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.You need to login to perform this action.
You will be redirected in
3 sec