- #1
Tianwu Zang
- 11
- 0
Hi all,
What is the potential generated by a electric loop? I have found two ways to sovle the problem. One is since the charge density does not change with time, we can write it as
[tex]\phi(\vec{r_{0}})[/tex]=[tex]\oint\frac{\rho_{static}}{|\vec{r}-\vec{r_{0}}|}d\vec{l}[/tex]. But what is [tex]\rho[/tex] in this loop? Is it should be zero? If the charge density is zero, than the electric potential is zero everywhere, thus there is no electric field. Is it true?
Another approach is that using the potential generated by the moving particles, therefore
[tex]\phi(\vec{r_{0}})[/tex]=[tex]\oint\frac{\rho_{total}}{|\vec{r}-\vec{r_{0}}|\times(1-\frac{\vec{v}\bullet(\vec{r}-\vec{r_{0}})}{c|\vec{r}-\vec{r_{0}}|})}d\vec{l}[/tex]. And this equation, which is always positive, is totally different from the former one, so which one is correct?
Thanks.
What is the potential generated by a electric loop? I have found two ways to sovle the problem. One is since the charge density does not change with time, we can write it as
[tex]\phi(\vec{r_{0}})[/tex]=[tex]\oint\frac{\rho_{static}}{|\vec{r}-\vec{r_{0}}|}d\vec{l}[/tex]. But what is [tex]\rho[/tex] in this loop? Is it should be zero? If the charge density is zero, than the electric potential is zero everywhere, thus there is no electric field. Is it true?
Another approach is that using the potential generated by the moving particles, therefore
[tex]\phi(\vec{r_{0}})[/tex]=[tex]\oint\frac{\rho_{total}}{|\vec{r}-\vec{r_{0}}|\times(1-\frac{\vec{v}\bullet(\vec{r}-\vec{r_{0}})}{c|\vec{r}-\vec{r_{0}}|})}d\vec{l}[/tex]. And this equation, which is always positive, is totally different from the former one, so which one is correct?
Thanks.
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