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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