Applying the last two equation with a rectangular loop that crosses the interface,shows that i×(E2-E1)=0, 元×(H2-H1)=K (39) where K is the surface current density (units:amp/m).(If H2 points up and n points to the right,K points out of the paper.) Specializing to electrostatics,we have that the tangential components of the electric field are continuous,and the discontinuity of the normal components of D=eE are proportional to the surface charge density.Since the electric field vanishes inside a conductor this gives a direct connection between the field just outside a conductor and the surface charge density. The field is normal to the conductor and of magnitude o/eo. 2.3 Uniqueness of electrostatic solutions,Green's theorem Suppose 1,2 are two solutions of Poisson's equation-V2=p/eo.Then =2-is a solution of Laplace's equation V2=0.One can find lots of solutions of Laplace's equation but they all have growing behavior at infinity.To see this consider a5n®=人r-(®r)=人T电电 (40) J8R R If the left side vanishes (either because -0 fast enough at infinity or because Dirichlet of Neumann boundary conditions apply on the boundary of R),then must be a constant. This is because the integrand on the right is positive definite.If the charge density p is localized and has finite total charge,will vanish faster than 1/r at infinity. More generally one has 7.(070-07b)=必72中-72沙 (41) which,upon integrating over a region R,yields Green's Theorem dv(UV26-0V2v)=p dS(vi.Vo-on.Vv) (42) R In the context of electrostatics,suppose o is the potential associated with charge density p=-V2co and =is associated with charge density p=-V2'co,and furthermore that the boundary of R is a conducting surface with surface charge density o =-nEeo =n.Voco in the first case and a'=n.Vo'eo in the second case.Then Green's theorem becomes the reciprocity theorem: dV(-o'p+op)= dS(p'o-oo） dvop'+ dSoo'= dW'p+φ dsoo (43) a JaR 10 ©2010 by Charles ThornApplying the last two equation with a rectangular loop that crosses the interface, shows that nˆ × (E2 − E1) = 0, nˆ × (H2 − H1) = K (39) where K is the surface current density (units: amp/m). (If H2 points up and nˆ points to the right, K points out of the paper.) Specializing to electrostatics, we have that the tangential components of the electric field are continuous, and the discontinuity of the normal components of D = E are proportional to the surface charge density. Since the electric field vanishes inside a conductor this gives a direct connection between the field just outside a conductor and the surface charge density. The field is normal to the conductor and of magnitude σ/0. 2.3 Uniqueness of electrostatic solutions, Green’s theorem Suppose φ1, φ2 are two solutions of Poisson’s equation −∇2φ = ρ/0. Then Φ = φ2 − φ1 is a solution of Laplace’s equation ∇2Φ = 0. One can find lots of solutions of Laplace’s equation but they all have growing behavior at infinity. To see this consider Z ∂R dSΦnˆ · ∇Φ = Z R dV ∇ · (Φ∇Φ) = Z R dV ∇Φ · ∇Φ (40) If the left side vanishes (either because Φ → 0 fast enough at infinity or because Dirichlet of Neumann boundary conditions apply on the boundary of R), then Φ must be a constant. This is because the integrand on the right is positive definite. If the charge density ρ is localized and has finite total charge, Φ will vanish faster than 1/r at infinity. More generally one has ∇ · (ψ∇φ − φ∇ψ) = ψ∇2φ − φ∇2ψ (41) which, upon integrating over a region R, yields Green’s Theorem Z R dV (ψ∇2φ − φ∇2ψ) = I ∂R dS(ψnˆ · ∇φ − φnˆ · ∇ψ) (42) In the context of electrostatics, suppose φ is the potential associated with charge density ρ = −∇2φ0 and ψ = φ 0 is associated with charge density ρ 0 = −∇2φ 0 0, and furthermore that the boundary of R is a conducting surface with surface charge density σ = −nˆ·E0 = nˆ·∇φ0 in the first case and σ 0 = nˆ · ∇φ 0 0 in the second case. Then Green’s theorem becomes the reciprocity theorem: Z R dV (−φ 0 ρ + φρ0 ) = I ∂R dS(φ 0σ − φσ0 ) Z R dV φρ0 + I ∂R dSφσ0 = Z R dV φ 0 ρ + I ∂R dSφ 0σ (43) 10 c 2010 by Charles Thorn