Capacitance with Non-Uniform Dielectrics

[Durin]

Homework Statement

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

Homework Equations

C = Q/V

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

P = X_eE

D = P + E = (1+X_e)E = εE

Where P is the polarization vector and X_e is electric susceptibility and D is the electric displacement
-grad(P) = ρ_bound

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

ε = ε₀(1+χ_e)

The Attempt at a Solution

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

So

P = ax/ε₀E

And I know that the E from the plates is σ_free/ε₀

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.

 

[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.

 

[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.