Page 146 - Electromagnetics

Figure 3.7: Electrostatic shielding by a conducting shell.

where S B is tangential to the inner surface of the shell and we have used ρ = 0 within
the shell. Because (r ) = 0 for all r on S B ,wehave

(r) = 0
everywhere in the region enclosed by the shell. This result is independent of the charge
outside the shell, and the interior region is “shielded” from the effects of that charge.
Conversely, consider a grounded conducting shell with charge contained inside. If we
surround the outside of the shell by a surface S 1 and let S B recede to infinity, then (3.71)
becomes

∂G D (r|r ) ∂G D (r|r )

(r) = lim (r ) dS + (r ) dS .
S B →∞ ∂n ∂n
S B S 1
Again there is no charge in V (since the charge lies completely inside the shell). The
contribution from S B vanishes. Since S 1 lies adjacent to the outer surface of the shell,

(r ) ≡ 0 on S 1 .Thus (r) = 0 for all points outside the conducting shell.
Example solution to Poisson’s equation: planar layered media. For simple
geometries Poisson’s equation may be solved as part of a boundary value problem (§ A.4).
Occasionally such a solution has an appealing interpretation as the superposition of
potentials produced by the physical charge and its “images.” We shall consider here the
case of planar media and subsequently use the results to predict the potential produced
by charge near a conducting sphere.
Consider a layered dielectric medium where various regions of space are separated by
planes at constant values of z. Material region i occupies volume region V i and has
permittivity i ; it may or may not contain source charge. The solution to Poisson’s
equation is given by (3.56). The contribution
ρ(r )

p
(r) = G(r|r ) dV
V
produced by sources within V is known as the primary potential. The term

s ∂G(r|r ) ∂ (r )
(r) = (r ) − G(r|r ) dS ,
S ∂n ∂n

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