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 $\log(\infty)$?
Best Answer
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".