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Introductory Physics Homework Help
Calculating Capacitance: A Derivation
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[QUOTE="raggle, post: 4603163, member: 486707"] [h2]Homework Statement [/h2] 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[SUB]r[/SUB] 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[h2]Homework Equations[/h2] Gauss's law ∫[B]D[/B].d[B]S[/B] = ρ [B]D[/B] = ε[SUB]0[/SUB](1+ε[SUB]r[/SUB])[B]E[/B] C = [itex]Q/V[/itex][h2]The Attempt at a Solution[/h2] 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[SUB]0[/SUB](1+ε[SUB]r[/SUB]) = σ/ε[SUB]0[/SUB](1+ε[SUB]r[/SUB]) 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[SUB]0[/SUB](1+ε[SUB]r[/SUB]) and another outside the dielectric with E = σ/2ε[SUB]0[/SUB]. 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! [/QUOTE]
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Calculating Capacitance: A Derivation
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