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Electrostatics - charges in stable equilibrium

by gandharva_23
Tags: electrostatics
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gandharva_23
#1
Mar8-06, 05:01 AM
P: 62
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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Galileo
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Mar8-06, 06:18 AM
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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
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Mar8-06, 10:27 AM
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PF Gold
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It is called Thomson's theorem.

Galileo
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Mar8-06, 10:56 AM
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Electrostatics - charges in stable equilibrium

Do you mean Earnshaw's theorem?
Meir Achuz
#5
Mar8-06, 11:51 AM
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PF Gold
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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
#6
Mar8-06, 09:48 PM
P: 62
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
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Mar9-06, 05:04 AM
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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].


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