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Electric field due to an uniformly charged infinite sheet at a distance $x$, if we calculate using Gauss Law, is ${\bf \frac{σ}{2ε_0}}$. But, How to calculate the electric field without using Gauss Law? Can it be done?

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  • $\begingroup$ You need Maxwell's equations in some form to do a computation In this case that means Gauss's law. $\endgroup$
    – my2cts
    May 11, 2021 at 10:42

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Consider a circular disc on the infinite sheet with radius $r$.

As an exercise, you can prove that electric field at a distance $a$ on the axis of a circular disc from its centre having uniform surface charge density $\sigma$ and radius $r$ is given as $\frac{\sigma}{2\epsilon_o}\Big[1-\frac{a}{\sqrt{a^2+r^2}}\Big]$
[Hint- Electric field due to a charged circular ring (having total charge $Q$ uniformly distributed) of a radius $r$ at a distance $a$ from its centre on its axis is given as $\frac{kQa}{(a^2+r^2)^\frac{3}{2}}$. Now take this elemental ring on the disc and use integration to find electric field.]

As the sheet is infinite so the disc which we have considered in the sheet, we have to take the limit of its radius to infinity.
So, electric field due to sheet is $E=\lim\limits_{r \to\infty} \frac{\sigma}{2\epsilon_o}\Big[1-\frac{a}{\sqrt{a^2+r^2}}\Big]$
So, $E=\frac{\sigma}{2\epsilon_o}$

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No you cannot do a calculation without Maxwell's equations in some form. In this case that means you need Gauss's law.

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