Solving for Vector & Scalar Potential of a Rotating Line Charge Density

[yungman]

Homework Statement

This is not a home work, I actually make this problem up and work on it. I want to verify whether I am correct in the step and I need help to solve the final integration.

The question is:

Given a plastic circular ring radius = a with line charge density glued on it. The charge density around the ring is given as \lambda = \lambda_0\left | sin \left ( \frac { \theta}{2}\right )\right |. Then the ring is place on the xy plane with the center at origin and spin at constant angular velocity \omega. I want to find the scalar potential V and vector potential A at P(x_0,0,z_0).

Homework Equations

\vec R \;\hbox { is position vector pointing to P. } \vec w(\phi) \;\hbox { is position vector pointing to a point of the ring.}

\phi = \theta + \omega t_r = \theta +\omega \left ( t-\frac {\eta}{c}\right ) \;,\; \vec R = \hat x_0 + \hat z z_0\;,\; \vec w (\phi) = \hat x a\;cos (\phi) + \hat y a\; sin(\phi) = \hat x a\;cos (\theta + \omega t_r ) + \hat y a\; sin(\theta + \omega t_r )

\vec v(\phi) = \vec w\;' (\phi) = -\hat x a\omega\;sin (\theta + \omega t_r ) + \hat y a\omega \; cos(\theta + \omega t_r )

V_{(\vec r,t)} = \frac {1}{4\pi \epsilon_0}\int \frac { \rho_{(\vec r\;',t_r)}}{\eta c -\vec{\eta} \cdot \vec v(\phi)} a d\phi \;\hbox { where }\;\vec {\eta} = \vec R - \vec w (\phi) = \hat x(x_0-acos\phi)-\hat y a sin\phi+\hat z z_0

\vec{\eta} \;\hbox { is the vector from a point of interest on the plastic ring to point P.}

\eta = \sqrt{(x_0-acos\phi)^2+a^2sin^2\phi + z_0^2} = \sqrt{(x_0^2+a^2 -2ax_0 cos\phi) + z_0^2} = c(t-t_r)

\vec {\eta} \cdot \vec v(\phi) = -a\omega sin\phi(x_0-acos\phi)-a^2\omega cos\phi sin\phi = -a\omega x_0sin\phi = -a\omega x_0 sin \left ( \theta + \omega \left ( t- \frac {\eta}{c}\right ) \right )

The Attempt at a Solution

V_{(\vec r,t)} = \frac {1}{4\pi \epsilon_0}\int \frac { \rho_{(\vec r\;',t_r)}}{\eta c -\vec{\eta} \cdot \vec v(\phi)} a d\phi = \frac {\lambda_0 c }{4\pi \epsilon_0}\int \frac {\left | sin \left ( \frac {\phi}{2}\right ) \right | }{c\sqrt{x_0^2 +a^2-2ax_0cos\phi+z^2} + a\omega x_0sin\phi} a d\phi

The next step is to substitude \phi = \theta + \omega \left ( t- \frac {\eta}{c}\right ) \;\hbox { and }\; \eta = c(t-t_r) \;\hbox { to solve for }\; t_r \;\hbox { and }\; \phi

I don't know an easy to solve for t_r \;\hbox { or } \phi as I have both t_r \;\hbox { and } \; cos \phi in the same equation.

Please check for me whether I am doing this correctly so far because I don't have the correct solution since I make up this problem. And also give me hints how to solve of t_r or \phi.


Thanks

Alan

 

[yungman]

Anyone?

I was told the method I use is correct. The solution is very hard to get because both sin\phi\;,\; cos \phi \hbox { and } t_r exist in the same equation. The solution might involve series or Bessel's equation and there is no easy way.