App L adder REVIEWED

This is the Title PhL 3.26.05 Here I write the last two sections of Appendix L as analogs of the first two sections. All installed. L.3 The potential of a line charge inside a thick dielectric cylindrical shell In this section, we repeat everything done in Section L.1 in the 2D world instead of the 3D world. The 3D Poisson propagator (1/4πε0)(1/r) becomes (1/2πε0)ln (1/r) as discussed in Appendix K. We reuse the same drawings, the first of which is Fig L.1' This is now a cross section of an infinite uniform cylindrical geometry. Quantity q is now a linear charge density with dimensions Coulombs/m. Rather that copy, paste and edit Section L.1, here we just show the altered equations and skip most of the words. The equation numbers are those of Section L.1 with a prime added. One difference encountered is that we must take b→R (a large value) rather than b→∞. As noted in Appendix K, a constant in a potential can be ignored even if it is infinite, and such constants do not appear in the field E = -φ. Ansatz potential forms: (to-be-determined constants are α, C, D) φ0(r) = (1/2πε0) q ln(1/r) Er0(r) = (1/2πε0) q/r region 0 φ1(r) = (1/2πα ) q ln(1/r) + C Er1(r) = (1/2πα ) q/r region 1 φ2(r) = (1/2πε0) q ln(1/r) + D Er2(r) = (1/2πε0) q/r region 2 (L.1.1)' Continuity of φ at r=a and b: (1/2πε0) q ln(1/b) = (1/2πα ) q ln(1/b) + C region 0/1 boundary, r = b (1/2πε0) q ln(1/a) + D = (1/2πα ) q ln(1/a) + C region 2/1 boundary, r = a (L.1.3)' Rule for E field normal components at a boundary: ε0Er0(b) = ε1Er1(b) region 0/1 boundary, r = b ε0Er2(a) = ε1Er1(a) region 2/1 boundary, r = a or ε0 q/ [2πε0b] = ε1 (1/2πα ) q/b => 1 = ε1/α ε0 q/ [2πε0a] = ε1 (1/2πα ) q/a => 1 = ε1/α . (L.1.4)' Restated continuity of φ with α = ε1: (1/2πε0) q ln(1/b) = (1/2πε1 ) q ln(1/b) + C region 0/1 boundary, r = b (1/2πε0) q ln(1/a) + D = (1/2πε1 ) q ln(1/a) + C region 2/1 boundary, r = a (L.1.5)' Second equation minus first above: D + (1/2πε0) q(ln(1/a)- ln(1/b)) = (1/2πε1)q (ln(1/a)- ln(1/b)) => D + (1/2πε0) q ln(b/a) = (1/2πε1) q ln(b/a) Solution for the three constants: α = ε C = q ln(1/b)(1/2π) (1/ε0-1/ε1) D = q ln(a/b)(1/2π) (1/ε0-1/ε1) (L.1.6)' The potentials in the three regions are then φ0(r) = (1/2πε0) q ln(1/r) φ1(r) = (1/2πε1) q ln(1/r) + (q/2π) ln(1/b) (1/ε0-1/ε1) φ2(r) = (1/2πε0) q ln(1/r) + (q/2π) ln(a/b) (1/ε0-1/ε1) (L.1.7)' while the fields are Er0 = (1/2πε0) q/r Er1 = (1/2πε1) q/r Er2 = (1/2πε0) q/r . (L.1.8)' The following Maple plots show the continuity of φ and the jumps in Er at the boundaries Fig L.2' What about the (now linear) bound charge densities at r = a and r = b? P = (ε-ε0)Er Pr1(r) = (ε1-ε0)Er1(r) θ(r>a)θ(r<b) // points radially outward since ε1 > ε0 (L.1.9)' From (1.1.11) the polarization charge density is then ρpol = - div P (1.1.11) so in cylindrical coordinates, ρpol(r) = - [r-1∂r(rPr) + r-1∂θPθ + ∂zPz] = - r-1∂r(rPr) = - r-1∂r(r[(ε1-ε0)Er1(r) θ(r>a)θ(r<b)]) = - r-1∂r(r[(ε1-ε0) (1/2πε1) q/r θ(r>a)θ(r<b)]) = - q(ε1-ε0) (1/2πε1)r-1∂r[θ(r-a)θ(b-r)] = - q(ε1-ε0) (1/2πε1)r-1 [ δ(r-a) θ(b-r) - θ(r-a)δ(r-b) ] = - q(ε1-ε0) (1/2πε1) [ δ(r-a)/a - δ(r-b)/b ] . (L.1.10)' We may then read off the bound linear charge densities at r = a and b σinner = - (ε1-ε0) (1/2πε1)(q/a) // Coulombs/m σouter = (ε1-ε0) (1/2πε1)(q/b) (L.1.11)' Qinner = ds σinner = - (ε1-ε0) (1/2πε1)(q/a) * 2πa = - (1-ε0/ε1) q Qouter = ds σouter = (ε1-ε0) (1/2πε1)(q/bb) * 2πb = (1-ε0/ε1) q (L.1.12)' Q = (1-ε0/ε1) q // exactly the same equation as in the 3D case (L.1.13)' Here Q is the total charge/m on the outer surface of the cylindrical shell at r = b, and -Q is the same thing at r = a. Recall that q is the charge/m of the central linear line charge. L.4 Limits of the Previous Problem (a) Line charge in an infinite cylindrical hole in a dielectric Fig L.3' The potentials and fields shown in (L.1.7)' and (L.1.8)' are then, taking b→R (some large value) φ1(r) = (1/2πε1) q ln(1/r) + (q/2π) ln(1/R) (1/ε0-1/ε1) φ2(r) = (1/4πε0) q/r - (q/4πa)(1/ε0-1/ε1) (L.2.1)' while the fields are Er1 = (1/2πε1) q/r Er2 = (1/2πε0) q/r . (L.2.2)' The induced bound charge density σ at r = a is still given by σ = - (ε1-ε0) (1/2πε1)(q/a) -Q = (1-ε0/ε1) q (L.2.3)' where -Q is the total bound charge on the r=a surface. (b) Line charge embedded in an infinite dielectric cylinder Fig L.5' The potentials and fields shown in (L.1.7)' and (L.1.8)' are then, taking a→0 φ0(r) = (1/2πε0) q ln(1/r) φ1(r) = (1/2πε1) q ln(1/r) + (q/2π) ln(1/b) (1/ε0-1/ε1) (L.2.4)' while the fields are Er0 = (1/2πε0) q/r Er1 = (1/2πε1) q/r = (1/2πε0) [q(ε0/ε1)] / r (L.2.5)' where the last expression shows the "shielded charge interpretation". (c) Line charge embedded in an infinite dielectric We now take b→R (a large value) in the previous limit to get: Fig L.6' There is only one region left and from (L.2.4)' and (L.2.5') we get φ1(r) = (1/2πε1) q ln(1/r) + (q/2π) ln(1/R) (1/ε0-1/ε1) (L.2.6)' while the field is Er1 = (1/2πε1) q/r = (1/2πε0) [q(ε0/ε1)] / r (L.2.7)' where the last expression shows the "shielded charge interpretation".