## Minggu, 26 September 2010

### Fisika untuk Universitas

Fisika untuk Universitas

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

# 5: Uniform Circular Motion

Today we will discuss what we call "uniform circular motion." What is uniform circular motion? An object goes around in a circle, has radius r and the object is here.This is the velocity.It's a vector, perpendicular. And later in time when the object is here the velocity has changed, but the speed has not changed. We introduce T, what we call the period-- of course it's in seconds-- which is the time to go around once.

We introduce the frequency, f, which we call the frequency which is the number of rotations per second. And so the units are either seconds minus one or, as most physicists will call it, "hertz" and so frequency is one divided by T. We also introduce angular velocity, omega which we call angular velocity.

Angular velocity means not how many meters per second but how many radians per second. So since there are two pi radians in one circumference-- in one full circle-- and it takes T seconds to go around once it is immediately obvious that omega equals two pi divided by T. This is something that I would like you to remember. Omega equals two pi divided by T-- two pi radians in capital T seconds.

The speed, v, is, of course, the circumference two pi r divided by the time to go around once but since two pi divided by T is omega you can also write for this "omega r." And this is also something that I want you to remember.

These two things you really want to remember. The speed is not changing, but the velocity vector is changing. Therefore there must be an acceleration. That is non-negotiable. You can derive what that acceleration must be in terms of magnitude and in terms of direction. It's about a five, six minutes derivation. You'll find it in your book. I have decided to give you the results so that you read up on the book so that we can more talk about the physics rather than on the derivation.

This acceleration that is necessary to make the change in the velocity vector is always pointing towards the center of the circle. We call it "centripetal acceleration." Centripetal, pointing towards the center. And here, also pointing towards the center. It's a vector.

And the magnitude of the centripetal acceleration equals v squared divided by r, which is this v and therefore it's also omega squared r. And so now we have three equations and those are the only three you really would like to remember. We can have a simple example.

Let's have a vacuum cleaner, which has a rotor inside which scoops the air out or in, whichever way you look at it. And let's assume that the vacuum cleaner these scoops have a radius r of about ten centimeters and that it goes around 600 revolutions per minute, 600 rpm.

600 rpm would translate into a frequency, f, of 10 Hz so it would translate into a period going around in one-tenth of a second. So omega, angular velocity, which is two pi divided by T is then approximately 63 radians per second and the speed, v, equals omega r is then roughly 6.3 meters per second.

The centripetal acceleration--

and that's really my goal--

the centripetal acceleration would be omega squared r or if you prefer, you can take v squared over r. You will get the same answer, of course, and you will find that that is about 400 meters per second squared.

And that is huge.

That is 40 times the acceleration due to gravity. It's a phenomenal acceleration, the simple vacuum cleaner. Notice that the acceleration, the centripetal acceleration is linear in r. Don't think that it is inversely proportional with r. That's a mistake, because v itself is a function of r.

If you were sitting here then your velocity would be lower.

Since omega is the same for the entire motion you really have to look at this equation and you see that the centripetal acceleration is proportional with r.

Therefore, if you were...

if this were a disc which was rotating and you were at the center of the disc the centripetal acceleration would be zero. And as you were to walk out, further out, it would increase. Now, the acceleration must be caused by something.

There is no such thing as a free lunch.

There is something that must be responsible for the change in this velocity and that something I will call either a pull or I will call it a push. In our next lecture, when we deal with Newton's laws we will introduce the word "force." Today we will only deal with the words "pull" and "push." So there must be a pull or a push. Imagine that this is a turntable and you are sitting here on the turntable on a chair. It's going around with angular velocity omega and your distance to the center, let's say, is little r.

You're sitting on this chair and you must experience--

that is non-negotiable--

centripetal acceleration A of c, which is omega squared times r.

Where do you get it from? Well, if your seat is bolted to the turntable then you will feel a push in your back so you're sitting on this thing, you're going around and you will feel that the seat is pushing you in your back and so you feel a push, and that gives the push out.

Yeah, I can give this a red color for now.

So you feel a push in your back.

That push, apparently, is necessary for the acceleration.

Alternatively, suppose you had in front of you a stick.

You're not sitting on a chair.

You don't get a push from your back.

But you hold onto the stick and now you can go around by holding onto the stick.

Now the stick is pulling on you in this same direction.

So now you would say, aha, someone is pulling on you.

Whether it is the pull or whether it is the push one of... either one of the two is necessary for you to go around in that circle on that turntable with that constant speed.

Now, the classic question comes up, which we often ask to people who have no scientific background.

If you were to go around like this and something is either pushing on you or is pulling on you to make this possible suppose you took that push out, all of a sudden.

The pull is gone.

[makes whooshing sound]

What is now the motion of the person who is sitting on the turntable? And many non-scientists say, "Well, it will do like this." That's sort of what your intuition says.

You go around in a circle, and all of a sudden you no longer have the pull or the push and you go around in a spiral and obviously, that is not the case.

What will happen is, if you have, at this moment in time a velocity in this direction and you take the pull or the push out you will start flying off in that direction and depending upon whether there is gravity or no gravity there may be a change, but if this were...

if there were no gravity you would just continue to go along that line and you would not make this crazy spiral motion.

I have here a disc, which we will rotate and at the end...

the edge of the disc here we have a little ball.

And the ball is attached to that disc with string.

So now this is vertical, and so this is going to go around with angular velocity omega.

And we have a string here and the string is attached to this ball and the whole thing is going around and so at one moment in time this has a velocity, like so.

And therefore there must be non-negotiable centripetal acceleration which in magnitude is omega squared r or, if you want to, v squared divided by r.

Now I cut it and that's like taking away the push and the pull. The string that you have here is providing the pull on this ball. This ball is feeling a pull from the string and that provides it with the centripetal acceleration. Cut the string and the pull is gone and the object will take off. And if there were gravity here, as there is in 26.100 it would become a parabola and it would end up here. If, however, I cut the ball exactly when it is here--

not the ball, but I cut the string--

then, of course, it would fly straight up gravity would act on it, it would come to a halt and it would come back. So it really would then go along a straight line. But you would clearly see, then that it's not going to do what many people think-- that it would start to swirl around.

It would just go...

[makes whooshing sound]

and comes back. Let's look at that. We have that here. So here is that ball. The string is behind here; you cannot see the string. I will rotate it, wait for it to pick up a little speed and the knife, that you can't see either, is behind here and when I push the knife in, I do it exactly here.

It cuts the string and it goes up. You ready for this? You sure you're ready? Three, two, one, zero. Wow! That was very high. So you see, it's nothing like this. It simply continued on in the direction that it was going. It wasn't going into a parabola because I was shooting it straight up. The string forms the connection between the rotating disc and the ball and therefore, the pull is responsible for the centripetal acceleration.

Planets go around the sun.

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

b. Prof. Bernd Surrow, Ph.D.

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

Terima Kasih Semoga Bermanfaat dan mohon Maaf apabila ada kesalahan.

## Sabtu, 25 September 2010

### Ikatan Keluarga Alumni Fisika UPI 07

IKA Fisika UPI 07

Visi

Misi

Program

Silaturahmi

## Minggu, 19 September 2010

### PUSTAKA FISIKA (PF)

1. Sebuah Visi Pengumpulan 100.000 Buah Buku yang terkait dg Fisika
2. Pengumpulan Data-Data Kefisikaan sebesar 1 Terra byte

Tempat Pengumpulan dan Pendataan Buku-buku fisika via Internet

Fisika Kuantum

• Introduction to Quantum Theory and Atomic Struncure
• New Developments in Quantum Field Theory
• Quantum Mechanics
• Quantum Physics
• Introduction to Quantum Mechanics
• Theory and Problems of Modern Physics
• Introduction to the Theory and Applications of Molecular and Quantum Mechanics

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Media Saling Berbagi Ilmu dan Informasi

## Sabtu, 18 September 2010

### Fisika untuk Universitas

Fisika untuk Universitas

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

# 4: The Motion of Projectiles

Today, we're going to take it quite easy.

I also have to take it a little easy because my voice may be petering out, if I'm not careful.

We're going to apply today what we have learned, so there is nothing new but its applications.

And that's important--

things that... you can let it sink in.

We have here a trajectory of a golf ball or a tennis ball in 26.100.

We shoot it up at an angle alpha.

The horizontal component in the x direction is v zero cosine alpha and the vertical component is v zero sine alpha.

It reaches the highest point at P and it returns to the ground at point S.

This is the increasing y direction and this is the increasing x direction.

We're going to use, very heavily, the equations that you see here that are so familiar with us.

These are the one-dimensional equations in x direction where there is no acceleration and the one-dimensional equations in the y direction where there is acceleration.

In order to use these equations we need all these constants--

x zero, v zero x and v zero y.

We have seen those last time.

I choose for x zero...

I choose zero arbitrarily.

Also for y zero.

The velocity in the x direction will never change.

This v zero x will always remain v zero cosine alpha.

The velocity in the y direction, however, in the beginning at t equals zero is v zero sine alpha.

And that one will change, because there is here this t and that's why the velocity is going to change.

This t will do it.

And the acceleration in the y direction--

since this is increasing value of y--

is going to be negative 9.8.

Since I call always 9.8 plus...

since I always call g "plus 9.8," this is minus g.

I now want to ask first the question that you may never have seen answered: what is the shape of this? Well, we can go to equation number three there and we can write down this equation number three: That y, as a function of time, equals v zero yt so it is v zero sine alpha times t minus one-half gt squared.

That's the equation in y.

I go to equation number one and I write down x--

at any moment in time--

equals v zero z times t so that is v zero cosine alpha times t.

Now I eliminate t, and the best way to do that is to do it here--

to write for t, x divided by v zero cosine alpha.

Now I can drop all subindexes t because we're now going to see x versus y.

We're going to eliminate t.

So this time here, I'm going to substitute in here and in there and so I'm going to get y equals...

There's a v zero here and there's a v zero there that cancels.

There's a sine alpha here and a cosine alpha there that makes it a tangent of alpha.

And then I have here the x and I get minus one-half g times this squared--

x squared divided by v zero cosine alpha squared.

And now look very carefully.

Y is a constant times x minus another constant times x squared.

That is a parabola.

It's a second-order equation in x, and is a parabola and a parabola has this shape.

So you so see now, by eliminating the time that we have here a parabola.

Now I want to massage this quite a bit further today.

I would like to know at what time the object here comes to a halt to its highest point.

It comes to a halt in the y direction.

It comes to a highest point and I want to know how high that is.

Well, the best way to do is to go to equation four and you say, to equation four, "When are you zero?" Because that is the moment that the velocity in the y direction becomes zero.

It must be at its highest point, then.

So in order to find, for us, the position of the highest point P we first ask ourselves the question from equation number four: when is the velocity in the y direction zero? And that then becomes v zero y which is v zero sine alpha minus gt and out pops that t at point P is going to be v zero sine alpha divided by g.

That's the time that it takes for the object to reach the highest point.

Where is it, then? What is the highest point above the ground? Well, now we have to go to equation number three and you have to substitute this time in there so that highest point h, which is y at the time t of P equals v zero yt--

that is v zero times the sine of alpha.

But you have to multiply it by this time and so I get another v zero sine alpha and I get a g here minus gt mi...

oh, no, no, this equa...

minus half dt squared minus one-half g times this one squared which is v zero sine alpha squared, divided by g...

divided by g squared because there is a g here, you see? So you square the whole thing if it's t squared.

You lose one g and you will find, then, that the highest point--

let's write it down here so that we don't block that blackboard--

the highest point in the sky equals v zero sine alpha squared divided by 2g.

That is the highest point.

Let's give that some color because we may want to keep that.

Is it reasonable that the point, the highest point in the sky gets higher when v zero is higher? Of course.

If I shoot it up at a higher speed of course it will get higher.

So that's completely intuitive that v zero is upstairs.

If I increase the angle from a small angle to larger and larger and larger is it reasonable that it will get higher? Of course.

You all feel in your stomach that the highest possible value you can get is when you make alpha 90 degrees for a given velocity.

That's the highest it will go in the sky.

So clearly, this is also very pleasing.

If you did the experiment on the moon with the same initial speed, it will go much higher so you are also happy to see that this g here is downstairs.

So that makes sense.

At what time will the object be at point S? Now, there are two ways that you can do that.

You either go to this equation, number three and you ask equation number three, "When are you zero?" It will give you two answers.

It will say, "I am zero here at this time" and "I am zero at that time." And those are the two times that you want and this is the one you pick.

That's perfectly fine.

I think there's a faster way to do it, and that's the following.

This is a parabola, so it's completely symmetric about the vertical, about P.

So to climb up from O to P must take the same amount of time as to go down from P to S and so I claim that the time to reach point S must be twice the time to reach point P and therefore it's going to be two v zero sine alpha divided by g.

But now we want to look again whether the v zeros and the sine alphas have the right place.

Indeed, if I increase the speed, I would expect it to take longer before it reaches S.

If I give it a larger speed, it will come out farther and obviously, the time will take longer.

If I do it at a higher angle, it will also take longer and if I do it on the moon, it will also take longer.

So this makes sense--

these equations are pleasing in terms of the rate that v zero and sine alpha appear in the equations.

But now comes an important point which I am going to use throughout this lecture.

I want to know what OS is.

The distance OS... I shoot it up and it hits the floor again What is that distance that it travels? Well, for that, I need equation number one.

It is v zero x times the time and v zero x is v zero cosine alpha.

We got v zero cosine alpha times the time to hit it--

that is two v zero sine alpha.

So I get a two here, I get a sine alpha, and I get a g here and I have another v zero there, and so the answer is a v zero squared times the sine of the double angle--

remember, two cosine alpha sine alpha is the sine of two alpha--

divided by g.

And this is OS, and I'm going to need this a lot.

This reminds me not to remove it.

Now, I sort of wonder, and you should too why is it that the highest point in the sky has a v zero squared and why is the farthest point also...

why does it also have a v zero squared? There must be a way that you can reason that.

Why is it not just v zero? Why is it v zero squared? Well, I'll let you argue about the highest points, and I'll give you a good reason for the distance, OS.

Don't look at the equations.

You simply...

Think for a change.

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

b. Prof. Bernd Surrow, Ph.D.

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

Terima Kasih Semoga Bermanfaat dan mohon Maaf apabila ada kesalahan.

## Jumat, 17 September 2010

### Belajar itu Asyik

Berikut ini adalah tips dan triks yang dapat menjadi masukan berharga dalam mempersiapkan diri dalam menghadapi ulangan atau ujian :

1. Belajar Kelompok

Belajar kelompok dapat menjadi kegiatan belajar menjadi lebih menyenangkan karena ditemani oleh teman dan berada di rumah sendiri sehingga dapat lebih santai. Namun sebaiknya tetap didampingi oleh orang dewasa seperti kakak, paman, bibi atau orang tua agar belajar tidak berubah menjadi bermain. Belajar kelompok ada baiknya mengajak teman yang pandai dan rajin belajar agar yang tidak pandai jadi ketularan pintar. Dalam belajar kelompok kegiatannya adalah membahas pelajaran yang belum dipahami oleh semua atau sebagian kelompok belajar baik yang sudah dijelaskan guru maupun belum dijelaskan guru.

2. Rajin Membuat Catatan Intisari Pelajaran

Bagian-bagian penting dari pelajaran sebaiknya dibuat catatan di kertas atau buku kecil yang dapat dibawa kemana-mana sehingga dapat dibaca di mana pun kita berada. Namun catatan tersebut jangan dijadikan media mencontek karena dapat merugikan kita sendiri.

3. Membuat Perencanaan Yang Baik

Untuk mencapai suatu tujuan biasanya diiringi oleh rencana yang baik. Oleh karena itu ada baiknya kita membuat rencana belajar dan rencana pencapaian nilai untuk mengetahui apakah kegiatan belajar yang kita lakukan telah maksimal atau perlu ditingkatkan. Sesuaikan target pencapaian dengan kemampuan yang kita miliki. Jangan menargetkan yang yang nomor satu jika saat ini kita masih di luar 10 besar di kelas. Buat rencana belajar yang diprioritaskan pada mata pelajaran yang lemah. Buatlah jadwal belajar yang baik.

4. Disiplin Dalam Belajar

Apabila kita telah membuat jadwal belajar maka harus dijalankan dengan baik. Contohnya seperti belajar tepat waktu dan serius tidak sambil main-main dengan konsentrasi penuh. Jika waktu makan, mandi, ibadah, dan sebagainya telah tiba maka jangan ditunda-tunda lagi. Lanjutkan belajar setelah melakukan kegiatan tersebut jika waktu belajar belum usai. Bermain dengan teman atau game dapat merusak konsentrasi belajar. Sebaiknya kegiatan bermain juga dijadwalkan dengan waktu yang cukup panjang namun tidak melelahkan jika dilakukan sebelum waktu belajar. Jika bermain video game sebaiknya pilih game yang mendidik dan tidak menimbulkan rasa penasaran yang tinggi ataupun rasa kekesalan yang tinggi jika kalah.

5. Menjadi Aktif Bertanya dan Ditanya

Jika ada hal yang belum jelas, maka tanyakan kepada guru, teman atau orang tua. Jika kita bertanya biasanya kita akan ingat jawabannya. Jika bertanya, bertanyalah secukupnya dan jangan bersifat menguji orang yang kita tanya. Tawarkanlah pada teman untuk bertanya kepada kita hal-hal yang belum dia pahami. Semakin banyak ditanya maka kita dapat semakin ingat dengan jawaban dan apabila kita juga tidak tahu jawaban yang benar, maka kita dapat membahasnya bersama-sama dengan teman. Selain itu

6. Belajar Dengan Serius dan Tekun

Ketika belajar di kelas dengarkan dan catat apa yang guru jelaskan. Catat yang penting karena bisa saja hal tersebut tidak ada di buku dan nanti akan keluar saat ulangan atau ujian. Ketika waktu luang baca kembali catatan yang telah dibuat tadi dan hapalkan sambil dimengerti. Jika kita sudah merasa mantap dengan suatu pelajaran maka ujilah diri sendiri dengan soal-soal. Setelah soal dikerjakan periksa jawaban dengan kunci jawaban. Pelajari kembali soal-soal yang salah dijawab.

7. Hindari Belajar Berlebihan

Jika waktu ujian atau ulangan sudah dekat biasanya kita akan panik jika belum siap. Jalan pintas yang sering dilakukan oleh pelajar yang belum siap adalah dengan belajar hingga larut malam / begadang atau membuat contekan. Sebaiknya ketika akan ujian tetap tidur tepat waktu karena jika bergadang semalaman akan membawa dampak yang buruk bagi kesehatan, terutama bagi anak-anak.

8. Jujur Dalam Mengerjakan Ulangan Dan Ujian

Hindari mencontek ketika sedang mengerjakan soal ulangan atau ujian. Mencontek dapat membuat sifat kita curang dan pembohong. Kebohongan bagaimanapun juga tidak dapat ditutup-tutupi terus-menerus dan cenderung untuk melakukan kebohongan selanjutnya untuk menutupi kebohongan selanjutnya. Anggaplah dengan nyontek pasti akan ketahuan guru dan memiliki masa depan sebagai penjahat apabila kita melakukan kecurangan.

Semoga tips cara belajar yang benar ini dapat memberikan manfaat untuk kita semua, amin.

Sources: Godam

## Jumat, 10 September 2010

### Fisika untuk Universitas

Fisika untuk Universitas

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

Kelistrikan dan Kemagnetan

Topics covered:

Electrostatic Potential
Electric Energy
eV
Conservative Field
Equipotential Surfaces

Instructor/speaker: Prof. Walter Lewin

### Video

We're going to talk about, again, some new concepts.

And that's the concept of electrostatic potential...

electrostatic potential energy.

For which we will use the symbol U and independently electric potential.

Which is very different, for which we will use the symbol V.

Imagine that I have a charge Q one here and that's plus, plus charge, and here I have a charge plus Q two and they have a distant, they're a distance R apart.

And that is point P.

It's very clear that in order to bring these charges at this distance from each other I had to do work to bring them there because they repel each other.

It's like pushing in a spring.

If you release the spring you get the energy back.

If they were -- they were connected with a little string, the string would be stretched, take scissors, cut the string they fly apart again.

So I have put work in there and that's what we call the electrostatic potential energy.

So let's work this out in some detail how much work I have to do.

Well, we first put Q one here, if space is empty, this doesn't take any work to place Q one here.

But now I come from very far away, we always think of it as infinitely far away, of course that's a little bit of exaggeration, and we bring this charge Q two from infinity to that point P.

And I, Walter Lewin, have to do work.

I have to push and push and push and the closer I get the harder I have to push and finally I reach that point P.

Suppose I am here and this separation is little R.

I've reached that point.

Then the force on me, the electric force, is outwards.

And so I have to overcome that force and so my force, F Walter Lewin, is in this direction.

And so you can see I do positive work, the force and the direction in which I'm moving are in the same direction.

I do positive work.

Now, the work that I do could be calculated.

The work that Walter Lewin is doing in going all the way from infinity to that location P is the integral going from in- infinity to radius R of the force of Walter Lewin dot dR.

But of course that work is exactly the same, either one is fine, to take the electric force in going from R to infinity dot dR.

Because the force, the electric force, and Walter Lewin's force are the same in magnitude but opposite direction, and so by flipping over, going from infinity to R, to R to infinity, this is the same.

This is one and the same thing.

Let's calculate this integral because that's a little easy.

We know what the electric force is, Coulomb's law, it's repelling, so the force and dR are now in the same direction, so the angle theta between them is zero, so the cosine of theta is one, so we can forget about all the vectors, and so we would get then that this equals Q one, Q two, divided by four pi epsilon zero.

And now I have downstairs here an R squared.

And so I have the integral now dR divided by R squared, from capital R to infinity.

And this integral is minus one over R.

Which I have to evaluate between R and infinity.

And when I do that that becomes plus one over capital R.

Right, the integral of dR over R squared I'm sure you can all do that is minus one over R.

I evaluate it between R and infinity and so you get plus one over R.

And so U, which is the energy that -- the work that I have to do to bring this charge at that position, that U is now Q one, times Q two divided by four pi epsilon zero.

Divided by that capital R.

And this of course this is scalar, that is work, it's a number of joules.

If Q one and Q two are both positive or both negative, I do positive work, you can see that, minus times minus is plus.

Because then they repel each other.

If one is positive and the other is negative, then I do negative work, and you see that that comes out as a sign sensitive, minus times plus is minus.

So I can do negative work.

If the two don't have the same polarity.

I want you to convince yourself that if I didn't come along a straight line from all the way from infinity, but I came in a very crooked way, finally ended up at point P, at that point, that the amount of work that I had to do is exactly the same.

You see the parallel with eight o one where we dealt with gravity.

Gravity is a conservative force and when you deal with conservative forces, the work that has to be done in going from one point to the other is independent of the path.

That is the definition of conservative force.

Electric forces are also conservative.

And so it doesn't make any difference whether I come along a straight line to this point or whether I do that in an extremely crooked way and finally end up here.

That's the same amount of work.

Now if we do have a collection of charges, so we have pluses and minus charges, some pluses, some minus, some pluses, minus, pluses, pluses, then you now can calculate the amount of work that I, Walter Lewin, have to do in assembling that.

You bring one from infinity to here, another one, another one, and you add up all that work, some work may be positive, some work may be negative.

Finally you arrive at the total amount of work that you have to do to assemble these charges.

And that is the meaning of capital U.

Now I turn to electric potential.

And for that I start off here with a charge which I now call plus capital Q.

It's located here.

And at a position P at a distance R away I place a test charge plus Q.

Make it positive for now, you can change it later to become a negative.

And so the electrostatic potential energy we -- we know already, we just calculated it, that would be Q times Q divided by four pi epsilon zero R.

That's exactly the same that we have.

So the electric potential, electrostatic potential energy, is the work that I have to do to bring this charge here.

Now I'm going to introduce electric potential.

Electric potential.

And that is the work per unit charge that I have to do to go from infinity to that position.

So Q doesn't enter into it anymore.

It is the work per unit charge to go from infinity to that location P.

And so if it is the work per unit charge, that means little Q disappears.

And so now we write down that V at that location P.

The potential, electric potential at that location P, is now only Q divided four pi epsilon zero R.

Little Q has disappeared.

It is also a scalar.

This has unit joules.

The units here is joules per coulombs.

I have divided out one charge.

It's work per unit charge.

No one would ever call this joules per coulombs.

We call this volts, called after the great Volta, who did a lot of research on this.

So we call this volts.

But it's the same as joules per coulombs.

If we have a very simple situation like we have here, that we only have one charge, then this is the potential anywhere, at any distance you want, from this charge.

If R goes up, if you're further away, the potential will become lower.

If this Q is positive, the potential is everywhere in space positive for a single charge.

If this Q is negative, everywhere in space the potential is negative.

Electro- electric static potential can be negative.

The work that I do per unit charge coming from infinity would be negative, if that's a negative charge.

And the potential when I'm infinitely far away, when this R becomes infinitely large, is zero.

So that's the way we define our zero.

So you can have positive potentials, near positive charge, negative potentials, near negative charge, and if you're very very far away, then potential is zero.

Let's now turn to our Vandegraaff.

It's a hollow sphere, has a radius R.

And I'm going to put on here plus ten microcoulombs.

It will distribute itself uniformly.

We will discuss that next time in detail.

Because it's a conductor.

We already discussed last lecture that the electric field inside the sphere is zero.

And that the electric field outside is not zero but that we can think of all the charge being at this point here, the plus ten microcoulombs is all here, as long as we want to know what the electric field outside is.

So you can forget the fact that it is a -- a sphere.

And so now I want to know what the electric potential is at any point in space.

I want to know what it is here and I want to know what it is here at point P, which is now a distance R from the center.

And I want to know what it is here.

At a distance little R from the center.

So let's first do the potential here.

The potential at point P is an integral going from R to infinity if I take the electric force divided by my test charge Q dot dR.

But this is the electric field, see, this force times distance is work, but it is work per unit charge, so I take my test charge out.

And so this is the integral in R to infinity of E dot dL -- dR, sorry.

And that's a very easy integral.

Because we know what E is.

The electric field we have done several times.

Follows immediately from Coulomb's law and so when you calculate this integral you get Q divided by four pi epsilon zero R which is no surprise because we already had that for a point charge.

So this is the situation if r, little r, is larger than capital R.

We can put in some numbers.

If you put in R equals R, which is ho point three meters, and you put in here the ten microcoulombs, and here the -- the thirty centimeters, then you'll find three hundred thousand volts.

So you get three times ten to the fifth volts.

If you take r equals sixty centimeters, you double it, if you double the distance, the potential goes down by a factor of two, it's one over R, so it would be a hundred and fifty kilovolts.

And if you go to three meters, then it is ten times smaller, then it is thirty kilovolts.

And if you go to infinity which for all practical purposes would be Lobby seven, if you go to Lobby seven, then the potential for all practical purposes is about zero.

Because R is so large that there is no potential left.

So if I, if I, Walter Lewin, march from infinity to this surface of the Vandegraaff, and I put a charge Q in my pocket, and I march to the Vandegraaff, by the time I reach that point I have done work, I multiply the charge now back to the potential, that gives you the work again, because potential was work per unit charge, and so the work that I have done then is the charge that I have in my pocket times the potential, in this case the potential of the Vandegraaff.

If I go all the way to this surface, which is three hundred thousand volts.

If I were a strong man then I would put one coulomb in my pocket.

That's a lot of charge.

Then I would have done three hundred thousand joules of work.

By just carrying the one coulomb from Lobby seven to the Vandegraaff.

That's about the same work I have to do to climb up the Empire State Building.

The famous MGH, my mass times G times the height that I have to climb.

So I know how the electric potential goes with distance.

It's a one over R relationship.

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

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

b. Prof. Bernd Surrow, Ph.D.

#### 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

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.

## Rabu, 01 September 2010

### Fisika untuk Universitas

Fisika untuk Universitas

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

Kelistrikan dan Kemagnetan

Topics covered:

Electric Flux
Gauss's Law
Examples

Instructor/speaker: Prof. Walter Lewin

### Video

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.

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

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

b. Prof. Bernd Surrow, Ph.D.

#### 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

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.