Electrostatics - charges in stable equilibrium

Click For Summary

Homework Help Overview

The discussion revolves around electrostatics, specifically focusing on the stability of charge distributions in equilibrium and the implications of potential functions in this context.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between electrostatic potential and equilibrium, questioning the conditions under which charges can be in stable equilibrium. There is a discussion on the implications of harmonic functions and theorems related to electrostatics.

Discussion Status

The conversation includes references to specific theorems and mathematical principles, with some participants clarifying terminology and correcting previous statements. There is ongoing exploration of the mathematical proof regarding the condition \nabla^2 V=0 and its implications.

Contextual Notes

Participants are navigating through the definitions and implications of electrostatic potentials, with some confusion about theorems and their correct attribution. The discussion reflects a need for clarity on the mathematical foundations of the concepts being discussed.

gandharva_23
Messages
61
Reaction score
0
We have a charge distribution in which all the charges are in equilibrium due to electrostatic forces . Can we prove that none of these charges will be in stable equilibrium ?
 
Physics news on Phys.org
Yes, since the potential due to the other charges satisfies [itex]\nabla^2 V=0[/itex]. V is harmonic with has the property that is has no local maxima or minima.
 
It is called Thomson's theorem.
 
Do you mean Earnshaw's theorem?
 
Whoops, Galileo is right again. I meant Earnshaw.
Thomson had several theorems (besides his home run), but not that one.
Sorry, and thank you.
 
Yes, since the potential due to the other charges satisfies [itex]\nabla^2 V=0[/itex]is being generated. Reload this page in a moment.. V is harmonic with has the property that is has no local maxima or minima.

thats what i wanted to ask . how can we prove that [itex]\nabla^2 V=0[/itex]
 
It's just Maxwell's (first) equation: [itex]\vec \nabla \cdot \vec E =\rho/\epsilon_0[/itex], or [itex]\nabla^2 V=-\rho/\epsilon_0[/itex].
In the region where there is no charge density you have [itex]\nabla^2 V=0[/itex].
 

Similar threads

Unstable or stable electrostatic equilibrium?
  • · Replies 69 ·
3
Replies
69
Views
7K
Conducting Sphere and Dipole Problem
  • · Replies 5 ·
Replies
5
Views
1K
Electric field of a neutral conductor just at the surface
  • · Replies 4 ·
Replies
4
Views
2K
High School Uniform charge distribution in a conductor
  • · Replies 5 ·
Replies
5
Views
3K
Electrostatic field of a sphere
  • · Replies 2 ·
Replies
2
Views
1K
Capacitors, equipotentials, and electric fields
  • · Replies 5 ·
Replies
5
Views
1K
Earnshaw's theorem and 'free space'
  • · Replies 1 ·
Replies
1
Views
2K
Potential energy of an electrostatic system in equilibrium
  • · Replies 14 ·
Replies
14
Views
2K
Equilibrium circular ring of uniform charge with point charge
  • · Replies 7 ·
Replies
7
Views
3K
Proving stable equilibrium: Rotating circular hoop
  • · Replies 6 ·
Replies
6
Views
3K