Direction of electric field

Question

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  ( considered point K between A  and B)of the ring is E. The electric field at the centre due to charge  on the  part ACDB of the ring is

A) E along KO

B ) 3E alongOK

C) 3E along KO

D) E along OK

Please explain through diagrams indicating  the direction of electric field.

Answer 1

D) E along OK

The field at O due to AC and BD cancel each other. The field due to CD is acting in the direction OK and equal magnitude to E due to AKB.

Answer 2

Correct option is (D) E along OK

Explanation:

 I think the figure is like this. If it is same as you are asking.

As per the question, the ring is conducting. So It is obvious that the total electric field at the centre is zero. 
E_total = 0
So, E_AKB + E_ACDB = 0

or, E_ACDB = - E_AKB
or, E_ACDB = - E (towards centre along KO)
_Or,  E_ACDB = E (Along OK)