Unstable or stable electrostatic equilibrium?

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Homework Help Overview

The discussion revolves around determining the stability of electrostatic equilibrium for a point charge at the center of a charged ring. Participants explore concepts related to potential energy, electric fields, and the implications of charge signs on stability.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants consider using Laplace's equation and the shell theorem to analyze the electric potential and field. There are discussions about the effects of charge signs and the nature of forces acting on the point charge when displaced from the center.

Discussion Status

The conversation is ongoing, with participants raising various approaches and questioning assumptions. Some have suggested calculating the electric field and potential in the vicinity of the center, while others are exploring the implications of symmetry and dimensionality in their analysis.

Contextual Notes

There are mentions of constraints such as the need to assume charge signs and the dimensionality of the problem, with some participants questioning whether to apply 1D or 2D equations in their calculations.

  • #61
Angela G said:
ok, so If the charges has oposite sign, the there will be an attractive force when the point charge is displaced a dz from the ring, and know we have to determine the stability on the plane of the ring.

Also, I wonder if you could take a look at post #50, and tell me if my reasoning is right or not, pls
I don't think you were supposed to get into detailed calculations. That's why the question said simply "motivate your answer". You're looking for a good approximation for the force on a particle that is perturbed off centre.
 
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  • #62
Angela G said:
ok, so If the charges has oposite sign, the there will be an attractive force when the point charge is displaced a dz from the ring, and know we have to determine the stability on the plane of the ring.
You have examined $$- \frac{\partial^2 V}{\partial z^2} = \frac{1}{\rho} \frac{\partial}{\partial \rho} \left( \rho \frac{\partial V}{\partial \rho} \right )$$ This says that:
1. If one side of the equation is positive (it doesn't matter which) the other is negative.
2. The positive side means stable equilibrium in that direction and the negative side means unstable equilibrium in that direction.

Is that observation going to change if the signs of the charged ring and the test particle are the same as opposed to different?

This thread has gone on for too long in my opinion. One can find more details about the potential and electric fields near the origin (for like charges) in post #2 here.
 
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  • #63
Angela G said:
ok, so If the charges has oposite sign, the there will be an attractive force when the point charge is displaced a dz from the ring, and know we have to determine the stability on the plane of the ring.

Also, I wonder if you could take a look at post #50, and tell me if my reasoning is right or not, pls
I think you already have said the way the ring field acts and you know which way it makes a charge with a small displacement go for each sign of ##q##.
Just spell it out.

It is interesting to compare the stability in the plane vs. out of the plane along the ##z## axis in both cases (of like or opposite charges).
 
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  • #64
PeroK said:
Which is an unjustified assumption. ##\lambda## and ##q## could have opposite signs.
It was used as a starting case, that's all. The possibility of different signs were discussed earlier.
 
  • #65
I think part ##a## is basically solved or at least enough was discussed to complete it, so what about part ##b##?
 
  • #66
could you please summarize a) because I'm little confused
 
  • #67
Angela G said:
could you please summarize a) because I'm little confused
You solved the stability for the ##z## axis above and you know the direction and therefore force in the plane (toward the center for positive ring) and so you know whether it is stable in the plane of the ring given the sign of the charges. If you want to be more detailed for example on the ##z## axis just compute the potential along the ##z## axis which is easy because of symmetry and take the second derivative of it.
 
  • #68
regarding to b) I think I already solved it could you please take a look of it

Edit: I missed a "q" in the third expression
 

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  • #69
Angela G said:
regarding to b) I think I already solved it could you please take a look of it

Edit: I missed a "q" in the third expression
Looks correct.
 
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  • #70
Thank you everyone that helped me in this problem I finally understood ☺️☺️☺️☺️
 
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