Extending the cpbc INSTALLED

Extending the cpbc for σd ≠ 0 PhL 6.2.14 Extending the Charge Pump Boundary Condition of Appendix D. This stuff has been installed as section D.9 (d). Question: If I allow the dielectric to conduct, how does that change the cpbc? It would then read Jr(a-α,θ) - Jr(a+α,θ) = (jω) n(θ) or σ Er(a-α,θ) - σdEr(a+α,θ) = (jω) n(θ) // this is div J = -jωρ But we know from (1.1.47) that [ points into medium 1 which for me is the dielectric ] [ε1En1 - ε2En2] = nfree // this is continuity of Dn or [εdEr1 - ε0Er2] = nfree or [εd Er(a+α,θ) - ε0 Er(a-α,θ)] = nfree So how does this relate to the cpbc?? It must be that nfree = n(θ) = the physical surface charge. so σ Er(a-α,θ) - σdEr(a+α,θ) = (jω) n(θ) εd Er(a+α,θ) - ε0 Er(a-α,θ) = n(θ) But should not these two equations be the same? No! They both appear below (1.1.47). You can eliminate n(θ) and end up with ξ1En1 = ξ2En2 or (ε1 + σ1/jω) En1 = (ε2 + σ2/jω) En2 or (εd + σd/jω) End = (ε0 + σ/jω) En or (εd + σd/jω) Erd ≈ (σ/jω) Er which then relates the two fields so (εd + σd/jω) Er(a+α,θ) ≈ (σ/jω) Er(a-α,θ) But this provides another "connection" between inside and outside that I forgot about! Ouch!!!! I cannot ignore this thread now, but it is 10AM pre Torrey. Summary: there are two independent conditions here relating inside and outside Er fields. I can take any pair of these three equations σ Er(a-α,θ) - σdEr(a+α,θ) = (jω) n(θ) // modified cpbc from div J = -jωρ εd Er(a+α,θ) - ε0 Er(a-α,θ) = n(θ) // div D = ρ straddle (εd + σd/jω) Er(a+α,θ) ≈ (σ/jω) Er(a-α,θ) // ξ1En1 = ξ2En2 NOTE: if ρ= 0, then div E = 0 is the same as div J = 0, but this is not true otherwise!!! For a vacuum dielectric with σd = 0 these three conditions read σ Er(a-α,θ) = (jω) n(θ) // cpbc from div J = -jωρ εd Er(a+α,θ) - ε0 Er(a-α,θ) = n(θ) // div D = ρ (**) (εd) Er(a+α,θ) ≈ (σ/jω) Er(a-α,θ) // ξ1En1 = ξ2En2 (*) The last one at least does say the field inside is "small" since σ is large. In the capacitor model of the transmission line, I assume that Er(inside) ≡ 0 and that is how I get n(θ). If I took (*) strictly, then I would be unable to solve the exterior problem using the capacitor method. Having ignored (*) to get n(θ) as an approximation, I suppose (*) is still reasonable if you want to know Er inside. Er(a-α,θ) ≈ [jω εd/σ] Er(a+α,θ) Now, to get n(θ) outside I had to say this: n(θ) = εdEr(a+α,θ) // from (**) But then (*) reads n(θ) ≈ (σ/jω) Er(a-α,θ) and this is the same as the cpbc. So if I decide to compute n(θ) as I do assuming Er ≈ 0 inside, then I don't really have two independent conditions, I have only one which is the cpbc. Now let's go back to the original question. ******************** continuing on June 1 *************************** Now let's go back to the original question which was: Question: If I allow the dielectric to conduct, how does that change the cpbc? In this case we have these three boundary conditions : 1 σ Er(a-α,θ) - σdEr(a+α,θ) = (jω) n(θ) // modified cpbc from div J = -jωρ 2 εd Er(a+α,θ) - ε0 Er(a-α,θ) = n(θ) // div D = ρ straddle 3 (εd + σd/jω) Er(a+α,θ) ≈ (σ/jω) Er(a-α,θ) // ξ1En1 = ξ2En2 Claim A: only two of these three BC's are independent. Proof: If we take -1 divided by jω we get 4 (σd/jω)Er(a+α,θ) - (σ/jω) Er(a-α,θ) = - n(θ) Now add this to 2 to get (εd + σd/jω) Er(a+α,θ) - (σ/jω) Er(a-α,θ) = 0 but this is BC#3, so only two are independent. I then have 2 equations in 3 unknowns which are the functions Er(a-α,θ), dEr(a+α,θ) and n(θ). Claim B. When I solve the capacitor problem, I am really using #2 with the approximation that Er(a-α,θ) ≈ 0 inside the conductor. This then gives a result for n(θ). What I am really assuming there is that Er(a-α,θ) << Er(a+α,θ) assuming εd and ε0 are about the same. This then "uses up" equation #2, and I have only one other equation left. It is unclear how to proceed from this point in order to answer the question : how is the cpbc altered by σd ≠ 0. I cannot simply drop the term σdEr(a+α,θ) in #1 without some justification. The problem with that is that, although Er(a-α,θ) is very small, σ is very large, so the size of product σ Er(a-α,θ) is hazy. ******************************** I am expecting involvement of this equation somehow, where ns is the surface charge, and nc is that other thing I discuss in Section 1.5 (c): ns(θ) = (εd/ξd) nc(θ) The ns is what appears in the capacitor problem, so my three BC equations are 1 σ Er(a-α,θ) - σdEr(a+α,θ) = (jω) ns(θ) // modified cpbc from div J = -jωρ 2 εd Er(a+α,θ) - ε0 Er(a-α,θ) = ns(θ) // div D = ρ straddle 3 ξdEr(a+α,θ) ≈ ξ Er(a-α,θ) // ξ1En1 = ξ2En2 So what might I do here? Suppose I eliminate ns in favor of nc : 1 σ Er(a-α,θ) - σdEr(a+α,θ) = (jω) (εd/ξd) nc(θ) 2 εd Er(a+α,θ) - ε0 Er(a-α,θ) = (εd/ξd) nc(θ) 3 ξdEr(a+α,θ) ≈ ξ Er(a-α,θ) How does this help? Rewrite as 1 σ (ξd/εd) Er(a-α,θ) - σd (ξd/εd) Er(a+α,θ) = (jω) nc(θ) 2 ξd Er(a+α,θ) - (ξd/εd) ε0 Er(a-α,θ) = nc(θ) 3 ξdEr(a+α,θ) ≈ ξ Er(a-α,θ) Now use #3 to eliminate Er(a+α,θ) in #1: σ (ξd/εd) Er(a-α,θ) - σd (ξd/εd) Er(a+α,θ) = (jω) nc(θ) or σ (ξd/εd) Er(a-α,θ) - σd (ξ/εd) Er(a-α,θ) = (jω) nc(θ) or [ σ (ξd/εd) - σd (ξ/εd) ] Er(a-α,θ) = (jω) nc(θ) or [ σ ξd - σd ξ ]/εd Er(a-α,θ) = (jω) nc(θ) or [ σ (εd + σd/jω) - σd (ε0 + σ/jω) ]/εd Er(a-α,θ) = (jω) nc(θ) or [ (σ εd - σdε0) + (σσd-σσd)/jω ]/εd Er(a-α,θ) = (jω) nc(θ) or [ (σ εd - σdε0)]/εd Er(a-α,θ) = (jω) nc(θ) or [ (σ - σd(ε0/εd) ] Er(a-α,θ) = (jω) nc(θ) Now finally: since the two ε are the same order, we drop the second term to get σ Er(a-α,θ) = (jω) nc(θ) and THIS then is the modified cpbc ! This what I was hoping for. The main idea is the cancellation shown in red above. Answer to the Question: the modified version of the cpbc is this σ Er(a-α,θ) = (jω) nc(θ) or σ Er(a-α,θ) = (jω) (ξd/εd) ns(θ) or σ Er(a-α,θ) = (jω) ([(εd + σd/jω)]/εd) ns(θ) or σ Er(a-α,θ) = (jω) (1 + σd/jωεd ) ns(θ) We can then ignore the σd/jωεd term here as long as σd/jωεd << 1 We already know that ωε << σ for ω < 1018, but this new condition is different. We have σd << ωεd At very low ω this won't be realized. Write as' σd << 2πf ε0 = 6.28 * 8.8541877 x 10-12 * f ≈ 54 * 10-12 ≈ 10-10 f ? So if f = 1 Hz, you only need σd << 10-10 to be OK. But when we use σeff, this probably won't be OK and the factor (ξd/εd) ns(θ) must be maintained. So everywhere you have a charge type factor, you now must add (ξd/εd). So how does that play out in the Appendix D results??? Take it in steps please. 1. We have to make these two replacements: n(θ) → (ξd/εd) n(θ) ≡ n'(θ) Nm → (ξd/εd) Nm ≡ N'm 2. The coefficient solutions then become am = (jω/2σ) 2m Nm → (ξd/εd) (jω/2σ) 2m Nm = (jω/2σ) 2m Nm' = a'm and similarly for Km, = (jω/2σ) Nm [ – ] → (ξd/εd) (jω/2σ) Nm [ – ] = (jω/2σ) Nm' [ – ] = K'm/2 = (jω/σ) N0 → (ξd/εd) (jω/σ) N0 = (jω/σ) N0' The coefficients get larger! I have just renamed them with a prime for the "new" situation. 3. The normalization condition (D.2.31) is then (ξd/εd)N0 = (βd/2πωa) I' = N0' where I' is the current in the new situation. So, or (βd/2πωa) I' = (ξd/εd)N0 Suppose we start with σd = 0 with some I and some N0 moment. Ignoring the change in βd, you would argue that I "increases" due to this factor (ξd/εd), since N0 does not change. That is, I' > I. 4. The new Ez field is this E'z(r,m) = -j(β'/βd) Jm(x) = -j(β'/βd) (jω/2σ) Nm' [ – ] Jm(x) = -j(β'/βd) (jω/2σ) (Nm'/N0') N'0 [ – ] Jm(x) = -j(β'/βd) (jω/2σ) ηm N'0 [ – ] Jm(x) Now use the fact that (βd/2πωa) I' = N0' to end up with E'z(r,m) = -j(β'/βd) (jω/2σ) ηm (βd/2πωa) I' [ – ] Jm(x) = (β') (1/4σ) ηm (1/πa) I' [ – ] Jm(x) = (1/4) ηm I' Rdc (aβ') [ - ] Now E'z(r,m) is different for two reasons. First, because I' is different from the old I. Second, βd is different.