Page 142 - Electromagnetics

This simple Green’s function is generally used to find the potential produced by charge
in unbounded space. Here N = 0 (no internal surfaces)and S B →∞.Thus

ρ(r ) ∂G(r|r ) ∂ (r )

(r) = G(r|r ) dV + lim (r ) − G(r|r ) dS .
V S B →∞ S B ∂n ∂n
We have seen that the Green’s function varies inversely with distance from the source,
and thus expect that, as a superposition of point-source potentials, (r) will also vary
inversely with distance from a source of finite extent as that distance becomes large with
respect to the size of the source. The normal derivatives then vary inversely with distance
squared. Thus, each term in the surface integrand will vary inversely with distance cubed,
while the surface area itself varies with distance squared. The result is that the surface
integral vanishes as the surface recedes to infinity, giving
ρ(r )

(r) = G(r|r ) dV .
V
By (3.60)we then have
1 ρ(r )

(r) = dV (3.61)
4π V |r − r |

where the integration is performed over all of space. Since
lim (r) = 0,
r→∞
points at infinity are a convenient reference for the absolute potential.
Later we shall need to know the amount of work required to move a charge Q from
infinity to a point P located at r. If a potential field is produced by charge located in
unbounded space, moving an additional charge into position requires the work
P
W 21 =−Q E · dl = Q[ (r) − (∞)] = Q (r). (3.62)
∞
Coulomb’s law. We can obtain E from (61)by direct differentiation. We have

1 ρ(r ) 1 1

E(r) =− ∇ dV =− ρ(r )∇ dV ,
4π V |r − r | 4π V |r − r |

hence
1 r − r

E(r) = ρ(r ) dV (3.63)
3
4π V |r − r |
by (3.57). So Coulomb’s law follows from the two fundamental postulates of electrostatics
(3.5)and (3.6).
Green’s function for unbounded space: two dimensions. We define the two-
dimensional Green’s function as the potential at a point r = ρ + ˆ zz produced by a
z-directed line source of constant density located at r = ρ . Perhaps the simplest way

to compute this is to first find E produced by a line source on the z-axis. By (3.63)we
have
1 r − r

E(r) = ρ l (z ) dl .
3
4π |r − r |
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