Insulators — Question 5

Question 5

Conduction loss and displacement current have different frequency dependence

A dielectric slab has $J_c=\sigma E$ and $J_d=\varepsilon\,dE/dt$. Apply $E=E_0\cos\omega t$. Define loss tangent as the amplitude ratio $\tan\delta=|J_c|/|J_d|$.

1. Derive the loss tangent and mean heating per volume.

2. Find the angular frequency where both current amplitudes are equal for $\sigma=10^{-6}\,\mathrm{S/m}$ and $\varepsilon=5.00\times10^{-11}\,\mathrm{F/m}$.

Question 5 - Solution

Conduction loss and displacement current have different frequency dependence

(a) Method and calculation

$$J_c=\sigma E_0\cos\omega t,\quad J_d=-\omega\varepsilon E_0\sin\omega t.$$$$\boxed{\tan\delta=\sigma/(\omega\varepsilon),\quad\langle p_v\rangle=\sigma E_0^2/2}.$$

(b) Interpretation and further calculation

$$\tan\delta=1\Rightarrow\omega=\sigma/\varepsilon.$$$$\boxed{\omega=2.00\times10^4\,\mathrm{s^{-1}},\quad f=\omega/(2\pi)=3.183\,\mathrm{kHz}}.$$

Physical check

The displacement current is in quadrature with the field and contributes no mean Joule heating in this ideal model. Polarization relaxation can add loss beyond the stated conductivity term.