# electrostatics

by gandharva_23
Tags: electrostatics
 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 ?
 Sci Advisor HW Helper P: 2,004 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.
 Sci Advisor HW Helper P: 1,930 It is called Thomson's theorem.
 P: 62 Yes, since the potential due to the other charges satisfies $\nabla^2 V=0$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 $\nabla^2 V=0$
 Sci Advisor HW Helper P: 2,004 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 $\nabla^2 V=0$.