# Electric field nearby an inifinite plane

Say we have an infinite plane with some surface charge density $$\sigma$$. If we build a Gaussian surface that encloses both of its sides, then the electric field is given by:

$$\oint_S \vec{E}\cdot d\vec{A}=\frac{\sigma \cdot A}{\epsilon_0} \Longrightarrow E=\frac{\sigma}{2\epsilon_0}$$

Because, since the Gaussian surface encloses both sides of the plane, then there is 2 equal areas through which electric field lines are going, which yields $$\oint_S dA = 2A$$.

Nonetheless, my textbook states the electric field immediately outside any charged conductor in electrostatic equilibrium is $$E=\frac{\sigma}{\epsilon_0}$$. How can I make sense of this? I've been doing some research and I think the electric field I calculated first is the electric field on the actual surface of the plane, whilst the latter is the electric field immediately outside the plane. Is this correct? If so, if we were to place a charge 3 metres away from our infinite plane, which of both electric fields would affect said charge?

• "Nonetheless, my textbook states the electric field immediately outside any charged conductor in electrostatic equilibrium is $E=\frac{\sigma}{\epsilon_0}$." We need to see this statement in context. Aug 7, 2023 at 20:10

That's because the electric field inside the conductor ("at electrostatic equilibrium") is zero. So only one side of your Gaussian surface has any flux through it.

Your first expression would be valid for the electric field either side of a charged, insulating surface.

If the context you are referring to is a capacitor then there is another plate with the same but opposite charge but now the total charge on the two plates is zero and on the outside, not between the plates, the field is essentially zero when the plates are much closer to each other than their linear dimensions. Now between the plates you get $$E=\tfrac{\sigma}{\epsilon_0}$$

if you load one side of a conductor, the other side will have the same charge, half of what you put on it. So you have 2*A for the $$\sigma$$ or you take your surface through the inner of the plate and get only the one E at one side.

This has been a puzzle to many students when they start to learn about Gauss's law.

If the conducting plane is assumed to be infinite in two dimensions but not the third then the charge distributes itself equally on both sides of the plane and the magnitude of the resulting electric field in each side is $$\dfrac {\sigma}{2\,\epsilon_0}$$.
then the field on one side is $$\dfrac \sigma {\epsilon_0}$$ and zero on the other side.