An infinitely long sheet of charge of width L lies in the xy-plane between x = - L{\rm{/}}2 and x = L/2. The surface charge density is \eta. Derive an expression for the electric field \vec E along the x-axis for points outside the sheet ({x > L/2}) for x >0.
Express your answer in terms of the variables \eta, x, L, unit vector \hat i, and appropriate constants.

The magnitude of the electric field of force between two charge particle is directly proportional to the charge and inversely proportional to the square of distance.

The expression for the electric field along the x-axis for points outside the sheet is,

[tex]\vec E=\dfrac{\eta}{\varepsilon_0} \left [tan_{-1}\dfrac{L}{2z} \right ]\\[/tex]

Given information-

An infinitely long sheet of charge of width L lies in the xy-plane between,

[tex]-\dfrac{L}{2}[/tex] and [tex]\dfrac{L}{2}[/tex].

The surface charge density is [tex]\eta[/tex].

The electric field is [tex]\vec E[/tex].

What is electric field?

The magnitude of the electric field of force between two charge particle is directly proportional to the charge and inversely proportional to the square of distance.

According to Guess law,

[tex]\oint \vec E \vec {ds}=\dfrac{dq}{2z\varepsilon_0 y }\\\oint \vec E \vec {ds}=\dfrac{\eta dx}{2z\varepsilon_0 \sqrt{x^2+y^2} }\\\int \vec E \vec {ds}\cos \theta=\dfrac{\eta dx}{2z\varepsilon_0 \sqrt{x^2+y^2} }\cos \theta[/tex]

Here, [tex]\cos \theta[/tex] is ratio of adjacent to the hypotenuse. From the attached figure by Pythagoras theorem,

[tex]\cos \theta =\dfrac{z}{\sqrt{x^2+z^2}}[/tex]

Put values,

[tex]\int dE\cos 0^o=\dfrac{\eta dx}{2z\varepsilon_0 \sqrt{x^2+z^2} } \times\dfrac{z}{\sqrt{x^2+z^2} }[/tex]

[tex]\int \vec dE=\dfrac{\eta dx}{2\varepsilon_0 ({x^2+z^2} )}[/tex]

Integrate,

[tex]\vec E=\int\limits^{L/2}_{-L/2} {\dfrac{\eta dx}{2\varepsilon_0 ({x^2+z^2} )}} \\\\vec E=\dfrac{\eta}{2\varepsilon_0} \left [tan_{-1}\dfrac{x}{z} \right ]^{L/2}_{-L/2}\\\vec E=\dfrac{\eta}{2\varepsilon_0} \left [tan_{-1}\dfrac{L}{2z} -tan_{-1}\dfrac{-L}{2z} \right ]\\\vec E=\dfrac{\eta}{2\varepsilon_0} \left [2tan_{-1}\dfrac{L}{2z} ]\\\vec E=\dfrac{\eta}{\varepsilon_0} \left [tan_{-1}\dfrac{L}{2z} \right ]\\[/tex]

Hence, the expression for the electric field along the x-axis for points outside the sheet is,

[tex]\vec E=\dfrac{\eta}{\varepsilon_0} \left [tan_{-1}\dfrac{L}{2z} \right ]\\[/tex]

Learn more about the electric field here;

https://brainly.com/question/12757739