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