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5.22: Dielectric material in an alternating electric field.

  • Page ID
    7791
  • 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.)

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