**Find the electric field caused by a disk of radius R with a uniform positive surface charge density $\sigma$ and total charge Q, at a point P. Point P lies a distance x away from the centre of the disk, on the axis through the centre of the disk.**

$$\sigma = \frac{Q}{\pi R^{2}}$$

To find dQ, we will need $dA$. Note that $dA = 2 \pi r \, dr$

$$\begin{aligned} dQ &= \sigma \times dA \\ &= 2 \pi r \sigma \, dr \end{aligned}$$

Note that due to the symmetry of the problem, there are no vertical component of the electric field at P. There is only the horizontal component.

Using the result from a ring of charge:

$$dE_{x} = \frac{x \, dQ}{4 \pi \epsilon_{0} (x^{2} + r^{2})^{\frac{3}{2}}}$$

Sub in dQ to obtain:

$$dE_{x} = \frac{\sigma}{2 \epsilon_{0}} \frac{xr \, dr}{(x^{2} + r^{2})^{\frac{3}{2}}}$$

$$E_{x} = \frac{\sigma x}{2 \epsilon_{0}} \int\limits_{0}^{R} \frac{r}{(x^{2} + r^{2})^{\frac{3}{2}}} \, dr$$

To evaluate the integral, you will need $\int\frac{x \, dx}{ \left( a^{2}+x^{2} \right)^{\frac{3}{2}}} = \, – \frac{1}{\sqrt{a^{2} + x^{2}}}$ from integration table.

$$\begin{aligned} E_{x} &= \frac{\sigma x}{2 \epsilon_{0}} \left( \frac{1}{x}- \frac{1}{\sqrt{x^{2} + R^{2}}} \right) \\ &= \frac{\sigma}{2 \epsilon_{0}} \left( 1 – \frac{1}{\sqrt{1 + \frac{R^{2}}{x^{2}}}} \right) \end{aligned}$$

The electric field produced by an infinite plane sheet of charge (which can be seen from the formula above as $r \rightarrow \infty$) is independent of the distance from the sheet.

$$E = \frac{\sigma}{2 \epsilon_{0}}$$

The electric field produced by an infinite plane sheet of charge can be found using Gauss’s Law as shown here.

Next: Electric Field Of Two Oppositely Charged Infinite Sheets

Minor typo. For page

https://www.miniphysics.com/uy1-electric-field-of-uniformly-charged-disk.html

your integral for Ex needs a “dr.”

Caught it. Thanks!