MIT OpenCourseWare http://ocw.mit.edu 8.02 Electricity and Magnetism, Spring 2002 Please use the following citation format: Lewin, Walter, 8.02 Electricity and Magnetism, Spring 2002 (Massachusetts Institute of Technology: MIT OpenCourseWare). http://ocw.mit.edu (accessed MM DD, YYYY). License: Creative Commons Attribution-Noncommercial-Share Alike. Note: Please use the actual date you accessed this material in your citation. For more information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/termsMIT OpenCourseWare http://ocw.mit.edu 8.02 Electricity and Magnetism, Spring 2002 Transcript – Lecture 3 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 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 …
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