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.


Shubhashish on Feb 03, 2017:
MT on Apr 06, 2017:
Thank you!
George E. Smith on Feb 28, 2018:
Earnshaw's theorem says there can be no static system subject to mutual inverse square law forces only. It's a mathematical theorem so there are no exceptions. It does NOT even apply to magnetic systems, since magnetism is a property of moving electric charges (currents). In the physical world, there are only two such forces outside of an atomic nucleus; gravity, and the Coulomb force. It can ONLY apply to the complete set of interacting entities, so you can't have masses or charges which are fixed by the use of other non-inverse square law forces.
It is also only valid in classical physics, since Heisenberg's principle does not permit any static system.
George E. Smith on Feb 28, 2018:
If Earnshaw's theorem were false, there would be a static system with some finite number (n) of entities subject to mutual inverse square law forces only, creating a force field with at least one other point at which the net vector force field was zero. Any additional entity could be placed there creating a static system of n+1 elements.
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.
Diego Assencio on Feb 28, 2018:
@George: Thank you very much for your comment!

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 ;-)
琼五大帝 on Mar 20, 2022:
Thank you!