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If we have two conducting plates, with charge $Q$ and $-Q$, why is the charge on the outer surfaces of each conductor zero?

I've been trying to wrap my head around the problem. Firstly, don't excess charges on a conductor spread out towards the surface, leading to a contradiction?

How can I use Gauss' law to prove this statement (even though it seems false to me)?

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  • $\begingroup$ Outer sides of plates do have charge. Who said that's not there? $\endgroup$ – Anubhav Goel Feb 13 '16 at 11:36
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Suppose you have an infinite thin sheet of charge with a certain charge density. By symmetry we know the electric field is perpendicular to the sheet. You can calculate the field using this method: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesht.html

$E = \dfrac{\sigma}{2\epsilon_0}$

This field is constant everywhere.

If we have a second sheet with opposite charge density its field is:

$E = -\dfrac{\sigma}{2\epsilon_0}$

If both plates exist, then we can use superposition and sum the two fields. The net field outside the sheets is 0. The net field between the sheets is

$E_{net} = \dfrac{\sigma}{\epsilon_0}$

Now suppose the two sheets are conductors. We know that at equilibrium, the electric field inside a conductor is 0. Otherwise we'd have moving charges. So we can assume the electric field inside these two sheets is 0.

Now look at a small volume that crosses the outersurface . Do gauss' law. We know the electric field outside and inside the outersurface is 0. So the charge within that small volume must be 0.

By contrast, look at the inner surface. Do gauss' law there. This time we know the field, and we're calculating the charge density:

$\dfrac{\sigma_{inner}A}{\epsilon_0} = E_{net}A$

$\dfrac{\sigma_{inner}A}{\epsilon_0} = \dfrac{\sigma A}{\epsilon_0}$

$\sigma_{inner} = \sigma$

This is an idealized situation. We're pretending the parallel plates are infinite sheets, when they're really not.

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I may possibly be very wrong, but wouldn't it be the same as when you get an electron and mesh it with a proton getting a neutron? If you have the same amount of an opposite charge, the charge will disappear..

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  • $\begingroup$ No, because you're actually making each plate charged. The question is why is the excess charge on the surface facing the other plate and not on the opposite (outside) surface. If you had a single conducting plate with excess charge, there would be charge distributed on both surfaces. $\endgroup$ – Bill N Feb 13 '16 at 4:42

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