5.22: Dielectric material in an alternating electric field.

We have seen that, when we put a dielectric material in an electric field, it becomes polarized, and the $\textbf{D}$ field is now $\epsilon\textbf{E}$ instead of merely $\epsilon_0\textbf{E}$. But how long does it take to become polarized? Does it happen instantaneously? In practice there is an enormous range in relaxation times. (We may define a relaxation time as the time taken for the material to reach a certain fraction – such as, perhaps $1-e^{-1}=63$ percent, or whatever fraction may be convenient in a particular context – of its final polarization.) The relaxation time may be practically instantaneous, or it may be many hours.

As a consequence of the finite relaxation time, if we put a dielectric material in oscillating electric field $E = \hat E \cos \omega t$ (e.g. if light passes through a piece of glass), there will be a phase lag of $D \text{ behind }E$. $D$ will vary as $D=\hat D \cos (\omega t-\delta)$. Stated another way, if the $E$-field is $E=\hat E e^{i\omega t}$, the $D$-field will be $D=\hat D e^{i(\omega t -\delta)}$. Then $\frac{D}{E}=\frac{\hat D}{\hat E}e^{-i\delta}=\epsilon (\cos \delta -i\sin \delta)$. This can be written

$$D=\epsilon ^*E,$$

where $\epsilon^* = \epsilon^\prime − i\epsilon^{\prime\prime} \text{ and }\epsilon^\prime= \epsilon \cos \delta \text{ and }\epsilon^{\prime\prime} = \epsilon \sin \delta$.

The complex permittivity is just a way of expressing the phase difference between $D \text{ and }E$. The magnitude, or modulus, of $\epsilon^* \text{ is }\epsilon$, the ordinary permittivity in a static field.

Let us imagine that we have a dielectric material between the plates of a capacitor, and that an alternating potential difference is being applied across the plates. At some instant the charge density $\sigma$ on the plates (which is equal to the $D$-field) is changing at a rate $\dot \sigma$, which is also equal to the rate of change $\dot D$ of the $D$-field), and the current in the circuit is $A\dot D$, where $A$ is the area of each plate. The potential difference across the plates, on the other hand, is $Ed$, where $d$ is the distance between the plates. The instantaneous rate of dissipation of energy in the material is $AdE\dot D$, or, let’s say, the instantaneous rate of dissipation of energy per unit volume of the material is $E\dot D$.

Suppose $E = \hat E \cos \omega t$ and that $D =\hat D \cos (\omega t − \delta )$ so that

$$\dot D = -\hat D \omega \sin (\omega t -\delta ) =-\hat D \omega (\sin \omega t \cos \delta -\cos \omega t \sin \delta ).\nonumber$$

The dissipation of energy, in unit volume, in a complete cycle (or period $2π/\omega$) is the integral, with respect to time, of $E\dot D$ from $0 \text{ to }2π/\omega$. That is,

$$\hat E \hat D \omega \int_0^{2\pi/\omega}\cos \omega t (\sin \omega t \cos \delta -\cos \omega t \sin \delta )\,dt.\nonumber$$

The first integral is zero, so the dissipation of energy per unit volume per cycle is

$$\nonumber \hat E \hat D \omega \sin \delta \int_0^{2\pi/\omega}\cos^2 \omega t\,dt = \pi \hat E \hat D \omega \sin \delta .$$

Since the energy loss per cycle is proportional to $\sin \delta,\, \sin \delta$ is called the loss factor. (Sometimes the loss factor is given as $\tan \delta$, although this is an approximation only for small loss angles.)