Infinitely long rod

1. Sep 29, 2008

germx3

The problem:

An infinitely long rod of radius R carries a uniform volume charge
density ρ (ρ > 0).

(a) Show how to use Gauss' Law to prove that the
electric field inside this rod points radially
outward and has magnitude:

$E = \rho r/2\epsilon_0$

(b) Integrate the electric field over an appropriate
displacement to find the potential difference from
the rod's surface to its axis. State explicitly
which of those two locations is at the higher
potential.

I solved for part (a) using a Gaussian surface symmetry and got this as my final answer.

$E(2 \pi rL) = \rho r/2 \epsilon_0$
$E = \rho r/2\epsilon_0$

I am having a hard time starting part (b). I am not sure where to start.

2. Sep 29, 2008

michalll

Electric potencial is a path integral of electric intensity from point A to point B

3. Sep 29, 2008

Defennder

More specifically the formula is $$V = -\int^0_{R} \mathbf{E} \cdot d\mathbf{r}$$. In this case since you've derived the expression in cylindrical coordinates you might as well make use of cylindrical basis vectors for the line integral. Just substitute all the values into the integration, find the dot product and integrate.