What is the Uniform E Field Inside the Capacitor?

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SUMMARY

The discussion focuses on the analysis of a circular cylindrical parallel plate capacitor filled with a linear and homogeneous material characterized by permittivity ε = e' - je'' and conductivity σ. Key calculations include determining the uniform electric field E inside the capacitor as E = σ/ε, the displacement current density Jd as Jd = I/(πa²hc), and the induced current density Ji as Ji = σ * E. The discussion also addresses the magnetic field H at the surface, given by H = I/(2πa), and concludes with the derivation of the complex impedance z of the capacitor, demonstrating that it behaves as an imperfect capacitor in parallel with a resistor.

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Electromagnetics applied to Capacitor

Homework Statement


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)]



Homework Equations





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!
 
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