Page 147 - Electromagnetics

on the other hand, involves an integral over the surface fields and is known as the sec-
ondary potential. This term is linked to effects outside V . Since the “sources” of s
s
(i.e., the surface fields)lie on the boundary of V , satisfies Laplace’s equation within
s
V . We may therefore use other, more convenient, representations of provided they
satisfy Laplace’s equation. However, as solutions to a homogeneous equation they are of
indefinite form until linked to appropriate boundary values.
Since the geometry is invariant in the x and y directions, we represent each potential
function in terms of a 2-D Fourier transform over these variables. We leave the z depen-
dence intact so that we may apply boundary conditions directly in the spatial domain.
The transform representations of the Green’s functions for the primary and secondary
potentials are derived in Appendix A. From (A.55)we see that the primary potential
within region V i can be written as

p p ρ(r )
(r) = G (r|r ) dV (3.76)
i
i
V i
where

1 1 ∞ e −k ρ |z−z |
p jk ρ ·(r−r ) 2

G (r|r ) = = e d k ρ (3.77)

4π|r − r | (2π) 2 2k ρ
−∞
2
is the primary Green’s function with k ρ = ˆ xk x + ˆ yk y , k ρ =|k ρ |, and d k ρ = dk x dk y .
We also find in (A.56)that a solution of Laplace’s equation can be written as
1 ∞
s k ρ z −k ρ z jk ρ ·r 2
(r) = 2 A(k ρ )e + B(k ρ )e e d k ρ (3.78)
(2π)
−∞
where A(k ρ ) and B(k ρ ) must be found by the application of appropriate boundary con-
ditions.
As a simple example, consider a charge distribution ρ(r) in free space above a grounded
conducting plane located at z = 0. We wish to find the potential in the region z > 0
using the Fourier transform representation of the potentials. The total potential is a sum
of primary and secondary terms:

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

(x, y, z) = 2 e d k ρ dV +
V (2π) −∞ 2k ρ 0
1 ∞ −k ρ z jk ρ ·r 2
+ B(k ρ )e e d k ρ ,
(2π) 2
−∞
where the integral is over the region z > 0. Here we have set A(k ρ ) = 0 because e k ρ z
grows with increasing z. Since the plane is grounded we must have (x, y, 0) = 0.

Because z < z when we apply this condition, we have |z − z |= z − z and thus

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

(x, y, 0) = e dV + B(k ρ ) e d k ρ = 0.
(2π) 2 V 0 2k ρ
−∞
Invoking the Fourier integral theorem we find
−k ρ z
ρ(r ) e − jk ρ ·r

B(k ρ ) =− e dV ,
V 0 2k ρ
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