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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.)