Capacitor with conducting dielectric REVIEWED

Capacitor with conducting dielectric PhL 10.16.13 I suddenly realize that I have never solved this problem, and one must solve it cleanly first before attempting a transmission line problem! 1. Parallel plate cap with large area A 1 2. Cap made of long cylinder-like conductors 2 1. Parallel plate cap with large area A Put a gaussian box which includes the charge on the positive plate. Apply this law div D = ρ ∫V ρ dV = ∫S D dA (1.1.13) The left side will be ndA. The right side we assume only contribution is the far surface of the box, and since D = εE have the RHS = ε E dA, and therefore n = εE and also E is independent of z falls out of this analysis, since z appears nowhere in this result. The total charge on the plate is Q = nA. 2. If the voltmeter voltage is the line integral of E between the plates, then V = Es = (n/ε)s = ns/ε = Qs/εA Since Q = CV, we say C = ε(A/s) and there is the capacitance, boosted by ε as expected. Function of geometry and ε. But something is missing here! Where is the leakage voltage drop? Well I think we know R = (1/σ)s/A = s/(σA) I = V/R = VσA/s J = Vσ/s E = V/s and this is consistent with that found above. This is a DC situation. So we just have our capacitor in which we have a leakage current, all is well. 2. Cap made of long cylinder-like conductors Imagine we solve this potential theory problem? Is Laplace valid with σ ? Very good question. If we use the potential with the Lorenz gauge, our potential equation here is ( 2 + β2) φ(x,ω) = - (1/ε) ρ(x,ω) (1.5.3) where β2 = ω2εμ does not see σ ! I don't think ρ is at all affected by σ, it is not in the simpler problem above. Maybe φ is the voltmeter voltage after all. Suppose yes. Then solve the above for φ(x) and we find V = Q/C where C is some function of the geometry. This result is not affected by σ. The new feature is this E(x) = -φ(x) + ∂tA(x) J(x) = σE(x) So you have already solved for E(x), the new feature is that there now exists some J. What is the total current? Put a surface around conductor 1 and integrate over it. E = n/ε just outside the surface. so the surface integral of EdA is Q/ε. ∫EdA = ∫(n(x)/ε)dA = Q/ε where A is the area of conductor C1 say. Then Ileak = ∫JdA = ∫σEdA = σ Q/ε Then it must be that R = V/Ileak = Vε/(σQ) = ε/(σC) The impedance of the capacitor is then Z = R + jXc = R + j (1/ωC) = ε/(σC) + j (1/ωC) = (1/C)(ε/σ +j/ω) The admittance is then Y = 1/Z = C (ε/σ +j/ω)-1 = Cεσ / (ε2σ2+1/ω2) - (Cj/ω) / (ε2σ2+1/ω2) Why is this so messy? Maybe I will say G = 1/R = conductance Y = G + jωC = (σC/ε) + jωC = jωC [ 1 + σ/jωε] = jωC[ ε + σ/jω]/ε = jωC(ξ/ε) = jωC' => C' = C + G/jω The idea here is this: Once you know the capacitance C, you also know the total Y without having to do any more work. [ This fact should be installed somewhere in lines doc ] Thus, I am able to get the conductance to appear at least in my Case 2 capacitor situation.