Potential and Electric Field

[Philosophaie]

Homework Statement

If you have an infinitely long straight conductor radius, a,with a current density ρ how do you find the Potential and Electric Field?

Homework Equations

Gauss Law $\phi = \int{\vec{E}.d\vec{a}} = 4*\pi*q_{enclosed}$

$q_{enclosed}=\int\int\int{\rho dz dy dx}$

$E=-grad(\phi)$

The Attempt at a Solution

I choose the conductor to be along the x-axis. The cross-sectional area is 2*$\pi$*a at a point $\sqrt{y^2+z^2}$ then the integral in x is taken from $\inf to -\inf$.

 

[vanhees71]

I'd use the local form of the electrostatic Maxwell equations. Best are cylinder coordiantes, here around the x axis of the Cartesian coordinate system
\vec{r}=(x,\rho \cos \phi,\rho \sin \phi), \quad x \in \mathbb{R}, \quad \rho>0, \quad \phi \in [0,2 \pi).
You have (in Gaussian units)
\vec{E}=-\vec{\nabla} \phi, \quad \Delta \phi=-\vec{\nabla} \cdot \vec{E}=-4 \pi \rho.
Then you make an ansatz for \phi accoding to the symmetries of the system.

Further note that \rho is the charge density. If it's a coductor, it should be a surface charge on the outer surface of the conductor!

 

[Philosophaie]

You used $\rho$ and $\phi$ for both cylindrical and charge density & Potential. It was a little confusing

Potential and Electric Field vary by the inverse of the distance which means it is weaker the further you get from the Charge Density. The infinitely long conductor has to be taken in consideration as well as surface charge of the outer shell of the conductor.

The divergence of E gives 4*$\pi*\rho$. How does that give the inverse of the distance?