Today we're going to um 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 NDA 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. And this is 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 SU 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. And 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 eight
oh two, 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 eight oh two. 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 -- 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 that if 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 have 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
with 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 eight oh one 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 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're 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 eight oh two. 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 ask 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 Ds 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
twenty-six one hundred 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 if 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 uh 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 strength of
the [inaudible] 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
uh 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. [laughter] 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 uh 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.
[laughter] OK. So look at the projection.
The uh 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 t- 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. 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.
Oh. [laughter] What happened?
I must have touched the side. There's no way around it.
I'll make sure that there is enough
charge on it. I'll charge it once more.
Discharge them. OK.
We'll do it again. Go inside.
I touch them.
Take it out. Take it out.
Nothing. 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 automatically 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 every
time 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,
along the field lines. 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 with 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 h- 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 c- 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.