A flat conducting sheet of area A has a charge q on each surface.....?
What is the electric field inside the sheet?
Use Gauss's law to show that the electric field just outside the sheet is E=q/(εoA)=σ/εo
- hfshawLv 71 decade agoFavorite Answer
The electric field inside any conductor at equilibrium is zero. If there were an electric field, the electrons in the material, which are free to move because it's a conductor, would move around until they reached equilibrium positions, at which point they would no longer move (neglecting thermal motion). The fact that they are no longer moving means they are not subject to a net force, hence there can be no electric field at equilibrium.
You should complain to your teacher/professor about the second part of this problem; it's poorly posed. The first part of the question implies that the sheet has a nonzero thickness, which means that the sheet is actually a thin rectangular prism. Each surface on this solid would then have a surface charge at equilibrium, not just the "large" surfaces; however, the wording of the second part of the problem (and the result you are supposed to obtain) really only hold for an infinite sheet with charge density σ on the two sides of the sheet. Nevertheless, let's proceed as though we have a finite sheet of area A that only has charge density on the "large" surfaces.
Gauss' Law states that the total "electric flux" through any closed surface that encloses a charge Q is given by Q/εo. The electric flux through a surface is defined as the dot product of the electric field vector and the normal to the surface (i.e., d(flux) = |E|*cos(theta)*dA, where d(flux) is the differential flux through differential area dA, |E| is the magnitude of the electric field, and theta is the angle between the field and the normal to dA).
One normally picks a "Gaussian surface" that simplifies the math. In this case, let's use a surface that has sides that are perpendicular to the plane of the sheet and that conform to the shape of the sheet. This surface has a top and bottom that have the same shape as the sheet, but that are displaced a small distance on either side of the sheet. (For instance, if the sheet were a rectangle, our surface would be a rectangular box bisected by the sheet. If the sheet were a circle, our surface would be a cylinder bisected by the sheet. If the sheet were a triangle, the surface would be a triangular prism, etc.)
We know that the electric field due to the sheet (so long as we are close enough that the sheet looks infinite in extent) is perpendicular to the sheet. It it weren't, the component of the field parallel to the surface would induce the charges in the sheet to move until they reached a equilibrium positions. (This would change the field in such a way as to remove any such parallel fields.)
The total charge contained by our surface is then 2*q (because each side of the sheet has charge q, and there are two sdes).
The electric flux through the sides of our surface, which are perpendicular to the sheet, is zero because the field is parallel to the sides of our surface.
The electric flux through the top surface is just E*A, and the flux through the bottom surface is also E*A, so the total flux through our surface is 2*E*A. By Gauss' Law, this equals the total charge enclosed by the surface (which we showed above is equal to 2Q) divided by the permittivity of free space:
2*E*A = 2*Q/εo
E = Q/(A*εo)
The surface charge density, σ, is defined as equal to Q/A, so
E = σ/εo