Question on a sinodal charge distribution regarding retarded potential.

[yungman]

The example is about finding the retarded scalar potential V at the origin from a plastic circular ring on xy plane center at origin spining at constant angular velocity $\omega$ with line charge density: [tex]\lambda = \lambda_0\left | sin \left (\frac {\theta}{2}\right ) \right |[/itex]<br /> <br /> <br /> <br /> <br /> For finding retarded potential at the origin,<br /> $$\hbox { Let }\; \theta = \phi - \omega t_r$$<br /> <br /> $$\lambda_{(\phi,t_r)} = \lambda_0\left | sin \left (\frac {\theta}{2}\right ) \right | = \lambda_0\left | sin \left (\frac {\phi -\omega t_r}{2}\right ) \right |$$<br /> <br /> Where $$V_{(\vec r,t)} = \frac 1 {4\pi \epsilon_0}\int \frac {\lambda_{(\phi,t_r)}}{a} dl' = \frac {\lambda_0} {4\pi \epsilon_0}\int_0^{2\pi} \frac {\left |sin \left (\frac {\phi - \omega t_r}{2}\right ) \right | }{a} ad\phi = \frac {\lambda_0} {4\pi \epsilon_0}\int_0^{2\pi} \left |sin \left (\frac {\theta}{2}\right ) \right | d\theta$$<br /> <br /> Notice the limit of the two integration still the same? My question is the book claim both $\phi \;\hbox { and }\; \theta$ are 0 to $2\pi$. But if you do the convensional substitution:<br /> <br /> $$\phi = 0 \Rightarrow \theta =-\frac { \omega t_r}{2} \hbox { and } \;\phi = 2\pi \Rightarrow \theta = 2\pi-\frac { \omega t_r}{2}$$<br /> <br /> This will change the limit. Please explain why I can do that.<br /> <br /> Thanks<br /> <br /> Alan[/tex]

 

[yungman]

Actually I figure out my original question. But I have a totally different question. I modified the original post #1. Please take a look and help me.

Thanks

Alan