# Electric Field between two large oppositely charged metal plates

1. Sep 25, 2010

### AlexChandler

1. The problem statement, all variables and given/known data
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.

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

3. 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

#### Attached Files:

• ###### Gauss.jpg
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2. Sep 25, 2010

### hikaru1221

The E-field at the surface of the conductor $$E=\sigma / \epsilon$$ 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.