Perfect conductors are not just dielectrics where the electric field is
completely zero inside. The electric field *is* exactly cancelled
on the interior by the induced surface charge. As we have seen, this
cancellation occurs close to the surface (within a few times the skin
depth). However, the induced currents *also* tend to expel the time
dependent magnetic field. We therefore have two modification of our
results from the previous section. The electric polarization will have
a different form, and there will be a contribution from the induced *magnetic* moment of the sphere as well.

Recall (from J2.5) that the induced dipole moment on a conducting sphere is
= 4&pi#pi;&epsi#epsilon;_0 a^3 _inc .
This is indeed the generalization of the result for **p** last time,
as you should be able to derive in a few minutes of work. Either review
that section or solve the boundary value problem where
is
discontinuous at the surface and
on the surface to
obtain:
&phis#phi;= -E_0 (r - a^3r^2 ) &thetas#theta;
from which we can easily extract this
.

**But**, the magnetic field is also varying, and it induces an EMF that
runs in loops around the magnetic field lines and opposes the change in
magnetic flux. Assuming that no field lines were trapped in the sphere
initially, the induced currents act to cancel component of the magnetic field
normal to the surface. The sphere thus behaves like a magnetically permeable
sphere (see e.g. section J5.10 and J5.11, equations J5.106, J5.107,
J5.115):
= 4&pi#pi;a^3/3 = 3 (&mu#mu;-
&mu#mu;_0&mu#mu;+ 2&mu#mu;_0)_inc
with
so that:
= - 2&pi#pi;a^3 _inc.
The derivation is again very similar to the derivation we performed last
time, with suitably chosen boundary conditions on
and
.

If we then repeat the reasoning and algebra for this case of the conducting
sphere (substituting *this*
and
into the expression we derived
for the differential cross-section), we get:
d&sigma#sigma;d&Omega#Omega; = k^4 a^6 ^&ast#ast;·_0 -
12(×^&ast#ast;) ·(_0 ×_0)^2 .

After much tedious but straightforward work, we can show (or rather you can
show for homework) that:
d&sigma#sigma;_&par#parallel;d&Omega#Omega; & = &k^4a^62 &thetas#theta;-
12^2

d&sigma#sigma;_&perp#perp;d&Omega#Omega; & = & k^4a^62 1 - 12
&thetas#theta;^2
so that the total differential cross section is:
d&sigma#sigma;d&Omega#Omega; = k^4 a^6 {58 (1+^2 &thetas#theta;) -
&thetas#theta;)}
and the polarization is:
&Pi#Pi;(&thetas#theta;) = 3 ^2 &thetas#theta;5(1 + ^2 &thetas#theta;) - 8 &thetas#theta;
Finally, integrating the differential cross section yields the total
cross-section:
&sigma#sigma;= 10 &pi#pi;k^4 a^63 = (4&pi#pi;a^2)(ka)^4 2.53
&sim#sim;&sigma#sigma;_dielectric
for
curiously enough.

What do these equations tell us? The cross-section is strongly peaked backwards. Waves are reflected backwards more than forwards (the sphere actually casts a ``shadow''. The scattered radiation is polarized qualitatively alike the radiation scattered from the dielectric sphere, but with a somewhat different angular distribution. It is completely polarized perpendicular to the scattering plane when scattered through an angle of 60 , not 90 .

We see that dipole scattering will always have a characteristic dependence. By know you should readily understand me when I say that this is the result of performing a multipolar expansion of the reaction field (essentially an expansion in powers of where is the characteristic maximum extent of the system) and keeping the first (dipole) term.

If one wishes to consider scattering from objects where or greater, one simply has to consider higher order multipoles (and one must consider the proper multipoles instead of simple expansions in powers of ). If (which is the case for light scattering from macroscopic objects, radar scattering from airplanes and incoming nuclear missiles, etc) then a whole different apparatus must be brought to bear. I could spend a semester (or a least a couple of weeks) just lecturing on the scattering of electromagnetic waves from spheres, let alone other shapes.

However, no useful purpose would be so served, so I won't. If you ever need to figure it out, you have the tools and can find and understand the necessary references.