How Does Gauss' Law Account for the Negative Plate in a Capacitor?

[kash25]

Hi,
I have a problem with electric fields in capacitors. Using Gauss' Law, we always choose the Gaussian surface that encloses the positive plate in the capacitor. When we go through field calculations, we get an equation for the electric field and we use that to find voltage, capacitance, and everything else. What I don't understand is how enclosing ONLY the positive plate takes into account the fact that the negative plate is present and that it amplifies the field between the two. In other words, how would our calculations or conclusions be any different if the negative plate were not even present?
Thank you very much for the help

 

[alphysicist]

Hi kash25,

I don't think that's correct. For example, for parallel plates we might use Gauss's law to find the electric field of just the positive plate (giving a result of E = \sigma/(2\epsilon_0) everywhere), but then we would double that to find the electric field of the capacitor between the plates (E=\sigma/\epsilon_0) and set the field to be zero outside the plates.

(Between the plates, the fields are in the same direction and so add; outside the plates they are in opposite directions and therefore cancel; also I'm making the usual approximation that the nonuniform field at the ends of the plate can be neglected, etc.)