Then Ji cos 01= J2 cos 02 by (3. 39), while o2J1 sin 01=01J2 sin e2 by(3.40). Hence It is interesting to consider the case of current incident from a conducting material onto material. If insulator, then J2n =J2=0: by (3.39)we ha Jin =0. But(3.40)does not require Jit =0; with 02=0 the right-hand side of (3.40) is indeterminate and thus JI, may be nonzero. In other words, when current moving through a conductor approaches an insulating surface, it bends and Hows tangential to the surface. This concept is useful in explaining how wires guide current Interestingly, (3.42 )shows that when o2<on we have 02-0: current passing fror onducting region into a slightly-conducting region does so normally. 3.2.3 Uniqueness of the electrostatic field In§ ic field is ur V when the tangential component of E is specified over the surrounding surface. Unfortunately this condition is not appropriate in the electrostatic case. We should remember that an additional requirement for uniqueness of solution to Maxwells equations is that the field be specified throughout V at some time to. For a static field this would completely determine e without need for the surface field! Let us determine conditions for uniqueness beginning with the static field equations Consider a region V surrounded by a surface S. Static charge may be located entirely or partially within V, or entirely outside V, and produces a field within V. The region may also contain any arrangement of conductors or other materials. Suppose 1,E1) and (D2, E) represent solutions to the static field equations within V with source p(r) We wish to find conditions that guarantee both El=E2 and DI=D Since V. DI= p and V. D2 p, the difference field Do= D2 - DI obeys the homogeneous equation Consider the quantity V·(DoΦ0)=Φo(VDo)+Do·(vΦo) here Eo=E2-E1=-Vφo=-V(Φ2-中1).We divergence theorem and ( 3.43)to obtain do[D·的dS=/Do·(vφo)dV D0·EodV 3.44 Now suppose that o=0 everywhere on S, or that f. Do=0 everywhere on S, or that po=0 over part of S and f Do =0 elsewhere on S. Then dv Since V is arbitrary, either Do =0 or Eo =0. Assuming E and D are linked by the constitutive relations, we have e= e and D,= D2 Hence the fields within V are unique provided that either the normal component of D, or some combination of the two, is specified over S. We often use a multiply connected surface to exclude conductors. By (3. 33)we see that specification of the ②2001 by CRC Press LLC
Then J1 cos θ1 = J2 cos θ2 by (3.39), while σ2 J1 sin θ1 = σ1 J2 sin θ2 by (3.40). Hence σ2 tan θ1 = σ1 tan θ2. (3.42) It is interesting to consider the case of current incident from a conducting material onto an insulating material. If region 2 is an insulator, then J2n = J2t = 0; by (3.39)we have J1n = 0. But (3.40)does not require J1t = 0; with σ2 = 0 the right-hand side of (3.40) is indeterminate and thus J1t may be nonzero. In other words, when current moving through a conductor approaches an insulating surface, it bends and flows tangential to the surface. This concept is useful in explaining how wires guide current. Interestingly, (3.42)shows that when σ2 σ1 we have θ2 → 0; current passing from a conducting region into a slightly-conducting region does so normally. 3.2.3 Uniqueness of the electrostatic field In § 2.2.1 we found that the electromagnetic field is unique within a region V when the tangential component of E is specified over the surrounding surface. Unfortunately, this condition is not appropriate in the electrostatic case. We should remember that an additional requirement for uniqueness of solution to Maxwell’s equations is that the field be specified throughout V at some time t0. For a static field this would completely determine E without need for the surface field! Let us determine conditions for uniqueness beginning with the static field equations. Consider a region V surrounded by a surface S. Static charge may be located entirely or partially within V, or entirely outside V, and produces a field within V. The region may also contain any arrangement of conductors or other materials. Suppose (D1,E1) and (D2,E2) represent solutions to the static field equations within V with source ρ(r). We wish to find conditions that guarantee both E1 = E2 and D1 = D2. Since ∇ · D1 = ρ and ∇ · D2 = ρ, the difference field D0 = D2 − D1 obeys the homogeneous equation ∇ · D0 = 0. (3.43) Consider the quantity ∇ · (D00) = 0(∇ · D0) + D0 · (∇0) where E0 = E2 − E1 = −∇0 = −∇(2 − 1). We integrate over V and use the divergence theorem and (3.43)to obtain S 0 [D0 · nˆ] d S = V D0 · (∇0) dV = − V D0 · E0 dV. (3.44) Now suppose that 0 = 0 everywhere on S, or that nˆ · D0 = 0 everywhere on S, or that 0 = 0 over part of S and nˆ · D0 = 0 elsewhere on S. Then V D0 · E0 dV = 0. (3.45) Since V is arbitrary, either D0 = 0 or E0 = 0. Assuming E and D are linked by the constitutive relations, we have E1 = E2 and D1 = D2. Hence the fields within V are unique provided that either , the normal component of D, or some combination of the two, is specified over S. We often use a multiplyconnected surface to exclude conductors. By (3.33)we see that specification of the
normal component of D on a conductor is equivalent to specification of the surface charge density. Thus we must specify the potential or surface charge density over all conducting surfaces One other condition results in zero on the left-hand side of (3.44). If S recedes to infinity and o and Do decrease sufficiently fast, then(3.45) still holds and uniqueness is guaranteed. IfD,E-1/r2 as r-00, then w 1/r and the surface integral in(3. 44) tends to zero since the area of an expanding sphere increases only as r. We shall find later in this section that for sources of finite extent the fields do indeed vary inversely with distance squared from the source, hence we may allow s to expand and encompass all space. For the case in which conducting bodies are immersed in an infinite homogeneous medium and the static fields must be determined throughout all space, a multiply connected surface is used with one part receding to infinity and the remaining parts surrounding the conductors. Here uniqueness is guaranteed by specifying the potentials or charges on the surfaces of the conducting bodies 3.2.4 Poisson's and Laplaces equations For computational purposes it is often convenient to deal with the differential versions E(r)=0, of the electrostatic field equations. We must supplement these with constitutive relations between E and D; at this point we focus our attention on linear, isotropic materials for which Dr)=∈(r)E(r) Using this in(3. 47)along with E=-Vqp Justified by(3.46)), we can write v·[e(r)VΦ(r)]=-p(r) (348) This is Poisson's equation. The corresponding homogeneous equation v·[e(r)VΦ(r)=0 holding at points r where p(r)=0, is Laplace's equation. Equations(3. 48)and (3.49) are valid for inhomogeneous media. By(B42) we can write vφ(r).V∈(r)+∈(r)V·[Vφ(r)]=-p(r) For a homogeneous medium,V∈=0; since v·(VΦ)≡V2Φ, we have vΦ(r)=-p(r)/e in such a medium. Correspondingly, V-(r=0 t points where p(r)=0. Poisson's and Laplace's equations can be solved by separation of variables, Fourier transformation, conformal mapping, and numerical techniques such as the finite difference and moment methods. In Appendix a we consider the separation of variables solution ②2001 by CRC Press LLC
normal component of D on a conductor is equivalent to specification of the surface charge density. Thus we must specify the potential or surface charge density over all conducting surfaces. One other condition results in zero on the left-hand side of (3.44). If S recedes to infinity and 0 and D0 decrease sufficiently fast, then (3.45)still holds and uniqueness is guaranteed. If D,E ∼ 1/r 2 as r → ∞, then ∼ 1/r and the surface integral in (3.44) tends to zero since the area of an expanding sphere increases only as r 2. We shall find later in this section that for sources of finite extent the fields do indeed vary inversely with distance squared from the source, hence we may allow S to expand and encompass all space. For the case in which conducting bodies are immersed in an infinite homogeneous medium and the static fields must be determined throughout all space, a multiplyconnected surface is used with one part receding to infinity and the remaining parts surrounding the conductors. Here uniqueness is guaranteed by specifying the potentials or charges on the surfaces of the conducting bodies. 3.2.4 Poisson’s and Laplace’s equations For computational purposes it is often convenient to deal with the differential versions ∇ × E(r) = 0, (3.46) ∇ · D(r) = ρ(r), (3.47) of the electrostatic field equations. We must supplement these with constitutive relations between E and D; at this point we focus our attention on linear, isotropic materials for which D(r) = (r)E(r). Using this in (3.47)along with E = −∇ (justified by (3.46)), we can write ∇ · [ (r)∇(r)] = −ρ(r). (3.48) This is Poisson’s equation. The corresponding homogeneous equation ∇ · [ (r)∇(r)] = 0, (3.49) holding at points r where ρ(r) = 0, is Laplace’s equation. Equations (3.48)and (3.49) are valid for inhomogeneous media. By (B.42)we can write ∇(r) · ∇ (r) + (r)∇ · [∇(r)] = −ρ(r). For a homogeneous medium, ∇ = 0; since ∇ · (∇) ≡ ∇2, we have ∇2 (r) = −ρ(r)/ (3.50) in such a medium. Correspondingly, ∇2 (r) = 0 at points where ρ(r) = 0. Poisson’s and Laplace’s equations can be solved by separation of variables, Fourier transformation, conformal mapping, and numerical techniques such as the finite difference and moment methods. In Appendix A we consider the separation of variables solution
's equation in three major coordinate systems for a variety of problems. For ction to numerical techniques the reader is referred to the books by Sadiku [ 162, Harrington [82 and Peterson et al. [146. Solution to Poisson's equation is often undertaken using the method of Green's functions. which we shall address later in this section. We shall also consider the solution to Laplace's equation for bodies immersed in an applied, or"impressed, field. Uniqueness of solution to Poisson's equation. Before attempting any solutions, ve must ask two very important questions. How do we know that solving the second-order differential equation produces the same values for E= -va as solving the first-order equations directly for E? A if these solutions are the same. what are the conditions for uniqueness of solution to Poisson's and Laplace's equations? To answer the first question, a sufficient condition is to have p twice differentiable. We shall not attempt to prove this, but shall instead show that the condition for uniqueness of the second-order equations is the same as that for the first-order equations Consider a region of space V surrounded by a surface S. Static charge may be located entirely or partially within V, or entirely outside V, and produces a field within V. This egion may also contain any arrangement of conductors or other materials. Now, assume that pI and p2 represent solutions to the static field equations within V with source P(r). We wish to find conditions under which pI=ap ince we have v·[e(r)VΦ1(r)=-p(r),V·[e(rvΦ2(r)=-p(r the difference fieldΦo=中2-中1 obeys V·[e(r)VΦo(r)]=0. (3.51) That is, o obeys Laplace's equation. Now consider the quantity V·(∈ po.o)=∈|Vdo2+dV·(eVdo) Integration over V and use of the divergence theorem and (3.51) gives go(r)le(r)vo(r)].ds= E(r)lvo(r)ldV As with the first order equations, we see that specifying either p(r)or E(r)v(r).nover S results in o(r)=0 throughout V, hence pI= 2. As before, specifying E(r)Vp(r)n for a conducting surface is equivalent to specifying the surface charge on S Integral solution to Poissons equation: the static Greens function. The method of Greens functions is one of the most useful techniques for solving Poissons equation. We seek a solution for a single point source, then use Green's second identity to write the solution for an arbitrary charge distribution in terms of a superposition We seek the solution to Poissons equation for a region of space V as shown in Figure 3.5. The region is assumed homogeneous with permittivity E, and its surface is multiply- connected, consisting of a bounding surface SB and any number of closed surfaces internal to V. We denote by S the composite surface consisting of Sg and the N internal surfaces Sn, n=l,.... N. The internal surfaces are used to exclude material bodies, such as the ②2001 by CRC Press LLC
to Laplace’s equation in three major coordinate systems for a variety of problems. For an introduction to numerical techniques the reader is referred to the books by Sadiku [162], Harrington [82], and Peterson et al. [146]. Solution to Poisson’s equation is often undertaken using the method of Green’s functions, which we shall address later in this section. We shall also consider the solution to Laplace’s equation for bodies immersed in an applied, or “impressed,” field. Uniqueness of solution to Poisson’s equation. Before attempting any solutions, we must ask two very important questions. How do we know that solving the second-order differential equation produces the same values for E = −∇ as solving the first-order equations directly for E? And, if these solutions are the same, what are the conditions for uniqueness of solution to Poisson’s and Laplace’s equations? To answer the first question, a sufficient condition is to have twice differentiable. We shall not attempt to prove this, but shall instead show that the condition for uniqueness of the second-order equations is the same as that for the first-order equations. Consider a region of space V surrounded by a surface S. Static charge may be located entirely or partially within V, or entirely outside V, and produces a field within V. This region may also contain any arrangement of conductors or other materials. Now, assume that 1 and 2 represent solutions to the static field equations within V with source ρ(r). We wish to find conditions under which 1 = 2. Since we have ∇ · [ (r)∇1(r)] = −ρ(r), ∇ · [ (r)∇2(r)] = −ρ(r), the difference field 0 = 2 − 1 obeys ∇ · [ (r)∇0(r)] = 0. (3.51) That is, 0 obeys Laplace’s equation. Now consider the quantity ∇ · ( 0∇0) = |∇0| 2 + 0∇ · ( ∇0). Integration over V and use of the divergence theorem and (3.51)gives S 0(r)[ (r)∇0(r)] · dS = V (r)|∇0(r)| 2 dV. As with the first order equations, we see that specifying either (r) or (r)∇(r)· nˆ over S results in 0(r) = 0 throughout V, hence 1 = 2. As before, specifying (r)∇(r)· nˆ for a conducting surface is equivalent to specifying the surface charge on S. Integral solution to Poisson’s equation: the static Green’s function. The method of Green’s functions is one of the most useful techniques for solving Poisson’s equation. We seek a solution for a single point source, then use Green’s second identity to write the solution for an arbitrary charge distribution in terms of a superposition integral. We seek the solution to Poisson’s equation for a region of space V as shown in Figure 3.5. The region is assumed homogeneous with permittivity , and its surface is multiplyconnected, consisting of a bounding surface SB and any number of closed surfaces internal to V. We denote by S the composite surface consisting of SB and the N internal surfaces Sn, n = 1,..., N. The internal surfaces are used to exclude material bodies, such as the
re 3.5: Computation of potential from known sources and values on bounding sur- plates of a capacitor, which may be charged and on which the potential is assumed to be known. To solve for p (r) within V we must know the potential produced by a point source. This potential, called the Greens function, is denoted G(rr); it has two rguments because it satisfies Poissons equation at r when the source is located at r VG(rr)=-s(r-r (3.52) Later we shall demonstrate that in all cases of interest to us the green's function is symmetric in its arguments G(rr)=g(rr) This property of G is known as reciprocity Our development rests on the mathematical result(B 30) known as Green's second identity. We can derive this by subtracting the identities V·(φVψ)=φv·(vψ)+(φ)·(Vy), v·(yVφ)=ψv·(φ)+(Vyψ)·(Vφ) to obtain V·(φVy-ψVφ)=φvy-ψV2φ Integrating this over a volume region V with respect to the dummy variable r and using the d ve obtain φ(r)vyr)-y(rvφ(r)dv"=-φ[φ(r)vv(r)-y(r)vφ(r)]ds The negative sign on the right-hand side occurs because f is an inward normal to V Finally, since ay(r)/an Vy(r), we have [φ(r)v"y(r)-yr)v"φ(r)dv′ 中(r) dy(r) a(r) ②2001 by CRC Press LLC
Figure 3.5: Computation of potential from known sources and values on bounding surfaces. plates of a capacitor, which may be charged and on which the potential is assumed to be known. To solve for (r) within V we must know the potential produced by a point source. This potential, called the Green’s function, is denoted G(r|r ); it has two arguments because it satisfies Poisson’s equation at r when the source is located at r : ∇2G(r|r ) = −δ(r − r ). (3.52) Later we shall demonstrate that in all cases of interest to us the Green’s function is symmetric in its arguments: G(r |r) = G(r|r ). (3.53) This property of G is known as reciprocity. Our development rests on the mathematical result (B.30)known as Green’s second identity. We can derive this by subtracting the identities ∇ · (φ∇ψ) = φ∇ · (∇ψ) + (∇φ) · (∇ψ), ∇ · (ψ∇φ) = ψ∇ · (∇φ) + (∇ψ) · (∇φ), to obtain ∇ · (φ∇ψ − ψ∇φ) = φ∇2 ψ − ψ∇2 φ. Integrating this over a volume region V with respect to the dummy variable r and using the divergence theorem, we obtain V [φ(r )∇2 ψ(r ) − ψ(r )∇2 φ(r )] dV = − S [φ(r )∇ ψ(r ) − ψ(r )∇ φ(r )] · dS . The negative sign on the right-hand side occurs because nˆ is an inward normal to V. Finally, since ∂ψ(r )/∂n = nˆ · ∇ ψ(r ), we have V [φ(r )∇2 ψ(r ) − ψ(r )∇2 φ(r )] dV = − S φ(r ) ∂ψ(r ) ∂n − ψ(r ) ∂φ(r ) ∂n d S
desired To solve for p in V we shall make some seemingly unmotivated substitutions into this dentity. First note that by (3. 52)and(3.53)we can write (rr)=-8(r-r) (3.54) We now set (r)=(r) and y(r)=G(rr)to obtain [p(r)V-G(rr)-G(rr)V-p(r)]dv aG(rr) (r) G(rr) (355) Φ(r)6(r-r)-G(rr ag(rr) dv op(r a(r rir ds By the sifting property of the Dirac delta G(rIrp(r) dv+ Φ(r) ag(rr) a(r) -Grr) ds+ gir)G(rr) a (r) With this we may compute the potential anywhere within V in terms of the charge density within V and the values of the potential and its normal derivative over S. We must simply determine G(rr) first Let us take a moment to specialize (3.56) to the case of unbounded space. Provided that the sources are of finite extent, as SB ->oo we shall find that p(r) dG(rr) a p(r) Φ(r)=G(rr)dv+ Φ(r) G(rr)- ds A useful derivative identity. Many differential operations on the displacement vector R=r-r occur in the study of electromagnetics. The identities VRE-VRERV 1 R R for example, follow from direct differentiation of the rectangular coordinate representa tion The identity (358) crucial to potential theory, is more difficult to establish. We shall prove the equivalent ②2001 by CRC Press LLC
as desired. To solve for in V we shall make some seemingly unmotivated substitutions into this identity. First note that by (3.52)and (3.53)we can write ∇2 G(r|r ) = −δ(r − r). (3.54) We now set φ(r ) = (r ) and ψ(r ) = G(r|r ) to obtain V [(r )∇2 G(r|r ) − G(r|r )∇2 (r )] dV = − S (r ) ∂G(r|r ) ∂n − G(r|r ) ∂(r ) ∂n d S , (3.55) hence V (r )δ(r − r) − G(r|r ) ρ(r ) dV = S (r ) ∂G(r|r ) ∂n − G(r|r ) ∂(r ) ∂n d S . By the sifting property of the Dirac delta (r) = V G(r|r ) ρ(r ) dV + SB (r ) ∂G(r|r ) ∂n − G(r|r ) ∂(r ) ∂n d S + + N n=1 Sn (r ) ∂G(r|r ) ∂n − G(r|r ) ∂(r ) ∂n d S . (3.56) With this we may compute the potential anywhere within V in terms of the charge density within V and the values of the potential and its normal derivative over S. We must simply determine G(r|r ) first. Let us take a moment to specialize (3.56)to the case of unbounded space. Provided that the sources are of finite extent, as SB → ∞ we shall find that (r) = V G(r|r ) ρ(r ) dV + N n=1 Sn (r ) ∂G(r|r ) ∂n − G(r|r ) ∂(r ) ∂n d S . A useful derivative identity. Many differential operations on the displacement vector R = r − r occur in the study of electromagnetics. The identities ∇ R = −∇ R = Rˆ , ∇ 1 R = −∇ 1 R = − Rˆ R2 , (3.57) for example, follow from direct differentiation of the rectangular coordinate representation R = xˆ(x − x ) + yˆ(y − y ) + zˆ(z − z ). The identity ∇2 1 R = −4πδ(r − r ), (3.58) crucial to potential theory, is more difficult to establish. We shall prove the equivalent version ∇2 1 R = −4πδ(r − r)