# How to calculate Electric field due to an uniformly charged infinite sheet without Gauss Law?

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?

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

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.]
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, $$E=\frac{\sigma}{2\epsilon_o}$$