A) 3E along KO
B) E along OK
C) E along KO
D) 3 E along OK
Correct Answer: B
Solution :
Key Idea: It is given that the ring is conducting. As the ring is conducting, so electric field at its centre is zero, ie, \[{{\overrightarrow{E}}_{total}}=0\]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.You need to login to perform this action.
You will be redirected in
3 sec