Fundamentals of Physics II

PHYS 201 - Lecture 2 - Electric Fields

Chapter 1: Review of Charges [00:00:00]

Professor Ramamurti Shankar: Okay, normally I would ask in a small class if there is something you didn’t follow from last time. I’m afraid to do that now, because it’s a big class and I don’t know how many things you follow or didn’t follow. What I will do first is write down a very quick summary of the main points from last time. So you should ask yourself, “Did I follow all those things?” And if your answer is yes, then you are fine. Because I talked about many, many things, but you don’t need all of that. So what I’m going to write down right now is the absolute essentials of last lecture, okay? That’s going to be needed for what I do next.

First point was: everything is made of atoms. You know that. And the atom has a nucleus. In the nucleus are some things called protons, some things called neutrons, and outside are some things called electrons. That’s all the atomic structure we need. Then we say certain entities have a property called electric charge. The symbol for electric charge is q, and you can put a subscript to say who you are talking about. So you can say q for the neutron is 0. A q for the electron is -1.6 times 10 to the -19, and it’s measured in coulombs. The q for the proton is really — let me put it this way, q for the proton is a positive number, so these minus signs cancel. Now, the importance of the coulomb is that if anything has some coulombs on it, it will interact with anything else that has some coulombs on it. So that if you have two entities and this one has a charge of q₁ coulombs, that one has a charge of q₂ coulombs, and the distance between them is r, then the force is q₁q₂over 4Πε₀r². I’m purposely not putting all the vector signs on F because it takes too long, but you all know what the answer is. Namely, if you want the force on 2, due to 1, will be repulsive if q₁ and q₂ are of the same sign and point to the direction joining them. Yep?

Student: Shouldn’t it be 1 over r²?

Professor Ramamurti Shankar: Yes, thank you. There is another force law which is r², 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 r²law. Newton is known for the 1 over r²law. 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 r². There is nothing I can do until some voice from the back says, “Hey, send it downstairs”, and I appreciate that a lot, okay? It’s good, so never hesitate to do that, plus sign, minus sign, symbols; anything that goes wrong. It also tells me that you’re following me. So for all those reasons you should not hesitate to correct anything, and you should not think that you don’t follow it because it’s your fault. Probably it is. Sometimes it’s my fault as was demonstrated now. Okay?

So this is the force law. You need only one other ingredient. That’s the superposition principle. You need that ingredient because we’re not going to be talking only about two charges. We’re going to be talking about many charges. And the question is: what will they do when they’re all present, and it’s a great blessing that we have the principle of superposition that says that if you’ve got, say, three charges, say 1, 2 and 3, you want to know the force on 3 due to 2 and 1. You can find the force that 1 would exert. You can find the force that 2 would exert. We have two vectors, and you can add the two vectors to get the net force. In other words, the interaction between pairs of charges is insensitive to the presence of any other charges. They go about their business exactly the same way. That is a principle that is not deduced by logic. You cannot say, “Of course, it had to be that way.” That’s not true. It doesn’t have to be that way. In fact it is not that way if you make supreme, I mean, extremely exquisite measurements, but they occur really at the really quantum level. For classical electromagnetic theory it is actually an experimental fact that you can superpose.

All right, if you combine those things you can calculate anything and everything that we will deal with for sometime, but that’s summary of — this is all I said last time, okay? Now, I didn’t do one thing, which is to emphasize to you that the force of gravitation divided by the electric force is some number like 10 to the -40. I put this twiddle here meaning not exactly. There are factors of 1 and 2 or 10 missing, but 10 to the -40 is roughly the magnitude of this ratio, and it was done by comparing the force between electron and a proton. It didn’t matter what the distance was because everything goes like 1 over r², so when they take the ratio that cancels out, but this was really the ratio of things like mass of the electron, mass of the proton, divided by 1 over 4Πε₀; q of the electron and q of the proton. If you put all these numbers in you’ve got that. I want to mention one thing. It may be interesting for you to think about. Look at this nucleus. Nucleus has a lot of protons in it, and according to Coulomb’s Law they all repel each other. And the neutrons, of course, don’t do anything because they have no electric charge. So you should ask yourself, “Why are all these protons staying together inside the nucleus if I don’t see any attraction between them?” I see why the electrons are hanging around, because they’re attracted to the nucleus, because nucleus is positively charged, electrons are negatively charged, and Mr. Coulomb tells you they’ll be attracted. What are the protons doing together in that tiny space? That space is like 10 to the -13 centimeters. So have you ever thought of that, or does anybody know why the protons are together? Yes?

Student: The strong nuclear force?

Professor Ramamurti Shankar: There is a strong nuclear force. Okay, that’s the answer. In other words, but that answer will lead to more questions, but I’ll first state the answer. There’s a force even stronger than the electric force. You see, this gives the impression electric force is very strong. It is much stronger than gravity, but there is a force even stronger than the electric force that is experienced by protons and by neutrons. In other words there is another charge which is not electric charge, which the protons are endowed with and the neutrons are endowed with. And the force due to that charge, if you like, it’s not called charge but it’s a similar thing, is much, much stronger. It had to be, because you have to beat the electrical repulsion. So then you can ask yourself, “Well, in that case, how did we ever find the electrical force,” because here’s another force even stronger than the electrical force, maybe a thousand times stronger. Then why wasn’t it completely overshadowing the electrical force? Yep?

Student: It’s irrelevant over larger distances.

Professor Ramamurti Shankar: 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 r². Other function looks like 10 times e to the -r over r₀ divided by r²; r₀ is some length, and the length is roughly 10 to the -13 centimeters, or 10 to -15 meters. Which force are you more impressed with is the question, okay? If r is much bigger than r₀, say 10 times bigger, you’ve got e to the -10 on the top. e to the -10 is a small number. e to the -3 is like 1 over 20. e to the -10 is 1 over 20 cubed. So then what will happen is, this force, even though there’s a number 10 in front of it, will be negligible compared to this one. Whereas if r is much smaller than r₀, say, r is .1 of r₀, so you can forget this number, e to the minus point is roughly 1, then this 10 will dominate the 1 over 100. So what happens is, if you’re sitting inside the nucleus, the nuclear force is numerically strong, not only because of the number in front of it, but also because this exponential factor has not kicked in. Therefore, the protons feel an attraction for each other due to the nuclear force that is stronger than the repulsion due to the coulomb force. And the neutrons attract the protons just as much as the protons attract the protons with respect to the nuclear force. So neutrons are actually good, because when you throw an extra neutron into an atom you don’t add to the coulomb repulsion, but you add to the overall attraction that they all feel for each other due to the nuclear force. So neutrons are like the glue. As you add more and more protons, you will find that you’ve got to add more and more neutrons to compensate the coulomb attraction.

But a time will come when the nucleus becomes so big, that even if you add enough neutrons, the repulsion between protons from one end of the nucleus to the other is now becoming comparable to the attraction due to the nuclear force, because nucleus has become so big this factor is no longer negligible. See, over long distances a coulomb force will always triumph, because no matter what the prefactors are, the exponential factor in front of the nuclear force will always weaken it. That’s why you cannot have nuclei beyond some size. If you make them any bigger, the nuclear electrical repulsion between the distant parts of the protons, due to the distant protons in the nucleus, cannot be compensated by the short-range attraction. So it’s the range of the interaction that is a significant factor here. So all the strong forces are strong, but at short distances the coulomb force is not that strong, but it falls like 1 over r². Now gravity, on the other hand, is exactly 1 over r²with a number that’s much, much, much smaller than this, but then we saw the other day why gravity managed to survive, because this is q₁q₂ and the q’s can be added algebraically and cancel each other. Whereas if you’ve got m₁m₂over r²there is no way to cancel the m.

This is how the different forces manage to survive for the different reasons, okay? Nuclear wins in the nuclear zone, but dies very quickly outside the size of a nucleus. Electrical forces falls more slowly at like 1 over r². They dominate atomic physics, but once you formed an atom you’ve got pretty much an electrical neutral thing, and once you’ve got many, many atoms making up our planet, then all that remains is the gravitational attraction between a planet and another planet. Okay, so these are examples of different forces and why they were found at various times, because they all dominate under different circumstances.

All right, so today I’m going to start with my new stuff. So all you need to know, really, if you want to do your problem sets and your homework, is Coulomb’s Law, you know, how to stick the numbers in the Coulomb’s Law. And the only thing I didn’t mention is this 1 over 4Πε₀ is 9 times 10 to the 9^th. So today we’re going to do something which is a part of a great abstraction and it goes as follows: So I’m gong to take two charges, a 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 are of all the arrows. This one is r₂ - r₁, that arrow there. You can check that I didn’t mess up anything, because r₁ + r₂ - r₁ should be r₂. So let’s write the coulomb force now as a vector. And you’ve got to say force on what, so I’m going to say force on 2, due to 1. Now, you’ve got to realize it’s a convention. So I use a convention: This is the force on this guy and the force due to this guy, with the second later.

Now, let’s write it out in detail. So that is q₁ over 4Πε₀. I’m going to put the q₂ here. Then here I want to write (r₁ - r₂)². That’s the 1 over r², but then I have to make it a vector. So for the vector part, once you have the magnitude of the vector, you should multiply by vector of length unit 1 going in the direction of this difference vector. So you can use any symbol you like. One is to say e₁₂. e’s always going to be unit vector going from 1 to 2, or if you’re inclined you can also write e₁₂ 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 = 7, because that doesn’t tell you which way it’s pointing. So invent the vector called i, which is a unit vector in the x direction and I multiply it by that. So 7i is a vector parallel to i and 7 times long. 7.3i is a vector parallel to i 7.3 times long, okay? So you need to add a vector of unit length to the magnitude, multiply it to get the actual vector. So that’s that formula.

Chapter 2: Electric Fields [00:16:35]

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, but you’ve got to add to the force due to 1. I’m not going to do that. I’m just taking two guys. Now, I’m going to formally write this as equal to the electric field at r₂ times q₂, and this is called a field. You see, if you look at this thing, all I’ve done is rewrite the expression as something that involves a charge of 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. In Coulomb’s Law you say that q₁ and q₂ exert a force on each other, okay? And the force depends on the charge and the distance between them. Now I’m going to say 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 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 to 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 a 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 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. If I go to you and say, “Find out the electric field at this point.” You say, “Okay, where?” I say, “Here.” What do you think you have to do to measure the field? Yes?

Student: Add up all the forces on that field from the different charges, or add up all of the fields at that point…

Professor Ramamurti Shankar: No, okay. Her answer was, “If I want the field here, I should find out where all the charges are,” right? “Find all the forces they will exert on this guy, on the unit charge here,” right? But I’m telling you, that’s correct. That’s the theoretical way to calculate the field at that point, but suppose you’re an experimentalist and you don’t want to know what produced it. You just want an answer to what’s the field here. What will you do? Yes?

Student: Place a test charge.

Professor Ramamurti Shankar: Okay, and then?

Student: And then see what happens to it.

Professor Ramamurti Shankar: By ‘see what happens’, you’ve got to be more — so his answer was: put a test charge, and see what happens. Now, you’ve got to be more precise. I think you know what you meant, but I want you to finish that sentence.

Student: Okay. Well, you can measure the force by measuring the acceleration on the charge.

Professor Ramamurti Shankar: Very good. See, that’s what I want. When you said, “See what happens,” does it get married? Does it have children? That’s not what I meant, okay? But you did give the right answer. The answer was, by ‘see what happens’, you put the charge there, and you see what acceleration it undergoes. Then that acceleration is the — that times the mass is the force it’s experiencing. That should be q that you place there times E. If q was 1 that force itself is equal to E. If q was 10 you’ve got to divide the force by 10 to get the field there. So, the field is like the sound of one hand clapping. People say one hand clapping is Zen concept, but the field is like that, because you don’t need two charges to have a field. You just need one charge.

So here is how we understand that. You put a charge q. If you go here something has happened there, see? You don’t need to put a second charge there to conclude something has happened. Something really is different at this location because the charge q is present. What is different? What is different is that when this guy was not here and I put a charge, it just sat there. Whereas when this guy is here and I put a charge, it experiences a force. So if you put a 1 coulomb here it experiences a force which is 1 times the field at that point, therefore, this charge has distorted the space around it, in fact, everywhere. Now, if you’ve got many, many, many charges, then they will all try to produce a force on a unit charge and you should, like you said, add up the vectors, which are the forces due to all the other charges on that one location, on a test charge on that location. So the field is the sum of the field due to all the charges at that point. Now, Coulomb’s Law doesn’t work when charges are moving. Why is that? Have you any idea why you cannot use it? Yes?

Student: The radius is changing?

Professor Ramamurti Shankar: Pardon me?

Student: r is changing?

Professor Ramamurti Shankar: r is changing so we’ll keep changing it as the charge moves, but it’s only an approximation when charges move. Do you know why? Yep?

Student: There’s a magnetic field?

Professor Ramamurti Shankar: True, but even the electric field — his answer was magnetic field, but even the electric field is not properly given. Yes?

Student: Is the electric field affected by something like the Doppler Effect?

Professor Ramamurti Shankar: Not the Doppler, something else. If Coulomb’s Law were exact, okay? Here is what I can do. You take a coulomb and you sit at the other end of the galaxy. I have a coulomb here. You know the force my coulomb exerts on yours because it’s pushing against you. You hold it. Now suddenly I move my coulomb away from you by a tiny amount. What will you feel? You will feel the force is reduced. Yes?

Student: Relativity is…

Professor Ramamurti Shankar: Yes, so the special theory does not allow that, because I have managed to communicate with you, arbitrarily far, instantaneously. The minute I move this charge, you know about it, because your charge moved away. For example, if your charge was connected to a spring, and it had been extended to some amount because of this charge, the minute I move it your spring will move. That instantaneous communication is forbidden by the special theory, so it does not happen. So the correct way to do it, it’ll maybe come later in the course, so probably not even at the end of this course, but the proper way to do that is to in the end realize that the electric field at this point is not only due to what the charges are doing now, but what they were doing in the past. Because if some charge at the other end of the galaxy did something it takes some time, namely traveling at the speed of light, to carry that information from there to here. So it’s the delayed response to all the motion in the charges that you’ve got to add to find the field here. That’s what makes the computation of the electric field much more complicated. But if you promise me charges never move then the location now is the location last year, the location a million years ago, then you can use Coulomb’s Law. Coulomb’s Law is good for electrostatics, but in real life charges are moving, so you cannot really use the formula. Now, in our room, if you put a charge here and another charge there, if you move this, that guy will move pretty much instantaneously. That’s because the time it takes a light signal to go from here to there is so small, you may treat it as instantaneous.

So Coulomb’s Law is used in electrical circuits and so on. You don’t worry about the time of transit because it’s too small, but over longer distances, where the time it takes for light to travel becomes non-negligible, you cannot use Coulomb’s Law. It’s not wrong. It is not appropriate when charges are moving. However, it will always be true that if you go to any one point the force on any charge q you put there is q times electric field at that point, okay? So the electric field notion survives because it doesn’t violate relativity. It says if the field here is so much q will experience the force q times E, but the complication is what is the field here? Well, it’s due to everybody else, and it’s not only due to everybody else right now, but everybody else from the dawn of time because things have been moving and shaking and sending signals to us. We collect all that and see what lands here at this instant. That decides the field here. So that’s the computation of the field, but the response to the field is very easy. You put a test charge; q times E is the answer. So in modern physics, in theories that are compatible with the special theory of relativity we break the force into two parts. Charges don’t immediately interact with other charges. Charges produce a field, and the field may even propagate outwards at the speed of light if you make motions, but another charge at the location of that particular point will respond to the field at that point. So it’s not responding to the charge right now. It’s responding to the field this charge produced at its location.

So all of electromagnetic theory is going to contain two parts. The first part is, find the field due to this charge configuration, that charge configuration, maybe due to various currents, and the second part is, given the field, find the response of charges to the field. So you understand charges play a double role. They are the producers of the field. They are also the ones who respond to the field. If you don’t have a charge you cannot produce field. If you don’t have a charge you cannot experience it, you cannot play that game. To gain membership into electrostatic interactions you’ve got to have charge. So neutrons cannot do that, but they can do other things. Like I said, they take part in nuclear force, in fact, just as well as protons do.

All right, so let’s go back now to the simplest problem in the world: the electric field due to one charge. The formula is very simple. Let’s put that charge at the origin. And the electric field due to 1 charge is q over 4Πε₀, 1 over r² if you’re here, and that’s your r vector. The field at this point is that magnitude, and I’m going to write e_rmeaning a vector of unit length in the radial direction. Once again, I will tell you if you want, you can write that as the position vector divided by the length of the position vector. They’re equivalent ways. If I write it the second way, you’ve got to be a little careful. It’ll look like qr over 4Πε₀r³. Don’t get fooled into thinking the field is falling at 1 over r³. It’s really 1 over r²because there’s an r on the top. See, if you write it as 1 over r² put unit vector. If you write it as 1 over r³ put the position vector. They’re all saying the same thing.

So here you have this formula. If you’re a person who likes to work with formulas this is all you need. You manipulate the stuff on paper, and you add different fields, but people like to visualize this. So how do I visualize this? That’s the real question. So here’s a very popular method for visualizing this formula. You know, you’ve got, for example — suppose someone asks you, what’s the height above ground level of a certain part of the United States? Okay, you’ve got some mountains. You’ve got some valleys. Well, somebody can give you a function that gives you the height at any point in the United States, but it’s more interesting to have some kind of a contour map that looks like this, right? And all these contours are different heights. If you’ve gone hiking, you can see those maps. They tell you pictorially what a certain function is trying to tell you. So you want a pictorial representation of this electric field. It’s very easy to write down the electric field at one point, namely you take that point, you draw an arrow there; E. That E is the electric field at that point. So we try to do our best by saying here is my test charge. I’m going to pick a few points, four points, maybe eight, and I’m going to tell you what the field is at those points. At this point it looks like that. At this point it looks like that. Here it looks like this. That’s already telling you something. You’ve got to be very careful on the interpretation. This arrow is not telling you what’s happening throughout the length of the arrow. It’s telling you what’s happening at the tip. You understand the arrow is in your mind. It’s a vector. It’s not really sticking out in space. It’s a property at a condition at that point, but we’ve got to draw it, so we draw it that way. It doesn’t tell you the state of affairs over its length, but only at its tip, at the starting point.

Then you can say, “Okay, what happens when I go further out?” When I go further out, say over here, it’s going to be still — if I put a test charge here it’s still going to be repelled radially, but a lot less. So I do that. So I draw arrows at other representative points and make them shorter. In fact, the length of the arrow will be 1 over r². So you can do this, okay? But now where all do you want to draw these arrows? It’s up to you. You pick a few points. You go to another radius, you draw more arrows. Someone had this clever idea of doing the following. You can probably guess. Their idea was when I join all these arrows like that, now if you go to a point like this, what have I gained and what have I lost? What more information have I got when I join the arrows? Yes.

Student: The more information is that now you know the direction that it will continue on forever if no other forces act on it, and what you’ve lost is the magnitude…

Professor Ramamurti Shankar: Okay, let me repeat that. That’s what you guys should have been thinking in your head. When I joined these lines — by the way, I do want you to anticipate what I’m going to say, because if I’m struck by lightning, another electromagnetic phenomenon, can you even complete my…

Student (Chorus): Sentence!

Professor Ramamurti Shankar: Sentence, right. Okay, now you should be able to go a little beyond if I’m doing a derivation. You’ve got to be following me, right? That’s very important. It’s got to be active. And I sat through a lecture yesterday for an hour. I know it’s a very long time. This is what, an hour and fifteen minutes? The only way you can survive this is if you somehow make it an active event. You’ve got to do something that keeps you awake during the process. One of them is to anticipate what I will do next in a calculation. That’ll make sure also that you’re on top of it, that’ll make sure that you catch mistakes.

Chapter 3: Electric Field Lines [00:33:56]

Okay. All right, so here are these lines. As she said quite correctly, previously I knew the field direction only at the chosen points, but now I know it throughout this line, but I’ve lost information on the magnitude of the field, because the arrows — there are no lengths of anything. These arrows don’t have any length. In fact, you can keep drawing more lines if you like. They go like that in all directions. It basically tells you, hey, the charge is pushing everything out radially no matter where you are. That’s the thrust of this picture. But, due to the miraculous property of the coulomb force, namely that it falls like 1 over r², there is information even on the strength of the electric field, and that information is contained in the density of electric field lines. And I’ll tell you precisely what I mean. So, here is the charge. Take a sphere of radius r and here are all these lines going. By density of lines, I mean the number of lines crossing a surface perpendicular to the lines, divided by the area of that surface. Because let’s make a convention that we will draw for every coulomb a certain number of lines, 32 lines per coulomb, 32 lines are going out. I draw a sphere of some radius, 32 lines cross that sphere. I draw a bigger sphere, 32 lines cross this sphere also, but they’re less dense, because the number of lines per area will be some number of lines per charge divided by the area of the sphere, which is 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 r². So these lines naturally diverge and spread out in space so that the density falls precisely as 1 over r².

That has to do also with the fact you’re living in three dimensions. Only in three dimensions where area goes like r²does this spreading of the density of lines coincide with the decline of the force. So these lines tell you more than simply the direction. They convey to you visually where the field is strong. Wherever the lines are dense, the field is strong. Wherever the lines are spread apart, the field is weak, and it’s a very precise statement. The only thing not precise is, how many lines do you want to draw per coulomb. That’s really up to you, but you’ve got to be consistent. Once you give 32 lines per coulomb, then if you’ve got a charge of 1 coulomb you should draw 32. If you’ve got two coulombs, you should draw 64 lines. As long as you do that, the number of lines crossing per unit area will be proportional to the field. But I’m going to make a certain choice that will make the number of lines per unit area exactly equal to the field, and here is the choice. It’s a choice that makes life simple.

Let us agree that 1 coulomb gets 1 over ε₀ lines. ε₀ is a number, right? 1 over 4Πε₀ is 9 times 10 to the 9^th. This is some number. Maybe 40 million, so one coulomb gets 40 million lines. Don’t quote the 40 million. It’s whatever this thing is. I don’t know what it is. It’s a definite number. Then, what’s the nice thing? If you’ve got q coulombs you will have q over ε₀ lines, and if you take a sphere of radius r you’ll get 1 over 4Πr²as 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 ε₀ you will always be measuring 2 times the electric field. The density will still convey the electric field, namely it’ll be proportional to it, but let’s make life easy by making it equal to it. This is just a convenience. Now, we are really set. If you draw pictures this way, you can go as far as you like from this charge. Simply take a unit area with you. Take a piece of wood 1 meter by 1 meter, put it there, see how many lines cross; that is equal to the electric field at that point.

Okay, so this is the way one likes to visualize field lines. So I’m going to give you some examples. For a single charge you just draw it that way. For two charges, let’s take two charges, a minus charge and a plus charge. Let’s say that one is -q, the other is +q. Then you are very near that charge. By the way, I’m not going to draw 1 over ε₀ lines per coulomb because it’s going to be too many lines, okay? I’m just going to draw a few so you get the picture. So I’m going to draw four lines right near the charge. You can forget about all other charges in drawing the lines. Why is that? Yep?

Student: The field only depends on the one charge.

Professor Ramamurti Shankar: Pardon me?

Student: Why does the field only depends on the one charge.

Professor Ramamurti Shankar: Why does it depend on the one charge? In principle it depends on every charge. Somebody had an answer back there? Yes?

Student: Since it falls as 1 over r²when you’re so much closer to one than the other then…

Chapter 4: Electric Dipoles [00:40:19]

Professor Ramamurti Shankar: Right, because the field is 1 over r²and the 1 over r for this guy is going to infinity. 1 over r for this guy is maybe 1 over 1 meter. It’s finite. So when you come arbitrarily close to a charge, it is going to dominate. Well, if it’s the only thing in the universe, we know the lines will look like this. At least they’ll start out this way, but soon, of course, it won’t go out this way forever, because you will realize there’s another charge. Likewise, it’s easy to draw the lines this way. Remember, the lines are coming in because if you put a test charge, it’ll be sucked into this. Test charge is always assumed to be unit positive charge, so the lines will be coming into a negative charge, and leaving, going away from a positive charge. Now we’ve just go to do what the agencies forgot to do, which is to connect the dots. You do this. You do this. You do this. You do this. Now, at some point you’ll have to think a little harder, because suppose I go here? How do I know I should draw the lines this way? If I take this guy here, it will repel it. This guy will attract it, and I add the two, and get a line in that direction, so you really have to do a lot of work. If you really want this picture to be exact, you have to compute the vector everywhere, but if you want a sketch, you’re allowed to guess, and things look like this. So this is called a dipole, and this is the field of a dipole.

So here’s another example. Both guys are plus. Now what do the lines look like? Again, they will start out this way near the charges, but now when you come to the midpoint here there should be no electric field right in the midpoint because it’s getting pushed equally from both sides. So the lines, if you think about how they will add, they will do something like this. Okay, look, I’m not going to do a good job for a variety of reasons, but you can look at your textbook or any other book to see what the lines will look like when you’ve two plus charges. If you go a mile away from these two plus charges, what do you think the lines would look like? Yep?

Student: Just like a point charge that’s twice as strong.

Professor Ramamurti Shankar: Right, so that’s the intuition you should keep in your mind. If you go very far from a charge distribution where you cannot look into the details, all you will see is some little dot that has the entire charge in it, and the lines will be coming out radially. So only when you zoom in you realize, hey, it’s not a single charge 2q, it is two guys of strength q. Finally, let’s take a case where this is charge 2q. This is charge q. Let’s say this is charge -q. Then some lines will go like this and some lines will run off to infinity. Here if you go very, very far away from the two charges you’ll again see radially outgoing lines, except there is a charge q at the center because 2q and -q give you a net of q. Okay? So this is the example of a dipole. If you’ve got more charges it gets more complicated, so people don’t usually draw them. There’s one example which is pretty interesting. If you’ve got one plate and another plate, this contains all positive charges, this contains all negative charges, then the field here will look like this, will go from the positive to the negative plate, because if you put a test charge between them, it’s getting repelled by the positive plate and attracted by the negative plate, so the lines will go from one to the other. Near the edges they may do something more complicated, but in the bulk they will look like this.

Okay, so now I’m going to do one calculation, which is: what is the actual electric field due to a dipole? In other words, not just the picture here, where is that? That picture on the left is a dipole. I’m going to do it quantitatively. So here’s my goal: I want to take a minus charge here, a plus charge here. This is at x = a. This is at x = -a. So I want to find the field everywhere. So today I’m not going to do the field everywhere, because later on I’ll show you a more effective way to calculate it, but I’m going to calculate it at a couple of interesting easy places. In other words, we all know the lines look like this, okay? But I want to go to some location and find the magnitude and direction of the field. But today I will only find it at two places, one along the axis at a point x, and one on the perpendicular bisector at a point (x = 0, y), just going to do those two. So let’s see what’s the field here, the field at this point? The field at that point, you agree, is going to be entirely in the x direction because this is pushing it, and that is pulling it. So E is going to be i(unit vector) times q over 4Πε₀ times that distance squared, which happens to be (x - a)². That’s the repulsion due to the charge that’s nearer to you. Then there’s the attraction due to the minus q, but it’s a little further away, so it looks like a minus sign but it is (x + a)². If a is equal to 0 you get 0. If a is equal to 0 the two guys are sitting exactly on top of each other. You will not see them. So you see them only because they’re not on top of each other. This whole thing fails to be 0 because this a is not 0, and you can understand why. The minute a is not 0 you’re closer to one of the two charges, so that they cannot really cancel each other.

So you’ve got to manipulate this expression, so I will do that now. q over 4Πε₀ and you find common denominator. I remind you (x + a) times (x - a) is (x² - a²)and everything is under squares. In the numerator you’ve got (x + a)² - (x - a)². So what does that give me? iq over 4Πε₀, (x² - a²)². And how about on the top, can you do that in your head? This is going to be x² + a² + 2xa. You’re going to subtract from it x² + a² - 2xa, and the only thing that will survive will be 4xa. Is everything okay?

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

Professor Ramamurti Shankar: It is. You are saying why is it not (x + a)² times (x - a)², right? It is, because this would be x + a times x - a, the whole thing squared.

Student: Oh!

Professor Ramamurti Shankar: And this guy is x² - a². So this is classified as a nice try, so. But I want you to keep doing this. This time I am right, but you never know, okay? I don’t want you to give up. There is nothing better than shooting me down, but this happened to be correct. You satisfied though?

Student: I knew you were right, I just couldn’t figure out why.

Professor Ramamurti Shankar: No, no, no, no. I don’t want to rush through this. Anybody have the same problem with this? Look, also I am doing it fast because this is 958th time I’m doing this calculation, so if you’re seeing it for the first time, I’ve got to slow down. So let’s see which part. Everybody okay with this, right? So I wrote that. Then this is really an x + a times another x + a, and an x - a and another x - a, but without these guys, I know it’s x² - a² because I’ve got two of everything. I squared everything. So this answer’s actually an exact formula of the electric field along the axis of the dipole. But normally what one is interested in is, when x is much bigger than a. When it’s much bigger than a, downstairs you’ve got x squared. This could be 1 kilometer squared, a squared was maybe 1 millimeter squared. So in the first approximation, it’s not an exact formula anymore, from now on it is approximate, in the limit x is much bigger than a. You can see it’s going to be i times q over 2Πε₀ times 2a divided by x³. So what did I do now? I took from the 4a. I borrowed a 2a to write this here and I canceled the 2 with the 4 here to get that. Then on the top I had an x, and the bottom had x⁴. I get x³. 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 2Πε₀ x³. So I will tell you what I’m doing.

So the final formula I had was electric field E = itimes p divided by 2Πε₀x³, where p — I’m sorry, itimes p, no arrow; p is equal to 2aq. So let’s delete some extra arrows I had. That’s right. You never should settle for something that looks like that. So that’s p. So p is called the dipole moment of this dipole, and it’s given by the product of the distance between the charges and the value of one of the charges, the plus charge. So whenever I give you two charges, call it dipole, you can associate with them a vector. And the vector is, if you’ve got a charge -q here and a charge +q there separated by a little vector r, then the dipole moment is q times the little vector r. In our example the little vector r was 2a in the x direction. So 2a times i times q is the dipole moment. So this means electric field is parallel to the dipole moment and falls like 1 over x³. That’s the most important part of the dipole. A single charge, the field falls like 1 over x² if you move a distance x. A dipole will always fall like a bigger power of x, because to go like 1 over x² you’ve got to have net charge. As long as the net charge is 0 the fact that there are two opposite charges that don’t quite cancel each other, it always comes from the fact that the distance between them is not 0. And the distance will appear in the numerator, and that must be the corresponding distance in the denominator, because the formula should have the same dimensions. That’s what turns the 1 over x² to 1 over x³. Yep?

Student: That only works if x is a lot bigger than a.

Professor Ramamurti Shankar: Yes. So this formula is good for all x; this formula is good only for x bigger than a. You’ll find whenever you’re working with dipoles, people will always ask you to find the field very far from the dipole. So here’s a second place where I can go and find the field. That’s going to be here, -a, +a, and I’m going to find the field there. So that’s at a distance y. So let’s look at this. That’s a +q here and -q here. A +q will repel it that way, and a -q will attract it this way, and their sum will be that. I’m going to compute that sum. So how do I do this? Let’s look at this guy here. We know that these arrows have equal magnitude because this distance is the same as that distance. Therefore, it’s the horizontal part that will remain. The vertical part will cancel. You see that? We’ve got two arrows of the same length with this angle and this angle equal. The horizontal part will be additive and the vertical part will be equal and opposite. So I’m only going to compute the horizontal part. So the electric field now will be -i (that’s to tell me it’s in the negative x direction) times q over 4Πε₀ times 1 over distance squared which is (y²+ a²). This is (y²+ a²).

That’s also (y²+ a²), 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 (y²+ a²)^½. See, you want to take that force and find this horizontal part. Then I’m going to put on another 2 because this is going to contribute an equal horizontal part. So the E, in the end, is equal to -iq2a over 4Πε₀ divided by (y²+ a²)^3/2. That is then -p divided by 4Πε₀ divided by y³, for y much bigger than a. I’m sorry, y³, y is really the distance. So, right, if y is 1 mile and a is 1 millimeter that’s essentially the distance. If you like you can call it -p divided by 4Πε₀r³. Where r is the distance, if you like, from the center of dipole. Look, the point of this exercise is twofold. One is to show you how to add vectorially the fields due to two guys. Another is, to have you understand, at least, how to do the computation at a few simple places, where the direction of the field is not so hard to calculate. Actually, one would like to compute it here, but it becomes quite nasty. The magnitude is not so hard, but the direction is hard to calculate, so we’ll find a shortcut. But at these two places, on the perpendicular bisector and on the axis, the mathematics is pretty simple. That’s the electric field. Yeah?

Student: Is p as a vector different from the p that you crossed out as a vector over there?

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 I’m going to do the second part of the problem, which is finding the response to E. This was all computing E. Of course there are more and more complicated examples, but we did a few simple ones. Next is going to be, if I give you E, can you find what will happen to the charge. So I’m going to do two examples. One is there are these two plates I mentioned to you. This one is all positively charged. This one is negatively charged. And I shoot a particle here, with some velocity v­_o in the x direction, and the field everywhere is down, and the electric field is some constant, -j times some number E₀. -j because i is this way and j is that way. So what will this do is the question, and where will it end up? Well, I think you can all tell that it’ll end up somewhere there. What we’re trying to find out is how much does it fall, and when it comes out, what’s the direction of this final velocity vector.

Well, the force on this charge is equal to -q times E₀ j. The acceleration will be -qE₀ over m times j in the y direction. So what’ll be the position? Position will be from lecture number one of your Physics 200. So let’s say the starting point r₀ is our origin. v₀ is whatever it was projected in with, plus ½ times qE₀ over m, t² j. Okay, so as a function of time this tells you where the position will be. At t = 0, you are at the origin. As t increases, it’s moving horizontally due to v₀ and it’s also dropping vertically due to the acceleration. Then it’s very easy from now on to compute anything you like. For example, when you want to go to that point, what will be the time? Anybody tell me what the time will be when I go to that point? How long will it be in the region between the plates? Yep?

Student: The distance of the length of the plate divided by initial velocity.

Professor Ramamurti Shankar: Which velocity should we take?

Student: The initial horizontal velocity.

Professor Ramamurti Shankar: That’s correct because t will be L over v₀ where v_0­’s the magnitude of the initial velocity because x velocity is never changing. Acceleration is in the y direction. So the time it takes to cross will be independent of the fact it’s falling in the y direction. So if you put t equal to all of this you will find out where it will end up. And that’s how you make pictures on the television. You’ve got a bunch of plates, and then you drive charges, and if you apply the right electric field the electron will land on a screen and make a little dot. The screen will look like this, and you’re looking at it from the other side. It’ll glow. Then you want the dot to move up and down, you can move the voltage. Then if you want to move it back and forth you’ve got to put another set of plates, not like this, but coming out of the blackboard. That way you can move the electron beam in all directions. That’s how you scan the television screen. You can also use magnetic fields, but this is one simple way using electric field.

All right, final thing to discuss is: what is the force of a uniform electric field on a dipole? So let’s take an electric field in the x direction like that. It’s got a magnitude E₀ and it’s in the x direction. And in this electric field I stick a dipole in like that. Here is the +q. Here is the -q. Let’s make that a. Let’s make that a. So the plus charge will experience a force like that. The minus charge will experience a force like that. This will be q times E₀. That’ll be -q times E₀. So dipole as a whole will not feel any net force, because the two parts are getting pulled by opposite amount. If the electric field were not uniform, namely if it were stronger here than here, then of course it will drift to the right, but I’m taking uniform electric field, and because the charges are equal and opposite, the net force on it is 0. But something is not 0. You know what something is that’s not 0? Yep?

Student: The torque.

Professor Ramamurti Shankar: The torque. There is a torque because there is a force here and the force there. You can imagine they’re trying to straighten out the dipole so it ends up looking like this. So let’s find the magnitude of that torque. Magnitude of the torque is the force times the perpendicular distance. So if this angle is θ here, and you want the perpendicular distance, it is asinθ. Then there’s another asinθ from that one. That’s the torque. But since 2q a is p, it’s pE₀sinθ. You can see this makes sense. If θ was 0, if the dipole was aligned with the field, the torque vanishes, because if the charges are like this there is no tendency to rotate. The biggest torque you get, if the charges are like this, then this gets rotated that way. That gets rotated that way. You get the maximum torque. Put θ equal to Π by 2, you get a torque of p times E₀. That’s also the reason it’s called a dipole moment.

Now, I’m going to write this as a cross product, and I’m assuming you guys are familiar with the cross product, right? You take two vectors, p and E, the cross product has a magnitude, which is p times E times sine of the angle between them, and a direction obtained by turning a screwdriver from p to E. So p is like this. E is like this. Turn a screwdriver from p to E, it goes into the board, and that’s the torque, and the dipole will then rotate until it lines up, or if you don’t want it to rotate you’ve got to provide a counter-torque of this magnitude. Now, if you take any dipole and leave it, you know it will like to become horizontal, so there is a certain restoring torque that tries to rotate it so it becomes horizontal, and it’s not very different from a spring, where if you pull it from equilibrium, there is a restoring force that brings you back to where you were. So this dipole is happiest when it’s horizontal. If you go away from horizontal the torque brings it back. So just like for a spring, if you’ve got a force which is -kx we can assign a potential energy U, which is ½ kx² so that the force is equal to -dU/dx. This is something I am assuming you guys know; the relation between potential and force. Relation between potential and force, the potential at 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. In other words, in rotational dynamics torque is to force — just like one of the SAT questions. Torque is to force as θ is to x, and work is just work. If you want to calculate the work done by the dipole, but if you like the potential energy when it’s at an angle θ and its potential energy when it’s at angle 0 is the integral of the torque dθ from 0 to θ. Now, what is the torque? The torque is -pEsinθ 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 sinθ is cosθ, so you will get pEcosθ - pEcos 0 which is -pE, and that is supposed to equal to U(θ) - U(0). By comparing the two expressions you can identify U(θ) to be — sorry, this is U(0) minus U(θ), therefore U(θ) is equal to -pEcosθ, or just -p dot E.

So that’s the final formula you have to remember, that the — can we bring it down here? When you have a dipole in an electric field, it has a potential energy associated with the angle which is -p dot E, and if you draw a picture of that as a function of θ it goes like this. θ = 0 is when the dipole is parallel to the field. That’s when it has the minimum energy, -pE. At 90 degrees, energy is 0. At 180 degrees, it’s maximum. And the torque is just -dU/dθ and you can see that the torque here and there are zero, but this is the point of stable equilibrium. That’s the point of unstable equilibrium. See, if this was a potential energy like a shape of the ground, if you left a marble there it’ll stay there, but if it moved a little bit, it would roll down hill. But if you left the marble here it’ll stay there, but if you move it, it’ll rattle back and forth. That’s a stable equilibrium. That’s unstable. So for the dipole when it’s parallel to the field you are here, and at anti-parallel to the field you are there. The difference is when you’re parallel and you move it a little bit, it’ll have stable oscillations, whereas if at anti-parallel if you move it a little bit it’ll flip over completely and come down here.

All right, so I’m going to summarize the main points so you can carry that with you, okay? We saw today that we should think in terms of electric field from now on, and we no longer talk about direct interaction between charges. We say charges produce fields, and fields act on charges to move them. The force of a field on a charge is just q times E. The field is found by adding the field due to all the charges in the universe, provided they’re all at rest, and you just add by Coulomb’s Law. So we found the field due to a dipole allowing the axis and perpendicular to the axis. We saw the notion of field lines as a good way to visualize what’s going on in the vicinity of the charges. The lines tell you the direction where the field is, and they tell you in your line density, lines cutting a unit area perpendicular to them, the strength of the field. Then I calculated for you the field of a dipole along the axis and perpendicular to the axis. There are a lot of formulas, but one thing you should carry in your head. When you’ve got two equal and opposite charges and you go very far, the field will go like 1 over r³. The 1 over r² part of them is canceled, okay? That’s the main point, and it goes like 1 over r³ times 2 in one place, and 1 over r³ times 1 in one place. It doesn’t matter. The main thing is it’s 1 over r³.

Finally, if you take a dipole and you put it in an electric field, it tends to line up because there’s a torque, p cross E, trying to line it up. With that torque you can associate a potential energy by the usual formula that the torque is minus a derivative of the potential energy. That potential energy is -p dot E. Some of these things may come in handy later on. So you don’t have to memorize them, but they’ll be involved later on. So I’m going to post the problem set today. You can bring them into class next week. There’s one change. I wanted you to give it before class. I realize that’s going to be too disruptive, so you can give it after class. The reason I wanted it before class is that I don’t want you doing your homework in class. I don’t know how many generations have perished trying to follow that strategy. It’s just not worth missing a lecture to get one more problem right, so don’t bother. Once you come to class follow the lecture and don’t do the problems.

[end of transcript]

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