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Infinitesimally small dielectric layers in a capacitor

  1. Oct 1, 2011 #1
    I have a parallel plate capacitor with a dielectric inside whose relative permittivity varies as εr = βexp([itex]\alpha[/itex]x) [a theoretical calculation, I have not measured this directly]

    Eventually I want to calculate the capacitance of the system. First I need to calculate the [itex]\overline{D}[/itex] field, to do this I am using [itex]\oint[/itex]S [itex]\overline{D}[/itex] . d[itex]\overline{s}[/itex] = qfree the result for a parallel plate capacitor is D = q/A with q being qfree again.

    Now it can be shown that
    [itex]\overline{D}[/itex] = εrε0[itex]\overline{E}[/itex] and from this [itex]\overline{E}[/itex] = D/ε0εr[*]

    And this would be fine if the permittivity was constant.

    My guess is to assume the dielectric is made up of many infinitesimally small dielectric's each with εr satisfying εr = βexp([itex]\alpha[/itex]x)

    I know that an integral will give the sum of many infinitesimal 'pieces' but here is seems as if I have the total (which is the full dielectric) and i need to find each infinitesimal piece of dielectric such that εr = βexp([itex]\alpha[/itex]x) is satisfied for each.

    Is that even possible? Will [*] then have an integral on the denominator in place of εr?

    I have done a similar calculation for 2 different dielectrics, this seems an extension of this where each dielectric is infinitesimally small.

    any help would be much appreciated
     
  2. jcsd
  3. Oct 1, 2011 #2
    Just add together the results of sheets as they would be connected as a series

    1/C=Ʃ_i Δx_i/ (ε_0 ε_r(x_i) A)

    Taking the limit Δx_i -> 0

    [itex] 1/C=\int_0^L dx/(ε_0 ε_r(x) A) [/itex]

    where Δx_i is the thickness of one sheet, L is the distance between the plates and A is the area. Finally, you need to integrate the second equation and that is all.
     
  4. Oct 2, 2011 #3
    Thanks a lot, I've derived an expression for the capacitance now. Any idea how i would go about calculating the surface polarisation charge density at the electrodes or the volume polarisation charge density through the dielectric?

    And also i have a feeling that the total polarisation charge density is zero, however I'm not sure how i can prove this.

    Thanks again :D
     
  5. Oct 2, 2011 #4
    It is good to start with determining the D vector from the Gauss theorem. D is constant inside the device

    D=Q/A,

    where Q is the charge on the metal plate of the capacitor. Then, the electric field is

    E(x)=D/( ε_0 ε_r(x) )

    The polarization vector is

    P(x)=ε_0 χ(x) E(x),

    where χ(x)=ε_r(x)-1 is the susceptibility. Finally, the polarization charge density is given by

    ρ_b(x)=-∂_x P(x)

    The total polarization charge can be obtained by integrating ρ_b(x) for the total volume inside the capacitor. In my opinion that is non-zero since ε_r(x) is not the same near the two plates of the capacitor.

    Good luck.
     
  6. Oct 3, 2011 #5
    If i use the definition [itex]\oint[/itex]P.ds = -qpol is that the same as what you have written here?
    I'm not sure if thats a partial derivative or another quantity.

    Is the surface polarisation charge density just the P vector?

    P = χε0E

    Thanks :)
     
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