Replacing V.J with -jop from the continuity equation and using the divergence theorem on the second term on the right-hand side. we then have 1 E. dV==p (2J'G+jop'v'g)dv (的·Jv"GdS Lastly we examine Jn·Vx(Gp) Use of J.V×(p)=Jn(VG×p)=p(Jn×VG)gi JmV=jp·/Jmxv We now substitute all terms into(6.5)and note that each term involves a dot product with p Since p is arbitrary we have axE×VG+(,EG-(x(dS+ dr.HV'G Jm x V'G+=V'G-jojJ' The electric field may be extracted from the above expression by letting the radius of the excluding volume Vs recede to zero. We first consider the surface integral over Ss Examining Figure 6.3 we see that R=rp-r=8,n=-R R VG(rrp) d VRER Assuming E is continuous at r=rp we can write [(×E)×VG+('EVG-joj(xmG]dS R R 40xE)×+(R.E-xm|82aB 1切~(,E+食RE+(RER]d2=E(r ②2001 by CRC Press LLCFigure 6.3: Geometry of surface integral used to extract E at rp. Replacing ∇ · J˜i with − jωρ˜i from the continuity equation and using the divergence theorem on the second term on the right-hand side, we then have  V−Vδ E˜ p · J˜i dV = 1 ˜ c p˜ ·  V−Vδ (k2 J˜i G + jωρ˜i ∇ G) dV − S+Sδ (nˆ · J˜i )∇ GdS  . Lastly we examine  V−Vδ H˜ p · J˜i m dV = jω  V−Vδ J˜i m · ∇ × (Gp˜) dV . Use of J˜i m · ∇ × (Gp˜) = J˜i m · (∇ G × p˜) = p˜ · (J˜i m × ∇ G) gives  V−Vδ H˜ p · J˜i m dV = jωp˜ ·  V−Vδ J˜i m × ∇ G dV . We now substitute all terms into (6.5) and note that each term involves a dot product with p˜. Since p˜ is arbitrary we have − S+Sδ (nˆ × E˜) × ∇ G + (nˆ · E˜)∇ G − jωµ(˜ nˆ × H˜ )G  d S + + 1 jω˜ c a+b (dl · H˜ )∇ G =  V−Vδ  −J˜i m × ∇ G + ρ˜i ˜ c ∇ G − jωµ˜ J˜i G  dV . The electric field may be extracted from the above expression by letting the radius of the excluding volume Vδ recede to zero. We first consider the surface integral over Sδ . Examining Figure 6.3 we see that R = |rp − r | = δ, nˆ = −Rˆ , and ∇ G(r |rp) = d d R e− jkR 4π R  ∇ R = Rˆ 1 + jkδ 4πδ2  e− jkδ ≈ Rˆ δ2 as δ → 0. Assuming E˜ is continuous at r = rp we can write − lim δ→0 Sδ (nˆ × E˜) × ∇ G + (nˆ · E˜)∇ G − jωµ(˜ nˆ × H˜ )G  d S = lim δ→0   1 4π  (Rˆ × E˜) × Rˆ δ2 + (Rˆ · E˜) Rˆ δ2 − jωµ(˜ Rˆ × H˜ ) 1 δ  δ2 d = lim δ→0   1 4π −(Rˆ · E˜)Rˆ + (Rˆ · Rˆ )E˜ + (Rˆ · E˜)Rˆ  d = E˜(rp).