The electric field due to a uniformly charged disc at a point very distant from the surface of the disc is given by
[$\sigma$ is the surface charge density on the disc]

Consider the disc of radius "$a$" with uniform charge density $\sigma$; $E$ be the electric field at a point along the axis of the disc at a distance $x$ from its center.

We can assume the charge distribution as a collection of concentric rings of charge. Consider one such ring of radius $r$ and charge $dq$.

Area element $dA=(2\pi r)dr=$ area of ring

$dq=\sigma dA=(2\pi \sigma r)dr=$ charge of that ring

Due to symmetry, there is no vertical component of electric field at point P. There is only the horizontal component so, using the result from a ring of charge;

$d{E}_x=\frac{k(dq)x}{(x^2+r^2{)}^{3/2}}$

$=\frac{k(2\pi \sigma r)drx}{(x^2+r^2{)}^{3/2}}$

$E_x={\int }_0^{a}d{E}_x={\int }_0^a\frac{(2\pi \sigma rkdr)x}{(x^2+r^2{)}^{3/2}}$

put $x^2+r^2=t^2$

Differentiating both sides, 

$2rdr=2tdt$

At $r=0;t=x$

At $r=a;t=\sqrt{x^2+a^2}$

So, $E_x=\frac{\sigma }{2{\epsilon }_0}{\int }_x^{\sqrt{x^2+a^2}}\frac{t}{t^3}dt$

$=\frac{\sigma }{2{\epsilon }_0}{\left[\frac{-1}{t}\right]}_x^{\sqrt{x^2+a^2}}$

$=\frac{\sigma }{2{\epsilon }_0}\left[\frac{1}{x}-\frac{1}{\sqrt{x^2+a^2}}\right]$ 

$=\frac{\sigma }{2{\epsilon }_0}\left[1-\frac{1}{\sqrt{\frac{a^2}{x^2}+1}}\right]$ is very small compared to 1.

So, $E_x=\frac{\sigma }{2{\epsilon }_0}$.