Minggu, 10 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:

High-Voltage Breakdown
Sparks - St. Elmo's Fire

Instructor/speaker: Prof. Walter Lewin

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Last time I mentioned to you that charge resides at the surface of solid conductors but that it's not uniformly distributed.

Perhaps you remember that, unless it happens to be a sphere.

And I want to pursue that today.

If I had a solid conductor which say had this shape and I'm going to convince you today that right here, the surface charge density will be higher than there.

Because the curvature is stronger than it is here.

And the way I want to approach that is as follows.

Suppose I have here a solid conductor A which has radius R of A and very very far away, maybe tens of meters away, I have a solid conductor B with radius R of B and they are connected through a conducting wire.

That's essential.

If they are connected through a conducting wire, then it's equipotential.

They all have the same potential.

I'm going to charge them up until I get a charge distribution QA here and I get QB there.

The potential of A is about the same that it would be if B were not there.

Because B is so far away that if I come with some charge from infinity in my pocket that the work that I have to do to reach A per unit charge is independent of whether B is there or not, because B is far away, tens of meters, if you can make it a mile if you want to.

And so the potential of A is then the charge on A divided by 4 pi epsilon 0 the radius of A.

But since it is an equipotential because it's all conducting, this must be also the potential of the sphere B, and that is the charge on B divided by 4 pi epsilon 0 R of B.

And so you see immediately that the Q, the charge on B, divided by the radius of B, is the charge on A divided by the radius on A.

And if the radius of B were for instance 5 times larger than the radius of A, there would be 5 times more charge on B than there would be on A.

But if B has a 5 times larger radius, then its surface area is 25 times larger and since surface charge density, sigma, is the charge on a sphere divided by the surface area of the sphere, it is now clear that if the radius of B is 5 times larger than A, it's true that the charge on B is 5 times the charge on A, but the surface charge density on B is now only one-fifth of the surface charge density of A because its area is 25 times larger and so you have this -- the highest surface charge density at A than you have at B.

5 times higher surface charge density here than there.

And I hope that convinces you that if we have a solid conductor like this, even though it's not ideal as we have here with these two spheres far apart, that the surface charge density here will be larger than there because it has a smaller radius.

It's basically the same idea.

And so you expect the highest surface charge density where the curvature is the highest, smallest radius, and that means that also the electric field will be stronger there.

That follows immediately from Gauss's law.

If this is the surface of a conductor, any conductor, a solid conductor, where the E field is 0 inside of the conductor, and there is surface charge here, what I'm going to do is I'm going to make a Gaussian pillbox, this surface is parallel to the conductor, I go in the conductor, and this now is my Gaussian surface, let this area be capital A, and let's assume that it is positive charge so that the electric field lines come out of the surface like so, perpendicular to the surface.

Always perpendicular to equipotential, so now if I apply Gauss's law which tells me that the surface integral of the electric flux throughout this whole surface, well, there's only flux coming out of this surface here, I can bring that surface as close to the surface as I want to.

I can almost make it touch the conductor.

So everything comes out only through this surface, and so what comes out is the surface area A times the electric field E.

The A and E are in the same direction, because remember E is perpendicular to the surface of the equipotentials.

And so this is all there is for the surface integral, and that is all the charge inside, well the charge inside is of course the surface charge density times the area A, divided by epsilon 0, this is Gauss's law.

And so you find immediately that the electric field is sigma divided by epsilon 0.

So whenever you have a conductor if you know the local surface charge density you always know the local electric field.

And since the surface charge density is going to be the highest here, even though the whole thing is an equipotential, the electric field will also be higher here than it will be there.

I can demonstrate this to you in a very simple way.

I have here a cooking pan and the cooking pan, I used to boil lobsters in there, it's a large pan.

The cooking pan I'm going to charge up and the cooking pan here has a radius, whatever it is, maybe 20 centimeters, but look here at the handle, how very small this radius is, so you could put charge on there and I'm going to convince you that I can scoop off more charge here where the radius is small than I can scoop off here.

I have here a small flat spoon and I'm going to put the spoon here on the surface here and on the surface there and we're going to see from where we can scoop off the most charge.

Still charged from the previous lecture.

So here, we see the electroscope that we have seen before.

I'm going to charge this cooking pan with my favorite technique which is the electrophorus.

So we have the cat fur and we have the glass plate.

I'm going to rub this first with the cat fur, put it on, put my finger on, get a little shock, charge up the pan, put my finger on, get another shock, charge up the pan, and another one, charge up the pan, make sure that I get enough charge on there, rub the glass again, put it on top, put my finger on, charge, once more, and once more.

Let's assume we have enough charge on there now.

Here is my little spoon.

I touch here the outside here of the can -- of the pan.

And go to the electroscope and you see a little charge.

It's very clear.

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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