The Figure 1 shows a disk of total charge \(q\). We are going to calculate the electric field due to the disk of charge at a point \(p\) which lies at perpendicular distance \(x\) from the centre of the disk. The origin of our coordinate system lies at the centre of the disk and the plane of the disk is perpendicular to the x-axis. To get our result we divide the charged disk into small rings of charge \(dq\) and get the the electric field at the point \(p\) by adding the electric fields produced by every ring element.

We already know the electric field of a ring of charge, it's easier to get our result for the disk of charge. As you already know that the total electric field at the point \(p\) is only due to the x-component of electric field of ring elements. Consider that the radius of one specific ring is \(r\) and thickness is \(dr\). So, the ring has area equal to \(2\pi rdr\) and the charge \(dq\) in the ring is \(dq=\sigma (2\pi rdr)\) where \(\sigma \) is the charge per unit area. The electric field \(dE\) of the ring of charge \(dq\) (you already know this) at point \(p\) is

\[dE = k\frac{{x\sigma (2\pi rdr)}}{{{{({x^2} + {r^2})}^{\frac{3}{2}}}}}\]

Now the electric field for the entire disk is calculated by integrating the above expression from \(0\) to \(R\).

\[E = 2k\pi \sigma x\int\limits_0^R {\frac{{rdr}}{{{{({x^2} + {r^2})}^{\frac{3}{2}}}}}} = 2k\pi \sigma x\left( {\frac{1}{x} - \frac{1}{{\sqrt {{x^2} + {R^2}} }}} \right)\]

\[{\rm{or,}}\quad E = 2k\pi \sigma \left( {1 - \frac{1}{{\sqrt {1 + {{\left( {\frac{R}{x}} \right)}^2}} }}} \right) \tag{1} \label{1}\]

In vector form you can write the above expression as

\[\vec E = 2k\pi \sigma \left( {1 - \frac{1}{{\sqrt {1 + {{\left( {\frac{R}{x}} \right)}^2}} }}} \right)\hat i \tag{2} \label{2}\]

where \(\hat i\) gives the direction of the electric field vector. Suppose the radius of the disk is large enough in comparison to the distance \(x\) so that \(R >> x\), the term \(1/{\sqrt{1+{{\left( \frac{R}{x} \right)}^{2}}}}\) is negligibly small and can be neglected. And the electric field in this case is

\[E=2k\pi \sigma \tag{3} \label{3}\]

So for the very large disk (or you can think as an infinite disk) the electric field is independent of the distance \(x\) and perpendicularly outward from the disk.