Electrostatics - charges in stable equilibrium

[gandharva_23]

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 ?

 

[Galileo]

Yes, since the potential due to the other charges satisfies \nabla^2 V=0. V is harmonic with has the property that is has no local maxima or minima.

 

[Meir Achuz]

It is called Thomson's theorem.

 

[Galileo]

Do you mean Earnshaw's theorem?

 

[Meir Achuz]

Whoops, Galileo is right again. I meant Earnshaw.
Thomson had several theorems (besides his home run), but not that one.
Sorry, and thank you.

 

[gandharva_23]

Yes, since the potential due to the other charges satisfies \nabla^2 V=0is 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 \nabla^2 V=0

 

[Galileo]

It's just Maxwell's (first) equation: \vec \nabla \cdot \vec E =\rho/\epsilon_0, or \nabla^2 V=-\rho/\epsilon_0.
In the region where there is no charge density you have \nabla^2 V=0.