Effect of negative electrostatic potential on infinite wire?

[Loonuh]

Homework Statement

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 Φ

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 in Eqn. (1).

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

Homework Equations

##
\begin{align}
B(r) = \frac{I\mu_0}{2\pi r} \\
\nabla \cdot \vec{E} = 0
\end{align}
##

The Attempt at a Solution

##
\begin{align*}
\nabla \cdot \vec{E} &= 0\\
\frac{d^2 V}{dr^2} &= 0\\
\therefore V &= Cr + D\\
\end{align*}
##

At r = 0, V = ##\phi##

##
\begin{align*}
V = Cr + \phi
\end{align*}
##

At r = ##\infty##, V = 0 ...?

This can't possibly be right now and it appears I made some mistake.

 

[vela]

Homework Statement

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 Φ

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 in Eqn. (1).

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

Homework Equations

##
\begin{align}
B(r) = \frac{I\mu_0}{2\pi r} \\
\nabla \cdot \vec{E} = 0
\end{align}
##

The Attempt at a Solution

\begin{align*}
\nabla \cdot \vec{E} &= 0\\
\frac{d^2 V}{dr^2} &= 0\\
\therefore V &= Cr + D\\
\end{align*}

That's not the correct expression for the Laplacian in cylindrical coordinates.

At r = 0, V = ##\phi##
\begin{align*}
V = Cr + \phi
\end{align*}

At r = ##\infty##, V = 0 ...?

This can't possibly be right now and it appears I made some mistake.

Are you supposed to take the wire as infinitely thin? Even after you get the proper expression for the potential, you'll run into math problems if you assume the wire doesn't have a finite radius.

 

[Loonuh]

Wow, that was a very obvious mistake, thanks for that correction. I believe that I am free to assume that the wire is of a finite radius.

Solving now where the radial derivative term of the Laplacian is expressed as:## \nabla^2 = \frac{1}{r} \frac{\partial}{\partial r} (r \frac{\partial}{\partial r}) ...##

We have
##
\begin{align*}
\frac{1}{r} \frac{\partial}{\partial r} (r \frac{\partial V}{\partial r}) = 0
\end{align*}
##

Let ##y = \dot{V}##, then:

##
\begin{align*}
\dot{y} + \frac{1}{r}y &= 0\\
\frac{\dot{y}}{y} &= \frac{-1}{r}\\
\frac{dy}{y} &= \frac{-dr}{r}\\
\ln(y) &= -\ln(r) + C\\
\\
\therefore y &= Cr^{-1}\\
\dot{V} &= Cr^{-1}\\
dV &= Cr^{-1}dr\\
\\
\therefore V(r) &= C_0\ln(r) + C_1\\
\end{align*}
##

Now how do I move forward with this new expression for the potential assuming that the wire has some finite radius ##r_0##?

How to proceed from here can be found here: http://physics.stackexchange.com/qu...al-of-infinite-wire-with-poisson-laplace-equa