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Review Exam 1 (Secret Top!)
Instructor/speaker: Prof. Walter Lewin
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You see here the topics the way I see them, you will get three problems to -- on the exam, and not all subjects can, of course, be represented on the exam.
Nor can I cover all of them in 50 minutes.
I will test some very basic ideas, the math will be utterly trivial, and if it becomes complicated, then you just know that you're on the wrong track.
If you get stuck, somehow, on a problem, my advice is, move on, don't stay with the problem, but move on and try some others first.
There is a reason why Gauss's Law there is in red, because Gauss's Law is, of course, extremely important in the early part of the course, the closed surface integral of E dot dA is the sum of the enclosed charge, divided by epsilon 0.
And that is so important that you can be sure that there will be one problem dealing with Gauss' Law.
Now, when you have Gauss' Law problems, there's always one of three.
You must have symmetry, you must have a special distribution of charges, because otherwise, Gauss' Law doesn't get you anywhere.
So we have spherical symmetry, we have cylindrical symmetry, and we have plane symmetry, and that's all there is.
So you're going to get one of those three.
I will do one now, you may choose.
We're going to have a vote.
One is a possibility, I do one on spherical symmetry, another one I do on cylindrical symmetry, or I do one on slab symmetry.
Who wants the spherical symmetry?
Hands.
Who wants the cylindrical symmetry?
Way more hands.
Who wants plane?
I think the cylinders have it.
But if you're clever, you can stay for the next lecture, and then you can try to get the other one.
We need a little bit of fun today.
And therefore, I want to introduce you first to something very special, which is close to my heart, it is a secret top, you're going to see it there, and that secret top, I'm going to spin, and if you're a believer in 8.01, which you should be by now, then we will -- should be able to predict that that stop cannot spin for very long, there is friction with the air and friction with the surface, and so chances are it will soon fall over.
We'll take a look at it later, again.
So let's now start our first problem, and that is a cylindrical symmetry.
Well, we have a cylinder -- and here is the cylinder -- it's very long, has radius R, and I have uniform charge distribution throughout the whole cylinder, and the density is rho coulombs per cubic meter.
Uniformly distributed through the cylinder.
I want to know what the electric field inside the cylinder is and outside the cylinder.
Let's first do outside the cylinder.
The gauss surface, clearly, is going to be itself a cylinder, there it goes -- you can give it any random length, L, cannot have any effect on the answer -- and so the end is flat, perpendicular to the axis of symmetry, and this front part is flat, and this is curved.
I give this a radius little r, and so I know that everywhere on the surface of that cylinder outside, that the electric field must be the same everywhere because the distance is the same, that's the symmetry argument.
Electric field cannot be any stronger here than it is there, if I'm on that surface.
Symmetry argument number one.
Symmetry argument number two is, given the fact that this is a cylinder, the electric field must everywhere be perpendicular to this axis, coming out -- I call it radially, but, of course, it is not radially, like a sphere -- it's radially coming out of this surface, always perpendicular to this axis of symmetry.
Nature could not decide any other way.
That's the second symmetry argument.
One you recognize that argument, the electric flux through this flat surface and through that flat surface must be 0.
Because then, the electric field and the local dA vector, which is the perpendicular to the surface, make angles of 90 degrees with each other, because E would be like this here, but dA is in the direction of the axis of symmetry.
So no flux can, therefore, get out here and get out here.
But only through this curved surface.
But on this curved surface, if it is a positive charge, then the E vector and the dA are in the same direction, if it is a negative charge, they are in opposite directions.
Later, you can change the sign of rho, let's just make it positive for now.
So if, now, I apply Gauss' Law, then I only have to take this surface into account and not these two end pieces.
And so I need to know, now, what this surface is, because E and dA are always in the same direction everywhere, thus the cosine of the angle between them is plus 1.
And so what is the surface area?
That is going to be L times 2 pi little r, and then the electric vector is everywhere, the same there, this was our symmetry argument, and that is now the charge inside this cylinder, divided by epsilon 0.
But, of course, the charge inside the cylinder, that's only the portion that is in this inner cylinder, and so that has also, then, length L.
The cross-section here is pi R squared, so this is the volume of the charge that I have inside my Gaussian surface, I must multiply by rho, that gives it a charge, and I divide by epsilon 0.
And of course, the L cancels, as it always does, and the pi cancels here, too, and so I find that the electric field equals R squared times rho divided by 2 epsilon 0 r, and if you want to see it vectorially, you can put an r roof there, r roof, then, would be a vector which is perpendicular to the axis -- I mentioned earlier, I called that radially outwards.
So this is the electric field outside the cylinder.
R squared rho 2 epsilon 0 r.
So it falls off as 1 over r.
So now I want to know what it is inside the cylinder.
So now I go to r, less than equal to R.
So it's clear that what I do now, I'm going to have a Gaussian surface which, again, is a cylinder, has length L, and it has, again, two flat pieces at the end, so no flux will go through those two pieces, so my first term of Gauss' Law is going to be the same, I have L times 2 pi little r, because the radius of this inner circle is also r, L 2 pi r times the electric field, the arguments are identical -- but now, there is less charge inside my Gaussian surface.
Uh, the volume is L, now times pi little r squared, and then I get rho to convert it to charge, divided by epsilon 0.
I lose my L, as I always do, my pi goes, and so now I get E equals -- there is an r here, and there is an r squared here, an so I only end up with one r, divided by 2 epsilon 0, and if you like that vector notation, you can always do this.
And of course, if rho were negative, then automatically, you see, if you put a negative charge density in here, then the E field flips over, so that's automatically taken into account both here and there.
So let's take a look at it, I'm quite happy with that.
If you substitute little r equals capital R, you are right at the surface of your cylinder, then you get the same answer in both cases.
Substitute R, capital R in here, then the magnitude of E -- don't worry about the direction now -- is rho capital R divided by 2 epsilon 0, and if you put in here for this little r, capital R, you find exactly the same answer.
So we can now make a plot of the electric field as a function of distance R, here being capital R, and here being the electric field strength.
It's a linear line, 0 here, it goes up to a certain maximum, and then it falls off as 1 over r.
And this value here is this value.
That's where little r is capital R.
It is obvious and pleasing that the electric field, on the axis itself, where little r is 0, that that electric field is 0, that is something that you could have predicted almost without any knowledge, because you have symmetry all around it, there is charge on the left, there is charge on the -- on the right, there's charge on north and south, and the electric fields right at the center, of course, all pair each other out, so you get no electric field right at the center.
If the charge, for some reason, would all be at the outer surface, if it were a solid conductor that would be the case, then the electric field would be 0 everywhere inside, and this would be unchanged, assuming then, that you have the same amount of charge on the outside per unit length as you now have on the inside.
So that is cylindrical symmetry.
I am dying to take a look at my top.
I am really -- and much to my shock, do I see that this top -- is still rotating.
So maybe I have to come to the conclusion that there is something wrong with 8.01.
.
There must be a layer deeper than 8.01 -- there is friction, and yet, this top doesn't come to a halt.
And so that layer deeper -- maybe that layer deeper is 8.02.
Give that some thought, it may add to your sleepless nights.
We'll visit it later, because maybe it will come to a stop.
Very well.
Let's now do something very different.
