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PHYS 201: Fundamentals of Physics II
Lecture 3
 Gauss's Law I
Overview
The electric field is discussed in greater detail and field due an infinite line charge is computed. The concepts of charge density and electric flux are introduced and Gauss’s Law, which relates the two, is derived. It is applied to the study of the electric field generated by a spherical charge distribution.
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htmlFundamentals of Physics IIPHYS 201  Lecture 3  Gauss's Law IChapter 1: Review of Electric Field Concepts [00:00:00]Professor Ramamurti Shankar: All right, class, I thought I’d start as usual by telling you what happened last time. Not the whole thing, but just the highlights so you can follow what’s happening today. The main things I did last time were the notion of an electric field, which is going to be with you from now till the end of the course. The idea of the electric field is that if you’ve got lots of charges, q_{1}, q_{2}, q_{3}, instead of worrying about the force they exert on each other, you ask yourself, at a generic point where there’s nothing, if I put 1 coulomb here, what will it experience? What force will it experience? You compute that. So in your mind, imagine a coulomb, and you find the force due to q_{1}, it looks like that. q_{2} may exert a force that way. q_{3} could be of opposite sign, so maybe it will exert a force that way. You add all these vectors, they add up to something. That something is called the electric field at that point. There is nothing there except the electric field, but it’s very real, because if you put something, something happens to it. So the electric field is everywhere. The charges are in a few places. Electric field is defined everywhere, except right on top of the charges, which is where it probably is infinite. Once you know the electric field anywhere, if you put another charge q, a real charge, it will experience a force equal to qE, because electric field was the force you would have had on the 1 coulomb, and if you put q coulombs, it will be qE. That’s the electric field. So you can imagine computing it for any given distribution of charges, because you know what each one of them does. Then I said there’s one nice way to visualize the electric field, which is to draw the field lines. You go to each point and you ask, if I put a charge here, a positive test charge, which way will it move? Then you follow that thing as it moves and you get that line and you get that line and you get that line. You can draw these lines. The lines give you one piece of information which is very obvious, namely, if you are here, the force is along that line. But I also pointed out to you, you get more than just the direction of the force. You also can understand the strength of the force. The strength of the force is contained in the density of these lines. Now density has to be defined carefully. It’s not like mass per unit volume. It’s an area density. That means if you get yourself 1 square meter, you know, meter by meter piece of wood, like a frame, and you hold it there, perpendicular to the lines and see how many lines go through, that’s called the area density of lines. That you can see will fall like 1/r^{2}, because if you draw a sphere of radius r, that area is 4Πr^{2}, and the same number of lines are going through that as any other sphere, so it will be proportional to 1/r^{2}. But so is the electrical field proportional to 1/r^{2}, and I said, let us agree that we will draw 1/ε_{0} lines per coulomb. This is a necessity; it’s just a convenience. It’s like saying I want to measure distances in inches and centimeters for daily life. You can measure them in parsecs and angstroms, but you’ll be dealing with nasty numbers. So it’s a convenience, and the convenience here is, let’s pick 1/ε_{0} lines per coulomb. You’ll see the advantage of that, because if you ask, what’s the density of lines here on a sphere of radius r, if there’s charge q at the center, I got that many lines, and the area of the sphere is 4Πr^{2}, so the density of lines per unit area, you see, is precisely equal in magnitude to the electric field at that point. If you had not drawn 1/ε_{0}, but maybe 5/ε_{0} lines, then the line density will be 5 times the electric field. It will still represent the electric field, but we don’t want to simply represent it. We want it to be the electric field. It makes it easier. Then I said, let’s take a slightly more complicated situation, two charges. This is called a dipole, one pole and another pole, and you can draw the field lines here. You can see if you put a test charge, it’ll run away from the plus to the minus. If you leave it somewhere here, it’ll go like that and loop around and come back, and you can calculate them. This is no longer guesswork. If you had enough time, I hope you all agree, you can go to any point you like and find the force of attraction due to this one, the force of repulsion due to that one, add them up and you will get an arrow that direction. So one can draw these lines, and the lines tell you a story. Then I said, let us find the field in an analytic expression due to the dipole. Yes? Student: I have a question about the dipole. If you put a test particle going in the positive direction of the x axis, would it also _________? Professor Ramamurti Shankar: No. She’s got a good point. If you put somebody here, it will never loop around. Can you see why? Because as it goes further away, this is trying to pull it back. It’s always going to be closer to this guy, so it’s never going to come back. This line will go like that and this line will go like this. But anything else at any other angle will loop around and come back. All right, now the field strength, if you want to calculate it, you can use the formula for E due to this one and E due to that one and add them, and I did that for you. I don’t want to go into the details, but I remind you in all cases, the electric field fell like 1/r^{3}, because the field of each charges goes like 1/r^{2}. And if these were on top of each other, they will completely cancel each other. So the reason you have a nonzero field is thanks to a. Therefore the answer has to contain an a in front of it, at least n the first approximation. But that a, from dimensional analysis, has to come with a 1/r so that the whole thing has the same dimension as before. That’s where you get a/r^{3}. That a times q times 2 and so on, that became the dipole moment. That was true here. We verified that’s true here. Later on, we’ll verify it everywhere, because there are easier ways to calculate than what I’m using. Then I said, forget about the field due to charges. Let’s look at what charges do when you put them in a field. So I took two examples. One was a very simple example. These are two parallel plates. They are not two lines; they are plates coming out of the blackboard. They’re filled with charge and this has got + charge on it. This has got  charge on it. And therefore, the electric field will look like this, right? Because if you leave a test charge, it will go away from the positive, towards the negative. Then I said, suppose there really is a particle here with some velocity, v_{0}, what will it do? You can see that the force on this guy is going to be q times E. E is pointing down. If you divide by the mass, that’s the acceleration, also pointing down, and it’s constant. So that’s like a particle in a gravitational field, except g is replaced by this number. So it will just curve like that, and you can calculate the trajectory. The final thing I did was, what happens when you put a dipole in a uniform field. Here as well, I think I was rushing near the end, and even I couldn’t read my stuff in the corner. So I will go over that. If there’s something that you didn’t follow, then I will be happy to repeat that part for you. But you should understand what the question is. There is an electric field which is pointing like this, as if you have two plates here, + charge is here,  charge is there. They’re producing a constant electric field in the horizontal direction. In that environment, I take an electric dipole whose  charge and + charge, +q and q, happen to be oriented like that. Question is, what will happen to this guy? If you want, you can imagine that it’s a little massless stick, and one end you glue q coulombs, other end you glue q coulombs, and your let it sit there. What will it do? First of all, it won’t feel any net force, because the force in this direction is q times E and the force that direction is also q times E, if you want, it’s q times E and they cancel. But that doesn’t mean it won’t react. It will react, because you can all see intuitively, it’s trying to straighten this guy out and applying a torque like that. You follow that? That’s what it will do. And the way to find the torque, the torque is the product of the force and the distance between the point of rotation and the force, and the sine of the angle between them, that is to say, sine of this angle. What the sine of the angle does is to take the component of the force perpendicular to this axis, because if you resolve the force into that part and that part, this part is no good for rotation. That’s trying to stretch the dipole along its own length. It’s the perpendicular part that’s going to rotate something, so you get that times sine theta. That you can write now as the vector equation p x E. Because p is equal to — I’m sorry. I need a 2 here. I forgot the 2, because this charge will have a torque and that charge will have a torque and the two torques are additive. They are both going the same way. Then 2q times a is p, and this is E and the sinθ comes in the cross product. I’m assuming all of you know about the cross product. Okay, final thing I did, which is, if you have a force, you can associate with that force a potential energy. Again, this is something you must have seen last time, but I will remind you. As long as it’s not a frictional force, you can say the force is connected to potential energy in this following fashion. Or the potential energy at x minus potential energy at some starting point x_{0 } — I’m sorry, x_{0}  x is the integral of the force from x_{0} to x. This is the relation between the force, as delivered to the potential, and the potential is the integral of the force. For example, for a spring, U is ½kx^{2}, and F = kx. If you go to this one, we tell you U(x_{0})  U(x) is equal to the integral of kx dx from x_{0} to x. So that gives you kx_{0}^{2}/2 − kx^{2}/2. And by comparison, you can see U(x) = ½kx^{2}. Actually, this is not the unique answer. Do you know why? Given this formula, can I immediately say this guy corresponds to that, this one corresponds to that one? Is there some latitude here? Yes? Student: You can always add in a constant. Professor Ramamurti Shankar: You can add a constant to both, because if I said that, that certainly works. If I add 92 to both, it still works, because the 92 extra doesn’t matter when you take the difference. So it’s conventional to simply pick the constant so that the formula looks simple. Coming to other expression, if you had a torque, which is pEsinθ, you can ask — that’s the torque — it’s minus, because it’s trying to reduce the angle θ — you can ask what U leads to that. And you can see it’s pEcosθ. See that, take the U and take the derivative, you’ll get the torque. But if you got two vectors, p and E and you see the cosθ, I hope you guys know that you can write it as a dot product. So that’s the end of what I did last time, okay? The potential energy is proportional to the dot product of p with E. The torque is equal to the cross product of p with E. And what does pEcosθ mean if you plot it as a function of angle? It will look like this. This is Π and this is 0. That means it likes to sit here and if you deviate a little bit and let it go there, it’ll rattle back and forth, just like a mass spring system. In fact, you can very easily show, near the bottom of the well, the potential energy is proportional to θ^{2}. That’s because cosθ can be written as 1 − θ^{2}/2+ θ^{4}/4!, etc. for small angles. If you just keep that term, you will find it looks like this. Not very different from U = kx^{2}. So what x does with forces, θ does with rotations. All right, so this is what we did last time. Now I’m going to do the new stuff. So new stuff is going to give you — I think it’s useful, because it tells you the level at which you should be able to do calculations. Chapter 2: Electric Field Due to an Infinite Line of Charge [00:15:02]So here’s a typical problem. You have an infinite line of charge, of which I will show that part, and somebody has sprinkled on it λ coulombs per meter. So it is not a discrete set of charges; it’s assumed to be continuous and it’s everywhere. I’m just showing a few of them, and if you cut out one meter, you’ll find there’s λ coulombs there. And you want to compute the electric field you will get due to this distribution, everywhere. So you want to go somewhere here and ask what’s the electric field. That’s what we’re going to do. Let’s go here. You will see why. Now first of all, you’ve got to have an intuition on which way the electric field will point. You have a feeling? Yes. It will point here, this way. Why not like that? Yeah? Student: [inaudible] Professor Ramamurti Shankar: Okay. She said the horizontal parts will cancel. That’s correct. Another argument from symmetry is that if anybody can give you a reason why it should tilt to the left, I can say, “Why don’t you use the same argument to say it will tilt to the right?” Because this is an infinitely long wire and things look the same if you look to the left and if you look to the right. And the field you get should have the same property. If this was a finite wire, I wouldn’t say that, because in a finite wire, that can be a tilt, somewhere here. But infinite wire, it cannot tilt to the left or to the right, because each point has the same symmetric situation to its right and to its left. In a finite wire, it’s not true. Life to the left is different from the life to the right, but for infinite wire, you know it cannot be biased one way or the other. It’s got to go straight up. Secondly, you can find the field here, here, here or at anywhere at the same distance, you’ve got to get the same answer. Again, because if you move two inches to the right, it doesn’t make any difference with an infinite wire. You’ve still got infinite wire on either side. So we’ll pick a typical point and calculate the field and we know that answer is going to be good throughout that line. So now I take this point here. I want the field here. I’m going to make that my origin. Then I take a piece of wire of length dx. dx is so small that I can treat it as a point. Now the dx I’ve drawn is not a point, but in the end, we’re going to make dx arbitrarily small so that it’s good enough. It’s like a point and it is at a distance x from the origin, and let’s say that distance is a. So let’s find the field due to just this guy, the shaded region. Think of a charge. How much charge is sitting here? I hope you all agree, the charge sitting in this region is just λ times dx. That’s just the definition. If you’ve got that many coulombs, imagine a test charge of 1 here. Well, it will push it this way, and the field due to that, I’m going to draw as infinitesimal, so I’m going to call it dE. You can call it E, but dE is to remind you, it’s a tiny field, due to a tiny section dx. Now that electric field is biased to the left, but for every such section you find here, I’ll find a section on the other side that’s precisely biased to the right by the same amount. Therefore the only part that’s going to survive due to this guy, combined with this, will be the portion here, which is the vertical part of that force. So let me find the contribution first, only from this one, then we will add the contribution for this one. For that one, you just find the vertical projection. So how much is that? Remember, the Coulomb’s law for the electric field is q/4Πε_{0}r^{2}. So this is the q. That’s the 4Πε_{0}. r^{2} is that distance squared, which is x^{2} + a^{2}. That’s really like the field of a point charge at that distance. But now this is the magnitude of the electric field vector at this angle, but I want the part along the y direction. So I’ve got to take cosine of that θ. You guys follow that? If you took this vector, this part is dEcosθ. That part is dEsinθ. But that angle and that angle are equal, and cosθ for this triangle, you can see, is a divided by (a^{2} + x^{2})^{½}. So this is the electric field in the y direction. I’m going to call it dE, in the y direction, due to the segment dx. The total electric field is obtained by adding all the dEs or adding all the contributions from all the segments on this line. And that goes from  infinity to + infinity. All right, so now it’s a matter of just doing this integral. So this gives me (λa/4Πε_{0}) dx/(x^{2} + a^{2})^{3/2}, integrated from  to + infinity. Now I can make life a little easier by saying that this function is an even function of x. That means when you change x to x, it doesn’t care. Therefore the contribution from a positive x region is the same as the contribution from a negative x region. That also makes sense in this picture here, because if you look at the field I’m computing, this section and this section give equal contributions in the y direction. But even if you did not know any of that background, as a mathematician, if you see this integral, you would say, “Hey, put a 0 there and put a 2 here” namely,integrate over half the region, because the second half is giving you the same answer. So you double the integral, but cut the region of integration in half. So at this point, you are free to look up a book, if it was an exam, but maybe not even if it was an exam at your level. You should be able to do this integral. So integrals have been around from the time of Newton and the question of an integral is, find the area of some graph with this particular functional form. And the answer to any integral is that function whose derivative is the integrand. So what you have to do is guess many answers until you get the right one. But people have been guessing for hundreds of years, and there’s big tables of integrals with all the integrals you want. But you should still be able to do some integrals from scratch and I’m going to tell you how to do this one. But before you do the integral, you’ve got to have some idea what the answer is going to look like. I want you to get some feeling about this. Answer depends on what, is the first question. What’s the answer going to depend on? Student: a. Professor Ramamurti Shankar: a, you understand? Whatever this is depends only on a because 0 and infinity are not going to be present as part of the answer. If the lower limit was 5, it can depend on 5, but it doesn’t depend on any other thing, other than a. Then from dimension analysis, I got a length squared to the 3/2, that’s length cubed, and a length on the top, so whole thing should look like something over length squared. The only length I have is a, so it’s going to look like 1 over a^{2} times a number. Once you got the number, you’re done. So I’m going to do all the work now to show you that the number is actually just 1. This number will turn out to be 1, in which case, you will find it’s λ/2Πε_{0}a. Well, let’s see how we get the number to be 1. So does anybody know what trick you use to do this integral? This is whatever, math 120 or — yes? Student: Use substitution? Professor Ramamurti Shankar: Yes. What substitution? Student: x^{2} + a^{2} = U. Professor Ramamurti Shankar: That won’t help you. Yes? Student: Can you use trigonometric substitution? Professor Ramamurti Shankar: Yes. Trigonometric substitution. Which one? Okay, look — no, no, I don’t blame you. I know the answer because I’ve seen it, but if I have to work on it, I’ll try for a while before I got it right. The whole idea is, we don’t like all these 3 _______ here. We want to turn that into something nice. So I’ll tell you what the answer is. You can all marvel at how wonderfully it works. So what we are going to do is to introduce an angle theta — nothing to do with the angle in the problem — so that x = tanθ. That means instead of going over all values of x, I’ll go with the suitable values of θ — I’m sorry, this would not even be correct dimensionally. x = atanθ. You can see that every x that I want, I can get by some choice of theta, because tanθ goes from 0 to infinity when θ goes from 0 to Π/2. You cannot say x = acosθ, for example. You are doomed. If x is acosθ, the biggest x you can get is a, whereas I want this x to go from 0 to infinity. So when you make the change of variables, you’ve got to make sure that for every x you want, there is some θ that will do it. Then the next thing you do, you say dx/dθ = a times derivative of tanθ which is sec^{2}θ. Then you write that as dx = that. What that means is an integral dx is related to an integral dθ by this factor. Therefore going to the integration here, I’m just doing that part, which is going to be a sec^{2}θ dθ. θ goes from 0 to Π/2. Now let’s look downstairs. Downstairs I’ve got x^{2} + a^{2}. x itself is atanθ, therefore a^{2} times 1 + tan^{2}θ. 1 + tan^{2}θ happens to be sec^{2}θ. That to the power 3/2, which is what I want, will give me an a^{3}sec^{3}θ. So what do we get? You can see as promised I get a 1/a^{2} and I get integral of 1/secθ, which is cosθdθ from 0 to Π/2. Yes. And integral of cosθ is sinθ from Π/2 to 0. That just happens to be 1. So the final answer is what I gave you here, E. Well, E is the vector. I’ve just shown you the magnitude, but we’ve all agreed what the direction is. The direction is away from the wire. So if you like, if you look at this wire from the end, the lines will look like this. If the infinite wire is coming out of the blackboard towards you and you look at it this way, if you go too close, you’ll poke your eyes out. Look from here, you’ll see the lines are going out radial everywhere. The question is, how do the fields get weak? How does it weaken with distance? It weakens like 1/a. That’s a big of a surprise, right? The field away from the wire doesn’t fall like 1 over distance squared, but falls like 1 over distance. The reason is that every individual portion of the wire has a contribution that does fall like 1 over distance squared, but it is an infinite wire. When you add it all up, the net answer goes like 1 over the distance. The field away from a wire falls like the distance from the wire, on the perpendicular from the wire and there’s pointing away from the wire. That’s it. Chapter 3: The Infinite Sheet and Charge Density [00:28:42]Okay, so that’s one calculation. Then I’m going to do one more and that’s going to be the end of the tough calculations. Second calculation is going to be an infinite sheet. On the infinite sheet, the appropriate quantity is called the charge density, which is coulombs per meter squared. That means if you cut out a tiny piece, the charge on it will be sigma times the area of that piece. So there is positive charge everywhere here, and the number of coulombs per unit area is called sigma. These are standard. λ is coulombs per meter, σ is used for coulombs per unit area. The question is, what’s going to be the electric field at some point away from that plane? Once again, I think we can all agree that the electric field at some point from the plane will not depend on where in front of the plane you are standing. Are you standing here or are you standing there? It doesn’t matter, because it’s an infinite plane. If I moved 1 inch — I’ll tell you why it won’t matter. If I moved 1 inch and the answer changed, I should get the same change if I didn’t move and somebody moved the sheet 1 inch the other way. But when I move an infinite sheet the other way by 1 inch, it looks exactly the same. It’s got to produce exactly the same field. So you can always ask, what will happen if I move to the left, the same as what will happen if the sheet moves to the right? The sheet moving to the right looks exactly like the sheet before. The answer won’t change, therefore the answer won’t change for you if you move to the left. I’ve got infinite plane below you. As long as you don’t change the distance from the plane, you navigate perpendicular to it, no matter where you are, you will get the same answer. Same answer, meaning same direction of the field, same magnitude. And that direction has to be perpendicular to the plane, again for symmetry reasons. If you tilt it in any one direction, you have no reason to do it. For example, if you tilt it this way, I can take the infinite plane and rotate it, then the tilt will be in some other direction, maybe like that, but the rotated infinite plane looks the same. In other words, if the cause does not change, the effect should not change. If I can do certain things to the infinite plane that leave it invariant, then I can do the same transformation to the location of the point, and that shouldn’t have a different answer. So the plane has the property that when you slide it up and down parallel to itself, or twist it and turn it, it looks the same, therefore the field pattern should have that property. Therefore the field has to be the same at all distances from the plane anywhere on top of the plane, and it’s going to point this way. But you can also find out in a minute — by the way, you don’t need any of the symmetry arguments. You just do the calculation by brute force, it will have these properties. But it’s good to know what to anticipate, because maybe you made a mistake somewhere. It’s good to know some broad features. So none of this is needed to calculate, even in that problem. Go ahead and find the electric field not where I found it, but 2 inches to the right. You’ll find the answer looks the same. So those symmetry properties will come out of the wash, but it’s good for you to anticipate that, and that’s where you should look at the symmetry of the source. For example, the source was a ball of charge. You know if you rotate the ball, when I’m not looking to rotate the ball, it’s going to look the same. That means the field pattern should have the property, when you rotate it, it looks the same, because the same cause should produce the same effect. Anyway, going to this problem now, let’s find the electric field here. Okay, now this is going to be a stretch for me to draw, so I’m going to try, but you’ll have to go look at some textbook if you want a really nice looking picture, but this is the best I can do. I take a ring of radius r and thickness dr. I take an annulus, and I ask, what will that ring do to this point? So let’s take a tiny part of that ring, this guy. Well, for that, you just did what you did, you draw the line here. You’ll produce a dE that looks like this. What is its magnitude? Magnitude is just given by Coulomb’s law. The q there is sigma times the tiny area, dA. Let’s call this dA. dA is the name for a small area. σ times dA is the name for a small charge. That charge will produce a force, 1/4Πε_{0}, square of the distance, r^{2} + a^{2}. Finally, here is where the symmetry comes in, can you see that for every section here, I can find an opposite section that will cancel everything but the part perpendicular to the plane? So I should only keep this portion of it. Namely, I should take the cosθ. The cosine of that θ is the distance a, just like in the other problem, a/(r^{2} + a^{2})^{½}. This is now dE. If you want, you can put this following symbol, dE_{perp}, meaning perpendicular to the plane. Yes? Student: Do you need to multiply by 2 again, because you’re __________? Professor Ramamurti Shankar: Let’s be careful. Her question was, should I multiply by 2, because of this guy here, right? In fact, I should multiply by all kinds of other numbers, because so far, I’ve found the field only due to this segment here. I’ve got to add the field due to that and that and that and so on, right? What will that contribution be? For every one of them, this factor, (r^{2} + a^{2})^{3/2} is the same. They all contribute to the same factor, so when I added the shaded region, I’ll just get the area of the shaded region. All these dAs, if you add them up, what will I get? It will be sigma over 4Πε_{0}. Now you’ve got to ask yourself, what’s the area of an annulus of radius r and thickness dr? So take that annulus, take a pair of scissors and you cut it, and you stretch it out like that, it’s going to look like this. This is dr and this is 2Πr. So the area of an annulus is just 2Πr dr. So the sum of all these areas is 2Πr dr and then I’ve got here (r^{2} + a^{2})^{3/2}. But now this is the dE, due to annulus of thickness of dr. Then I’ve got to integrate over all values of r, but r goes from 0 to infinity. So I have here σ/2ε_{0} times rdr, divided by (r^{2} + a^{2})^{3/2}, 0 to infinity. Student: [inaudible] Professor Ramamurti Shankar: Did I miss a pi? Student: Shouldn’t it be sigma over ____ pi? Professor Ramamurti Shankar: There is a 2 pi here. Student: Oh, okay, yeah. Professor Ramamurti Shankar: 2ε_{0}. So do you understand what I did? I broke the plane into concentric rings and I took one ring and looking head on at that ring, I took a portion of that ring and see what field is produces. And I know that even though the field due to that is at an angle, the only part that’s going to remain is the part perpendicular to the plane, because the counterpart to this one on the other side will produce a similar field with the opposite angle here that will cancel, so only the part perpendicular will survive. Then I found out that the contribution from every dA had exactly these factors. They all had the same r and they all had the same a, so some of all the dAs, all I have to add is 2Πr dr. And that’s the contribution from this annulus, then I still have to look at annulus of every radius, so that’s the integral over dr. Yes? Student: What happened to the a _______? I thought it was a over — Professor Ramamurti Shankar: Oh, I’m sorry. It’s there. Thank you. There is an a still here. Yeah, I would have caught that guy in a while, but I’m always happy when you do that. That’s correct. Okay, so now about how this integral. Do you have any idea what you might do now? Yes? Student: Use substitution. Professor Ramamurti Shankar: Right. What substitution? “Use substitution” is a pretty safe answer, but you’ve got to go a little beyond that. Student: Substitute r^{2} + a^{2} for the ________. Professor Ramamurti Shankar: Yes. You can do that in this problem, because there’s an r on the top. If you didn’t have the r, you couldn’t do that, but now you can. I’ll tell you how it works. First of all, you can always do that tanθ substitution even here. It will always work. The tanθ substitution, if you put it here, it will still work. You can go home and verify that, but I will do it a different way now. I will say, let w = r^{2}, then dw is equal to 2r dr. So if I come here, I can write it as aσ/4ε_{0}. I borrow a 2 top and bottom to make it d. w also goes from 0 to infinity, but now I get (w + a^{2})^{3/2}. Now this is simple integral, dx/x + some number to some power is x^{n+1}/(n+1), but n is now 3 over 2. So you get aσ/4ε_{0}, divided by (w+a^{2})^{½}, divided by ½, which is 2 on the top, and that goes from infinity to 0. So I’m not going to do this much slower than this. This is the kind of integral that you can see right away, or you can go and work out the details. This is something you should do. If you have trouble with such integrals, then you should work harder than people who don’t have trouble with such integrals, because you should be able to do x^{n+1}/(n+1), and know that n+1 is ½, and when it comes upstairs, it becomes 2. Now if you look at this integral, in the upper limit omega’s infinity, you get 1 over infinity, which is 0. The lower limit when omega is 0, you get 1/a, and that will cancel the a here, and you will get σ/2ε_{0}. So that’s the final answer. So the electric field of this infinite plane, if you look at it from the side, looks like this. The σ/2ε_{0}. So what do you notice about this one that’s interesting? Student: It doesn’t depend on the distance. Professor Ramamurti Shankar: It doesn’t fall with distance. No matter how far you go from this infinite plane, the field is the same. Again, each part of it makes a contribution that falls like 1 over distance squared. As you go further and further out, you might think the field should get weaker, right? How could it not get weaker? They’re moving away from everything. At least with the line charge, it didn’t go weaker like 1/r^{2}, but it did get weaker. How can you go further and further from a plane? You are going further away from everybody? How could it not matter? Yes, any ideas? For example, if you go very close to the plane, what happens? If you go really close to the plane what happens is, the field in each section here looks like this. Therefore the part that’s useful is very small. If you go further away, you get things like that. Maybe it’s a little weaker, but the part that’s useful, this one, is getting bigger. So by magic, these tendencies cancel in the end. It doesn’t depend on distance. Now unless you do the integral, you will not know it doesn’t depend on the distance, because you can give arguments for why it’ll get weaker, arguments for why it’ll get stronger. The fact that it’ll precisely be independent of distance, you have to do the calculation. Yes. Student: What’s the negative sign? Professor Ramamurti Shankar: Negative sign where, here? Student: Yes. Professor Ramamurti Shankar: 2 is there, but the upper limit is infinity. Student: Oh, okay. Professor Ramamurti Shankar: All right. Now here’s the third problem, and the good news is, I’m not going to solve it for you, but I’ll tell you what it is. Here is a solid ball of charge. It’s got some charge density ρ coulombs per meter cubed. So ρ is the standard name. You use density for mass over volume and you use the same symbol rho for charge per unit volume. So somebody’s assembled a blob of electrical charge, and 1 cubic meter of that has ρ coulombs. You want to find the field due to this one. Now when you do a similar problem in gravitation, it is generally assumed that when you’re outside the sphere, the whole sphere acts like a point charge with the entire charge sitting at the center. But you actually have to prove that. That’s what took Newton a long time to prove. He knew it was true but he couldn’t prove it, because for that, you’ve got to be able to do integral calculus. And even today, to find the field due to a sphere using integration is quite difficult. Think about what you have to do. You want to sit somewhere here. First of all, for a sphere, we know the field is going to be radial. It doesn’t matter where you pick, everything looks the same. You can decide to be horizontally here at that point. Then you’ve got to divide the sphere into tiny pieces, tiny little cubes, each with some charge rho times the volume of the cube, and that will exert a force like this. And you’ve got to integrate over the volume of the sphere, but each portion is at a different distance and a different angle. Chapter 4: Gauss’s Law [00:44:29]You’ve got to add it all up. That’s why it’s a tough problem. So to solve that tough problem, we’re now going to use a very powerful trick and that trick is called Gauss’s law. So we’re going to learn today about the Gauss’s law. Now a prelude to that, you need a little more mathematical definitions, but they’re not bad. I just have to tell you what the definition is. Suppose I have in three dimensions a tiny little area, like a snowflake, but it’s flat and it’s rectangular, let’s say. I want to tell you everything about it. I want you to be able to visualize the area. What can I do to specify this little thing? First I have to tell you how big it is. If it’s a tiny area, let this area be dA meters squared, but that doesn’t tell me the orientation of this area, because that area could be like this, it could be like this, it could be tilted in many ways. So I want to tell you it’s an area in a certain plane, what should I do? How do I nail down the plane in which the area is located? Yeah? Student: Define the vector that’s perpendicular to that surface. Professor Ramamurti Shankar: Define a vector normal to that surface, because if you draw that vector, then there’s only one plane perpendicular to it. Then we can follow, we can then form a vector, dA. It’s a tiny vector whose direction is perpendicular to its area and whose magnitude is the value of the area itself. So areas can be associated with vectors. You may not have thought about it that way, but you can by this process. I’ve told you, there’s only one ambiguity even now. Do you know what that one — yes? Student: Which direction. Professor Ramamurti Shankar: There are two normal’s you can draw to an area, right? We’ve got an area like this, it can come out towards you or go away from you. Therefore simply drawing that rectangular patch is not enough to nail that. That’s like saying, “Here is a vector.” That’s not enough. Where is the head and where is the tail? That’s not a vector. That is a vector. Similarly, this area has to be specified some more and here is what you’re supposed to do. You take that area here, I’m just drawing it another place, draw some arrows, then circulate around it in one sense or another. I picked a particular sense in which they’re going around. Then use the famous right hand rule, where your fingers curl along the arrow and your thumb points in some direction, that is the direction, the area of the vector. If the arrows are running round the opposite way, then your thumb will point into the blackboard. So an area like this is like a vector without a head. Area like this is a signed area. It’s an area that’s got a magnitude and unique direction. So get used to the notion that a tiny planar area can be represented as a vector. Another way to see that is, if you took any area, a rectangular area like this — square is a special case — if you took two vectors A and B that form the two edges, then A x B is just double the area. It’s the fact that given two vectors, I can find the third vector perpendicular to them, up to a sign, is what makes a cross product possible, only in three dimensions. You cannot have a cross product of two vectors and four dimensions because in four dimensions, if I pick two vectors, they’ll be two other directions perpendicular to that plane. Only in 3D, there’s only one direction left. The question is, is it in or out? That you pick a sign in the cross product, A x B is something that goes from A to B. Or for an area, you draw arrows around the edge in a certain direction. So area is a vector. You have to get used to that notion, along with all the other vectors you know. Now I’ll tell you why that becomes useful. So we’re going to take — let’s see — there is a rectangular tube which has got a height h and a width w, and some fluid is flowing along the tube with the velocity v along the length of the tube. You got that? It’s like an air duct. Stuff is going through that tube. It’s got a rectangular cross section. The cross section area is hw. If the fluid is going velocity V along the length of the tube, what is the flow rate, which is equal to meter cubed of stuff flowing per second. I’m going to denote it by the symbol Φ. If I wait 1 second and I watch all the fluid go by me, past any cross section, how much stuff goes by? I think you can all see, if I wait 1 second, the fluid whose front was here would have advanced to here, and the volume here will be v times 1, because in 1 second, it goes a distance v. So the flow rate will be hwv or area times v. That makes sense? The faster it’s going, the more stuff you get. The bigger the cross section, the more stuff you get. This doesn’t depend on a rectangular cross section. It can be cylindrical pipe carrying oil. Again, the flow rate is area times velocity. But I’m going to write this in terms of vectors, because I know the velocity is a vector. But now I have also learned area is a vector, because area vector here, you can draw a vector perpendicular to this area, and I’ll draw up this convention that it’s area vectors along this way, rather than the other way. Then I can write this v⋅A. Let’s check that v⋅Ais correct. v⋅A is the length of v, the size of v, the size of Aand cosine of the angle between them. Here we’ve got to be very careful. The velocity vector is like this. It’s perpendicular to the plane but don’t say it’s cosine of 90 degrees. The area vector is perpendicular to the area itself. Do you understand that? When I take the dot product and I will ask for the angle between the area vector and the velocity vector, that angle is 0. For the area, there’s a little confusion. The vector representing it happens to be perpendicular to the plane of the area itself. So if you remember that, then v⋅Ais vA times cos 0, so this indeed is one way to write the flow. This flow is also called a flux. But now let us do the following. Let’s take the same problem, and I have this area here. Let me now take a tilted area like this. It also goes from the ceiling to the floor but still turned at an angle θ. So it’s a bigger area than the original one. How much bigger? That area prime, I claim, is equal to the base w, times this side. This side is h/cosθ. θ is the angle between these two planes. It’s a bigger area, but you all know that just because it’s a bigger area, it doesn’t intercept more fluid per second. Any stuff crossing this guy also crosses this at the same rate. So how am I going to get the same rate? The flux is not going to be v times A’. It’s going to be vA’cosθ. But θ is the angle between the area vector and the velocity vector, which is the same angle here. So the moral of the story is, v⋅A’, or v⋅A in general is the flux or the flow, off of any vector across an area. If it’s a fluid that’s flowing, then v⋅A is the fluid flow past that area. If you need the dot product, you need the cosine of the angle, because the area, if it’s not perpendicular to the flow, it’s not useful. In fact, you can take a huge area parallel to the flow and nothing goes through it. So area is most effective if the plane of the area is normal to the fluid, or the area vector is parallel to the velocity of the fluid. So that’s that lesson. Okay, this had nothing to do with the electric field, but we’re going to come to the electric field now. This is just a warm up. Let’s come to the electric field and see what’s going on. So I take a charge q and I draw the lines coming out of it. How many lines do I get crossing a sphere S? Well, we know that we have agreed to draw 1/ε_{0} lines per coulomb, so this q here, that many lines cross the sphere. I’m now going to relate it to something I can do with the electric field as follows. I’m going to say that if I go to that sphere, I look at the electric field. Electric field is in that direction, E. And any portion of the sphere — and I want this is where you got to __________ imagine — take a tiny part of the sphere. There’s a tiny area that’s got a size dA. What is its direction? Direction of the area vector is radial. You understand? The area is on the surface and normal to that is the radial vector, which I always denote by e_{r}. The electric field is equal to q/4Πε_{0}(1/r^{2})E_{r}. Therefore E⋅dA = q/4Πε_{0}(1/r^{2})dAe_{r}⋅e_{r}. This is a dot product of the area vector with the electric field vector. The two unit vectors are parallel. The dot product of that guy with itself is just 1. So this is the number of lines crossing the tiny area, because electric field numerically is equal to the line density. So those lines crossing this area, this is the number of lines crossing that patch, which is given by E⋅dA. If you add up all the lines, you must add up all the dAs. Sum of all the dAs on a sphere of radius r is just q/4Πε_{0}r^{2} times sum of all the dAs on a sphere, that is just 4Πr^{2} . So you get q/ε_{0}. In other words, either you can draw the picture and it’s immediately obvious to you the lines crossing is q/ε_{0} by construction, or you can remember, “Hey, the electric field is a direct measure of the number of lines per unit area,” and the electric field times area times the cosine of the angle between the area and the electric field will count the lines going through a tiny area. If I add them all up, I’d better get q/ε_{0}, and indeed you do. So the moral of this little exercise is that the surface integral, let’s call it the surface integral of the electric field, on a surface is equal to q/ε_{0} where this was a sphere. In other words, even if you’ve never heard of field lines, just take the electric field and do the surface integral, you get this. So surface integral is a new concept. You probably have not done that before and I’ve got to remind you how it is done at least operationally. If you’ve got a computer, you can find the surface integral of anything as follows. Take the surface over which you’re doing an integration. Divide it into tiny pieces, each is a little area dA. Take the dot product of the dA with the electric field there, and sum over all the patches covering the sphere. Then take the limit in which every patch gets vanishingly small. That is called the surface integral of the electric field and we see that if you do that, you get the charge inside divided by ε_{0}. But that was on a sphere, and the interesting thing was, the answer was independent of the radius of the sphere, because the 1/r^{2} in the field canceled the r^{2} in the area. But it is even better than that, because it’s clear to you know, without any calculation, that if I took some crazy surface like this, the lines crossing that is also the same. Because the lines leave the origin, they go radially outwards. They don’t terminate on anything. They’re only supposed to terminate on other charges, therefore you can count them anywhere you like. You can take a census here, you can take a census there or there. You’re always going to get the same number of lines. Therefore q/ε_{0 }is also going to be equal to the line count on some weird surface enclosing the charge. So how am I going to count the lines on a weird surface? I take the surface, divide it into little pieces, but now the area vector and the electric field vector may not be parallel, because it’s not a sphere. But it is still true that if you’re trying to count the lines going through, just like in the velocity, you must take the dot product of these. Therefore you will again find E⋅dA for any surface, any closed surface, is equal to q/ε_{0 }. If you got lost in the mathematics, the physical picture is very clear. If you want to count how many lines leave the charge, you can surround it with any surface you like. If it’s a sphere, it’s very easy for you to do the check. If it’s a crazy surface, it’s harder but they all come from the fact that E⋅dA, count the number of lines and that’s independent of the surface at stake. So now that it what is called Gauss’s law, so I’m going to write it down here. What I’ve shown you is the following. Here is some strange closed surface that’s a charge q inside. Then E⋅dAon that surface is equal to q_{inside}/ε_{0}. This is not yet the theorem, but this is the case for 1 charge. If the charge were outside, suppose the charge were here, then the lines would go like this, and the total number leaving the surface would actually be 0. You might say, “How is that? I see all these lines penetrating the surface,” but you’ve got to remember, if you take an area vector here, with the definition for a closed surface, the area vector is the outward pointing normal. For a closed surface, every area is a vector pointing out from the closed surface. Then you can see that in this surface, lines are coming out; on this surface, lines are going in. It’s very clear what’s going in is coming out, because nothing is terminating. Therefore if you took a surface that did not enclose the charge, this answer would be 0, but if it enclosed the charge, it will be the charge inside, but it won’t matter where it is. You can also see that you can move this charge around anywhere you like, you don’t change the number of lines coming out. Then the most important generalization is, if instead of 1 charge, I have 2 charges, what do you do? If you had 2 charges, let us take the total E⋅dAon a surface. The total electric field is the electric field due to the first charge on that surface, + the electric field due to the second charge on the surface, thanks to superposition. It’s the fact that the electric field is additive over charges. But this one is q_{1}/ε_{0}. This is q_{2}/ε_{0}. Therefore, we can now write, generalizing to any number, whether one draws a double integral with this thing, meaning it’s a closed surface. You can also have an open surface, take the skin off an orange and cut it in half. That hemispherical skin is also a surface, but to signify it’s a closed surface, we do it like this. On a closed surface of E⋅dA= q_{enclosed}/ε_{0}, which means sum of all the q’s. And sometimes we write this as follows. Suppose inside you don’t have a discrete set of charges that you can count, but a continuous blob of charge, and the charge has a certain density, ρ is the charge density. What is the q_{enclosed}? I claim the answer is due to a volume integral of ρ or any point r inside, times dxdydz inside that volume. You know, if you want to say how many coulombs are there, well divide the volume into tiny cubes of size dxdydz. That times the ρ of r, meaning ρ at x, y and z, is the charge inside the cube. You add it all up, you get the charge enclosed in that funny shaped object. So this is the final form of Gauss’s law that we’re going to use. I’m going to use the symbol, which is very useful. This is the theorem of Gauss. And S = DV. DV is the boundary of V. In other words, V is like a potato, the skin of the potato is DV. Therefore if the potato is full of charge, it will emit some electric flux and the surface integral electric flux over the skin of the potato is the charge enclosed inside it. The charge enclosed inside it is not simply some constant density times volume, if the charge density varies from point to point. In each neighborhood inside their volume, you take a tiny section, a tiny little cube of size dxdydz, see how much is in there, as the function of x, y and z, and you do the integral over the volume. Now these integrals may be hard to do, but you should know at least what I’m saying. This is just a way of counting the total charge inside a volume when the charge is continuously distributed. So I’m going to show you one application of this and we’ll come back to more next time. And the one application I’m going to show you is to find the electric field due to a solid ball, solid ball of charge. So I’m going to use Gauss’s theorem to do that. This is true for any surface you pick. It doesn’t matter what surface you pick. So if I want to find the electric field here, I’m going to pick a surface of radius little r. Let the sphere have a radius big R. I want to find the field here and I’m going to use Gauss’s law. So on the left hand side and the right hand side are two different things. On the right hand side, what is the charge enclosed? That region. Well, the charge enclosed is just some number q. This whole thing is q, q spread over a sphere of radius r, over ε_{0}. That’s going to be the surface integral of the electric field on that sphere. E⋅dA on that sphere. Now normally, if you knew the value of an integral over a region, you cannot deduce anything but the integrand, unless what? There’s one exception where if you know the integral of a function, you can find the integrand. You know when that might be? Here’s a function. I tell you its integral from here to here, but I don’t show you the picture. What’s the integrand? You don’t know. But if I also tell you the function is a constant, and I tell you the integral, and you know the width of this region, you can find the integrand. So you’re going to cook this up so that this whole integral is a constant times the area of integration. And we argue that if you’ve got a spherical charge density, the electric field must be radial everywhere. You’ve not proven this, but you argue that because it’s the only distribution with lines coming out everywhere, invariant under every possible rotation of the sphere. Because if you rotate the sphere, I won’t know you rotated it, so the field pattern cannot look different. If the field pattern looked different when you rotated it, then you have a problem because the cause looks the same, but the effect looks different. That’s not allowed. The only allowed this is the radially outgoing electric field. Therefore the electric field is radial and also constant throughout the sphere, because all points on the sphere are equivalent. There’s no reason why this is any better than that. The sphere looks the same for all directions. Therefore this whole integral is going to be 4Πr^{2} times the electric field at that radius r, because the area vector and the electric field vector are both parallel, so in this dot product, forget the dot product. It’s just E times dA. There’s no cosθ, just 1. You can pull the E out of the integral, because E is constant in magnitude on the sphere. And the integral of the dA is just 4Πr^{2}. Therefore you deduce E(r) is q/4Πε_{0}r^{2}. It’s a very profound result. It looks very simple, but it is true for not a point charge, but for the spherical distribution of charge. That the field goes like that of a point charge sitting at the origin, is a consequence of Gauss’s law. If the charge inside was not uniform, suppose it’s a charge q, but there’s more stuff here, less stuff here, this theorem would still be true up to this point, but you can never deduce that E is a constant on a sphere, because even though a sphere is nice and symmetrical, the charge distribution is not. It could be big here, it could be small here. You know something about the integral over a surface of a varying function, then Gauss’s law is no good, not useful. True, but not useful. Gauss’s law is useful only when in a given problem, there’s only one number you don’t know. That number here happens to be, what’s the strength of E at a radius r? I know it’s direction is radial, I know it’s magnitude is constant on the sphere by symmetry, but what is the number? You can trade that one number for this one number on the left hand side, q/ε_{0}, you can calculate it. I’ll come back and do more examples for you guys next time. Okay, so see you next Wednesday. [end of transcript] Back to Top 
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