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So, I know $\oint E\centerdot dA = 4\pi Q_{enc}$

I'm trying to solve for a TEM mode with two concentric (infinite) cylindrical wave guides of radius a and b, $a<b$. I know that for TEM modes, I can solve by assuming that the outside and inside are at two different potentials, $\pm V$.

I'm told the solution is $\vec E=a\sqrt {\mu/\epsilon}H_0\hat r/r$. It seems to me that the solutions found $\vec H$ first, and then found $\vec E$. I should be able to do the reverse, and end up with the same answer. So, my question is, how can I apply Gauss' Law in this situation? Or, is there simply a better way to solve for $\vec E $ and $\vec H$?

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There's no charge enclosed between the cylinders, and since it's a TEM mode you know that the electric field component along the waveguide axis is zero. So it essentially boils down to solving for the electric field for, say, a cylindrical shell capacitor (which I would guess you have seen before). That will also allow you to express your electric field in terms of V, rather than just throwing in some E0 factor. –  user758556 Jun 12 '12 at 5:30
That can't be the case. There is an electric field between the two cylinders (since it is, after all, a TEM mode) and thus there must be a charge on the inner cylinder in order for the electric field to exist at all. You can easily show this by considering boundary conditions and using the uniqueness theorem. –  Teofrostus Jun 12 '12 at 10:40
Sorry, meant to say no charge density between the cylinders. –  user758556 Jun 12 '12 at 17:31
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