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Homework Help: Electromag applied to Capacitor

  1. Feb 22, 2010 #1
    Electromagnetics applied to Capacitor

    1. The problem statement, all variables and given/known data
    consider the circular cylindrical parallel plate capacitor shown. The cap has height hc, diameter 2a, and its interior volume is filled with a linear and homogeneous material with permittivity epsilon=e'-je'' and conductivity sigma. A time harmonic source of voltage V0 is applied between the top and bottom perfectly conducting plates of the capacitor such that at t=0, the top plate is at its max voltage. The capacitor can be considered small relative to the wavelength associated with the source frequency and its interior volume v is enclosed by a snug surface S. Assuming V0,hc,2a, e, sigma are known and neglecting all field fringe effects, provide answers to the items below:
    a) What is the uniform E field inside the cap
    b) what is the displacement current density Jd inside the cap
    c) what is the induced current density Jc inside the cap
    d) what is the electric current I, flowing on the wire that connects the source to the cap
    e) what is the uniform mag field H on the surface(at r=a)
    f) starting from E and H, using poynting theorem over the cylindrical surface S enclosing the cap, determine power P provided to the cap
    g) from item f, determine the complex impedance z of the cap
    h) from result of item g, show that cap is always imperfect; it is actually perfect cap in parallel with a perfect resistor, where

    C= [tex]\epsilon[/tex]{ (pi a^2)/hc}
    and
    R= [1/([tex]\sigma[/tex]+[tex]\omega[/tex]e'')] [ (hc/(pi*a^2)]



    2. Relevant equations



    3. The attempt at a solution

    a) E field inside a cap is [tex]\sigma[/tex]/[tex]\epsilon[/tex]

    b) displacement current Jd is I/Area = I/pi a^2 hc

    c) induced current Ji= [tex]\sigma[/tex] * E field

    d) Not sure. I = Jd+Ji or is it I=C.dv/dt

    e) H = I/2pi*a (by using [tex]\oint[/tex](B.dl) = [tex]\mu[/tex]*I

    f) I did [tex]\oint[/tex] ExH.ds and got {[tex]\sigma[/tex]/[tex]\epsilon[/tex]}*I/2pi*a
    Not sure if my answer is right.

    g) and h) unable to solve. Help!
     
    Last edited: Feb 22, 2010
  2. jcsd
  3. Feb 22, 2010 #2
    Anybody?
     
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