# Potential and Electric Field

1. Nov 11, 2013

### Philosophaie

1. The problem statement, all variables and given/known data
If you have an infinitely long straight conductor radius, a,with a current density ρ how do you find the Potential and Electric Field?

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

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

$E=-grad(\phi)$

3. 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 $\inf to -\inf$.

2. Nov 11, 2013

### 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!

3. Nov 11, 2013

### 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?