Today we're going to work on a whole new concept
and that is the concept of electric flux. We've come a long way. We started out with Coulomb's law. We got electric field lines. And now we have electric flux. Suppose I have an electric field
which is like so-- and I bring in that electric field a surface,
an open surface like a handkerchief
or a piece of paper. And so here it is. Something like that. And I carve this surface up in very small
surface elements, each with size dA, that's the area, teeny weeny little area
and let this be the normal, N roof, the normal on that surface. So now the local electric field say at that
location would be for instance this. It's a vector. The electric flux d phi that goes through
this little surface now-- is defined as the dot product of E and the vector perpendicular to this element which has this, as a magnitude, dA. Now our book will always write for N dA simply
dA. So I will do that also although I don't like
it but I will follow the notation of the book. So this vector dA is always perpendicular
to that little element dA and it has the magnitude dA. And so this since it is a dot product is the
magnitude of E times the area dA times the cosine of the angle between these two vectors, theta. It is a scalar. The number can be larger than zero, smaller
than zero, and it can be zero. And I can calculate the flux through the entire
surface by doing an integral over that whole surface. The unit of flux follows immediately from
the definition. That is newtons per coulombs, for the units
of this flux, is newtons per coulombs times square meters. But no one ever thinks of it that way. Just SI units. I can give you a, some intuition for this flux
by comparing it first with an airflow. These red arrows that you see there represent
the velocity of air and you see there a black rectangle three times. In the first case notice that the normal to
the surface of that area is parallel to the velocity vector of the air and so if you want to know now
what the amount of air is in terms of cubic meters per second
going through this rectangle it would be V times A, it's very simple. However, if you rotate this rectangle ninety
degrees so that the normal to that rectangle is perpendicular to the velocity vector, nothing
goes through that rectangle and so it's zero. So now the flux-- the air flux is zero,
and if the angle is sixty degrees then it is of course V times A times the cosine of
sixty degrees. Now think of these red vectors as electric
fields. So now the electric flux going in the first case
through that surface is now simply E times A. In the second case it's zero. And in the last case it is EA times the cosine
of sixty degrees. So you can sometimes think of this as airflows. We also saw that when we dealt with field
lines that can come in sometimes very handy. I now take a surface which is not open as
this one is, this is an open surface. Can come in from both sides. But now I choose one that is completely closed. Like a potato bag or a balloon. I'll draw, put this line in here to give you
a feeling there's a completely closed surface. So you can only get inside if you penetrate
that surface from the outside. And so now I can put up here and here these
normals, dA and there's another normal here,
maybe in this direction, dA. In this case by convention the normal
to the surface, locally to the surface, is always from the inside of
the surface to the outside world. It's uniquely determined because it's a closed
surface. Here it was not uniquely determined. I arbitrarily chose this one but I could have
flipped it over a hundred eighty degrees since it's an open surface, it's ill-defined. Here it's never ill-defined. So the normal is always chosen to go from
the inside to the outside. And now I can calculate the total flux going
through this closed surface. Locally multiplying E with dA,
dot product over the whole surface, out comes a certain number,
and that is now therefore the integral of E dot dA integrated over that closed surface and since it is a closed surface we put a
circle here, to remind us that it is a closed integral and here in this case it is
a closed surface. And this now is the total flux
through that surface. It could be larger than zero. It could be smaller than zero. It's a scalar, it's not a vector. It could be equal to zero. If it's equal to zero then you can think of
it whatever flows in, if you think of it as air, also flows out. If more flows out than flows in
then it is positive. If more flows in than flows out
it is negative. So let's now calculate the flux for a very
simple case where I have a point charge. So here I have a point charge and I'm going
to put a bag around this point charge and the bag is a sphere. It is a sphere and the sphere has radius capital
R. And let this charge be plus Q. Just for simplicity. Well, I pick a small element dA here. And at element dA is radially outward... dA... This is the normal to that surface so that
this radial... The electric field at that point is also radial. We have dealt with that before. So dA and E not only here but anywhere on
the surface of this sphere are parallel. For the cosine of the angle equals one. I can also introduce here the unit vector
R roof which is the unit vector going from capital Q to that element where I evaluate
the teeny weeny little amount of flux. So if now I want to know what the total flux
is through this sphere that's very easy because since this is a sphere the E vector
in magnitude is everywhere the same because the radius is the same,
the same distance to this charge, and dA and E are parallel. So it's simply the surface four pi R squared
of that sphere, times E. And so now I have that the total flux
through that closed surface is simply four pi R squared times E. Well, what is E?
The electric field at this distance R-- equals Q divided by four pi epsilon zero R squared
times R roof. That gives me the direction. And so,
if I know that the flux is four pi R squared times E, I put the four pi R squared here,
I lose the four pi R squared and I find that the E vector,
at least the magnitude of the electric field-- uh excuse me, that the flux phi,
that's what I want to calculate, I multiply this by E,
equals Q divided by epsilon zero. And this is independent of the distance R. And that's not so surprising because if you
think of it as air flowing out then all the air has to come out somehow
whether I make the sphere this big or whether I make the sphere this big. So the flux, being independent of the size
of my sphere, the flux is given by the charge which is right here
at the center divided by epsilon zero. Now if I had chosen some other shape,
not a sphere, but I have dented it like this, it's clear that the air that flows out would
be exactly the same. And so I don't have to take a sphere
to find this result. I could have taken any type of strange closed
surface around this point charge and I would have found exactly the same result. And if I put more than one charge inside this
potato bag then clearly since I know that electric fields from different charges can be added, should be added, vectorially, it is clear that the relation should also hold
for any collection of charges inside the bag and therefore we now arrive at our first milestone
in 802, which we call Gauss's law. And Gauss's law says that the flux,
the electric flux, going through a closed surface,
being the closed surface of E dot dA, is the sum of all charges Q which are inside
the bag, that you may choose at any time
you pick that bag, divided by epsilon zero. And this is the first of four equations of
Maxwell which are at the heart of this course. So the electric flux through any closed surface
is always the charge inside that closed surface divided by epsilon zero. And if that flux happens to be zero, it means
there is no net charge inside the bag. There could be positive, there could be negative
charges, but the net is zero. Gauss's law always holds. No matter how weird the charge distribution
inside the bag. No matter how weird the shape of this bag. It always holds. But Gauss's law won't help you very much
if you don't have a situation whereby the charges are distributed
in a very symmetric way. Gauss's law holds but it doesn't do you any
good if you want to calculate the electric field. In order to calculate successfully the electric
field you do need forms of symmetry and there are three forms of symmetry
that we will deal with in 802. One is of course spherical symmetry. Another one is cylindrical symmetry. And a third one is flat planes with uniformly
charged distributions. Then we also have situations of symmetry. And so now I would like to, as a first example,
use an application of Gauss's law and I will start with a situation
of spherical symmetry. And I use a thin shell, a hollow sphere,
which is thin and so this radius is R and I put charge Q on here
but it is uniformly distributed. That's crucial. If it's not uniformly distributed I have no
symmetry, I can't do the problem. So it's uniformly distributed. We will learn later in the course,
that it's very easy to do this, because any conductor of this shape,
if you bring charge on it, will automatically distribute itself uniformly. So we have the charge plus Q on there
uniformly distributed, that's a must, and I would like to know now
what is the electric field here-- at a distance R from the center and what is the electric field here
at a distance R from the center. In other words I want to know what is the
electric field everywhere in space. Just due to this charged, uniformly charged
sphere. And with Gauss's law it just goes like that. You now have to choose your Gauss surface. And if you don't choose it in a clever way
you get nowhere. In a case like this,
I would think it is rather obvious, that the Gauss surface that you would choose, are themselves spheres,
concentric spheres. If you want to know what the electric field
is at this point you choose a sphere with this radius R going through that point and
if you want to know what it here is you choose a sphere going through that point. All the way enclosed. It's a concentric sphere. And now you have to use symmetry arguments. And the symmetry arguments are the following. Since this is spherically symmetric, this problem,
if you were here, whatever the electric field is here in magnitude
must be the same as it is there and it must be the same as it is there. Because of the symmetry of the problem,
it couldn't be any larger here than it could be here. That's obvious. That's a symmetry argument because the charge
here is uniformly distributed. That's symmetry argument number one. Now comes another symmetry argument
and that is-- the electric field, if there is an electric field,
must be either radially pointing outwards or radially pointing inwards. So either it has to be like this or it has
to be like this and here the same. Either like this or like this. Because we already know if this is a positive
charge then it's going to be pointing outward. It cannot go like this or like this
because nature could not decide, in this spherically symmetric problem,
to go like this or like this. It can only go radially. That's the second symmetry argument. So now if we go to this sphere now
and we know that E is radially outwards apart from a plus or a minus sign,
apart from the fact that the angle between dA and E
could either be zero degrees or a hundred eighty degrees,
we know now that the surface area of that sphere,
which is four pi R squared, times the magnitude of the E vector right here, I can do that now because DA and E are either
parallel or antiparallel, that must be equal to Q inside divided by
epsilon zero. There is no Q inside, so E must be zero. That's an amazing result. You say well uh there's no charge inside. Still an amazing result. Because it means that anywhere inside here
no matter what radius you choose the electric field equals exactly zero. And it means there is some crazy conspiracy
of all these charges, that are uniformly distributed here, which each individually contribute to
the electric field inside through Coulomb's law, that all those together can for through
a conspiracy make the E field everywhere inside zero. It's a nontrivial result. All right. So now we know that the E field inside is
zero. So this is for r smaller than R. Let's now go r larger than R. Everything I told you holds for the sphere
which is outside this hollow sphere. Everything holds. The E field here must be the same everywhere
on the surface. dA and E are either parallel or antiparallel. So I can write down again that four pi r squared
which is the surface area times the electric field vector must be the Q inside divided by epsilon zero. But this Q is that Q. It's not zero. There is charge inside. And so now I know that the electric field E,
in terms of its magnitude, is Q divided by four pi r squared epsilon zero. And we know the direction
if it is positive of course it is radially outwards and if this is negative it's radially inwards. And this is a nontrivial result. We have seen this before. If I had put all the charge right here at
the middle at the center we would have gotten exactly the same answer. We've seen that before. In other words whether the charge is uniformly
distributed over a sphere or whether the charge is, all of it,
exactly at the center of the sphere, that makes no difference for the E field
as long as you're outside the sphere. If you plot the electric field as a function of R
and if here is capital R and if this is the -- the field strength,
then you would get that the electric field is zero inside, jumps to a maximum value and this falls off
as one over R squared proportional to one over R squared. If I go back to the situation that the charge,
that the electric field inside is zero, you may say: "Isn't that a little bit of a cheat?". "Because yeah there is no charge inside.". "But have you really used the charge outside.". "And if you have used it how did you use it.". Well I have used it. I use it through my symmetry arguments. The symmetry arguments take into account
that the charge is uniformly distributed. If the charge on the sphere had not been
uniformly distributed, I could not have used the symmetry argument, and therefore the electric field inside
would in fact not have been zero. If there is more charge on the sphere here
than there is there, the field inside the sphere is not zero. So I have used all that charge by using my
symmetry argument. Gauss's law and Coulomb's law in a way are
the same law. They both link the electric field with the
charge Q. Key is the fact that the electric force falls
off as one over R square. If the electric field strength did not fall
off as one over R square Gauss's law would not even hold. And the electric field inside this uniformly
charged sphere would not be zero. So it is the immediate consequence of the fact that electric forces fall off as one over R squared. Gravitational forces also fall off as one
over R squared. Therefore if you take a planet if it existed
which is a hollow spherically spherical planet but hollow inside it means there would be
no gravitational field inside that hollow planet. So if you were there, there would be no
gravitational force on you. If it is spherical. If that planet were a cubical planet then
the gravitational field inside would not be zero. You say well big deal, with 801 we
always take a planet and then it's not as far as we're outside the planet we put all the mass
and we consider it as a point. Yeah indeed. It's not a big deal for you
and it is not a a big deal for me but it was a big deal for Newton. He intuitively sensed that it was correct
that if you have a planet of uniform mass distribution that you can
consider it as a point mass as long as you are outside the planet. But it took him twenty years to prove it
and he finally published his results. It would take us now thirty seconds. He didn't have access to Gauss's law. Came about a hundred years later. But the net result is that you see here in
front of you that if you have uniformly charge distribution
and you can draw the parallel with gravity, that it's you get the same electric field outside that you would have gotten if
all the charge is at one location. At the center. This is spherical symmetry, number one. That's the easiest symmetry that we have in
802. Now I will present you with a second form
of symmetry which is a flat horizontal plane. And I want you to work out most of it but
I'll help you a little bit to set it up. Suppose we have a plane which is very very
large. Think of it for now as infinitely large. That doesn't exist of course infinitely large. And I put on this plane charge. And I put a certain amount of charge density
which I call sigma. Sigma is an amount of charge Q per area A. So it is a certain number of coulombs per
square meter. And it's uniformly distributed,
so the whole plane everywhere has the same number of coulombs
per square meter. Or microcoulombs or nanocoulombs,
whatever you prefer. And this is a plane which is huge
and you are being asked: "What is the electric field anywhere in space?" Just like we before, we asked,
what is the electric field anywhere inside the sphere and anywhere outside the sphere? Now I want to know what it is anywhere in
the vicinity of this plane. If now you pick a clever Gauss surface
the answer pops out very quickly. If you would choose a sphere as a Gauss surface
you're dead in the waters, you get nowhere because there's no spherical symmetry. I will define for you that Gauss surface
but I want you to work out at home how you get the electric field. Suppose I want to know what the electric field
is at a distance d above the plane. What I do now is I choose this as my Gauss
surface. Watch me closely. This is the intersection with the plane. This is my Gauss surface. It is a closed surface. Three conditions have to be met for you to
be able to calculate what the E vector is at that location D. The first one is that this is a flat plane
here and this is the same flat plane. Must be parallel to this plane. That's a must. If you don't do that you can't use Gauss's
law. The second one is that these vertical walls,
that you have here, are indeed perpendicular to that plane. In other words these are parallel and these
are ver- exactly vertical. If you don't make them vertical if you do
this you're dead in the waters. Can't use Gauss's law very effectively. And then the third argument
which is very important, that this flat surface is a distance d
above the plane and that this flat surface is exactly
the same distance below the plane. And you can already smell why that is important. Because if you ever want to use a symmetry
argument if this plane is uniformly charged the electric field vector here in terms of
magnitude obviously must be the same as there in terms of magnitude,
maybe not in terms of direction, as long as this d is the same as that d. So that's why it's important that the two
d-s are the same. And the only charge that you have inside
when you apply Gauss's law is the charge which is of course here. That's the only charge inside that closed
box. If you work this out at home you will find
an amazing result. You will find that the electric flux through
these vertical walls is zero. Nothing comes out through the vertical walls. Think about it. Why that is. Use symmetry arguments. But something comes out here or comes in here
if it is a negative charge and something goes out here. And so you only have two contributions from
those two end plates. You'll work on that and you will find perhaps
to your amazing result that the electric field equals sigma divided by two epsilon zero and
that it is independent of how far you are from that plane. Whether you're very far away or whether you're
close it's the same. So if this is that plane
and if the plane is positively charged then E would be like this here
and E would be like this here and it would be independent of distance
and if it is negatively charged E would be like so and it would be like,
like so pointing towards the plane and in all cases would the magnitude be
sigma divided by two epsilon zero. Does it mean if I go very far away from that
plane that it is still independent of the distance? Yeah, if that plane is infinitely large. But if the plane is only as large as the lecture
hall here then clearly it would hold very accurately as long as I stay relatively close
to the plane. In other words if my distance to the plane
is small compared to the linear size of the plane. But if I go miles away, well of course then
that plane is charged looks like a point charge if I'm five miles away
from 26-100. if the plane is only as large as this lecture hall then it looks like a point charge
and obviously the electric field will then fall off as one over R squared. So when I say the E field doesn't change
with distance, it means of course that you have to be
relatively close to the surface, relative to the linear size of that surface. So you are going to prove this
and I'm going to use this now to calculate for you
a much more complicated configuration of two charged planes. But I use that result. That's very important. And suppose I have here a, a plate,
very large, nothing is infinitely large of course, and it has a surface charge density plus sigma
and I have here a plate which has surface charge density minus sigma and the separation between these
two plates happens to be d. And the question now is what is the electric
field anywhere in space. Here, here and here. And we'll think of them as being infinitely
large, each plate. And I now use the superposition principle. I say to myself aha! This plate alone, forget this one, this plate
alone would give me an E vector, oh, stick to my colors,
give me an E vector like so and that is sigma divided by two epsilon zero, this one is also pointing away from this,
sigma divided by two epsilon zero and here it's also sigma divided by two epsilon zero because it's independent of the distance to this plate. What is the negative charge doing? Well, the negative charge has
E vectors pointing towards it. So here I have an E vector which is sigma
divided by two epsilon zero. Here I have one that is sigma divided by
two epsilon zero and I have one that is pointing towards the plate,
which is sigma divided by two epsilon zero. I use the superposition principle,
I can add electric vectors, and when I do that I find that
these two cancel each other out. So the electric field here is zero. The electric field here is sigma
divided by epsilon zero. The two support each other. They are both in the same direction. And the electric field here is again zero. And that is an amazing result. Of course it's only accurate if these plates
are extraordinarily large and so I have to draw the field lines
in the situation like this then the field lines would be like so. If the upper plate is positive and the field
in here would be the same everywhere, would be outside zero and outside zero here. Now clearly this cannot be true
if you get into this area here where you are near the end
of these plates. That is not possible. Why not? Well you can't use your symmetry
arguments so Gauss's law is not going to help you if you get anywhere near this area. And it is very difficult to calculate the
electric field configuration when you are near the edges,
which we call the... the fringe field. Maxwell of course was a clever man
and he knew how to do that. Today we can also do that very easily
with computers. But I'll show you from Maxwell's original
publications that in a situation like that,
he was already perfectly capable of calculating these electric field lines
and you have these two horizontal plates, which one is plus and which one is minus
doesn't matter, he doesn't put arrows in there
and what you see is an extremely strong field inside the two plates
and remember that the density of field lines tells you something about the very strong field
but when you get near the edge the field is not really zero. The field strength drops very rapidly
because look the density is very low. But it is not zero. And the electric field is not zero here either
and is not zero there. In our assumption, in our simplification,
we have however assumed that the plate is so large that we don't have to worry about any end effects and in that case the electric field is only existent
in between the plates but not anywhere else. I now want to demonstrate to you some of the
things that we have learned today. And the first thing that I want to demonstrate
is that the electric field outside a large plane is more or less constant. Doesn't matter how far away you are. Now the way I'm going to do that is of course
I don't have an infinite large plane, the plane that you're going to see only a few
square meters in size. And so with only something like one by one
meter, then it would only be true that the electric field is very close to constant
if I stay very close to that plane. The moment that I go out as far as a meter
of course it's no longer true. So it's very qualitative, what I'm going to
show you. But you're going to see very shortly there
a very large plane. I'm going to get it in a few minutes. And let's assume that we look at that plane
edge on. So here is that plane. Look at it from edge on,
it will be put here. It will block your view,
that's why we don't have it up now. And what I will do now is I will connect that
with the VandeGraaff which is behind it. If you wait a few minutes then class will
pay attention to me and not to you. Uh here is the VandeGraaff,
we're going to attach it to the VandeGraaff and then we use this interesting fishing rod
which is a small Mylar balloon which we will charge with the same charge
as the VandeGraaff, the same charge as the plate
and we will hold that in front of the plane. And then of course there will be a force. So here is my glass rod. This is the vertical. And because there will be a repelling force
on this air-filled balloon, there will be an angle. There's an electric force on it because the
two have the same charge. And this is the angle theta that I will show
you projected on that wall. And when I move this away from this plane
you will see that the angle theta becomes smaller. Yes of course because look how small that
plane is. No matter what I do if I go from twenty to
forty centimeters you can't really say that the plane is infinitely large compared to
forty centimeters. But you will see that the angle of theta will
change very slowly. And then we will remove that plane and then
we will do exactly the same experiment but we will use only the VandeGraaff
which produces now an electric field. And that electric field now falls off as one
over R squared. It's not constant as a function of distance
but it falls off as one over R squared. This is a hollow sphere. So you can think of it as all the charge right
at the center. As we demonstrated, it's on the blackboard
still here. You know, you get that amazing result. And so now if I place this,
if I place this fishing rod, this balloon, near the spherical VandeGraaff
you will see that this angle theta drops very fast when I start moving my hand away. Extraordinarily fast. If I double the distance to the center,
the force on that little object will become four times smaller. It's inverse R square. So let's first do the plane and then we'll
try to do the... the single VandeGraaff. And we'll try to optimize the light conditions. We have a projection here. There's a carbon arc. Which will hopefully produce some light in
that direction. If the carbon arc works. [chuckles] Marcos, oh I forgot to turn on the
power. Thank you. So this carbon arc is coming on now and you'll
see there the shadows on that wall. See my hand, here is that plane, and it is
far from infinitely large, that plane. If I were this far away from it, four centimeters,
very close approximation, it would be infinitely large. But if I'm here and there and that's where
I will be, of course it is not infinitely large anymore. So let's start the VandeGraaff. You can see that I turned it on. It's rotating now. I have to put charge on here so I'll touch
it with the VandeGraaff and so this is now charged. It has the same charge as the plane. The plane is being charged. And here you see the angle. Try to remember that angle. It's hard to estimate, maybe fifteen degrees. You see the vertical and if now I--
it's about um thirty centimeters away from the plane. And if now I go back to fifty centimeters,
which is where I am now, you see the angle hasn't changed
very much. If I go further out, to sixty centimeters,
yeah, the angle goes down a little. Of course it does. But not very much. And if I go far away,
all the way to Mass Avenue, of course the force on this little object would be inversely R squared because then the whole plane would behave
like a point source. So I've shown you that very close to this
plane the electric field stays approximately constant. So if now we remove this, Marcos if you can
yeah, you'll have to take this also off. Thank you very much. So now we have the VandeGraaff alone. So now we know that the electric field falls
off as one over R squared. It's a very good approximation now. We can think of the charge as being right
at the center. I will give it a little bit of charge. Oh, it is already charged. [chuckles] OK. So look at the projection. The eh, the balloon is now uh oh maybe thirty
centimeters away from the center, maybe forty. Boy, the angle is almost forty-five degrees. And now I go, I double the distance,
I go to about ninety centimeters and look at that angle theta. The angle theta is now down to oh maybe ten
degrees. I will go back where I was. This angle is about forty degrees. And now it's very small and when I go here
which is about a meter-and-a-half you can hardly see that there is any angle. It's only a few degrees. And so I've shown you only qualitatively that
the electric field falls off very rapidly. In the vicinity of a hollow uniformly charged
sphere. And that it doesn't fall off very fast
if you are in the near vicinity of a plane. The second thing I want to show you
has to do with the fact that the electric field inside a uniformly charged sphere is zero. Here I have a sphere which is not entirely
closed. I can't make it closed because I want to demonstrate
to you that there is no electric field inside when I charge it uniformly. And since I have to get inside I need an opening. There's nothing I can do about it. Since there is an opening, the electric field
is not exactly zero inside. It's only true if this is a complete closed
surface and if the charge is uniformly distributed. But it's a good approximation. The opening is quite small. And what I'm going to do is I'm going to charge
this sphere. I'm putting charge outside. I use a device that we have not used before,
but that's not so important, but here is now that hollow sphere. I'm going to put charge on there. Let's suppose it is positive charge. So this will be positively charged. Since it is a -- a conductor, as we will learn
I think the next lecture or at least this week, that the charge will automatically distribute uniformly, only does that on a conductor and now to demonstrate to you that there is
an electric field here, I will use induction. I have two Ping Pong balls
painted with conducting paint. They touch each other. Under influence of this electric field
this one will become negative and this one will become positive,
we have discussed that last time, you create a dipole. Not important that it is a dipole. I separate the two. I have negative charge here
and positive charge there. I will touch any one of these two balls,
it doesn't matter which one, with the electroscope and you will see
that there is charge there. So I have demonstrated then that there is
an electric field outside that sphere. Now I will do exactly the same demonstration,
but now I put these two conducting balls inside, so here they are. I touch them, you just have to trust me
that I really will touch them and then I will take them out. And if I didn't make a mistake,
if I didn't touch the rim by accident, then I will show you that there is
no electric field inside, it means there is no induction,
so these balls did not pick up charge. I show you with the electroscope that indeed
there is no charge on it. So that is the way I want to do this. So there is the electroscope. Here is the sphere. And the way I'm going to charge it,
has a nice name, it's called electrophorus,
hard to pronounce. I first rub a glass plate with cat fur. Then I take a metal plate. I put on top. And I touch it with my finger. And now I transfer charge
and you think about it why that is. I put it on here again
touch it again with my finger. I'm again charging it. Put it on top, touch it again with my finger. I want a little bit more charge. So I'm rubbing this again. Put this on top. Touch it with my finger. Every time I do that I feel a little shock. Put it on there. Touch it with my finger. OK, let's hope that's enough. So now comes demonstration number one. These two spheres conducting completely discharged,
I bring them close to this sphere. There they are. I separate them. And now they must have picked up charge. Shall I use this one or this one to touch
the electroscope? The same to me. My right hand or my left hand? Who wants right? Who wants left? The right ones have it. And there's the charge. So I've shown you that there is an electric
field there. Through induction I have created
charge on here. Now I'll do the same inside. It's always tricky because if I hit--
if I hit the rim then it's not zero. This one has to go in first
because the opening is too small. Then the second one has to come in. Now I have to touch them
and I really do. I wouldn't cheat on you. Not this time. They are now in contact with each other. And now I take one out. And I take the other out. Which one shall I touch it, there shouldn't
be any charge on either one of them. We had left before or we had right before?
Well let's do this one. This one? Who is for left? Who is for right? The lefts have it. Nothing. Maybe a teeny weeny little bit,
well, the electric field inside is not necessarily exactly zero. But it's extremely close. The last thing I want to show you
has to do with the fringe field that we have seen here. I have here two parallel plates
which I'm going to charge with an instrument
that we have not seen before which is called the uh
-- a Wimshurst. If I rotate this crank I can produce positive
and negative charge. And this plate becomes positively charged
and the other plate automaticly becomes
negatively charged. And I'm going to show this to you
right there. That's the idea. Yeah, we will make it uh... So there you see these two plates. And you see a Ping Pong ball. And this Ping Pong ball is a conductor,
we put conducting paint on it. And remember when I did the demonstration
with the balloon which bounced between my head
and the VandeGraaff and everytime that it bounced
on the VandeGraaff it took all the charge of the VandeGraaff
and when it bounced on my head it took my charge,
and so it went back and forth and that is what I want to show you now. That this Ping Pong ball will start to probe
that field first outside the capacitor, or I shouldn't use the word capacitor but these
plates and then I will bring the Ping Pong ball inside and then you will see that the
field is much stronger there. So let's first get some charge on there. And listen to the sounds. Every time that it hits it bangs. So it's following almost those field lines. And in doing that it's actually transferring
charge every time from one plate to the other. It's nicely going around in an arc
the way you see it there. So it's clear that there is an electric field
outside. I've proven that to you. Otherwise it would never do what it's doing. So the electric field outside is not exactly
zero, of course not. This plate is not infinitely large. And now I will bring this Ping Pong ball inside,
I have to open up the -- the gap a little and I will bring it inside. And you see the field is much stronger. Now it's going back and forth between those
very high-density field lines, very strong electric field, going back
and forth each time that it hits the plate, it changes polarity and this is not too different from
the experiment I did with the balloon when I bounced it back from the VandeGraaff
to my head and back to the VandeGraaff. OK. Start working on that assignment. It's not an easy assignment this week. See you Wednesday.
