# What is the electric field produced by a conductor “plate”?

I know that the electric field is zero inside a conductor. However, a plate does not have some kind of "interior space." So, if there is a infinitely large conductor plate with a uniform charge density, what is the electric field made by the plate? Is it just the same as a plate with uniform charge distribution? It is so much confusing...

• An infinitely large plate with a finite net change has a charge density of zero and no electric field. Maybe change it to a charge density of $\sigma$ instead of a net charge of $Q$. – Ben51 Jan 17 '18 at 13:30
• Ok I will edit it. What is the answer then? Could you tell me? – Keith Jan 17 '18 at 13:31
• The field only cares where the charges are, not whether they’re in a conductor or not. So it’s the same. But in general, you’re not allowed to specify the distribution of charge on a conductor: it will rearrange itself to eliminate the field inside and make the field just outside normal to the surface. – Ben51 Jan 17 '18 at 13:36

The electric field from an infinite single plane of charge is given by $$\vec{E}=\frac{\sigma}{2\epsilon_0}\hat{n},$$ where $\sigma$ is the area charge density and $\hat{n}$ is the unit vector normal to the plane and away from the plane on both sides.
A conducting plate necessarily has 2 planes of charge because to be a conductor it must be material object with 2 different sides. The charge will rearrange until there is a plane of charge on each side of the plate. At some distance away from the surface, the net electric field will be the sum of the fields which will be $$\vec{E}=\frac{\sigma}{\epsilon_0}\hat{n}.$$