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"Particles", i.e. atoms or
molecules in a liquid or solid are basking in electrical fields - the external
field that we apply from the outside is not necessarily all they
"see" in terms of fields. |
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First, of course, there is a tremendous
electrical field inside any atom. We have
after all, positive charges and negative charges separated by a distance
roughly given by the diameter of the atom. |
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Second, we have fields between atoms, quite evident for ionic crystals, but
also possible for other cases of bonding. |
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However, if you look at the
materials at a scale somewhat larger than the atomic scale, all these fields
must average to zero. Only then do we have
a field-free interior as we always assume in
electrical engineering
("no electrical field can penetrate a metal"). |
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Here, however, we are looking at the
effect an external field has on atoms and
molecules, and it would be
preposterous to assume that
what an atom "sees" as local
electrical field is identical to what we apply from the outside.
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Since all our equations obtained so far always
concerned the local electrical field - even
if we did not point that out in detail before - we now must find a relation
between the external field and the local field, if we want to use the insights
we gained for understanding the behavior of dielectrics on a macroscopic
scale. |
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We define the
local field
Eloc to be the field felt by one particle (mostly an atom) of the material at its
position (x,y,z). |
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Since the superposition principal for fields
always holds, we may express Eloc as a superposition
of the external field Eex and some field
Emat introduced by the surrounding material. We thus
have |
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All electrical fields can, in principle, be
calculated from looking at the charge distribution r(x, y, z) in the material, and
then solving the
Poisson
equation (which you should know). The Poisson equation couples the charge
distribution and the potential V(x, y, z) as
follows: |
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| D = Delta
operator = |
¶2V
¶x2 |
+ |
¶2V
¶y2 |
+ |
¶2V
¶z2 |
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The electrical field then is just the (negative)
gradient of the potential; E = ÑV. |
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Doing this is pretty tricky, however. We can
obtain usable results in a good approximation in a much simpler way, by using
the time-honored Lorentz approach or the Lorentz model. |
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In this approach we decompose the total field
into four components. |
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For doing this, we imagine that we remove a small sphere containing a
few 10 atoms from the material. We want to know the local field in the
center of this sphere while it is still in the material; this is the local
field Emat we are after. We define that field
by the force it exerts on a charge at the center of the sphere that acts as a
"probe". |
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The essential trick is to calculate
the field produced from the atoms inside the sphere and the field inside the
now empty sphere in the material. The total local field then is simple the sum
of both. |
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Like always, we do not consider the
charge of the "probe" in computing the field that it probes. The cut-out sphere thus must not contain the charge we use as
the field probe! |
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The cut-out material, in general, could produce
an electrical field at its center since it is composed of charges. This is the
1st component of the field,
Enear which takes into account the
contributions of the atoms or ions inside the sphere. We will consider that
field in an approximation where we average over the volume of the small sphere.
To make things clear, we look at an ionic crystal where we definitely have
charges in our sphere. |
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Enear, however, is
not the only field that acts on our probe.
We must include the field that all the other atoms of the crystal produce
in the hollow sphere left after we cut out
some material. This field now fills the "empty" void left by taking
out our sphere. |
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This field is called
EL (the "L" stands for
Lorentz); it compensates for the cut-out part - and that provides our
2nd component. |
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Now we only have to add the
"macroscopic" fields from 1. the polarization of the material
and 2. the external field that causes everything: |
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The field Epol is
induced by the macroscopic polarization (i.e. by area charges equal to the
polarization); it is the 3rd
component. |
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The external field Eex =
U/d from the applied voltage at our capacitor which supplies
the 4th component. |
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In a visualization, this looks like this: |
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The blue "sphere" cuts
through the lattice (this is hard to draw). The yellow "point" is
where we consider the local field; we have to omit the contribution of the
charged atom there. We now have |
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| Eloc = Eex +
Epol + EL +
Enear |
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How large are those fields? We know the external
field and also the field from the polarization (always assuming that the
material completely fills the space inside the capacitor). |
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We do not know the other two fields, and it is not
all that easy to find out how large they are. The results one obtains, however,
are quite simple: |
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Lorentz showed that Enear = 0 for
isotropic materials, which is easy to
imagine. Thus for cubic crystals (or polycrystals, or amorphous materials), we
only have to calculate EL. |
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EL needs some thought. It
is, however, a standard problem from electrostatics in a slightly different
form. |
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In the standard problem one calculates the field in a
materials with a DK given by er that does not fill a rectangular capacitor totally, but is in
the shape of an ellipsoid including the extreme cases of a pure sphere, a thin
plate or a thin needle. The
result is always |
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| Eellipse = NP · |
P
er · eo |
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In words: The field inside a dielectric in the
shape of an ellipsoid (of any shape whatsoever) that is put between the
parallel plates of a typical capacitor arrangement, is whatever it would be if
the dielectric fills the space between the plates completely times a number NP, the value of
which depends on the geometry. |
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NP is the so-called
depolarization factor, a pure number, that only depends on the shape of the
ellipsoid. For the extreme cases of the ellipsoid we have fixed and well-known
depolarization factors:
- Thin plate: N = 1
- Needle: N = 0
- Sphere: N = 1/3
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Our case consists of having a sphere with er = 1. We thus obtain |
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We have now all components and
obtain |
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| Eloc |
= |
U
d |
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P
e0 |
+ |
P
3eo |
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U/d P/e0 is just the field we would use in the
Maxwell equations, we call it E0. It is the homogeneous field averaged over the whole volume of
the homogeneous material |
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The local
field finally becomes
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This seems a bit odd? How can
the local field be different from the
average field? |
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This is one of the tougher questions one can
ask. The answer, not extremely satisfying, comes from the basic fact that
all dipoles contribute to
E0, whereas for the local
field you discount the effect of one charge - the charge you use for probing the
field (the field of which must not be added to
the rest!). |
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If you feel somewhat uneasy about this, you
are perfectly right. What we are excluding here is the action of a charge on
itself. While we may do that because that was one way of defining electrical fields (the other one
is Maxwells equation defining a field as directly resulting from charges), we
can not so easily do away with the energy
contained in the field of a single charge. And if we look at this, the whole
theory of electromagnetism blows up! If the charge is a point charge, we get
infinite energy, and if it is not a point charge, we get other major
contradictions. |
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Not that it matters in everyday aspects - it
is more like a philosophical aspect. If you want to know more about this, read
chapter 28 in the "Feynman lectures, Vol.
2" |
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But do
not get confused now! The relation given above is perfectly valid
for everyday circumstances and ordinary matter. Don't worry - be happy that a
relatively complex issue has such a simple final formula! |
© H. Föll (Advanced Materials B, part 1 - script)