Physics Help Forum Position and velocity vector for the charge in finding retarded potential.

 May 5th 2011, 12:41 AM #1 Member   Join Date: Nov 2010 Posts: 40 Position and velocity vector for the charge in finding retarded potential. I want to verify that I am correct in setting up the two cases specifically where the charges is moving in a circle radius = a with constant velocity \omega. Case 1: A point charge q. The position vector w depends on a single variable t_r. The position and velocity vectors are: $\vec w(\phi) = \vec w (t_r) = \hat x a cos ( \omega t_r) +\hat y a sin ( \omega t_r) \;\hbox { and } \vec v(\phi) = \vec w'(\phi)= -\hat x a\omega sin ( \omega t_r) +\hat y a\omega cos ( \omega t_r) \;\hbox { where } \; \phi = \omega t_r$ Case 2: Charge distribution of line charge density $\lambda =\lambda_0 \left |sin \left (\frac{\theta } {2}\right ) \right |$ The position vector depends on both \theta and t. So the position and velocity vectors are: $\vec w(\phi) = \vec w (\theta,t_r) = \hat x a cos (\theta + \omega t_r) +\hat y a sin ( \theta +\omega t_r) \;\hbox { and } \vec v(\phi) = \vec w'(\phi)= -\hat x a\omega sin (\theta + \omega t_r) +\hat y a\omega cos (\theta + \omega t_r) \;\hbox { where } \; \phi = \theta + \omega t_r$ The reason I claim this is because in the first case, it is a point charge and I don't need \theta. \phi is function of t only. But in the second case where charge density is function of \theta, the position vector w is function of both t and \theta. Last edited by yungman; May 5th 2011 at 01:08 AM.
May 5th 2011, 04:22 AM   #2

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 Originally Posted by yungman I want to verify that I am correct in setting up the two cases specifically where the charges is moving in a circle radius = a with constant velocity \omega. Case 1: A point charge q. The position vector w depends on a single variable t_r. The position and velocity vectors are: $\vec w(\phi) = \vec w (t_r) = \hat x a cos ( \omega t_r) +\hat y a sin ( \omega t_r) \;\hbox { and } \vec v(\phi) = \vec w'(\phi)= -\hat x a\omega sin ( \omega t_r) +\hat y a\omega cos ( \omega t_r) \;\hbox { where } \; \phi = \omega t_r$ Case 2: Charge distribution of line charge density $\lambda =\lambda_0 \left |sin \left (\frac{\theta } {2}\right ) \right |$ The position vector depends on both \theta and t. So the position and velocity vectors are: $\vec w(\phi) = \vec w (\theta,t_r) = \hat x a cos (\theta + \omega t_r) +\hat y a sin ( \theta +\omega t_r) \;\hbox { and } \vec v(\phi) = \vec w'(\phi)= -\hat x a\omega sin (\theta + \omega t_r) +\hat y a\omega cos (\theta + \omega t_r) \;\hbox { where } \; \phi = \theta + \omega t_r$ The reason I claim this is because in the first case, it is a point charge and I don't need \theta. \phi is function of t only. But in the second case where charge density is function of \theta, the position vector w is function of both t and \theta.
I'm sorry but I really don't follow what's going on here at all. For case 1 all I can tell you is that you have correctly written the displacement and velocity vectors of an object moving in a circle of radius a about the origin at constant angular speed (omega). For case 2, is the line charge bent into a circle? And we are talking about the motion of what?.

-Dan
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May 5th 2011, 09:10 AM   #3
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 Originally Posted by topsquark I'm sorry but I really don't follow what's going on here at all. For case 1 all I can tell you is that you have correctly written the displacement and velocity vectors of an object moving in a circle of radius a about the origin at constant angular speed (omega). For case 2, is the line charge bent into a circle? And we are talking about the motion of what?. -Dan
Thanks so much for you help.

In case 2, consider the line charge is gluded onto a plastic ring according to the condition given initially. Then the whole ring is spinning at a constant angular velocity \omega.

What I am trying to establish is that for any line charge distributions that spinning in a circle, I have to present the Position vector (which implies the Velocity vector) as function of \phi, which in turn \phi can be function of other independent variables eg.

Case 1 for point charge q spinning at \omega:

$\vec w(\phi) = \vec w (t_r)$

Case 2: Charge distribution of GLUDED line charge density $\lambda =\lambda_0 \left |sin \left (\frac{\theta } {2}\right ) \right |$ spinning at \omega

$\vec w(\phi) = \vec w (\theta,t_r)$

Last edited by yungman; May 5th 2011 at 09:21 AM.

May 5th 2011, 09:50 AM   #4

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 Originally Posted by yungman Thanks so much for you help. In case 2, consider the line charge is gluded onto a plastic ring according to the condition given initially. Then the whole ring is spinning at a constant angular velocity \omega. What I am trying to establish is that for any line charge distributions that spinning in a circle, I have to present the Position vector (which implies the Velocity vector) as function of \phi, which in turn \phi can be function of other independent variables eg. Case 1 for point charge q spinning at \omega: $\vec w(\phi) = \vec w (t_r)$ Case 2: Charge distribution of GLUDED line charge density $\lambda =\lambda_0 \left |sin \left (\frac{\theta } {2}\right ) \right |$ spinning at \omega $\vec w(\phi) = \vec w (\theta,t_r)$
Okay. Yes for case 1, and basically yes for case 2. I can't see how else to characterize the problem for case 2 any differently than you did. The slight change you might consider is instead of using q for case 2 you might want to use dq. It depends on what further work you are going to do.

-Dan
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 May 5th 2011, 10:33 AM #5 Member   Join Date: Nov 2010 Posts: 40 Thanks for you help. The ultimate thing why I am being so particular is because I am looking in terms coordinates transformation specifically from rectangular to cylindrical and visa versa. $\hat x = (\hat r cos \phi -\hat {\phi} sin\phi) \;\hbox { and } \hat y = (\hat r sin\phi +\hat{\phi} cos\phi)$ $\vec w(\phi)= (\hat r cos \phi -\hat {\phi} sin\phi) a cos (\phi) +(\hat r sin\phi +\hat{\phi} cos\phi) = \hat r a$ $\vec v(\phi) = \vec w'(\phi)= -(\hat r cos \phi -\hat {\phi} sin\phi) a \frac {d\phi}{dt} sin (\phi) +(\hat r sin\phi +\hat{\phi} cos\phi) a\frac {d\phi}{dt}cos (\phi) = \hat{\phi} a \frac {d\phi}{dt}\;\hbox { where } \; \phi = \theta + \omega t_r$ $\frac {d\phi}{dt}=\omega \;\hbox{ for both case 1 and case 2.}$ Using \phi like this is the only way to make it consistant during coordinates transformations. Please verify I am correct. Thanks Alan Last edited by yungman; May 5th 2011 at 10:35 AM.

 Tags charge, finding, position, potential, retarded, vector, velocity

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