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I want to calculate the electric potential of a uniformly charged wire with infinite length. The problem I run into is that one boundary of the integral is $\infty$. That’s what I have so far:

Given the uniform charge density $\lambda$ and $E(r) = \frac{\lambda}{2\pi r \epsilon_0}$.

$$ V(r) := \int_r^\infty \vec E \cdot \vec dr = \frac{\lambda}{2\pi\epsilon_0}[\log(r)]_r^\infty $$

How should I evaluate $\log(\infty)$?

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  • $\begingroup$ Use angles instead of length. Say that the angle between the line joining the $\vec{dl}$ element of wire and the point where you want to find the potential and the line drawn perpendicularly from that point on the wire be $\theta$. Then you could tend the angle of $\frac{\pi}{2}$. $\endgroup$ – Mitchell Jul 26 '17 at 16:00
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By default we usually suppose that the electric field vanishes in infinity, since for a point charge it is KQ/r². We usually stablish V=0 at infinity in order to cancel one of the terms of the integral.

However, when you do have electric charges in the infinity (and that's the case if the wire is infinite), then you cannot say that the potential in the infinite is 0, and so that cannot be your origin of potentials anymore.

On the contrary, you must stablish a new point as reference. That means you have to choose any point you want and set it as V(there)=0. Then, the integral limits must be "from that reference point to the generic r position".

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  • $\begingroup$ Yup, that's pretty much it. Note that you can use LaTeX notation (so e.g. $KQ/r^2$ will render as $KQ/r^2$). $\endgroup$ – Emilio Pisanty Jul 26 '17 at 15:23
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The expression you use assumes $V(\infty)=0$, which is the same as assuming there is no charge at $\infty$. This is clearly not the case for your setup since your uniformly charged wire is infinitely long.

In the specific case you have the reference potential, i.e. the location where $V=0$, is usually taken to be at $r=0$. In this way you can keep your expression for the potential, which then simply becomes $$ V(r)= \frac{\lambda}{2\pi\epsilon_0}(\log(r)-\log(1))=\frac{\lambda}{2\pi\epsilon_0}\log(r)\, . $$ Note that the $\log$ behavior is typical of problems with cylindrical symmetry.

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