Electric Field Of Charged Disk MCQ - Practice Questions with Answers

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) $\sigma$ 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 $r+d r$. Its area $d A$ is approximately equal to its width d r times its circumference $(2 \pi r)$ so, $d A=2 \pi r \cdot d r$. The charge per unit area is $\sigma=d Q / d A$, from this - the charge of ring is $d Q=\sigma(2 \pi r d r)$. The field component $d E_x$ at point $P$ due to charge $d Q$ of a ring of radius $r$ is -

$
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., $E_y=E_z=\overline{0}$

Special case -
1) When $R \gg x$, then $E_x=\frac{\sigma}{2 \varepsilon_0} \quad$ Note that, this equation is independent of ' $x$ '

$
E_x=\frac{\sigma}{2 \varepsilon_0}
$