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. 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 802 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 call 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 give 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, an "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,
it takes both into account. So you can write down now,
that E equals the free electric field, 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 surface integral
of E dot dA 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 induced 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. 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 than the substances that I just discussed,
which themselves, do not have dipoles. Water is an example,
an 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 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 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 PS. 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 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. And 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 eye on the 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 now 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 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 dL,
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. [crggk] 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, if C goes down 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 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 I expect the amp meter to go to the left. Every time that I give it a little jerk,
I do it now, it went to the left. I do 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. When 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 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 no 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. The only advice I have for you,
when you're dealing with these cases whereby dielectric goes in, dielectric goes out,
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 air 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, 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 go 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? [laughter] One hundred... [student] micro [Lewin] microfarad How many volts? [student] Forty. [Lewin] 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 see it 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 Wimshurst 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. [spark] 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 VandeGraaff. And that has to do with the potential
that we can achieve. Remember the large VandeGraaff? 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 VandeGraaff 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,
with 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 VandeGraaff. There will be an electric field between
this charged object and the VandeGraaff 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 VandeGraaff. I go back to my power supply,
I touch again the few thousand volts, and I keep spooning charge
on the VandeGraaff. Will I be able to get the VandeGraaff 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 VandeGraaff 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 VandeGraaff, 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 would you do? And here comes the breakthrough
by Professor Van de Graaff 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, 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 VanderGraaff, and now the charge will run
on the outer shell. And I can keep doing that. Inside, touch. 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 VanderGraaff, and that is when the VanderGraaff
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 VanderGraaff. So how does the VanderGraaff work? Uh, we have a belt,
which is run by a motor-- here is the VanderGraaff--
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 VandeGraaff. And the VandeGraaff 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 VandeGraaff there. Uh, that is a very crude measure for the potential
on the VandeGraaff, 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, 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 Graaff. But then came
Professor van der Graaff! 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! [laughter] Ten thousand,
scream when you see ten thousand! [crowd roars] Scream when you see
fifteen thousand! Scream when you see
fifteen thousand! [crowd roars] Very good,
keep an eye on it, tell me when
you see twenty thousand. 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? [laughter] 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? [students] Yeah! [sound of a shock] [students scream] OK, see you next week.
