electric field outside two parallel conducting plates 
Hi, I'm confused about this question. I watched a video, and it said that the electric field in region 1 equals the electric field in region 5, and I wanted to know why. Is it just a property of two parallel conducting plates that the electric field outside either plate is the same? I would understand that if the two metal slabs were 4 non-conducting sheets of charge, the fields in regions 1 and 5 would be equal since the fields from each sheet are additive and don't depend on distance; however, the question says that they are slabs of metal, so you have to use properties of conductors, and you don't find out until later that the charge densities on the outside surface of each plate is the same. Any help is appreciated. Thanks!
 A: Revised Answer
The field in regions 1 and 5 has the same constant magnitude (opposite in direction), independent of distance from the plates (provided this distance is small compared with the width of the plates). This occurs because the plates are parallel and the electric field from each is uniform, independent of distance from the plate. It is true for any number of parallel planes of uniform charge density, and does not depend on them being conductors/insulators.
The electric field from each face of the plates is uniform and points away from that face. Suppose the charge on each face is +ve. Then in regions 1 and 5 the electric fields are all equal and constant, and all pointing in the same direction (all up in region 1, all down in region 5), so they add up to the same value in region 1 as in region 5.
The fact that the plates are conductors makes no difference. The excess charge will be distributed evenly over each face, probably with a different surface charge density on each. Each conducting plate is then equivalent to 2 infinitely thin planes of charged insulating material. 
A: Actually, you are quite right in worrying, why the magnitude of the electric field is the same in areas 1 and 5.  To show that unambiguously, additional assumption of no "external electric field" is required.  That is, in the analysis in the previous answer, if som external field is present, it will add positive flux through one of the bases of the Gaussian Surface cylinder, and negative flux of equal magnitude through the other base, leaving the Gauss theorem unchanged.  If you assume that there is no external field (and that is quite often an implicit assumption in most of the electrostatic problems), then you'll get to the conclusion that was given in the video.
