Lecture on Atomic Structure and Electric Forces

Professor Ramamurti Shankar M.D.: 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 I'm going to 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're 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. Professor Ramamurti Shankar M.S.: 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, Professor Ramamurti Shankar 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're talking about. So you can say q for the neutron is 0, q for the electron is minus 1.6 times 10 to the minus 19, and you can measure it in coulombs. The Q for the proton is really, let me put it this way. Q for the proton is a positive number, so it's really 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 q1 coulombs, that one has a charge of q2 coulombs, and the distance between them is r, then the force So the answer is q1q2 over 4Πε 0 R squared. 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, it'll be repulsive if q1 and q2 are of the same sign and point in the direction joining them. Yep? Shouldn't it be 1 over Professor Ramamurti Shankar Yes, thank you. There's another force law, which is r squared, 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 squared law. Newton is known for the 1 over r squared 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 squared. There is nothing I can do until some voice from the back says, hey, isn't 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, then you're going to have a superposition that's going to be superposition. 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. They are 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 extremely exquisite measurements, but they occur 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 some time. But that's a summary of, this is all I said last time. Okay? Now, I did 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 minus 40. I put this twiddle here, meaning not exactly. There are factors of 1 and 2 or 10. the time missing, but 10 to the minus 40 is roughly the magnitude of this ratio. And it was done by comparing the force between an electron and a proton. It didn't matter what the distance was, because everything goes like 1 over r squared, so when you 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 4Πε0, Q of the electron and Q of the proton. If you put all these numbers in, you got that. I want to mention one thing. This 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 repelling each other? 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 will be attracted. What are the protons doing together in that tiny space? That space is like 10 to the minus 13 centimeters. So have you ever thought of that, or does anybody know why the protons are together? Yes? There's a strong nuclear force? 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's another charge, which is not an 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 1000 times stronger, then why wasn't it completely overshadowing the electrical force? Yes? It's irrelevant to this? 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 squared. Other function looks like 10 times e to the minus r over r ℏ divided by r squared. r ℏ is some length, and the length is roughly 10 to the minus 13 centimeters, or 10 to the minus 15 meters. Which force are you more impressed with is the question, okay? If r is much bigger than r say 10 times bigger, you've got e to the minus 10 on the top. e to the minus 10 is a small number. e to the minus 3 is like 1 over 20. e to the minus 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 0.1 of r-0, so you can forget this number, e to the minus 0.1 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 you've got to add more and more neutrons to compensate the 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 the nucleus has become so big, this factor is no longer negligible. See, over long distances the Coulomb force will always triumph, because no matter what the pre-factors are, the exponential factor in front of the nuclear force will always 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 the 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 squared. So it's a little bit more of a gravity. gravity, on the other hand, is exactly 1 over r squared with a number that's much, much, much smaller than this. But then we saw the other day why gravity manages to survive, because this is q1,q2, and the q's can be added algebraically and cancel each other, whereas if you've got m1, m2 over r squared, there is no way to cancel the end. This is how the different forces manage to survive for the different reasons. Okay? So here wins. It's in the nuclear zone, but dies very quickly outside the size of a nucleus. Electrical forces fall more slowly at like 1 over r squared. They dominate atomic physics. But once you've formed an atom, you've got pretty much an electrically neutral thing. And once you've got many, many atoms making up our planet, then all that remains is a gravitational attraction between a planet and another planet. Okay, so these are examples of different forces and why they were founded 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, how to stick the numbers in to Coulomb's Law. And the only thing I didn't mention is this 1 over 4 times 9 times. 10 to the 9th. So today we are going to do something which is part of a great abstraction, and it goes as follows. So I'm going to take two charges of Q1 here and Q2 here, and I'm going to give them some locations. So let's say this guy is at vector r1, This one is at vector r2. Now I will really take care of all the arrows. This one is r2-r1, that arrow there. You can check that I didn't mess up anything, because r1, plus r2-r1, should be r2. Professor Ramamurti Shankar So let's write the Coulomb force now as a vector. Now you've got to say force on what? I'm going to say force on 2 due to 1. Now you've got to realize it's a convention. So I use the convention, this is the force on this guy and the force due to this guy. It's a second label. Now let's write it out in detail. The force on that is Q1 over 4Πε 0. I'm going to put the Q2 here. Then here I want to write R1-R2 squared. That's the 1 over r squared, but then I have to make it a vector. So for that vector part, once you have the magnitude of the vector, you should multiply it by a 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 E12. E is always going to be unit vector going from 1 to 2. Or if you have, if you're inclined, As r2-r1 divided by the length of r2-r1. 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 equal to 7 because that doesn't tell you which way it's pointing, so invent a vector called i, which is a unit vector in the x direction, and then 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 Q2, which you've got to add to the force due to 1. But 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 R2 times Q2, and this is called the field. Professor Ramamurti Shankar You see, if you look at this thing, all I've done is rewrite the expression as something that involves a charge of q2 and everything else that involves the q1 and the distance from q1 and q2. 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. Here's the problem's law. You say that q1 and q2 exert a force on each other, and the force depends on the charge and the distance between them. Now I'm going to say q1 produces an electric field at the location of q2 given by this vector, and when I multiply by q2 it gives me the force on q2. Now what's the importance of the electric field? Whereas Q1 and Q2 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 r2. So the picture we have is that Q1 produces an electric field all over space, and Q2 experiences that field, and that's what we're going to do here. the force gets repelled as a result, and the force it feels is the field at that point times q2. 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 understand this. The charges are only in two places in our example. but the field due to q1 is everywhere. At every point in space, I can compute the field due to q1. 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 a 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 q2 is equal to 1, then numerically, E is exactly equal to the force. So 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? No, okay. Her answer was, if I want the field here, I should find out where all the charges are, find all the forces they will exert on this guy, on a 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? Raise the test charge. Okay, and then? 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 when you say, I think you know what you meant, but can you, I want you to finish that sentence. Very good. See, that's what I want. When you say, see what happens, does it get married? Does it have children? That's not what I meant. Okay? But you also, 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. 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 a 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's 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, charge, 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 I put a one coulomb here, it experiences a force, which is one 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, 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? The radius is changing. Pardon me? R is changing. R is changing, so we'll keep changing it as the charge moves. But it's only an approximation when charge is moved. Do you know why? Yes? True, but even the electric field, his answer was magnetic field, but even the electric field is not properly given. Yes? Not the Doppler, something else. If Coulomb's Law were exact, okay, here's 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? Relativity. 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 is 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. It does not happen. So the correct way to do it, it'll maybe come later in the course, or 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, It's 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 cable 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 the that if you go to any one point, the force on any charge Q you put there is Q times the 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 a 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. 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. 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're 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 in the electric field. You do one chart, it's q over 4Πε 0, 1 over r squared. If you're here and that's your r vector, the field at this point is that magnitude, and I'm going to write ER, 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 will look like qr over 4Πε 0r cubed. Don't get fooled into thinking the force field is falling at 1 over r cubed. It's really 1 over r squared, because there's an r on the top. If you write it as 1 over r squared, put a unit vector. If you write it as 1 over r cubed, 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 this 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. 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? There are all these contours of different height. 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, 4 points, maybe 8, and I'm going to tell you what the field is at those points. At this point, it looks like that. This point, it looks like that. This point looks like this. It'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 or 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 squared. So you can do this, okay? Professor Ramamurti Shankar M.D.: 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. And someone had this clever idea of doing the following. You can probably guess. Their idea was, why not 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? Professor Ramamurti Shankar Okay, let me repeat that. That's what you guys should have been thinking in your head. When I join 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 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 15 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 you catch mistakes. Okay. All right, so here are these lines. As you 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. the length of the arrows. 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 falls like 1 over r squared, There is information even on the strength of the electric field. 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 area of that surface. Because the area of the surface is perpendicular to the area of the surface, the area of the surface is perpendicular to the area of the surface. Professor Ramamurti Shankar, Professor R.C. M.D.: 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 that sphere also. but they are 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 4Πr². 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 squared. So these lines naturally diverge and spread out in space so that the density falls precisely as 1 over r squared. That has to do also with the fact you're living in three dimensions. Only in three dimensions where area goes like r squared does the 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 That's really up to you, but you've got to be consistent. Once you get 32 lines per coulomb, then if you've got a charge of 1 coulomb, you should draw 32. If you've got 2 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. Please agree that... One coulomb gets 1 over ε0 lines. You know, ε0 is some number, right? 1 over 4 by ε0 is 9 times 10 to the 9. 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. lines, and if you take a sphere of radius r, you'll get 1 over 4Πr² 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, you will always be measuring 2 times the electric field. The density will still convey the electric field, namely it will be proportional to it, but let's make life easy by making it equal to it. This is just a convenience. Now you 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, 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 one is minus Q, other is plus Q. When you're very near the charge, by the way, I'm not going to draw 1 over ε, zero lines per coulomb, because it's going to be too many lines. 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? Yeah? Why does it depend on the one charge? In principle, it depends on every charge. Somebody had an answer back there? Yes? Since it falls as one of the two. Right? Because the field is 1 over r squared, 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 will start out this way, but soon, of course, it won't go out this way forever, because you'll 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 you've just got 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 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 got 2 charges. Professor Ramamurti Shankar If you go a mile away from these two plus charges, what do you think the lines will look like? Yeah? Just like a point charge that's twice as strong. Professor Ramamurti Shankar Right. 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. 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 minus 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, as if there was a charge q at the center, because 2q and minus q give you a net of q. Professor Ramamurti Shankar 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 chart 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 that 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 equal to A. This is at x equal to minus 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. 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 equal to 0y. I'm 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, 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 squared. 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 squared. 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. You can understand why. The minute a is not 0, you're closer to one of the two charges, so they cannot really cancel each other. So, you've got to manipulate this expression. So, I will do that now. Q over 4Πε0, and you find common denominator. I remind you, x plus a times x minus a is x squared minus a squared, and everything is under squares. In the numerator, you've got x A squared minus x A squared. So what does that give me? Iq over 4Πε 0, x squared minus A squared squared. And how about on the top? You can do that in your head. This is going to be x squared A squared 2xA. You're going to subtract from it x squared A squared minus 2xA. The only thing that will survive will be 4xA. Everything okay? It is. You're saying, why is it not x A squared times x A squared, right? It is, because this would be x A times x A, the whole thing squared. And this guy is x squared minus A squared. So this is classified as a nice try. Professor Ramamurti Shankar But I want you to keep doing this. This time I'm right, but you never know, okay? I don't want you to give up. There's nothing better than shooting me down. But this happened to be correct. You satisfied though? No, no, no, no. I don't want to rush through this. Anybody have the same problem with this? Look, also I'm doing it fast because this is the 958th time I'm doing this calculation. So, if you're seeing it for the first time, I've got to slow down. Professor Ramamurti Shankar So let's see which part. Everybody okay with this, right? So I wrote that. Then this is really an x plus a times another x plus a and then x minus a and another x minus a. But without these guys, I know it's x squared minus a squared, because I got two of everything, I squared everything. So this answer is actually an exact formula for 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 x², which could be 1 km², a² which may be 1 mm². So in the first approximation, it's not an exact formula anymore. From now on, it is approximate. If the limit x is much bigger than a, you can see it's going to be q 2A over 2Πε 0 times 2A divided by x cubed. 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 a 4 here to get that. Then on the top I had an x, and the bottom I had x to the 4th. I get x cubed. 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 2Πε0x cubed. So I will tell you what I'm doing. So the final formula I had was electric field E equal to I times P divided by 2Πε 0x cubed, where P, I'm sorry, I times P, no arrow, P is equal to 2AQ. Professor Ramamurti 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. The value of the charge is given by the product of the distance between the charges and the value of one of the charges, the plus charge. Whenever I give you two charges, called a dipole, you can associate with them a vector. And the vector is, If you've got a charge minus q here and a charge plus 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. Professor Ramamurti Shankar So, this means electric field is parallel to the dipole moment and falls like 1 over x cubed. That's the most important part of the dipole. A single charge, the field falls like 1 over x squared, if you move a distance x. A dipole will always fall like a bigger power of x, because to go like 1 over x squared, you've got to have net charge. As long as the net charge is zero, the fact that there are two charges that don't quite cancel each other, It always comes from the fact that the distance between them is not 0. The distance will appear in the numerator, and there must be a corresponding distance in the denominator, because the formula should have the same dimensions. That's what turns a 1 over x squared to 1 over x cubed. Yeah? So the formula works if x is a lot bigger than A. 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 the second place where I'm going to find the field. That's going to be here, and I'm going to find the field there. So that's at a distance y. So let's look at this. There's a q here and a q here. q will ripple it that way, and the q will attract it this way, and the 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. the horizontal part. 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 minus I. That should tell me it's in the right direction. times q over 4Πε 0 times 1 over distance squared, which is y squared, plus a squared. This is y squared plus a squared. That's also y squared plus a squared. 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 squared. I'm going to put a 2, plus a squared to the 1 half. See, you want to take that force and find a horizontal part. Then I'm going to put 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 squared plus A squared, which is Iq2A over 2. That is then minus P divided by 4Πε 0 divided by y cubed, for y much bigger than A. I'm sorry, y cubed. y is really the distance. Professor Ramamurti Shankar So, right? If y is 1 mile and a is 1 millimeter, that's essentially the distance. If you like, you can call it minus p divided by 4Πε0r cubed. But r is the distance, if you like, from the center of the dipole. Professor Ramamurti Shankar M.D.: Look, the point of this exercise is twofold. One is to show you how to add vectorially the fields due to two guys. Professor Ramamurti Shankar And other 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? Is a vector different from the p that you crossed out as a vector? Professor Ramamurti Shankar 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, Professor Ramamurti Shankar I'm going to do the second part of the problem, which is finding the response to E. This was all computing E. Professor Ramamurti Shankar M. Professor Ramamurti Shankar M. Professor Ramamurti Shankar M. Professor Ramamurti Shankar M. Professor Ramamurti Shankar M. Professor Ramamurti Shankar M. Professor Ramamurti Shankar M. Professor Ramamurti Shankar M. Professor Ramamurti Shankar M. Professor Ramamurti Shankar M. Professor Ramamurti Shankar M. Professor Ramamurti Shankar M. Professor Ramamurti Shankar M. Professor Ramamurti Shankar M. Professor Ramamurti Shankar M. Professor Ramamurti Shankar M. Professor Ramamurti Shankar M. Professor Ramamurti Shankar M. Professor Ramamurti Shankar M. Professor Ramamurti Shankar M. and I shoot a particle here with some velocity v-0 in the x direction, and the field everywhere is down, and the electric field is some constant, a minus j times some number e-0. Minus j because i is this way and j is that way. So what will this do is the question, 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 The acceleration will be minus qE₀ over m times j in the y direction. So what will be the position? The position will be from lecture number 1 of your Physics 200. Professor Ramamurti Shankar So, let's say the starting point, r naught, is our origin. v naught is whatever it was projected in with, plus times qE0 over Professor Ramamurti Shankar Okay. So, as a function of time, this tells you where the position will be. At t equal to 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, Professor Ramamurti Shankar M.D.: 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? Yes? The distance. Professor Ramamurti Shankar M.D.: Which velocity should we take? The initial. Professor Ramamurti Shankar M.D.: That's correct, because T is the distance. the velocity of the acceleration will be L over V0, where V0 is 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 factors falling in the y direction. So, if you put t equal to all of this, where it will end up. And that's how you make pictures on the television. You've got a bunch of plates, then you drive charges, and if you apply the right electric field, the electron will land on a screen and make a little dot, then you will, well, 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, and you're going to have a 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 naught and it's in the x direction. And in this electric field, I stick a dipole in, like that. Here is the plus Q, here is the minus 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 naught. That will be minus q times E naught. So the 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 was stronger here than here, then of course it'll 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? Yes? The torque. The torque. 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. The 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 A sin θ. Then there's another a sin θ from that one. That's the torque. Since 2Qa is p, it's pE₀ sin θ. You can see this makes sense. If θ was 0, if the dipole is 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 way, that gets rotated that way, you get the maximum torque. At θ equal to π by 2, you get a torque of P times E zero. That's also the reason it's called a dipole moment. Now, I'm going to write this as a cross product. I'm assuming you guys are familiar with the cross product. 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. 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 would like to become horizontal. So, there is a certain restoring torque that tries to rotate it so it becomes horizontal. the force is equal to the force of the force. And it's not very different from a spring, where if you pull it from equilibrium, that's the 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 minus kx, we can assign a potential energy u, which is kx squared, so that the force is equal to There's something I'm assuming you guys know, the relation between potential and force. The relation between potential and force, the potential at x1 minus the potential at x2 is the integral of the force from x1 to x2. 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. Professor Ramamurti Shankar M.D.: In the 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, or if you like, the potential energy when it's at an angle θ, when it's potential energy when it's at an angle 0, is the integral of the torque dθ from 0 to θ. Professor Ramamurti Shankar Now what is the torque? The torque is minus pE sin θ dθ 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 sine θ is cosine θ, so you will get pEcosθ minus pEcos 0, which is minus pE. That is supposed to be equal to u of θ minus u of 0. By comparing the two expressions, you can identify u of θ to be, sorry, this is u of 0 minus u of θ. Therefore, U is equal to pEcosθ, which is p.e. So that's the final formula you have to remember, that the Let me bring it down here. When you have a dipole in an electric field, it has a potential energy associated with the angle, which is minus p dot e. If you draw a picture of that as a function of θ, it goes like this. θ equals 0 is when the dipole is parallel to the field. That's when it has the minimum energy, minus Pe. At 90 degrees, energy is 0. At 180 degrees, it's maximum. And the torque is just minus du dθ, and you can see that the torque here and there are 0, 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'll roll downhill. But if you left a marble here, it'll stay there, but if you moved 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. When you're anti-parallel to the field, you are there. So what happens is, when you're parallel, and you move it a little bit, it'll have stable oscillations, whereas if you're 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 point so you can carry that with you. Okay? We saw today that we should think in terms of electric field from now on. 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 along the axis and perpendicular to the axis. We saw the notion of field lines. It's a good way to visualize what's going on in the vicinity of charges. The lines tell you in the and they tell you in your light 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 cubed. 1 over r squared, part of them is canceled. That's the main point. And it goes like 1 over r cubed times 2 in one place is 1 over r cubed times. 1 in one place, it doesn't matter. The main thing is it's 1 over r cubed. 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 x E, trying to line it up. And with that torque, you can associate a potential energy. Professor Ramamurti Shankar M.D.: By the usual formula, that the torque is minus a derivative of the potential energy. That potential energy is minus 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 invoked later on.

What I will do first is I'm going to 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're 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. Professor Ramamurti Shankar M.S.: 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, Professor Ramamurti Shankar 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're talking about.

So you can say q for the neutron is 0, q for the electron is minus 1.6 times 10 to the minus 19, and you can measure it in coulombs. The Q for the proton is really, let me put it this way. Q for the proton is a positive number, so it's really 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 q1 coulombs, that one has a charge of q2 coulombs, and the distance between them is r, then the force So the answer is q1q2 over 4Πε 0 R squared.

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, it'll be repulsive if q1 and q2 are of the same sign and point in the direction joining them. Yep?

Shouldn't it be 1 over Professor Ramamurti Shankar Yes, thank you. There's another force law, which is r squared, 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 squared law.

Newton is known for the 1 over r squared 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 squared.

There is nothing I can do until some voice from the back says, hey, isn't 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, then you're going to have a superposition that's going to be superposition. 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.

They are 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 extremely exquisite measurements, but they occur 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 some time. But that's a summary of, this is all I said last time. Okay?

Now, I did 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 minus 40. I put this twiddle here, meaning not exactly. There are factors of 1 and 2 or 10. the time missing, but 10 to the minus 40 is roughly the magnitude of this ratio. And it was done by comparing the force between an electron and a proton. It didn't matter what the distance was, because everything goes like 1 over r squared, so when you 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 4Πε0, Q of the electron and Q of the proton.

If you put all these numbers in, you got that. I want to mention one thing. This 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 repelling each other? 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 will be attracted.

What are the protons doing together in that tiny space? That space is like 10 to the minus 13 centimeters. So have you ever thought of that, or does anybody know why the protons are together?

Yes? There's a strong nuclear force? 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's another charge, which is not an 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 1000 times stronger, then why wasn't it completely overshadowing the electrical force? Yes?

It's irrelevant to this? 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 squared. Other function looks like 10 times e to the minus r over r ℏ divided by r squared.

r ℏ is some length, and the length is roughly 10 to the minus 13 centimeters, or 10 to the minus 15 meters. Which force are you more impressed with is the question, okay? If r is much bigger than r say 10 times bigger, you've got e to the minus 10 on the top. e to the minus 10 is a small number.

e to the minus 3 is like 1 over 20. e to the minus 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 0.1 of r-0, so you can forget this number, e to the minus 0.1 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 you've got to add more and more neutrons to compensate the 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 the nucleus has become so big, this factor is no longer negligible. See, over long distances the Coulomb force will always triumph, because no matter what the pre-factors are, the exponential factor in front of the nuclear force will always 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 the 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 squared.

So it's a little bit more of a gravity. gravity, on the other hand, is exactly 1 over r squared with a number that's much, much, much smaller than this. But then we saw the other day why gravity manages to survive, because this is q1,q2, and the q's can be added algebraically and cancel each other, whereas if you've got m1, m2 over r squared, there is no way to cancel the end. This is how the different forces manage to survive for the different reasons.

Okay? So here wins. It's in the nuclear zone, but dies very quickly outside the size of a nucleus. Electrical forces fall more slowly at like 1 over r squared. They dominate atomic physics.

But once you've formed an atom, you've got pretty much an electrically neutral thing. And once you've got many, many atoms making up our planet, then all that remains is a gravitational attraction between a planet and another planet. Okay, so these are examples of different forces and why they were founded 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, how to stick the numbers in to Coulomb's Law. And the only thing I didn't mention is this 1 over 4 times 9 times. 10 to the 9th. So today we are going to do something which is part of a great abstraction, and it goes as follows.

So I'm going to take two charges of Q1 here and Q2 here, and I'm going to give them some locations. So let's say this guy is at vector r1, This one is at vector r2. Now I will really take care of all the arrows. This one is r2-r1, that arrow there. You can check that I didn't mess up anything, because r1, plus r2-r1, should be r2.

Professor Ramamurti Shankar So let's write the Coulomb force now as a vector. Now you've got to say force on what? I'm going to say force on 2 due to 1. Now you've got to realize it's a convention.

So I use the convention, this is the force on this guy and the force due to this guy. It's a second label. Now let's write it out in detail.

The force on that is Q1 over 4Πε 0. I'm going to put the Q2 here. Then here I want to write R1-R2 squared. That's the 1 over r squared, but then I have to make it a vector. So for that vector part, once you have the magnitude of the vector, you should multiply it by a 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 E12. E is always going to be unit vector going from 1 to 2. Or if you have, if you're inclined, As r2-r1 divided by the length of r2-r1. 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 equal to 7 because that doesn't tell you which way it's pointing, so invent a vector called i, which is a unit vector in the x direction, and then 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 Q2, which you've got to add to the force due to 1. But 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 R2 times Q2, and this is called the field. Professor Ramamurti Shankar You see, if you look at this thing, all I've done is rewrite the expression as something that involves a charge of q2 and everything else that involves the q1 and the distance from q1 and q2. 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.

Here's the problem's law. You say that q1 and q2 exert a force on each other, and the force depends on the charge and the distance between them. Now I'm going to say q1 produces an electric field at the location of q2 given by this vector, and when I multiply by q2 it gives me the force on q2. Now what's the importance of the electric field?

Whereas Q1 and Q2 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 r2. So the picture we have is that Q1 produces an electric field all over space, and Q2 experiences that field, and that's what we're going to do here.

the force gets repelled as a result, and the force it feels is the field at that point times q2. 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 understand this.

The charges are only in two places in our example. but the field due to q1 is everywhere. At every point in space, I can compute the field due to q1. 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 a 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 q2 is equal to 1, then numerically, E is exactly equal to the force.

So 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?

No, okay. Her answer was, if I want the field here, I should find out where all the charges are, find all the forces they will exert on this guy, on a 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? Raise the test charge. Okay, and then? 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 when you say, I think you know what you meant, but can you, I want you to finish that sentence.

Very good. See, that's what I want. When you say, see what happens, does it get married? Does it have children?

That's not what I meant. Okay? But you also, 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. 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 a 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's 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, charge, 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 I put a one coulomb here, it experiences a force, which is one 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, 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? The radius is changing.

Pardon me? R is changing. R is changing, so we'll keep changing it as the charge moves.

But it's only an approximation when charge is moved. Do you know why? Yes? True, but even the electric field, his answer was magnetic field, but even the electric field is not properly given.

Yes? Not the Doppler, something else. If Coulomb's Law were exact, okay, here's 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?

Relativity. 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 is 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. It does not happen.

So the correct way to do it, it'll maybe come later in the course, or 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, It's 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 cable 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 the that if you go to any one point, the force on any charge Q you put there is Q times the 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 a 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.

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. 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're 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 in the electric field. You do one chart, it's q over 4Πε 0, 1 over r squared. If you're here and that's your r vector, the field at this point is that magnitude, and I'm going to write ER, 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 will look like qr over 4Πε 0r cubed.

Don't get fooled into thinking the force field is falling at 1 over r cubed. It's really 1 over r squared, because there's an r on the top. If you write it as 1 over r squared, put a unit vector.

If you write it as 1 over r cubed, 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 this 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. 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?

There are all these contours of different height. 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, 4 points, maybe 8, and I'm going to tell you what the field is at those points.

At this point, it looks like that. This point, it looks like that. This point looks like this. It'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 or 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 squared.

So you can do this, okay? Professor Ramamurti Shankar M.D.: 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. And someone had this clever idea of doing the following.

You can probably guess. Their idea was, why not 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? Professor Ramamurti Shankar Okay, let me repeat that. That's what you guys should have been thinking in your head.

When I join 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 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 15 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 you catch mistakes. Okay. All right, so here are these lines.

As you 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. the length of the arrows. 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 falls like 1 over r squared, There is information even on the strength of the electric field. 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 area of that surface. Because the area of the surface is perpendicular to the area of the surface, the area of the surface is perpendicular to the area of the surface. Professor Ramamurti Shankar, Professor R.C. M.D.: 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 that sphere also. but they are 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 4Πr². 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 squared. So these lines naturally diverge and spread out in space so that the density falls precisely as 1 over r squared. That has to do also with the fact you're living in three dimensions. Only in three dimensions where area goes like r squared does the 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 That's really up to you, but you've got to be consistent.

Once you get 32 lines per coulomb, then if you've got a charge of 1 coulomb, you should draw 32. If you've got 2 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. Please agree that... One coulomb gets 1 over ε0 lines.

You know, ε0 is some number, right? 1 over 4 by ε0 is 9 times 10 to the 9. 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. lines, and if you take a sphere of radius r, you'll get 1 over 4Πr² 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, you will always be measuring 2 times the electric field. The density will still convey the electric field, namely it will be proportional to it, but let's make life easy by making it equal to it.

This is just a convenience. Now you 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, 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 one is minus Q, other is plus Q. When you're very near the charge, by the way, I'm not going to draw 1 over ε, zero lines per coulomb, because it's going to be too many lines.

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? Yeah?

Why does it depend on the one charge? In principle, it depends on every charge. Somebody had an answer back there?

Yes? Since it falls as one of the two. Right?

Because the field is 1 over r squared, 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 will start out this way, but soon, of course, it won't go out this way forever, because you'll 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 you've just got 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 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 got 2 charges. Professor Ramamurti Shankar If you go a mile away from these two plus charges, what do you think the lines will look like?

Yeah? Just like a point charge that's twice as strong. Professor Ramamurti Shankar Right. 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. 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 minus 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, as if there was a charge q at the center, because 2q and minus q give you a net of q. Professor Ramamurti Shankar 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 chart 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 that 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 equal to A. This is at x equal to minus 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. 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 equal to 0y.

I'm 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, 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 squared.

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 squared. 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. You can understand why.

The minute a is not 0, you're closer to one of the two charges, so they cannot really cancel each other. So, you've got to manipulate this expression. So, I will do that now.

Q over 4Πε0, and you find common denominator. I remind you, x plus a times x minus a is x squared minus a squared, and everything is under squares. In the numerator, you've got x A squared minus x A squared.

So what does that give me? Iq over 4Πε 0, x squared minus A squared squared. And how about on the top? You can do that in your head.

This is going to be x squared A squared 2xA. You're going to subtract from it x squared A squared minus 2xA. The only thing that will survive will be 4xA. Everything okay? It is.

You're saying, why is it not x A squared times x A squared, right? It is, because this would be x A times x A, the whole thing squared. And this guy is x squared minus A squared.

So this is classified as a nice try. Professor Ramamurti Shankar But I want you to keep doing this. This time I'm right, but you never know, okay? I don't want you to give up.

There's nothing better than shooting me down. But this happened to be correct. You satisfied though? No, no, no, no.

I don't want to rush through this. Anybody have the same problem with this? Look, also I'm doing it fast because this is the 958th time I'm doing this calculation. So, if you're seeing it for the first time, I've got to slow down. Professor Ramamurti Shankar So let's see which part.

Everybody okay with this, right? So I wrote that. Then this is really an x plus a times another x plus a and then x minus a and another x minus a.

But without these guys, I know it's x squared minus a squared, because I got two of everything, I squared everything. So this answer is actually an exact formula for 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 x², which could be 1 km², a² which may be 1 mm². So in the first approximation, it's not an exact formula anymore.

From now on, it is approximate. If the limit x is much bigger than a, you can see it's going to be q 2A over 2Πε 0 times 2A divided by x cubed. 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 a 4 here to get that. Then on the top I had an x, and the bottom I had x to the 4th. I get x cubed.

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 2Πε0x cubed.

So I will tell you what I'm doing. So the final formula I had was electric field E equal to I times P divided by 2Πε 0x cubed, where P, I'm sorry, I times P, no arrow, P is equal to 2AQ. Professor Ramamurti 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.

The value of the charge is given by the product of the distance between the charges and the value of one of the charges, the plus charge. Whenever I give you two charges, called a dipole, you can associate with them a vector. And the vector is, If you've got a charge minus q here and a charge plus 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.

Professor Ramamurti Shankar So, this means electric field is parallel to the dipole moment and falls like 1 over x cubed. That's the most important part of the dipole. A single charge, the field falls like 1 over x squared, if you move a distance x. A dipole will always fall like a bigger power of x, because to go like 1 over x squared, you've got to have net charge. As long as the net charge is zero, the fact that there are two charges that don't quite cancel each other, It always comes from the fact that the distance between them is not 0. The distance will appear in the numerator, and there must be a corresponding distance in the denominator, because the formula should have the same dimensions.

That's what turns a 1 over x squared to 1 over x cubed. Yeah? So the formula works if x is a lot bigger than A. 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 the second place where I'm going to find the field. That's going to be here, and I'm going to find the field there. So that's at a distance y.

So let's look at this. There's a q here and a q here. q will ripple it that way, and the q will attract it this way, and the 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.

the horizontal part. 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 minus I. That should tell me it's in the right direction.

times q over 4Πε 0 times 1 over distance squared, which is y squared, plus a squared. This is y squared plus a squared. That's also y squared plus a squared.

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 squared. I'm going to put a 2, plus a squared to the 1 half.

See, you want to take that force and find a horizontal part. Then I'm going to put 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 squared plus A squared, which is Iq2A over 2. That is then minus P divided by 4Πε 0 divided by y cubed, for y much bigger than A.

I'm sorry, y cubed. y is really the distance. Professor Ramamurti Shankar So, right? If y is 1 mile and a is 1 millimeter, that's essentially the distance. If you like, you can call it minus p divided by 4Πε0r cubed.

But r is the distance, if you like, from the center of the dipole. Professor Ramamurti Shankar M.D.: Look, the point of this exercise is twofold. One is to show you how to add vectorially the fields due to two guys.

Professor Ramamurti Shankar And other 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? Is a vector different from the p that you crossed out as a vector?

Professor Ramamurti Shankar 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, Professor Ramamurti Shankar I'm going to do the second part of the problem, which is finding the response to E. This was all computing E.

Professor Ramamurti Shankar M. Professor Ramamurti Shankar M. Professor Ramamurti Shankar M. Professor Ramamurti Shankar M. Professor Ramamurti Shankar M. Professor Ramamurti Shankar M. Professor Ramamurti Shankar M. Professor Ramamurti Shankar M. Professor Ramamurti Shankar M. Professor Ramamurti Shankar M. Professor Ramamurti Shankar M. Professor Ramamurti Shankar M. Professor Ramamurti Shankar M. Professor Ramamurti Shankar M. Professor Ramamurti Shankar M. Professor Ramamurti Shankar M. Professor Ramamurti Shankar M. Professor Ramamurti Shankar M. Professor Ramamurti Shankar M. Professor Ramamurti Shankar M. and I shoot a particle here with some velocity v-0 in the x direction, and the field everywhere is down, and the electric field is some constant, a minus j times some number e-0. Minus j because i is this way and j is that way.

So what will this do is the question, 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 The acceleration will be minus qE₀ over m times j in the y direction. So what will be the position?

The position will be from lecture number 1 of your Physics 200. Professor Ramamurti Shankar So, let's say the starting point, r naught, is our origin. v naught is whatever it was projected in with, plus times qE0 over Professor Ramamurti Shankar Okay. So, as a function of time, this tells you where the position will be. At t equal to 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, Professor Ramamurti Shankar M.D.: 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? Yes? The distance.

Professor Ramamurti Shankar M.D.: Which velocity should we take? The initial. Professor Ramamurti Shankar M.D.: That's correct, because T is the distance.

the velocity of the acceleration will be L over V0, where V0 is 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 factors falling in the y direction.

So, if you put t equal to all of this, where it will end up. And that's how you make pictures on the television. You've got a bunch of plates, then you drive charges, and if you apply the right electric field, the electron will land on a screen and make a little dot, then you will, well, 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, and you're going to have a 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 naught and it's in the x direction. And in this electric field, I stick a dipole in, like that.

Here is the plus Q, here is the minus 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 naught. That will be minus q times E naught. So the 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 was stronger here than here, then of course it'll 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? Yes? The torque. The torque.

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. The 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 A sin θ. Then there's another a sin θ from that one. That's the torque. Since 2Qa is p, it's pE₀ sin θ. You can see this makes sense.

If θ was 0, if the dipole is 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 way, that gets rotated that way, you get the maximum torque. At θ equal to π by 2, you get a torque of P times E zero. That's also the reason it's called a dipole moment.

Now, I'm going to write this as a cross product. I'm assuming you guys are familiar with the cross product. 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. 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 would like to become horizontal. So, there is a certain restoring torque that tries to rotate it so it becomes horizontal. the force is equal to the force of the force. And it's not very different from a spring, where if you pull it from equilibrium, that's the 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 minus kx, we can assign a potential energy u, which is kx squared, so that the force is equal to There's something I'm assuming you guys know, the relation between potential and force. The relation between potential and force, the potential at x1 minus the potential at x2 is the integral of the force from x1 to x2.

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.

Professor Ramamurti Shankar M.D.: In the 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, or if you like, the potential energy when it's at an angle θ, when it's potential energy when it's at an angle 0, is the integral of the torque dθ from 0 to θ. Professor Ramamurti Shankar Now what is the torque? The torque is minus pE sin θ dθ 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 sine θ is cosine θ, so you will get pEcosθ minus pEcos 0, which is minus pE.

That is supposed to be equal to u of θ minus u of 0. By comparing the two expressions, you can identify u of θ to be, sorry, this is u of 0 minus u of θ. Therefore, U is equal to pEcosθ, which is p.e. So that's the final formula you have to remember, that the Let me bring it down here. When you have a dipole in an electric field, it has a potential energy associated with the angle, which is minus p dot e.

If you draw a picture of that as a function of θ, it goes like this. θ equals 0 is when the dipole is parallel to the field. That's when it has the minimum energy, minus Pe. At 90 degrees, energy is 0. At 180 degrees, it's maximum.

And the torque is just minus du dθ, and you can see that the torque here and there are 0, 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'll roll downhill.

But if you left a marble here, it'll stay there, but if you moved 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. When you're anti-parallel to the field, you are there. So what happens is, when you're parallel, and you move it a little bit, it'll have stable oscillations, whereas if you're 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 point so you can carry that with you. Okay? We saw today that we should think in terms of electric field from now on.

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 along the axis and perpendicular to the axis.

We saw the notion of field lines. It's a good way to visualize what's going on in the vicinity of charges. The lines tell you in the and they tell you in your light 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 cubed. 1 over r squared, part of them is canceled. That's the main point. And it goes like 1 over r cubed times 2 in one place is 1 over r cubed times.

1 in one place, it doesn't matter. The main thing is it's 1 over r cubed. 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 x E, trying to line it up.

And with that torque, you can associate a potential energy. Professor Ramamurti Shankar M.D.: By the usual formula, that the torque is minus a derivative of the potential energy. That potential energy is minus 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 invoked later on.

Transcript for:Lecture on Atomic Structure and Electric Forces

Transcript for:
Lecture on Atomic Structure and Electric Forces