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Intro. to Modern Optics Question

  1. Feb 6, 2012 #1
    1. The problem statement, all variables and given/known data
    Show that the displacement current density which flows when a parallel plate capacitor gets charged is ε(dE/dt) where ε is the permittivity of the medium between the capacitor. What is the energy density stored in the capacitor and the force per unit area (pressure) on the plates?


    2. Relevant equations
    D=εoE + p= εoE +εoχE=εE
    p=[itex]\alpha[/itex]E (mircoscopic)
    p=χE (marcoscopic)
    (Constitute of Relations) p is the polarization of the medium

    3. The attempt at a solution
    (a) I am really confused as to how to show the first part, which I think is needed for the whole problem.
    (b) For the work, from my notes I have[itex]\tau\tau[/itex]=[itex]\sigma[/itex]E which means it behaves equivallently to Ohms Law V=IR so the Voltage which is the work done per unit charge is equal to that expression.
    (c) For the energy density of the capacitor Ue=1/2 εo E^2
    And the Total energy density is Uto/2 *E^2 +B^2/2μo which reduced to 2εoE^2
    (d) P=1/2 εoE^2 +B^2/μ
    P=εo E^2
    P=s/c (s is the Poynting vector)

    I am just really confused but this is what my notes and external sites have led me to. Any help to make sense of this is appreciated (I'll be talking to my professor today too as well so hopefully I can understand this better)
     
  2. jcsd
  3. Feb 6, 2012 #2
    tau is the displacement current, and sigma should be the conductivity? I think
     
  4. Feb 6, 2012 #3
    I didn't need the equation for D in the first part, All I needed was C=eA/d for capactiance and q=CV
     
  5. Feb 6, 2012 #4
    So If i take the time derivative of q=cv I get the current I = e A/d *dV/dt
     
  6. Feb 6, 2012 #5
    Nvm, I figured most of it now.
     
  7. Feb 6, 2012 #6

    vela

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    Oh, I thought you had figured it out five hours ago.
     
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