# Capacitance with Non-Uniform Dielectrics

1. Oct 13, 2011

### Durin

1. The problem statement, all variables and given/known data

1. A parallel plate capacitor, with plates of area A, and spacing d, is filled with a non-uniform
dielectric, with a permitivity that varies as

ε = ε0 + ax

where a is a constant, and x is the distance from one plate

2. Relevant equations

C = Q/V

I'm assuming I'm allowed to assume this is a linear dielectric so

P = XeE

D = P + E = (1+Xe)E = εE

Where P is the polarization vector and Xe is electric susceptibility and D is the electric displacement

P dot n = σbound
Where n is a unit vector normal to the surface

ε = ε0(1+χe)

3. The attempt at a solution

I am able to solve for Xe and sub it back into the equation for P

So

P = ax/ε0E

And I know that the E from the plates is σfree0

And that C = Q/V and that the sum of the bound charges should be zero because otherwise conservation of charge would be violated so V should be changing.

But I am really unsure what to do to figure out the total electric field.

Last edited: Oct 13, 2011
2. Oct 17, 2011

### runnergirl

Are you trying to solve for the capacitance or the E-field? I'm assuming the capacitance since that was one of your relevant equations and typically for a parallel plate capacitor, that's what you're looking for.

A few equations to get you started:
$E_n*\epsilon(x) = \rho_s$ where $E_n$ is the normal component of the E-field and $\rho_s$ is the surface charge density on the plates. If you use $\rho_s$ and substitute in for the charge and area, then solve for the potential V using:
$V = -\int_{0}^{d}E\cdot dl$
You should be able to get the capacitance from there.

3. Oct 17, 2011

### Durin

I atually ended up figuring it out, but it was really crazily easy. Basically I was supposed to use the electric field from a parallel-plate capacitor and substitute in ε for ε0 then integrate to find the voltage.