Solution to Exercise 3.2-1

	Sometimes, a question can be more tricky than originally intended. That is the case here - lets see why.
		First lets get the e.s.u out of the way. It means "electrostatic units" which are sub-units of the old c.g.s (centimeter-gram-second) system, and still much in use.
		Few things are more confusing than converting electric or magnetic c.g.s. units into the SI (Standard International) kilogram- meter- second-Ampère system. If you are not somewhat familiar with that, read up the basic modules accessible by the links to this topic.
	In the case given here, you have to multiply with \|c\|/10 = 3,3356 · 10^–10 (c = vacuum speed of light) to obtain the charge in [C] (The magnitude signs \| \| simply mean hat you only take the number!); and since the dipole moment is charge times distance, the distance in e.s.u units must be cm.
		We obtain µ_water = 1,87 · 10^–18. 3,3356 ·10^–10 C · cm = 6,24 · 10^–28 C · cm
		Lets see if that is reasonable: A water molecule carries about one elementary charge = 1,6 · 10^–19 C at the end of the dipole, and the distance will be about 1 Å = 10^–8 cm. This would give a dipole moment of 1,6 · 10^–27 C·cm, so the number we got should be correct

Now to the tricky part. First it is important to realize that:

A material with completely oriented natural dipoles does not have a dielectric constant e_r or dielectric susceptibility c = e_r – 1 anymore!

Consider: c was the proportionality factor between the external field E and the induced polarization P

P = e · ₀c · E

If the field doubles, the polarization, and thus the degree of orientation into the field doubles.

However, if all dipoles are fully aligned, the polarization is at a maximum and will not respond to the field anymore; c looses its meaning.

Nevertheless, we could take this fully polarized material, stick it into a plate capacitor, and just measure how the capacitance C changes . This would give us a value for e_r simply by computing C_after/C_before. Lets see if we can do this.

For the capacity before we use our fully polarized dielectric we have with some applied voltage U and some corresponding charge Q₀

C_before	=	Q₀ U

For the capacity after we use our fully polarized dielectric we have
C_after = (Q₀ + Q_pol)/U,
and this gives us

C_after C_before	=	e_r	=	Q₀ + Q_pol Q₀

This does not help, however, because we do not know Q₀. Lets try a different approach and look at C_after – C_before.

We obtain .

C_after – C_before = e_r · C_before – C_before = C_before · (e_r – 1) = C_before · c =	Q₀ + Q_pol U	–	Q₀ U	=	Q_pol U

c	=	Q_pol U · C_before	=	Q_pol Q₀

This looks better, but it is still not useful - we do not know Q₀. We still have the same problem: The changes are not proportional to what we had before the introduction of the dielectric, but absolute - we are, in effect, adding a fixed charge and thus switching a second capacitor in series.

Lets try a different approach. We know that c(H₂O) » 80. The polarization that goes with this value increases steadily as the field strength inducing the polarization increases - as long as we have P = c · E

For large field strength, however, this "law" must break down - we reach the absolute limit of polarization sooner or later.

So lets compute in a first approximation the field strength needed (within the simple law) to induce the maximum polarization and compare the value obtained to field strengths usually encountered.

First, we compute the maximum polarization P_max. This is simply the the charge q_H₂O on one end of the water dipole times the distance of the charges d_H₂O divided by the (area) density of the dipoles, i.e. the (area density) of water.

The dipole moment of water is given by

µ_water = q_H₂O · d_H₂O = 1,87 · 10^–18 · 3,3356 · 10^–10 C · cm

We need d_H₂O to compute q_H₂O; from the picture in the question we find it to be d_H₂O = 0,0958 nm · cos(104,45^o/2) = 0,0586 nm.

The (effective) charge q_H₂O at the end of a dipole thus is

q_H₂O = q_H₂O = 6,24 · 10^–28 C·cm / 0,0586 ·10^–7 cm = 1,065 · 10^–19 C

about 2/3 of an elementary charge.

The density of water is r_H₂O = 1 kg/l = 1g/cm³ by definition.

One mol of water is 1+ 1+ 16 = 18 g which tells us that we have 1 mol = 6.022 · 10²³ water molecules in 18 cm³.

The areal density r_arealof dipoles is therefore

r_areal	=	6.022 · 10²³ · 0.0586 nm 18 cm³	= 1,96 · 10¹⁴ dipoles/cm²

Converting volume densities to areal or surface densities may appear tricky. If you are not sure about how it is done, consult the link.

The maximum polarization P_max thus is.

P_max	=	1,065 10^–19 · 1,96 · 10¹⁴C /cm² = 2,087 · 10^–5 C /cm²

If we want to generate this polarization with an electrical field and a susceptibility c = 80, we need a saturation field strength E_sat of

E_sat	=	P_max/80 · e₀ = 2,087 · 10^–5/80 · 8,854 · 10^–12(C/cm²) · (Vm/C) = 2,946 ·10⁶ V/cm

OK, that is a definite result. Now we have to ask ourselves, how we must compare a field strength of about 3 ·10⁶ V/cm to "normal" field strengths.

To some extent, we do that in sub-chapter 3.5.1, but common sense tells us that we would certainly use 1mm or more of a dielectric to insulate a wire carrying 1000 V, for example. This translates to a "typical" field strength of 10.000V/cm.

Many materials will be destroyed at field strengths of very roughly 100.000 V/cm, so 3 · 10⁶V/cm is very large, indeed.

However, dielectrics in integrated circuits must be able to operate at field strength of this order of magnitude. Take 3 V and a thickness of the dielectric of 10 nm - a not atypical combination - and you have a field strength of 3 · 10⁶V/cm, just what we calculated.

Anyway, if we take 100.000 V/cm as a "normal value, we realize that the only 3,4% of the dipoles need to be oriented in field direction, whereas the rest could be oriented at random. (1 · 10⁵/2,946 ·10⁶ = 0,034).

This is not the physical reality, of course. A more physical interpretation is that all dipoles change whatever orientation they happen to have by about 3,4 % in field direction. What that means precisely, we will leave open, the general meaning, however, is clear:

The effect of polarization would hardly be noticeable by just looking at the distribution of the dipoles. It is a rather small effect, even for a material with a comparatively very large dielectric susceptibility.

With frame

Gauss law or integral theorem

Exercise 3.2-1

Solution to 3.2-2