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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?

Loonuh
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1 Answers1

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We would have to know the rest of the problem to be sure, but most likely it means what you say. A wire connected to a DC source will in fact have a current going through it and it will be at some nearly uniform potential. If we wanted to solve Laplace's equation to find the potential everywhere, the wire would work as a boundary condition.

Javier
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  • 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 03 '15 at 00:07
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    @Loonuh: That would be one way to do it, though you still can't impose a boundary condition at infinity. The problem here (see this question for more details) is that the potential for this wire is not well defined. If the problem asks you to find the potential everywhere, then it's a badly-posed problem. – Javier Jun 03 '15 at 00:42