Calculating Capacitance: A Derivation

[raggle]

Homework Statement

A parallel plate capacitor consists of two plates, each of area A, separated by a small distance d. in this gap, a dielectric of relative permittivity ε_r and thickness d/2 is fitted tight against one of the plates, leaving an air gap of thickness d/2 between it and the other plate. Calculate the capacitance of the capacitor

Homework Equations

Gauss's law ∫D.dS = ρ

D = ε₀(1+ε_r)E

C = $Q/V$

The Attempt at a Solution

First I said the plates have a charge density σ. By using Gauss's law in the dielectric I got D = σ, and then the second equation gives

E = D/ε₀(1+ε_r) = σ/ε₀(1+ε_r)

Then (this is where I'm worried I start going wrong) I use this to figure out the potential between the plates, and I split the integral up into two integrals, one inside the dielectric with E = σ/ε₀(1+ε_r) and another outside the dielectric with E = σ/2ε₀. Altogether this ends up giving:

V = -($\int_{0}^{d/2} \frac{\sigma dl}{2\epsilon_0 (1+\epsilon)} + \int_{d/2}^{d} \frac{\sigma dl}{2\epsilon_0}$)

and going through the integrals gives

V = $\frac{(\epsilon - 1)d\sigma}{4\epsilon_0}$

Finally, putting Q = Aσ, I ended up with

C = $4A \epsilon_0/(\epsilon -1)d$

Could someone tell me if I made a mistake somewhere? I'm quite bad at calculating capacitance.
Also is it possible to do this problem by thinking of the capacitor as two capacitors connected in series? Because when I try the problem that way I end up getting a d in the numerator, so I think I've slipped up somewhere.

Thanks!

 

[xophergrunge]

D = ε₀(1+ε_r)E

This should be D = ε₀ε_rE

Then (this is where I'm worried I start going wrong) I use this to figure out the potential between the plates, and I split the integral up into two integrals, one inside the dielectric with E = σ/ε₀(1+ε_r) and another outside the dielectric with E = σ/2ε₀. Altogether this ends up giving:

Where are those 2's coming from?