# Why can't I contain a charged particle electrostatically in stable equilibrium? Referring to Earnshaw's Theorem

Contrary to the Earnshaw's theorem, I have an example. Consider 6 positively charged particles of equal magnitudes places at the corners of a regular hexagon. Naturally, the field at the centre of the hexagon is zero. Now, place another positively charged particle at centre of the hexagon. This particle will be in STABLE EQUILIBRIUM. Which goes against the Earnshaw's theorem. Please help me with this contradiction, and where it's going wrong.

• Could you explain why you think the equilibrium the particle is in is "stable"? What physical quantity would you use to define equilibrium? – Philip Cherian Aug 13 '17 at 5:25
• possible duplicate of physics.stackexchange.com/questions/351567/… – ZeroTheHero Aug 13 '17 at 6:13
• This arrangement is not stable. The particle in the center is in an unstable equilibrium in the direction perpendicular to the hexagon. – Johnathan Gross Aug 13 '17 at 8:54

In order for a charged particle to be in stable equilibrium electrostatically, you must have that all the electric field lines at some point (the point where you want to place your particle) be pointing towards (away from) the particle. This means that the divergence of the electric field in this region would be negative (positive).

However, Gauss' Law tells us that in free space (such as at this point we're considering):

$$\nabla\cdot E = 0,$$

which means that we cannot have such a stable configuration, and so there must be instability in some direction. This point that naively seems stable is actually a saddle point, which isn't stable.

You seem to be assuming that the six particles at the corners are held in place. But Earnshaw's theorem applies to all the particles, not just the one in the center, and no other forces are allowed (including whatever is holding those particles in place). If the six particles are allowed to move freely, they will all move away from each other.

• If I've understood you correctly, you're saying that such a 'violation' occurs since we're assuming the corner particles to be fixed. However, even if they were fixed, the theorem would still hold: the particle at the center would not be in stable equilibrium. – Philip Cherian Aug 13 '17 at 5:28
• Good point. Jared's answer is better. – A. Newell Aug 13 '17 at 5:35
• However, my answer is not wrong, so it seems a bit harsh to give it a negative vote. – A. Newell Aug 19 '17 at 21:00
• (Wasn't me! :) However, I believe that a downvote cannot be removed until the post has been edited.) – Philip Cherian Aug 20 '17 at 6:43
• Oh, I figured editing was for tweaking the answer. I can just write a whole new response? – A. Newell Aug 20 '17 at 14:50