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# Electric Field Of Charged Disk - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

## Quick Facts

• Electric field due to uniformly charged disc is considered one the most difficult concept.

• 7 Questions around this concept.

## Solve by difficulty

A thin disc of radius b=2a has a concentric hole of radius 'a' in it (see figure). It carries uniform surface charge on it. If the electric field on its axis at height' h ' (h<<a) from its centre is given as 'Ch' then value of 'C' is :

## Concepts Covered - 1

Electric field due to uniformly charged disc

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) $\sigma$. 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 dQ, 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, $dA=2 \pi r .d r$. 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 dQ 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., $E_y=E_z=0$

Special case -

1) When $R>>x$then   $E_{x}=\frac{\sigma}{2 \varepsilon_{0}}$      Note that this equation is independent of 'x'

2)  When $x\rightarrow 0$ (i.e very near to disc), then $E_{x}=\frac{\sigma}{2 \varepsilon_{0}}$

## Study it with Videos

Electric field due to uniformly charged disc

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