electrostatics

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 [itex]\nabla^2 V=0[/itex]. 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 [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]

Galileo

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].