Electric potential generated by a electric loop

AI Thread Summary
The discussion centers on the electric potential generated by an electric loop, with two approaches presented for solving the problem. The first method questions the charge density in the loop, suggesting that if it is zero, the electric potential and field would also be zero. The second approach considers the potential from moving particles, leading to a different equation that remains positive. A key point raised is that the electric potential is a scalar quantity, challenging the validity of the first expression. The thread highlights the complexity of determining electric potential in dynamic versus static scenarios.
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
\phi(\vec{r_{0}})=\oint\frac{\rho_{static}}{|\vec{r}-\vec{r_{0}}|}d\vec{l}. But what is \rho 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
\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}. 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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