Well, we have a current going through a wire,
like so. And we look at the magnetic
field in the vicinity of this wire , then we know from
experiment that if you put pieces of magnetite around
the wire that they line up in a circle.
Put around like this. If that circle has a radius R,
then the magnetic fields, that's an experimental fact,
is proportional with the current I and is inversely
proportional with the radius of that circle.
By convention, the direction of the magnetic
fields given by the right-hand corkscrew,
rotate this way, the current goes up.
You've seen before, with electric charges,
when you have a wire which is uniformly distributed say with
positive charge, you've also seen that electric
fields in the vicinity of that straight wire falls off as one
over R, whereas the direction is different than the magnetic
field but it also falls off as one over R, and the reason is
that electric monopoles, individual charges,
the electric fields fall off as one over R squared.
And so when you integrate that out over a straight wire you get
the one over R field. So by analogy,
it would be very plausible that if you took magnetic monopoles
that the magnetic field would also fall off as one over R
squared, but magnetic monopoles as far as we know don't exist.
In principle they could exist, but we've never seen one,
and if any one of you ever find one, that would certainly be a
Nobel Prize. It's by no means impossible.
And so the simple fact that the magnetic field around a current
wire falls off as one over R, sort of suggests that if you
carve this wire up in little elements DL, that each one of
those elements contributes to the magnetic fields
in inverse R-squared law, and by integrating out over the
whole wire you'd then get the one over R fields.
And this behind the idea of the formalism by Biot-Savart,
who introduced the idea that if you have a little current
element DL, and the current is in this direction,
and you want to know what the magnetic field is,
This is small contribution DB to
that little current element, and the distance is R,
and the unit vector from the element DL to the point where
you want to know the magnetic field is R roof.
Then the idea here is DB, it's a little bit of curr-
little bit of magnetic fields. In this case it would be in the
blackboards because of the right-hand corkscrew rule.
The current is in this direction, so these little
elements would contribute to magnetic fields in
this direction perpendicular to the blackboard.
It's some constant, proportional to the current no
doubt, and then it's proportional to the lengths of
that little element DL, if it's longer than the
magnetic field is larger, and in order to get the
direction right perpendicular to the blackboard you take the
cross product with the unit vector R.
The unit vector R has length one so you only do that in order
to get the direction right. And this, and that inversely
proportional to R squared. That's of course key.
And this is, the formalism by Biot-Savart
and you can do experiments and measure the magnetic field in
the vicinity of wires and this formalism works,
so you then calculate the individual contributions of all
these little elements DL and then you do an integration and
this formalism works. You can then also measure what
C is, in SI units. C is ten to the minus seven.
But we write for C something quite peculiar.
We write for C mu zero divided by four pi, and we call this mu
zero the permeability of free space.
You've seen earlier with Coulomb's law that this constant
nine times ten to the ninth, we call that one over four pi
epsilon zero. What is in the name?
And so here we call this mu zero divided by four pi.
So now you can apply Biot-Savart's Law and you can go
to a straight wire and you have a current I, and suppose you
want to know what the magnetic field at that location
is at a distant capital R, and so what you now have to do,
if you carve this up, in an infinite number of small
elements DL, and this distance is R, and the unit vector is
then like so, and you calculate the small
amount of magnetic field due to this little element and you
integrate this over the whole wire.
It's mathematics. You've done it.
You've done it before, where we had uniformly electric
charge on the wire. So I'm not going to do this
again for you. It's a very straightforward
piece of mathematics. The magnetic field by the way,
in this case, would come out of the
blackboards. Thus it's the right-hand
corkscrew rule. And what you find when you do
this, we will find that B equals mu zero times I,
divided by two pi R, this being R,
and so you indeed see that the inverse one over R comes out.
And so if you, for instance,
take a radius of oh point one meters, ten centimeters,
and you have a current through the wire of about a hundred
amperes, then you would end up with a B field.
You use this equation two times ten to the minus four Tesla.
That is about two Gauss. Hundred amperes.
Ten centimeter distance is only two Gauss.
Think about it. The Earth's magnetic field is
half a Gauss. So if you go one meter away
from the wire, so we have a magnetic field
which is ten times lower [inaudible] is one over R,
than the magnetic field of the Earth already dominates
substantially. So you need very high current,
actually, when you do these experiments.
It's nice to see that out of Biot-Savart's formalism the one
over R pops out, but of course you must realize
that Biot-Savart knew that the magnetic field falls off as one
over R. That was an experimental fact.
So the fact that it falls out is logical, because it was
cooked into that formalism. If you think about it,
it all goes back to Newton. Newton was the one who first
suggested that the gravitational field falls off as one
over R squared. And then later a logical
extension was that the electric fields would fall off as one
over R squared and out of that came the idea that the fields of
magnetic monopole, if they only existed,
would fall off as one over R squared, and that's all behind
this and so the person who really deserves most of the
credit for all this in my book is Newton.
Using Biot-Savart, we can calculate now easily the
magnetic fields at the center of a current loop.
Let this be a wire circle and let the current go in this
direction, and I would ask you what is the magnetic field right
at the center. Well, the magnetic field right
at the center of course is pointing upwards.
Each little element along the line here, DL,
each little element will contribute a little bit
magnetic field at that point right in this direction.
And if this radius is R, with Biot-Savart now,
we can calculate quite easily the total field that you would
get at this location, because that total field is
then the integral of DB factorially over the entire wire
so the entire loop... So if you go there,
so you would get your mu zero, divided by four pi,
you get your current and you get your one over R squared,
and now we have to do an integral over that DL cross R.
Well, R is of course always perpendicular to DL.
Any element DL that you choose, the unit vector R is exactly
perpendicular to the element DL, because that's characteristic
of a circle.
And so the sine of the angle between DL and R is one,
and so all we have to do is do an integral over DL,
which is the integral of the circle, which is the
circumference of the circle, and that is two pi R.
And so now you find, lose a pi, you lose an R,
so we find mu zero times I divided by two R.
Just to show you an example, how in this case how easy it is
to use Biot-Savart and calculate the magnetic field right at the
center. If you were asked what the
magnetic field was here or there, that would be also
relatively easy. You've done that.
I've given you a problem earlier where we had point
charges uniformly distributed on a wire and I asked you what the
electric field was here. So that can also be done now
with magnetic fields. If I ever asked you what the
magnetic fields would be here, that of course is an
impossibility to do that with Biot-Savart, practically an
impossibility. I wouldn't know how to do that.
But in principle it could be done and certainly with a
computer you can do it. So we can go to our same
situation, we can take a hundred amperes for I and you can take R
oh point one meters and then the B field, the strength of the B
field right at the center of this loop that I found is then
six times ten to the minus four Tesla.
And that would be six Gauss. It's clear that if you want to
put in some field lines, magnetic field lines,
as a result of this current going around in a circle,
that the- through the center there would be a field line like
so. If you're very close to the
wire here, which goes into the blackboards, I want you to see
this three-dimensionally, then the magnetic fields would
go like this, clockwise.
Here the current comes to you, so would be counterclockwise.
If the magnetic field line is here like so,
and here it is curled up, then clearly I expect them to
be here, sort of like so, and like so,
and like so. This is the kind of magnetic
field line configuration that I would expect,
then, in the vicinity of such a current loop.
And I want to show this to you in a little bit more detail.
I have here a transparency, and you see there on the right
side, current goes into the paper and here it comes out of
the paper. There is a circular loop.
And you see here the field line configuration.
It's not too different from what I have on the blackboard
there. Very close to the wires,
of course, you get circles because the one over R dominates
there. It's so close to the wire that
the one over R relationship makes it come out like circles
and here too, but then if you're farther away
you get configurations like I have there.
When you're very far away from a current loop,
the magnetic field configuration is very similar to
that of an electric dipole. I can show you that in the
following way. Let's first look at the
electric dipole that you see up there.
This is a positive charge, this is a negative charge.
Don't look anywhere near the charges.
Don't look in between the charges.
Look far away. Here you see electric field
lines and you see them here. Now look at your current loop
here. The current is going into the
paper here, coming out of the paper.
There is a loop. And look, you see the same
configuration, field lines,
field lines. This goes like so.
This one goes like so. Here, the electric field lines
coming in, magnetic field lines are coming in.
Electric field lines are going out.
Magnetic field lines are going out.
They look very similar. Gauss's Law tells me that the
closed surface integral of the electric flux is the charge
inside the box divided by epsilon zero,
and so if you have a closed surface
here, it looks like a line but I meant it to be a surface,
then that closed surface integral of the electric flux is
now zero because there's a charge inside the box.
No matter where in the magnetic field you make a closed surface
there is never any magnetic flux going through that surface.
Never, unless you come into twenty-six one hundred and show
me a magnetic monopole. Only then will there be
magnetic flux coming out of a closed
surface, if we put the magnetic monopole inside.
And this, now, brings us to the second of four
Maxwell's equations, the first one being Gauss's
Law, the second one is that the closed surface- closed surface
integral of B dot DA is always zero, unless you come with a
magnetic monopole.
So we now have two of Maxwell's four equations in
place. Historic day.
I want to show you the magnetic field in the vicinity of a wire
like this. I have to use a few hundred
amperes through that wire. I told you why,
because magnetic field falls off
quite rapidly, and I do that with iron file
which I will sprinkle around the wire and these magnetites will
orient themselves in that magnetic field and then I will
try to also make you see this field configuration by having
one wire going into the paper and one coming out of the paper.
So let's first look at the single current wire.
It's coming towards you and it goes
into of course. It is a wire that goes like
this and the reason why you see it here is of course we have to
get the current in somehow. But it really is a wire like
this, and this platform I'm going to show you.
I'm going to put some iron file around this.
We've got magnetites and when they see magnetic fields,
they will try to orient themselves in the direction of
the magnetic fields, so you're going to see these
circles. But appreciate the fact that
you need huge currents for this. Hundreds of amperes we do.
That's why we have car battery here, remember,
that was capable of delivering many hundreds of amperes.
So I close the current now. I tap, you can see,
these circular configurations. I hope you can see that.
Looks like circles. I want to do the same now.
This is a little bit more exciting, perhaps,
with one wire going into the plane and the other
one coming out. Even though it's not a circle,
the idea is that you get a field configuration very much
like you see there. Boy, it's already hot.
These cables are already hot. They don't like the two hundred
amperes. OK, let's do it this way.
So you're going to see a field configuration similar to what I
have on the blackboard, similar to what you have here.
Oh, not, not this one. Similar to this.
Yeah, actually, it was also the one that I
have, but this is nicer way to look at it.
So we have one wire going in and one wire coming out of the
plane. All right, so let's first put
some iron file on. All right,
and now a few hundred amperes through it, tap it,
and you can see close to the wire that goes in here and
here, you see circles. Those are the one over R
relationship. They dominate the magnetic
fields, but look in between here.
With a little bit of imagination you can see these
field lines going like this, just like I have today on the
blackboard. All right.
We don't need it anymore, and we don't need that anymore.
So this gives you a little bit of insight into the magnetic
field configurations that we have about these wires when we
run a current through. I will return to magnetic
fields next lecture and we will expand on it.
We will learn techniques to calculate magnetic fields in a
way which is highly superior to Biot-Savart's law.
In fact, that wil get us to the third Maxwell equation,
almost. But I now want you to relax a
little bit because this may already have been a little rough
on you, so now I want to discuss something entirely different,
something very practical, and it has to do with the
transport of electrical energy. So we have a power station
somewhere at location A, and they deliver electric
power, electric energy to Boston.
Here is Boston, location B.
And A is the location of the power station.
This could be a thousand miles, the distance.
There's a cable going from the power station to us,
and the potential here in this line, say, is B,
a field B. Here's a cable for the return
current, so the current goes like this, and the return
current is in this direction, and, and here you use this
energy. You hook up your computer,
you hook up your hairdryer, your heaters,
electric toothbrush and what have you, your TV station,
everything. And so you are the consumer
here. You take the energy that is
provided by this power station. I will call the potential of
this line zero and so this is V of A higher than this line here.
Well, according to Ohm's Law, VA minus VB is the current
times capital R which is now not radius, but that is resistance
in the wire, this wire. This cable, however thick it
may be, has a finite resistance. And so VB, that is the
potential that we receive in Boston,
equals VA minus IR. So if there's no current going
through the wire, no one is using any electric
energy, then VB is the same as VA.
Now I want to know if we consume energy,
so we're dealing now with power, so the power that we take
off at Boston is I times V of B. That's the number of joules per
second that we are consuming. So that then equals VA times I
minus I square R. That's fine.
What is this? This is the energy per second
that we are consuming. What is this?
This is the energy per second that the power station is
delivering to us. What is this?
That's lost energy. It's the I square R that is the
heat produced in this cable that goes into the universe.
It's gone. So the economy demands that we
try to make this as small as possible.
Uh, this is the power that is available but you get a loss of
power in terms of heats, the minus sign here,
so you get less in Boston. And so how can you make this I
square R low? Well, what is the resistance of
a wire that is rho, which is the resistivity,
times the length of the wire divided by the cross-section of
the wire. So we have several options.
You could make A very large, a very thick copper wire,
and that's expensive. You could also make the wires
out of gold, which has a lower resistivity than copper.
That's also expensive. People are thinking of making
these transmission wires of superconducting material.
They have to cool them at very low temperatures.
That's outrageously expensive but that's a way you could get
the resistance down. Let's now look at the current.
What can we do with the current?
Suppose we consume a hundred megawatts.
Not an unreasonable number, so we are consuming a hundred
megawatts, and just for the sake of the argument,
suppose at VB the potential is hundred volts,
so V of B is a hundred volts. What now is the current?
Well, current times potential gives me power,
and so my current is now a million amperes.
Alternatively, suppose that the potential at B
in the wire is a hundred thousand volts,
a thousand times higher. Now the current is only a
thousand amperes, gives me the same power.
In both cases, am I consuming at a rate of a
hundred million joules per second.
But I square R, the heat loss on the way from
the power station to me, is a million times lower in
this case than in that case, because I is a thousand times
lower, and the heat loss goes with I squared,
and so now you understand why electricity, when it is
transported from one place to another, why this is done at the
highest voltage possible. When you get to Boston,
you've obviously got to do something about this enormous
potential, because if you were to deliver a
hundred thousand potential difference there,
then half the population in Boston would electrocute itself,
so now you've got to come down in voltage, which you do with
transformers. We will talk about that later
in the course. And so you bring it down to a
comfortable voltage, which is in the United States
about a hundred and ten volts. In Europe it's two twenty.
Now comes the question, how high can you make V of A.
The higher you could make it, the less loss there would be
along the way.
Well, you've got to stay away from the breakdown electric
field, which is the three million volts per meter.
If at the surface of these cables you get three million
volts per meter, you get corona discharge.
That's a big loss and you want to stay away from that,
and so typical cables have about a radius R.
This is now, R is the radius of the cable of
about two centimeters that gives him a cross-sectional area I
think of about ten to the minus three
meters squared, that's correct.
And the potential V at A is roughly three hundred kilovolts,
and with that configuration you stay comfortably below the
electric field of three million volts per meter,
but you don't get the corona discharge.
If the length of that cable, L, if that were something like
thousand kilometers, not an unreasonable number,
thousand kilometer distance from if we get our electricity
from Niagara Falls to Boston, not an unreasonable number,
you can calculate now what the resistance of that cable would
be, because that resistance R equals rho times L divided by A.
If you take copper, that has a resistivity of two
times ten to the minus eight SI units.
We have a length of ten to the sixth meters of the cable and we
have a cross-sectional area of ten to the minus three square
meters so that thousand kilometer cable would only have
a resistance of twenty ohms. And to make the numbers a
little easy, if we have a current say of three hundred
amperes, then the power that the power
station produces, if you take the three hundred
kilovolts for now, that power would be the three
hundred kilovolts times the three hundred amperes and that
is about ninety megawatts. That's close to my hundred that
I had in mind earlier. So you can calculate now what
the loss is. The loss is I squared R.
You know R is twenty ohms, you know I, three hundred
amperes, and so you'll find now that you have about two megawatt
loss. That's not bad.
Two out of ninety. So we have about two percent
energy loss in transportation. You can also calculate now what
the difference is in potential between the power station and
Boston, VA minus VB is IR. You know that I is three
hundred amperes and you know that R is
twenty ohms, so VA minus VB is about six
kilovolts. In other words,
if the power station puts it on the line at three hundred
kilovolts, then you would get it here with only four six
kilovolts less. So it's not a very unreasonable
situation. I told you that these power
lines have to stay away from the three
million volts per meter electric field because then you
get corona discharge and when there is thunderstorms in the
area it can actually push up the electric field on the wire and
you can get corona discharge. I have seen that several times,
not only seen it at night, with my naked eyes that you see
the power lines glow, but I've also heard it,
because you can here this cracking noise of corona
discharge. It's very fascinating,
actually. I have a slide here
which shows that. So here you see a high voltage
power line, transmission line, and you clearly see the glowing
of the corona discharge. They compare that with what's
called a [inaudible] kite string.
Kite strings at night when you fly them near thunderstorms can
also produce corona discharge and light.
That's why the call [inaudible] kite string candles.
Benjamin Franklin did quite a bit of
experiments at night with, uh, kites near thunderstorms.
Dangerous business by the way. So you see that these
high-voltage power lines can go into a corona discharge.
I now want to revisit our Leyden jar.
We had a truly absurd situation whereby we did an experiment
with a jar which you still see here,
and I will redo the demonstration,
but it's crying for an explanation because it looked
like there was something wrong with physics,
and I want to refresh your memory of what we have seen
before and what this Leyden jar is all about.
The Leyden jar is nothing but a capacitor, which is a dielectric
between two conductors in the form,
in the shape of a- of a bottle, the jar.
And so the inner portion, which is glass,
say has this shape. You see it there,
you're going to see it shortly there, and then we have
conductors, the conducting beaker around it here,
and we have a conducting beaker on the inside.
And we charge these up with the Wimshurst.
This is the Wimshurst machine. And when we do that,
we get free charge on the conductor, and so you get sigma
free right here and you get it here, opposite signs of course,
and you get sigma induced on the dielectric.
Why do you get it? Because the dielectric sees an
external field due to this sigma free, so it begins to polarize,
and so you get here induced charge and you get there induced
charge. If this side is positive,
then the induced charge here will be negative,
and vice-versa. We have here a metal hook,
so that we can lift out the inner portion.
And so what we did, we charged it up with the
Wimshurst, and then there is a certain amount of energy.
The electrostatic potential energy of this configuration is
one-half Q free times the potential difference,
and the Q free is the charge which is on the outer conductor.
And what I then did, I disassembled it very
carefully after we had charged it up, took the inner portion
out, took the glass out, the outer portion,
and I took all the free charge off.
I touched the conductors and discharged them,
so there is no Q free left. It's gone.
The moment that that is gone, the induced charge must also go
away because the induced charge is only there because of the
free charge. Remember the induced charge
density is one minus one over kappa times sigma free.
So if glass has a kappa of five,
then the induced surface charge density is about oh point eight
times the free surface charge density.
The moment that the free one goes, the induced one goes.
And then I assembled it again, and much to our surprise when I
short out the inner portion with the outer portion,
we saw a huge spark. That means there was energy
left and that is very puzzling. There can not be any energy
left unless there is something wrong with physics.
So I first want to show that again, to remind you of what you
have seen before, and then I will make a proposal
for a- for an explanation. Let me check my- my light
configuration. Ah, we're going to make it all
dark. I can charge it up while you're
seeing it, actually. Yeah.
Make it dark now. And now I will disassemble it.
I take the inner portion out. The glass is a good insulator,
so I don't mind touching that with my hand.
OK, and so now I take the inner conductor, touch it,
lick it, kiss it, take all the charge off.
There it is. Do the same with the outer
conductor, have it in my hand. For sure, there is no Q free
left on anymore, on that anymore.
I put the glass back in again, and then I put the inner
[inaudible], and now I'll make it a little darker
for you because I want you to see that when I short this out
that you're going to see a spark, so I'm going to turn the
lights down, so look very closely.
I'm going to tell you when I'm going to do it.
Three, two, one, I'm approaching it now,
zero. And there's a huge spark.
That means energy, and that's crazy.
There shouldn't be any. And so some of you must have
had sleepless nights not being able to explain this.
Some of uh- some of you actually wrote me e-mail.
Uh, you didn't have sleepless nights, I can tell that.
So what's going on? There's only one possibility
and that is there must be free charge on the glass.
How did it get there? Well, corona discharge.
That's the only way it can get there.
Keep in mind that the electric field in air,
in air, can be no larger than three times ten to the six volts
per meter. If it gets larger,
you get corona discharge. In glass, by the way,
it's a bit higher. It's ten to the seventh volts
per meter. I will now make some
calculations based on certain assumptions.
And those assumptions may not be exactly accurate because I
don't know the exact dimensions of this system,
but the exact dimensions don't matter.
What matters is the idea behind it, why it does such crazy
things. So first of all,
I will assume that this capacitor is just two parallel
plain plates. That's a simplifying situation
because it has the shape of a bottle.
I will assume that the Wimshurst produces about thirty
kilovolts. I know that it's approximately
right, but it may be twenty-five.
It may be thirty-five. I will assume that the air gap
between the outer conductor and the glass is one millimeter,
that both are one millimeter, and I would assume that the
glass thickness is three millimeters, and I take kappa
equal five for the glass. So that is the basis of my
calculations. So now I have here the
conductor on the outside. This is the glass,
and this is the conductor on the inside, so this is one
millimeter thick. This is the glass and this is
the conductor on the inside, so this is one millimeter
thick, this is three millimeters thick, and this is one
millimeter thick, and the potential difference
over the whole thing is going to be thirty kilovolts.
But I know that the potential difference is always E times D.
That holds for this gap the local E times this D,
the local E times this D, and the local E times that D.
I also know that E glass is the same as E in the air divided by
that kappa, which is five. And I know that the total
potential difference between here and here must be thirty
thousand volts. And so this allows me now to
calculate in a very straightforward way the electric
field here in the air, the electric field
here in the air, and the electric field in the
glass because I get simple equation with one unknown and
that's the following. I first go over this gap and so
I get E in the air times the distance D, which is one
millimeter. But of course later on I have
to go through this gap again so I'll multiply it by two now.
And now I have to add the electric
field in the glass, which is the same as air
divided by five, times its distance which is
three millimeters, and that must now be thirty
thousand, because that's the potential difference between
here and there. That's one equation with one
unknown. And I can calculate the
electric field in the air gap. And the electric field in the
air gap turns out to be eleven point five times ten to the six
volts per meter. It's here the same,
of course, eleven point five times ten to the sixth,
and here it is five times smaller, so I find here two
point three times ten to the sixth.
To show you that I did my homework correctly,
the potential difference here is now eleven point five
kilovolts. I put it here in kilovolts.
The potential difference here is now about seven kilovolts and
the potential difference here is
the same, is eleven point five kilovolts.
And if you add them up, you get thirty.
If you look at this, this can not be,
because a field of eleven point five times ten to the sixth is
way above the breakdown electric field.
And so what are you going to- what's going to happen,
you're going to get corona discharge from the conductor to
the glass, and so what you're doing is
you're spraying charge on the glass, and that's the key to the
solution of this bizarre behavior.
And so when you later disassemble it and you take the
free charge of the conductors. There is still this free charge
which you have sprayed on the glass.
I never remove that. It's very hard to remove
because the glass is an insulator.
It's very difficult to take charge off an insulator.
It's easy to take it off a conductor but I've never even
attempted that because I made you believe,
as I believed myself for years, that once you take the Q free
of the conductor that there can be no charge on the glass.
That is wrong, because there's corona
discharge. So now comes the question,
before we disassemble, what now is the configuration
of the electric fields and the potentials.
I'll make a drawing here and so now I draw again the conductor,
the glass, and the conductor. I resume now that after the
corona discharge this field here is three times ten to the sixth.
Maybe a little lower, but that's the maximum that it
can be, and so the field here in the air will be also three times
ten to the sixth volts per meter.
But since I know that the potential difference between
here and here must be thirty kilovolts, I can now immediately
conclude that the electric field here is now eight
times ten to the sixth volts per meter.
That is the only way that it adds up to thirty kilovolts,
because the three million volts per meter gives me here three
kilovolts. This here gives me three
kilovolts. So now I need a potential
difference here of twenty-four kilovolts, and that over three
millimeters requires this field. Look, this field is stronger
than that field, whereas earlier we made the
assumption that this field was five times lower than that
field. Yah, why is it now higher?
Because we have sprayed right here on this surface,
we have sprayed free charge. It's no longer the field that
is dictated by the external field and then the induced
charges. That's no longer the case.
It carries now itself free charge.
You now have all the tools, maybe not the courage,
to calculate how much free charge there is right
here on the glass to get these fields.
It's a very straightforward calculation.
And you will find that there is twelve times more free charge
here on the glass than there is here on the conductor,
twelve times more, and so if I disassemble and
remove the free charge on the conductors, I have almost done
nothing because most of the free charge is on the glass,
and I have not touched that. So now, if I reassemble,
I have almost all energy left. I have not lost much.
Lost some, but not lost much. And so what I should have
really have done, I should also have discharged
the inner glass. That's not easy,
but I will try that today. It's not easy because it's very
hard to take that charge off, but I will try that.
And then there shouldn't be much energy left if we
reassemble it again and try to get a spark out of it.
So I go through the same routine and
I'm going to charge it up now. OK, take this cable off,
take that cable off. I take it apart,
do everything that I did before the same way,
gone, whole charge gone, whatever there was.
Whole charge gone. Now, this is more difficult.
This is not enough if I do this.
I have to get in there. Ooh, I could actually feel it.
It's really, it's a great feeling.
I can feel a sort of corona discharge with my -- I have to
really get everything out and that's not easy.
It's not. In fact, when I rub in with my
shirt, I may even make it worse. I may be charging it up through
friction. But I'll do the best I can.
It's very indecent, what I'm doing.
OK. So I try to get a charge now,
something that we ignored completely before.
This is really where the energy was, and I'm trying to kill that
now, and so now I'm going to reassemble it,
and I'll go through the same routine [clink] [whoop].
Good thing I didn't break it. Put it back in again.
I'll make it dark, so that you can see where the-
perhaps there may be a little bit of spark,
if I didn't succeed to remove all the charges from the glass.
So I'm going to short it out again, three,
two, one, zero, and I saw a teeny-weeny little
spark. You may not even have seen it,
so we have to conclude now that the physics behind this lies in
the glass and physics works even when it sometimes surprises us.
See you Wednesday.