- #1

raggle

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## 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**.d

**S**= ρ

**D**= ε

_{0}(1+ε

_{r})

**E**

C = [itex]Q/V[/itex]

## 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/ε

_{0}(1+ε

_{r}) = σ/ε

_{0}(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/ε

_{0}(1+ε

_{r}) and another outside the dielectric with E = σ/2ε

_{0}. Altogether this ends up giving:

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

and going through the integrals gives

V = [itex]\frac{(\epsilon - 1)d\sigma}{4\epsilon_0}[/itex]

Finally, putting Q = Aσ, I ended up with

C = [itex]4A \epsilon_0/(\epsilon -1)d[/itex]

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!