# Infinite, straight, current-carrying wire uniformly charged to a negative electrostatic potential

I am working on a problem that states the following:

Imagine an infinite straight wire carrying a current $I$ and uniformly charged to a negative electrostatic potential $\phi$

I know here that the current $I$ will set up a magnetic field around the wire that abides to the right hand rule with magnitude:

$B(r) = \frac{I\mu_0}{2\pi r}.$

However, what is the importance of there being a negative electrostatic potential $\phi$? Does this mean that the wire sets up an electrostatic $\vec{E}$ field in addition to the magnetic field?

• I found that the potential should be of the form: $V(r) = C_1 \ln(r) + C_0$ but then how am I supposed to apply the boundary condition that at $r=0, V(r) = \phi$ ? Should I assume instead that at some $r_0 \ne 0$ that $V(r_0) = \phi$, meaning that the wire has some finite width? – Loonuh Jun 3 '15 at 0:07