Electric fields can induce dipoles in insulators.
Electrons and 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 become 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 eight oh two 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 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 a 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 zero,
and E-induced is the induced surface charge density,
divided by epsilon zero.
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
strengths of the E fields created by the so-called free
charge, minus the strengths 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 one. If B were point one,
it means that sigma-induced would be ten percent of sigma
free, that's the meaning of B. So clearly, if this is the
case, then, also, E of I must also me 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 one minus B, and that one minus B,
now, we call one over kappa. I call it one over kappa,
our book calls it one 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 one minus B is one 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 five.
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' Law still hold? And the answer is,
yes, of course, Gauss' Law will still hold.
Gauss' Law tells me that the closed loop -- closed surface,
I should say, not closed loop -- the closed
surface integral of E dot D A is one 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' Law simply means that you have to take both into
account, and so, therefore, you can write down
one over epsilon zero, 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 mex- uh,
Gauss' Law very easily by this factor of kappa.
Dielectric constant is dimensionless,
as I mentioned already, it is one, in vacuum,
by definition. One atmosphere gases typically
have dielectric constant just a hair larger than one.
We will, most of the time, assume that it is one.
Plastic has a dielectric constant of three,
and glass, which is an extremely good insulator,
has a dielectric constant of five.
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
that 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 eighty. That's enormous.
And if you go down to lower temperature, if you take ice of
minus forty degrees, it is even higher,
then the dielectric constant is one hundred.
I'm now going to massage you through four demonstrations,
four experiments. One of them,
you have already seen. And try to follow them as
closely as you can, because if you miss
one small step, then you miss,
perhaps, a lot. I have two parallel plates
which a- are on this table, as you have seen last time,
and I have, here, a current meter,
I put it -- an A on there, that means amp meter.
And the plates have a certain separation D.
I'm going to charge this capacitor up by connecting these
ends to a power supply, and I'm going to connect them
to fifteen hundred volts. I'm -- I'm already going to set
my light, because that's where you're going to see it very
shortly. I'm going to start off with a
distance D -- so this is going to be my experiment one -- with
a distance D of one millimeter. And the voltage V always means
the voltage the -- the -- the potential difference between the
plates is going to be fifteen hundred volts.
Forgive me for the two Vs, I can't help that.
This means, here, the potential difference,
and this is the unit in volts. Once I have charged them,
I disconnect -- this is very important -- I disconnect the
power supply, for which I write P S.
That's it. So the charge is now trapped.
As I charge it, as you saw last time,
you will see that the amp meter shows a short surge of current,
because, as I put charge on the plates, the charge has to go
from the power supply to the plates, and you will see a short
surge of current which will make the handle -- the hand of the
power su- of the amp meter, as you will see on the -- on
the wall there -- go to the right side, just briefly,
and then come back. This indicates that you are
charging the plates. Now, I'm going to open up the
gap -- so this is my initial condition, there is no
dielectric -- and now I'm going to go D to
seven millimeters. And this is what I did last
time. The reason why I do it again,
because I need this for my next demonstration.
If I make the distance seven millimeters, then the charge,
which I call now, Q free, but it is really the
charge on the plates, is not going to be -- is not
going to change, it is trapped.
So there can be no change when I open up the gap.
That means the amp meter will do
nothing, you will not see any charge flow.
The electric field E is unchanged, because E is sigma
divided by epsilon zero. If sig- if Q free is not
changing, sigma cannot change. So, no change in the electric
field. But the potential V is now
going to go up by a factor of seven, because V equals E times
D. E remains constant,
D goes up, V has to go up. And this is what I want to show
you first, even though you have already seen this.
And I need the new conditions for my demonstration that comes
afterwards. I'm going from fifteen hundred
volts to about ten thousand volts, it goes up by a factor of
seven. And you're going to see that
there. There you see your amp meter.
I'm going to -- you see the, um, this is this propeller volt
meter that we discussed last time, and here you see the --
the plates. They're one millimeter apart
now, very close. And I'm going to charge the
plates, I will count down, so you keep your amp meter,
three, two, one, zero, and you saw a current
surge. So I charged the capacitor.
It is charged now. The volt meter doesn't show
very much, fifteen hundred volts.
Maybe it went up a little, but not very much,
but now I'm going to increase the gap to ten -- to seven
millimeters, and look that the amp meter is not doing anything,
the charge is trapped, so there is no charge going to
the plates, but look what the volt meter is doing.
It's increasing the voltage, it's not approaching almost ten
thousand volts, although this is not very
quantitative, and now I have a gap of about
seven millimeters, and that's what I wanted.
We've seen that the plates on the left side here are now
farther apart than they were before.
So that is my demonstration number one, a repeat of what we
did last time. So now comes number two.
So now my initial conditions are that V is now ten kilovolts,
so that's the potential difference between the plates
that I have now, and D is now seven millimeters,
and I'm not going to change that.
At this moment, kappa is one.
But now, I'm going to insert the dielectric.
So I take a piece of glass, and I'll just put it into that
gap. Q free cannot go anywhere,
because I have disconnected the power supply.
So Q free, no change. If there is no chee- no change
in the free charge, the amp meter will do nothing.
So as I plunge in this dielectric, you will not see any
reading on the amp meter. But, as we discussed at length
now, the electric field, which is the net electric
field, will go down by that factor kappa.
That's what the whole discussion was all about.
That's going to be a factor of five.
And since the potential equals electric field times D -- but I
keep D at seven millimeters, I'm not going to change it --
if E goes down by a factor kappa, then clearly,
the potential will also go down by a factor kappa.
So now you're going to see the second part, and that is I'm
going -- as it is now,
I'm going to plunge in this glass, the seven millimeters
thick, I put it in there, you expect to see no change on
the amp meter, but you expect the voltage
difference over the plates to go down by a factor of five,
so you will see that -- that the propeller volt meter will
have a smaller deflection. You ready for this?
There we go. Now you have a smaller
potential difference, but there was no current
flowing through the plates or from the plates.
When I take it out again, the potential difference comes
back to the ten thousand volts. So that's demonstration number
two. Now we go to number three.
But before we go to number three, I want to ask myself the
question, what actually happened with the capacitance when I
bring the dielectric between those plates?
Well, the capacitance is defined as the free charge
divided by the potential difference over the plates.
That's the definition of capacitance.
And since, in this experiment, as you have seen,
the voltage went down by a factor of kappa,
the capacitance goes up by a factor of kappa,
because Q free was not changing.
And so, since the capacitance, as we derived this last time
for plane -- plate capacitors, I still remember,
it was the area times epsilon zero divided by the separation D
-- since we now know that with the
glass in place, that's -- the capacitance is
higher by a factor of kappa, this is now the amendment we
have to make. To calculate capacitance,
we simply have to multiply, now, by the dielectric constant
of the thin layer that separates the two conductors,
the layer that has thickness D that is in between the two
plates. In our case,
I brought in glass. I could write down a few
equations now that you can always hold on
to in your life, and you can also use them in
the two demonstrations that follow.
And one is that E -- which is always the net E,
when I write E it's always the net one -- equals sigma free
divided by epsilon zero times kappa.
There comes that kappa that we discussed today.
Let's call that equation number one.
The second one is that the potential
difference over the plates is always the electric field
between the plates times D, because the integral of E dot D
L, over a certain pass, is the potential difference.
That's not going to change. And then the third one that may
come in handy is the one that I have already there,
C equals Q free divided by the potential difference,
which, in terms of the plate area, is A times epsilon zero,
divided by D, times kappa.
Let's call this equation number three.
Now comes my third experiment. In the third demonstration,
I am not going to disconnect my power supply.
So now, in number three, I start out with fifteen
hundred volts, just like we did with number
one, but the power supply will stay in there throughout,
never take it off. We start with D equals one
millimeter, just like we did in experiment one.
No glass. I'm going to charge it up,
just like I did with number one, and, of course,
I will see that the amp meter will show this charge.
[clk]. See a surge of current.
Now I'm going to increase D to seven
millimeters. Now something very different
will happen from what we saw in the first experiment.
The reason is that the potential difference is going to
be fixed, because the power supply is not disconnected,
the power supply stays in place.
Look, now, at equation number two.
If that V cannot change, and if I increase D by a factor
of seven, now the electric field must come down by a factor of
seven. And so now the electric field
will come down by that factor of seven, because I go from one
millimeter to seven millimeters. So now the electric field
changes, because D goes up. In case you were interested in
the capacitance, the capacitance will also go
down by a factor of seven, because, if you look at this
equation, kappa is one. If I make D go up by a factor
of seven, C goes down by a factor of seven.
Just look at this, simple as that.
So C must also go down by a factor of seven.
Nothing to do with dielectric. Nothing.
And so Q free must now also go down by a factor of seven,
because if the potential difference doesn't change,
but if Q free goes down a factor of seven -- or by -- if C
goes down by a factor of seven, Q free must go down by a factor
of seven.
This goes down by a factor of seven, this doesn't change.
So the free charge goes down by a factor of seven.
And what does that mean? That means charge will flow
from the plates, away from the plates,
and so my amp meter will now -- will tell me that charge is
flowing from the plates, and so that handle -- that hand
there will go [wssshhht] to the left.
And so, as I open up, depending upon how fast I can
do that, charge will flow from the
plates, in the other direction, it -- the charge will flow off
the plates, and that current meter will show you,
every time that I open it a little bit [klk],
it will go to this direction. So let's do that first,
no dielectric involved, simply keeping the power supply
connected. So I have to go back,
first, to one millimeter, which is what I'm doing now,
I have here this thin sheet to make sure that I don't short
them out, it's about one millimeter, and I am going to
now connect the fifteen hundred volts, and keep it on,
and as I charge it, you will see the current meter
surge to the right, right?
That always means we charge the plates.
So there we go, did you see it?
I didn't see it because I had to concentrate.
Did it go like this? Good.
So now it's charged. We don't take this connection
off, it's connected with the power supply all the time.
And now I'm going to open up, and as I'm going to open up,
the potential remains the same, so this volt meter doesn't give
a damn, it will stay exactly where it is, because fifteen
hundred volts remains fifteen hundred volts,
but now, we go -- as we open up, we're going to take charge
off the plates and so this, I expect to go to the left.
Every time that I give it a little jerk, I do it now,
it went to the left. I go it now,
again, I go to two millimeters, go to three millimeters,
go to four millimeters, make it five millimeters,
five millimeters, six millimeters,
and I finally end up at seven millimeters.
And every time that I made it larger, you saw the hand go to
the left. Every time I took some charge
off. So that is demonstration number
three. Why did I go to seven
millimeters? You've guessed it!
Now I want to plunge in the dielectric.
So my experiment number four, I start with fifteen hundred
volts, I start with D equals seven millimeters,
and I'm not going to change that.
There's no dielectric in place, but now, I put a dielectric in.
So kappa goes in. What now is going to happen?
Well, for sure, V is unchanged,
because it's connected with the power supply,
so that cannot change. What happens with Q free?
Look at this equation. W hen I put in the dielectric,
I know that the capacitance goes up by a factor of kappa.
C will go up by a factor of kappa.
If C goes up with a factor of kappa, and if V is not changing,
then Q free must go up by a factor of kappa.
Follows immediately from equation three.
So this must go up by a factor of kappa.
What does that mean? That the charge will flow
through the plates. I increase the charge on the
plates, and so my amp meter will tell me that.
And so my amp meter will say, "Aha!
I have to put charge on the plates," and so my amp meter
will now do this. And that's what I want to show
you. The remarkable thing,
now, is that the electric field E, the net electric field E,
will not change. And you may say,
"But you put in a dielectric!" Sure, I put in a dielectric.
But I kept the potential difference constant,
and I kept the D constant.
And since V is always E times D, if I keep this at fifteen
hundred volts, and I keep the seven millimeter
seven millimeters, then the net electric field
cannot change, it's exactly what it was
before. That is the reason why Q free
has to change, think about that.
Because you do introduce -- induce charges on the
dielectric, and you have to compensate for that to keep the
E field constant, and the only way that nature
can com- compensate for that is to
increase the charge on the plates, the free charge.
And so that's what I want to show you now,
which is the last part. So I'm going now to put in the
dielectric, and what you will see, then, is that current will
flow onto the plates, so the propeller will do
nothing, will sit there, and you will see this one go
klunk when I bring in the glass. And then it goes back,
of course. There's only a little charge
that comes off, and
then it will go back. So as I plunge it in,
you will see charge flowing onto the plates.
There we go, you're ready for it?
Three, two, one, zero.
And you saw a charge flowing onto the plates.
When I remove the glass, of course, then the charge goes
off the plates again, and you see that now.
I've shown you four demonstrations.
None of this is intuitive. Not for you,
and not for me. Whenever I do these things,
I have to very carefully sit down and think,
what actually is changing and what is not changing?
I have not gut feeling for that.
There is not something in me that says, "Oh yes,
of course that's going to happen.
Not at all. And I don't expect that from
you, either. Then only advice I have for
you, when you're dealing with these cases whereby dielectric
goes in, dielectric goes in, plates separate,
plates not separate, power supply connected,
power supply not connected, approach it in a very
cold-blooded way, a real classical MIT way,
very cold-blooded. Think about what is not
changing, and then pick it up from there, and see what the
consequences would be. How can I build a very large
capacitor, one that has a very large capacitance?
Well, capacitance, C, is the area,
times epsilon zero, divided by D,
times kappa, which your book calls K.
So give K -- make K large, make A large,
and make D as small as you possibly can.
Ah, but you have a limit for D. If you make D too small,
you may get sparks between the conductors, because you may
exceed the electric field, the breakdown electric field.
So you must always stay below that breakdown field,
which in A, it would be three million volts per meter.
If you want a very large kappa, you would say,
"Well, why don't you make the layer
water, in between, that has a kappa of eighty."
Ah, the problem is that water has a very low breakdown
electric field, so you don't want water.
If you take polyethylene -- I'll just call it poly here,
to se- as abbreviation -- polyethylene has a breakdown
electric field of eighteen million volts per meter,
and it has a kappa, I believe of three.
Many capacitors are made whereby the layer in between is
polyethylene, although mica would be really
superior. Be that as it may,
I want to evaluate, now, with you,
two capacitors, which each have the same
capacitance of one hundred microfarads.
But one of them, the manufacturer says,
that you could put a maximum potential difference of four
thousand volts over it, that's this baby.
And the other, I got to Radio Shack,
and it says you cannot exceed the potential difference,
not more than forty volts. Well, if I have polyethylene in
between the layers of the conductors, then I can calculate
what the thickness D should be before I get breakdown.
That's very easy, because V equals E D,
and so I put in here, eighteen million volts per
meter, and I go to four thousand volts,
and then I see what I unintelligible D.
And it turns out that the minimum value for D,
you cannot go any thinner, is then two hundred and twenty
microns, and so for this one, it is only two point two
microns. You can make it much thinner,
because the potential difference is hundred times
lower. So you can make the layer a
hundred times thinner before you get electric breakdown.
I want the two capacitors to have the same capacitance.
That means, since they have the same kappa,
and they have the same epsilon zero, it means that A over D has
to be the same for both capacitors.
So A divided by D, for this one,
must be the same as A divided by D for that one.
But if D here is a hundred times larger than this one,
then this A must also be hundred times larger,
because A over D is constant. So if A here is hundred,
then A is here one. But now, think about it.
What determines the volume of a capacitor?
That's really the area of the plates, times the thickness.
And if I ignore, for now, the thickness of the
conducting plates, then the volume of a capacitor
clearly is the product between the area and the thickness,
and so it tells me, then, that this capacitor,
which has a hundred times larger area, is hundred times
thicker, will have a ten thousand times larger volume
than this capacitor. And this baby is four thousand
volts, hundred microfarads, it has a length of about thirty
centimeters, ten centimeters like this, twenty centimeters
high, that is about ten thousand cubic centimeters.
Ten thousand cubic centimeters. You go to Radio Shack,
and you buy yourself a forty-volt capacitor,
hundred microfarads, which will be ten thousand
times smaller in volume. It will be only one cubic
centimeter. And if I had one of them behind
my ear, you wouldn't even notice
that, would you? Could you tell me what it says
here? One hundred micro microfarad.
How many volts? Forty.
Forty volts. That's small.
Compared to this one, which can handle four thousand
volts. But the capacitance is the
same. So you see now,
the connection with area and with thickness,
by no means trivial. All this has been very rough on
you. I realize that.
It takes time to digest it, that you have to go over your
notes. And therefore,
for the remaining time -- we have quite some time left -- I
will try to entertain you with something which is a little bit
easier. A little nicer to digest.
Professor Musschenbroek in the Netherlands, invented -- yes,
you can say he invented the -- the capacitor.
It was an accidental discovery. He called them a Leyden jar,
because he worked in Leyden. And a Leyden jar is the
following. This is a glass bottle,
so all this is glass, that's an insulator,
and he has outside the insulator, he has two conducting
plates, so that's a beaker outside, and there's a beaker
inside, conducting. That's a capacitor.
Although he didn't call it a capacitor.
And so he charged these up, and so you can have plus charge
here, and minus Q on the inside, and he did experiments with
that. The, um, the energy stored in a
capacitor -- we discussed that last time -- equals one-half
times the free charge times the potential
difference, if you prefer one-half C V squared,
that's the same thing, I have no problem with that,
because the C is Q free divided by V, so it's the same thing.
What I'm going to do, I'm going to put a certain
potential difference over a Leyden jar, I will show you the
Leyden jar that we have -- you'll see there -- and once I
have put in -- put on some potential difference,
put on some charge on the outer surface and on the inner
surface -- you can see the outer surface there,
the inner one is harder to see, but I will show that later to
you. So here you see the glass,
and here you see the outer conductor, and there's an inner
one, too, which you can't see very well.
Once I have done that, I will disassemble it.
So I first charge it up so there is energy in there,
this much energy. And then I will take the glass
out, I will put the, um, the outside conductor here
, the inside conductor here,
I will discharge them completely.
I will hold them in my hands, I will touch them with my face,
I will lick them, I will do anything to get all
the charge off. And then I will reassemble
them. Well, if I get all the charge
off, all this Q free [wssshhh] goes away, there's no longer any
potential difference. When I reassemble that baby,
then, clearly, there couldn't be any energy
left. And the best way to demonstrate
that, then, to you, is, to take these prongs,
which I have here, conducting prongs,
and see whether I can still draw a spark by connecting the
inner part with the outer part. And you would not expect to see
anything. So it is something that is not
going to be too exciting. But let's do it anyhow.
So here is this Leyden jar, and I'm turning the wind
unintelligible to charge it up. I'm going to remove this
connection, remove this connection,
take this out, take this out,
come on -- believe me, no charge on it any more.
This one. It's all gone.
Believe me. There we go.
And now let's see what happens when I short out the outer
conductor with the inner conductor.
Watch it. That is amazing.
There shouldn't be any energy on that capacitor.
Nothing. And I saw a huge spark,
not even a small one. When I saw this first,
and I'm not joking, I was totally baffled.
And I was thinking about it, and I couldn't sleep all night.
I couldn't think of any reasonable explanation.
And so my charter for you is, to also have a few sleepless
nights, and to try to come up, why this is happening.
How is it possible that I first bring charge on these two
plates, disassemble them, totally take all the charge
off, and nevertheless, when I reassembled them again,
there is a huge potential difference between the two
plates, otherwise, you wouldn't have seen the
spark. So give that some thought,
and later in the course, I will make an attempt to
explain this. At least, that's the
explanation that I came up with, it may not be the best one,
but it's the only one that I could come up with.
In the remaining eight minutes, I want to tell you the last
secret, which I owe you, of the van der Graf.
And that has to do with the potential that we can achieve.
Remember the large van der Graf?
We could get it up to about three hundred thousand volts.
How do we charge a conducting sphere?
Well, let's start off with a -- with this hollow sphere,
which is what the con- the van der Graf is -- and suppose I
have here a voltage supply, with a few kilovolts.
I can buy that. And I have a sphere,
and I touch with this sphere, which an insulating rod,
I touch the output of the kilo- the few kilovolt
supply, and I bring this -- so there's positive charge on here,
say -- and I bring it close to the van der Graf,
there will be an electric field between this charged object and
the van der Graf, and the closer I get,
the stronger that electric field will be.
And when I touch the outer shell, then the charge will flow
in the van der Graf. I go back to my power supply,
I touch again the few thousand volts,
and I keep spooning charge on the van der Graf.
Will I be able to get the van der Graf up to three hundred
thousand volts? No way, because there comes a
time that the potential of this object -- which comes from my
power supply -- is the same electric potential as the van
der Graf, and then you can no longer exchange charge.
What it comes down to is that when you come with this
conductor and you approach the van der Graf,
there will be no longer any electric fields between the two.
So there will be no longer any potential difference.
So you can't transfer any more charge.
So you run very quickly into a situation which will freeze.
You cannot get it above a few thousand volts.
So now what do you do? And here comes the breakthrough
by Professor van der Graf from MIT, who now said,
"Ah. I don't have to bring the
charge on this way, but I can bring the charge in
this way." So now you go to your power supply,
a few thousand volt, and you bring it inside this
sphere, where there was no electric
field to start with. When you charge the outside,
there's going to be an electric field from this object,
and there's going to be an electric field from this object,
the net result will be zero in between.
There was no electric field inside.
If I now bring the positively charged sphere there,
I'm going to get E field lines like this, problem two one,
and so now there is a potential difference between this object
and the sphere. What I have done by moving it
from here to the inside, I have done positive work
without having realized it, and therefore,
I have brought this potential higher than the sphere.
Now I touch the inside of the van der Graf,
and now the charge will run on the outer shell.
And I can keep doing that. Inside, touch.
Inside, touch.
And every time I come in here, there is no electric field in
there. So I can do that until I'm
green in the face. Well, there comes a time that I
can no longer increase the potential of the van der Graf,
and that is when the van der Graf goes into electric
breakdown. When I reach my three hundred
thousand volts, it's all over.
I can try to bring the potential up,
but it's going to lose charge to the air.
And so that is the -- ultimately the limit of the
potential of the van der Graf. So how does the van der Graf
work? Uh, we have a belt,
which is run by a motor -- here is the van der Graf -- and right
here, through corona discharge, we put charge
on the belt. They're very sharp points,
and we get a corona discharge at a relatively low potential
difference, it goes on the belt, the belt goes here,
and right here, there are two sharp points,
which through corona discharge take the charge off.
On the inside, that's the key.
And then it goes through the dome, and then it charges up,
up to the point that you begin to hear the sparks,
and that you have breakdown. And I can demonstrate that to
you. I built my own van der Graf.
And the van der Graf that I built to you is this paint can.
I'm going to charge that paint can by touching it repeatedly
with a conductor, and the conductor has a -- is
going to be -- yes, I'm going to touch the
conductor with a few thousand volt power supply every time --
this is the power supply, turning it on now -- and you're
going to see the potential of the
van der Graf there. Uh, that is a very crude
measure for the potential on the van der Graf,
but very crudely, when it reads one,
I have about ten thousand volts -- this is the probe that I'm
using for that -- two, it's twenty thousand volts.
My power supply is only a few thousand volts.
But that's not very good. Well, I will first start
charging it on the outside to demonstrate to you that I very
quickly run into the wall that I just described.
That if they have the same potential, then I can no longer
transfer a charge. But then I'm going to change my
tactics and then I go inside. And then you will see that it
will go up further. So let's first see what happens
if I now bring charge on the outside.
There it goes. It's about a thousand volts,
about two thousand volts, two thousand volts,
keep an eye on it, two thousand volts,
it's heading for three thousand volts, three thousand volts,
three thousand volts, three thousand volts,
not getting anywhere,
I'm beginning to reach the saturation, maybe three and a
half thousand volts, three and half,
it's slowly going to four, let's see whether we can get it
much higher than four, I don't think we can.
So this is the end of the story before Professor van der Graf.
But then came Professor van der Graf.
And he said, "Look, man, you've got to go
inside. Now watch it.
Now I have to concentrate on this scooping,
so I would like you to tell me when we reach five thousand,
you just scream. Oh, man, we already passed the
five thousand, you dummies!
Ten thousand, scream when you see ten
thousand. [crowd roars].
Scream when you see fifteen thousand.
[crowd roars]. Very good, keep an eye on it,
tell me when you see twenty thousand.
[noise] I don't hear anything! [crowd roars] Now I want you
tell me every one thousand, because I think we're going to
run into the wall very quickly. Twenty one?
I want to hear twenty two. [crowd roars].
Already at twenty three. So I expect that very s- very
quickly now -- [crowd roars] -- the can will go into discharge,
you won't see that, but you get corona discharge,
and then, no matter how hard I work, I will not be able to
bring the potential up. But let's keep going.
Are we already at twenty five hundred?
Twenty five thousand, sorry, twenty five thousand?
Twenty five thousand volts. Twenty five six.
Twenty seven. Twenty seven.
Twenty eight. Twenty eight.
It looks like we are beginning to get into the corona
discharge. Twenty eight!
Boy, twenty eight! That's a record.
Twenty-eight, keep an eye on it.
Twenty nine? Twenty nine?
Whew. You realize I'm doing all this
work. Well, I get paid for it,
I -- I think I've reached the limit.
I've reached my own limit and I've reached the limit of the
charging. Now, we have thirty thousand
volts, and we started off with only a few thousand volts.
Originally, it wasn't a very dangerous object.
But now, thirty thousand volts -- shall I?
OK, see you next week.