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 ?
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.
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].