App D for par plates REVIEWED

Appendix D for a Parallel Plate Transmission line PhL 3.18.14 Review 5.15.14. Basically I replicate pretty much all of Appendix D replacing the round wire as part of a twin-lead transmission line with fat rectangular conductors with a gap between them, the "parallel plate transmission line". My motivation is to see what happens in terms of Jz asymmetry in this case for ω→0. Although I don't say so, I assume that the two plates are infinitely fat, and that the gap is small and the plates are also "high" relative to the gap size (I do say this). In this geometry, Jz asymmetry means that Jz is larger in both conductors near the gap (I am only studying one conductor I guess). I find that Jz(x) = (1+j) (vd/δ) n exp( - x/δ) exp( -j x/δ) which of course is just a usual skin effect situation. You see the current "asymmetry" exp( - x/δ). I cheer below and say that as ω→0, we have δ → ∞ and then "the asymmetry goes away at ω = 0". But one could also say that for any ω > 0, no matter how small, there is asymmetry. But it is different from the twin lead case where it seems that that asymmetry pattern is constant as ω → 0 since it tracks n(θ). This geometry doesn't really have an n(θ) because the fat conductors have only a surface facing the gap where I presume we find that n(θ) = constant. So although this was a valiant effort, I am not sure if the conclusion is very helpful. D.1 Partial Wave Expansion 1 (a) The General Method 2 (b) Partial Wave Expansions 2 (c) The Vector Laplacian 2 (d) The three Helmholtz equations and div E = 0 2 D.2 Solutions for Ez,Ex and Ey 3 (a) The Ez Solution 3 (b) The Ex Solution 4 (c) The Ey Solution 4 (d) The Charge Pumping Boundary Condition 5 (e) Application of the Boundary Conditions 6 Examination of the solution found for Ez(x) 7 Verify the E field solutions in Maple 8 Motivation: I cannot obtain constant Jz at ω = 0 in Appendix D. Maybe PP is a simpler geometry which will shed light on this ongoing problem. Here is the picture, and we shall be interested in the fields inside the right conductor, so we put its left edge at x = 0 D.1 Partial Wave Expansion (a) The General Method (2 - με ∂t2 - μσ∂t)E(x,yz,t) = 0 . (1.3.36) E(x,yz,t) = ej(ωt-βz) E(x,y) . (D.1.1) (2 + β2) E(x,yz,t) = 0 (D.1.2a) or [2D2 + (β2-βd2) ] E(x,y = 0 2 = 2D2 + ∂z2 (D.1.2b) (b) Partial Wave Expansions Skip the partial wave expansion! (c) The Vector Laplacian The vector Laplacian is this (2E)x = 2Ex (2E)y = 2Ey (2E)z = 2Ez (D.1.12) (d) The three Helmholtz equations and div E = 0 1. The z equation. [∂x2 + ∂y2 + ∂z2 + β2 ] ej(ωt-βz) Ez(x,y) (1) Now assume that the parallel plate capacitor is very tall and so we expect fields away from the edge effect regions to be independent of y. Then we can simplify the above to read [∂x2 + (β2-βd2) ] Ez(x) (D.1.15) Conversion Rules: ∂z → -jβd (D.1.16) 2. The x equation. [∂x2 + (β2-βd2) ] Ex(x) = 0 (D.1.17) 3. The y equation. [∂x2 + (β2-βd2) ] Ey(x) = 0 (D.1.18) 4. The z equation [∂x2 + (β2-βd2) ] Ez(x) = 0 (D.1.15) 5. The divE = 0 equation ∂xEx(x) + ∂yEy(x) -jβdEz(x) = 0 or ∂xEx(x) -jβdEz(x) = 0 (D.1.19) The Three Helmholtz Equations and the div E = 0 equation (in partial waves) (D.1.20) [∂x2 + (β2-βd2) ] Ez(x) = 0 (D.1.15) [∂x2 + (β2-βd2) ] Ex(x) = 0 (D.1.17) [∂x2 + (β2-βd2) ] Ey(x) = 0 (D.1.18) ∂xEx(x) - jβdEz(x) = 0 (D.1.19) D.2 Solutions for Ez,Ex and Ey (a) The Ez Solution Equation of interest is [∂x2 + β'2 ] Ez(x) = 0 β'2 = β2 - βd2 (D.1.15) Still claim | | >> 1 as before so we then have β' ≈ β = β = (j - 1) = [(j-1)/] = [(j-1)/] (/δ) = (j-1) (1/δ) The solution to this equation we take to be Ez(x) = cz exp( j β' x) = cz exp( j [(j-1)/] (/δ) x) = cz exp( j j/ (/δ) x)exp( j (-1)/ (/δ) x) = cz exp( - x/δ) exp( -j x/δ) = cz exp [ -(1+j) x/δ ] We select this sign for the right conductor in order to get exponential decay going into the conductor. This then is our skin effect, very simple. Constant cz is TBD. (b) The Ex Solution Equation of interest is [∂x2 + (β2-βd2) ] Ex(x) = 0 (D.1.17) Solution is Ex(x) = cx exp( - x/δ) exp( -j x/δ) Same solution, different constant. (c) The Ey Solution Equation of interest is [∂x2 + (β2-βd2) ] Ey(x) = 0 (D.1.17) Solution is Ey(x) = cy exp( - x/δ) exp( -j x/δ) Same solution, different constant. Now what does div E = 0 add to the story? ∂xEx(x) - jβdEz(x) = 0 (D.1.19) so ∂x{ cx exp [ -(1+j) x/δ ] } - jβd{ cz exp [ -(1+j) x/δ ] }= 0 or [-(1+j)/δ * cx exp [ -(1+j) x/δ ] - jβd{ cz exp [ -(1+j) x/δ ] }= 0 or [-(1+j)/δ * cx - jβd{ cz }= 0 or -(1+j) cx /δ * - jβdcz= 0 or -(1+j) cx = j(βdδ)cz which relates cx and cz, leaving us with only two independent constants, same as with Km and am First summary of the E field solutions (D.2.21) Ez(x) = cz exp( - x/δ) exp( -j x/δ) Ex(x) = cx exp( - x/δ) exp( -j x/δ) Ey(x) = cy exp( - x/δ) exp( -j x/δ) -(1+j) cx = j(βdδ)cz // from div E = 0 Notice that last time we had two constants Km and am (d) The Charge Pumping Boundary Condition div J = - jωρ -jω[∫V ρ dV] = ∫S J dS . (D.2.22) Here is our picture, piece of the above ∫S J dS = Jx dS -jω[∫V ρ dV] = -jω n(y) dS Therefore Jx(0+ε) = -jω n(y) where Jx is just below the surface. But we know n(y) does not vary with y so we have Jx(x=0+ε) = -jωn and this is the charge pumping boundary condition. The other boundary condition is that Ey(x=0) = 0. This is because the E field is perp to the surface. So: We have two boundary conditions to impose: Ex(x=0+ε) = -(jω/σ)n // just inside conductor (D.2.26) Ey(x=0) = 0 (D.2.27) We know Ey is continuous through the boundary like Eθ in Appendix D, so no ±ε needed. (e) Application of the Boundary Conditions The second boundary condition when applied says: Ey(x) = cy exp( - 0/δ) exp( -j 0/δ) = 0 or cy = 0 That was not too hard! Next we have Ex(x=0+ε) = cx exp( - 0/δ) exp( -j 0/δ) = -(jω/σ)n Therefore cx = -(jω/σ)n Then we bring in our div E = 0 condition, -(1+j) cx = j(βdδ)cz or -(1+j) {-(jω/σ)n} = j(βdδ)cz or (1+j) {(jω/σ)n} = j(βdδ)cz or (j-1) (ω/σ)n = j(βdδ)cz or (1+j) (ω/σ)n = (βdδ)cz cz = (βdδ)-1 (1+j) (ω/σ)n To summarize: cy = 0 (D.2.28) cx = -(jω/σ)n cz = (βdδ)-1 (1+j) (ω/σ)n Installing these constants into (D.2.21) above we get Second summary of the E field solutions : Rdc = β'2 = β2 - βd2 (D.2.33) Ez(x) = (βdδ)-1 (1+j) (ω/σ) n exp( - x/δ) exp( -j x/δ) δ ≡ Ex(x) = -(jω/σ)n exp( - x/δ) exp( -j x/δ) Ey(x) = 0 So far, things have gone well with this parallel plate case, or so it would seem. Examination of the solution found for Ez(x) To follow Appendix D, the next step would be to show that these solutions solve the three Helm equations and solve the BC's. I will do that in a moment in a new Maple file. But for the moment, let's examine Ez as ω → 0 which is my main interest right now. to do this note first that (βdδ)-1 (ω/σ) = (1/βd) (ω/σ)(1/δ) = (vd/ω) (ω/σ)(1/δ) = (vd/1) (1/σ)(1/δ) = (vd/σδ) so then Ez(x) = (1+j) (vd/σδ) n exp( - x/δ) exp( -j x/δ) Jz(x) = (1+j) (vd/δ) n exp( - x/δ) exp( -j x/δ) Now let's assume the right conductor is infinitely wide and has some large height Y. Then I = Y * !Syntax Error, I Jz(x) dx = Y (1+j) (vd/δ) n!Syntax Error, I exp [ -(1+j) x/δ ] dx But we know that the integral !Syntax Error, Idx e-ax = 1/a , and in our case we have a = (1+j)/δ so then I = Y (1+j) (vd/δ) n δ/(1+j) = Y (vd/1) n = Y n/vd = total volume current = finite Now look at our main result here, which integrates to the above finite value for I, Jz(x) = (1+j) (vd/δ) n exp( - x/δ) exp( -j x/δ) As ω→ 0, we know that δ → ∞. In this limit we get Jz(x) = (1+j) (vd/δ) n but δ → ∞ We have to have Jz(x) → 0 because it is going to a constant value and we have an infinite x integral! But here is the point. Conclusion: For ω very small but not zero, Jz(x) has a very slight decay going into the right conductor from its surface at x = 0. But as ω→0, Jz(x) becomes a constant value, spread over the entire conductor. And each conductor carries I = Y n/vd which is a finite current. So we don't get any residual Jz "asymmetry" in the limit ω → 0. This seems to conflict my conclusion with the two cylinders! In the parallel plate case we have several advantages: (1) no need for partial waves (2) Cartesian coordinates make equations simpler (3) the solutions are just exponentials, not special functions, so don't need fancy limits. Verify the E field solutions in Maple Now finally let's check that our solutions satisfy the three Helm and the two BC's : Maple file: par plates 1.mws Here are the four expressions we ant to be 0: Here are the solution fields in terms of their constants and here are the constants as found above in the main body, We now show that the four expressions above all evaluate to 0 (they were inspected) And finally we show that the boundary conditions are met: So everything checks out.