Electric Field Of Charged Disk - Practice Questions & MCQ

Electric field due to uniformly charged disk

Let us take a disk of radius R with a uniform positive surface charge density (charge per unit area) . Our aim is to find an electric field at a point on the axis of the disk at a distance x from its centre. 

From the figure, we can see that the we have taken a typical ring has charge $d Q$, inner radius $r$ and outer radius $\mathrm{r}+\mathrm{d} \mathrm{r}$. Its area $\mathrm{d} A$

$
d E_x=\frac{1}{4 \pi \varepsilon_0} \frac{(2 \pi \sigma r d r) x}{\left(x^2+r^2\right)^{3 / 2}}
$

If we integrate from 0 to $R$, we will get the total field -

$
E_x=\int d E_x=\int_0^R d E_x=\int_0^R \frac{1}{4 \pi \varepsilon_0} \frac{(2 \pi \sigma r d r) x}{\left(x^2+r^2\right)^{3 / 2}}
$

Here, ' $x$ ' is constant and 'r' is the variable. After integration we get -

$
E_x=\frac{\sigma x}{2 \varepsilon_0}\left[-\frac{1}{\sqrt{x^2+R^2}}+\frac{1}{x}\right]=\frac{\sigma}{2 \varepsilon_0}\left[1-\frac{x}{\sqrt{x^2+R^2}}\right]
$

As this disc is symmetric to $x$-axis, so the field in the rest of the component is zero i.e., $E_y=E_z=0$

Special case -
1) When $R \gg>x$, then $\quad E_x=\frac{\sigma}{2 \varepsilon_0} \quad$ Note that this equation is independent of ' x '
2) When $x \rightarrow 0$ (i.e very near to disc), then $E_x=\frac{\sigma}{2 \varepsilon_0}$