The maximum electric field intensity on the axis of a uniformly charged ring of charge q and radius R will be.

let the total charge on the ring is Q then electric field at the axis of a charged ring is given by

E=kQx/(R^{2} + x^{2})^{3/2}

to find maximum electric field we can use the concept of maxima and minima…

dE/dx = d/dx of {kQx/(R^{2} + x^{2})^{3/2} }

=kQ{ 1 .(R^{2} +x^{2} )^{3/2} - 3/2 (R^{2} +x^{3/2} )^{1/2} . x }/(R^{2} +x2)^{3} (by applying quotient rule)

now on putting dE/dx =0 ,we get

x^{2} -3x/2+R^{2} =0

x^{2} -3x/2 +9/16 -9/16 +R^{2} =0

(x-3/4)^{2} = 9/16-R^{2}

x={(9-16R^{2} )^{1/2} +3 }/4 units

therefore at a distance x from center of the ring electric field is maximum