**Topics covered:**

Polarization

Dielectrics

The Van de Graaff

More on Capacitors

**Instructor/speaker:** Prof. Walter Lewin

## Free Downloads

### Video

» Download this transcript (PDF)

Electric fields can induce dipoles in insulators.

Electrons in insulators are bound to the atoms and to the molecules, unlike conductors, where they can freely move, and when I apply an external field -- for instance, a field in this direction, then even though the molecules or the atoms may be completely spherical, they will become a little bit elongated in the sense that the electrons will spend a little bit more time there than they used to, and so this part becomes negatively charged and this part becomes positively charged, and that creates a dipole.

I discussed that with you, already, during the first lecture, because there's something quite remarkable about this, that if you have an insulator -- notice the pluses and the minuses indicate neutral atoms -- and if now, I apply an electric field, which comes down from the top, then, you see a slight shift of the electrons, they spend a little bit more time up than down, and what you see now is, you see a layer of negative charge being created at the top, and a layer of positive charge being created at the bottom.

That's the result of induction, we call that also, sometimes, polarization.

You are polarizing, in a way, the electric charge.

Uh, substances that do this, we call them dielectrics, and today, we will talk quite a bit about dielectrics.

The first part of my lecture is on the web, uh, if you go to 8.02 web, you will see there a document which describes, in great detail, what I'm going to tell you right now.

Suppose we have a plane capacitor -- two planes which I charge with a certain potential, and I have on here, say, a charge plus sigma and here I have a charge minus sigma.

I'm going to call this free -- you will see, very shortly why I call this free -- and this is minus free.

So there's a potential difference between the plate, charge flows on there, it has an area A, and sigma free is the charge density, how much charge per unit area.

So we're going to get an electric field, which runs in this direction, and I call that E free.

And the distance between the plates, say, is D.

So this is given.

I now remove the power supply that I used to give it a certain potential difference.

I completely take it away.

So that means that this charge here is trapped, can not change.

But now I move in a dielectric.

I move in one of those substances.

And what you're going to see here, now, at the top, you're going to see a negative-induced layer, and at the bottom, you're going to see a positive-induced layer.

I called it plus sigma induced, and I call this minus sigma induced.

And the only reason why I call the other free, is to distinguish them from the induced charge.

This induced charge, which I have in green, will produce an electric field which is in the opposite I- direction, and I call that E-induced.

And clearly, E free is, of course, the surface charge density divided by epsilon 0, and E induced is the induced surface charge density, divided by epsilon 0.

And so the net E field is the vectorial sum of the two, so E net -- I gave it a vector -- is E free plus E induced, vectorially added.

Since I'm interested -- I know the direction already -- since I'm interested in magnitudes, therefore the strength of the net E field is going to be the strength of the E fields created by the so-called free charge, minus the strength of the E fields created by the induced charge, minus -- because this E vector is down, and this one is in the up direction.

And so, if I now make the assumption that a certain fraction of the free charge is induced, so I make the assumption that sigma induced is some fraction B times sigma free, I just write, now, and I for induced and an F for free.

B is smaller than 1.

If B were .1, it means that sigma-induced would be 10% of sigma free, that's the meaning of B.

So clearly, if this is the case, then, also, E of I must also be B times E of F.

You can tell immediately, they are connected.

And so now I can write down, for E net, I can also write down E free times 1 minus B, and that 1 minus B, now, we call 1 over kappa.

I call it 1 over kappa, our book calls it 1 over K.

But I'm so used to kappa that I decided to still hold on to kappa.

And that K, or that kappa, whichever you want to call it, is called the dielectric constant.

It's a dimensionless number.

And so I can write down, now, in general, that E -- and I drop the word net, now, from now on, whenever I write E, throughout this lecture, it's always the net electric field, takes both into account.

So you can write down, now, that E equals the free electric fields, divided by kappa, because 1 minus B is 1 over kappa.

And so you see, in this experiment that I did in my head, first, bringing charge on the plate, certain potential difference, removing the power supply, shoving in the dielectric that an E field will go down by a factor kappa.

Kappa, for glass, is about 5.

That will be a major reduction, I will show you that later.

If the electric field goes down, in this particular experiment, it is clear that the potential difference between the plates will also go down, because the potential difference between the plates, V is always the electric field between the plates times D.

And so, if this one goes down, by a factor of kappa, if I just shove in the dielectric, not changing D, then, of course, the potential between the plates is also going down.

None of this is so intuitive, but I will demonstrate that later.

The question now arises, does Gauss's Law still hold?

And the answer is, yes, of course, Gauss's Law will still hold.

Gauss's Law tells me that the closed loop -- closed surface, I should say, not closed loop -- the closed surface integral of E dot dA is 1 over epsilon times the sum of all the charges inside my box.

All the charges! The net charges, that must take into account both the induced charge, as well as the free charge.

And so let me write down here, net, to remind you that.

But Q net is, of course, Q free plus Q induced.

And I want to remind you that this is minus, and this was plus.

The free charge, positive there, is plus, and at that same plate, if you have your Gaussian surface at the top, you have the negative charged Q induced.

And so therefore, Gauss's Law simply means that you have to take both into account, and so, therefore, you can write down 1 over epsilon 0, times the sum of Q free, but now you have to make sure that you take the induced charge into account, and therefore, you divide the whole thing by kappa.

Then you have automatically taken the induced charge into account.

So you can amend Gauss's Law very easily by this factor of kappa.

Dielectric constant is dimensionless, as I mentioned already, it is 1, in vacuum, by definition.

1 atmosphere gases typically have dielectric constant just a hair larger than 1.

We will, most of the time, assume that it is 1.

Plastic has a dielectric constant of 3, and glass, which is an extremely good insulator, has a dielectric constant of 5.

If you have an external field, that can induce dipoles in molecules -- but there are substances, however, which themselves are already dipoles, even in the absence of an electric field.

If you apply, now, an external field, these dipoles will start to align along the electric field, we did an experiment once, with some grass seeds, perhaps you remember that.

And as they align in the direction of the electric field, they will strengthen the electric field.

On the other hand, because of the temperature of the substance, these dipoles, these molecules which are now dipoles by themselves, through chaotic motion, will try to disalign, temperature is trying to disalign them.

So it is going to be a competition, on the one hand, between the electric field which tries to align them and the temperature which tries to disalign them.

But if the electric field is strong, you can get a substantial amount of alignment.

Uh, permanent dipoles, as a rule, are way stronger than any dipole that you can induce by ordinary means in a laboratory, and so the substances which are natural dipoles, they have a much higher value for kappa, a much higher dielectric constant that the substances that I just discussed, which themselves, do not have dipoles.

Water is an example, extremely good example.

The electrons spend a little bit more time near the oxygen than near the hydrogen, and water has a dielectric constant of 80.

That's enormous.

## Tidak ada komentar:

Posting Komentar