}}_{ACDB}}=0\]

or         \[{{\overrightarrow{E}}_{ACDB}}=-{{\overrightarrow{E}}_{AKB}}\]

or         \[{{\overrightarrow{E}}_{ACDB}}=-\overrightarrow{E}\]  (along KO)

Therefore, the electric field at the centre due to the charge on the part ACDB of the ring is E along OK.

Alternative:

The fields at O due to AC and BD cancel each other.

The field due to CD is acting in the direction OK and equal in magnitude to E due to AKB.