Does a Charge on a Closed Surface Contribute to Electric Flux?

  • Context: Undergrad 
  • Thread starter Thread starter lugita15
  • Start date Start date
  • Tags Tags
    Gauss's law Law
Click For Summary

Discussion Overview

The discussion centers around the application of Gauss's Law in relation to charges located on a closed surface. Participants explore whether a charge on the surface contributes to the electric flux through that surface, examining the implications of such a scenario within the context of theoretical physics.

Discussion Character

  • Debate/contested
  • Conceptual clarification

Main Points Raised

  • Some participants assert that according to Gauss's Law, only charges enclosed by the surface contribute to electric flux, implying that charges on the surface do not count as enclosed.
  • One participant emphasizes that by definition, all charge must be enclosed, suggesting that there cannot be a surface charge contributing to flux.
  • Another participant notes that a Gaussian surface is a mathematical construct, and for Gauss's theorem to apply, charges must be either inside or outside the surface, but not on it.
  • It is proposed that if a charge is on the surface, the electric flux would be undefined at that point due to the electric field being undefined at the location of the charge itself.

Areas of Agreement / Disagreement

Participants express differing views on the treatment of charges on a closed surface, with no consensus reached regarding their contribution to electric flux.

Contextual Notes

The discussion highlights the ambiguity surrounding the definition of "enclosed" charges and the implications of surface charges on electric flux, which may depend on specific interpretations of Gauss's Law.

lugita15
Messages
1,553
Reaction score
15
According to Gauss's Law, the electric flux through a closed surface is equal to the charge enclosed by the surface divided by epsilon_0. This obviously excludes charges outside the surface. But what about *on* the surface? That is, it lies neither inside nor outside. Does this charge count as being "enclosed" in the sense of Gauss's Law? In other words, does the electric field of a charge "on" a closed surface create a nonzero flux through the surface?

Any help would be greatly appreciated.
Thank You in Advance.
 
Physics news on Phys.org
lugita15 said:
According to Gauss's Law, the electric flux through a closed surface is equal to the charge enclosed by the surface divided by epsilon_0. This obviously excludes charges outside the surface. But what about *on* the surface? That is, it lies neither inside nor outside. Does this charge count as being "enclosed" in the sense of Gauss's Law? In other words, does the electric field of a charge "on" a closed surface create a nonzero flux through the surface?

Any help would be greatly appreciated.
Thank You in Advance.

I believe by definition all charge must be enclosed, therefore there is no surface charge.
 
A Gaussian surface is a mathematical surface you decide on.
To apply Gauss's theorem, you must choose a surface that has charge either within or without, but not on your Gaussian surfce.
 
If a charge was on the surface, the flux would be undefined at that point (since the field itself is undefined at the charge itself)
 

Similar threads

  • · Replies 83 ·
3
Replies
83
Views
5K
  • · Replies 7 ·
Replies
7
Views
890
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K