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Capacitor with dielectric

  • Thread starter 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!
 

Answers and Replies

  • #2
D = ε0(1+εr)E
This should be D = ε0εrE


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:
Where are those 2's coming from?
 

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