Electric Field Created by 2 Infinite Plates

Click For Summary
SUMMARY

The discussion centers on the electric field generated by two infinite parallel plates, utilizing Gauss's law for analysis. The electric field due to a single plate is defined as E = σ/(2ε₀), where σ represents surface charge density and ε₀ is the permittivity of free space. When considering two plates with equal charge, the electric fields between the plates cancel each other out, resulting in a net electric field of zero. Conversely, outside the plates, the fields reinforce each other, yielding a total electric field of E = σ/ε₀. This conclusion is rooted in the principles of superposition and the linearity of Maxwell's equations.

PREREQUISITES
  • Understanding of Gauss's Law
  • Familiarity with electric field concepts
  • Knowledge of vector addition in physics
  • Basic principles of Maxwell's equations
NEXT STEPS
  • Study the application of Gauss's Law in various geometries
  • Explore the concept of electric field superposition
  • Learn about the implications of Maxwell's equations in electrostatics
  • Investigate the behavior of electric fields in different configurations of charged plates
USEFUL FOR

Physics students, electrical engineers, and anyone interested in understanding electrostatics and electric field behavior in parallel plate configurations.

Heisenberg7
Messages
101
Reaction score
18
Today, I watched a video about electric field created by an infinite plate by Khan Academy. They were talking about the clever application of the Gauss's law in this case (the cylinder method), so I wondered if I could apply the same thing but to 2 plates. For example, let's say that the plates are parallel. In this case the electric field created by one plate is ##E = \frac {\sigma}{2\epsilon_o}##. Since electric field is a vector quantity we can vectorially add up the electric field created by both plates. Between the plates the electric field created by one plate is opposite and equal to the electric field created by the other, thus if we vectorially add them up, we get 0. But on the left and right side, it's different. They have the same direction and magnitude at each point in space, thus the electric field at any point is ##E = \frac {\sigma}{\epsilon_o}##. Is this the correct way to think about this problem? (both plates have the same charge ##q##)
 
Physics news on Phys.org
Yes, electric fields generated from different sources add up to a superposition of the individual contributions. However, note that this is a result of the linearity of the governing differential equation, ie Gauss’ law or ultimately Maxwell’s equations. It does not follow solely from being a vector field.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 30 ·
2
Replies
30
Views
4K
  • · Replies 7 ·
Replies
7
Views
20K
Replies
8
Views
2K