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  • Prof: Okay, normally I would ask in a small

  • class if there is something you didn't follow from last time.

  • I'm afraid to do that now, because it's a big class and I

  • don't know how many things you follow or didn't follow.

  • What I will do first is write down a very quick summary of the

  • main points from last time.

  • So you should ask yourself, "Did I follow all those

  • things?"

  • And if your answer is yes, then you are fine.

  • Because I talked about many, many things,

  • but you don't need all of that.

  • So what I'm going to write down right now is the absolute

  • essentials of last lecture, okay?

  • That's going to be needed for what I do next.

  • First point was: everything is made of atoms.

  • You know that.

  • And the atom has a nucleus.

  • In the nucleus are some things called protons,

  • some things called neutrons, and outside are some things

  • called electrons.

  • That's all the atomic structure we need.

  • Then we say certain entities have a property called electric

  • charge.

  • The symbol for electric charge is q, and you can put a

  • subscript to say who you are talking about.

  • So you can say q for the neutron is 0.

  • A q for the electron is -1.6 times 10 to the -19,

  • and it's measured in coulombs.

  • The q for the proton is really--let me put it this way,

  • q for the proton is a positive number,

  • so these minus signs cancel.

  • Now, the importance of the coulomb is that if anything has

  • some coulombs on it, it will interact with anything

  • else that has some coulombs on it.

  • So that if you have two entities and this one has a

  • charge of q_1 coulombs,

  • that one has a charge of q_2 coulombs,

  • and the distance between them is r,

  • then the force is q_1q

  • _2_ over 4Πε

  • _0r^(2).

  • I'm purposely not putting all the vector signs on F

  • because it takes too long, but you all know what the

  • answer is.

  • Namely, if you want the force on 2,

  • due to 1, will be repulsive if q_1 and

  • q_2 are of the same sign and point to the

  • direction joining them.

  • Yep?

  • Student: Shouldn't it be 1 over r^(2)?

  • Prof: Yes, thank you.

  • There is another force law which is r^(2),

  • not so famous, called Hooke's Law,

  • but you're absolutely right.

  • That's the difference between being Newton and being Hooke.

  • Hooke is known for the r^(2)^( )law.

  • Newton is known for the 1 over r^(2)^( )law.

  • So this is a very important thing.

  • If you see anything wrong you should stop me,

  • because when I go home, they're going to play the video

  • for me to watch, and I'm going to see that

  • r^(2).

  • There is nothing I can do until some voice from the back says,

  • "Hey, send it downstairs",

  • and I appreciate that a lot, okay?

  • It's good, so never hesitate to do that, plus sign,

  • minus sign, symbols; anything that goes wrong.

  • It also tells me that you're following me.

  • So for all those reasons you should not hesitate to correct

  • anything, and you should not think that you don't follow it

  • because it's your fault.

  • Probably it is.

  • Sometimes it's my fault as was demonstrated now.

  • Okay?

  • So this is the force law.

  • You need only one other ingredient.

  • That's the superposition principle.

  • You need that ingredient because we're not going to be

  • talking only about two charges.

  • We're going to be talking about many charges.

  • And the question is: what will they do when they're

  • all present, and it's a great blessing that

  • we have the principle of superposition that says that if

  • you've got, say, three charges,

  • say 1,2 and 3, you want to know the force on 3

  • due to 2 and 1.

  • You can find the force that 1 would exert.

  • You can find the force that 2 would exert.

  • We have two vectors, and you can add the two vectors

  • to get the net force.

  • In other words, the interaction between pairs

  • of charges is insensitive to the presence of any other charges.

  • They go about their business exactly the same way.

  • That is a principle that is not deduced by logic.

  • You cannot say, "Of course,

  • it had to be that way."

  • That's not true.

  • It doesn't have to be that way.

  • In fact it is not that way if you make supreme,

  • I mean, extremely exquisite measurements,

  • but they occur really at the really quantum level.

  • For classical electromagnetic theory it is actually an

  • experimental fact that you can superpose.

  • All right, if you combine those things you can calculate

  • anything and everything that we will deal with for sometime,

  • but that's summary of--this is all I said last time,

  • okay?

  • Now, I didn't do one thing, which is to emphasize to you

  • that the force of gravitation divided by the electric force is

  • some number like 10 to the -40.

  • I put this twiddle here meaning not exactly.

  • There are factors of 1 and 2 or 10 missing,

  • but 10 to the -40 is roughly the magnitude of this ratio,

  • and it was done by comparing the force between electron and a

  • proton.

  • It didn't matter what the distance was because everything

  • goes like 1 over r^(2), so when they take the ratio

  • that cancels out, but this was really the ratio

  • of things like mass of the electron,

  • mass of the proton, divided by 1 over

  • 4Πε _0;

  • q of the electron and q of the proton.

  • If you put all these numbers in you've got that.

  • I want to mention one thing.

  • It may be interesting for you to think about.

  • Look at this nucleus.

  • Nucleus has a lot of protons in it, and according to Coulomb's

  • Law they all repel each other.

  • And the neutrons, of course, don't do anything

  • because they have no electric charge.

  • So you should ask yourself, "Why are all these protons

  • staying together inside the nucleus if I don't see any

  • attraction between them?"

  • I see why the electrons are hanging around,

  • because they're attracted to the nucleus,

  • because nucleus is positively charged,

  • electrons are negatively charged, and Mr. Coulomb tells

  • you they'll be attracted.

  • What are the protons doing together in that tiny space?

  • That space is like 10 to the -13 centimeters.

  • So have you ever thought of that, or does anybody know why

  • the protons are together?

  • Yes?

  • Student: The strong nuclear force?

  • Prof: There is a strong nuclear force.

  • Okay, that's the answer.

  • In other words, but that answer will lead to

  • more questions, but I'll first state the

  • answer.

  • There's a force even stronger than the electric force.

  • You see, this gives the impression electric force is

  • very strong.

  • It is much stronger than gravity, but there is a force

  • even stronger than the electric force that is experienced by

  • protons and by neutrons.

  • In other words there is another charge which is not electric

  • charge, which the protons are endowed with and the neutrons

  • are endowed with.

  • And the force due to that charge, if you like,

  • it's not called charge but it's a similar thing,

  • is much, much stronger.

  • It had to be, because you have to beat the

  • electrical repulsion.

  • So then you can ask yourself, "Well,

  • in that case, how did we ever find the

  • electrical force," because here's another force

  • even stronger than the electrical force,

  • maybe a thousand times stronger.

  • Then why wasn't it completely overshadowing the electrical

  • force?

  • Yep?

  • Student: It's irrelevant over larger

  • distances.

  • Prof: Yes, so let me repeat what he said.

  • It has to do with what's called the range of the force.

  • In other words, look at two different

  • functions, okay?

  • One function looks like 1 over 137 times 1 over r^(2).

  • Other function looks like 10 times e to the -r

  • over r_0 divided by r^(2);

  • r_0 is some length, and the length is

  • roughly 10 to the -13 centimeters, or 10 to -15

  • meters.

  • Which force are you more impressed with is the question,

  • okay?

  • If r is much bigger than r_0,

  • say 10 times bigger, you've got e to the -10

  • on the top.

  • e to the -10 is a small number.

  • e to the -3 is like 1 over 20.

  • e to the -10 is 1 over 20 cubed.

  • So then what will happen is, this force, even though there's

  • a number 10 in front of it, will be negligible compared to

  • this one.

  • Whereas if r is much smaller than

  • r_0, say, r is .1 of

  • r_0, so you can forget this number,

  • e to the minus point is roughly 1,

  • then this 10 will dominate the 1 over 100.

  • So what happens is, if you're sitting inside the

  • nucleus, the nuclear force is

  • numerically strong, not only because of the number

  • in front of it, but also because this

  • exponential factor has not kicked in.

  • Therefore, the protons feel an attraction for each other due to

  • the nuclear force that is stronger than the repulsion due

  • to the coulomb force.

  • And the neutrons attract the protons just as much as the

  • protons attract the protons with respect to the nuclear force.

  • So neutrons are actually good, because when you throw an extra

  • neutron into an atom you don't add to the coulomb repulsion,

  • but you add to the overall attraction that they all feel

  • for each other due to the nuclear force.

  • So neutrons are like the glue.

  • As you add more and more protons, you will find that

  • you've got to add more and more neutrons to compensate the

  • coulomb attraction.

  • But a time will come when the nucleus becomes so big,

  • that even if you add enough neutrons,

  • the repulsion between protons from one end of the nucleus to

  • the other is now becoming comparable to the attraction due

  • to the nuclear force, because nucleus has become so

  • big this factor is no longer negligible.

  • See, over long distances a coulomb force will always

  • triumph, because no matter what the

  • prefactors are, the exponential factor in front

  • of the nuclear force will always weaken it.

  • That's why you cannot have nuclei beyond some size.

  • If you make them any bigger, the nuclear electrical

  • repulsion between the distant parts of the protons,

  • due to the distant protons in the nucleus,

  • cannot be compensated by the short-range attraction.

  • So it's the range of the interaction that is a

  • significant factor here.

  • So all the strong forces are strong, but at short distances

  • the coulomb force is not that strong, but it falls like 1 over

  • r^(2).

  • Now gravity, on the other hand,

  • is exactly 1 over r^(2)^( )with a number

  • that's much, much, much smaller than this,

  • but then we saw the other day why gravity managed to survive,

  • because this is q_1q

  • _2 and the q's can be added

  • algebraically and cancel each other.

  • Whereas if you've got m_1m

  • _2_ over r^(2)^( )there is no way

  • to cancel the m.

  • This is how the different forces manage to survive for the

  • different reasons, okay?

  • Nuclear wins in the nuclear zone, but dies very quickly

  • outside the size of a nucleus.

  • Electrical forces falls more slowly at like 1 over

  • r^(2).

  • They dominate atomic physics, but once you formed an atom

  • you've got pretty much an electrical neutral thing,

  • and once you've got many, many atoms making up our

  • planet, then all that remains is the

  • gravitational attraction between a planet and another planet.

  • Okay, so these are examples of different forces and why they

  • were found at various times, because they all dominate under

  • different circumstances.

  • All right, so today I'm going to start with my new stuff.

  • So all you need to know, really, if you want to do your

  • problem sets and your homework, is Coulomb's Law,

  • you know, how to stick the numbers in the Coulomb's Law.

  • And the only thing I didn't mention is this 1 over

  • 4Πε _0 is 9 times 10

  • to the 9^(th).

  • So today we're going to do something which is a part of a

  • great abstraction and it goes as follows: So I'm gong to take two

  • charges, a q_1 here and

  • q_2 here, and I'm going to give them some

  • locations.

  • So let's say this guy is at vector r_1,

  • this one is at vector r_2.

  • Now, I will really take are of all the arrows.

  • This one is r_2 -

  • r_1, that arrow there.

  • You can check that I didn't mess up anything,

  • because r_1 r_2 -

  • r_1 should be r_2.

  • So let's write the coulomb force now as a vector.

  • And you've got to say force on what, so I'm going to say force

  • on 2, due to 1.

  • Now, you've got to realize it's a convention.

  • So I use a convention: This is the force on this guy

  • and the force due to this guy, with the second later.

  • Now, let's write it out in detail.

  • So that is q_1 over

  • 4Πε _0.

  • I'm going to put the q_2 here.

  • Then here I want to write (r_1 -

  • r_2)^(2).

  • That's the 1 over r^(2), but then I have to make it a

  • vector.

  • So for the vector part, once you have the magnitude of

  • the vector, you should multiply by vector

  • of length unit 1 going in the direction of this difference

  • vector.

  • So you can use any symbol you like.

  • One is to say e_12.

  • e's always going to be unit vector going from 1 to 2,

  • or if you're inclined you can also write

  • e_12 as r_2 -

  • r_1 divided by the length of

  • r_2 - r_1.

  • They're all unit vectors.

  • Now, do you follow that?

  • I mean, are you having trouble with unit vectors?

  • Anytime I have a vector pointing from here to there,

  • I want to give a magnitude and direction.

  • The magnitude in this case is 1 over the distance squared,

  • but you have to append to it a vector of unit length in that

  • direction.

  • That's what makes it into a vector.

  • For example, suppose I want to describe that

  • vector r, and it is 7 meters long?

  • I cannot write r = 7, because that doesn't tell you

  • which way it's pointing.

  • So invent the vector called i, which is a unit vector

  • in the x direction and I multiply it by that.

  • So 7i is a vector parallel to i and 7 times

  • long.

  • 7.3i is a vector parallel to i 7.3 times

  • long, okay?

  • So you need to add a vector of unit length to the magnitude,

  • multiply it to get the actual vector.

  • So that's that formula.

  • Then if you have many charges, I'm not going to do that now,

  • say one more here, it'll exert a force on

  • q_2, but you've got to add to the

  • force due to 1.

  • I'm not going to do that.

  • I'm just taking two guys.

  • Now, I'm going to formally write this as equal to the

  • electric field at r_2 times

  • q_2, and this is called a field.

  • You see, if you look at this thing,

  • all I've done is rewrite the expression as something that

  • involves a charge of q�_2 and

  • everything else that involves the q_1 and the

  • distance from q_1 and

  • q_2.

  • So it looks like there's no real content to giving this

  • object a name, but it's a very profound

  • notion, so I've got to tell you the story that goes with it.

  • In Coulomb's Law you say that q_1 and

  • q_2 exert a force on each other,

  • okay?

  • And the force depends on the charge and the distance between

  • them.

  • Now I'm going to say q_1 produces an

  • electric field at the location of q_2 given by

  • this vector, and when I multiply by

  • q_2 it gives me the force on

  • q_2.

  • Now, what's the importance of the electric field?

  • Whereas q_1 and q_2 exist

  • only at these two places, the electric field can be

  • defined everywhere.

  • It doesn't require a second charge because you see,

  • this number you can compute for any value of

  • r_2.

  • So the picture we have is that q_1 produces an

  • electric field all over space, and q_2

  • experiences that field and gets repelled as a result,

  • and the force it feels is the field at that point times

  • q_2.

  • In general, we say there's an electric field at this point in

  • space.

  • If you put a charge q there it should experience a

  • force q times the electric field.

  • So to understand this, the charges are only in two

  • places in our example, but the field due to

  • q_1 is everywhere.

  • At every point in space I can compute a field due to

  • q_1.

  • So the field, you can see,

  • will turn into a force if you multiply it by a charge you put

  • at the location of the field.

  • So if you've got one charge here, I claim there's a field

  • here, there's field there; there's a field everywhere due

  • to this guy.

  • How do I know, because if you put a test

  • charge it begins to feel a force.

  • So one way to say it, is that the field is the force

  • on a unit charge you put at that location,

  • unit charge because if q_2 is equal to

  • 1 then numerically E is exactly equal to the force.

  • If I go to you and say, "Find out the electric

  • field at this point."

  • You say, "Okay, where?"

  • I say, "Here."

  • What do you think you have to do to measure the field?

  • Yes?

  • Student: Add up all the forces on that field from the

  • different charges, or add up all of the fields at

  • that point...

  • Prof: No, okay.

  • Her answer was, "If I want the field here,

  • I should find out where all the charges are,"

  • right?

  • "Find all the forces they will exert on this guy,

  • on the unit charge here," right?

  • But I'm telling you, that's correct.

  • That's the theoretical way to calculate the field at that

  • point, but suppose you're an experimentalist and you don't

  • want to know what produced it.

  • You just want an answer to what's the field here.

  • What will you do?

  • Yes?

  • Student: Place a test charge.

  • Prof: Okay, and then?

  • Student: And then see what happens to it.

  • Prof: By `see what happens', you've got to be

  • more--so his answer was: put a test charge,

  • and see what happens.

  • Now, you've got to be more precise.

  • I think you know what you meant, but I want you to finish

  • that sentence.

  • Student: Okay.

  • Well, you can measure the force by measuring the acceleration on

  • the charge.

  • Prof: Very good.

  • See, that's what I want.

  • When you said, "See what happens,"

  • does it get married?

  • Does it have children?

  • That's not what I meant, okay?

  • But you did give the right answer.

  • The answer was, by `see what happens',

  • you put the charge there, and you see what acceleration

  • it undergoes.

  • Then that acceleration is the--that times the mass is the

  • force it's experiencing.

  • That should be q that you place there times E.

  • If q was 1 that force itself is equal to E.

  • If q was 10 you've got to divide the force by 10 to get

  • the field there.

  • So, the field is like the sound of one hand clapping.

  • People say one hand clapping is Zen concept, but the field is

  • like that, because you don't need two charges to have a

  • field.

  • You just need one charge.

  • So here is how we understand that.

  • You put a charge q.

  • If you go here something has happened there,

  • see?

  • You don't need to put a second charge there to conclude

  • something has happened.

  • Something really is different at this location because the

  • charge q is present.

  • What is different?

  • What is different is that when this guy was not here and I put

  • a charge, it just sat there.

  • Whereas when this guy is here and I put a charge,

  • it experiences a force.

  • So if you put a 1 coulomb here it experiences a force which is

  • 1 times the field at that point, therefore, this charge has

  • distorted the space around it, in fact, everywhere.

  • Now, if you've got many, many, many charges,

  • then they will all try to produce a force on a unit charge

  • and you should, like you said,

  • add up the vectors, which are the forces due to all

  • the other charges on that one location,

  • on a test charge on that location.

  • So the field is the sum of the field due to all the charges at

  • that point.

  • Now, Coulomb's Law doesn't work when charges are moving.

  • Why is that?

  • Have you any idea why you cannot use it?

  • Yes?

  • Student: The radius is changing?

  • Prof: Pardon me?

  • Student: r is changing?

  • Prof: r is changing so we'll keep changing

  • it as the charge moves, but it's only an approximation

  • when charges move.

  • Do you know why?

  • Yep?

  • Student: There's a magnetic field?

  • Prof: True, but even the electric

  • field--his answer was magnetic field,

  • but even the electric field is not properly given.

  • Yes?

  • Student: Is the electric field affected by

  • something like the Doppler Effect?

  • Prof: Not the Doppler, something else.

  • If Coulomb's Law were exact, okay?

  • Here is what I can do.

  • You take a coulomb and you sit at the other end of the galaxy.

  • I have a coulomb here.

  • You know the force my coulomb exerts on yours because it's

  • pushing against you.

  • You hold it.

  • Now suddenly I move my coulomb away from you by a tiny amount.

  • What will you feel?

  • You will feel the force is reduced.

  • Yes?

  • Student: Relativity is...

  • Prof: Yes, so the special theory does not

  • allow that, because I have managed to

  • communicate with you, arbitrarily far,

  • instantaneously.

  • The minute I move this charge, you know about it,

  • because your charge moved away.

  • For example, if your charge was connected to

  • a spring, and it had been extended to

  • some amount because of this charge,

  • the minute I move it your spring will move.

  • That instantaneous communication is forbidden by

  • the special theory, so it does not happen.

  • So the correct way to do it, it'll maybe come later in the

  • course, so probably not even at the end

  • of this course, but the proper way to do that

  • is to in the end realize that the electric field at this point

  • is not only due to what the charges are doing now,

  • but what they were doing in the past.

  • Because if some charge at the other end of the galaxy did

  • something it takes some time, namely traveling at the speed

  • of light, to carry that information from

  • there to here.

  • So it's the delayed response to all the motion in the charges

  • that you've got to add to find the field here.

  • That's what makes the computation of the electric

  • field much more complicated.

  • But if you promise me charges never move then the location now

  • is the location last year, the location a million years

  • ago, then you can use Coulomb's Law.

  • Coulomb's Law is good for electrostatics,

  • but in real life charges are moving, so you cannot really use

  • the formula.

  • Now, in our room, if you put a charge here and

  • another charge there, if you move this,

  • that guy will move pretty much instantaneously.

  • That's because the time it takes a light signal to go from

  • here to there is so small, you may treat it as

  • instantaneous.

  • So Coulomb's Law is used in electrical circuits and so on.

  • You don't worry about the time of transit because it's too

  • small, but over longer distances,

  • where the time it takes for light to travel becomes

  • non-negligible, you cannot use Coulomb's Law.

  • It's not wrong.

  • It is not appropriate when charges are moving.

  • However, it will always be true that if you go to any one point

  • the force on any charge q you put there is q times

  • electric field at that point, okay?

  • So the electric field notion survives because it doesn't

  • violate relativity.

  • It says if the field here is so much q will experience

  • the force q times E, but the complication

  • is what is the field here?

  • Well, it's due to everybody else, and it's not only due to

  • everybody else right now, but everybody else from the

  • dawn of time because things have been moving and shaking and

  • sending signals to us.

  • We collect all that and see what lands here at this instant.

  • That decides the field here.

  • So that's the computation of the field, but the response to

  • the field is very easy.

  • You put a test charge; q times E is the

  • answer.

  • So in modern physics, in theories that are compatible

  • with the special theory of relativity we break the force

  • into two parts.

  • Charges don't immediately interact with other charges.

  • Charges produce a field, and the field may even

  • propagate outwards at the speed of light if you make motions,

  • but another charge at the location of that particular

  • point will respond to the field at that point.

  • So it's not responding to the charge right now.

  • It's responding to the field this charge produced at its

  • location.

  • So all of electromagnetic theory is going to contain two

  • parts.

  • The first part is, find the field due to this

  • charge configuration, that charge configuration,

  • maybe due to various currents, and the second part is,

  • given the field, find the response of charges to

  • the field.

  • So you understand charges play a double role.

  • They are the producers of the field.

  • They are also the ones who respond to the field.

  • If you don't have a charge you cannot produce field.

  • If you don't have a charge you cannot experience it,

  • you cannot play that game.

  • To gain membership into electrostatic interactions

  • you've got to have charge.

  • So neutrons cannot do that, but they can do other things.

  • Like I said, they take part in nuclear

  • force, in fact, just as well as protons do.

  • All right, so let's go back now to the simplest problem in the

  • world: the electric field due to one charge.

  • The formula is very simple.

  • Let's put that charge at the origin.

  • And the electric field due to 1 charge is q over

  • 4Πε _0,

  • 1 over r^(2) if you're here,

  • and that's your r vector.

  • The field at this point is that magnitude,

  • and I'm going to write e_r_

  • meaning a vector of unit length in the radial direction.

  • Once again, I will tell you if you want, you can write that as

  • the position vector divided by the length of the position

  • vector.

  • They're equivalent ways.

  • If I write it the second way, you've got to be a little

  • careful.

  • It'll look like qr over 4Πε

  • _0r^(3).

  • Don't get fooled into thinking the field is falling at 1 over

  • r^(3).

  • It's really 1 over r^(2)^( )because there's

  • an r on the top.

  • See, if you write it as 1 over r^(2) put unit vector.

  • If you write it as 1 over r^(3 )put the position

  • vector.

  • They're all saying the same thing.

  • So here you have this formula.

  • If you're a person who likes to work with formulas this is all

  • you need.

  • You manipulate the stuff on paper, and you add different

  • fields, but people like to visualize this.

  • So how do I visualize this?

  • That's the real question.

  • So here's a very popular method for visualizing this formula.

  • You know, you've got, for example--suppose someone

  • asks you, what's the height above ground level of a certain

  • part of the United States?

  • Okay, you've got some mountains.

  • You've got some valleys.

  • Well, somebody can give you a function that gives you the

  • height at any point in the United States,

  • but it's more interesting to have some kind of a contour map

  • that looks like this, right?

  • And all these contours are different heights.

  • If you've gone hiking, you can see those maps.

  • They tell you pictorially what a certain function is trying to

  • tell you.

  • So you want a pictorial representation of this electric

  • field.

  • It's very easy to write down the electric field at one point,

  • namely you take that point, you draw an arrow there;

  • E.

  • That E is the electric field at that point.

  • So we try to do our best by saying here is my test charge.

  • I'm going to pick a few points, four points,

  • maybe eight, and I'm going to tell you what

  • the field is at those points.

  • At this point it looks like that.

  • At this point it looks like that.

  • Here it looks like this.

  • That's already telling you something.

  • You've got to be very careful on the interpretation.

  • This arrow is not telling you what's happening throughout the

  • length of the arrow.

  • It's telling you what's happening at the tip.

  • You understand the arrow is in your mind.

  • It's a vector.

  • It's not really sticking out in space.

  • It's a property at a condition at that point,

  • but we've got to draw it, so we draw it that way.

  • It doesn't tell you the state of affairs over its length,

  • but only at its tip, at the starting point.

  • Then you can say, "Okay, what happens when I

  • go further out?"

  • When I go further out, say over here,

  • it's going to be still--if I put a test charge here it's

  • still going to be repelled radially, but a lot less.

  • So I do that.

  • So I draw arrows at other representative points and make

  • them shorter.

  • In fact, the length of the arrow will be 1 over

  • r^(2).

  • So you can do this, okay?

  • But now where all do you want to draw these arrows?

  • It's up to you.

  • You pick a few points.

  • You go to another radius, you draw more arrows.

  • Someone had this clever idea of doing the following.

  • You can probably guess.

  • Their idea was when I join all these arrows like that,

  • now if you go to a point like this, what have I gained and

  • what have I lost?

  • What more information have I got when I join the arrows?

  • Yes.

  • Student: The more information is that now you know

  • the direction that it will continue on forever if no other

  • forces act on it, and what you've lost is the

  • magnitude...

  • Prof: Okay, let me repeat that.

  • That's what you guys should have been thinking in your head.

  • When I joined these lines--by the way,

  • I do want you to anticipate what I'm going to say,

  • because if I'm struck by lightning,

  • another electromagnetic phenomenon,

  • can you even complete my...

  • Student (Chorus): Sentence!

  • Prof: Sentence, right.

  • Okay, now you should be able to go a little beyond if I'm doing

  • a derivation.

  • You've got to be following me, right?

  • That's very important.

  • It's got to be active.

  • And I sat through a lecture yesterday for an hour.

  • I know it's a very long time.

  • This is what, an hour and fifteen minutes?

  • The only way you can survive this is if you somehow make it

  • an active event.

  • You've got to do something that keeps you awake during the

  • process.

  • One of them is to anticipate what I will do next in a

  • calculation.

  • That'll make sure also that you're on top of it,

  • that'll make sure that you catch mistakes.

  • Okay.

  • All right, so here are these lines.

  • As she said quite correctly, previously I knew the field

  • direction only at the chosen points,

  • but now I know it throughout this line,

  • but I've lost information on the magnitude of the field,

  • because the arrows--there are no lengths of anything.

  • These arrows don't have any length.

  • In fact, you can keep drawing more lines if you like.

  • They go like that in all directions.

  • It basically tells you, hey, the charge is pushing

  • everything out radially no matter where you are.

  • That's the thrust of this picture.

  • But, due to the miraculous property of the coulomb force,

  • namely that it falls like 1 over r^(2),

  • there is information even on the strength of the electric

  • field, and that information is

  • contained in the density of electric field lines.

  • And I'll tell you precisely what I mean.

  • So, here is the charge.

  • Take a sphere of radius r and here are all these

  • lines going.

  • By density of lines, I mean the number of lines

  • crossing a surface perpendicular to the lines,

  • divided by the area of that surface.

  • Because let's make a convention that we will draw for every

  • coulomb a certain number of lines, 32 lines per coulomb,

  • 32 lines are going out.

  • I draw a sphere of some radius, 32 lines cross that sphere.

  • I draw a bigger sphere, 32 lines cross this sphere

  • also, but they're less dense,

  • because the number of lines per area will be some number of

  • lines per charge divided by the area of the sphere,

  • which is 4Πr^(2).

  • Do you follow that?

  • If you take a sphere, first of all,

  • every portion of the sphere the area that you have is

  • perpendicular to the lines.

  • So the area intercepts the lines perpendicularly.

  • That's the agreement here.

  • And you see how many are crossing per unit area.

  • That is going like 1 over r^(2).

  • So these lines naturally diverge and spread out in space

  • so that the density falls precisely as 1 over

  • r^(2).

  • That has to do also with the fact you're living in three

  • dimensions.

  • Only in three dimensions where area goes like r^(2)^(

  • )does this spreading of the density of lines coincide with

  • the decline of the force.

  • So these lines tell you more than simply the direction.

  • They convey to you visually where the field is strong.

  • Wherever the lines are dense, the field is strong.

  • Wherever the lines are spread apart, the field is weak,

  • and it's a very precise statement.

  • The only thing not precise is, how many lines do you want to

  • draw per coulomb.

  • That's really up to you, but you've got to be

  • consistent.

  • Once you give 32 lines per coulomb, then if you've got a

  • charge of 1 coulomb you should draw 32.

  • If you've got two coulombs, you should draw 64 lines.

  • As long as you do that, the number of lines crossing

  • per unit area will be proportional to the field.

  • But I'm going to make a certain choice that will make the number

  • of lines per unit area exactly equal to the field,

  • and here is the choice.

  • It's a choice that makes life simple.

  • Let us agree that 1 coulomb gets 1 over

  • ε_0 lines.

  • ε_0 is a number, right?

  • 1 over 4Πε _0 is 9 times 10

  • to the 9^(th).

  • This is some number.

  • Maybe 40 million, so one coulomb gets 40 million

  • lines.

  • Don't quote the 40 million.

  • It's whatever this thing is.

  • I don't know what it is.

  • It's a definite number.

  • Then, what's the nice thing?

  • If you've got q coulombs you will have q over

  • ε_0 lines,

  • and if you take a sphere of radius r you'll get 1

  • over 4Πr^(2)^( )as the line density,

  • namely lines per unit area, but that is exactly equal to

  • the strength of the electric field.

  • If you picked a different number like 2 over

  • ε_0 you will always be measuring 2 times

  • the electric field.

  • The density will still convey the electric field,

  • namely it'll be proportional to it, but let's make life easy by

  • making it equal to it.

  • This is just a convenience.

  • Now, we are really set.

  • If you draw pictures this way, you can go as far as you like

  • from this charge.

  • Simply take a unit area with you.

  • Take a piece of wood 1 meter by 1 meter, put it there,

  • see how many lines cross; that is equal to the electric

  • field at that point.

  • Okay, so this is the way one likes to visualize field lines.

  • So I'm going to give you some examples.

  • For a single charge you just draw it that way.

  • For two charges, let's take two charges,

  • a minus charge and a plus charge.

  • Let's say that one is -q, the other is

  • q.

  • Then you are very near that charge.

  • By the way, I'm not going to draw 1 over

  • ε_0 lines per coulomb because it's

  • going to be too many lines, okay?

  • I'm just going to draw a few so you get the picture.

  • So I'm going to draw four lines right near the charge.

  • You can forget about all other charges in drawing the lines.

  • Why is that?

  • Yep?

  • Student: The field only depends on the one charge.

  • Prof: Pardon me?

  • Student: Why does the field only depends on the one

  • charge.

  • Prof: Why does it depend on the one charge?

  • In principle it depends on every charge.

  • Somebody had an answer back there?

  • Yes?

  • Student: Since it falls as 1 over r^(2)^( )when

  • you're so much closer to one than the other then...

  • Prof: Right, because the field is 1 over

  • r^(2)^( )and the 1 over r for this guy is going

  • to infinity.

  • 1 over r for this guy is maybe 1 over 1 meter.

  • It's finite.

  • So when you come arbitrarily close to a charge,

  • it is going to dominate.

  • Well, if it's the only thing in the universe,

  • we know the lines will look like this.

  • At least they'll start out this way, but soon,

  • of course, it won't go out this way forever, because you will

  • realize there's another charge.

  • Likewise, it's easy to draw the lines this way.

  • Remember, the lines are coming in because if you put a test

  • charge, it'll be sucked into this.

  • Test charge is always assumed to be unit positive charge,

  • so the lines will be coming into a negative charge,

  • and leaving, going away from a positive

  • charge.

  • Now we've just go to do what the agencies forgot to do,

  • which is to connect the dots.

  • You do this.

  • You do this.

  • You do this.

  • You do this.

  • Now, at some point you'll have to think a little harder,

  • because suppose I go here?

  • How do I know I should draw the lines this way?

  • If I take this guy here, it will repel it.

  • This guy will attract it, and I add the two,

  • and get a line in that direction, so you really have to

  • do a lot of work.

  • If you really want this picture to be exact,

  • you have to compute the vector everywhere,

  • but if you want a sketch, you're allowed to guess,

  • and things look like this.

  • So this is called a dipole, and this is the field of a

  • dipole.

  • So here's another example.

  • Both guys are plus.

  • Now what do the lines look like?

  • Again, they will start out this way near the charges,

  • but now when you come to the midpoint here there should be no

  • electric field right in the midpoint because it's getting

  • pushed equally from both sides.

  • So the lines, if you think about how they

  • will add, they will do something like this.

  • Okay, look, I'm not going to do a good job for a variety of

  • reasons, but you can look at your

  • textbook or any other book to see what the lines will look

  • like when you've two plus charges.

  • If you go a mile away from these two plus charges,

  • what do you think the lines would look like?

  • Yep?

  • Student: Just like a point charge that's twice as

  • strong.

  • Prof: Right, so that's the intuition you

  • should keep in your mind.

  • If you go very far from a charge distribution where you

  • cannot look into the details, all you will see is some little

  • dot that has the entire charge in it,

  • and the lines will be coming out radially.

  • So only when you zoom in you realize, hey,

  • it's not a single charge 2q, it is two guys of

  • strength q.

  • Finally, let's take a case where this is charge 2q.

  • This is charge q.

  • Let's say this is charge -q.

  • Then some lines will go like this and some lines will run off

  • to infinity.

  • Here if you go very, very far away from the two

  • charges you'll again see radially outgoing lines,

  • except there is a charge q at the center because

  • 2q and -q give you a net of q.

  • Okay?

  • So this is the example of a dipole.

  • If you've got more charges it gets more complicated,

  • so people don't usually draw them.

  • There's one example which is pretty interesting.

  • If you've got one plate and another plate,

  • this contains all positive charges,

  • this contains all negative charges,

  • then the field here will look like this,

  • will go from the positive to the negative plate,

  • because if you put a test charge between them,

  • it's getting repelled by the positive plate and attracted by

  • the negative plate, so the lines will go from one

  • to the other.

  • Near the edges they may do something more complicated,

  • but in the bulk they will look like this.

  • Okay, so now I'm going to do one calculation,

  • which is: what is the actual electric field due to a dipole?

  • In other words, not just the picture here,

  • where is that?

  • That picture on the left is a dipole.

  • I'm going to do it quantitatively.

  • So here's my goal: I want to take a minus charge

  • here, a plus charge here.

  • This is at x = a.

  • This is at x = -a.

  • So I want to find the field everywhere.

  • So today I'm not going to do the field everywhere,

  • because later on I'll show you a more effective way to

  • calculate it, but I'm going to calculate it

  • at a couple of interesting easy places.

  • In other words, we all know the lines look like

  • this, okay?

  • But I want to go to some location and find the magnitude

  • and direction of the field.

  • But today I will only find it at two places,

  • one along the axis at a point x,

  • and one on the perpendicular bisector at a point (x =

  • 0, y), just going to do those two.

  • So let's see what's the field here, the field at this point?

  • The field at that point, you agree, is going to be

  • entirely in the x direction because this is pushing it,

  • and that is pulling it.

  • So E is going to be i (unit vector) times

  • q over 4Πε

  • _0 times that distance squared,

  • which happens to be (x - a)^(2).

  • That's the repulsion due to the charge that's nearer to you.

  • Then there's the attraction due to the minus q,

  • but it's a little further away, so it looks like a minus sign

  • but it is (x a)^(2).

  • If a is equal to 0 you get 0.

  • If a is equal to 0 the two guys are sitting exactly on

  • top of each other.

  • You will not see them.

  • So you see them only because they're not on top of each

  • other.

  • This whole thing fails to be 0 because this a is not 0,

  • and you can understand why.

  • The minute a is not 0 you're closer to one of the two

  • charges, so that they cannot really cancel each other.

  • So you've got to manipulate this expression,

  • so I will do that now.

  • q over 4Πε

  • _0 and you find common denominator.

  • I remind you (x a) times (x -

  • a) is (x^(2 )- a^(2))^( )and everything

  • is under squares.

  • In the numerator you've got (x a)^(2) -

  • (x - a)^(2).

  • So what does that give me?

  • iq over 4Πε

  • _0, (x^(2 )-

  • a^(2))^(2).

  • And how about on the top, can you do that in your head?

  • This is going to be x^(2) a^(2)

  • 2xa.

  • You're going to subtract from it x^(2) a^(2) -

  • 2xa, and the only thing that will survive will be

  • 4xa.

  • Is everything okay?

  • Student: Why is it not x - a squared times x a squared?

  • Prof: It is.

  • You are saying why is it not (x a)^(2) times

  • (x - a)^(2), right?

  • It is, because this would be x a times x

  • - a, the whole thing squared.

  • Student: Oh!

  • Prof: And this guy is x^(2 )- a^(2).

  • So this is classified as a nice try, so.

  • But I want you to keep doing this.

  • This time I am right, but you never know,

  • okay?

  • I don't want you to give up.

  • There is nothing better than shooting me down,

  • but this happened to be correct.

  • You satisfied though?

  • Student: I knew you were right, I just couldn't

  • figure out why.

  • Prof: No, no, no, no.

  • I don't want to rush through this.

  • Anybody have the same problem with this?

  • Look, also I am doing it fast because this is 958th time I'm

  • doing this calculation, so if you're seeing it for the

  • first time, I've got to slow down.

  • So let's see which part.

  • Everybody okay with this, right?

  • So I wrote that.

  • Then this is really an x a times another x

  • a, and an x - a and

  • another x - a, but without these guys,

  • I know it's x^(2 )- a^(2 )because I've got

  • two of everything.

  • I squared everything.

  • So this answer's actually an exact formula of the electric

  • field along the axis of the dipole.

  • But normally what one is interested in is,

  • when x is much bigger than a.

  • When it's much bigger than a, downstairs you've got

  • x squared.

  • This could be 1 kilometer squared, a squared was

  • maybe 1 millimeter squared.

  • So in the first approximation, it's not an exact formula

  • anymore, from now on it is approximate, in the limit

  • x is much bigger than a.

  • You can see it's going to be i times q over

  • 2Πε _0 times

  • 2a divided by x^(3).

  • So what did I do now?

  • I took from the 4a.

  • I borrowed a 2a to write this here and I canceled the 2

  • with the 4 here to get that.

  • Then on the top I had an x, and the bottom had

  • x^(4).

  • I get x^(3).

  • So let me write--can everybody see this thing from wherever you

  • are; the last formula here?

  • So I'm going to write it as i times p divided

  • by 2Πε _0 x^(3).

  • So I will tell you what I'm doing.

  • So the final formula I had was electric field E = i

  • times p divided by 2Πε

  • _0x^(3), where p--I'm sorry,

  • i times p, no arrow;

  • p is equal to 2aq.

  • So let's delete some extra arrows I had.

  • That's right.

  • You never should settle for something that looks like that.

  • So that's p.

  • So p is called the dipole moment of this dipole,

  • and it's given by the product of the distance between the

  • charges and the value of one of the charges,

  • the plus charge.

  • So whenever I give you two charges, call it dipole,

  • you can associate with them a vector.

  • And the vector is, if you've got a charge

  • -q here and a charge q there separated by a

  • little vector r, then the dipole moment is q times

  • the little vector r.

  • In our example the little vector r was 2a in

  • the x direction.

  • So 2a times i times q is the dipole

  • moment.

  • So this means electric field is parallel to the dipole moment

  • and falls like 1 over x^(3).

  • That's the most important part of the dipole.

  • A single charge, the field falls like 1 over

  • x^(2) if you move a distance x.

  • A dipole will always fall like a bigger power of x,

  • because to go like 1 over x^(2) you've got to have

  • net charge.

  • As long as the net charge is 0 the fact that there are two

  • opposite charges that don't quite cancel each other,

  • it always comes from the fact that the distance between them

  • is not 0.

  • And the distance will appear in the numerator,

  • and that must be the corresponding distance in the

  • denominator, because the formula should have

  • the same dimensions.

  • That's what turns the 1 over x^(2) to 1 over

  • x^(3).

  • Yep?

  • Student: That only works if x is a lot

  • bigger than a.

  • Prof: Yes.

  • So this formula is good for all x;

  • this formula is good only for x bigger than a.

  • You'll find whenever you're working with dipoles,

  • people will always ask you to find the field very far from the

  • dipole.

  • So here's a second place where I can go and find the field.

  • That's going to be here, -a, a,

  • and I'm going to find the field there.

  • So that's at a distance y.

  • So let's look at this.

  • That's a q here and -q here.

  • A q will repel it that way, and a -q will

  • attract it this way, and their sum will be that.

  • I'm going to compute that sum.

  • So how do I do this?

  • Let's look at this guy here.

  • We know that these arrows have equal magnitude because this

  • distance is the same as that distance.

  • Therefore, it's the horizontal part that will remain.

  • The vertical part will cancel.

  • You see that?

  • We've got two arrows of the same length with this angle and

  • this angle equal.

  • The horizontal part will be additive and the vertical part

  • will be equal and opposite.

  • So I'm only going to compute the horizontal part.

  • So the electric field now will be -

  • i (that's to tell me it's in the negative x

  • direction) times q over 4Πε

  • _0 times 1 over distance squared which is

  • (y^(2)^( ) a^(2)).

  • This is (y^(2)^( ) a^(2)).

  • That's also (y^(2)^( ) a^(2)),

  • but I want the horizontal part of this,

  • so I want the cosine θ.

  • That is the same as this one.

  • The cosine θ is a divided by

  • (y^(2)^( ) a^(2))^(½).

  • See, you want to take that force and find this horizontal

  • part.

  • Then I'm going to put on another 2 because this is going

  • to contribute an equal horizontal part.

  • So the E, in the end, is equal to

  • -iq2a over 4Πε

  • _0 divided by (y^(2)^( )

  • a^(2))^(3/2).

  • That is then -p divided by 4Πε

  • _0 divided by y^(3),

  • for y much bigger than a.

  • I'm sorry, y^(3), y is really the

  • distance.

  • So, right, if y is 1 mile and a is 1

  • millimeter that's essentially the distance.

  • If you like you can call it -p divided by

  • 4Πε _0r^(3).

  • Where r is the distance, if you like,

  • from the center of dipole.

  • Look, the point of this exercise is twofold.

  • One is to show you how to add vectorially the fields due to

  • two guys.

  • Another is, to have you understand,

  • at least, how to do the computation at a few simple

  • places, where the direction of the

  • field is not so hard to calculate.

  • Actually, one would like to compute it here,

  • but it becomes quite nasty.

  • The magnitude is not so hard, but the direction is hard to

  • calculate, so we'll find a shortcut.

  • But at these two places, on the perpendicular bisector

  • and on the axis, the mathematics is pretty

  • simple.

  • That's the electric field.

  • Yeah?

  • Student: Is p as a vector different from the

  • p that you crossed out as a vector over there?

  • Prof: No.

  • It's the same p.

  • So p as a vector in our example will be the charge at

  • either end times the distance between them.

  • The vector difference is 2a times i.

  • So you can write the formula in terms of the dipole moment.

  • Okay, so now I'm going to do the second part of the problem,

  • which is finding the response to E.

  • This was all computing E.

  • Of course there are more and more complicated examples,

  • but we did a few simple ones.

  • Next is going to be, if I give you E, can you

  • find what will happen to the charge.

  • So I'm going to do two examples.

  • One is there are these two plates I mentioned to you.

  • This one is all positively charged.

  • This one is negatively charged.

  • And I shoot a particle here, with some velocity

  • v�_o in the x direction,

  • and the field everywhere is down, and the electric field is

  • some constant, -j times some number

  • E_0.

  • -j because i is this way and j is that

  • way.

  • So what will this do is the question, and where will it end

  • up?

  • Well, I think you can all tell that it'll end up somewhere

  • there.

  • What we're trying to find out is how much does it fall,

  • and when it comes out, what's the direction of this

  • final velocity vector.

  • Well, the force on this charge is equal to -q times

  • E_0 j.

  • The acceleration will be -qE_0 over

  • m times j in the y direction.

  • So what'll be the position?

  • Position will be from lecture number one of your Physics 200.

  • So let's say the starting point r_0 is our

  • origin.

  • v_0 is whatever it was projected in

  • with, plus ½

  • times qE_0 over m,

  • t^(2) j.

  • Okay, so as a function of time this tells you where the

  • position will be.

  • At t = 0, you are at the origin.

  • As t increases, it's moving horizontally due to

  • v_0 and it's also dropping vertically due to

  • the acceleration.

  • Then it's very easy from now on to compute anything you like.

  • For example, when you want to go to that

  • point, what will be the time?

  • Anybody tell me what the time will be when I go to that point?

  • How long will it be in the region between the plates?

  • Yep?

  • Student: The distance of the length of the plate

  • divided by initial velocity.

  • Prof: Which velocity should we take?

  • Student: The initial horizontal velocity.

  • Prof: That's correct because t will be

  • L over v_0 where

  • v_0�'s the magnitude of the initial

  • velocity because x velocity is never changing.

  • Acceleration is in the y direction.

  • So the time it takes to cross will be independent of the fact

  • it's falling in the y direction.

  • So if you put t equal to all of this you will find out

  • where it will end up.

  • And that's how you make pictures on the television.

  • You've got a bunch of plates, and then you drive charges,

  • and if you apply the right electric field the electron will

  • land on a screen and make a little dot.

  • The screen will look like this, and you're looking at it from

  • the other side.

  • It'll glow.

  • Then you want the dot to move up and down, you can move the

  • voltage.

  • Then if you want to move it back and forth you've got to put

  • another set of plates, not like this,

  • but coming out of the blackboard.

  • That way you can move the electron beam in all directions.

  • That's how you scan the television screen.

  • You can also use magnetic fields, but this is one simple

  • way using electric field.

  • All right, final thing to discuss is: what is the force of

  • a uniform electric field on a dipole?

  • So let's take an electric field in the x direction like that.

  • It's got a magnitude E_0 and it's in

  • the x direction.

  • And in this electric field I stick a dipole in like that.

  • Here is the q.

  • Here is the -q.

  • Let's make that a.

  • Let's make that a.

  • So the plus charge will experience a force like that.

  • The minus charge will experience a force like that.

  • This will be q times E_0.

  • That'll be -q times E_0.

  • So dipole as a whole will not feel any net force,

  • because the two parts are getting pulled by opposite

  • amount.

  • If the electric field were not uniform,

  • namely if it were stronger here than here,

  • then of course it will drift to the right,

  • but I'm taking uniform electric field,

  • and because the charges are equal and opposite,

  • the net force on it is 0.

  • But something is not 0.

  • You know what something is that's not 0?

  • Yep?

  • Student: The torque.

  • Prof: The torque.

  • There is a torque because there is a force here and the force

  • there.

  • You can imagine they're trying to straighten out the dipole so

  • it ends up looking like this.

  • So let's find the magnitude of that torque.

  • Magnitude of the torque is the force times the perpendicular

  • distance.

  • So if this angle is θ here,

  • and you want the perpendicular distance, it is

  • asinθ.

  • Then there's another asinθ from

  • that one.

  • That's the torque.

  • But since 2q a is p, it's pE_0

  • sinθ.

  • You can see this makes sense.

  • If θ was 0, if the dipole was aligned with

  • the field, the torque vanishes,

  • because if the charges are like this there is no tendency to

  • rotate.

  • The biggest torque you get, if the charges are like this,

  • then this gets rotated that way.

  • That gets rotated that way.

  • You get the maximum torque.

  • Put θ equal to Π by 2,

  • you get a torque of p times E_0.

  • That's also the reason it's called a dipole moment.

  • Now, I'm going to write this as a cross product,

  • and I'm assuming you guys are familiar with the cross product,

  • right?

  • You take two vectors, p and E,

  • the cross product has a magnitude, which is p

  • times E times sine of the angle between them,

  • and a direction obtained by turning a screwdriver from

  • p to E.

  • So p is like this.

  • E is like this.

  • Turn a screwdriver from p to E,

  • it goes into the board, and that's the torque,

  • and the dipole will then rotate until it lines up,

  • or if you don't want it to rotate you've got to provide a

  • counter-torque of this magnitude.

  • Now, if you take any dipole and leave it,

  • you know it will like to become horizontal,

  • so there is a certain restoring torque that tries to rotate it

  • so it becomes horizontal, and it's not very different

  • from a spring, where if you pull it from

  • equilibrium, there is a restoring force that

  • brings you back to where you were.

  • So this dipole is happiest when it's horizontal.

  • If you go away from horizontal the torque brings it back.

  • So just like for a spring, if you've got a force which is

  • -kx we can assign a potential energy U,

  • which is ½ kx^(2) so that the force

  • is equal to -dU/dx.

  • This is something I am assuming you guys know;

  • the relation between potential and force.

  • Relation between potential and force,

  • the potential at x_1 minus the

  • potential at x_2 is the

  • integral of the force from x_1 to

  • x_2.

  • This is how potential is defined.

  • This is the reason why, if you knew the potential minus

  • the derivative gives the force, and if you knew the force,

  • its integral will give you the change in potential.

  • All right, so now when you do rotations,

  • whatever you had for force you had for torque,

  • and whatever played the role of x is played by

  • θ.

  • In other words, in rotational dynamics torque

  • is to force--just like one of the SAT questions.

  • Torque is to force as θ is to x,

  • and work is just work.

  • If you want to calculate the work done by the dipole,

  • but if you like the potential energy when it's at an angle

  • θ and its potential energy when it's at angle 0 is

  • the integral of the torquefrom 0 to

  • θ.

  • Now, what is the torque?

  • The torque is -pEsinθ

  • from 0 to θ.

  • There's a minus sign because if θ tries to

  • increase, the torque tries to decrease it.

  • That's why it has got a minus sign.

  • Now, integral of minus sinθ is

  • cosθ, so you will get

  • pEcosθ - pEcos 0 which is

  • -pE, and that is supposed to equal

  • to U(θ) - U(0).

  • By comparing the two expressions you can identify

  • U(θ) to be-- sorry, this is

  • U(0) minus U(θ),

  • therefore U(θ) is

  • equal to -pEcosθ,

  • or just -p dot E.

  • So that's the final formula you have to remember,

  • that the--can we bring it down here?

  • When you have a dipole in an electric field,

  • it has a potential energy associated with the angle which

  • is -p dot E, and if you draw a picture of

  • that as a function of θ it goes like

  • this.

  • θ = 0 is when the dipole is parallel to the field.

  • That's when it has the minimum energy, -pE.

  • At 90 degrees, energy is 0.

  • At 180 degrees, it's maximum.

  • And the torque is just -dU/dθ and

  • you can see that the torque here and there are zero,

  • but this is the point of stable equilibrium.

  • That's the point of unstable equilibrium.

  • See, if this was a potential energy like a shape of the

  • ground, if you left a marble there

  • it'll stay there, but if it moved a little bit,

  • it would roll down hill.

  • But if you left the marble here it'll stay there,

  • but if you move it, it'll rattle back and forth.

  • That's a stable equilibrium.

  • That's unstable.

  • So for the dipole when it's parallel to the field you are

  • here, and at anti-parallel to the field you are there.

  • The difference is when you're parallel and you move it a

  • little bit, it'll have stable oscillations,

  • whereas if at anti-parallel if you move it a little bit it'll

  • flip over completely and come down here.

  • All right, so I'm going to summarize the main points so you

  • can carry that with you, okay?

  • We saw today that we should think in terms of electric field

  • from now on, and we no longer talk about direct interaction

  • between charges.

  • We say charges produce fields, and fields act on charges to

  • move them.

  • The force of a field on a charge is just q times

  • E.

  • The field is found by adding the field due to all the charges

  • in the universe, provided they're all at rest,

  • and you just add by Coulomb's Law.

  • So we found the field due to a dipole allowing the axis and

  • perpendicular to the axis.

  • We saw the notion of field lines as a good way to visualize

  • what's going on in the vicinity of the charges.

  • The lines tell you the direction where the field is,

  • and they tell you in your line density,

  • lines cutting a unit area perpendicular to them,

  • the strength of the field.

  • Then I calculated for you the field of a dipole along the axis

  • and perpendicular to the axis.

  • There are a lot of formulas, but one thing you should carry

  • in your head.

  • When you've got two equal and opposite charges and you go very

  • far, the field will go like 1 over r^(3).

  • The 1 over r^(2) part of them is canceled,

  • okay?

  • That's the main point, and it goes like 1 over

  • r^(3) times 2 in one place, and 1 over r^(3)

  • times 1 in one place.

  • It doesn't matter.

  • The main thing is it's 1 over r^(3).

  • Finally, if you take a dipole and you put it in an electric

  • field, it tends to line up because

  • there's a torque, p cross E,

  • trying to line it up.

  • With that torque you can associate a potential energy by

  • the usual formula that the torque is minus a derivative of

  • the potential energy.

  • That potential energy is -p dot E.

  • Some of these things may come in handy later on.

  • So you don't have to memorize them, but they'll be involved

  • later on.

Prof: Okay, normally I would ask in a small

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