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First Paradox of the Day for 9_19_14 REVIEWED

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Short note by Phil dated 9/19/14 and reviewed 9/23/14, stored with Appendix D of his transmission lines work. It compares I = (ω/k)CV, derived from the surface charge density N0, with I = V/Z0 and shows they agree because k/Z0 = ωC when G = 0. It then repeats the check for G > 0 by replacing C with C' using the factor ξd/εd, so the paradox disappears.

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First Paradox of the Day for 9_19_14 PhL 9.19.14 This relates to the relation between I and N0 ending up with I = (ω/k) CV which is a relatively new equation in lines doc. I added it just now as (D.31.2c) so it will be locatable. I use the result in Chapter 7 when taking ω→ 0 where I want to see k and ω clearly indicated. I also use it in (D.9.25) where I turn on G in Appendix D. This started as a little paradox, but that soon went away. This relates directly to Appendix D, so it gets stored there. This review done 9/23/14. 1. In Appendix D I claim that I = 2πω (a/k) N0 = 2πa (ω/k) N0 But I also claim that N0 = <n(θ)> = q/(2πa) = CV/(2πa) This I can write I = 2πa (ω/k) N0 = 2πa (ω/k) CV/(2πa) = (ω/k) CV. 2. On the other hand, I have always claimed that I = V/Z0 3. Comparing these two expressions for the same quantity I requires that (ω/k) CV = V/Z0 or (ω/k) C = 1/Z0 Cancelling V, this says ωC = k/Z0 Is this true? We know that k = -j Z0 = Therefore k/Z0 = -j * = -j (G+jωC) But in App D for the CP BC I assume that G = 0, so then k/Z0 = -j (0+jωC) = -j2ωC = ωC and so there is no paradox! 4. Let's redo this where we allow G > 0 using the App D generalization to this condition. The rule as I show in Section D.9 (?) is to replace Nm → (ξd/εd) Nm . Thus, the above discussion would say N0(new) = CV/(2πa) * (ξd/εd) = C' V/(2πa) I = 2πω (a/k) N0(new) = 2πa (ω/k) C' V/(2πa) = (ω/k) C' V and I = V/Z0 so we then want to show that ωC' = k/Z0 where jωC' = (ξd/εd)jωC = G + jωC (4.11.24) Then we find that k/Z0 = -j (G+jωC) = -j (jωC') = ωC' and result is reconfirmed.