Unstable or stable electrostatic equilibrium?

[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

 

[PeroK]

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

That's true if the charges are of the same sign. But, if the charges are of opposite sign, there there is stable equilibrium in the z-direction.

 

[PeroK]

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

In general, a line charge of $\lambda$ may be positive or negative; likewise $q$ may be positive or negative.

 

[Angela G]

In general, a line charge of λ may be positive or negative; likewise q may be positive or negative.

exacly, we assumed that both charges are positive

 

[PeroK]

exacly, we assumed that both charges are positive

Which is an unjustified assumption. $\lambda$ and $q$ could have opposite signs.

 

[Angela G]

Which is an unjustified assumption. λ and q could have opposite signs

that's true, how can we know the signs?

 

[PeroK]

that's true, how can we know the signs?

You have to analyse both cases. Same sign and opposite sign.

 

[Angela G]

You have to analyse both cases. Same sign and opposite sign

ok, thanks I'll keep that in mind

 

[PhDeezNutz]

In general, a line charge of $\lambda$ may be positive or negative; likewise $q$ may be positive or negative.

I think we should assume opposite signs now that I think about it.

if it were the same signs the point charge wouldn’t experience a restoring force to the center of the ring when displaced up and down the axis of symmetry. It would be repelled. In which case the problem would be a trivial one liner.

I guess for completeness the OP should address this case.

your thoughts?

 

[PeroK]

I think we should assume opposite signs now that I think about it.

if it were the same signs the point charge wouldn’t experience a restoring force to the center of the ring when displaced up and down the axis of symmetry. It would be repelled. In which case the problem would be a trivial one liner.

I guess for completeness the OP should address this case.

your thoughts?

Yes, in 3D the instability of the system for the same sign can be seen simply. That leaves the opposite sign case to be determined.

 

[Angela G]

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