The correct option is C 14πε02q(3√3)R2
REF.Image.
The diagram is :-
The electric field due to the ring an it &
axis at a distance x is given by:-
E=kqx(x2+R2)3/2
To find maximum electric field, we will use the concept of maximum and minimum :-
dEdx=kq(x2+R2)3/2−3/2(x2+R2)1/2.2x2(x2+R2)
Now, dEdx=0⇒(x2+R2)3/2=32×2x2(x2+R2)1/2
⇒x2+R2=3x2
⇒2x2=R2
⇒x2=R22
⇒x=±R√2
So Emax=k.q.R√2(R22+R2)3/2=k.q.2R3√3R3
Emax=14πE02q3√3R2