Question

derivation of internal electric field of capacitor

Answer 1

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.^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^E=σϵ0n^ and zero everywhere else. Here, σσ is the surface charge density on a single side of the plate, or Q/2AQ/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 dd denote the distance between the plates, then we must have

limd→0E=2σϵ0n^limd→0E=2σϵ0n^