Electric Field between two large oppositely charged metal plates

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AlexChandler
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Homework Statement


I am trying to calculate the electric field between these two plates... and I am using two different methods to do so. One gives me the correct solution, and the other does not. I am curious as to why the second method is wrong.



Homework Equations


Gauss's Law Phi = Q/e0
E field near conductor = Charge density/ E0
Superposition of electric fields E=E1+E2


The Attempt at a Solution


The first method: Using gauss' law I am able to come up with the correct solution (I used a rectangular gaussian surface containing an area A2 of plate 1 reaching out halfway between the plates. The only face with flux through it is the one parallel to both plates located halfway between them.. The solution comes out as the charge density of the plates devided by the electric constant.
The second method: Using the superposition principle along with the equation for the electric field near a conductor I get twice the answer that I had calculated using Gauss' Law. I have attached a picture of my work in hope that it will help to illustrate my problem. Thanks and hope to hear back soon,
Alex
 
Attachments
  • Handwritten worksheet titled What is E between plates? with a sketch of two parallel charged plates labeled sigma and minus sigma, and two solutions: a Gauss's law derivation giving E = -sigma/epsilon0 marked right, and a superposition attempt giving E-middle = -2 sigma/epsilon0 marked wrong, asking why.
    Gauss.jpg
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The E-field at the surface of the conductor [tex]E=\sigma / \epsilon[/tex] is already taken into account the presence of both plates. That means, what you get is E-field due to both plates on the surface on each plate, which is actually E-field between the plates.
However, in order to apply the superposition principle, you have to find the E-field due to ONE "plate", i.e. one layer of charges on one plate, assuming that the other layer is not there.