Unstable or stable electrostatic equilibrium?

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Homework Statement:
Consider a circular ring with radius R and uniform longitudinal charge density λ
a) Assume that a charge of magnitude q is in the center of the circle. The power of it then becomes zero, i.e. we have an equilibrium position. Is this stable or unstable? Motivate the answer!

b) Suppose now that we take off half of the ring and that the long charge density remains the same. (The ring is not conductive.) Determine the magnitude of the force acting on the charge q.
Relevant Equations:
## V(0) = \frac{\lambda}{2\epsilon_0} ##
## \nabla^2 V = 0 ##
## U = - \vec p \cdot \vec E ##
## W = q V ##
I wonder if you could help me with both I'm stuck, I know that in order to see if the electrostatic equilibrium is stable or not at the center of the ring , the potential energy has to be minimum there. I was going to use Laplace eq. but it allows neither minimum nor maximum. Then I also thought that if you move the point charge to either the right or left, it should experience a force that takes the point charge back to the center of the ring (for stable equilibrium). But since we have only got the magnitude of charges, I guess I have to make assumptions (if both the point and line circles have the same sign or if they have different signs of charges). Another alternative that I thought of was to use that U = - p * E, where if the electric dipole is parallel to the electric field we have the smallest potential energy, but again the problems arise about the signs of the charges and now it is also an external electric field involved, which we do not have. So I'm wondering if you could guide me. I would like to solve in with advanced physics, please.

I determined the electric potential at the center of the ring without the charge it was $$ V(0) = \frac{\lambda}{2\epsilon_0} $$

for b) I was thinking that I can determine the electric field cause we dont hade longer circular simetry and then the force by columbs law
 
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  • #2
PeroK
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For b) you should be able to calculate the electric field, using the remaining symmetry to simplify the integration.

For a) if we assume that ##\lambda## is positive have you thought about what happens if ##q## is positive or negative?
 
  • #3
PeroK
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PS another thought about a). If it were a sphere, rather than a ring, then the shell theorem would apply. Do you know an intuitive way to deduce the shell theorem? What happens if you apply that here?
 
  • #4
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For a) if we assume that λ is positive have you thought about what happens if q is positive or negative?
The charge will experience either a attractive or a repulsive force depending if it is negative or positive.
 
  • #5
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PS another thought about a). If it were a sphere, rather than a ring, then the shell theorem would apply. Do you know an intuitive way to deduce the shell theorem? What happens if you apply that here?
I dont know, I just used the shell theorem once and it was long ago or Im just confused which shell theorem do you mean?
 
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PeroK
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The charge will experience either a attractive or a repulsive force depending if it is negative or positive.
I was hoping for a little more thought than that. The central charge is surrounded by a ring of charge. The equilibrium is only achieved by repulsive (or attractive) forces in different directions cancelling out.

You may need to calculate the forces in this case, assuming there is a small displacement off centre.

PS The shell theorem (for electrostatics) can be found easily enough online.
 
  • #7
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ok, so for a charged shell we will have that the electric field will be ## \vec E = 0 ## due to the simmetry and when ## r> R## the electric field will be as for a point charge. I think this is the shell theorem or this is what I found.

You may need to calculate the forces in this case, assuming there is a small displacement off centre.
how can I do that? Think I can use the coloumb law and then I have to determine the electric field at that point?
 
  • #8
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The power of it then becomes zero, i.e. we have an equilibrium position
I found that there will be a " force" instead for "power"
 
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PeroK
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how can I do that? Think I can use the coloumb law and then I have to determine the electric field at that point?
Yes, you have to approximate the field for a small perturbation from centre.
 
  • #10
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I was also thinking how I can see if the equilibrium is stable or not by looking at the variation in the potential around the center. And I was wondering if we can also use laplace equition, to solve it. Since we have from a) what V is at r = 0 and we do the assumtion that ## V = 0 ## at ## R = \infty ##
 
  • #11
PeroK
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I was also thinking how I can see if the equilibrium is stable or not by looking at the variation in the potential around the center. And I was wondering if we can also use laplace equition, to solve it. Since we have from a) what V is at r = 0 and we do the assumtion that ## V = 0 ## at ## R = \infty ##
You could calculate the potential in a neighbourhood of the centre, yes.
 
  • #12
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could you please give me som advice for how to solve the laplacian in this case? Should I use the 2d or 1d equation?
 
  • #13
bob012345
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could you please give me some advice for how to solve the laplacian in this case? Should I use the 2d or 1d equation?
The ring is a 2d object so you are concerned with the 2d plane. I would suggest using polar coordinates but you can use symmetry arguments to simplify it.
 
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  • #14
kuruman
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could you please give me som advice for how to solve the laplacian in this case? Should I use the 2d or 1d equation?
As far as the stability of the equilibrium point at the center is concerned, just look at Laplace's equation in cylindrical coordinates,$$\frac{1}{\rho}\frac{\partial }{\partial\rho}\left(\rho\frac{\partial V}{\partial \rho}\right)+\frac{\partial^2 V}{\partial z^2}=0.$$For the sum of the two terms to be zero, one must be the negative of the other everywhere in space. What does this say about stability for small displacements about the origin or anywhere else?
 
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  • #15
bob012345
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As far as the stability of the equilibrium point at the center is concerned, just look at the Laplacian in cylindrical coordinates.$$\frac{1}{\rho}\frac{\partial }{\partial\rho}\left(\rho\frac{\partial V}{\partial \rho}\right)+\frac{\partial^2 V}{\partial z^2}=0.$$For the sum of the two terms to be zero, one must be the negative of the other everywhere in space. What does this say about stability for small displacements about the origin or anywhere else?
Isn't this a flat 2D ring? I thought the OP concerned a small displacement from the center in the plane of the ring? I was thinking polar coordinates for 2D.
 
  • #16
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I thought the OP concerned a small displacement from the center in the plane of the ring? I was thinking polar coordinates for 2D.
I thought it too
 
  • #17
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What does this say about stability for small displacements about the origin or anywhere else?
As you said we will have that $$
\frac{\partial^2 V}{\partial z^2}= -\frac{1}{\rho}\frac{\partial }{\partial\rho}\left(\rho\frac{\partial V}{\partial \rho}\right) $$
I think that we have a maximun since the second derivative is negative and thus unstability, but if the laplacian applies we will not have any maximun or minimun in this area
 
  • #18
kuruman
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Isn't this a flat 2D ring? I thought the OP concerned a small displacement from the center in the plane of the ring? I was thinking polar coordinates for 2D.
It is a flat ring indeed, but the electric potential it generates is 3-dimensional. Stable equilibrium means that there is a restoring force if the test particle is displaced in any direction in 3-d space. Of course, if the particle were somehow constrained from moving off the plane of the flat ring, then one would consider equilibrium in 2-d space.
 
  • #19
kuruman
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As you said we will have that $$
\frac{\partial^2 V}{\partial z^2}= -\frac{1}{\rho}\frac{\partial }{\partial\rho}\left(\rho\frac{\partial V}{\partial \rho}\right) $$
I think that we have a maximun since the second derivative is negative and thus unstability, but if the laplacian applies we will not have any maximun or minimun in this area
Which second derivative is negative? All that says is they have opposite signs and that's enough. Laplace's equation applies because there is no charge density in the area of interest.
 
  • #20
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how can I relate the Laplacian with the stability of the equilibrium?
 
  • #21
bob012345
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It is a flat ring indeed, but the electric potential it generates is 3-dimensional. Stable equilibrium means that there is a restoring force if the test particle is displaced in any direction in 3-d space. Of course, if the particle were somehow constrained from moving off the plane of the flat ring, then one would consider equilibrium in 2-d space.
True indeed. I can't say more without saying too much on this point. Just that it is not clear whether the problem was meant as a 2D only problem or not from the description in the OP.
 
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  • #22
kuruman
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how can I relate the Laplacian with the stability of the equilibrium?
Think about what equilibrium means in terms of forces. They must be restoring in both directions ##\rho## and ##z##. Here is a simple exercise.

Consider two potential energies ##U_1=\frac{1}{2}kx^2## and ##U_2=-\frac{1}{2}kx^2 ~~ (k>0)##.
1. Which one gives rise to a restoring force, which one does not and why?
2. Can you deduce a general rule for answering the previous question that is applicable to any potential energy ##U(x)##?
 
  • #23
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So equilibrium means that the net force is zero. If we derive the equations you gave me we will have the force of a oxcillation so ## F_1 = \frac{\partial U_1 }{\partial x} = kx## and the ##F_2= \frac{\partial U_2 }{\partial x} = -kx ##. So the restoring force has to be negative, ## F_2##, cause the restoring force has to take the particle back to the equilibrium point and thus it most be against the motion direction.
 
  • #24
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I'm thinking about , If we derivate again the ##U_2## we will have ## \frac{\partial^2U}{\partial x^2} = - k##, where k is a constant, and if we compare it with the laplacian ##
\frac{\partial^2 V}{\partial z^2}= -\frac{1}{\rho}\frac{\partial }{\partial\rho}\left(\rho\frac{\partial V}{\partial \rho}\right) ## we can put that ## \frac{1}{\rho}\frac{\partial }{\partial\rho}\left(\rho\frac{\partial V}{\partial \rho}\right)## = constant, but I dont know how to continue or if my reasoning is right
 
  • #25
kuruman
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So equilibrium means that the net force is zero. If we derive the equations you gave me we will have the force of a oxcillation so ## F_1 = \frac{\partial U_1 }{\partial x} = kx## and the ##F_2= \frac{\partial U_2 }{\partial x} = -kx ##. So the restoring force has to be negative, ## F_2##, cause the restoring force has to take the particle back to the equilibrium point and thus it most be against the motion direction.
A force is derived from a potential using ##F=-\dfrac{\partial U}{\partial x}##. The negative sign is very important.
 

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