The theorm states: "You can't trap a charged particle in a charge-free electrostatic field; there is no point of stable equilibrium.' R. Good, p284. Is this theorem trying to say that if you are ever able to trap a charge x inside any region than that region must already contain a charge y (although that y was not trapped in the region beforehand). I suppose any region of interest must have electric fields through it so any charge placed inside it will have a force on it but with divF=0. Hence the froces cannot be directed in a way so as to trap the particle. What about a region without any electric fields than if the charge is placed inside it with 0 intitial velocity than it will stay stationary. But a region like this does not apply to the theorem because it assumes an electrostatic field, not a vacuum.
I believe what it is saying is that any distribution of charges will result in local minima of the potential (ie, points of stable equilibrium) only at points where the charge density is non-zero. Local maxima (which are points of unstable equilibrium) however, can exist at points where the charge density is zero. For instance, if you have two equal negative charges separated by some distance, the midpoint of the line joining the charges is a point of unstable equilibrium for both positive and negative charges placed there. However, the positions of the charges themselves are points of stable equilibrium for a positive charge. If you take a positive charge and place it at the midpoint, it will stay there so long as there are no perturbations. The slightest distaurbance will likely cause it to collapse into one of the two neegative charges.
This is the case of unstable equilibrium so the theorem does not apply. What do you think of the validity of 'This theorem is trying to say that if you are ever able to trap a charge x inside any region than that region must already contain a charge y.' where x and y are nonzero?
That's not true. For the described situation, there exists one point of unstable equilibrium, and two points of stable equilibrium (ignoring points at infinity). The goal of the illustration was to demonstrate that, in accordance with the theorem, the only point of equilibrium where the charge density is zero is not a stable one. Yes, that works.
For this situation to have a charge density 0, it means the mid charge must contain twice the charge of one of the end charges. i.e (-1C) (+2C) (-1C) C=coulomb From the electrostatics of the situtation, its true that the midcharge will be in unstable equilbrium but the two end charges will not be in any equilibirum and wiill be attracted toward the centre positive charge. Even without the centre positive charge, how can the two negative charges be in equillibrium in the first place because they will move to get as far from each other as possible. Or are you assuming they are bound somehow such as being inside a container?
This would not work anyway, because technically you are not trapping the particle. Some energy minimum must exist in order for the particle to exist in a true 'trapped' state. As Gokul as pointed out, it is not enough that it exist in an unstable equilibrium, because the particle is not trapped any more that it would be sitting in a vacuum at infinity because the slightest imbalance would cause it to accelerate away from its initial position. Claude.
If we imagined a true vacuum existed than everywhere has a minimum energy. If we give the particle no initial kinetic energy than would it be trapped?
If there's no potential, then there's no confining potential, and so you can't have a local minimum or whatnot because there's nothing there. It won't move, but what you've just described is about the most uninteresting problem in physics. Having zero potential everywhere just means that the kinetic energy and momentum are both conserved, and there is nothing to study at this point.
That is a good point because Earnshaw's theorem only concerns stable equilibrium. There is still the question in post 5 to be settled.