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

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SUMMARY

This discussion focuses on solving for the scalar potential V and vector potential A of a rotating line charge density on a circular ring with a specific charge density function, λ = λ₀ |sin(θ/2)|. The problem involves complex integration due to the presence of both time and angular variables, specifically tₗ and φ, in the equations. The participant, Alan, is seeking verification of his approach and hints for solving the integral, which may involve advanced techniques such as series expansions or Bessel's equations.

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yungman
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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 [tex]\lambda = \lambda_0\left | sin \left ( \frac { \theta}{2}\right )\right |[/tex]. Then the ring is place on the xy plane with the center at origin and spin at constant angular velocity [itex]\omega[/itex]. I want to find the scalar potential V and vector potential A at P(x_0,0,z_0).

Homework Equations



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

[tex]\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 )[/tex]

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

[tex]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[/tex]

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

[tex]\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)[/tex]

[tex]\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 )[/tex]


The Attempt at a Solution



[tex]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[/tex]

The next step is to substitude [tex]\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[/tex]

I don't know an easy to solve for [itex]t_r \;\hbox { or } \phi[/itex] as I have both [itex]t_r \;\hbox { and } \; cos \phi[/itex] 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
 
Last edited:
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Anyone?

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

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