Consider a uniformly charged disc of radius R. If the field at the centre of the disc is E₀, then the field along the axis at a distance R from the centre is nearly:

Let us subdivide the disc into thin concentric rings of width dr each. Charge on one such element ring of radius r is dq $=\sigma \left(2\pi rdr\right)$

Field at P due to this element ring is

$\begin{array}{l}dE=\frac{1}{4\pi {\epsilon }_0}\frac{xdq}{{\left(x^2+r^2\right)}^{\frac{3}{2}}}=\frac{\sigma x}{2{\epsilon }_o}\frac{rdr}{{\left(x^2+r^2\right)}^{\frac{3}{2}}}\\ \Rightarrow E=\frac{\sigma x}{2{\epsilon }_0}{\int }_0^R \frac{rdr}{{\left(x^2+r^2\right)}^{\frac{3}{2}}}\end{array}$

$=\frac{\sigma x}{4{\epsilon }_o}{\left|\frac{-2}{\sqrt{x^2+r^2}}\right|}_0^R\left[\text{ Putting }{x}^2+r^2=t^2\right]$

$=\frac{\sigma x}{2{\epsilon }_o}\left(\frac{1}{x}-\frac{1}{\sqrt{R^2+x^2}}\right)=\frac{\sigma }{2{\epsilon }_o}\left(1-\frac{x}{\sqrt{R^2+x^2}}\right)$

Putting x = 0 gives the field at the centre of the disc , 

$E_o=\frac{\sigma }{2{\epsilon }_o}$

Hence, $E\left(R\right)=E_o\left(1-\frac{1}{\sqrt{2}}\right)\simeq 0.3E_o$