91影视

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

role="math" localid="1657466514484" $E=\frac{1}{4蟺{蔚}_0}\frac{4位az}{\left(z^2+\frac{a^2}{4}\right)\sqrt{z^2+\frac{a^2}{2}}}\widehat{z}$

Here, $E$is the electric filed, $位$is the linear charge density, ${蔚}_0$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.

$d位=蟽\left(\frac{da}{2}\right)dE=\frac{1}{4蟺{蔚}_0}\frac{4az}{\left(z^2+\frac{a^2}{4}\right)\sqrt{z^2+\frac{a^2}{2}}}d位$

Here, $蟽$is the charge density.

Differentiating the equation $\left(1\right)$on both sides,

Substitute $蟽\left(\frac{da}{2}\right)$for $d位$.

width="233">dE=14蟺蔚04az蟽da2z2+a24z2+a22

$=\frac{1}{4蟺{蔚}_0}\left(\frac{4蟽z}{2}\right)\frac{ada}{\left(z^2+\frac{a^2}{4}\right)\sqrt{z^2+\frac{a^2}{2}}}$

$=\frac{蟽z}{2蟺{蔚}_0}\frac{ada}{\left(z^2+\frac{a^2}{4}\right)\sqrt{z^2+\frac{a^2}{2}}}$

Thus, the differential equation solution is$=\frac{蟽z}{2蟺{蔚}_0}\frac{ada}{\left(z^2+\frac{a^2}{4}\right)\sqrt{z^2+\frac{a^2}{2}}}$

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

$E=\frac{蟽z}{2蟺{蔚}_0}{鈭�}_0^a鈥�\frac{a}{\left(z^2+\frac{a^2}{4}\right)\sqrt{z^2+\frac{a^2}{2}}}da$

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

$\begin{array}{r}E=\frac{蟽z}{2蟺{蔚}_0}{鈭�}_0^{\frac{a^2}{4}}鈥�\frac{a}{\left(z^2+t\right)\sqrt{\left(z^2+2t\right)}}\frac{2}{a}dt\end{array}$

$=\frac{蟽z}{蟺{蔚}_0}{鈭�}_0^{\frac{a^2}{4}}鈥�\frac{1}{\left(z^2+t\right)\sqrt{\left(z^2+2t\right)}}dt$

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

Solving further based on the above tangential formula,

$E=\frac{蟽z}{蟺{蔚}_0}{\left[\frac{2}{z}{\tan }^{鈭�1}鈦�\left(\frac{\sqrt{\left(z^2+2t\right)}}{z}\right)\right]}_0^{\frac{a^2}{4}}$
$=\frac{2蟽}{蟺{蔚}_0}\left[{\tan }^{鈭�1}鈦�\left(\frac{\sqrt{z^2+2\left(\frac{a^2}{4}\right)}}{z}\right)鈭�{\tan }^{鈭�1}鈦�\left(\frac{\sqrt{z^2+2\left(0\right)}}{z}\right)\right]$

$=\frac{2蟽}{蟺{蔚}_0}\left[{\tan }^{鈭�1}鈦�\left(\frac{\sqrt{z^2+\frac{a^2}{2}}}{z}\right)鈭�{\tan }^{鈭�1}鈦�\left(1\right)\right]$

$=\frac{2蟽}{蟺{蔚}_0}\left[{\tan }^{鈭�1}鈦�\left(\frac{\sqrt{z^2+\frac{a^2}{2}}}{z}\right)鈭�{\tan }^{鈭�1}鈦�\left(\tan 鈦�\frac{蟺}{4}\right)\right]$

Simplifying the expression further,

$\begin{array}{r}E=\frac{2蟽}{蟺{蔚}_0}\left[{\tan }^{鈭�1}鈦�\left(\frac{\left.\left.\sqrt{z^2+\frac{a^2}{2}}\right)鈭�\frac{蟺}{4}\right]}{z}\right]鈭�1\right]\end{array}$

$=\frac{2蟽}{蟺{蔚}_0}\frac{蟺}{4}\left[\left[\frac{4}{蟺}{\tan }^{鈭�1}\right]\left(\frac{\sqrt{z^2+\frac{a^2}{2}}}{z}\right]鈭�1\right]$

localid="1657466675640" $=\frac{蟽}{2蟺{蔚}_0}\left[\frac{4}{蟺}{\tan }^{鈭�1}鈦�\left(\sqrt{1+\left(\frac{a^2}{2z^2}\right.}\right)-1\right]$

Thus, the electric filed is$\begin{array}{r}E=\frac{蟽}{2蟺{蔚}_0}\left[\frac{4}{蟺}{\tan }^{鈭�1}鈦�\left(\sqrt{1+\left(\frac{a^2}{2z^2}\right.}\right)-1\right]\end{array}$

If$a鈫�鈭�$then the electric field square plate is,

$E=\frac{2蟽}{蟺{蔚}_0}\left[{\tan }^{鈭�1}鈦�\left(\mathrm{鈭�}\right)鈭�\frac{蟺}{4}\right]$

Since, $\tan \frac{蟺}{2}=鈭�$

$E=\frac{2蟽}{蟺{蔚}_0}\left[{\tan }^{鈭�1}\left(\tan \frac{蟺}{2}\right)鈭�\frac{蟺}{4}\right]$

$=\frac{2蟽}{蟺{蔚}_0}\left[\frac{蟺}{2}鈭�\frac{蟺}{4}\right]$

$=\frac{蟽}{2{蔚}_0}$

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$a鈫�\mathrm{鈭�}$is$E=\frac{蟽}{2{蔚}_0}$

Now, let鈥檚 consider that,$f\left(x\right)={\tan }^{-1}\sqrt{1+x}-\frac{x}{4}$and$x=\frac{a^2}{2z^2}$. If$z>>a$, then$\frac{a^2}{2z^2}<<1$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,

role="math" localid="1657468740239" $f\left(x\right)=f\left(0\right)+xf\text{'}\left(0\right)+\frac{1}{2}{x}^{2}f\text{'}\text{'}\left(x\right)+...$

$鈮�f\left(0\right)+xf\text{'}\left(0\right)$

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

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

$=\frac{x}{4}$

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,

$E=\frac{2蟽}{蟺{蔚}_0}\left[\frac{a^2}{8z^2}\right]$

$=\frac{蟽a^2}{4蟺{蔚}_0{z}^2}$

$=\frac{1}{4蟺{蔚}_0}\frac{q}{z^2}$

Thus from above result it is clear that, the square sheet acts as a point change whenrole="math" localid="1657470504883" $z>>a$.

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