Electrostatics - charges in stable equilibrium

AI Thread Summary
In electrostatics, a charge distribution in equilibrium cannot achieve stable equilibrium due to the properties of the electric potential, which satisfies the equation ∇²V=0. This indicates that the potential is harmonic, lacking local maxima or minima, as stated in Earnshaw's theorem. The discussion clarifies that the relationship between electric field and potential is derived from Maxwell's equations, specifically highlighting that in regions without charge density, the potential also satisfies ∇²V=0. The conversation briefly corrects a misattribution of the theorem to Thomson instead of Earnshaw. Overall, the key takeaway is that stable equilibrium for charges is impossible in electrostatic systems.
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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 ?
 
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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.
 
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 \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
 
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.
 
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