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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 εr 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+εr)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/ε0(1+εr) = σ/ε0(1+εr)
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+εr) and another outside the dielectric with E = σ/2ε0. 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!