I've been learning about particle traps and Earnshaw's theorem. When dealing purely with electrostatic forces the theorem makes intuitive sense to me. But does it really apply to systems involving gravity AND electrostatic forces? According to Earnshaw proof it does, but I have trouble understanding it. For example, if I imagine a system consisting of a positively charged metal coffee can (a cylinder with a closed bottom and open top), sitting on the surface of Earth with the open top pointed skyward, and now I drop in a small particle of positively charged dust (large enough that it has non-negligible mass), couldn't I find a stable equilibrium for that dust grain where it sits just inside the the coffee can, repelled from all the walls but not allowed to exit through the top because of gravity?
1 Answer
The said theorem is a direct consequence of the "maximum principle (theorem)". It states that a (twice continuously differentiable) function $\phi$ satisfying $\Delta \phi=0$ cannot have local minima. Stable equilibrium configurations for a particle correspond to local minimum points of the total potential energy $\phi$ due to external sources. Hence $\phi$ satisfies the said equation where it has (should have) a local minimum. It does not matter if the potential energy is due to the electric field, gravitational field or both. If the sources also occupy the candidate equilibrium configurations everything changes. The theorem (in the version I know) refers to a charge immersed in the field due to other charges. It seems to me that you are considering more complicated situations where the theorem applies as well.
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$\begingroup$ "The theorem (in the version I know) refers to a charge immersed in the field due to other charges. It seems to me that you are considering more complicated situations." To clarify, I'm asking about the equilibrium point of the massive charged dust/test particle, as allowed by the electric field of the open cylinder and the gravity field of the Earth (I'm not considering how the test particle itself will affect these fields.) And to make sure I understand: your answer would suggest that you could not trap dust particles in a charged "cup" open at the top but in a gravity field? $\endgroup$ Commented Apr 12, 2021 at 19:24
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$\begingroup$ The theorem applies to that situation too. In the space of the box where the particle moves the total potential energy $\phi(x) = -qV(X) + m\varphi(x)$ satisfies $\Delta \phi=0$ so that this function cannot have a local minimum. The only possibility for an equilibrium configuration is that the particle touches the boundary of the box... $\endgroup$ Commented Apr 12, 2021 at 19:47