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the capacitance matrix stuff REVIEWED

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Working notes on the two-conductor capacitor problem in the 2D transmission line setting, noting that the material is already installed in the lines document. They write the potential as a sum of Green's function normal-derivative integrals over the two conductor contours, derive surface charge densities and the capacitance matrix c_ij, and invert it to get the elastance matrix s. The single capacitance is recovered as C = 4πε/K with K = 4πε[s11+s22-2s12]. The second half repeats the derivation with a parameter K in the Green's function.

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This is all installed in lines doc in "the capacitor problem" section of Chapter 5. φt(x,y) = – C1 ds' K1 ∂ng(x,y|x',y') – C2 ds' K2 ∂ng(x,y|x',y') = K1 F1(x,y) + K2F2(x,y) En(x,y) = - ∂n1φ = - ∂n1φt(x,y) = { K1 G11(x) + K2 G12(x) } x on C1 En(x,y) = - ∂n1φ = - ∂n2φt(x,y) = {K1 G21(x) + K2 G22(x) } x on C2 n1(x,y) = εEn(x,y) = εK1 G11(x) + εK2 G12(x) x on C1 n2(x,y) = εEn(x,y) = εK1 G21(x) + εK2 G22(x) x on C2 q1 = C1 ds' n1(x',y') = [εK1H11 + εK2H22] = [ c11V1 + c12V2 ] q2 = C2 ds' n2(x',y') = [εK1H21 + εK2H22] = [ c21V1 + c22V2 ] = or q = c K . = or K = sq = = 4πε K1 = 4πε (s11- s12) K2 = 4πε (s21- s22) K = K1 - K2 = 4πε [s11+ s22 - 2s12] . // since s12 = s21 as noted above = 4πε Then the same capacitance shown in *** is recovered from. C = 4πε/K = ************************************************************* φt(x,y) = – C1 ds' K1 ∂ngK(x,y|x',y') – C2 ds' K2 ∂ngK(x,y|x',y') = K1 F1(x,y,K) + K2F2(x,y,K) From this point on, every function and constant acquires and argument K: Gij(x,K), Hij(K) and then cij(K). We end up then with K = 4πε En(x,y) = - ∂n1φ = - ∂n1φt(x,y) = { K1 G11(x) + K2 G12(x) } x on C1 En(x,y) = - ∂n1φ = - ∂n2φt(x,y) = {K1 G21(x) + K2 G22(x) } x on C2 n1(x,y) = εEn(x,y) = εK1 G11(x) + εK2 G12(x) x on C1 n2(x,y) = εEn(x,y) = εK1 G21(x) + εK2 G22(x) x on C2 q1 = C1 ds' n1(x',y') = [εK1H11 + εK2H22] = [ c11V1 + c12V2 ] q2 = C2 ds' n2(x',y') = [εK1H21 + εK2H22] = [ c21V1 + c22V2 ] = or q = c K . = or K = sq = = 4πε K1 = 4πε (s11- s12) K2 = 4πε (s21- s22) K = K1 - K2 = 4πε [s11+ s22 - 2s12] . // since s12 = s21 as noted above = 4πε Then the same capacitance shown in *** is recovered from. C = 4πε/K =