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.
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, 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. 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.
