Page 140 - Electromagnetics

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

2 2
[ (r )∇ G(r|r ) − G(r|r )∇ (r )] dV =
V
∂G(r|r ) ∂ (r )

− (r ) − G(r|r ) dS , (3.55)
S ∂n ∂n
hence

ρ(r ) ∂G(r|r ) ∂ (r )

(r )δ(r − r) − G(r|r ) dV = (r ) − G(r|r ) dS .
V S ∂n ∂n
By the sifting property of the Dirac delta
ρ(r ) ∂G(r|r ) ∂ (r )

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

∂G(r|r ) ∂ (r )

+ (r ) − G(r|r ) dS . (3.56)
∂n ∂n
n=1 S n
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 S B →∞ we shall find that

N
ρ(r )
∂G(r|r ) ∂ (r )

(r) = G(r|r ) dV + (r ) − G(r|r ) dS .
∂n ∂n
V n=1 S n
A useful derivative identity. Many differential operations on the displacement vector
R = r − r occur in the study of electromagnetics. The identities

ˆ
R
1
1
ˆ

∇ R =−∇ R = R, ∇ =−∇ =− , (3.57)
R R R 2
for example, follow from direct differentiation of the rectangular coordinate representa-
tion

R = ˆ x(x − x ) + ˆ y(y − y ) + ˆ z(z − z ).
The identity

1
2
∇ =−4πδ(r − r ), (3.58)
R
crucial to potential theory, is more difficult to establish. We shall prove the equivalent
version

2 1
∇ =−4πδ(r − r)
R
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