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 $dQ$, inner radius $r$ and outer radius $r+dr$. Its area $dA$ is approximately equal to its width d r times its circumference $(2\pi r)$ so, $dA=2\pi r\cdot dr$. The charge per unit area is $\sigma =dQ/dA$, from this - the charge of ring is $dQ=\sigma (2\pi rdr)$. The field component $d{E}_x$ at point $P$ due to charge $dQ$ of a ring of radius $r$ is -

$d{E}_x=\frac{1}{4\pi {\epsilon }_0}\frac{(2\pi \sigma rdr)x}{{(x^2+r^2)}^{3/2}}$

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

$E_x=\int d{E}_x={\int }_0^{R}d{E}_x={\int }_0^R\frac{1}{4\pi {\epsilon }_0}\frac{(2\pi \sigma rdr)x}{{(x^2+r^2)}^{3/2}}$

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

$E_x=\frac{\sigma x}{2{\epsilon }_0}[-\frac{1}{\sqrt{x^2+R^2}}+\frac{1}{x}]=\frac{\sigma }{2{\epsilon }_0}[1-\frac{x}{\sqrt{x^2+R^2}}]$
 

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

Special case -
1) When $R\gg x$, then $E_x=\frac{\sigma }{2{\epsilon }_0}$ Note that, this equation is independent of ' $x$ '

$E_x=\frac{\sigma }{2{\epsilon }_0}$