
#1
Sep2306, 09:49 PM

P: 2,268

The theorm states:
"You can't trap a charged particle in a chargefree 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. 



#2
Sep2306, 11:27 PM

Emeritus
Sci Advisor
PF Gold
P: 11,154

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. 



#3
Sep2406, 12:29 AM

P: 2,268

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? 



#4
Sep2406, 07:49 AM

Emeritus
Sci Advisor
PF Gold
P: 11,154

Earnshaw's Theorem 



#5
Sep2406, 07:35 PM

P: 2,268

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? 



#6
Sep2406, 08:56 PM

Sci Advisor
P: 1,465

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. 



#7
Sep2506, 03:40 AM

P: 2,268





#8
Sep2506, 12:33 PM

P: 223

Having zero potential everywhere just means that the kinetic energy and momentum are both conserved, and there is nothing to study at this point. 



#9
Sep2506, 06:49 PM

Sci Advisor
P: 1,465

Claude. 



#10
Sep2606, 02:15 AM

P: 2,268

There is still the question in post 5 to be settled. 


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