# Electrostatics - charges in stable equilibrium

by gandharva_23
Tags: electrostatics
 Sci Advisor HW Helper P: 2,002 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.
 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: 2,002 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 $\nabla^2 V=0$.