Originally Posted by
wslgx1024 The original problem is
In electrostatics, a charge distribution over a curve $\displaystyle \mathcal{C}\subset \textbf{R}^{3}$ induces an electric potential in $\displaystyle $\textbf{R}^{3}\backslash\mathcal{C}$.
$
The charge distribution is given by the charge density $\displaystyle $\rho: \mathcal{C} \rightarrow \textbf{R}$$, and the induced electric potential at $\displaystyle $p \in \textbf{R}^{3}\backslash\mathcal{C}$ $
is given by
$\displaystyle
\begin{center}
$V(p)=\frac{1}{4\pi\varepsilon}\int_{\mathcal{C}} \frac{\rho}{\mathrm{dist(p,\cdot)}}$
\end{center}
$
where $\displaystyle $\mathrm{dist(p,\cdot)} = \|p-x\|$$ denotes the distance between $\displaystyle $p\in \textbf{R}^{3}\backslash\mathcal{C}$$ and $\displaystyle $x\in \mathcal{C}$$.
Consider a uniform charge density $\displaystyle $\rho$$ on the set
$\displaystyle
\begin{center}
$\mathcal{C}=\{(x,y,z)\in \textbf{R}^{3}:x^{2}+y^{2}=1,z=0\}$.
\end{center}
$
Find the potential at points on the x-axis, |x|>1.
And by using the parametrization $\displaystyle $x=\cos(\theta)$$ and $\displaystyle $y=\sin(\theta)$
I got the previous integral.
And how could the integral in the link you gave solve this integral?