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    Polarization, Relaxation and Loss

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    Concepts of relaxation, polarization, and dielectric loss. Why (not a mathematical argument) does a polarization following (in phase with) a field generate the highest loss? Why wouldn't slow polarization mechanisms lead to loss (is this because they're too slow to polarize in the first place)?

    © BrainMass Inc. brainmass.com September 27, 2022, 10:17 am ad1c9bdddf
    https://brainmass.com/physics/polarisation/polarization-relaxation-loss-560149

    SOLUTION This solution is FREE courtesy of BrainMass!

    In layman words.

    When external field is applied the orientation of dipole changes and it aligns with external field.

    Now if an electric field changes the directions (reverses), the orientation of a dipole also changes (reverses) accordingly. This process of the reorientation of a dipole causes work to be done against friction (and in-turn heat energy loss. Obviously more time taken in reorientation (larger delay or relaxation time) means friction is high, loss will be more. Less time taken, means friction is less and loss will also be less.

    If relaxation time << frequency: friction is low, loss will be less in re-orientation per cycle of wave.

    If relaxation time == frequency: Sufficient large time of re-orientation and and it properly reorients within a wave time period and hence maximum loss.

    If relaxation time >> frequency: Because high relaxation (or delay) can not reorient in sync with the external field, it will not be able to re-orient, and in-turn loss will be less (in-fact negligible).

    Note: Energy storage will be there. But loss will be as per the frequency.

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    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com September 27, 2022, 10:17 am ad1c9bdddf>
    https://brainmass.com/physics/polarisation/polarization-relaxation-loss-560149

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