Electric field from metal rod with surface charge

Question

I'm not sure if I'm thinking about this problem right.

If you have a metal solid rod of length $L$, radius $a$ and charge density $\sigma$ and want to know the electric field outside and inside the rod, I've done the following using Gauss' Law:

Outside the rod use a Gaussian surface that is a rod of radius $R$, so $R>a$. Then We know from Gauss' Law that $\int \vec{E}\dot{}d\vec{A}=q_{enc}/\epsilon_0$ and $q_{enc}=\sigma 2\pi aL$. We also know that the electric field will always be perpendicular to the Gaussian surface, so we can take that out of the integral, so $E\int dA=(2\pi aL+2\pi a^2)/\epsilon_0$ and $\int dA=2\pi aL$ so $E=\sigma /\epsilon_0$. The ends of the rod cancel out from symmetry so that's why I'm ignoring them here. Inside the rod, there is no enclosed charge since the charge is all on the surface, so $E_{inside}=0$. Is my understanding of this correct?

Answer 1

Charge density can have different meanings. Since you define $q_{enc}=σ2πaL$, I will assume that $\sigma$ is the surface charge density of the metal rod.

One thing is incorrect. $\int E ~dA \neq E (2 \pi a L)$. Since your gaussian surface is a cylindrical of radius $R$ and length $L$, this equation should be $$\int E ~dA = E (2 \pi R L)$$, yielding $ E = \frac{\sigma a}{\epsilon_0 R}$ for $R > a$.

On the other hand, you are right about $E_{\text{inside}} = 0$. Since metal is ideally a conductor, the net enclosed charge is zero.