Understanding the Contradiction in Gauss Law for Planar Capacitors

Click For Summary

Discussion Overview

The discussion revolves around the application of Gauss's Law to planar capacitors, particularly addressing the perceived contradiction in the electric field calculations between conducting and non-conducting sheets. Participants explore the implications of charge distribution and the resulting electric fields in different scenarios.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant states that the electric field between the plates of a planar capacitor is σ/ε, but questions why it seems to imply E=2σ/ε due to both plates contributing to the field.
  • Another participant recalls that a sheet of charge produces an electric field of E = σ/2ε, providing a derivation based on Gauss's Law.
  • A third participant discusses the application of Gauss's Law to non-conducting sheets and notes that for conducting sheets, the electric field inside the conductor is zero, leading to an electric field of E=σ/ε outside the conductor.
  • This participant further elaborates that for a charged conducting sheet, the charge is evenly distributed, and when considering a Gaussian surface, only half the charge contributes to the enclosed charge, leading to the conclusion that the electric field behaves consistently with the earlier derivation.
  • Another participant expresses concern about the implications of an infinitesimally thin conducting sheet and the potential contradiction that arises if the electric field were to double in this limit.

Areas of Agreement / Disagreement

Participants express differing views on the application of Gauss's Law to conducting versus non-conducting sheets, with some agreeing on the derivation of the electric field for sheets of charge while others raise concerns about the implications of charge distribution and thickness. The discussion remains unresolved regarding the apparent contradiction in the electric field calculations.

Contextual Notes

Participants highlight the importance of charge distribution and the behavior of electric fields in different materials, noting that assumptions about the thickness of sheets and the nature of charge distribution may affect the conclusions drawn.

hokhani
Messages
601
Reaction score
22
we know that the electric field between the planes of planar capacitor is σ/ε (according to gauss law)
however we have two conducting plate that each plate produe this electric field that are in the same direction therefore we must have E=2σ/ε
what is the reason for this contradiction.
 
Physics news on Phys.org
It seems to be my recollection that a sheet of charge has an electric field of

[tex]E = \frac{\sigma}{2\epsilon}[/tex]

It's trivial to derive this by hand.

[tex]\oint \mathbf{E}\cdot d\mathbf{a} = \frac{1}{\epsilon} Q = \frac{\sigma A}{\epsilon}[/tex]
[tex]\oint \mathbf{E}\cdot d\mathbf{a} = 2A \left| \mathbf{E} \right| = \frac{\sigma A}{\epsilon}[/tex]
[tex]\mathbf{E} = \frac{\sigma}{2\epsilon} \hat{n}[/tex]
where \hat{n} is the normal direction of the sheet.
 
Ok
If we have non conducting sheet we can take a cylindrical gauss surface around two sides of the sheet and you are right but if we have conducting sheet and take a cylindrical gauss surface in such that on of the circular surface of gauss surface take between two sides of the sheet(as shown in the picture), the electric field will be:
E=σ/ε
Because the electric field will be zero in the conductor
 

Attachments

hokhani said:
Ok
If we have non conducting sheet we can take a cylindrical gauss surface around two sides of the sheet and you are right but if we have conducting sheet and take a cylindrical gauss surface in such that on of the circular surface of gauss surface take between two sides of the sheet(as shown in the picture), the electric field will be:
E=σ/ε
Because the electric field will be zero in the conductor

Ahh, I got it. However, if we have a charged conducting sheet then the charge is spread evenly across its surface. Neglecting the charge on the edges (assuming a thin sheet) then the top side will have half the charge and the bottom side would have the other half. Thus, when you take your Guassian surface that has one surface terminating inside the volume then the enclosed charge is actually

[tex]Q = \sigma A[/tex]
[tex]Q_{upper} = \frac{1}{2}Q = \frac{\sigma A}{2}[/tex]

since you are only enclosing one surface of the conductor. So it still works out ok. If it didn't work out the same, then we come to a problem where we now have a problem. If we have an infinitesimally thin sheet of charge we get the factor of 1/2. If a conducting sheet gave us twice the E-field then what happens when we take the limit of the sheet to have zero thickness? We arrive at a contradiction.
 
thank you very very much.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 29 ·
Replies
29
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
3K
  • · Replies 12 ·
Replies
12
Views
4K