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Electric energy of a dielectric sphere

  1. Aug 29, 2005 #1
    Electric energy of a dielectric sphere!!

    Hi there, I have attempted a question, I was just seeing if some one can go over it and check if it is ok!!

    Q. The Electric field E inside a dielectric sphere placed between the plates of a large parallel-plate capacitor is uniform. Given that the sphere has a relative dielectric constant "epsilon(r)" and radius r, find an expression for the total electric energy of the sphere.

    My working out is as attached to this message:

    If there is more that can be done then can you help aswell!!!

    Thanks alot

    Attached Files:

    Last edited: Aug 29, 2005
  2. jcsd
  3. Aug 29, 2005 #2


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    Homework Helper

    It looks like you've answered the question as if the charge on the plates is known. I would have presumed, from the wording of the question, that the Electric Field is known. This would make the Electric field energy within the sphere obvious (epsilon E^2 Volume), but the total Electric (potential) Energy *of* the sphere would not be trivial to calculate. In my view, the surface charges are what actually belong to the sphere; the external E-field belongs more to the environment than to the sphere.
    I would've found the surface charge density first, then the Electric PE due to that surface charge density by surface integration. That way you end up *showing* the utility of the new Energy density approach (which I suppose is the "new topic").
    of course, I've been known to misinterpret questions before ... .
  4. Aug 30, 2005 #3
    I don't fully understand, do i have to use maxwells equations to find the charge density, and if so how does this apply to a dielectric between 2 capacitor plates. I thought that the E field would be E(0)/EPSILON(r) where E(0) is the Electric field of the capacitors without the dielectric.

    using DIV(E)=Charge desnsity/epsilon(0) I think I'm right in saying this is one of maxwells equations.

    Hope you can help clarify this for me. Thanks alot for you help

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