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Parallel plate capcitor displacement current magnitude

  1. Apr 26, 2007 #1

    neu

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    1. The problem statement, all variables and given/known data

    Use an integral form of the Gauss law to show how the magnitude of D inside the capacitor depends on the charge on the plates of a parallel plate capacitor and their area.

    2. Relevant equations

    Gauss
    [tex]\int{E.nda}=\frac{\rho}{\epsilon_{0}}[/tex]

    [tex]\nabla.D=\rho[/tex]

    where div, D, E all vectors


    3. The attempt at a solution

    so [tex]\epsilon\int{D.nda}=\frac{\rho}{\epsilon_{0}}[/tex]

    where integrals are all closed loops.

    so integral over area = 2 pi r^2 : [tex] D.2.\pi.r^2=\epsilon_{r}\rho[/tex]

    that right?
     
  2. jcsd
  3. Apr 26, 2007 #2
    First of all, lets get the definition straightened out

    [tex]\int \vec{D} \bullet d\vec{a} = Q_{f,enc}[/tex]

    So the (closed) surface integral of [itex]D[/itex] depends only on the free charge enclosed by the volume that is bounded by this surface.

    I don't understand your last question, but before we get to that, I'll try to explain:

    Now, in the region between the plates, you have no free charge: there is bound charge (induced charge) on the dielectric surfaces facing the conducting capacitor plates, but thats all.

    Don't write something like [itex]D = \epsilon E[/itex] right away. Work out the expression for D in the different regions and then get E.

    I can argue by symmetry that D inside is normal to the capacitor plates (neglecting fringing) at all points. Further, D is zero inside the conductor (as P and E are both zero inside a conductor and [itex]D = \epsilon_{0} E + P[/itex]). Now, if [itex]\sigma[/itex] is the free surface charge density (=Q/A where A is the plate area and Q is the charge on the capacitor plate), then can you show that [itex]D = \sigma[/itex] everywhere?

    Once you have D, you can easily get E.

    Now to your last equation. If you want to use the integral form of Gauss's Law with E (rather than D) then please understand that the charge you write on the right hand side is the sum of the bound charge (the induced charge in the dielectric..whatever part of it that contributes to the flux through your chosen Gaussians surface) and the free charge (capacitor charge). So you have to be careful to include the bound charge in this expression as well.

    Generally, since we control free charge (you charge the capacitor) and bound charge readjusts along the way to satisfy boundary conditions in local regions, its easier to work out D first and then E.

    While this post might not have solved your problem completely, I hope you can take it from here with the hints given. But please feel free to reply back esp if you are stuck or want to discuss this further. Include your solution with your reasoning so we can see where the problem is.

    Best wishes
     
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