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E field outside wire REVEIWED

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Short note by Phil, dated 10.5.13, asking how Appendix D changes when the field is outside the wire, where the Helmholtz equation uses the dielectric propagation constant βd. It collects the component equations for Ez, Er and Eφ plus div E = 0, then solves the Ez radial equation. He identifies it as an Euler equation with solutions A r^m + B r^-m, or A + B ln r for m = 0. It says the material is already installed in Appendix D.

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E fields outside the round wire? PhL 10.5.13 This has now been installed in appendix D, no need to keep long term. How is Appendix D different if you are thinking outside the wire? Just page through the appendix. First of all, the β appearing in the Helmholtz is then βd since we are "in the dielectric". So we have ( 2 + βd2 ) E(r,φ z, t) = 0 . (D.1.1) Assume the same form E(r,φz,t) = ej(ωt-βz) E(r,φ) . (D.1.2) As I read through the Helmholtz section, what comes out at the far end is this: Summary of Examples: [2E]z + β2 Ez = 0 : [r2∂r2 + r ∂r - m2] Ez(r,m) = 0 (D.1.16) [2E]r + β2 Er = 0 : [r2∂r2 + r∂r - (m2+1)] Er(r,m) - 2jm Eφ(r,m) = 0 (D.1.17) [2E]φ + β2 Eφ = 0 : [r2∂r2 + r∂r - (m2+1)] Eφ(r,m) + 2jmEr(r,m) = 0 (D.1.18) div E = 0 : ∂r [r Er(r,m)] + jmEφ(r,m) -jβd r Ez(r,m) = 0 (D.1.19) So basically you set β' = 0 it seems to me. Now down to the Ez solution. [r2∂r2 + r ∂r - m2] Ez(r,m) = 0 [∂r2 + (1/r) ∂r - m2/r2] Ez(r,m) = 0 [r∂r2 + ∂r - m2/r] Ez(r,m) = 0 This is not Bessel's equation! Ouch, what is it? r∂r(r∂rf(r)) = m2f(r) How do I "look up" an ODE. Polyanin is so massive. Here is an offering he has [r2∂r2 + r ∂r - m2] Ez(r,m) = 0 a = 1 and b = -m2 Then μ = 1/2*| 0 - 4(-m2)|1/2 = (1/2) | 4m2| = (1/2)2m = m. The if condition is 0 > 4b = -4m2 so Ez(r,m) = C1|r|m + C2|r|-m = Arm + B r-m Yes, this is the radial equation for the 2D Laplace equation 2u = 0 as on Stak II page 90, equation (7.8) page 92. I never knew it by the name Euler Equation. For m = 0 it is the second case and the solutions are A + Bln r as I now well remember.