Why does charge accumulate at the sharpest point?

  • Thread starter Thread starter jdstokes
  • Start date Start date
  • Tags Tags
    Charge Point
Click For Summary

Homework Help Overview

The discussion revolves around the phenomenon of charge accumulation at sharp points on conductors, exploring the underlying principles of charge distribution and electric fields in conductive materials.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between charge density and the geometry of conductors, particularly focusing on the effects of radius of curvature. Questions are raised about the implications of charge distribution on potential and electric fields at sharp points and corners.

Discussion Status

Several participants are actively engaging with the concepts, questioning the reasoning behind observed phenomena. There is a mix of models and theoretical considerations being explored, with no clear consensus yet on the simplest explanation for the charge distribution.

Contextual Notes

Some participants reference specific models involving conducting spheres and the implications of electric field behavior at corners, indicating a deeper inquiry into the assumptions about charge distribution in conductive materials.

jdstokes
Messages
520
Reaction score
1
I'm sure I used to have a really simple answer to this. But I've long since forgotten.

BTW This is not my homework.

James
 
Physics news on Phys.org
I believe the idea is that at least for a conductor, the charge exists evenly on the surface. So when you have a sharp point, you may have charges that are very close to each other in space, even though they may be equally spaced on the surface.
 
The local charge density varies inversely as the square of the radius of curvature.
 
maverick280857 said:
The local charge density varies inversely as the square of the radius of curvature.

Yes, of course. But why??
 
jdstokes said:
Yes, of course. But why??
So that the potential may satisfy the Laplacian outside the conductor.
 
Here's a simple model.
Consider two conducting spheres, with unequal radii A and B, connected by a long thin conducting wire... so the spheres are at the same potential. How do free charges distribute themselves on the two spheres?
 
robphy said:
Here's a simple model.
Consider two conducting spheres, with unequal radii A and B, connected by a long thin conducting wire... so the spheres are at the same potential. How do free charges distribute themselves on the two spheres?

[itex] \begin{align*}<br /> V_A & = \frac{\sigma_A 4\pi A^2}{4\pi\epsilon_0 A} \\<br /> & = \frac{\sigma_A A}{\epsilon_0} \\<br /> V_B & = \frac{\sigma_B B}{\epsilon_0} \\<br /> \sigma_A A & = \sigma_B B<br /> \end{align*}[/itex]

This suggests that the charge density is higher on the smaller sphere. Quite nice, but I think there is probably a simpler explanation.
 
jdstokes, have you thought about what would happen at a true corner in a conductor? You know the electric field is always normal to the surface of a conductor and continuous outside the conductor, but what happens at a corner? Imagine you sit just outside the corner: if you move a little to the left you get a field parallel to one surface normal, move a little to the right you get a field parallel to the other surface normal (the surfaces meet at an edge so imagine two planes intersecting). The field seems discontinuous just outside the corner. Of course there are physical limits, but the point is that the electric field is varying really fast, so what does this tell you about the charge density?
 
Last edited:

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
5
Views
3K
  • · Replies 13 ·
Replies
13
Views
6K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
36
Views
4K
Replies
4
Views
4K
Replies
16
Views
5K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
3
Views
3K