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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)

##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}##

##\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}##

##\vec{F} = \frac{\lambda q}{2 \epsilon_0} \left(\frac{r}{R^2} \right) \hat{r}##

##\vec{F} = -\frac {\lambda q}{2 \epsilon_0} \left(\frac{r}{R^2} \right) \hat{r}##

##\vec{F} = -\frac{\lambda q}{2 \epsilon_0} \left(\frac{r}{R^2} \right) \hat{r}##

##\vec{F} = \frac {\lambda q}{2 \epsilon_0} \left(\frac{r}{R^2} \right) \hat{r}##

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

Thoughts?