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• 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