Page 80 - Electromagnetics Handbook
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as the boundary condition appropriate to a surface of field discontinuity containing a
                        magnetic surface current.
                          We can also postulate boundary conditions on the normal fields to make Gauss’s laws
                        valid for surfaces of discontinuous fields. Integrating (2.147) over the regions V 1 and V 2
                        and adding, we obtain

                                        D · ˆ n dS −  D 1 · ˆ n 10 dS −  D 2 · ˆ n 20 dS =  ρ dV.
                                    S 1 +S 2      S 10           S 20           V 1 +V 2
                        As δ → 0 this becomes

                                             D · ˆ n dS −  ρ dV =  (D 1 − D 2 ) · ˆ n 12 dS.  (2.191)
                                            S          V        S 10
                        If we integrate Gauss’s law over the entire region V , including the surface of discontinuity,
                        we get

                                                  D · ˆ n dS =  ρ dV +  ρ s dS.               (2.192)
                                                 S          V         S 10
                        In order to get identical answers from (2.191) and (2.192), we must have

                                                      (D 1 − D 2 ) · ˆ n 12 = ρ s
                        as the boundary condition appropriate to a surface of field discontinuity containing an
                        electric surface charge. Similarly, we must postulate

                                                     (B 1 − B 2 ) · ˆ n 12 = ρ ms
                        as the condition appropriate to a surface of field discontinuity containing a magnetic
                        surface charge.
                          We can determine an appropriate boundary condition on current by using the large-
                        scale form of the continuity equation. Applying (2.10) over each of the volume regions
                        of Figure 2.6 and adding the results, we have
                                                                                    ∂ρ

                                       J · ˆ n dS −  J 1 · ˆ n 10 dS −  J 2 · ˆ n 20 dS =−  dV.
                                                                                    ∂t
                                   S 1 +S 2      S 10          S 20             V 1 +V 2
                        As δ → 0 we have
                                                                             ∂ρ

                                            J · ˆ n dS −  (J 1 − J 2 ) · ˆ n 12 dS =−  dV.    (2.193)
                                          S          S 10                   V ∂t
                        Applying the continuity equation over the entire region V and allowing it to intersect
                        the discontinuity surface, we get
                                                                   ∂ρ          ∂ρ s

                                          J · ˆ n dS +  J s · ˆ m dl =−  dV −     dS.
                                         S                       V ∂t       S 10  ∂t
                        By the two-dimensional divergence theorem (B.20) we can write this as

                                                                    ∂ρ          ∂ρ s
                                         J · ˆ n dS +  ∇ s · J s dS =−  dV −       dS.
                                        S          S 10           V ∂t       S 10  ∂t
                        In order for this expression to produce the same values of the integrals over S and V as
                        in (2.193) we require
                                                                        ∂ρ s
                                                ∇ s · J s =−ˆ n 12 · (J 1 − J 2 ) −  ,
                                                                         ∂t



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