When 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
field is given by the right-hand corkscrew. Rotate this way,
the current goes up. You've seen before,
with electric charges, when you have a--
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 field falls 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 and 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, it's a 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 is that dB, it's a little bit
of curr- little bit of magnetic fields. In this case it would be in the blackboard
because of the right-hand corkscrew rule. The current is in this direction,
so this little element 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 length
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 p 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 blackboard. 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, look it goes with 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
quite easily the magnetic field 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,
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 blackboard, 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 not 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 26-100
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. These are magnetites
and when they see a magnetic field, 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 and 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 few 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 of the one over R relationship. They dominate the magnetic field,
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 this 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 and so now I want to discuss something
entirely different, something very practical and it has to do with the transport of
electric 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,
is V of B. And 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 heat, 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 electrity, 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 hundred and ten volts. In Europe it's two twenty. So 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 we 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 hear 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 Romas' kite string. Kite strings at night when you fly them
near thunderstorms can also produce corona discharge
and light. That's why they're called
Romas' 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-- 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,
with 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,
a 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] again, 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 light down,
so look very closely. I'll 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,
E 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 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 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 much. And so what I should 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 all-- 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 [sound of glass falling]
whoa! 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.
