Page 149 - Electromagnetics

To determine A and B we impose (3.36)and (3.37). By (3.36)we have

1 ∞ ρ(r ) e −k ρ z − jk ρ ·r jk ρ ·r 2

e dV + B(k ρ ) − A(k ρ ) e d k ρ = 0,

(2π) 2 V 1 2k ρ
−∞
hence
ρ(r ) e − jk ρ ·r
−k ρ z
e dV + B(k ρ ) − A(k ρ ) = 0

V 1 2k ρ
by the Fourier integral theorem. Applying (3.37)at z = 0 with ˆ n 12 = ˆ z, and noting that
there is no excess surface charge, we find
e − jk ρ ·r
−k ρ z

ρ(r ) e dV − 1 B(k ρ ) − 2 A(k ρ ) = 0.
V 2k ρ
The solutions
−k ρ z
2 1 ρ(r ) e − jk ρ ·r

A(k ρ ) = e dV ,
1 + 2 V 1 2k ρ
−k ρ z
1 − 2 ρ(r ) e − jk ρ ·r

B(k ρ ) = e dV ,
1 + 2 V 1 2k ρ
are then substituted into (3.81)and (3.82)to give

e
∞ e −k ρ |z−z | 1 − 2 −k ρ (z+z )
1 + jk ρ ·(r−r ) 2 ρ(r )

1 (r) = 2 1 + 2 e d k ρ dV
V (2π) −∞ 2k ρ 1

ρ(r )

= G 1 (r|r ) dV ,
V 1

1 2 2 e jk ρ ·(r−r ) 2 ρ(r )
∞ −k ρ (z −z)

2 (r) = 2 e d k ρ dV
V (2π) −∞ 1 + 2 2k ρ 2
ρ(r )

= G 2 (r|r ) dV .
V 2
Since z > z for all points in region 2, we can replace z − z by |z − z | in the formula for

2 .
As with the previous example, let us compare the result to the form of the primary
Green’s function (3.77). We see that
1 1 − 2 1

G 1 (r|r ) = + ,

4π|r − r | 1 + 2 4π|r − r |
1
2 2 1

G 2 (r|r ) = ,

1 + 2 4π|r − r |
2

where r = ˆ xx + ˆ yy − ˆ zz and r = ˆ xx + ˆ yy + ˆ zz . So we can also write

1 2

1 1 1 − 2 1 ρ(r )

1 (r) = + dV ,

4π V |r − r | 1 + 2 |r − r | 1

1
1 2 2 1 ρ(r )

2 (r) = dV .

4π V 1 + 2 |r − r | 2
2
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