Exercise 5.2.3: Leaky capacitor

$E=V/d$, with amplitude $E_0=V_0/d=10^{5}V/m$. Then

with $\epsilon \omega E_0=2.3\times 8.842\times 10^{-12}\times 2\pi \times 10^3\times 10^5=1.28\times 10^{-2}A/{\mathrm{m}}^2$. Multiplying by $A=10^{-2}{\mathrm{m}}^2$:

The total current through the dielectric is $i=i_c+i_d=\sigma AE+\epsilon AdE/dt=V/R+CdV/dt$ with

$i=I_0\cos (\omega t+\phi )$ with $\tan \phi =\omega CR=i_{d0}/i_{c0}=128$, so $\phi =89.6^{\circ }$: the current leads the voltage by almost $90^{\circ }$. The loss angle is $\delta =90^{\circ }-\phi =0.45^{\circ }$, with $\tan \delta =\sigma /(\omega \epsilon )=7.8\times 10^{-3}$.

Units: $d/(\sigma A)$ is $m/(S/m\cdot {\mathrm{m}}^2)=1/\mathrm{S}=\Omega$. The same two terms of $\nabla \times H=J+\partial D/\partial t$ that appear side by side in a material appear as the parallel $R$ and $C$ of the equivalent circuit: $\tan \delta$ is the "dissipation factor" printed in capacitor data sheets, and $RC=\epsilon /\sigma$ is the relaxation time of Exercise 5.4.1 ($RC=10^8\times 203\times 10^{-12}=20\mathrm{ms}$ here).