- #1
yungman
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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: \lambda = \lambda_0\left | sin \left (\frac {\theta}{2}\right ) \right |[/itex]<br />
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For finding retarded potential at the origin,<br />
\hbox { Let }\; \theta = \phi - \omega t_r<br />
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\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 />
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Where V_{(\vec r,t)} = \frac 1 {4\pi \epsilon_0}\int \frac {\lambda_{(\phi,t_r)}}{a} dl&#039; = \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 />
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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 />
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\phi = 0 \Rightarrow \theta =-\frac { \omega t_r}{2} \hbox { and } \;\phi = 2\pi \Rightarrow \theta = 2\pi-\frac { \omega t_r}{2}<br />
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This will change the limit. Please explain why I can do that.<br />
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Thanks<br />
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Alan
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