Question

When we find the electric field between the plates of a parallel plate capacitor we assume that the electric field from both plates is

E=σ2ϵ0n.^The factor of two in the denominator comes from the fact that there is a surface charge density on both sides of the (very thin) plates. This result can be obtained easily for each plate. Therefore when we put them together the net field between the plates is E=σϵ0n^ and zero everywhere else. Here, σ is the surface charge density on a single side of the plate, or Q/2A , since half the charge will be on each side.
But in a real capacitor the plates are conducting, and the surface charge density will change on each plate when the other plate is brought closer to it. That is, in the limit that the two plates get brought closer together, all of the charge of each plate must be on a single side. If we let d denote the distance between the plates, then we must havelimd→0E=2σϵ0n^ which disagrees with the above equation. Where is the mistake in this reasoning?
Or more likely, do our textbook authors commonly assume that we are in this limit, and that this is why the conductor behaves like a perfectly thin charged sheet?

Answer 1

The mistake in the reasoning comes from assuming that in a real capacitor, all of the charge of each plate must be on a single side when the plates are brought closer together. The net electric field between the plates of a parallel plate capacitor is not doubled when the distance between the plates is reduced to zero.

Step-by-step explanation:

The mistake in the reasoning comes from assuming that in a real capacitor, all of the charge of each plate must be on a single side when the plates are brought closer together. This assumption is not correct when considering the behavior of conducting plates. In a real capacitor, as the plates get closer together, the charge will distribute itself on both sides of each plate.

Therefore, the net electric field between the plates of a parallel plate capacitor is not doubled when the distance between the plates is reduced to zero. The electric field remains as E=σϵ0n^ and zero everywhere else, just like in the case where the plates are not brought closer together.