= Solution
Use peak <phasors> with time dependence $e^{-i\omega t}$, real positive <electrical conductivity> $\sigma$, and absolute permeability $\mu_a=\mu\mu_0$. <Ampère-Maxwell equation> and Faraday equations give
$$
\nabla\times\mathbf H=(\sigma-i\omega\epsilon\epsilon_0)\mathbf E=-i\omega\epsilon_0\epsilon'\mathbf E,\qquad \nabla\times\mathbf E=i\omega\mu_a\mathbf H,
$$
so \b[$\epsilon'=\epsilon+i\sigma/(\omega\epsilon_0)$]. The sign of the imaginary term follows from the stated time convention. For a transverse <plane wave> in a homogeneous conductor,
$$
k^2=\mu_a\epsilon\epsilon_0\omega^2+i\mu_a\sigma\omega.
$$
In the good-conductor limit, choose the root that decays into $z>0$:
$$
\boxed{k=\frac{1+i}{\delta},\qquad \delta=\sqrt{\frac2{\sigma\mu_a\omega}}.}
$$
The electric <phasor> and physical field are consequently $\widetilde E_x=E_0e^{(i-1)z/\delta}$ and $E_x=\operatorname{Re}[E_0e^{i(z/\delta-\omega t)-z/\delta}]$, as required. Tangential boundary matching sets $E_0$; it need not equal the incident <amplitude>.
<Faraday's law> fixes the <magnetic field>, including its phase:
$$
\boxed{\widetilde{\mathbf H}=\frac{1+i}{\omega\mu_a\delta}E_0e^{(i-1)z/\delta}\widehat{\mathbf y}.}
$$
For real $E_0$ and $\psi=z/\delta-\omega t$, this means $H_y=E_0e^{-z/\delta}(\cos\psi-\sin\psi)/(\omega\mu_a\delta)$, with other components zero. The time-averaged <Poynting vector> is
$$
\boxed{\langle\mathbf N(z)\rangle=\frac12\operatorname{Re}(\widetilde{\mathbf E}\times\widetilde{\mathbf H}^{*})=\frac{|E_0|^2}{2\omega\mu_a\delta}e^{-2z/\delta}\widehat{\mathbf z}.}
$$
The <energy> entering a thin slab exceeds that leaving it by $-d\langle N_z\rangle/dz$ per unit <volume>, giving
$$
p_{\rm loss}=\frac{|E_0|^2}{\omega\mu_a\delta^2}e^{-2z/\delta}=\boxed{\frac{\sigma|E_0|^2}{2}e^{-2z/\delta}.}
$$
Directly, $\mathbf J=\sigma\mathbf E$ and $\mathbf J\cdot\mathbf E=J^2/\sigma$ at each time. Therefore $\langle J^2/\sigma\rangle=\sigma|\widetilde E_x|^2/2$ is exactly this loss. The factors one half use peak rather than <root mean square> <phasors>; <amplitude> decays over $\delta$, while intensity decays over $\delta/2$.
For a cold collisionless unmagnetized <plasma>, <electron> inertia gives $\widetilde\sigma=i\epsilon_0\omega_p^2/\omega$ and real $\epsilon'=1-\omega_p^2/\omega^2$. Above the <plasma frequency> the <wave> propagates without this resistive attenuation. Below it, a semi-infinite <plasma> supports an evanescent field with electric and magnetic components in quadrature and \b[zero mean inward <energy> flux]: the incident <energy> is reflected, not absorbed as <Joule heating>. <electron> <kinetic energy> is reversibly stored, so the nondispersive <dielectric> field-energy formula alone is incomplete. Collisions introduce a real dissipative <electrical conductivity> and can produce absorption.<Oscillations and Waves by Richard Fitzpatrick>[Section 9.7, equations 9.36–9.53, evanescence, phase quadrature and unit reflectance]{ref}
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