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Dielectric constant, metals, frequency a bit confused

  1. Mar 27, 2007 #1
    A metal is not a good dielectric. A vacuum is a decent dielectric. There are much better dielectrics than a vacuum. A high dielectric constant, k, means the material exhibits good dielectric behaviour. A vacuum has k = 1, better dielectrics have k > 1.

    What is the dielectric constant of metals? k = 0? K < 1 ?

    Doesn't the dielectric constant of a material depend on the frequency of the EM waves pasing through it?

    Where can I find an expression for the dielectric constant as a function of frequency?

    I'm confused, please try to unconfuse me. thanks.
  2. jcsd
  3. Mar 27, 2007 #2
    Assuming that a metal is a perfect conductor, there is never any electric field inside. As far as I know, we generally don't define a dielectric constant for metals.
  4. Mar 28, 2007 #3
    Thank you for answering one of my questions. :)
  5. Mar 28, 2007 #4
    The permittivity of a metal is negative. (for a perfect conductor it is -Inf)

    I am thinking here in the high frequency regime (but below the 'plasma frequency'): an electromagnetic wave cannot penetrate the metal and the wave is evanascent (exponential decay) below the surface of the metal. Having a negative dielectric conductivity leads to evanescence.

    The permittivity for "real metals" can be more generally a complex number. The imaginary part will represent the absorption of a part of the wave energy.

    As an exercice, consider a metal with a finite resistivity, and try to show the link between resistivity and permittivity. Remember, the permittivity will be a complex number. Then you can explore the limit of perfect metals as well as the limit of static fields (!).

    Also consider that perfect conductors do not exist in nature. The laws of motions apply for the electric charges and they cannot react instanteneously to an electric field. Therefore, the question is rather unrealistic. The next step would be to consider a model for a conductor and establish a formula for its permittivity. In the high frequency domain, for the simplest model, the electron dynamics leads to a permitivity that changes its sign: below the 'plasma frequency' waves cannot propagate, while they can above this frequency. This simplest model shows already the limits of the idea of a perfect conductor.
    Last edited: Mar 28, 2007
  6. Mar 28, 2007 #5
    Thank you lalbatros. Why can't EM waves propagate through the material if they are below the plasma frequency? Or rather why can EM waves propagate if they are at or above the plasma frequency?

    EDIT: Am I right in thinking this is because higher frequencies can be in resonance with the plasma frequency and lower frequencies cannot? Or is that rather the plasma frequency is the lowest energy oscillation that will make the electrons oscillate? I'm inclined to believe the latter is true.
    Last edited: Mar 28, 2007
  7. Mar 28, 2007 #6
    More or less, the plasma frequency is that frequency above which the conduction band electrons can no longer "follow" the oscillation of the incident EM wave. As a consequence, above this frequency, EM waves are no longer reflected off the material but pass through it. Below this frequency, incident EM waves are reflected by the material because of this oscillating behaviour of the conduction electrons.

    I answered a similar question here : plasma frequency

  8. Mar 28, 2007 #7
    Below the plasma frequency, the electrons have enough time to react to the electric field.
    The nearly cancel it, as a consequence.

    For very high frequencies, it is clear that the electrons inertia, as small as it is, is too large and they are not able to react to the electric field. The electric field propagates just as in the vacuum.

    In between these two cases, the permitivity changes from negative (-Inf) to positive (up to +1).

    Now another question:

    When can the permitivity be larger than 1 ?​


    The amplitude of the electric field, therefore its energy, does not play any role here.
    Only the frequency is involved.
    The Maxwell's equation together with the equations of motion for the charges is an homogeneous linear system of differential equations. Therefore the amplitude of the fields (related to the energy) does not come into play.
    If the field is twice higer, the charge displacements will be twice larger.

    However, this linearity can have some limitations. When the fields become very large, the response of the charged particle can be influenced by the non-linear contributions. This results simply from the equations of motion. In this case, the permitivity loses its usual meaning. In the simplest point of view it would become not only a function of frequency but also a function of the field amplitude. There is a lot of interresting and challenging physics there. The nonlinear Landau damping is one of the (many) nice related pieces of physics.
    Last edited: Mar 28, 2007
  9. Mar 28, 2007 #8
    Thanks for the replies.

    Doesn't a high frequency EM wave have a higher energy than a lower frequency EM wave of the same amplitude?

    OK so lets see if I can understand:

    Dielectrics/insulators absorb all frequencies and remit this as infrared waves. Conductors/mirrors/metals absorb all frequencies but re-emit waves at the frequency of absorbtion. Below the plasma frequency the electrons in a metal absorb the energy in terms of standard electron excitation, and thus re-emit this energy at the frequency of absorption. Above the plasma frequency the metal absorbs the energy in terms of lattice energy... Why the difference? The only reason I can think of is that higher frequency waves, as in the photoelectric effect, have the ability to increase the electron kinetic energy independently of atomic structure...
  10. Mar 29, 2007 #9
    No, frequency has no relation to energy, in classical physics as we discuss here.
    (in quantum physics there is a relation, but forget about it for the moment)
    For your initial question, think low frequencies and consider the static limit.
    The behaviour of materials for the whole frequancy spectrum is a quite larger and difficult subject, I know a little bit about plasmas.
    The physics of the permitivity is related to the response of charged particles within the material to the excitation by electric field (or electromagnetic fields for high frequencies). In materials electrons can be free to move from one place to another, like in metals. In some materials the electrons are essentially attached to an ion. And there are intermediate case, semiconductors.
  11. Mar 29, 2007 #10
    Nope, the energy of EM waves (ie the Poynting vector) does not depend on frequency because energy is averaged out. Check hyperphysics for the expression of the Poynting vector : check this

    All the properties of these materials are determined by the fact that you have a lot of atoms sitting together. This is the world of solid state physics. We all know that sending EM waves to an atom gives a discrete emission spectrum because of the discrete electronic energy levels (eg absorb a photon and kick out an electron). But when many atoms start sitting together, like in a crystal, they exhibit new properties because of the many new neigbors. For example, the valence electronorbitals of atoms in metals will overlap to form the conduction band.

    Not only the electron energy levels will give rise to bands but also the atoms that make up the crystal have their new stuff going on. These atoms are connected to each other, eg through ionic or valence bounds. This atomic lattice can vibrate, which is described in terms of phonons (ie the particles associated with the quantisation of these lattice vibrations). The phonons spectrum is continuous. Sending EM radiation to a crystal means that this radiation can interact with the electrons and with the phonons (ie making the lattice vibrate). The continuous phonon energy spectrum explains why we observe continuous EM emission spectra in the case of crystals, while these spectra were discrete in the case of single atoms.

    This is, ofcourse, a generalized picture, but i just want to point out what parameters are important here.

    Last edited: Mar 29, 2007
  12. Aug 31, 2009 #11
    How effective a dielectric is at allowing a capacitor to store more charge depends on the material the dielectric is made from. Every material has a dielectric constant k. This is the ratio of the field without the dielectric (Eo) to the net field (E) with the dielectric:
    k = Eo/E
    E is always less than or equal to Eo, so the dielectric constant is greater than or equal to 1. The larger the dielectric constant, the more charge can be stored.

    If a metal was used for the dielectric instead of an insulator the field inside the metal would be zero, corresponding to an infinite dielectric constant. The dielectric usually fills the entire space between the capacitor plates, however, and if a metal did that it would short out the capacitor - that's why insulators are used instead.
  13. Sep 8, 2009 #12
    Its a very complicated answer would you like to explain it a little bit more
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