Find the electric field a distance $\mathit{z}$above the center of a flat circular disk of radius $\mathit{R}$(Fig. 2.1 0) that carries a uniform surface charge $\mathit{a}$.What does your formula give in the limit $\mathit{R}\mathbf{\to }\mathbf{\infty }$? Also check the case localid="1654687175238" $\mathit{z}\mathbf{\gg }\mathit{R}$.

The radius of the circular disk is $R$.

The charge is$a$.

The uniform surface charge is$\sigma$.

Electric field due to charge $\mathit{q}$at a distance $\mathit{r}$is proportional to the charge $\mathit{q}$and inversely proportional to the square of the distance $\mathit{r}$

$\mathit{E}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\epsilon }}_{\mathbf{0}}}\frac{\mathbf{q}}{{\mathbf{r}}^{\mathbf{2}}}$

The center line of the flat circular disk of radius $R$is drawn. At a point z on the center line the edges of the loop make the angle$\theta$, as shown below:

From the above, using Pythagoras theorem in left right triangle, we can write,

$\cos \theta =\frac{z}{\sqrt{r^2+z^2}}$

It can be considered that the flat circular disk is made up of symmetrical elements of small lengths $dx$, having charge $dq$then the differential field at the point P due to $dq$can be written as

$dE=\frac{1}{4{\epsilon }_0}\frac{dq}{\sqrt{z^2+x^2}}\cos \theta$

The surface charge density $\sigma$is the ratio of differential charge $dq$to the differential surface$2\pi xdx$, written as,

$\sigma =\frac{dq}{2\pi xdx}$

Rearrange the above equation as,

$$\begin{gathered} \sigma =\frac{dq}{2\pi xdx} \\ dq=\sigma 2\pi xdx \end{gathered}$$

Substitute $\frac{z}{\sqrt{r^2+z^2}}$$\cos \theta and$localid="1654689936947" $\sigma 2\pi xdxforintodE=\frac{1}{4\pi {\epsilon }_0}\frac{dq}{R^2}\cos \theta as,$

localid="1654690248345" $$\begin{gathered} dE=\frac{1}{4\pi {\epsilon }_0}\frac{dq}{r^2+z^2}\left(\frac{z}{\sqrt{r^2+z^2}}\right) \\ =\frac{1}{4\pi {\epsilon }_0}\frac{\sigma 2\pi xzdx}{{\left(r^2+z^2\right)}^{3/2}} \end{gathered}$$

Integrate above differential integral as,

$$\begin{gathered} E=\int dE \\ ={\int }_0^R\frac{1}{4\pi {\epsilon }_0}\frac{\sigma 2\pi xzdx}{{\left(r^2+z^2\right)}^{3/2}} \\ =\frac{\sigma z}{2{\epsilon }_0}{\int }_0^R\frac{xdx}{{\left(x^2+z^2\right)}^{3/2}} \\ =\frac{\sigma z}{2{\epsilon }_0}{{\left(\frac{-1}{\sqrt{x^2+z^2}}\right)}_0}^R \end{gathered}$$

Simplify further as,

localid="1654691498152" $$\begin{gathered} E=\frac{\sigma z}{2{\epsilon }_0}{{\left(\frac{-1}{\sqrt{x^2+z^2}}\right)}_0}^R \\ =\frac{\sigma z}{2{\epsilon }_0}{{\left(\frac{-1}{\sqrt{x^2+z^2}}\right)}_0}^R \\ =\frac{\sigma z}{2{\epsilon }_0}\left[1-\frac{z}{\sqrt{x^2+z^2}}\right] \end{gathered}$$

Thus, the electric field at a distance zabove the center of a flat circular disk is

localid="1654691528997" $E=\frac{\sigma z}{2{\epsilon }_0}\left[\frac{z}{\sqrt{z^2+R^2}}\right]$

Substitute $\infty$for $R$, into $E=\frac{\sigma z}{2{\epsilon }_0}\left[-1\frac{z}{\sqrt{z^2+R^2}}\right]$

$$\begin{gathered} E=\frac{\sigma z}{2{\epsilon }_0}\left[-1\frac{z}{\sqrt{z^2+{\left(\infty \right)}^2}}\right] \\ =\frac{\sigma z}{2{\epsilon }_0}\left[1-0\right] \\ =\frac{\sigma z}{2{\epsilon }_0} \end{gathered}$$

Thus as $R\to \infty$,$E=\frac{\sigma }{2{\epsilon }_0}$.

Apply binomial distribution localid="1654692341834" ${\left(1+x\right)}^{-n}=1-nx+\frac{n\left(n+1\right)}{2!}{x}^2-.....$, in the result

$E=\frac{\sigma z}{2{\epsilon }_0}\left[1-\frac{z}{\sqrt{z^2+R^2}}\right]$

$E=\frac{\sigma z}{2{\epsilon }_0}\left[1-\left\{1-\left(\frac{R_2}{z_2}\right)+...\right\}\right]$

As $z\gg R$, the higher order terms of $\frac{R}{z^2}$for $n>2$can be neglected. Thus the above expression becomes

$$\begin{gathered} E=\frac{\sigma }{2{\epsilon }_0}\left[1-\left\{1-\left(\frac{1}{2}\left(\frac{R_2}{z_2}\right)\right)\right\}\right] \\ =\frac{\sigma }{2{\epsilon }_0}\left[\left(\frac{1}{2}\left(\frac{R^2}{z^2}\right)\right)\right] \\ =\frac{\sigma }{4{\epsilon }_0}\frac{R^2}{z^2} \end{gathered}$$

Thus, for$z\gg R$, the electric field is obtained as$E=\frac{\sigma }{4{\epsilon }_0}\frac{R^2}{z^2}$