Electric Field between two large oppositely charged metal plates

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SUMMARY

The discussion focuses on calculating the electric field between two large oppositely charged metal plates using two methods: Gauss's Law and the superposition principle. The first method, applying Gauss's Law, yields the correct electric field value, while the second method results in an incorrect value due to a misunderstanding of the superposition principle. Specifically, the error arises from not isolating the electric field contribution from a single plate when using superposition, leading to a doubled result. The correct approach requires considering the electric field from one plate at a time.

PREREQUISITES
  • Understanding of Gauss's Law and its application in electrostatics
  • Familiarity with the concept of electric field and charge density
  • Knowledge of the superposition principle in electric fields
  • Basic principles of conductors in electrostatic equilibrium
NEXT STEPS
  • Study the application of Gauss's Law in different geometries, including cylindrical and spherical cases
  • Learn about the concept of electric field lines and their relationship to charge distributions
  • Explore the implications of superposition in electric fields with multiple charge sources
  • Investigate the behavior of electric fields in conductors and insulators under various conditions
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Students studying electromagnetism, physics educators, and anyone interested in understanding electric fields in electrostatic systems.

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
 

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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.
 

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