Equilibrium circular ring of uniform charge with point charge

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Homework Statement:
A circular ring of uniform line charge (##\lambda## and radius ##R##) is placed in the ##xy##-plane with it's center at the origin. A charge ##q## is placed at the center of the ring. Is the point charge ##q## in stable or unstable equilibrium?

Broken into parts

a) What is the potential on the ##z-axis## of the circular ring of uniform line charge? Moreover what is the potential in the limit that ## z \ll R##? What is the potential at ##z = 0##?

b) Using the previous answer what is the general off-axis potential (for an axis-symmetric potential) provided that we are in the regime ##r \ll R##?

c) Now consider a point charge ##q## placed at the origin. Is ##q## in stable or unstable equilibrium? (Consider the cases where ##q## and ##\lambda## have same and opposite signs)
Relevant Equations:
The general formula for the potential off-axis for an axis symmetric potential in the "interior region" is

##\sum_{n=0}^{\infty} A_n r^n P_n \left( \cos \theta \right)##

Where

##\sum_{n=0}^{\infty} A_n r^n## corresponds to the on-axis potential because when ##\theta = 0## we have ##\cos \theta = 1## and ##P_n \left( 1 \right) = 1## for all ##n##

I will eventually use the Taylor Expansion approximation ##\frac{1}{\sqrt{1 + \left(\frac{r}{R}\right)^2}} \approx 1 - \left(\frac{r}{R} \right)^2## to quadratic order

For good measure let us list the first few Legendre Polynomials in

##P_0 \left( \cos \theta \right) = 1##

##P_1 \left( \cos \theta \right) = \cos \theta ##

##P_2 \left( \cos \theta \right) = \frac{1}{2} \left( 3 \cos^2 \theta -1 \right)##
This is an offshoot of @Angela G 's thread. I don't want to hijack her thread so I decided to create a new one. Original thread https://www.physicsforums.com/threads/unstable-or-stable-electrostatic-equilibrium.1007881/

@kuruman @PeroK @bob012345 If you have the time I'd appreciate your input.

a) What is the potential on the ##z-axis## of the circular ring of uniform line charge? Moreover what is the potential in the limit that ## z \ll R##? What is the potential at ##z = 0##?

General potential on the ##z##-axis

##\Phi \left( z \right) = \frac{1}{4 \pi \epsilon_0} \int \frac{\lambda}{\sqrt{R^2 + z^2}} R \, d \phi' = \frac{\lambda}{2 \epsilon_0} \frac{R}{\sqrt{R^2 + z^2}}##

In the limit that ## z \ll R##

##\Phi \left( z \ll R \right) = \frac{\lambda}{2 \epsilon_0} \frac{1}{\sqrt{1 + \left(\frac{z}{R}\right)^2}}##

Moreover potential at the origin ## \left(x,y,z\right) = \left( 0,0,0 \right) ## is

##\Phi_{origin} = \frac{\lambda}{2 \epsilon_0}##

b) Using the previous answer what is the general off-axis potential (for an axis-symmetric potential) provided that we are in the regime ##r \ll R##?

Recalll that the general form of off_axis potential (for an axis symmetric potential) in the limit ##r // R## is

##\Phi_{off-axis} \left( r , \theta \right) = \sum_{n=0}^{\infty} A_n r^n P_n \left( \cos \theta \right)##

On axis ##\theta = 0## and ## \cos \left( \theta = 0 \right) = 1##, Recall ##P_n \left( 1 \right) = 1## for all ##n##

so ##\sum_{n=0}^{\infty} A_n r^n ## is the on_axis potential

The on_axis potential

##\sum_{n=0}^{\infty} A_n r^n = \frac{\lambda}{2 \epsilon_0} \frac{1}{\sqrt{1 + \left(\frac{z}{R}\right)^2}} \approx \frac{\lambda}{2 \epsilon_0} \left(1 - \left(\frac{r}{R} \right)^2 \right)##

Therefore ##A_0 = 1## and ##A_1 = 0## and ##A_2 = -\frac{1}{R^2}##

Therefore general off_axis potential in the limit ##r \ll R## is

##\Phi \left(r \ll R \right) = \frac{\lambda}{2 \epsilon_0} - \frac{\lambda}{4 \epsilon_0} \left( \frac{r}{R}\right)^2 \left(3 \cos^2 \theta - 1 \right)##


(c) Now consider a point charge ##\left(q\right)## placed at the origin (either same sign or opposite sign of ##\lambda##). Is this point charge in stable or unstable equilibrium?

Recall ##U = q \Phi## and ## \vec{F} = - \nabla U = - \frac{\partial U}{\partial r} \hat{r} - \frac{1}{r}\frac{\partial U}{\partial \theta} \hat{\theta}## (No azimuthal dependence)



Same Sign

##U = \frac{\lambda q}{2 \epsilon_0} - \frac{\lambda q}{4 \epsilon_0} \left( \frac{r}{R}\right)^2 \left( 3 \cos^2 \theta - 1\right)##

##\vec{F} = \frac{\lambda q}{4 \epsilon_0} \left(\frac{r}{R^2} \right) \left( 3 \cos^2 \theta - 1\right) \hat{r} - \frac{3 \lambda q}{2 \epsilon_0} \left(\frac{r}{R^2} \right) \cos \theta \sin \theta \hat{\theta}##

Opposite Sign

##\vec{F} =- \frac{\lambda q}{4 \epsilon_0} \left(\frac{r}{R^2} \right) \left( 3 \cos^2 \theta - 1\right) \hat{r} + \frac{3 \lambda q}{2 \epsilon_0} \left(\frac{r}{R^2} \right) \cos \theta \sin \theta \hat{\theta}##


Displaced Along z-axis SAME SIGN ##\theta = 0##

##\vec{F} = \frac{\lambda q}{2 \epsilon_0} \left(\frac{r}{R^2} \right) \hat{r}## Points away from origin so UNSTABLE in that direction

Displaced Along z-axis OPPOSITE SIGN ##\theta = 0##


##\vec{F} = -\frac {\lambda q}{2 \epsilon_0} \left(\frac{r}{R^2} \right) \hat{r}## Points towards origin so STABLE in that direction


Displaced Along radial direction in the plane SAME SIGN
##\theta = 0##

##\vec{F} = -\frac{\lambda q}{2 \epsilon_0} \left(\frac{r}{R^2} \right) \hat{r}## Points towards origin so STABLE in that direction

Displaced Along z-axis OPPOSITE SIGN ##\theta = 0##


##\vec{F} = \frac {\lambda q}{2 \epsilon_0} \left(\frac{r}{R^2} \right) \hat{r}## Points away from origin so UNSTABLE in that direction


Given that both cases of sign (opposite and same) are stable in one direction and unstable in the other

I conclude that the center is a saddle point in both cases (i.e. unstable)

Thoughts?
 

Answers and Replies

  • #2
bob012345
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I agree on the stability issue. Regarding the small ##r## displacement in the plane of the ring, why would there be a ##\theta## dependence?

Also, It would be instructive to for you to do the stability formally by taking the second derivative of the potential energy like we discussed.
 
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  • #3
vela
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In the limit that ## z \ll R##

##\Phi \left( z \ll R \right) = \frac{\lambda}{2 \epsilon_0} \frac{1}{\sqrt{1 + \left(\frac{z}{R}\right)^2}}##
Note that this holds true for all values of ##z##. You haven't used the assumption that ##z \ll R##.

##\sum_{n=0}^{\infty} A_n r^n = \frac{\lambda}{2 \epsilon_0} \frac{1}{\sqrt{1 + \left(\frac{z}{R}\right)^2}} \approx \frac{\lambda}{2 \epsilon_0} \left(1 - \left(\frac{r}{R} \right)^2 \right)##
You're missing a factor of 1/2 on the second term.
 
  • #4
kuruman
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I agree with you and if you read my posts in Angela's thread, you will see that I consistently tried to direct the OP to the conclusions you have above in bold oversize letters. In post #38 I even mention "saddle point". The reasoning is simple and does not require Legendre polynomials to support it.
1. The two second derivatives add up to zero according to Laplace's equation.
2. This means that they have opposite signs so that's a saddle point. If the equilibrium is stable in the radial direction, it will be unstable in the axial direction and vice versa.

This much answers the stability question. To figure out what condition must be true for, say, the axial direction to have stability, one has to calculate the axial second derivative of the potential at the center of the ring.$$\varphi(0,0,z)=\frac{kQ}{(z^2+R^2)^{1/2}}\implies \left. \frac{\partial^2 \varphi(0,0,z)}{\partial z^2}\right |_{z=0}=-\frac{kQ}{R^3}.$$Here ##Q## is the total charge on the ring. From this, we conclude that
  • When you place a like charge ##q## at the center, the derivative of the potential energy is negative; equilibrium is unstable in the axial direction and stable in the radial direction.
  • When you place an unlike charge ##q## at the center, the derivative of the potential energy is positive; equilibrium is stable in the axial direction and unstable in the radial direction.
 
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  • #5
vela
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Displaced Along z-axis SAME SIGN ##\theta = 0##

Displaced Along radial direction in the plane SAME SIGN ##\theta = 0##
Why ##\theta = 0## when you're in the plane of the ring? Was this a typo?
 
  • #6
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Why ##\theta = 0## when you're in the plane of the ring? Was this a typo?
Yes meant ##\theta = \frac{\pi}{2}## Guess I have to redo the calculations real quick but I think my conclusion would hold.

@kuruman you're several levels higher than me when it comes to these type of problems. I'm still at the stage where I rely on brute computation.

Brb redoing problem to incorporate the missing factor and the correct angle.
 
  • #7
bob012345
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@PhDeezNutz, yes your basic conclusions were correct but it would be good to refine your treatment of how you estimated the radial off-axis potential. How did you get what you got?
 
  • #8
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@kuruman I just took the time to digest your post and it is quite insightful. I need to get that level where I no longer rely on brute (or semi-brute) computation to come to conclusions.
 

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