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Energy in a Capacator with a Diaelectric

  1. May 14, 2009 #1

    r16

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    1. The problem statement, all variables and given/known data
    I was attemping to solve problem 10.13 in Purcell's E&M Book: By considering how the introduction of a dielectric changes the energy stored in a capacitor, show that the correct expression for the energy density in a dielectric be [tex]\frac{\epsilon E^2}{8\pi}[/tex]


    2. Relevant equations
    1 [tex] U = 1/2 C V^2[/tex]
    2 [tex] Q_0 = Q + Q' = \epsilon Q_0 + Q'[/tex]
    3 [tex] U = \frac{1}{8 \pi} \int_V E^2 d^3 x [/tex]
    4 [tex] E_{tot} = E_{vac} - E_{pol} = \epsilon E - 4 \pi P[/tex]

    3. The attempt at a solution
    It is not hard to derive Eq. 3 from Eq. 1. This is the energy stored in a vacuum capacitor.

    Q_0 and E_0 are the charge and field in a vacuum capacitor before the dielectric is added

    I attempted to use Eq 2 to derive the electric field inside a capacitor with a dielectric. The total electric field inside the capacitor (Eq 4)will be the superposition of the electric field of a vacuum capacitor of strength [tex]\epsilon E_0[/tex] and the field of a polarized dielectric, [tex] 4 \pi P[/tex]. Since P can be equated to the charge density per area, [tex]\sigma[/tex], you can solve equation 2 for Q', which is the surface charge density on the polarized dielectric's surface and plug that in for P in Eq. 4. You can also do this for the field of the vacuum capacitor:
    [tex] E = \epsilon E_0 - 4\pi P = \frac{4 \pi \epsilon Q_0}{ A } - \frac{4 \pi Q_0(1-\epsilon)}{A} [/tex]
    factoring out a 4 pi Q_0 /A yeilds E_0:
    [tex]E = E_0 ( \epsilon - 1 + \epsilon ) = E_0(2 \epsilon - 1)[/tex]
    plugging this into equation 2 after calculating the voltage from E yeilds
    [tex]U = \frac{1}{8 \pi}E_0^2 (4 \epsilon^2 - 4\epsilon -1)[/tex]
    this is not what was expected in the problem.

    The only thing I can think of that would make this not work is equating P = Q' / A.

    Please advise
     
  2. jcsd
  3. May 14, 2009 #2
    Ok, you seem to be confused about the electric field in the dielectric. If you assume the material to be a linear dielectric, then you can write:

    [tex]D = \epsilon E[/tex]

    Which will make your life much easier. And the energy density is written:

    [tex]U = \frac{1}{2}E\cdot D[/tex]

    Of course you might have a [tex]\pi[/tex] in there depending on what units you use.
     
  4. May 16, 2009 #3

    r16

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    Ok, I think I remember seeing that equation some where, however it was not in my book. Is there a link to a good derivation of this equation? With this, the answer is trivial.

    Where I learn the most, however, is having someone explain to me how what I was doing was incorrect. Could anyone tell me why my original approach to this problem is erroneous?

    On another hand, even though [tex]U = 1/2 E \cdot D[/tex] gives the right answer, how does it apply to the context of considering the addition of a dielectric to a vacuum capacitor?
     
  5. May 16, 2009 #4
    You can solve it easily for a parallel plate capacitor knowing...

    [tex]U = \frac{1}{2}CV^2[/tex]

    But to solve it for any capacitor will take some work. You start with...

    [tex]U = \frac{1}{2}CV^2 = \frac{1}{2}QV[/tex]

    but now you make it a charge density...

    [tex]U = \frac{1}{2}\int \rho V d^3x=\frac{1}{2}\int V\nabla \dot Dd^3x[/tex]

    Do integration by parts...

    [tex]U = \frac{1}{2}\int D\cdot\nabla V d^3x = \frac{1}{2}\int D\cdot E d^3x[/tex]

    Tada... I may have made a mistake in the math, but you can fix it if you see any...
     
  6. May 16, 2009 #5

    r16

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    wow thats quite an elegant proof, i really like it.

    by [tex]\rho[/tex] I assume you mean the free charge, since, [tex]\nabla \mathbf{D} = 4 \pi \rho_f[/tex]

    i guess this is what is confusing me: why are you able to ignore the effects of the bound charge? The energy stored in the capacator is stored in terms of the net electric field E. Shouldn't I be able to find an expression for [tex]\mathbf{E}=\epsilon \mathbf{E} - 4 \pi \mathbf{P}[/tex] and stick it into [tex]U=1/2 C V^2[/tex]? I think I may see how this works, I still need to play around with it a little bit.

    I guess I expect to find that the electric field inside is [tex]\sqrt{\epsilon}[/tex] so when I solve for [tex]V=\int_s E ds[/tex] and square it in U=1/2CV^2 the equation comes out nicely as 1/8 pi E^2. The fact that in a standard electromagnetic plane wave [tex]B=\sqrt{\epsilon}E[/tex] put this idea in my head.
     
  7. May 16, 2009 #6
    A more elegant proof is given here:

    http://farside.ph.utexas.edu/teaching/jk1/lectures/node41.html

    Or you can find it in the E&M book by Jackson where it explains what happens when you have a polarizing material, it also explains that the equation given only works for linear dielectrics. So it will be completely different if the dielectric behaves nonlinear. My proof was sloppy and quick.
     
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