Potential due to two infinite parallel planes

Click For Summary
SUMMARY

The discussion focuses on calculating the electric potential due to two infinite parallel planes of charge, one located in the yz plane and the other at a distance x = a. For equal positive charge densities (+σ), the potential is defined as zero at x = 0. In the case of opposite charge densities, the potential is determined to be V = -σa/ε₀ at x = 0, as the potential difference between the planes is established. The user initially struggled with the concept but later clarified the relationship between electric field and potential.

PREREQUISITES
  • Understanding of electric fields and potentials
  • Familiarity with Gauss's Law
  • Knowledge of charge density concepts
  • Basic calculus for potential calculations
NEXT STEPS
  • Study the application of Gauss's Law in electrostatics
  • Learn about electric field calculations for infinite planes of charge
  • Explore the relationship between electric field and electric potential
  • Investigate the effects of varying charge densities on potential
USEFUL FOR

This discussion is beneficial for physics students, electrical engineers, and anyone studying electrostatics, particularly those interested in understanding electric potentials due to charged planes.

Lisa...
Messages
189
Reaction score
0
Consider two infinite parallel planes of charge, one in the yz plane and the other at distance x = a. (a) Find
the potential everywhere in space when V = 0 at x = 0 if the planes carry equal positive charge densities + σ. (b)
Repeat the problem with charge densities equal and opposite, and the charge in the yz plane positive.


I haven't had any problems with part a. As for part b, E=0 if x > a (the field lines of both planes cancel) , so I figured the potential needs to be 0 too, but the answer should be V= [tex]\frac{- \sigma a}{\epsilon_0}.[/tex] Why is that?

EDIT: I just figured it out by myself :)
 
Last edited:
Physics news on Phys.org
The potential difference between the planes is V = \frac{- \sigma a}{\epsilon_0}, so the potential at x=0 must be - \frac{ \sigma a}{\epsilon_0}.
 

Similar threads

Replies
9
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
4
Views
5K
Replies
26
Views
2K
  • · Replies 58 ·
2
Replies
58
Views
6K
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
569
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K