Unstable or stable electrostatic equilibrium?

[Angela G]

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 don't hade longer circular simetry and then the force by columbs law

 

[PeroK]

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?

 

[PeroK]

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?

 

[Angela G]

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.

 

[Angela G]

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 don't know, I just used the shell theorem once and it was long ago or I am just confused which shell theorem do you mean?

 

[PeroK]

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.

 

[Angela G]

ok, so for a charged shell we will have that the electric field will be $\vec E = 0$ due to the symmetry 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?

 

[Angela G]

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"

 

[PeroK]

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.

 

[Angela G]

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 $R = \infty$

 

[PeroK]

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.

 

[Angela G]

could you please give me som advice for how to solve the laplacian in this case? Should I use the 2d or 1d equation?

 

[bob012345]

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.

 

[kuruman]

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?

 

[bob012345]

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.

 

[Angela G]

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

 

[Angela G]

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

 

[kuruman]

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.

 

[kuruman]

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.

 

[Angela G]

how can I relate the Laplacian with the stability of the equilibrium?

 

[bob012345]

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.

 

[kuruman]

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 $U(x)$?

 

[Angela G]

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.

 

[Angela G]

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 don't know how to continue or if my reasoning is right

 

[kuruman]

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.

 

[kuruman]

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 don't know how to continue or if my reasoning is right

Answer question 2 in post #22 and you will see.

 

[Angela G]

I think that you already answer it, it is that $- \nabla V = F$?

 

[kuruman]

I think that you already answer it, it is that $- \nabla V = F$?

No, the questions in post #22 have not been answered. You are given two potential energies, $U_1=\frac{1}{2}kx^2$ and $U_2=-\frac{1}{2}kx^2.$ These give rise to forces $\vec F_1=-\vec\nabla U_1$ and $\vec F_2=-\vec\nabla U_2.$ The two questions for you to answer are:
1. Which one of the two forces, $\vec F_1$, $\vec F_2$ is restoring, which one is not and why?
2. Can you figure out a rule for distinguishing the restoring from the non-restoring forces given any potential $U(x)$, not just the ones I gave you?

If you can figure out the answers (we are here to help), you will be able to understand what's going on here.

 

[bob012345]

This video might help.

 

[PhDeezNutz]

I'm willing to bet it's unstable. To prove stable equilibrium you need to prove it's stable in EVERY direction at the origin. To prove unstable you only need to prove it's unstable in one direction at the origin. That should be your goal.

Use symmetry considerations.

Also as someone said look at the field and see if it points back or away from the suggested equilibrium point.

I believe we have already covered the potential on the $z-axis$ and concluded that the field pointed back towards the origin so it's stable in that direction of displacement.

Consider instead the potential in the plane of the ring and inside the ring

I'll get started and maybe others can chime in
$V = \frac{1}{4 \pi \epsilon_0} \int \frac{\lambda}{\left|\vec{R} - \vec{r} \right|} \, dl' = \frac{\lambda R}{4 \pi \epsilon_0}\int \frac{1}{ \sqrt{R^2 + r^2 - 2 \vec{R} \cdot \vec{r}} } \, d \phi'$

$\vec{R} = \left( R \cos \phi', R \sin \phi', 0\right)$
$\vec{r} = \left( r \cos \phi, r \sin \phi, 0\right)$

Make $\phi = 0$, you can do this arbitrarily due to cylindrical symmetry

And you get

$V = \frac{\lambda R}{4 \pi \epsilon_0}\int \frac{1}{ \sqrt{R^2 + r^2 - 2Rr \cos \phi' }} \, d \phi '$Can you invoke a smallness parameter and get rid of that middle term?

Can you take a (negative) derivative with respect to $r$ (you can move this inside the integral) to find the field?

Can you then evaluate the integral and find out which direction the field points? (u-substitution)

 

[Angela G]

Also as someone said look at the field and see if it points back or away from the suggested equilibrium point.

But isn't the field at this point zero? should I do some assumption for the charge line density? because I didnt get the sign of it , and if so I will choose a positive charge line

 

[Angela G]

I believe we have already covered the potential on the z−axis and concluded that the field pointed back towards the origin so it's stable in that direction of displacement.

where did we that ?

 

[Angela G]

I'm sorry, I'm feeling lost in this problem

 

[PhDeezNutz]

But isn't the field at this point zero? should I do some assumption for the charge line density? because I didnt get the sign of it , and if so I will choose a positive charge line

If I wasn't given a sign for the charge I'd assume it was positive.

Yes the field is zero at the middle of the ring, but I'm not talking about the middle of the ring. I'm talking about a point ever so slightly off the center of the ring but still in the plane of the ring. If the field there points away from the origin then you have unstable equilibrium. I would think anyway.

Let me get back to you in like half an hour to an hour and try working it out. Let me make sure what I'm saying makes sense.

 

[bob012345]

I'm sorry, I'm feeling lost in this problem

I'm willing to bet it's unstable. To prove stable equilibrium you need to prove it's stable in EVERY direction at the origin. To prove unstable you only need to prove it's unstable in one direction at the origin. That should be your goal.

Use symmetry considerations.

Also as someone said look at the field and see if it points back or away from the suggested equilibrium point.

I believe we have already covered the potential on the $z-axis$ and concluded that the field pointed back towards the origin so it's stable in that direction of displacement.

Consider instead the potential in the plane of the ring and inside the ring

I'll get started and maybe others can chime in
$V = \frac{1}{4 \pi \epsilon_0} \int \frac{\lambda}{\left|\vec{R} - \vec{r} \right|} \, dl' = \frac{\lambda R}{4 \pi \epsilon_0}\int \frac{1}{ \sqrt{R^2 + r^2 - 2 \vec{R} \cdot \vec{r}} } \, d \phi'$

$\vec{R} = \left( R \cos \phi', R \sin \phi', 0\right)$
$\vec{r} = \left( r \cos \phi, r \sin \phi, 0\right)$

Make $\phi = 0$, you can do this arbitrarily due to cylindrical symmetry

And you get

$V = \frac{\lambda R}{4 \pi \epsilon_0}\int \frac{1}{ \sqrt{R^2 + r^2 - 2Rr \cos \phi' }} \, d \phi '$Can you invoke a smallness parameter and get rid of that middle term?

Can you take a (negative) derivative with respect to $r$ (you can move this inside the integral) to find the field?

Can you then evaluate the integral and find out which direction the field points? (u-substitution)

Is it legal to reduce the integral by using a smallness assumption before the integration? Also, I know what all these terms mean but it may not be obvious to everyone what you mean by $r, R, \phi'$ ect.

 

[bob012345]

I'm sorry, I'm feeling lost in this problem

Understandable. I suggest taking a step back. Assuming the signs of the ring charge and the test charge in the center are all positive, use logic and symmetry to determine which way the forces must act and thus whether it is stable in the plane of the ring and along the $Z$ axis for a small displacement. Then, if you need to solve for an approximate potential in the plane of the ring near the center you can use that to confirm your suspicions. Computing the actual potential off axis in the plane is very difficult and that's why I say approximate the solution for a small displacement. For a $Z$ displacement it is a bit easier because it still has symmetry around the ring and you can get that by hand. That is also true for part b.

 

[PhDeezNutz]

Is it legal to reduce the integral by using a smallness assumption before the integration? Also, I know what all these terms mean but it may not be obvious to everyone what you mean by $r, R, \phi'$ ect.

I didn't think anything of it when I did it out of instinct. Now I'm second guessing myself. Thanks for pointing it out.

Also I think to keep this problem from being trivial we have to assume charges of different signs. Your thoughts?

 

[kuruman]

Is it legal to reduce the integral by using a smallness assumption before the integration? Also, I know what all these terms mean but it may not be obvious to everyone what you mean by $r, R, \phi'$ ect.

I agree. Whether legal or illegal, all that integration stuff is unnecessary and perhaps confusing. The potential at the center of the ring has a saddle point and that can be shown quite easily from Laplace's equation which is where I tried to lead the OP in #22 and #28.

 

[bob012345]

I didn't think anything of it when I did it out of instinct. Now I'm second guessing myself. Thanks for pointing it out.

Also I think to keep this problem from being trivial we have to assume charges of different signs. Your thoughts?

Assume a positive ring charge and allow the test $q$ to be either sign. Once you know one you know the other.

 

[Angela G]

I wonder if we could start again from the begining.

So we have a circle with line charge density $\lambda$ and at the centre of it we have a point charge q. We assume that both charges are positive and thus the electric field due to the line charge $\lambda$ will point towards to the center. Due to the symmetry, the electric field at the center of the circle will be zero, and from the problem we know that the system is in equilibrium, it means that there is no resultant force, $\sum \vec F_{res} = 0$.

Since we are in vacuum the Laplace equation applies around the centre. From a previous exercise I determined the potential at the center of the ring without the point charge. I got $$ V (0) = \frac{\lambda}{2\epsilon_0} $$

we know that if the second derivative of the potential energy $\nabla ^2 U > 0$ we have a stable equilibrium. But I thinkt this is not usable because the potential acoording to the Laplace equ. has no maxima or minima. So we can go another way, and it was what we disscused earlier. That the negative gradient of the potential is equal to the restoring force. $$ - \nabla U = F $$

 

[kuruman]

I wonder if we could start again from the begining.

So we have a circle with line charge density $\lambda$ and at the centre of it we have a point charge q. We assume that both charges are positive and thus the electric field due to the line charge $\lambda$ will point towards to the center. Due to the symmetry, the electric field at the center of the circle will be zero, and from the problem we know that the system is in equilibrium, it means that there is no resultant force, $\sum \vec F_{res} = 0$.

Since we are in vacuum the Laplace equation applies around the centre. From a previous exercise I determined the potential at the center of the ring without the point charge. I got $$ V (0) = \frac{\lambda}{2\epsilon_0} $$

we know that if the second derivative of the potential energy $\nabla ^2 U > 0$ we have a stable equilibrium. But I thinkt this is not usable because the potential acoording to the Laplace equ. has no maxima or minima. So we can go another way, and it was what we disscused earlier. That the negative gradient of the potential is equal to the restoring force. $$ - \nabla U = F $$

How do you know that the potential according to Laplace's equation has no maxima or minima? You are very close to figuring this out. If the potential has a minimum in both radial and axial directions near the origin, then the origin is a stable equilibrium point. What do you need to show in order to prove or disprove that?

 

[bob012345]

I wonder if we could start again from the begining.

So we have a circle with line charge density $\lambda$ and at the centre of it we have a point charge q. We assume that both charges are positive and thus the electric field due to the line charge $\lambda$ will point towards to the center. Due to the symmetry, the electric field at the center of the circle will be zero, and from the problem we know that the system is in equilibrium, it means that there is no resultant force, $\sum \vec F_{res} = 0$.

Since we are in vacuum the Laplace equation applies around the centre. From a previous exercise I determined the potential at the center of the ring without the point charge. I got $$ V (0) = \frac{\lambda}{2\epsilon_0} $$

we know that if the second derivative of the potential energy $\nabla ^2 U > 0$ we have a stable equilibrium. But I thinkt this is not usable because the potential acoording to the Laplace equ. has no maxima or minima. So we can go another way, and it was what we disscused earlier. That the negative gradient of the potential is equal to the restoring force. $$ - \nabla U = F $$

So just knowing that, what happens if the charge $q$ is slightly displaced off-center? Does it go back to the center or fly away towards the ring? Also, the potential you got is only valid at the center.

 

[Angela G]

How do you know that the potential according to Laplace's equation has no maxima or minima?

I think its a propiety of Laplace's equation

if we move the point charge slightly off the center the potential will increase

 

[Angela G]

So just knowing that, what happens if the charge $q$ is slightly displaced off-center? Does it go back to the center or fly away towards the ring? Also, the potential you got is only valid at the center.

I'm not sure, I tried to draw the electric field to get a visual picture. when we displace the charge off the center to the right , there will be a resultant field, but still not sure in which direction.

 

[Angela G]

What do you need to show in order to prove or disprove that?

I think that I need to see if the second derivative of the potential is positive or negative

 

[bob012345]

I'm not sure, I tried to draw the electric field to get a visual picture. when we displace the charge off the center to the right , there will be a resultant field, but still not sure in which direction.

I see the confusion. The charge is acted on by the field of the ring and not the field of the ring plus its own field. That should make things easier.

 

[kuruman]

I think that I need to see if the second derivative of the potential is positive or negative

Read post #14 carefully, very carefully, especially the last sentence. Do you understand what it is saying to you?

 

[kuruman]

I see the confusion. The charge is acted on by the field of the ring and not the field of the ring plus its own field. That should make things easier.

I think by "resultant" OP means "non-zero".

 

[bob012345]

I think by "resultant" OP means "non-zero".

Perhaps but it was the phrase but still not sure in which direction which led me to think there was confusion because it was already stated the ring field points toward the center.

 

[Angela G]

I looked at what happened if we displace the point charge in the z- axis, we call the displacement dz. So by symmetry all the element charges of the ring will act on the charge q. But they will not longer cancel each other. Know we will have a non- zero field,$dE_z$, acting on a charge at a distance dz above the center of ring . The non-zero field is in the positive z-direction . Then the force, $dF_z$ due to the ring will point away from the ring and thus there is no restoring force on the z-direction . It means that the equilibrium is not statble in the z- direction

And I was thinking how to connect $$ - \frac{\partial^2 V}{\partial z^2} = \frac{1}{\rho} \frac{\partial}{\partial \rho} \left( \rho \frac{\partial V}{\partial \rho} \right ) $$ To this reasoning, As disscused above we know that the force is $- \nabla V = F$ so I was thinking if we can do
$$ - \frac{\partial^2 V}{\partial z^2} = \frac{1}{\rho} \frac{\partial}{\partial \rho} \left( \rho \frac{\partial V}{\partial \rho} \right ) \iff \frac{\partial}{\partial z} \left(- \frac{\partial V}{\partial z} \right ) = \frac{1}{\rho} \frac{\partial}{\partial \rho} \left( \rho \frac{\partial V}{\partial \rho} \right ) $$ $$ \iff \partial \left(- \frac{\partial V}{\partial z} \right ) = \frac{1}{\rho} \frac{\partial}{\partial \rho} \left( \rho \frac{\partial V}{\partial \rho}\right ) \partial z \iff \partial F_z = \frac{1}{\rho} \frac{\partial}{\partial \rho} \left( \rho \frac{\partial V}{\partial \rho}\right ) \partial z $$ Here can we say that an element of the force is positive and thus it is repulsive $\iff$ the equilibrium is not stable in the z- direction. But I'm not sure if we can do that