Today I'm going to work with you on a new
concept and that is the concept of what we call electric field. We spend the whole lecture on electric fields. If I have a -- a charge, I just choose Q,
capital Q and plus at a particular location and at another location I have another charge
little Q, I think of that as my test charge. And there is a separation between the two
which is R. The unit vector from capital Q to li- little
Q is this vector. And so now I know that the two charges if
they were positive -- let's suppose that little Q is positive, they would repel each other. Little Q is negative they would attract each
other. And let this force be F and last time we introduced
Coulomb's law that force equals little Q times capital Q times Coulomb's constant divided
by R squared in the direction of R roof. The two have the same sign. It's in this direction. If they have opposite sign it's in the other
direction. And now I introduce the idea of electric field
for which we write the symbol capital E. And capital E at that location P where I have
my test charge little Q at that location P is simply the force that a test charge experienced
divided by that test charge. So I eliminate the test charge. So I get something that looks quite similar
but it doesn't have the little Q in it anymore. And it is also a vector. And by convention, we choose the force such
that if this is a positive test charge then we say the E field is away from Q
if Q is positive, if Q is negative the force is in the other direction, and
therefore E is in the other direction. So we adopt the convention that the E field
is always in the direction that the force is on a positive test charge. What you have gained now is that you have
taken out the little Q. In other words the force here depends on little
Q. Electric field does not. The electric field is a representation for
what happens around the charge plus Q. This could be a very complicated charge configuration. An electric field tells you something about
that charge configuration. The unit for electric field you can see is
newtons divided by coulombs, in SI units-- and normally we won't even indicate
the unit, we just leave that as it is. Now we have graphical representations for
the electric field. Electric field is a vector. So you expect arrows and I have here an example
of a charge plus three. So by convention the arrows are pointing away
from the charge in the same direction that a positive test charge
would experience the force. And you notice that very close to the charge
the arrows are larger than farther away. That it that sort of represents,
is trying to represent, the inverse R square relationship. Of course it cannot be very quantitative. But the basic idea is this is of course spherically
symmetric, if this is a point charge. The basic idea is here you see the field vectors and the direction of the arrow tells you in
which direction the force would be. If it is a positive test charge. And the length of the vector give you an idea
of the magnitude. And here I have another charge minus one. Doesn't matter whether it is minus one coulomb
or minus microcoulomb. Just it's a relative representation. And you see now that the E field vectors are
reversed in direction. They're pointing towards the minus charge
by convention. And when you go further out they are smaller
and you have to go all the way to infinity of course for the field to become zero. Because the one over R square field falls
off and you have to be infinitely far away for you to not experience at least in principle
any effect from the charge. What do we do now when we have more than one
charge? Well, if we have several charges-- here we have Q one and here we have Q two
and here we have Q three and let's say here we have
Q of I, we have I charges. And now we want to know what is the electric
field at point P. So it's independent of the test charge that
I put here. You can think of it if you want to as the
the force per unit charge. You've divided out the charge. So now I can say what is the E field due to
Q one alone? Well, that would be, if Q one were positive, then this might be a representation for E one. If Q two were negative,
this might be a representation for E two, pointing to the negative charge. And if this one were negative, then I would
have here a contribution E three, and so on. And now we use the superposition principle
as we did last time with Coulomb's law, that the net electric field at point P as a vector
is E one, influence of charge Q one, plus the vector E two plus E three and so on
and if you have I charges it is the sum over all I charges
of the individual E vectors. Is it obvious that the superposition principle
works? No. Does it work? Yes. How do we know it works? Because it's consistent
with all our experimental results. So we take the superposition principle for
granted and that is acceptable. But it's not obvious. If you tell me what the electric field at
this point is which is the vectorial sum of the individual E field vectors then I can
always tell you what the force will be if I bring a charge at that location. I take any charge that I always would carry
in my pocket, I take it out of my pocket and I put it at that location. And the charge that I have in my pocket is
little Q. Then the force on that charge is always Q
times E. Doesn't matter whether Q is positive. Then it will be in the same direction as E. If it is negative it will be in the opposite
direction as E. If Q is large the force will be large. If Q is small the force will be small. So once you know the E field it could be the
result of very complicated charge configurations, the real secret behind the concept of an
E field is that you bring any charge at that location and you know what force acts
at that point on that charge. If we try to be a little bit more quantitative,
suppose I had here a charge plus three and here I had a charge minus one. Here's minus one. And I want to know what the field configuration
is as a result of these two charges. So you can go to any particular point. You get an E vector which is going away from
the plus three, you get one that goes to minus one and
you have to vectorially add the two. If you are very close to minus one, it's very
clear because of the inverse R square relationship that the minus one is probably going to win. Let's in our mind take a plus test charge
now. And we put a plus test charge very close to
minus one, say we put it here, even though plus three is trying to push it out,
clearly minus one is most likely to win. And so there will probably be a force on my
test charge in this direction. The net result of the effects of the two. Suppose I take the same positive test charge
and I put it here, very far away, much farther away than this separation. What do you think now is the direction of
the force on my plus charge? Very far away. Excuse me? Why do you think it's to the left? Do you
think minus one wins? [student reply is inaudible] So you really think the minus one is stronger
than the plus three because the plus three will push it out and
the minus one tries to lure it in, right? If the test charge is positive. [inaudible] Imagine you are all the way on Mass Avenue. You think that this distance matters? Who thinks the force is in this direction? Who thinks it is in this direction? Very good, you've helped him really. The force is obviously in that direction because if you're very far away, the field will be the same as if you just had a plus three and a minus one somewhere here which is plus two. So if you're far away from a configuration
like this, even if you were here, or if you were there, or if you're way there,
clearly the field is like a plus two charge. And falls off as one over R squared. So therefore if you're far away the force
is in this direction. And now look, what is very interesting. Here if you're close to the minus one,
the force is in this direction. Here when you're very far away, maybe I should
be all the way here, it's in that direction. So that means there must be somewhere here
the point where the E field is zero. Because if the force is here in this direction
but ultimately turns over in that direction, there must be somewhere a point where E is
zero. And that is part of your assignment. I want you to find that point for a particular
charge configuration. So let's now go to some graphical representations
of a situation which is actually plus three minus one. Try to improve on the light situation. And let's see how these electric vectors,
how they show up-- in the vicinity of these two charges. So here you see the plus three and the minus
one, relative units, and let's take a look at this in some detail. First of all the length of the arrows again
indicates the strength. It gives you a feeling for the strength. It's not very quantitative of course. And so let's first look at the plus three,
which is very powerful. You see that these arrows all go away from
the plus three and when you're closer to the plus three, they're stronger, which is a representation
of the inverse R square field. If you're very close to the minus one, ah
the arrows are pointing in towards the minus one, because the one over R square, the minus
one wins. And so you see they're clearly going into
the direction of the minus one. Well, if you're in between the plus and the
minus on this line, always the E field will be pointing from the plus to the minus, because the plus is pushing out and the minus
is sucking in. So the two support each other. But now if you go very far away from this
charge configuration, anywhere, but very far away, much farther than the distance between
the two charges, so somewhere here, or somewhere there, or somewhere there, or here, notice that always the arrows are pointing away. And the reason is that plus three and minus
one is as good as a plus two if you're very, very far away. But of course when you're very close in, then the field configuration can be very very complicated. But you see very clearly that these arrows
are all pointing outwards. None of them come back to the minus one. None of them point to the minus one direction. And that's because the plus three is more
powerful and then there is here this point and only point whereby the electric field is zero,
if you put a positive test charge here the minus will attract it, the plus will repel it,
and therefore there comes a point where the two cancel each other exactly. Now there is another way of electric field
representation which is more organized. And we call these field lines. So you see again the plus three and you see
there the minus one. If I release right here or I place here
a positive test charge all I know is-- that the force will be tangential to the field lines. That is the meaning of these lines. So if I'm here the force will be in this direction. If I put a positive test charge here, the
force will be in this direction, and of course, if it's a negative charge the force flips
over. So the meaning of the field lines are that
it always tells you in which direction a charge experiences a force. A force a positive charge always in the direction
of the arrows, tangentially to the field lines and a negative charge in the opposite direction. How many field lines are there in space? Well
of course there are an infinite number. Just like these little arrows that we had
before, we only sprinkled in a few, but of course in every single point there is an
electric field and so you can put in an infinite number
of field lines and that would make this a
representation of course useless. So we always limit ourselves to a certain
number. If you look very close to the minus one notice
that all the field lines come in on the minus one. We understand that of course because a positive
charge would want to go to the minus one. If you're very close to the plus and they
all go away from the plus because they're being repelled. You can sort of think as these field lines
if you want to imagine the configuration that the plus charges blow out air like a hairdryer, and that the minus suck in air like a vacuum cleaner, and then you get a feeling for there
is on this left side here this hairdryer which wants to blow out stuff and then there is that little sucker
that wants to suck something in and it succeeds to some degree, it's not
as powerful as the plus three, though. Have we lost all information about field strength?
We had earlier with these arrows, we had the length of the arrow, the magnitude of the
field was represented. Yeah, you have lost that, but there is still
some information on field strength. If the lines are closer together, if the density
of the lines is high, the electric field is stronger than when the density becomes low. So if you look for instance here, look how
many lines there are per few millimeters, and when you go further out these lines spread
out, that tells you the E field is going down, the strength of the E field is going down. It's the one over R square field of course. If you want to make these drawings what you
could do to make them look good, you can make three times more field lines going out from the plus in this case than return to the minus one. So the field lines are very powerful and we
will often think in terms of electric fields and the line configurations and you will have
several homework problems that deal with electric fields and with the electric field lines. If an electric field line is straight, so
I have electric fields, get some red chalk, say we have fields that are like this, straight
E field lines, and I release a charge there, for instance a positive charge, then the positive
charge would experience a force exactly in the same direction as the field lines, because the tangential now is in the direction
of the field line, it would become accelerated in this direction
and would always stay on the field lines. If I release it with zero speed, start to accelerate
and it would stay on the field lines. In a similar way, if we think of the earth
as having a gravitational field, with 801 we may never have used that word, gravitational field, but in physics we think ot the-- of gravity also being a field. If I have here a piece of chalk the field lines,
the gravitational field lines, here in 26-100, nicely parallel and straight
and if I release this piece of chalk at zero speed it will begin to move
in the direction of the field lines, and it will stay on the field lines. So now you can ask yourself the question if
I release a charge would it always follow the field lines? And the answer is no. Only in this very special case. But suppose now that the field lines are curved. So here are field lines as you have seen in
those configurations. It's very common. If now I release a -- a charge in here, say
I have a point charge here, it will experience a force in this direction. So it will get an acceleration in this direction,
so it will immediately abandon that field line. And so if now you ask me what is the trajectory
of that charge, well, it could become very complicated,
I really don't know. Maybe it's going like this, and by the time
it reaches this point, what I do know that then the force will be tangential to this field line, so will be in this direction. And so as it marches out and picks up speed,
locally it will experience forces representative of those field lines and so the
trajectory can be rather complicated. So field lines are not trajectories,
and not even when you release a charge with zero speed. Only in case that the field lines are straight
lines. Let's now look at a field configuration which
Maxwell himself, the great maestro, in some of his publications put there. And it's a ratio one to four and whether it is
plus four plus one or minus four minus one is not important because that's just a
matter of the direction of the arrows. But... Maxwell didn't put arrows in. So I leave it up to you. If it's plus four and plus one you have to
put arrows going outwards. And what you see now here is this airblower
effect. Think of them as both being positive. So there is the plus four trying to blow air
out like a hairdryer and the plus one is trying to do its own thing
and so you get a field configuration, field lines which are sort of, not perhaps easy, but you can sort of imagine why it has this peculiar shape. If you... put a plus test charge in between
the one and the four, then the four will repel it but the one will also repel it
and so there's going to be a point somewhere, probably close to one, whereby the two forces
exactly cancel out. Therefore E will be zero there. In a similar way between the moon and the
earth, there is a point, not too far away from the moon, where the gravitational attraction from
the earth and the gravitational attraction from the moon
exactly cancel each other out. That's not too dissimilar from this situation. So when you have charges of the same polarity,
you always find in between somewhere a point where the electric field is zero. Let's now go to a very special case whereby
I make the two charges equal in magnitude but opposite in sign and we have a name for
that, we call that a dipole. The plus charge is here and the minus charge
is there. Situation is extremely symmetric, as you would
expect, because they have equal power. There's one airblower upstairs and one vacuum
cleaner downstairs. If you're close to the plus charge notice that
all the field lines go away from the plus. And if you're close to the minus, notice that
all the field lines come in on the minus. You expect that. If you are far away from this dipole,
now you have a problem. Before we had a plus three and a minus one
and when you're far away the plus three wins. So it's like having a plus two charge. If you're far away you always expect the electric
field then to be pointing away from the-- equivalent charge of plus two. But if you add up plus and minus and they
have equal magnitude, let's say plus one and minus one, you get zero, so neither one wins if you're far away and notice carefully if you're very far away, indeed you do not see arrows either pointing out nor pointing in. Nature cannot decide, there isn't one that
is stronger than the other. And that makes dipole fields very very special. In the case of the plus three and the minus one,
if you're very far away, it's like having a plus two charge and the E field,
when you go further and further out, will fall off as one over R squared. With a dipole your intuition sort of tells
you that it will probably fall off faster than one over R squared. And that is part of a homework assignment
that you have this week. In fact I can already give you the answer. You have to prove it. If you're far away from an electric dipole the
electric field falls off as one over R cubed. It goes faster than one over R squared. There is not a single point in space where
the electric field is zero. And you can think about that why that is the
case. So these field configurations can be rather
complicated and can be very interesting and each one has its own applications. Are dipoles rare in physics? Not at all. In fact, they're extremely common. You cannot avoid them. Remember last time, I told you if you have
a spherical atom or you have a spherical molecule, and you bring that close to a charge-- let's now think of it you bring it in an electric field,
it's another way of saying the same thing. So we have a nice spherical atom or a nice
spherical molecule and we bring it in an electric field. The electrons want to go upstream,
the electric field vectors, they go against the direction of the electric field. And the positive charge wants to go in the
direction, wants to go downstream. And so what are you going to do? The electrons
will spend a little bit more time on one side of the nucleus than they would in the absence
of that electric field. And therefore you are through induction turning
that atom, turning that molecule, in becoming a dipole. If you have a little bit more charge on this
side, averaged over time, you have the same amount of extra charge plus
on that side averaged over time. So you make dipoles very often whether you
like it or not. And later in this course we will learn more
about the polarization of atoms and molecules creating dipoles
when we talk about dielectrics. And you will see that it will have... can have an enormous impact on the properties of the material. Could I make you a dipole here in class?
Oh yeah, that's very easy. To make one of nonconductors
is not so easy in class. To make one of conductors is very easy. And I'm going to do that with these two spheres
that you have. Look at these two metal spheres. Conductors. Free electrons. It's very easy for them to move. And I'm going to bring this rubber rod, which I will rub and becomes I think negatively
charged if I remember correctly, and I will bring that... say close to these two
which are touching each other. So here is this one metal sphere and here is the other
metal sphere and here comes the rubber. Negatively charged. Ah! What's going to happen? The electrons
want to go away, so this becomes negatively charged and therefore this remains a little
bit positively charged. For every one electron that has is excess here,
when I start it's neutral, there will be a positive excess there because charge
is conserved. You can't create charge out of nothing. And now what I do, while this rubber is still here,
while that rubber rod is there, I separate them, so here the- they're in contact with
each other first, they have to be in contact. Wow, we get some visitors. [VISITORS]
[clap hands twice] Don't be late, that means you, pretty boy. [LEWIN] I'm impressed. Thank you. Um... so what I do now is while this rubber rod
is still in place, I take them apart, and when I take them apart this negative charge
is trapped and this positive charge is trapped. And so I have thereby created negative charge
on this one. Positive on this one and it's equal in magnitude,
so I have a dipole. What I want to demonstrate to you is that
indeed I have positive charged here and negative here, that there is a difference in polarity
between these two, and that's the way that I will do the experiment. I will not show you that the amount of charge
is exactly the same on each, which of course it has to be. So let me give you some better light, or we
have to get the view graph off, the overhead. You see there for the first time an electroscope,
we discussed it last time, it is a piece of aluminum foil, very thin, with a metal rod
next to it, and when I put charge on the rod it will also go into the aluminum foil, and they will repel each other, and so the,
the aluminum tinsel will go to the right and the more charge there is on it,
the farther it goes to the right. So let me first put these two together,
make sure they are completely discharged, and now I'm going to bring these two into an electric
field which is produced by this rubber rod, I have to rub with the cat fur, and I believe
it was negative. But if you... You'd never have to remember whether it's
negative or positive of course, that is not so important,
what is in a name, after all. But it did happen to be negative. OK, so now we go here. I bring it here, I hope that no sparks will
fly over because that ruins the demonstration. And now notice what I do. While the -- while the rod is here I separate
them. So as I was holding it there, things were going
on in there that you and I couldn't see but electrons, the rubber rod is negative,
electrons were shifting in this direction and this is now positive and that is now negative. If I take this one and I touch it with the
electroscope, you clearly see that there is charge on this. How can I show you now that there is charge
of different polarity on the other one? Well, the way I will do that is I will approach
this electroscope by bringing this sphere very close to it. And if this charge is different than the charge
that is on it-- the electroscope will--
the reading will become smaller. And why is that? Why will the reading become
smaller? Well, here is the situation of the electroscope now. And here is that ball that you see on top,
this is upside-down there, if this is all negative, that's why it is apart,
if now I approach this here with an object which is positively charged and I claim that this one
now is positively charged, because this one was negatively charged,
the electrons love the positive charge, so the electrons want to come to the positive charge,
so these electrons drift down again, and so if they come down, fewer will be here
and so you will see this. If however I put here a negative rod,
then the electrons which are here want to go further away, they will stream up
and therefore the reading will become larger. So you can always through induction test
what the polarity is of your charge. Let's hope that this one is still holding
its charge while I was talking. So I claim now that if this polarity is different
and if it's still there when I approach the electroscope, come very close, that the reading should become a little smaller
without even touching it. Let's see whether that works. You see, it goes down. You see, it goes down. Goes down. So through induction I have demonstrated that
this has indeed a different polarity from this one. If I approached it with this one, it would
go further out, unless it already is at the maximum, let's try that,
you see, it goes further out. So not only have I demonstrated that I created
a dipole, but you've also seen... that... by means of induction
that you can demonstrate... that... there's a difference in polarity
between the two spheres. If I create a dipole and I put that dipole
in a... in an electric field, the dipole will start to rotate. Let's first talk about it.
Why it rotates. And then I will try to demonstrate that by
making a dipole, a big one, this big, right in front of you, almost as big as the
one there. So let's have a... an electric field like
so. And I bring in this electric field a dipole,
a biggie, here, this is the one I'm going to use for this demonstration. Ping Pong balls on either side, they are conducting,
and they are connected with a rod which is not conducting. And so here is this dipole. So this rod is not conducting. And this is a conducting and this is a conductor. And let's suppose this is positive and this
is negative for now. And I'll show you how we get the charge on
it. Well, the positive charge will experience
a force in this direction, always in the direction of the electric field
and the negative charge will experience a force always upstream. And now there is a torque on this and there
is a torque on this dipole. It will start to rotate clockwise. And of course if it overshoots the field lines
when it is in this direction the torque will reverse.
It's very easy to see. And so what you will see, it's going to oscillate
and if there is enough damping it will come to a halt more or less
in the direction of the field lines. And this is something that I can demonstrate. First I have to make a dipole of this kind
and the way I will do that is the following. This is a metal bar, it's this insulator and here
is this... are these two Ping Pong balls, the one on this side has a yellow marker,
the one on that side has an orange marker, and I'm going to attach them holding them
up against this metal bar. In other words, here is this dipole,
it's not a dipole yet, metal, metal and here is a metal bar, this is a conductor
which connects them. I'm going to turn on the VandeGraaff here,
and the VandeGraaff creates an electric field, so we have the VandeGraaff here. And let's suppose that this VandeGraaff creates
a positive charge. Sometimes a VandeGraaff creates
positive charge on the dome, others can be designed to create
negative charge on the dome. And remember for now I assume that's positive. What will happen now? Electrons want to go
in this direction. So this becomes negative. Protons, positive charge, stays behind. So that becomes through induction a dipole. Because I have them connected. I have them connected with this metal bar. So these electrons can flow through this bar
and end up here. Now I remove the bar. And so when I remove the bar I have created
now a dipole. I have here an insulating thread
and I have a fishing rod and at the end of my fishing rod
I have now a permanent dipole. With that permanent dipole I'm now going to
probe the electric field around this VandeGraaff. I could have chosen the same VandeGraaff
but there's a reason why I picked this one and as I walk around this VandeGraaff
you will see that this fishing rod at the end is this dipole,
that the dipole always wants to go radially inwards or outwards depending
on how you look at it, of this field. So I can probe this field
and make you see for the first time that there is indeed somewhere
here a strong radial field going in or out of the VandeGraaff. And now comes something very interesting,
which I found out this morning for the first time when I did this experiment. If the other VandeGraaff there is also positive
when I run it, how do you think this dipole is going to align then
if I walk into it? Will the negative ball be closer to the VandeGraaff or will the positive one go closer to the
VandeGraaff? So I give you thirty seconds to think about it. So I make the dipole as it is here,
let's assume this one is positive, this VandeGraaff. So this side becomes minus, I call that A,
and this side become positive, that's B. I now walk with this dipole, I bring it in
this field. And let's assume that one is also positive. We don't know that yet. How will the dipole align now? Will A go inwards
or will A go outwards? Who thinks A goes inwards? Very good. Who says A goes outwards? OK. A will go inwards. If the two VandeGraaffs have the same polarity. So if that doesn't happen... that... doesn't
mean that physics doesn't work, it means the two VandeGraaffs have different polarities. And we'll see what happens. So let me first then create a dipole. So here is the... the dipole. It's shorted out now. I turn on the... this VandeGraaff. So induction takes place. Remember that the yellow is pointing towards
the VandeGraaff and that the orange is away from the VandeGraaff. OK. So I induce a dipole. Oh! I really should redo that. I don't know what happens when it-- I have
to remove the field first. OK. The yellow was inside, right, was that the
way it was? OK. Yellow inside. There we go. So now it's creating a dipole through this
metal bar. And I break contact and this should now be
a dipole. Now I turn on the field of the... so if the
polarity is the same, yellow will go in. I will try to swing it a little. Notice two things. It's going to line up beautifully radially
but the yellow is not in, the yellow is out. So the two VandeGraaffs have different polarities. But you will see they rotate nicely. And they end up beautifully radial and when
I go all the way around here... again... they may swing a little, they may oscillate a little,
but through damping they will come to a halt and look, the field is indeed beautifully
radial and the yellow is pointing outwards. The two VandeGraafs have different polarities. So you see how we can create a dipole and
you've also seen how we often can make statements about... the specific polarity. I can probe an electric field using grass
seeds in oil. Grass seeds are elongated and when I put a
grass seed in the electric field it will become polarized,
there's nothing you can do about it. Here is a grass seed and
the electric field is like so. And so the electrons want to go as far away
in this direction as they can through induction. And so this side remains positive, and so
what is this grass seed going to do? It's going to rotate. It's going to line up with the electric field. And this is the way that I'm going to show
you now field configurations in the vicinity of a dipole. And I will also show you then field configurations
in a vicinity of two charges which have equal polarity. You may have seen this in high school with
magnetic fields, with iron files, that's kid stuff. That's the easiest thing to do. This is the real thing, this is the electric
fields, I bet you you've never seen electric fields which are traced
by these mysterious seeds. So I'll give you some light that may optimize
the demonstration. These seeds first have to be oriented in a
way so that it is chaos. The first thing you see is I'm going to make
this-- I believe it's going to be a dipole first.
Almost certain. So I'm going to charge one positive and charge
the other negative. And then we'll see how these grass seeds will
form each other. Watch closely. There you go. My goodness. That is a wonderful dipole field. Of course we don't know which one is plus
or minus because the grass seeds have no arrows on them. But you clearly see these incredible lines-- radially inwards or outwards on each one of
the charges and then you see these nice arcs
in between. Who could see it easily? OK, you got something
worth for your tuition. Put a little bit more charge on maybe. Very clear. And now which is perhaps more interesting,
I'd like to show you the field surrounding two charges but now the charges are both the
same polarity. So we have to undo the, the memory of the
grass seeds. OK, now we'll try to make them both the same
polarity. And then watch this hairblower effect that
I told you about. Aaah, look at that. Great. Now you really cle- clearly see these -- these
field lines, and you see in between how the two airblowers are competing with each other. Very impressive. All right, so that's the way you see field
lines now, electric field lines. And some of you may have seen with iron files...
magnetic field lines. If I have the VandeGraaff and I have the VandeGraaff
here-- and let's suppose the VandeGraaff is positive, I don't know whether it's positive or negative, let's suppose it is, I'm going to use the one over there, and I'm going to stand here,
on the ground, Walter Lewin, what is going to happen with me? Through induction,
the electrons being sucked out of the earth and coming up because they want to go close
to the positive charge. So I will become negatively charged. What will the field lines do? Oh, they will
be extremely complicated. Very complicated. But something like this maybe. Maybe something like this. Uh some may come out here. Some may end up on my neck here. Some may go here. Like so. Very complicated field configurations. But I want to probe that field. Somehow, a little. Get a feeling for how that -- what that field
is like. And the way I'm going to do that is I'm going
to put a charged balloon. There you see the balloon. It's a conductor. I'm going to put a charged balloon and put
it here say. Well if it is a positively charged balloon
it will take off in that direction. Right? The force is always tangential to the
field lines. It will abandon the field lines, it won't
stay on the field lines, there's a lot of damping on the balloon, that's why I chose the balloon, so it will move relatively slowly, and it'll ultimately maybe end up on my head,
right here. Once it ends up on my head there,
so it comes in maybe like this, now it will get the negative charge from my head
and so it will become immediately negatively charged and so the force now will reverse
and will be in this direction, tangential to this field line, and so it will go back. When it hits the VandeGraaff again it will
get positive charge, reverse its polarity, and it will go back. And so it will bounce back and forth between
me and the VandeGraaff and it gives you some rough feeling of what this field configuration is about, although I want to remind you that the charge
does not follow exactly field lines. So I'm going to sit here and I will be part of this,
so that's probably going to be positive, I will automatically become negative,
there's nothing I have to do, all I turn is the VandeGraaff and we have to put a little bit of charge
on that balloon. It will probably do that by itself,
but I can always give it a little kick so that it goes to the VandeGraaff,
there it goes. [laughter] Oh [inaudible] because my,
my glasses are a good insulator, so I better take my glasses off,
so that every time it hits me. [laughter] Changes polarity. [laughter] So this is a way you can do physics
and have fun at the same time. See you Monday. [applause]
