Find the electric field at a height zabove the center of a square sheet (side a) carrying a uniform surface charge $\mathit{\sigma }$.Check your result for the limitingcases $\mathit{a}\mathbf{\to }\mathbf{\infty }$and z >> a.

The expression for the electric filed at a distance z above the center of the square loop carrying uniform line charge $\lambda$is,

$\mathit{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{\mathbf{4}\mathbf{\lambda }\mathbf{a}\mathbf{z}}{\mathbf{(}{\mathbf{z}}^{\mathbf{2}}\mathbf{+}{\displaystyle \frac{a^2}{4}}\mathbf{)}\sqrt{{\mathbf{z}}^{\mathbf{2}}\mathbf{+}{\displaystyle \frac{{\mathbf{a}}^{\mathbf{2}}}{\mathbf{2}}}}}\hat{\mathbf{z}}$

.......(1)

Here, E is the electric filed, $\lambda$is the linear charge density, ${\epsilon }_0$is the permittivity for the free space, is the length of each side of the square sheet.

The square sheet is shown in below figure.

Write the expression for linear charge density for the above square loop.

$\mathit{d}\mathit{\lambda }\mathbf{=}\mathit{\sigma }\mathbf{\left(}\frac{\mathbf{d}\mathbf{a}}{\mathbf{2}}\mathbf{\right)}dE=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{4\mathrm{az}}{({\mathrm{z}}^2+{\displaystyle \frac{{\mathrm{a}}^2}{4}})\sqrt{{\mathrm{z}}^2+{\displaystyle \frac{{\mathrm{a}}^2}{2}}}}d\lambda$

Here, $\sigma$is the charge density.

Differentiating the equation (1) on both sides,

Substitute$\sigma \left(\frac{da}{2}\right)$for$d\lambda$.

$$\begin{gathered} dE=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{4\mathrm{az}\sigma \left({\displaystyle \frac{da}{2}}\right)}{({\mathrm{z}}^2+{\displaystyle \frac{{\mathrm{a}}^2}{4}})\sqrt{{\mathrm{z}}^2+{\displaystyle \frac{{\mathrm{a}}^2}{2}}}} \\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{4\sigma z}{2}\right)\frac{ada}{({\mathrm{z}}^2+{\displaystyle \frac{{\mathrm{a}}^2}{4}})\sqrt{{\mathrm{z}}^2+{\displaystyle \frac{{\mathrm{a}}^2}{2}}}} \\ =\frac{\sigma z}{2{\mathrm{\pi \epsilon }}_0}\frac{ada}{({\mathrm{z}}^2+{\displaystyle \frac{{\mathrm{a}}^2}{4}})\sqrt{{\mathrm{z}}^2+{\displaystyle \frac{{\mathrm{a}}^2}{2}}}} \end{gathered}$$

Thus, the differential equation solution is $\frac{\sigma z}{2{\mathrm{\pi \epsilon }}_0}\frac{ada}{({\mathrm{z}}^2+{\displaystyle \frac{{\mathrm{a}}^2}{4}})\sqrt{{\mathrm{z}}^2+{\displaystyle \frac{{\mathrm{a}}^2}{2}}}}$.

Now, integrate the equation (1) with limits from 0 to a and solve for the electric filed due to the square sheet at a height z above its center.

$E=\frac{\sigma z}{2{\mathrm{\pi \epsilon }}_0}{\int }_0^a\frac{a}{({\mathrm{z}}^2+{\displaystyle \frac{{\mathrm{a}}^2}{4}})\sqrt{{\mathrm{z}}^2+{\displaystyle \frac{{\mathrm{a}}^2}{2}}}}da$ ………(2)

Let $a^2=4t$, then $da=\frac{2dt}{a}$. The limits of t are 0 and $\frac{a^2}{4}$. Then the equation becomes,

$$\begin{gathered} E=\frac{\sigma z}{2{\mathrm{\pi \epsilon }}_0}{\int }_0^{\frac{a^2}{4}}\frac{a}{({\mathrm{z}}^2+t)\sqrt{{\displaystyle \left({\mathrm{z}}^2+2t\right)}}}\frac{2}{a}da \\ =\frac{\sigma z}{2{\mathrm{\pi \epsilon }}_0}{\int }_0^{\frac{a^2}{4}}\frac{1}{({\mathrm{z}}^2+t)\sqrt{{\displaystyle \left({\mathrm{z}}^2+2t\right)}}}da \end{gathered}$$

As,$\left(a^2+b\right)\sqrt{\left(a^2+2b\right)}={\tan }^{-1}\left(\frac{\sqrt{\left(a^2+2b\right)}}{a}\right)$

Solving further based on the above tangential formula,

$$\begin{gathered} E=\frac{\sigma z}{{\mathrm{\pi \epsilon }}_0}{\left[\frac{2}{z}{\tan }^{-1}\left(\frac{\sqrt{z^2+2t}}{\begin{array}{c}z\\ \end{array}}\right)\right]}_0^{\frac{a^2}{4}} \\ =\frac{2\sigma }{{\mathrm{\pi \epsilon }}_0}\left[{\tan }^{-1}\left(\frac{\sqrt{z^2+2\left({\displaystyle \frac{a^2}{4}}\right)}}{\begin{array}{c}z\\ \end{array}}\right)-{\tan }^{-1}\left(\frac{\sqrt{z^2+2\left(0\right)}}{z}\right)\right] \\ =\frac{2\sigma }{{\mathrm{\pi \epsilon }}_0}\left[{\tan }^{-1}\left(\frac{\sqrt{z^2+{\displaystyle \frac{a^2}{2}}}}{\begin{array}{c}z\\ \end{array}}\right)-{\tan }^{-1}\left(1\right)\right] \\ =\frac{2\sigma }{{\mathrm{\pi \epsilon }}_0}\left[{\tan }^{-1}\left(\frac{\sqrt{z^2+{\displaystyle \frac{a^2}{2}}}}{\begin{array}{c}z\\ \end{array}}\right)-{\tan }^{-1}\left(\tan \frac{\mathrm{\pi }}{4}\right)\right] \end{gathered}$$

Simplifying the expression further,

$$\begin{gathered} E=\frac{\sigma z}{{\mathrm{\pi \epsilon }}_0}\left[{\tan }^{-1}\left(\frac{\sqrt{z^2+{\displaystyle \frac{a^2}{2}}}}{\begin{array}{c}z\\ \end{array}}\right)-\frac{\mathrm{\pi }}{4}\right] \\ =\frac{2\sigma }{{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{\pi }}{4}\left[\frac{4}{\mathrm{\pi }}{\tan }^{-1}\left(\frac{\sqrt{z^2+{\displaystyle \frac{a^2}{2}}}}{\begin{array}{c}z\\ \end{array}}\right)-1\right] \\ =\frac{\sigma }{2{\mathrm{\pi \epsilon }}_0}\left[\frac{4}{\mathrm{\pi }}{\tan }^{-1}\left(\sqrt{1+\left(\frac{a^2}{2z^2}\right)}\right)-1\right] \end{gathered}$$

Thus, the electric filed is $E=\frac{\sigma }{2{\mathrm{\pi \epsilon }}_0}\left[\frac{4}{\mathrm{\pi }}{\tan }^{-1}\left(\sqrt{1+\left(\frac{a^2}{2z^2}\right)}\right)-1\right]$.

If $a\to \infty$then the electric field square plate is,

$E=\frac{2\sigma }{{\mathrm{\pi \epsilon }}_0}\left[{\tan }^{-1}\left(\infty \right)-\frac{\mathrm{\pi }}{4}\right]$

Since, $\tan \frac{\mathrm{\pi }}{2}=\infty$

$$\begin{gathered} E=\frac{2\sigma }{{\mathrm{\pi \epsilon }}_0}\left[{\tan }^{-1}\left(\tan \left(\frac{\mathrm{\pi }}{2}\right)\right)-\frac{\mathrm{\pi }}{4}\right] \\ =\frac{2\sigma }{{\mathrm{\pi \epsilon }}_0}\left[\frac{\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{4}\right] \\ =\frac{\sigma }{2{\epsilon }_0} \end{gathered}$$

From the above equation, it is clear that the square sheet act as square plane. Thus the electric filed is due to square plate when $a\to \infty$is $E=\frac{\sigma }{2{\epsilon }_0}$.

Now, let’s consider that, $f\left(x\right)={\tan }^{-1}\sqrt{1+x}-\frac{\mathrm{\pi }}{4}$and $x=\frac{a^2}{2z^2}$. If$z\gg a$ , then $\frac{a^2}{2z^2}\ll 1$and $f\left(0\right)=0$. Then the value of $f\text{'}\left(x\right)$is,

$f\text{'}\left(x\right)=\frac{1}{1+\left(1+x\right)}\frac{1}{2}\frac{1}{\sqrt{1+x}}$

Then the value of $f\text{'}\left(0\right)$by substituting 0 for x is,

$f\text{'}\left(0\right)=\frac{1}{4}$

Applying Taylor series and solving,

$$\begin{gathered} f\left(x\right)=f\left(0\right)+xf\text{'}\left(0\right)+\frac{1}{2}{x}^{2}f\text{'}\text{'}\left(x\right)+... \\ \approx f\left(0\right)+xf\text{'}\left(0\right) \end{gathered}$$

Substitute 0 for $f\left(0\right)$and $\frac{1}{4}$for$f\text{'}\left(0\right)$

$$\begin{gathered} f\left(x\right)=0+\frac{x}{4} \\ =\frac{x}{4} \end{gathered}$$

Substitute $\frac{a^2}{2z^2}$for x

$f\left(x\right)=\frac{a^2}{8z^2}$

Therefore, the electric filed due to square plate when z >> a is,

$$\begin{gathered} E=\frac{2\sigma }{{\mathrm{\pi \epsilon }}_0}\left[\frac{a^2}{8z^2}\right] \\ =\frac{\sigma a^2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{z}}^2} \\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q}{z^2} \end{gathered}$$

Thus from above result it is clear that, the square sheet acts as a point change when z >> a.

Therefore, the electric field due to square plate z >> a is $E=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q}{z^2}$.