Page 144 - Electromagnetics

A good choice for G(r|r ) will minimize the effort required to evaluate (r). Examining
(3.56)we notice two possibilities. If we demand that

G(r|r ) = 0 for all r ∈ S (3.70)

then the surface integral terms in (3.56)involving ∂ /∂n will vanish. The Green’s

function satisfying (3.70)is known as the Dirichlet Green’s function. Let us designate it
by G D and use reciprocity to write (3.70)as

G D (r|r ) = 0 for all r ∈ S.
The resulting specialization of (3.56),

ρ(r ) ∂G D (r|r )

(r) = G D (r|r ) dV + (r ) dS +
V S B ∂n
N

∂G D (r|r )

+ (r ) dS , (3.71)
∂n
n=1 S n
requires the specification of (but not its normal derivative)over the boundary surfaces.
In case S B and S n surround and are adjacent to perfect conductors, the Dirichlet bound-
ary condition has an important physical meaning. The corresponding Green’s function is

the potential at point r produced by a point source at r in the presence of the conductors
when the conductors are grounded — i.e., held at zero potential. Then we must specify
the actual constant potentials on the conductors to determine everywhere within V

using (3.71). The additional term F(r|r ) in (3.69)accounts for the potential produced
by surface charges on the grounded conductors.
By analogy with (3.70)it is tempting to try to define another electrostatic Green’s
function according to

∂G(r|r )
= 0 for all r ∈ S. (3.72)

∂n
But this choice is not permissible if V is a finite-sized region. Let us integrate (3.54)over
V and employ the divergence theorem and the sifting property to get
∂G(r|r )

dS =−1; (3.73)
∂n
S
in conjunction with this, equation (3.72)would imply the false statement 0 =−1. Sup-
pose instead that we introduce a Green’s function according to
∂G(r|r ) 1

=− for all r ∈ S. (3.74)
∂n A
where A is the total area of S. This choice avoids a contradiction in (3.73); it does not
nullify any terms in (3.56), but does reduce the surface integral terms involving to
constants. Taken together, these terms all comprise a single additive constant on the
right-hand side; although the corresponding potential (r) is thereby determined only
to within this additive constant, the value of E(r) =−∇ (r) will be unaffected. By
reciprocity we can rewrite (3.74)as

∂G N (r|r ) 1
=− for all r ∈ S. (3.75)
∂n A

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