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.

## Comments

It is also only valid in classical physics, since Heisenberg's principle does not permit any static system.

since the system is static, there must be a net zero field at each of the entities, due to all the others, so any sized entity can be placed there including a zero size. So any element M( i ) can be removed leaving a static n-1 system. Repeating the process, all of the elements can be removed so there is no minimum size such system, which is absurd. So our conjecture that Earnshaw's theorem is false cannot be correct.

I wrote this article based on how Earnshaw's theorem was presented to me when I was a Physics student. To be frank with you, I am not surprised with the fact that what the theorem actually states (in a mathematical sense) is not 100% in accordance with what I've learned in a Physics class ;-)