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A question about the fields D and H for fields in matter

  1. Jan 12, 2015 #1
    Let me give you an example of a reasoning I made in a simple case. Afterwards comes the question:

    Start of reasoning.

    Consider two standard ideal capacitor plates with a dielectric material in between them. Let's call the external field caused by the plates ##\vec{E_{ext}}## and the average macroscopic internal field caused by polarisation ##\vec{E_{int}}## The total field at any point is then on average ##\vec{E_{tot}}=\vec{E_{ext}}+\vec{E_{int}}##

    One can prove that for homogenous polarisation ##\vec{E_{int}}=-\frac{\vec{P}}{\epsilon_{0}}## and thus ##\vec{E_{tot}}=\vec{E_{ext}}-\frac{\vec{P}}{\epsilon_{0}}##

    Let's now look at the ##\vec{D}## field at any point in the material. The definition is ##\vec{D}=\epsilon_{0}\vec{E_{tot}}+\vec{P}##. Plugging the previous expression into D, results in:

    ##\vec{D}=\epsilon_{0} \vec{E_{ext}}##

    End of reasoning.

    QUESTION:

    1) So in this case ##\vec{D}## seems to be not only independent of polarization but really equal to the external field up to a constant. Is this the correct way to interpret the ##\vec{D}##-field, as something that is equal to the E field that is caused by sources of free charges and neglect any E fields caused by polarization?

    2) If I replace D with H and E with B , is the same interpretation correct?
     
  2. jcsd
  3. Jan 12, 2015 #2

    mfb

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    The polarization changes E, and D is defined specifically to be invariant of that (as long as the setup is as nice as yours).
    D with B and E with H. Yes, that is not intuitive.

    Edit: removed weird sentence fragment
     
    Last edited: Jan 12, 2015
  4. Jan 12, 2015 #3
    May I ask what you mean by your last line? Is H no longer invariant in nice setups?
     
  5. Jan 12, 2015 #4

    mfb

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    B takes the place of D, and H the one of E, so yes.
     
  6. Jan 12, 2015 #5

    TSny

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    In general, the "source" of D is not just the free charges.

    The polarization P will also have an effect on the what the value of D will be. For certain simple cases with lots of symmetry, D does end up being the same as if you treated only the free charges as "producing" D. But, in general, this is not true. Examples where D depends on the properties of the dielectric as well as the free charges can be found in standard textbooks.

    It is true that the divergence of D is given by ##\vec{\nabla} \cdot \vec{D} = \rho_{free}##, but that equation is not sufficient to find D except for simple configurations with sufficient symmetry. In general, a vector field is determined by both its divergence and curl. You can see that the curl of D depends on the curl of P. So, P can act as a "source" for the curl of D.
     
  7. Jan 12, 2015 #6
    Thanks. Same reasoning applies to B and H right?
     
  8. Jan 12, 2015 #7

    TSny

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    Yes, that's right.
     
  9. Jan 13, 2015 #8
    May I then ask about the formula (assuming no E field is changing) ''closed loop integral of H'' = sum of macroscopic currents through the loop.

    This formula seems to only care about what the macro current is through the loop. For example, I take a toroid-shaped solenoid and put material inside this solenoid and then take the loop integral of H. I will find that the field H does not depend on the material inside the solenoid and only on the properties of the solenoid itself. The previous poster seemed to imply this was the opposite case for H fields. Am I misunderstanding something?
     
  10. Jan 13, 2015 #9

    mfb

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    You have a different setup now, a solenoid is not a capacitor.
    H does not change in that case, but B does.
     
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