# Electrostatics - charges in stable equilibrium

1. Mar 8, 2006

### gandharva_23

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 ?

2. Mar 8, 2006

### Galileo

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.

3. Mar 8, 2006

### Meir Achuz

It is called Thomson's theorem.

4. Mar 8, 2006

### Galileo

Do you mean Earnshaw's theorem?

5. Mar 8, 2006

### Meir Achuz

Whoops, Galileo is right again. I meant Earnshaw.
Thomson had several theorems (besides his home run), but not that one.
Sorry, and thank you.

6. Mar 8, 2006

### gandharva_23

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 $\nabla^2 V=0$

7. Mar 9, 2006

### Galileo

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 $\nabla^2 V=0$.