# 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 :)