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Inserting dielectric into isolated capacitor

  1. Sep 22, 2012 #1
    U = (1/2)CV^2 = (1/2)Q^2 /C

    Increasing C while keeping Q constant (isolated capacitor) will decrease the energy contained in the capacitor by and amount X
    Where does this X energy come from?

    Case 1 (slab polarizes immediately when it crosses the electric field of the plates)

    The capacitor tugs on the slab, so does work on it,
    If no friction, slab oscillates back and forth with constant
    transferring between kinetic and potential energy.

    Case 2 (slab takes a very long time to polarize)

    The plate can be moved into position without the electric field of the capacitor doing any work. Once in position after along time the slab gets polarized. Thus all of the energy X must go into polarizing the slab.

    In case 1, energy is used to pull in the slab as well as polarize the slab. There will be a point in case 1 where the slab is centred. At this point
    X would have been converted to: Ke + Polarization energy (stored in the induced electric field)

    Where as in case 2
    X would have been converted to: Polarization energy only
    Does this mean the case 2 slab will be more polarised (in the steady state)
    Last edited: Sep 23, 2012
  2. jcsd
  3. Sep 22, 2012 #2

    Simon Bridge

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    It goes into polarizing the dielectric.
    If you'd been holding the slab, you would, indeed, have felt it.

    In the case where the polarization takes a long time, the effect would just be initially as if the material were not a dielectric at all and the dielectric constant would change over time as the slab polarized.
  4. Sep 22, 2012 #3
    Does this mean that in the first case there will be less polarization since when the slab in centered it will have some kinetic energy as well being polarized
  5. Sep 22, 2012 #4

    Simon Bridge

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    You can get really tangled up with dielectrics. This is why your exercises and examples are all steady-state and static. Transient behavior can be very complicated.

    As the dielectric falls between the plates, some of the dielectric will be polarized, some will be in the process of becoming polarized, and some will be unpolarized. There are no sharp boundaries, and the dipoles will affect their neighbors.

    In the case of a very long polarization time, it is possible for the dielectric to fall completely through the plates without becoming polarized at all.

    It is good that you are thinking about this because there are a lot of pmm proposals based on misunderstanding how dielectrics work.
  6. Sep 23, 2012 #5
    Thanks for the reply, i updated my question a little.

    What i am still confused about, is how there could be different levels of polarization, which is dependent on how the dielectric is introduced into the system. (case 1 and 2)

    - based on energy conservation
  7. Sep 23, 2012 #6

    Simon Bridge

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    The steady-state polarization should be the same in each case.
    In the second case, you'd just have to wait longer for this to happen.
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