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Kelistrikan dan Kemagnetan

Topics covered:

Review Exam 1 (Secret Top!)

Instructor/speaker: Prof. Walter Lewin

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» Download this transcript (PDF)

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.

Pengembangan Perkuliahan

1. Buatlah sebuah Esai mengenai materi perkuliahan ini

2. Buatlah sebuah kelompok berjumlah 5 orang untuk menganalisis materi perkuliahan ini

3. Lakukan Penelitian Sederhana dengan kelompok tersebut

4. Hasilkan sebuah produk yang dapat digunakan oleh masyarakat

5. Kembangkan produk tersebut dengan senantiasa meningkatkan kualitasnya

Ucapan Terima Kasih Kepada:

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

(http://web.mit.edu/physics/people/faculty/lewin_walter.html)

b. Prof. Bernd Surrow, Ph.D.

(http://web.mit.edu/physics/people/faculty/surrow_bernd.html)

Staff

Visualizations:
Prof. John Belcher

Instructors:
Dr. Peter Dourmashkin
Prof. Bruce Knuteson
Prof. Gunther Roland
Prof. Bolek Wyslouch
Dr. Brian Wecht
Prof. Eric Katsavounidis
Prof. Robert Simcoe
Prof. Joseph Formaggio

Course Co-Administrators:
Dr. Peter Dourmashkin
Prof. Robert Redwine

Technical Instructors:
Andy Neely
Matthew Strafuss

Course Material:
Dr. Peter Dourmashkin
Prof. Eric Hudson
Dr. Sen-Ben Liao

Acknowledgements

The TEAL project is supported by The Alex and Brit d'Arbeloff Fund for Excellence in MIT Education, MIT iCampus, the Davis Educational Foundation, the National Science Foundation, the Class of 1960 Endowment for Innovation in Education, the Class of 1951 Fund for Excellence in Education, the Class of 1955 Fund for Excellence in Teaching, and the Helena Foundation. Many people have contributed to the development of the course materials. (PDF)

2. Para Dosen Pendidikan Fisika, FPMIPA, Universitas Pendidikan Indonesia.

Terima Kasih Semoga Bermanfaat dan mohon Maaf apabila ada kesalahan.

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3. Photonic Crystals Molding the Flow of Light
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5. Silicon Photonics
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6. Coarse Wavelength Division Multiplexing
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7. Silicon Photonics
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8. Integrated Photonics Fundamentals
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9: Review of Lectures 1 through 5

» Download this transcript (PDF)

The exam on Wednesday will cover our first five lectures and the first two homework assignments.

And so I list here the topics the way we discussed them.

Of course, it is not possible to discuss all of them today but I will make a selection.

I recall that we discussed scaling and we used the interesting example of Galileo Galilei--

an animal, and the animal has legs.

And we defined the overall size of the animal as yea big--

we called that "s." And then we said, well, there is here the femur and the femur has length l and thickness d.

It was completely reasonable to say well... that l will have to be proportional to S.

If an animal is ten times larger than another its legs will be typically ten times longer.

Since the mass of the animal must be proportional to its size to the power three it will also be proportional to the length of the femur to the power three, and then came in this key argument--

namely, you don't want the bones to be crushed.

Which is called "yielding" in physics.

If I take a piece of concrete, a block of concrete, and I put too much pressure on it, it starts to crumble.

And that's what Galileo Galilei may have had in mind.

And in order to protect animals who get bigger and bigger and bigger against this crushing, we argued--

and I will not go through that argument now anymore--

that the mass will have to be proportional to d squared, which is the cross-section of the femur.

And so, you see immediately that d squared has to be proportional to l to the third so d must be proportional to the length of the femur to the power one and a half.

So this would mean that if you compare an elephant with a mouse the elephant's overall size is about 100 times larger than a mouse.

You would expect the femur to be about 100 times larger, which is true.

But you would then expect the femur to be about 1,000 times thicker and that turns out to be not true, as we have seen.

In fact, the femur of the elephant is only 100 times thicker, so it scales just as the size.

And the answer lies in the fact that nature doesn't have to protect against crumbling of the bones.

There is a much larger danger, which we call "buckling." And buckling is the phenomenon that the bones do this and if now you put too much pressure on it the bones will break.

And if that's the case, you remember that, in fact, all you have to do is you have to scale d proportional to l, which is not intuitive--

that's not so easy to derive--

but that's the case.

And so the danger, then, that nature protects animals against is this buckling, and when the buckling becomes too much then, I would imagine, the bones, at some point in time--

well, these are tough bones, aren't they?--

[snaps]

will break, and that's what nature tries to prevent.

So that was a scaling argument.

And let's now talk about dot products.

If I look there...

I scan it a little bit in a random way over my topics, so let's now talk about dot products.

I have a vector A...

Ax times x roof, which is the unit vector in the x direction, plus Ay y roof plus Az Z roof.

So these are the three unit vectors in the x, y and z direction.

And these are the x components, y and the z component of the vector A.

I have another vector, B.

B of x, x roof, B of y, y roof, B of z, z roof.

Now, the dot product...

A dot B--

also called the scalar product--

is the same as B dot A and it is defined as Ax Bx plus Ay By plus Az Bz.

And it's a number.

It is a scalar, it is a simple number.

And so this number can be larger than zero--

it can be positive--

it can be equal to zero, it can also be smaller than zero.

They're just dumb numbers.

Ucapan Terima Kasih Kepada:

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

(http://web.mit.edu/physics/people/faculty/lewin_walter.html)

b. Prof. Bernd Surrow, Ph.D.

(http://web.mit.edu/physics/people/faculty/surrow_bernd.html)

2. Para Dosen Pendidikan Fisika, FPMIPA, Universitas Pendidikan Indonesia.

Terima Kasih Semoga Bermanfaat dan mohon Maaf apabila ada kesalahan.

Pusat Penelitian Pendidikan Fisika Indonesia

Indonesian Physics Education Research Group

Visi

Fisika untuk Sekolah

Misi

Perbaikan Terus Menerus dalam bidang Pengajaran Fisika

Program

1. Sains Fisika untuk Anak-anak dan Sekolah Dasar

2. Fisika untuk Sekolah Menengah Pertama

3. Fisika untuk Sekolah Menengah Atas

External links

• Paradigms in Physics, Physics Today
• Paradigms in Physics, Oregon State University
• Modeling Instruction Program (Arizona State University)
• Matter and Interactions (Ruth Chabay and Bruce Sherwood)
• Socratic Dialogue Inducing Labs (Richard Hake)
• American Association of Physics Teachers
• Harvard - Mazur Group
• Physics Education Resources
• The Physics Teacher
• PER Central - a clearinghouse of physics education research and results
• University of Maryland Physics Education Research Group
• University of Colorado - Physics Education Research at Colorado
• University of Massachusetts-Amherst Physics Education Research Group
• University of Minnesota Physics Education Research and Development
• University of Washington Physics Education Group
• Kansas State University Physics Education Research Group
• Physics Education journal
• Rutgers Physics & Astronomy Education Research Group
• Florida International University (Physics Education Research Group)
• Problem Based Learning for College Physics (CCDMD)
• HyperPhysics site at Georgia State University
• Physics Education Research Group at the University of Illinois at Urbana-Champaign

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Categories: Physics education