Electric potential generated by a electric loop

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SUMMARY

The discussion focuses on the electric potential generated by an electric loop, specifically examining two approaches to calculate it. The first approach uses the static charge density formula, represented as \(\phi(\vec{r_{0}})=\oint\frac{\rho_{static}}{|\vec{r}-\vec{r_{0}}|}d\vec{l}\), questioning the validity of \(\rho\) in this context. The second approach considers the potential from moving charges, expressed as \(\phi(\vec{r_{0}})=\oint\frac{\rho_{total}}{|\vec{r}-\vec{r_{0}}|\times(1-\frac{\vec{v}\bullet(\vec{r}-\vec{r_{0}})}{c|\vec{r}-\vec{r_{0}}|})}d\vec{l}\). The discussion concludes that the first expression is incorrect as it misrepresents the scalar nature of electric potential.

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Tianwu Zang
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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.
 
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"We" cannot write the electric potential as$$\phi(\vec{r_{0}})=\oint\frac{\rho_{static}}{|\vec{r}-\vec{r_{0}}|}d\vec{l}$$because the electric potential is a scalar quantity and your expression implies that it s a vector quantity.
 

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