The maximum electric field intensity on the axis of a uniformly charged ring of charge q

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/(R2 + x2)3/2
to find maximum electric field we can use the concept of maxima and minima…

dE/dx = d/dx of {kQx/(R2 + x2)3/2 }

=kQ{ 1 .(R2 +x2 )3/2 - 3/2 (R2 +x3/2 )1/2 . x }/(R2 +x2)3 (by applying quotient rule)

now on putting dE/dx =0 ,we get

x2 -3x/2+R2 =0
x2 -3x/2 +9/16 -9/16 +R2 =0
(x-3/4)2 = 9/16-R2
x={(9-16R2 )1/2 +3 }/4 units

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