# Why is electric field constant at any point due to infinite plane of charge while a finite plane of charge can give the same result ?

We know that the electric field is constant at any point irrespective of the radius due to a plane sheet of charge of infinite extent and can be proved by Gauss theorem however I don't understand why do we need an infinite length because we can apply the same gauss theorem to a finite length of plane sheet of charge and we get the same result. Also incase of capacitors, the electric field is taken constant even when the sheets are finite.

I have another question which is that the electric field due to an infinite linear charge distribution at any point is inversely proportional to the radius but we can apply the Gauss theorem there too and prove that electric field can be constant at any point irrespective of the radius by drawing a cuboid perpendicular to the linear charge.

• If the sheet of charge is finite then the electric field magnitude (and vector direction) will depend upon location with respect to the finite sheet of charge. So no, the answer will NOT be the same. Commented Jan 3, 2018 at 14:20
• But for capacitors we consider the electric field to be constant even tho it's finite. And we can apply Gauss law on any finite charge and draw a cylinder across it and say that the electric field is constant Commented Jan 3, 2018 at 14:21
• That is an approximation used when you have two very closely spaced (with respect to the size of the parallel plates) sheets of charge held apart by mechanical forces and one uses a dielectric to contain the fields between. This is very different than an isolated, finite sheet of charge. Commented Jan 3, 2018 at 14:24
• But like I don't understand why we only need an infinite plane sheet of charge, because if we take a finite sheet of charge and apply Gauss theorem to it then we will also get the same result. Is it because the electric field lines will not be perpendicular to the surface of the plane if it's finitie? Apart from that I don't find any reason. And what about a infinitely long uniformly linear charged rod, it also produces electric field but the electric field gets reduced/increased proportional to the radius. But here too we can apply Gauss law and say electric field is constant ? Commented Jan 3, 2018 at 14:40
• If you are on the axis of symmetry for the finite sheet or not on the wire in the 2nd case, the electric field should be one dimensional. The problem with the finite sheet is that the electric field is not constant with distance. Eventually if you are far enough way, the finite sheet of charge can be approximated as a point charge. Does that make sense? Commented Jan 3, 2018 at 15:05