Jumat, 15 Oktober 2010

Fisika untuk Universitas

Fisika untuk Universitas

Ditujukan untuk meningkatkan kualitas proses dan hasil perkuliahan Fisika di tingkat Universitas

Kelistrikan dan Kemagnetan

Topics covered:

Field Energy

Instructor/speaker: Prof. Walter Lewin

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...assemble charges, I have to do work, we discussed that earlier.

And we call that electrostatic potential energy.

Today, I will look at this energy concept in a different way, and I will evaluate the energy in terms of the electric field.

Suppose I have two parallel plates, and I charge this one with positive charge, which is the surface charge density times the area of the plate, and this one, negative charge, which is the surface charge density negative times the area of the plate.

And let's assume that the separation between these two is h, and so we have an electric field, which is approximately constant, and the electric field here is sigma divided by epsilon 0.

And now, I'm going to take the upper plate, and I'm going to move it up.

And so as I do that, I have to apply a force, because these two plates attract each other, so I have to do work.

And as I move this up, and I will move it up over distance X, I am creating here, electric field that wasn't there before.

And the electric field that I'm creating has exactly the same strength as this, because the charge on the plates is not changing when I am moving, the surface charge density is not changing, all I do is, I increase the distance.

And so I am creating electric field in here.

And for that, I have to do work, that's another way of looking at it.

How much work do I have to do?

What is the work that Walter Lewin has to do in moving this plate over the distance X?

Well, that is the force that I have to apply over the distance X.

The force is constant, and so I can simply multiply the force times the distance, that will give me work.

And so the question now is, what is the force that I have to apply to move this plate up?

And your first guess would be that the force would be the charge on the plate times the electric field strength, a completely reasonable guess, because, you would argue, "Well, if we have an electric field E, and we bring a charge Q in there, then the electric force is Q times E, I have to overcome that force, so my force is Q times E." Yes, that holds most of the time.

But not in this case.

It's a little bit more subtle.

Let me take this plate here, and enlarge that plate.

So here is the plate.

So you see the thickness of the plate, now, this is one plate.

We all agree that the plus charge is at the surface, well, but, of course, it has to be in the plate.

And so there is here this layer of charge Q, which is at the bottom of the plate.

And the thickness of that layer may only be one atomic thickness.

But it's not 0.

And on this side of the plate, is that electric field, which is sigma divided by epsilon 0.

But inside the plate, which is a conductor, the electric field is 0.

And therefore, the electric field is, in this charge Q, is the average between the two.

And so the force on this charge, in this layer, is not Q times E, but is one-half Q times E.

So I take the average between these E fields, and this E field is then this value.

And so now I can calculate the work that I have to do, the work that I have to do is now my force, which is one-half Q times E, and I move that over a distance X.

And so what I can do now is replace Q by sigma A, so I get one-half sigma A times E times X, and I multiply upstairs and downstairs by epsilon 0, so that's multiplied by 1.

And the reason why I do that is, because then I get another sigma divided by epsilon 0 here -- divided by epsilon 0, and that is E, and therefore, I now have that the total work that I, Walter Lewin have to do -- has to do is one-half epsilon 0 E-squared times A times X.

And look at this.

A X is the new volume that I have created, it is the new volume in which I have created electric field.

And this, now, calls for a work done by Walter Lewin.

Per unit volume, and that, now, equals one-half epsilon 0 times E squared.

This is the work that I have done per unit volume.

And since this work created electric field, we called it "field energy density".

And it is in joules per cubic meter.

And it can be shown that, in general, the electric field energy density is one-half epsilon 0 E squared, not only for this particular charge configuration, but for any charge configuration.

And so, now, we have a new way of looking at the energy that it takes to assemble charges.

Earlier, we calculated the work that we have to do to put the charges in place, now, if it is more convenient, we could calculate that the electrostatic potential energy, is the integral of one-half epsilon 0 E-squared, over all space -- if necessary, you have to go all the way down to infinity -- and here, I have now, dV, this is volume.

This has nothing to do with potential, this V, in physics, we often run out of symbols, V is sometimes potential, in this case, it is volume.

And the only reason why I chose H there is I already have a D here, so I didn't want 2 Ds.

Normally, we take D as the separation between plates.

And so this, now, is another way of looking at electrostatic potential energy.

We look at it now only from the point of view of all the energy being in the electric field, and we no longer think of it, perhaps, as the work that you have done to assemble these charges.

I will demonstrate later today that as I separate the two plates from these charged planes, that indeed, I have to do work.

I will convince you that by creating electric fields that, indeed, I will be doing work.

So, from now on, uh, we have the choice.

If you want to calculate what the electrostatic potential energy is, you either calculate the work that you have to do to bring all these charges in place, or, if it is easier, you can take the electric field everywhere in space, if you know that, and do an integration over all space.

We could do that, for instance, for these two parallel plates now, and we can ask what is now the total energy in these plates -- uh, in the field.

And at home, I would advise you, to do that the way that it's done in your book, whereby you actually assemble the charges minus Q at the bottom and plus Q at the top, and you calculate how much work you have to do.

That's one approach.

I will now choose the other approach, and that is, by simply saying that the total energy in the field of these plane-parallel plates, is the integral of one-half epsilon 0 E-squared, over the entire volume of these two plates.

And since the electric field is outside 0, everywhere, it's a very easy integral, because I know the volume.

The volume that I have, if the separation is H -- so we still have them H apart -- this volume that I have is simply A times H, and the electric field is constant, and so I get here that this is one-half epsilon 0.

For E, if I want to, I can write sigma divided by epsilon 0, I can square that, and dV, in doing the integral over all space, means simply I get A times H, it is the volume of that box.

So I get A times H.

And so this is now the total energy that I have, I lose one epsilon here, I have an epsilon 0 squared and I have an epsilon.

I also remember that the charge Q on the plate is A times sigma, and that the potential difference V, this now is not volume, it's the potential difference between the plates, is the electric field times H.

The electric field is constant, it can go from one plate to the other, the integral E dot dL in going from one plate to the other, gives me the potential difference.

And so I can substitute that now in here, I can take for A, sigma, I can put in the Q, and you can also show that this is one-half Q V.

V being, now, the potential difference between the plates.

And so this is a rather fast way that you can calculate what the total energy is in the field, or, say, the same thing, the total work you have to do to assemble these charges.

Or, to say it differently, the total work you have to do to create electric fields.

You have created electric fields that were not there before.

I now will introduce something that we haven't had before, that is the word "capacitance".

I will define the capacitance of an object to be the charge of that object divided by the potential of that object.

And so the unit is coulombs per volt, this V is volt, now, it's potential.

Uh, but we never say that it is coulombs per volt in physics, we write for that a capital F, which is Farad, we call that, 1 farad is the unit of capacitance, undoubtedly called after the great maestro Faraday, we will learn more about Faraday later in this course.

So let us go to, um, a sphere which has a radius R, and let us calculate what the capacitance is of this sphere.

Think of it as being a conductor, and we bring a certain charge Q on this conductor, it will then get a potential V, which we know is Q divided by 4 pi epsilon 0 R.

We've seen this many times, and so, by definition, the capacitance now is Q divided by the potential, and therefore, this becomes 4 pi epsilon 0 R.

So that is the capacitance of a single sphere.

And so we can now look at the values as a function of R.

I have here some numbers, I calculated it for the Van de Graaff, and I calculated it for the Earth.

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.

b. Prof. Bernd Surrow, Ph.D.


Prof. John Belcher

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


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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