# Electric potential of uniformly charged wire

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 $\log(\infty)$?

• 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 $\frac{\pi}{2}$. – Mitchell Jul 26 '17 at 16:00

• Yup, that's pretty much it. Note that you can use LaTeX notation (so e.g. $KQ/r^2$ will render as $KQ/r^2$). – Emilio Pisanty Jul 26 '17 at 15:23
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.