Sectiion 3.4 rewrite

Section 3.4 rewrite PhL 10.11.14 3.4 Size of E fields in conductor and dielectric; conservation of total current at a boundary We know from (1.1.48) that the following E field condition applies at a boundary between two media, where n refers to the normal component, ξ1En1 = ξ2En2 // frequency domain (1.1.48) or (ε1 + σ1/jω) En1 = (ε2 + σ2/jω) En2 (3.4.1) or = = = . (3.4.2) Since most dielectrics have a small imaginary part for ε1, we must interpret σ1 as being the σeff shown in (3.3.4). σ1 ≈ ( σ1DCωε1 tanL) . (3.4.3) Let 1 = dielectric = polyethylene and 2 = conductor = copper with these assumed parameters ε1 = 2.3 ε0 ε2 = ε0 ε0 = 8.85 x 10-12 σ1DC = 10-15 σ2 = 5.81 x 107 . tanL = .0002 (3.4.4) Then rat ≡ = = (3.4.5) Looking at this ratio, for f in the range ( 10 Hz, 1016 Hz) we can approximate the ratio as rat = = ≈ = .45 x 1018 / f . (3.4.6) Then for the "practical" range (10 Hz, 500 GHz) this ratio varies from ~1017 to 106. We conclude: Fact 1: At a boundary between a good dielectric and a good conductor, the normal E field is at least 1 million times larger in the dielectric than it is in the conductor for frequencies under 500 GHz. (3.4.7) Fact 2: This large jump in En at the boundary must be supported by a significant surface charge density ns on the boundary since, according to (1.1.47), ns = ε1En1 - ε2En2 ≈ ε1En1. (3.4.8) Using Ji = σiEi we can restate the above facts in terms of current densities, (jωε1 + σ1) En1 = (jωε2 + σ2) En2 . (3.4.1) (jωε1 + σ1) (Jn1/σ1) = (jωε2 + σ2) (Jn2/σ2) . [1 + jω(ε1/σ1)] Jn1 = [1 + jω(ε2/σ2)] Jn2 (3.4.9) [ 1 + ] Jn1 = [1 + jω(ε2/σ2)] Jn2 // using (3.4.3) [ 1 + ] Jn1 = [1 + j2πf(ε2/σ2)] Jn2 // ω = 2πf (3.4.10) For f >> σ1DC/(2πε1tanL) = .04 Hz, one can ignore σ1DC on the left to get dielectric conductor [ 1 + j/tanL] Jn1 = [1 + j2πf (ε2/σ2)] Jn2 f >> σ1DC/(2πε1tanL) (3.4.11) cond disp cond disp where we have now labeled the conduction current and displacement current terms. Using the numbers above one finds, dielectric conductor [ 1 + 5000 j ] Jn1 = [ 1 + 10-18 f ] Jn2 (3.4.12) cond disp cond disp For f in the range (10 Hz, 1016 Hz) it is clear that in the dielectric, essentially all the current is displacement current, while in the conductor it is essentially all conduction current. We conclude: Fact 3: For any practical frequency and good dielectric, the total current in the dielectric is almost all displacement current, while that in the conductor it is almost all conduction current. (3.4.13) Imagine now a tiny patch of area (bordered in red) on the surface between a conductor and a dielectric, Fig 3.2 Defining a total current Jtot,n ≡ jωεEn + σEn, as in (2.2.1), we have shown that this total current flows right through the patch but changes its nature from nearly all conduction current in the conductor to nearly all displacement current in the dielectric. In the next section, we identify the normal direction with the local radial direction. Then the total current passing through a tiny square patch like that in Fig 3.2 can be regarded as being "fed" by the radial conduction current Jr just inside the conductor where Jr = σEr. This current feeds the surface charge on the boundary which creates a large E field and thus a large displacement current in the dielectric. We sometimes refer to this mechanism as "charge pumping". Recall now some results from Section 1.5, Jn1 = ns(σ1/ε1) (1.5.15) J2n = (ξ1/ε1) (jω) ns = [ jω + (σ1/ε1) ] ns . (1.5.16) It follows that (remember that these J's are conduction currents) J2n = jωns + Jn1 . (3.4.14) We interpret this to say that during the charge pumping process, some of J2n is used to feed the change in the surface charge ns (think ∂tns), and the rest flows through into the dielectric as Jn1. This last equation is just an application of continuity (1.1.35) at the boundary.