# Confusion about surface charge density and electric field intensity of an infinite plate

consider a plate with some thickness. If both surface of the plate have a total of q charge and the area of each surface is A then would the charge density be q/A or q/2A ?

And what would be the electric field intensity of the plate . would it be Charge density/ 2 × epsilon or charge density/ epsilon

I actually have come across this confusion in my textbook. It states the above two different formula of intensity for an infinite plate. So I thought there might be some confusion regarding charge density

If both the surfaces are having a charge q, to avoid confusion use this:-

Consider a point somewhere. The charge on one side of the sheet is q and on the other side is also q. The electric field due to first side at that point is q/2epsilon and due to the other side also it is q/2epsilon. Net electric field will be q/2epsilon + q/2epsilon = q/epsilon. similarly the charge density on each side is q/2A and total charge density is q/A

If both surface of the plate have a total of q charge and the area of each surface is A then would the charge density be q/A or q/2A?

The field of a charged plate, in vacuum, will always be $E=\frac \sigma {\epsilon_0}$, where $\sigma$ is the surface charge density on the side of the plate where the field is measured. But, for a given $q$ and $A$, the charge density will depend on the distribution of charge between two surfaces of the plate.

In the absence of other nearby conductive objects or an external field, in other words, in a free standing conductive plate, the charge will be evenly distributed between the two surfaces. Correspondingly, the charge density on each surface will be $\frac q {2A}$ and the field on either side of the plate will be $E=\frac q {2\epsilon_0}$

On the other hand, if the plate is one of the plates of a capacitor (plate A on the diagram below), most of its charge, and, in an infinitely large capacitor, all of its charge, will migrate to one, inner, side (attracted by the charge of the opposite sign on the second plate, B). Correspondingly, the charge density on the inner surface of the plate will be close to $\frac q A$ with the field close to $E=\frac q {\epsilon_0}$ and the charge density on the outer surface of the plate will be close to zero with the field close to zero as well. 