## A proof of Earnshaw's theorem

Posted by Diego Assencio on 2014.09.19 under Physics (Electromagnetism)

Consider a set of $n$ fixed point charges $q_1, q_2, \ldots, q_n$ in vacuum and let ${\bf E}({\bf x})$ be the electric field produced by these charges (see figure 1). Can we find a point ${\bf z}$ in space such that if a charge $q$ is placed there, it will remain in stable equilibrium?

 Fig. 1: A set of $n$ fixed charges and a very small Gaussian surface $S$ centered at a point ${\bf z}$. For ${\bf z}$ to be a point of stable equilibrium (assuming we would place a positive charge $q$ there), ${\bf E}({\bf z})$ must be zero and ${\bf E}({\bf x})$ must point towards ${\bf z}$ at all points ${\bf x}$ of $S$.

The answer is no, and this fact is referred to as the Earnshaw's theorem. We will prove this assuming $q \gt 0$, but the proof is similar for $q \lt 0$.

Let's first think about what do we mean by stable equilibrium for a charge $q$ placed at ${\bf z}$:

 1 the electric force $q{\bf E}({\bf z})$ acting on $q$ must be zero, i.e, ${\bf E}({\bf z}) = {\bf 0}$ 2 for a small displacement $\delta{\bf z}$ of $q$ around ${\bf z}$, $q{\bf E}({\bf z} + \delta{\bf z})$ must point towards ${\bf z}$.

The second property means the electric force on $q$ must be restoring, meaning if we move $q$ a little bit away from ${\bf z}$, this force will will tend to bring $q$ back to ${\bf z}$. Since we are assuming $q \gt 0$, ${\bf E}({\bf z} + \delta{\bf z})$ itself must then point towards ${\bf z}$ for the electric force on $q$ to be restoring.

Consider now a very small Gaussian surface $S$ shaped like a sphere of radius $r$ and centered at ${\bf z}$ (as shown in figure 1). Since ${\bf E}({\bf x})$ points towards $q$ (i.e., inwards) for any point ${\bf x}$ on the surface of this sphere, we have that: $$\int_{S} {\bf E}\cdot d{\bf A} \lt 0 \label{post_bc04395b103021d338b4e30a061bfc74_surface_int}$$ where $d{\bf A}$ is an outward-oriented infinitesimal surface area element. But using the divergence theorem and Gauss's law on equation \eqref{post_bc04395b103021d338b4e30a061bfc74_surface_int} implies we must have a negative charge $Q_{\textrm{enc}}$ enclosed inside our sphere: $$\int_{S} {\bf E}\cdot d{\bf A} = \int_{V}\nabla\cdot{\bf E}\,dV = \displaystyle\frac{Q_{\textrm{enc}}}{\epsilon_0} \lt 0$$ This goes against our initial assumption that ${\bf z}$ is a point in vacuum where no charge is placed ($q$ itself does not count since it does not contribute to the electric field ${\bf E}$).

One of the consequences of Earnshaw's theorem is the fact that a classical model for the atom cannot be stable if it is based on a static (fixed) distribution of charges. For such an atom, the charge configuration would not be in stable equilibrium and therefore unable to resist even the slightest disturbance.