Question

Question: Find the electric field at a height z above the center of a square sheet (side a) carrying a uniform surface charge $\mathit{\sigma }$. Check your result for the limiting

cases $\mathit{a}\mathbf{\to }\mathbf{\infty }$and $\mathit{z}\mathbf{>}\mathbf{>}\mathit{a}$.

Short answer

Answer

The electric filed is due to square plate is when $\mathit{a}\mathbf{\to }\mathbf{\infty }$is $\mathit{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{\mathbf{q}}{{\mathbf{z}}^{\mathbf{2}}}$.

The electric field due to square plate $\mathit{z}\mathbf{>}\mathbf{>}\mathit{a}$is $\mathit{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{\mathbf{q}}{{\mathbf{z}}^{\mathbf{2}}}$.

Step-by-step solution

Define functions

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

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

Here, E is the electric filed, $\mathit{\lambda }$is the linear charge density, ${\mathit{\epsilon }}_{\mathbf{0}}$is the permittivity for the free space, a 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.

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

Here, $\mathit{\sigma }$ is the charge density.

Determine linear charge density

Differentiating the equation (1) on both sides,

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

$$\begin{gathered} \mathit{d}\mathit{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{\mathbf{4}\mathbf{a}\mathbf{z}\mathbf{\sigma }\left({\displaystyle \frac{da}{2}}\right)}{\left(z^2+{\displaystyle \frac{a^2}{4}}\right)\sqrt{{\mathbf{z}}^{\mathbf{2}}\mathbf{+}{\displaystyle \frac{{\mathbf{a}}^{\mathbf{2}}}{\mathbf{2}}}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\left(\frac{4\sigma z}{2}\right)\frac{\mathbf{a}\mathbf{d}\mathbf{a}}{\left(z^2+{\displaystyle \frac{a^2}{4}}\right)\sqrt{{\mathbf{z}}^{\mathbf{2}}\mathbf{+}{\displaystyle \frac{{\mathbf{a}}^{\mathbf{2}}}{\mathbf{2}}}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{\sigma }\mathbf{z}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\frac{\mathbf{a}\mathbf{d}\mathbf{a}}{\left(z^2+{\displaystyle \frac{a^2}{4}}\right)\sqrt{{\mathbf{z}}^{\mathbf{2}}\mathbf{+}{\displaystyle \frac{a^2}{2}}}} \end{gathered}$$

Thus, the differential equation solution is $\frac{\mathbf{\sigma z}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{\mathbf{ada}}{\mathbf{(}{\mathbf{z}}^{\mathbf{2}}\mathbf{+}{\displaystyle \frac{a^2}{4}}\mathbf{)}\sqrt{{\mathbf{z}}^{\mathbf{2}}\mathbf{+}{\displaystyle \frac{a^2}{2}}}}$.

Determine Electric field

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

$\mathit{E}\mathbf{=}\frac{\mathbf{\sigma }\mathbf{z}}{\mathbf{2}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\underset{\mathbf{0}}{\overset{\mathbf{a}}{\mathbf{\int }}}\frac{\mathbf{a}}{\left(z^2+{\displaystyle \frac{a^2}{4}}\right)\sqrt{{\mathbf{z}}^{\mathbf{2}}\mathbf{+}{\displaystyle \frac{{\mathbf{a}}^{\mathbf{2}}}{\mathbf{2}}}}}\mathit{d}\mathit{a}$ ………(2)

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

$$\begin{gathered} \mathit{E}\mathbf{=}\frac{\mathbf{\sigma }\mathbf{z}}{\mathbf{2}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\underset{\mathbf{0}}{\overset{\frac{{\mathbf{a}}^{\mathbf{2}}}{\mathbf{4}}}{\mathbf{\int }}}\frac{\mathbf{a}}{\left(z^2+t\right)\sqrt{\left(z^2+2t\right)}}\frac{\mathbf{2}}{\mathbf{a}}\mathit{d}\mathit{t} \\ \mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{\sigma }\mathbf{z}}{\mathbf{2}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\underset{\mathbf{0}}{\overset{\frac{{\mathbf{a}}^{\mathbf{2}}}{\mathbf{4}}}{\mathbf{\int }}}\frac{\mathbf{1}}{\left(z^2+t\right)\sqrt{\left(z^2+t\right)\sqrt{{\mathbf{z}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{t}}}}\mathit{d}\mathit{t} \end{gathered}$$

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

Solving further based on the above tangential formula,

$$\begin{gathered} \mathit{E}\mathbf{=}\frac{\mathbf{\sigma }\mathbf{z}}{\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}{\left[\frac{2}{z}ta{n}^{-1}\left(\frac{\sqrt{\left(z^2+2t\right)}}{z}\right)\right]}_{\mathbf{0}}^{\frac{{\mathbf{a}}^{\mathbf{2}}}{\mathbf{4}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{2}\mathbf{\sigma }}{\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\left[ta{n}^{-1}\left(\frac{\sqrt{z^2+2\left({\displaystyle \frac{a^2}{4}}\right)}}{z}\right)-ta{n}^{-1}\left(\frac{\sqrt{z^2+2\left(0\right)}}{z}\right)\right] \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{2}\mathbf{\sigma }}{\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\left[ta{n}^{-1}\left(\frac{\sqrt{z^2+{\displaystyle \frac{a^2}{2}}}}{z}\right)-ta{n}^{-1}\left(1\right)\right] \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{2}\mathbf{\sigma }}{\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\left[ta{n}^{-1}\left(\frac{\sqrt{z^2+{\displaystyle \frac{a^2}{2}}}}{z}\right)-ta{n}^{-1}\left(tan\frac{\pi }{4}\right)\right] \end{gathered}$$

Simplifying the expression further,

$$\begin{gathered} \mathit{E}\mathbf{=}\frac{\mathbf{2}\mathbf{\sigma }}{{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathbf{[}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left(\frac{\sqrt{z^2+{\displaystyle \frac{a^2}{2}}}}{z}\right)\mathbf{-}\frac{\mathbf{\pi }}{\mathbf{4}}\mathbf{]} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{2}\mathbf{\sigma }}{{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{\mathbf{\pi }}{\mathbf{4}}\mathbf{[}\left[\frac{4}{\pi }{\tan }^{-1}\right]\left(\frac{\sqrt{z^2+{\displaystyle \frac{a^2}{2}}}}{z}\right)\mathbf{-}\mathbf{1}\mathbf{]} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{\sigma }}{\mathbf{2}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathbf{[}\frac{\mathbf{4}}{\mathbf{\pi }}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left(\sqrt{1+\left(\frac{a^2}{2z^2}\right)}\right)\mathbf{-}\mathbf{1}\mathbf{]} \end{gathered}$$

Thus, the electric filed is $\mathit{E}\mathbf{=}\frac{\mathbf{\sigma }}{\mathbf{2}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathbf{[}\frac{\mathbf{4}}{\mathbf{\pi }}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\sqrt{\mathbf{1}\mathbf{+}\mathbf{\left(}\frac{{\mathbf{a}}^{\mathbf{2}}}{\mathbf{2}{\mathbf{z}}^{\mathbf{2}}}\mathbf{\right)}}\mathbf{\right)}\mathbf{-}\mathbf{1}\mathbf{]}$.

If $\mathit{a}\mathbf{\to }\mathbf{\infty }$then the electric field square plate is,

$\mathit{E}\mathbf{=}\frac{\mathbf{\sigma }}{\mathbf{2}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathbf{[}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\mathbf{(}\mathbf{\infty }\mathbf{)}\mathbf{-}\frac{\mathbf{\pi }}{\mathbf{4}}\mathbf{]}$

Since, $\mathit{t}\mathit{a}\mathit{n}\frac{\mathbf{\pi }}{\mathbf{2}}\mathbf{-}\mathbf{\infty }$

$$\begin{gathered} \mathit{E}\mathbf{=}\frac{\mathbf{\sigma }}{\mathbf{2}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathbf{[}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathbf{tan}\frac{\mathbf{\pi }}{\mathbf{2}}\mathbf{\right)}\mathbf{-}\frac{\mathbf{\pi }}{\mathbf{4}}\mathbf{]} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{\sigma }}{\mathbf{2}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathbf{[}\frac{\mathbf{\pi }}{\mathbf{2}}\mathbf{-}\frac{\mathbf{\pi }}{\mathbf{4}}\mathbf{]} \end{gathered}$$

$\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{\sigma }}{\mathbf{2}{\mathbf{\epsilon }}_{\mathbf{0}}}$

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 $\mathit{a}\mathbf{\to }\mathbf{\infty }$is $\mathit{E}\mathbf{=}\frac{\mathbf{\sigma }}{\mathbf{2}{\mathbf{\epsilon }}_{\mathbf{0}}}$.

Determine Electric field due to z>>a

Now, let’s consider that, $\mathit{f}\left(x\right)\mathbf{=}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\sqrt{\mathbf{1}\mathbf{+}\mathbf{x}}\mathbf{-}\frac{\mathbf{\pi }}{\mathbf{4}}$and $\mathit{x}\mathbf{=}\frac{{\mathbf{a}}^{\mathbf{2}}}{\mathbf{2}{\mathbf{z}}^{\mathbf{2}}}$. If $\mathit{z}\mathbf{>}\mathbf{>}\mathit{a}$, then $\frac{{\mathbf{a}}^{\mathbf{2}}}{\mathbf{2}{\mathbf{z}}^{\mathbf{2}}}\mathbf{<}\mathbf{<}\mathbf{1}$and $\mathit{f}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$. Then the value of $\mathit{f}\mathbf{\text{'}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$is,

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

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

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

Applying Taylor series and solving,

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

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

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

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

Therefore, the electric filed due to square plate when $\mathit{z}\mathbf{>}\mathbf{>}\mathit{a}$is,

$$\begin{gathered} \mathit{E}\mathbf{=}\frac{\mathbf{2}\mathbf{\sigma }}{\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\left[\frac{a^2}{8z^2}\right] \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{\sigma }{\mathbf{a}}^{\mathbf{2}}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}{\mathbf{z}}^{\mathbf{2}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\frac{\mathbf{q}}{{\mathbf{z}}^{\mathbf{2}}} \end{gathered}$$

Thus from above result it is clear that, the square sheet acts as a point change when $\mathit{z}\mathbf{>}\mathbf{>}\mathit{a}$.

Therefore, the electric field due to square plate $\mathit{z}\mathbf{>}\mathbf{>}\mathit{a}$is $\mathit{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{\mathbf{q}}{{\mathbf{z}}^{\mathbf{2}}}$.