Electric Field Of Charged Disk

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 we have taken a typical ring that has charge $dQ$, inner radius $r$ and outer radius $r+dr$. Its area $dA$

$d{E}_x=\frac{1}{4\pi {\epsilon }_0}\frac{(2\pi \sigma rdr)x}{{(x^2+r^2)}^{3/2}}$
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 the x -axis, the field in the rest of the component is zero i.e., $E_y=E_z=0$

Special case -
1) When $R>>x$, then $E_x=\frac{\sigma }{2{\epsilon }_0}$ Note that this equation is independent of ' $x$ '
2) When $x\to 0$ (i.e very near to the disc), then $E_x=\frac{\sigma }{2{\epsilon }_0}$

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Some Solved Examples:

Example 1: A thin disc of radius b=2a has a concentric hole of radius 'a' in it (see figure). It carries a uniform surface charge $'\sigma '$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 :

$\begin{array}{rl}& \frac{\sigma }{\frac{\sigma }{a{ϵ}_0}}\\ & \frac{\sigma }{2{)}^{2a{ϵ}_0}}\end{array}$

3) $\frac{\sigma }{4a{ϵ}_0}$

$\frac{\sigma }{8{)}^{8a{ϵ}_0}}$
Solution:

As we discussed in

Uniformly charged disc -

$\begin{array}{rl}& E=\frac{\sigma }{2{ϵ}_0}[1-\frac{x}{{(x^2+R^2)}^{\frac{1}{2}}}]\\ & V=\frac{\sigma }{2{ϵ}_0}[\sqrt{x^2+R^2}-x]\end{array}$
- wherein

Electric Field due to complete disc $(R=2a)$ at distance $x$

$\begin{array}{rl}& E_1=\frac{\sigma }{2{ϵ}_o}[1-\frac{x}{{(R^2+x^2)}^{\frac{1}{2}}}][h=x;2a=R]\\ & E_1=\frac{\sigma }{2{ϵ}_o}[1-\frac{h}{{(4a^2+h^2)}^{\frac{1}{2}}}]=\frac{\sigma }{2{ϵ}_o}[1-\frac{h}{2a}]\end{array}$
Similarly, the electric field due to disc $(R=a)$

$E_2=\frac{\sigma }{2{ϵ}_o}[1-\frac{h}{a}]$
Now $E=E_1-E_2=\frac{\sigma }{2{ϵ}_o}[1-\frac{h}{2a}]-\frac{\sigma }{2{ϵ}_o}[1-\frac{h}{a}]=\frac{\sigma h}{4{ϵ}_{o}a}$
Hence $C=\frac{\sigma }{4{ϵ}_{o}a}$

Example 2: What will be the electric field due to a uniformly charged disc At a distance $x$ from centre $O$ on its axis if $X\to 0$

$\begin{array}{rl}E& =\frac{\sigma }{4{\epsilon }_0}\\ E& \simeq \frac{\sigma }{2{\epsilon }_0}\\ E& =\frac{2K\lambda }{R}=\frac{Q}{2{\pi }^2{\epsilon }_0{R}^2}\end{array}$

4)0

Solution:
As we learned
Uniformly charged disc -

$\begin{array}{rl}& \text{If}x\to 0\\ & E\simeq \frac{\sigma }{2{ϵ}_0}\end{array}$

wherein

i.e. Point situated near the disc it behaves as an infinite sheet of charge.

$E=\frac{\sigma }{2{\epsilon }_0}[1-\frac{x}{\sqrt{x^2+R^2}}]\text{Putx}=0$

Example 3 The surface charge density of a thin charged disc of radius R is $\sigma .$ The value of the electric field at the centre of the disc is $\frac{\sigma }{2{ϵ}_0\underset{―}{\underset{―}{y}}}$ With respect to the field at the centre, the electric field along the axis at a distance R from the centre of the disc :
1)reduces by $70.7\mathrm{\%}$
2)reduces by $29.3\mathrm{\%}$
3)reduces by $9.7\mathrm{\%}$
4)reduces by $14.6\mathrm{\%}$

Solution:
Electric field intensity at the centre of the disc

$\Rightarrow \mathrm{E}=\frac{\sigma }{2{ϵ}_0}(\text{given})$
Electric field along the axis at any distance $x$ from the centre of the disc

${\mathrm{E}}^{\prime }=\frac{\sigma }{2{ϵ}_0}(1-\frac{\mathrm{x}}{\sqrt{{\mathrm{x}}^2-{\mathrm{R}}^2}})$
From question, $x=R$ (radius of disc)

$\therefore {\mathrm{E}}^{\prime }=\frac{\sigma }{2{ϵ}_0}(1-\frac{\mathrm{R}}{\sqrt{{\mathrm{R}}^2+{\mathrm{R}}^2}})=\frac{\sigma }{2{ϵ}_0}\left(\frac{\sqrt{2}\mathrm{R}-\mathrm{R}}{\sqrt{2}\mathrm{R}}\right)=\frac{4}{14}\mathrm{E}$
So reduction in the value of electric field

$=\frac{(\mathrm{E}-\frac{4}{14}\mathrm{E})\times 100}{\mathrm{E}}=\frac{1000}{14}\mathrm{\%}=70.7\mathrm{\%}$

Example 4:Find out the surface charge density at the intersection of point $\mathrm{x}=3\mathrm{m}$ plane and the x -axis in the region of uniform line charge of $8\mathrm{nC}/\mathrm{m}$ along the z -axis in free space.
1) $47.88\mathrm{C}/\mathrm{m}$
2) $0.424\mathrm{nC}{\mathrm{m}}^{-2}$
3) $0.07\mathrm{nC}{\mathrm{m}}^{-2}$
4) $4.0\mathrm{nC}{\mathrm{m}}^{-2}$

Solution:
Electric field due to uniformly charged rod- $E=\frac{2K\lambda }{r}$

Electric field due to uniformly charged disk-

$\begin{array}{rl}& E=\frac{\sigma }{2{\epsilon }_0}\\ & \frac{2\mathrm{K}\lambda }{\mathrm{r}}=\frac{\sigma }{{\epsilon }_0}\\ & \sigma =\frac{2K\lambda {\epsilon }_0}{r}=\frac{\lambda }{2\pi r}=\frac{8\times {10}^{-9}}{2\pi \times 3}\\ & \sigma =0.424\times {10}^{-9}\frac{\mathrm{C}}{{\mathrm{m}}^2}\end{array}$

Summary:

To compute the electric field at a point along the axis of a uniformly charged disk, the disk is treated as a series of infinitesimal rings of charge. Each ring contributes to the net electric field, with its components integrated to find the total field. For a point close to the disk, the field behaves like that of a charged plane, while at far distances, it resembles the field of a point charge. The formula derived depends on the radius of the disk, surface charge density, and the distance from the point of interest to the disk’s centre.