# Doubt regarding proof of Earnshaw's Theorem using Gauss's theorem

While proving Earnshaw's theorem using Gauss's theorem, we consider a small sphere surrounding our test charge, and apply Gauss law on this sphere, stating that field from all external charges must point either inward or outward, for stable equilibrium. However, then many sources conclude that a contradiction arises because the enclosed charge in the sphere is zero, and thus since the LHS of Gauss's law is violated. I also find many sources claim that the sign of the test charge contradicts the LHS of Gauss's law (I can follow the second reasoning).

I think the first explanation is incorrect as there is a net flux in the sphere precisely due to the presence of the test charge, so we cannot ignore it.

My question is if the first reasoning is wrong or am I really missing something?

Source-

Competitive physics by Jinhui Wang $$-$$

Show that a collection of point charges cannot be maintained in a stable equilibrium solely by their electrostatic interactions.

Firstly, we have to recall the meaning of a stable equilibrium — it requires that once a particle is displaced from its equilibrium position, there is a cor- recting force exerted on it that tends to return it to its equilibrium position. In the current context, if a certain charge rests in a stable equilibrium state, the electric ﬁeld due to all other charges in the region around this equilibrium position must point towards (if this particular charge is positive) and away from (if the charge is negative) the equilibrium position. However, applying Gauss’ law to an inﬁnitesimal surface (such as a small sphere) surrounding the mentioned equilibrium position, the total electric ﬂux through this sur- face would be non-zero (as the ﬁelds at all points on the surface either all point inwards or outwards) even though the enclosed charge is zero (note that the charge in the equilibrium position does not count as we are looking at the electric ﬁeld due to all other charges) — leading to a contradiction. Therefore, a stable equilibrium cannot exist in a system purely held by electrostatic forces. Only an unstable equilibrium, along at least one direction, or a neutral equilibrium can occur.

Purcell and Morin gives both reasonings$$-$$

First, suppose we have an electric field in which, contrary to the theorem, there is a point P at which a positively charged particle would be in stable equilibrium. That means that any small displacement of the particle from P must bring it to a place where an electric field acts to push it back toward P. But that means that a little sphere around P must have E pointing inward everywhere on its surface, which in turn means that there is a net inward flux through the sphere. This contradicts Gauss’s law, for there is no negative source charge within the region. (Our charged test particle doesn’t count; besides, it’s positive.)

No, the flux arises from the electric field of the other charges. From basic electrostatics, the force on a test charge $$q$$ in an electric field $$\mathbf{E}$$ is $$q\mathbf{E}$$. Here, $$\mathbf{E}$$ is the electric field considered without the test charge, i.e. it is the field produced by all other charges. That's why it is called a test charge, because we do not imagine it changing $$\mathbf{E}$$ itself. Here, we are considering the electric force on such a test charge, which is proportional to the electric field produced by the charge configuration (without the test charge itself). In other words, $$\rho = 0$$ at the point of consideration.
The statement is essentially saying that where $$\rho = 0$$, there cannot be a net electric flux through a closed surface.
• @Eisenstein That is exactly the point of the contradiction. A stable equilibrium at a point means that the electric field must always point inward in a sphere around it. But this would mean a net flux through the sphere, which is a contradiction because it must be zero. Therefore there cannot be a point of stable equilibrium. This is equivalent to saying that the electric potential (which obeys Laplace's equation where $\rho=0$) cannot have local minima/maxima. Commented Apr 4 at 18:21