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An infinitely long sheet of charge of width lies in the x y-plane between $-\frac{L}{2}$to $\frac{L}{2}$.The surface charge density $畏$.

The electric field due to small element in the sheet is,

$dE=\frac{畏dx}{2蟺{蔚}_0\sqrt{\left(x^2+z^2\right)}}\cos 胃\hat{k}\left(\mathrm{l}\right)$

${蔚}_0$ is the vacuum permittivity.

$畏$is the surface charge density.

d E is the electric field due to small element.

In trianglelocalid="1650104006638" $\mathrm{ABC}$

localid="1650104011399" $\cos 胃=\frac{z}{\sqrt{\left(x^2+z^2\right)}}$

Substitute localid="1650104022042" $\frac{z}{\sqrt{\left(x^2+z^2\right)}}$for$\cos 胃$in equation (I)

localid="1650104015937" $dE=\frac{畏dx}{2蟺{蔚}_0\sqrt{\left(x^2+z^2\right)}}\left(\frac{z}{\sqrt{\left(x^2+z^2\right)}}\right)\hat{k}$

localid="1650104026083" $=\frac{畏z}{2蟺{蔚}_0}\frac{dx}{\left(x^2+z^2\right)}\hat{k}$

The total electric field due to infinite sheet is,

localid="1650104030518" $E={鈭�}_{\frac{-L}{2}}^{\frac{L}{2}}\frac{畏z}{2蟺{蔚}_0}\frac{dx}{\left(x^2+z^2\right)}\hat{k}$

localid="1650104034469" $=\frac{畏z}{2蟺{蔚}_0}\frac{1}{z}{\left[{\tan }^{-1}\left(\frac{x}{z}\right)\right]}_{\frac{-L}{2}}^{\frac{L}{2}}\hat{k}$

Given info: An infinitely long sheet of charge of width L lies in the

x y-plane between $-\frac{L}{2}$to $\frac{L}{2}$. The surface charge density $畏.$

The total electric field expression is,

$E=\frac{畏}{蟺{蔚}_0}{\tan }^{-1}\left(\frac{L}{2z}\right)\hat{k}$

if $z>>L$ the value of ${\tan }^{-1}\left(\frac{L}{2z}\right)$is tends to infinity. So the value of total electric field is also tends to infinity.

If $z<<L$the value of ${\tan }^{-1}\left(\frac{L}{2z}\right)$ is tends to zero. So the value of total electric field is also tends to zero.

An infinitely long sheet of charge of width L lies in the x y-plane between $-\frac{L}{2}$to $\frac{L}{2}$. The surface charge density $畏.$