Relating the Poynting Vector to Resistive Power Dissipation

I'm trying to show that for a electromagnetic plane wave incident and persistent within a linear, homogeneous, and isotropic material that obeys Ohm's law (all part of a spherical cow - just kidding :-), the power dissipated within a region of the material is given by the rate at which energy is flowing into the region. For simplicity, I've chosen a small rectangular prism of infinite length with one end at $z=0$ and the other end negligibly far away. The electric field within the material is given by

$$\mathbf{E}=E_0e^{-\kappa z}cos(\eta ) \hat{x}$$

Where $1/\kappa$ is the penetration depth and $\eta=kz-\omega t$. To calculate the average power dissipated within the material, I compute the power per unit volume (i.e. per $dq$ of charge).

\begin{align*} P_{dq}&=\mathbf{F}_{dq}\cdot\mathbf{v}\\ &=\mathbf{E}\cdot\rho\mathbf{v}\,dV\\ &=\mathbf{E}\cdot\mathbf{J}\,dV\\ &=\sigma E^2\, dV \end{align*}

where I got the last line by Ohm's law. The average power per unit area (i.e. intensity) then is easily evaluated as

\begin{align*} \left<I\right>=\int_0^{\infty} \left<P_{dq}\right> dz & = \int_0^{\infty} \sigma \left<E^2\right> dv\\ &=\sigma E_0^2 \int_0^{\infty} e^{-2\kappa z}\left<\cos^2 (\eta)\right> dz\\ &=\frac{1}{2}\sigma E_0^2 \int_0^{\infty} e^{-2\kappa z} dz \\ &=\boxed{\frac{1}{4\kappa}\sigma E_0^2} \end{align*}

Now for the the average power input per unit area, I can just calculate the time-average of the Poynting vector, $\left<\mathbf{S}\right>$, and dot it with the unit vector normal to the surface it's incident on. This is a bit trickier though, because $\mathbf{S}=(1/\mu)\mathbf{E}\times\mathbf{B}$, and $\mathbf{B}$ is both out of phase and proportional to the real part of the wavenumber.

\begin{align*} \left<\mathbf{S}\right>\cdot\hat{n}&=\left<\frac{1}{\mu}\mathbf{E}\times\mathbf{B}\right>\cdot\hat{n}\\ &=\frac{1}{\mu}E_0B_0e^{-2\kappa z} \left<\cos (\eta )\cos (\eta + \phi )\right>\\ &\vdots \\ &=\frac{1}{2\mu}E_0^2e^{-2\kappa z} \frac{k}{\omega}\\ &=\boxed{\frac{1}{2\mu}E_0^2\frac{k}{\omega}}\,\,\,(z=0) \end{align*}

I have left out the details of the last calculation because (1) I'm fairly certain they are correct, and (2) I'm pretty sure if I've made an error, it will be in one of my base assumptions rather than the details of my calculations. However, I can reproduce them here if there's sufficient demand.

What have I done wrong here?

• Just a few thoughts: Ohm's law ignores the magnetic field because they are always in phase; but in this situation they are out of phase -- so how is this power dissipation justified? Secondly, if it is valid, your results develop a relationship between the constants; test this relationship for consistancy of units and of magnitude for some test cases. – Peter Diehr Mar 9 '16 at 12:34
• @PeterDiehr Thanks for the advice. The power dissipation term is a specific case of the more general formula for the net power done on the charges in any system, $$P=\int \mathbf{E}\cdot\mathbf{J}\,dV$$ There is no magnetic field term in here because they do no work on charges. In the special case of Ohm's law, $\mathbf{J}=\sigma \mathbf{E}$, we can recover the standard $P=IV$. – Arturo don Juan Mar 9 '16 at 13:23
• yes, I know the general formula; my real doubt is the practicability of Ohm's Law here. I've "derived" Ohm's law for students, using statistical mechanics and the Drude model, but I don't see how this would extend to waves in metals -- time to look at Jackson's Classical Electrodynamics! – Peter Diehr Mar 9 '16 at 13:50