Infinitesimally small dielectric layers in a capacitor

1. Oct 1, 2011

mitch_1211

I have a parallel plate capacitor with a dielectric inside whose relative permittivity varies as εr = βexp($\alpha$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 $\overline{D}$ field, to do this I am using $\oint$S $\overline{D}$ . d$\overline{s}$ = qfree the result for a parallel plate capacitor is D = q/A with q being qfree again.

Now it can be shown that
$\overline{D}$ = εrε0$\overline{E}$ and from this $\overline{E}$ = 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($\alpha$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($\alpha$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. Oct 1, 2011

Pi01

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

$1/C=\int_0^L dx/(ε_0 ε_r(x) A)$

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.

3. Oct 2, 2011

mitch_1211

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

4. Oct 2, 2011

Pi01

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.

5. Oct 3, 2011

mitch_1211

If i use the definition $\oint$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 :)