6. Capacitors
Shankar, Ramamurti
The electric potential is defined for the electric field. It is introduced as an integral of the electric field making the field the derivative of the potential. After discussing the ideas of electric potential and field as presented in the previous lecture, the concept of capacitance is introduced as a means of storing charge and energy.
 Okay, let's turn to the subject proper. I will start by doing the usual stuff of just refreshing your memory on what happened last time. I mean, last time was pretty heavy, a lot of mathematical machinery which I like to work with. Now, you might say, you know, I'm going to be a psychologist or art historian and I don't need all those details. Just tell me what I need to get through this course. So, that's a very reasonable question. Probably would have the same attitude if I took a course in your field. So, here it is. What I've told you so far is that the electric field, which we understand as basically Coulomb's forces, allows you to define a potential because it's a conservative force. That means the integral of the electric field on any closed loop is zero, and remember that's what it takes to define a potential. And the electric force, which is Q times E, will also satisfy the condition. That means we can define for this problem a potential v, which is called the potential, so that 1⁄2mv1 squared. This is a little v. This is big v. Capital V at the point r1 is equal to 1⁄2mv2 squared plus capital V at the point r2. That is our goal. Ah, missed one mistake already. Did you know what's missing here? Pardon me, yes? What's missing? Yes, the charge. So, this should really be Q times v, and this should be Q times v. So, let me remind you why it is. The force on any charge is really not just the electric field. It's Q times the electric field, and therefore the potential difference is associated with the work done by the electric field times Q. So, you've got to remember Q times v is the real energy. So, v is like an electrical height. For example, if this is the ground, there's a mountain like that, you give me the height. The potential energy is not just the height, it's mg times h. So, let me say you give me g times h. That's like the potential. That tells you the whole story, because if you want to lift a kilogram mass to that height, you do 1 times g h, 5 kilograms, 5 times g h. So, depending on the mass you want to lift, the work done will be different, or the potential energy will be different. But this doesn't depend on the mass you're planning to carry around. So, you take out the mass, you can always put it back. Similarly, the potential concerns itself with the unit charge. If Q were 1, nothing I wrote is wrong. But in general, Q is not 1. So, this is the actual energy. This is the old potential energy, U1. So, the potential energy is Q times the potential. So, think of V as the electrical height at a point, and Q is like the mass, and Q times V is like mg h. Okay, we got a formula for V. The general formula for V is that V of R2 minus V of R1 is the line integral of E dot dr from R1 to R2. This minus sign has the following meaning. The electric field exerts the force E on the unit charge, but if you want to drag the charge against the field from R1 to R2, you apply force minus E to compensate that. So, this really worked done by you, this really worked done by you, to move it from R1 to R2. Let me give you the gravity example. Gravitational force acts this way. But if you want to lift something from here to here, you oppose it, namely it's doing minus mg, your force mg, you move it to height h. That's the work you do, and that you know is the potential at the top minus potential at the bottom. So, that's the origin of the minus sign. The electric field wants to go one way, and if you want to take it against that force and raise it electrically speaking, the force you apply is precisely minus E. That dot dr is the work done by you. That's what you invest. That's what you will get back if it rolls downhill again. Now, if you want V at some point R, minus V at infinity, that integral will be minus, starting from infinity to the point, and then we all vector to the point R of E.dr. We will now take the convention that V of infinity is equal to 0. Remember, potentials are defined only by the differences. So, you can put the 0 anywhere you like. We will choose the 0 to be at infinity. Then in this integral, it's going to be 0, and if you do this integral from infinity to R, you will find for one charge, it is q over 4 pi epsilon 0 times R for single charge, and R is the distance from that charge. It looks like Coulomb's Law, but a couple of differences. There's no 1 over R squared and there is no vector. The potential due to a charge here at the point here is a number given by this. So, if you look at lots of charges, let's just find V due to many charges is sum over all those charges, the value of each charge before pi epsilon 0 times the length from where that charge is to where you want the potential. That's the principle of superposition. The Coulomb potential due to one charge is 1 over R due to many charges is each q times its own 1 over R times its own 1 over 4 pi epsilon. And finally, if you want the potential difference due to the field, you integrate it, and if you want the field from the potential, you take the derivative called the gradient, which is identical to minus i x partial derivative minus j y partial derivative. So, you expect two derivatives because the force you're looking for now is a vector. You're living in 2D. So, you cannot say the force is the derivative of the potential in what direction? If you want the force in the x direction, take the x derivative. If you want in the y direction, take the y derivative. And you put them together into a single vector, and that's the electric field. So, you go from potential to the field by taking the derivative of the gradient, and you go from the field to the potential by doing the integral. So, the subtleties, the integral is of a dot product of E with the displacement because everything is a vector now. It used to be fx dx in the old days, but f has become a vector, dx has become a vector dr, so you need the dot product. Okay, so this is a summary, but I want to make two course corrections from last time. First one was, when I tried to prove to you that the electric field is conservative, I did one thing which I realized was a swindle. I didn't realize it myself, and I wasn't waiting for anyone to correct me. I did not think about it hard enough. So, here is the mistake I made. Remember what I said? I said that's a potential due to point charge. I'm trying to look at it. These are different lines of the electric field. Let's say you want to go from here to here. You want to do some work, E dot dr along this line. The claim was that's the same work you do if I take another path that looks like this. Go this way, go that way, and go that way. Same work, and that's still correct statement. The same work because here dr is angular, E is radial, the dot product is 0. Likewise, 0. Here at every segment you have here, E is radial, dr is radial, the dot product is the field E times that length. Here is the field, here times that length, the field is constant at a given radius, so they're also equal. Still, no wrong statements. Then I said, what if I do this? A little radial, a little angular, and so on, and that's also okay. Then I finally said, what if I had some nice smooth path which is not made up of radial and angular, and my claim was, well, you take for this segment, for example, you take a radial segment, then you take an angular segment, then you may take another radial segment, another angular segment. So you approximate the smooth path by a bunch of radial angular, radial angular, and from a long distance it all goes the same, so it should be the same amount of work. That's wrong. I'll give you an example of why such arguments can be wrong. Let's take a triangle. This side is 1 and this side is 1. I want to walk from here to there. So you know I go a distance square root of 2. But then you say, I'm going to take a different path, I'm going to go horizontal, vertical, horizontal, vertical, equal amounts like that. So these are very tiny differences. You can say it's the same thing. But you know this path is of length 2. Understand why? All the horizontal parts add up to 1, all the vertical parts add up to 1. So going zigzag is longer by a factor root 2. So just because things are approximately the same, doesn't mean the answer will be the same, because if you make an infinite number of infinitesimal mistakes, you can add up to something finite. Likewise here, you've got to make sure that you can deform your path from what was really given to you to what you did, which is radial and angular. The work done should be the same. The path length is clearly not the same in this example, and it can all add up, but the work will be the same. That's what I did not show to you. So let me take a segment here where that's the radial part and an angular part, but the person really wanted to take a segment along some straight line joining the two. What I want to show you is that not only does that approximate this from long distance, the work done on that plus that is the work done here. Okay? So for your convenience, I'm going to blow up that picture here so you can all see what I'm trying to do. This should be the radial direction. This is the radial direction. This is E. This is some perpendicular direction, and this is the direction in which the move dr was made. That's the actual path which is going like that, and you manage to just deform it. So what's the line integral on the radial path? Well, it's the magnitude of the electric field at that distance times the difference in radius between this end and this end. Let me call it delta r. Delta r is the spacing here between that curve and that curve. You agree? The dot product is trivial. E is this way. Displacement is this way. No cos theta is to worry about. That's the work done. Work done here is zero because perpendicular to the field. So that's the total work done going like that. How about going along this line? It's going to be E at r dot dr. When I say E at r, r is of course not one point. It's going to spread out, but this is an infinitesimal region, so you can talk about the E at one point and not worry about the variation. So don't forget it's an infinitesimal triangle. E dot dr will be E at that distance r, the length of dr, and the cosine of the angle between them, which is this angle. But length of dr times cosine is precisely that distance. Delta r, and therefore you do the same amount of work. So basically I'm saying as you wander around, you may not choose to go radial angular, radial angular. But if you go at a slant to the radial, the work you do really depends on how much radial distance you cover. That's what E dot dr does for you. That's why you can deform the path anyway and get the same answer. That's the first thing. The second thing is I forgot to tell you anything about units. The electric field is forced on a unit charge, so the unit for that is Newton's per coulomb. The electric potential looks like the work done on a unit charge, so the unit for that is Joules per coulomb. And the name for Joules per coulomb is a volt. It means if there's a voltage difference between two points of the two poles, it's going to take you 12 Joules of work to lug a coulomb from the lower point to the higher point. So you guys have to get your units right, okay? Now you might say, why don't you use units? Right? You know why? Because I have tenure. When you have tenure, you don't have to use units. I have tenure, I have a car next to the fire hydrant, I got tenure, I tear up the tickets, I don't do duty, duty, I don't pay taxes. You don't have to do anything when you reach. You know, Allen was going to tell you the joys of being a graduate student. The joys of being tenure prof, hard to describe, okay? Priceless. In fact, one of my sons, when he saw my lifestyle, said it reminded him of a jellyfish deep in the ocean. He saw a rock, he attaches himself to a rock, and then never moves and eats his own brain for food, because he has no further use for it. So apparently that's what I reminded him of. So you can, if you work hard, you can be like me. You can do that. Anyway, look, so you guys have to get the units because seriously speaking, if you don't write any units, we don't know if you write the rock, right? The problem answered is 19, in some units, right? So if you don't give the unit, the number is not worth anything. Another advantage of units is you can keep track of your calculation. You multiply one quantity by another quantity. Everything has units. You cross out the lengths and the meters. You've got to make sure the stuff you get has the right units on the two sides. That helps you also keep track of calculations. Okay, so let's go back now. This is all stuff left over from before. So now for the new stuff. I'm going to tell you the first part of this lecture is going to be what are the advantages of V? I will just mention them, then I will elaborate on them. The first advantage is the conservation of energy. The second advantage is the computation of E. The third advantage is a lot of good visual pictures, which are very helpful. So you won't know what these mean, but this is what we're going to do, so you know where we're going. First thing is conservation of energy, there's no need to belabor at this point. I mean, that's why we did the whole thing. The whole conservative forces, gradient, this, that, all that was to make sure in the end you can get a potential energy out of it. And as you know, from the roller coaster problem, you can do kinetic plus potential equals kinetic plus potential, it saves you a lot of trouble. In other words, in this room right now there's a lot of electric fields. A lot of charges everywhere applying a field here, and I want to carry a charge from here, and I want to drag it over here. Or maybe the charge was shot by a cyclotron or something came flying out at some speed here, and it comes here a little later, and I want to know how fast it's moving. One way to do that is to follow the charge as it moves at the same time. You can, for instance, find the force on it, the acceleration from that, change in velocity, add up all the changes, you'll get the final velocity. But as you know, with the law of conservation of energy, you can skip all the intermediate stuff. You just have to know the initial kinetic energy, initial potential energy, final kinetic energy, final potential energy. They are equal if you sum them. So you can find one number from it, one of the numbers you can find. In particular, if you didn't know the final kinetic energy, you can find it. That's the energy. The second thing is that it makes life easy when you want to compute the electric field. So this is an example I did again near the end of the class, and I think it was needlessly complicated. So I will try to repeat the derivation a little more quickly, because it's a lot easier than I gave the impression. So we are trying to find, let us say, the electric field is a simple minus q and plus q separated by a distance 2a. If you did it with forces, with forces I didn't even try to do it everywhere. I remember I did it here and I did it there. The force here was like this, the force there was like that. Then I said if you do a lot of work, you'll get all these lines. What makes life difficult is that if you go to this point here, this I explained to you last time, this guy will repel it with some force, this guy will attract it, and you've got to add the two arrows and get a new arrow. That's the electric field at that point. So you've got to add arrows, and that's more difficult than adding the potential, which is just a number from this guy and that guy, then once you've got the potential, you find the field by taking derivatives. You want to take the derivative of ex, take the derivative of the minus sign, you want ey, you take minus dv dy. So what's the e here? You remember there's q over 4πε0 times 1 over r+, r+, at that length, minus 1 over r-, that's it. No arrow is nothing. That's the total v. This is r-, let's define r to be the distance from the center. And I want you to imagine this is long way off. This picture is not drawn to scale. You should imagine these two points are like this and I'm somewhere there. That's when the dipole approximation is working. This is not an approximation, this is exact. So let's do the following then. We combine the denominators, you get r+, times r-, and the denominator is r-, minus r+. Now you can see that r+, and r-, differ from r because of this little guy, a. If you use the law of cosines, you know, it's that squared plus that squared minus 2a r cosine theta or something. In the bottom, we ignore that difference and we write everything as 4πε0r². Because r+, I approximate with r, r-, I approximate with r. So this is not the exact answer. There are corrections here that look like 1+, maybe a number a divided by r and also a² divided by r². But these are all negligible. In the numerator, you can again say, why don't you replace this by r and this by r, then you get 0. It is true. If you replace everything by r, then you do get 0, and that's because in the difference the big guy cancels and what's left over. And the difference is what's going to keep this from vanishing. So we have to find the difference in that distance and that distance. So for that what you do, you draw a line here. Now you cannot draw a real perpendicular here, but if these things are very far and these lines are almost parallel, you can draw a perpendicular there. And if that angle is theta, theta is the angle, then it's very easy to show that this distance is simply 2a cos theta. Because if theta is this angle, then you can say cosine of this angle here will be, by whatever complementary angle, is 2a cos theta. So if I draw a line here, I have 2a times cosine theta, and that I will write as dipole moment P, cosine theta over 4πε0 r². That's what I was trying to get. Yep? Student is asked which angle is theta? Theta is the angle measured from here, let me draw this. You can, one can worry about, is it this angle or that angle? But I'm saying draw the angle from the middle, okay? That angle is theta. And roughly speaking, that angle is also theta. Now drop a perpendicular here, and this side is 2a, and this side is 2a, and this side is 2a. So this difference is 2a cos theta. So now I have v in terms of r and theta, but I can write it in terms of x and y very easily. It is P over 4πε0 r², and cosine theta is x over r. So the whole thing is x over r cubed, and it's the same as x over x² plus y² to the 3 over 2. Right? r is the square root and r cubed is three powers of that. Now you can easily take the derivative of this guy with respect to x, or with respect to y. That's very, derivative is very easy to take. Integrals are hard to do, and that'll give you the electric energy, and you'll get the ex and ey everywhere. So ex is equal to minus dv dx, and ey is minus dv dy. So one of the homework problems is to really calculate this. And if you calculate it and you plot it, you will see this thing emerging. Okay. In other problems, most of the time it's easy to find b and then to find e. For example, if I have a ring of charge, and I want to find the b here, then let's say the ring has charge q and some radius a. Then a little guy here produces a little potential dv, which is the charge dq. dq is not any distance, it's like the differential, divided by 4πε0. And what's the distance? If that is z, then it's simply root of z² plus a². So that's the potential of the x², due to this little charge. But every unit of charge on this ring is equidistant from this point. You understand it's a ring and I'm perpendicular to the ring. So they all contribute the same number and some of all the dq is just the q. So that's the potential on the axis. Now you can take a z derivative to get the field. Okay? Generally, it is easier to do the v. Either it's a sum or integral over the charges producing the v, then take derivatives. But I want to give you one example in which the opposite is true. But in fact, it's easier to get e than to get v. And that is, there are many such examples, I'm just giving you one. So I say that's a hollow conducting sphere. It's got charge q. This guy's hollow and conducting. I want to find the potential everywhere due to this. And the charge q is uniformly spread on the surface. So what's going to be the potential? First of all, by symmetry, you can see that if you pick any point, anywhere, the answer is going to depend on the radial distance from the center, because of spherical symmetry. But you still have a lot of hard work to do, because you can divide the sphere into little rings like this, slice it at different latitudes, and find the potential due to each ring using this formula I just did somewhere, then integrate over all the rings, and you will get an answer. But that's a decent amount of calculus. But you don't have to do that, because in this problem, you can't do that. Because it's vertically symmetric, and there's a charge q here, we know from Gauss's law the electric field is going to be purely radial, and will look like q over 4 point epsilon 0 r square, as long as r is outside the sphere. So here's one problem, but it's easier to get E, because E is pretty trivial. You just need to know the magnitude. Gauss's law tells you the magnitude. You put the unit vector back and you get this. So as long as you're outside the sphere, it's as if q were sitting at the origin, and the potential V of r, which is minus q plus q over 4 pi epsilon 0 r, outside that sphere. So that's because remember V of r minus V of infinity is equal to integral of E dot dr from infinity to r, the minus sign. And if you just take the field of a point charge and do the integral, you will get the potential of a point charge, which is this. So if I draw the potential as a function of radius, it will look exactly like that of a point charge, and I'm going to draw the potential of radius over r. Now what about for r less than r? How do I continue this graph? Pardon me? You can use Gauss's law. If you use Gauss's law, inside what will you get for E? E is 0. Integral of E dot 0 is dr is 0, so potential does not change inside the sphere. But don't think potential is 0. If it does not change, it's stuck at this value. So throughout the hollow sphere, there's only one potential. Electric field is a different story. Electric field is in fact 0 inside, then falls like 1 over r squared. Electric potential is a constant. So these are, there are certain problems like a cylinder of charge surrounded by another cylinder of charge, where again it's easier to get E from symmetry arguments than E to get the potential. Okay, so this is my second point, namely, barring a few exceptions like this one with high symmetry, the potential as the advantage that it lets us compute the electric field by first doing a sum of scalar quantities rather than vectors, then taking derivatives. Now third advantage has to do with being able to visualize the potential. It gives you a landscape, and that's intuitively very helpful. So here's one example. Let's take two parallel plates, put a lot of plus charge here, and equal amount of minus charge here. What will the electric field look like in this region? Let's say both have the same density sigma for unit area and that infinite. I'm just showing you a finite portion. So this guy will produce sigma over 2ε0 here and here. This one, because it's negative charge, will do this, sigma over 2ε0 on this side. In the region between them, you can see the two arrows are additive. So the field will look like this. In the region outside, the two arrows are opposite. This guy is coming and that guy is going out. So the field doesn't diminish with distance due to either plate so they can completely cancel each other. So the electric field is trapped between the two plates. And as the strength, which is double what you would have on either side if you had only one sheet, that's the electric field E. That's my electric field. We can ask what's the potential difference? Or what's the potential at various points? This whole thing being metallic is at one potential. I say v is equal to 0. Throughout this line, let's say v is 0. In fact, it's very clear v is 0 because if I move horizontally on this line, I don't do any work because the electric field is perpendicular to me. So if v was 0 at one point, it is 0 throughout this plane. Yes, I should qualify the statement. You cannot pick v equal to 0 over an extended region. You have freedom to pick it to be 0 at one point. I'm telling you, if you pick the one point in this plane, you can navigate throughout the plane without changing v, because you move perpendicular to E. So v is 0 on the lower plate, but it won't be 0 anywhere else, but you can see that may be v equal to 1, that may be v equal to 2, v equal to 3, upper plate may be v equal to 4 volts. So lines of E are vertical and lines of constant v are vertical. This is really like the gravitation problem. An electric field raining down on you is just like the force of gravity. This might as well be gh. This might as well be g. You multiply a mass, you get mg. You multiply by the q, you get the force on a charge. It's straight down and the potential just increases with height. Yep? Student is asking how do you do the cutters to get the constant voltage? Oh, okay. Think of it as a plate coming outside the blackboard. Student is asking for the hollow sphere. Oh, the hollow sphere. The potential is always the work done to go from infinity to wherever you are, right? So when I come from here, what do I think is at the origin? I think there's a point charge at the origin. So I push up against that. I come here. So I do that amount of work. Once I'm in here, there is no electric field inside. So e.dr is 0 from then on. But don't forget the e, the v that you got up to that point. You don't drop it. You add, keep on adding 0 to it. So it is 0 everywhere. You add 0 to all points here because you can go from here to here, no electric field. So the whole sphere has the same potential as the surface. So now, here I got these potentials, and it gives you a picture that you're running uphill, and these are the steps you've got to climb to go uphill. And if you've got four volts here, if you take a cool arm and you let it drop, it will pick up four joules by the time it hits the slower plate. That's the meaning. So the work done by the electric field, just like gravity, is the potential difference, which is the work done on unit charge times the charge, which is one cool arm here. If it is ten cool arms, you do 40 joules of work. If an electron is released, you've got to be a little careful. By definition, the charge of an electron is negative. So if an electron is released here, in fact, it will go like this. You don't have any analogs in gravity. Gravity, everything falls down. So what was something said negative mass and something said positive mass? All guys with negative mass, you've got to nail them to the ground, otherwise they'll just float up. That happens in electricity, so any electron you leave here will go up. So the force on the electron is minus q times e, so everything's got a minus sign. So an electron going from here to here will in fact lose energy. It will lose potential energy, and it will turn into kinetic energy. So if an electron falls through 1 volt, the work done is charge of an electron times 1 volt, which is 1.6 times 10 to the minus 19 joules, that happens to be called an electron volt, just a convenient thing to use. When the charges you're pushing around are not cool arms, but this tiny electron is good to know in electron volts. For example, in the hydrogen atom, the cool arm potential is negative because the charge at the origin is a proton plus q. Electron has got some minus q divided by 4πε 0r. That's the potential. And in quantum mechanics, only at certain radius you can have electrons. So the lowest electron is somewhere here, and they say it is minus whatever some number of electron, 13 points something electron volts. That means if you want to drag an electron from this hole and liberate it from the atom, you've got to do 13 times 1.6 times 10 to the minus 19 joules of work. So as if this were 13 volts down compared to infinity. It's a hole in the ground, 13 points, and it's got a 4 or 6 volts deep. So if you want to liberate that electron, you have to furnace that amount of energy. Okay, now going on with visualization, let me say let's take another case we can all do, which is a single point charge. Here's a point charge. The electric field due to that looks like this, and it is million times. Now what about the electric potential? For a point charge, the potential is 1 over r. That means it's a constant on a circle of radius r. That means if you pick any circle or a sphere, I'm sorry, v is one number here, v is a different number there, a different number there. So that's the whole surface surrounding the charge on which v is some fixed number. It depends on that radius. You go further out, it's a different number depending on that radius, and it falls like 1 over r. You notice once again that the lines of constant potential are perpendicular to the lines of the field. That was true there, that is true here. In fact, let me do one more example. If you take a dipole, negative and positive, and the field lines go like this, if you compute the potential, if you go very, very close to this guy, I think you all seem to be in the same position. If you're very near one charge, forget about the rest of the world. It dominates everything. Potentially, it will look like circles centered here, but as you go further out, the shape will change and it will look like this. A similar thing on the other side, then there will be this huge infinite sphere that becomes planar, that goes all the way to infinity. These are contours of constant v. If you really plot them, you'll again find contours of constant v. If you plot them, you'll find that they're perpendicular to the electric field lines, perpendicular to e. So I want to explain to you that this is a very general result. This is not something true in these isolated cases of nice symmetry. Why is that? So let's understand why that is true. The v2 minus v1 is minus integral of e.dr from 1 to 2. So let's say v2 and v1 are 2 and 1 are very close. In that case, let's call it delta v is equal to minus e.dr. There's no integral to do because I've picked two points which are nearby an infinitesimal amount dr, and e.dr is equal to delta v, which is minus of v2 minus v1. What does that mean? That means if you start at some point and you move a distance dr, you take a step dr and you find the change in potential, the answer depends not only on the length of your step, but on which direction you move. Because of this dot product, the change in potential will be the magnitude of the electric field at that point, the magnitude of the step length, and the cosine of the angle between them. So if electric field, if the electric potential is replaced by height in two dimensions, you know, top of sleeping giants and the hill is going up and down, up and down, you're at some point. There's a height function h of x and y instead of v of x and y, and I want to take one step and I can ask what are the consequences to my height? The answer is if the electric field is pointing in some direction and you take a step dr in a generally different direction, this is the change you get. So you can ask, how do I get the biggest drop in height, or the biggest drop in potential, when cosine theta is 1? Right? Because then it's this, that times 1 with a minus sign. So the electric field points in the direction of the greatest rate of change of potential. It drops that direction. If you release a marble in a gravitational field, you'll roll along the gravitational field, which is obtained the same way from the gravitational potential. If you want to climb up the mountain, suppose there's a tsunami coming, and you want to get to the top as quickly as possible, don't panic, compute the gradient, and move along the gradient. Okay? Then stop, think again, recompute the gradient, and in that manner you will get to the top in the quickest time if you can compute quickly enough. Okay? But that's the process. At every point, that's the vector pointing in the direction of greatest increase, the electric field happens to be the direction of greatest decrease because of this minus sign in our convention. Now, let us say you are in this mountain, and you don't want to go down, you don't want to go up, but you still want to take a walk. Can you do that? Yes? Or are you stuck at one place? What do you have to do, guys? Yes? You move perpendicular to the field. To move perpendicular to the field, the cos theta is 0. That's why, now going back to this, this is called Mount Kula. It's coming out of the blackboard, really tall, and you are here. If you want to go downhill, you go along the field, you run away from the hill, all the way to the valley, to the plains. If you want to go to the top, you go opposite to the field, and you go along this equipotential at every point, you do not do any extra work, so your potential does not change. So it's clear from this that the equipotential is always, in fact, in general, the change in any potential is gradient of the potential.dr, except for the minus sign. It's the same story that if dr is parallel to grad v or v parallel to e, you get the biggest change in voltage. All right. So this is what you should understand in general. What is good about the potential is you can take a whole bunch of charges, put them here, there, there, there, and you can draw little contours, maybe of constant v. It's just like the graph you have of the weather, of the temperature, for example, or if you have a topographic map of the mountain, there are some valleys and there are plains and there are some mountain tops, and these lines of constant height are like constant v. So it helps you visualize. And then if you're told you really something here, and it rolled down and came here, you can find the speed here, given the difference of these two numbers. You don't have to go through the intermediate step. So these are the advantages. Okay. So now I'm going to return to the conductors briefly to tell you something, one other thing that is new compared to what we did before. Remember what I told you about a conductor? A conductor is one in which the electrons do not belong to any one parent. They run around to the whole sample, but they cannot leave the metal, or the boundary of the metal that a force is bringing them back. Inside, they're free to move. Therefore, if you take a conductor and you stick it into a metal, initially an electric field will penetrate the conductor. The charges, because they're free to move, will move. Negative charges will move. One way positive charges will move. The other way, until they set up a back reaction to this, that blocks the field inside the metal, until the field inside is zero. So here's a generic conductor. If you put it in an external field, it will have some charges on its surface. It will produce a field this way that will oppose the external field in which you put it, until at the end there is no field inside the metal. Or if you did not put it in the field, it just took this material and dumped some charge on it, threw some charge on it, and asked, where does it go? Well, the charges are all plus, and they don't like each other. They want to move away from each other, and they move as far as they can until they come to the boundary, and they cannot go any further. That's where they stop. That we saw before. So we know all the charges in a metal are in the boundary. There's no electric field inside because of Gauss's law. If I want to find the charge here, I do a tiny sphere, do integral e.da, I get zero, and that's a charge inside. And I can move this little microscope all over the inside, I keep getting zero. That's why I know there's no charge inside. There's no charge inside, there's no field inside. Q equal to zero inside metal. This is a solid metal, and e equal to zero inside solid. The other variation you can have is, suppose I make a hole in this. There is no electric field anywhere here because it's a metal. The whole logic is, if you've got a field, you've got to find a static situation. Let the charges move. They'll keep moving till they have no reason to move. That reason will stop when the field vanishes. It's quite amazing. They can always find an arrangement in which they kill the external field. That's the perfect metal. So there's no field here. So if you do a Gaussian surface like this, the charge inside that is zero. But that does not preclude the possibility that there are maybe two plus charges there and some minus charges there. Last time I told you that if that happens, then the pluses and minuses will run around and meet and destroy each other and utilize. But that's a little subtle because maybe if they were the only two things in the world, I can understand that, but there's also these charges outside. Maybe they'll impede their motion. But here's one way to prove that there can be really no charge even on the inner surface. If there were some plus charges here and minus charges there, remember q is zero inside, so you can only have equal number, then electric field lines will have to leave the plus charge and terminate on the minus charge. So there'll be an electric field inside the cavity. And I'm saying that's not allowed because of the following. If there is field inside the cavity, let's take that path and another path, which are two ways to go from here to here. One path, I get line integral of e to be not 0. Second one, I get line integral to be 0 because there is no e. The line integral cannot vary with the path in an electrostatic situation. Therefore, there cannot be any electric field inside the cavity due to anything you put outside. You can have electric field inside the cavity if you go into the cavity and put by hand some extra charge there. Then what will happen is that guy will produce a field and the electric field lines will come to the boundary and terminate on some negative charges. Again, there'll be no field here. But a hollow cavity with no charges inside cannot have charges, no charges inside the hole, will not have charges on the edge either. So now what else can we say about conductors? Today I use the fact that the line integral of e is 0 to tell you that in a hollow cavity you cannot have any charges in the wall. The second thing is every conductor is in equipotential. It's in equipotential. Why is that? Can you tell me why that is true? If it is not equipotential, the whole lump. Yep? Student is asking, can there be an equipotential in a hollow cavity? And how do we know that? That's right. In other words, if you take the gradient, namely taking derivatives, you'll get a non-zero derivative because I'm telling you v is not constant. In some direction it's got to change. There's an electric field in that direction, at least. But that's not allowed. So v is a constant. So take any metal shaped like anything. You can associate a single potential to all of it. Okay? If you've got charge on it, that's the same. And what it means, if you start at infinity, the work doesn't come there, the work doesn't come there, the work doesn't come there, all the same because they're all measuring the potential. The whole metal is at one potential. That allows you to do certain tricks. So here's one trick you do. There are more tricks, but this is the only one we can discuss in this class. So here's an infinite conducting plane, and you've got a ground attached to the ground. And over this guy I put a charge Q. And I ask you, find the electric field everywhere now. That's your talents. Without this guy, fields are just radial. So radial, radial. But the fields come near this metal. They cannot penetrate. They've got to stop at the surface of the metal. So the field lines will do that. They will all terminate on the metal. And to terminate, you need some negative charge, which the metal will suck out of the ground. That's the reason for grounding it. So the question is, what kind of charge distribution do you get on the metal? How much charge is there on the metal? And because there's positive charge and negative charge here, there's the force of attraction between this charge and the plane. In general, that's a very difficult problem. If I give you a potato, aluminum potato, put a charge next to it, it will have a force, but it's not easy to calculate. But this infinite sheet turns out very easy to calculate. The reason is that we can solve this problem by borrowing from another problem. The problem I borrow from is the following. Take this plane, continue these lines, and think of the problem of a dipole with an equal negative charge here. So the lines from them go like this, right? And if you slice it in the perpendicular bisector of this distance between them, you will hit every line perpendicularly. You guys remember that part about the dipole field, it's somewhere there. You cut it right down the middle. So this plane was an equipotential at v equal to 0. If it's an equipotential, there is nothing there, it was just a mathematical surface on which v was constant and equal to 0. If v is constant, you can slide in a conductor there and nobody will know, because the field is 0 along the surface of the conductor, nothing happens. So you can reconcile the fact that conductors at equipotentials and solve problems in which a charge is in front of a conductor, if you're lucky enough to find a situation where that's an equipotential surface of the same shape as your conductor, then you can slip your conductor there. So in this example, what happens is that the electric field here, throughout the plane, is exactly what you would get due to a dipole minus and plus, you can easily calculate at all points on the plane. Once you've got the electric field, you remember, σ over ε0 is the electric field, so you can find σ, the charge density. And if you integrate the charge density, you will get exactly minus Q, because the number of lines coming from here have to terminate on the plane rather than on this, and that'll be minus Q. So finally, what's the force of attraction between the plane and this charge? Can you make a guess? What's the force with which the plane attracts the charge? I'm asking you to guess, so it cannot be too difficult. Same as the force that this negative charge would exert on this positive charge. And I'll tell you why. So this guy is looking around, okay, seeing some forces. It's due to these negative charges on the plane, but they are identical to the field created by this guy. So it cannot tell the difference, and therefore it will move, if you're releasing it, it'll move the same way in this problem as in the other problem. So it'll be drawn to the plate with an attractive force, and Q times minus Q divided by square of that distance. So the summary of what I did now is that if you solve any problem, let's say the dipole problem, and you get some shape, it's an equi-potential, you're allowed to then replace that with a chunk of aluminum and keep the field lines the way they are, because that aluminum will be an equi-potential and it'll be a sit-on equi-potential, because there are no forces on the surface of the conductor moving the charges. So the model of the story is you can sneak in a solid object bounded by the surface of an equi-potential for any problem. So if you can solve certain problems and the equi-potentials have nice shapes, you can put conductors of that shape in that precise location. That'll answer a different problem in which the second charge is absent. You only have the first charge. It's like a phantom charge, an image charge. It's like this charge looks at the mirror and sees the image. The image is minus of itself and it's drawn to it, because the field produced by this guy and this guy is the same as the field produced by all of these and this. All right, so now we've got a final problem of energetics. Let's see. The question is, I want to bring a whole bunch of charges, Q1, Q2, Q3, from infinity and bring them to this arrangement. So when they're infinitely far, they don't even know about each other. They don't feel any force. So here's my goal. I've got to do some work, bring everybody together. They all be positive. They may not want to be in the same region, but I'm going to force them. The question is, what work do I have to do? We're going to calculate that. We do that by saying, first let's take charge 1, let it be wherever it has to be. Then I bring charge 2 from infinity and put it here. The work done for that is by definition Q2 times the potential here. The potential there is Q1 over 4πε0 R12. R12 is the distance between 1 and 2. Now I nailed the two charges. Nobody's allowed to move, so I don't do further work. Then I bring a third charge from infinity to its location here. How much work should I do? The work that I have to do is the potential energy from the two charges I've calculated here due to these two charges. That means the work I'll have to do is equal to Q1Q3 over 4πε0 R13 plus Q2 Q3 divided by 4πε0 R23. You follow that? That's the work I have to do to bring the first guy, and Q2 had to fight only this one. The third guy has to fight these two, and the fourth guy will have to fight these three, and so on. I'm just going up to three charges. From this, it should not be too hard to guess. In general, if you've got many, many charges, you do Qi Qj over 4πε0 Rij, then Qi goes from 1 to N, where j goes from 1 to N, but i cannot be equal to j, and you divide by 2. If you're not able to read this, I'll tell you the summation is 1 over 2, i from 1 to N, j from 1 to N, but with a condition that i should not be equal to j times all these. You can see that's valid here. Yes? This is the, oh, I'm sorry. You're absolutely right. This is not the potential, it's the potential energy, it should be called U. Did I say, yeah, that's right. Because Q over 4πε0 is the potential, but when I multiply by Q2, that is actually an energy. These are all energies I denote by U. Now if you have a continuous set of charges, you must replace a sum by the integral. Do you see what's going on here? Why there's a half? Because this one counts Q1 Q3, then again it counts Q3 Q1, but there's only one such thing, so I divide by 2, and I have nothing like Q1 Q1, so you don't allow that. That's the energy to assemble a whole bunch of charges together. So that's the work you do. If you do that work and you take your hands off, you can fly apart and give you back that energy in the form of kinetic energy. Now let me do another simple example. I want to take a sphere of radius r, and I want to put Q coulombs on it, and I want to know how much work I have to do for that one. What work do I have to do to charge this sphere with Q coulombs? Well, first a couple of coulombs come in, they don't run into anything. I want to put some charge here. That makes it harder to bring the next set of coulombs, the next set of coulombs. At some intermediate stage, when the charge on this sphere is Q, and I want to bring in a charge dQ, how much work do I have to do? When the charge is Q, the potential of the surface of the sphere is Q over 4π ε0 over r. The whole sphere is at that potential, and I'm trying to bring in a tiny more dQ from all sides and smear it on this. The work for that will be that times dQ. Then the total work I do, when I go all the way from 0 to Q, will be Q squared over 2 times 1 over 4π ε0 r. Sorry, R. Again, you see why you get the half, because the last coulomb is much harder to bring than the first one. The first one is free. It's like a spring. When you pull it, the first inches are easier, the other ones are harder, because the force is Kx. The work done is integral of x that becomes x squared over 2. Similarly, to bring in an extra charge, you've got to worry about how much charge is already there on the sphere. That's opposing you with this potential. You work against that by bringing dQ from infinity, and when you integrate it, you get Q squared over 2. So this is a good problem to explain the difference between potential and energy. This is the potential. This is the charge I brought from infinity, and that together is du, and integral of du is my total u. So one can ask, what's the energy to produce any changement of charges, and you should be able to either do the sum or do the integral. One homework problem will be how to assemble a solid ball of charge, not a sphere solid ball. It's got no radius, and slowly it grows up to some size, and you can do similar integrals where you slap on more and more spherical shells on this onion to build it up. Again, if you knew the potential at any stage, you could find the work to add a thin layer to it and integrate. Okay, so final topic is one of capacitors. So capacitor is like a way of storing charge and storing energy. In other words, if I want to store gravitational energy, I can do the following. I build this tank, and maybe there's some water in the tank, and I make a bake here at the foot of the tank, and I go to the top and go to the top and empty little buckets until I got some water at a big height. Then I've done some work, and it'll pay me back when I open the tap and let it all rush down. Then it may turn a turbine blade and give me back a thermoelectric power. You want to do similar things electrically. You want to invest with some work, and it'll give you back. What I can do is I can take a plus and minus charge, which love to be next to each other and separate them. Now I've done some work, but I cannot walk around like this. I've got other things to do. I've got to find a better way to store energy. How do you shake hands with people? So you have to find another way. So here's what you do. You take one conductor and you take another conductor, any shape you like. It doesn't matter. They're both electrically neutral. They're both electrically neutral. So you take a conductor, so you take a tiny charge from this guy and put it there. Then take more and more charge and start putting it there. At any stage, when you stop, you will have a lot of positive charges on this conductor and a lot of negative charges on this conductor. This conductor will be at one potential because it's a conductor. That conductor will be at another potential because it's a conductor. But both have unique potentials. They're not the same, but this whole conductor has one potential. This is another potential v'. And because it's harder and harder to pump the coulombs, you can imagine that there is a potential difference between them, because you've got to go uphill to pump even more charge into this already charged object. So we define the capacitance as the charge that you transferred divided by the voltage difference you got. So let me call this zero by convention. Let's call it the capacitance. That's the ability of that system to hold charge. If there are very, very tiny metal spheres, if someone want to work, if there are huge metal spheres, you can transfer a lot of charge. Okay? That's the definition. It's measured in Farad's. Again, it's got some units, coulomb for volt, but again it's got a different name, named after Mr. Faraday, measured in Farad's. So usually you'll have micro-Farad's in a lab. Farad's a huge unit. So let's find the capacitance of certain simple arrangements. Simplest arrangement is the parallel plate capacitor. Here it is. One plate is a second plate. The area of the plate is A. The distance between them is D. And I put plus charges on this one, and I got them from this one. Do you understand that? You don't have to bring your own charge. You remove charge from one conductor, you therefore leave it negatively charged, and you dump it on the other conductor, thereby leaving it positively charged. So you don't need to bring charge from the outside world. You just redistribute the neutrality so that one is now negative and one is positive. So q stands for the magnitude of charge that you transfer. So suppose I transfer some charge q. The only job is to find q divided by v, and that's my capacitance. Okay, so how am I going to find the voltage difference between these two? I think you guys should know by now voltage difference is the electric field times the distance between the plates. There's no line integral. E dot dr is just E times D. E, if you remember, is sigma over epsilon 0 between the plates. So sigma is q over A. Charge density is a total charge over that. That is by definition q over C. Therefore, the capacitance of this parallel plate capacitor is epsilon 0A over D. That's one of the easier calculations to do. So anytime someone gives you two objects and says, find for me the capacitance, you will take charge from one, you will put it on the other one, and then you will, and you will get the same voltage. Where is all these black holes? Bad choice. So you guys took all this down? Yes? Oh, yes. Sigma is q over A. This is what comes from knowing the answer. This is what you guys do in the exam. You turn that page and all kinds of stuff happens. Signs change, numbers change, and the right answer appears on the next page. That's really what I did, because I knew this formula. You're getting there no matter what, but you're quite right. I didn't put the A in the right place. So in a way, it's not good to know the answer every time, because you're not that alert doing this. Anyway, this is the correct answer. So here's the capacitance of this parallel plate. So another problem you could do. These are also common. Spherical capacitors are one sphere of radius A surrounded by a second sphere of radius B, and your shove charges may be plus on this one and equal number minus on that one. So what's my goal? My goal is to find the electric field in the region between them and integrate it from here to here to give the potential difference and divide by the charge. So if I put a charge Q here, you can see that the electric field in this region is just Q over 4.0r2, because by Gauss's law, if I draw a sphere here, that encloses charge Q, and it's isotropic and uniform, and it's like that of a point charge. Therefore, the potential difference between these two points will be like the potential difference between these two points with a point charge of the origin. So V2 minus V1 will be Q over 4p epsilon 0 times 1 over A minus 1 over B. So this is what we have done, because this is what I call V. So capacitance is Q over B, and that's going to be 4p epsilon 0 divided by 1 over A minus 1 over B, which is 4p epsilon 0AB divided by B minus A. So in every geometry, if they want capacitance, move charge from one guy to the other guy and find the field between them, then integrate it to find the potential, then the potential will always be proportional to the charge, take the ratio, and that's capacitance. I'm going to put this guy to a test. Let me write down clearly the answer here. 4p epsilon 0AB divided by B minus A. What test can I do if I want to compare it to a parallel plate capacitor? Can you see how I can relate it to a parallel plate capacitor? Yep? Student is asking, make the A. A what? Student is asking, make the B minus A. Right. You can take B minus A very close, but it still may not look like a parallel plate capacitor. So here's what I say we do. We take B and A astronomical, size of the galaxy, okay, and B minus A is just 1 inch. Then in that case, it will really look like two infinite planes. The fact that it gets wrapped around over the length of the galaxies, just like saying the Earth is round, which is not obvious from short excursions. In that case, we can write 4p epsilon 0, A and B are really equal to some common number r, and B minus A is the distance between the inner shell and the outer shell. That you can see. Wait. I'm sorry, 4pi r squared is the area times epsilon 0 over d. Right? So that's one test. It doesn't mean it's right, but if this test failed, it would definitely be wrong. Okay. Now, how much energy does it take to charge up a capacitor? It's what I did earlier, but let me do it again. If you take two plates or any two surfaces, and there's a certain amount of charge q on it, and I transfer delta q, the work I do is q, the first a q times the voltage at that stage, voltage at that stage is equal to q over C, which is the q you have at that point. Therefore, the total work you do is 1 over 2 q square over 2, q square over C. So, when you want to charge a capacitor with a charge q on it, either you can write it that way, or you can also do the following. You can write as 1 over 2 C v square, because q over C is a v multiplied top and bottom by C. You can write the energy in a capacitor, in terms of the charge stored, or in terms of the voltage between the plates. We'll resume this next time.
 Okay, let's turn to the subject proper.  I will start by doing the usual stuff of just refreshing your  memory on what happened last time.  I mean, last time was pretty heavy, a lot of mathematical  machinery which I like to work with.  Now, you might say, you know, I'm going to be a  psychologist or art historian and I don't need all those  details.  Just tell me what I need to get through this course.  So, that's a very reasonable question.  Probably would have the same attitude if I took a course in  your field.  So, here it is.  What I've told you so far is that the electric field,  which we understand as basically Coulomb's forces,  allows you to define a potential because it's a  conservative force.  That means the integral of the electric field on any closed  loop is zero, and remember that's what it takes to  define a potential.  And the electric force, which is Q times E,  will also satisfy the condition.  That means we can define for this problem a potential v,  which is called the potential, so that 1⁄2mv1  squared.  This is a little v.  This is big v.  Capital V at the point r1 is equal to 1⁄2mv2  squared plus capital V at the point r2.  That is our goal.  Ah, missed one mistake already.  Did you know what's missing here?  Pardon me, yes?  What's missing?  Yes, the charge.  So, this should really be Q times v,  and this should be Q times v.  So, let me remind you why it is.  The force on any charge is really not just the electric  field.  It's Q times the electric field, and therefore the  potential difference is associated with the work done  by the electric field times Q.  So, you've got to remember Q times v is the real energy.  So, v is like an electrical height.  For example, if this is the ground,  there's a mountain like that, you give me the height.  The potential energy is not just the height,  it's mg times h.  So, let me say you give me g times h.  That's like the potential.  That tells you the whole story, because if you want to lift  a kilogram mass to that height, you do 1 times g  h, 5 kilograms, 5 times g h.  So, depending on the mass you want to lift,  the work done will be different,  or the potential energy will be different.  But this doesn't depend on the mass you're planning to carry  around.  So, you take out the mass, you can always put it back.  Similarly, the potential concerns itself with the unit  charge.  If Q were 1, nothing I wrote is wrong.  But in general, Q is not 1.  So, this is the actual energy.  This is the old potential energy, U1.  So, the potential energy is Q times the potential.  So, think of V as the electrical height at a point,  and Q is like the mass, and Q times V is like mg h.  Okay, we got a formula for V.  The general formula for V is that V of R2  minus V of R1 is the line integral of E dot dr  from R1 to R2.  This minus sign has the following meaning.  The electric field exerts the force E on the unit charge,  but if you want to drag the charge against the field from R1  to R2, you apply force minus E to compensate that.  So, this really worked done by you, this really worked done by  you, to move it from R1 to R2.  Let me give you the gravity example.  Gravitational force acts this way.  But if you want to lift something from here to here,  you oppose it, namely it's doing minus mg,  your force mg, you move it to height h.  That's the work you do, and that you know is the  potential at the top minus potential at the bottom.  So, that's the origin of the minus sign.  The electric field wants to go one way,  and if you want to take it against that force and raise it  electrically speaking, the force you apply is  precisely minus E.  That dot dr is the work done by you.  That's what you invest.  That's what you will get back if it rolls downhill again.  Now, if you want V at some point R,  minus V at infinity, that integral will be minus,  starting from infinity to the point,  and then we all vector to the point R of E.dr.  We will now take the convention that V of infinity is equal to  0.  Remember, potentials are defined only by the  differences.  So, you can put the 0 anywhere you like.  We will choose the 0 to be at infinity.  Then in this integral, it's going to be 0,  and if you do this integral from infinity to R,  you will find for one charge, it is q over 4  pi epsilon 0 times R for single charge,  and R is the distance from that charge.  It looks like Coulomb's Law, but a couple of differences.  There's no 1 over R squared and there is no vector.  The potential due to a charge here at the point here is a  number given by this.  So, if you look at lots of charges, let's just find V  due to many charges is sum over all those charges,  the value of each charge before pi epsilon 0 times the length  from where that charge is to where you want the potential.  That's the principle of superposition.  The Coulomb potential due to one charge is 1 over R  due to many charges is each q times its own 1 over R times  its own 1 over 4 pi epsilon.  And finally, if you want the potential difference due to the  field, you integrate it, and if you want the field  from the potential, you take the derivative called  the gradient, which is identical to  minus i x partial derivative minus j y partial derivative.  So, you expect two derivatives because the force you're  looking for now is a vector.  You're living in 2D.  So, you cannot say the force is the derivative of the  potential in what direction?  If you want the force in the x direction,  take the x derivative.  If you want in the y direction, take the y derivative.  And you put them together into a single vector,  and that's the electric field.  So, you go from potential to the field by taking the  derivative of the gradient, and you go from the field to the  potential by doing the integral.  So, the subtleties, the integral is of a dot  product of E with the displacement because everything  is a vector now.  It used to be fx dx in the old days,  but f has become a vector, dx has become a vector dr,  so you need the dot product.  Okay, so this is a summary, but I want to make two course  corrections from last time.  First one was, when I tried to prove to you  that the electric field is conservative,  I did one thing which I realized was a swindle.  I didn't realize it myself, and I wasn't waiting for anyone  to correct me.  I did not think about it hard enough.  So, here is the mistake I made.  Remember what I said?  I said that's a potential due to point charge.  I'm trying to look at it.  These are different lines of the electric field.  Let's say you want to go from here to here.  You want to do some work, E dot dr along this line.  The claim was that's the same work you do if I take another  path that looks like this.  Go this way, go that way, and go that way.  Same work, and that's still correct statement.  The same work because here dr is angular,  E is radial, the dot product is 0.  Likewise, 0.  Here at every segment you have here, E is radial,  dr is radial, the dot product is the field E times  that length.  Here is the field, here times that length,  the field is constant at a given radius,  so they're also equal.  Still, no wrong statements.  Then I said, what if I do this?  A little radial, a little angular,  and so on, and that's also okay.  Then I finally said, what if I had some nice  smooth path which is not made up of radial and angular,  and my claim was, well, you take for this segment,  for example, you take a radial segment,  then you take an angular segment, then you may take another  radial segment, another angular segment.  So you approximate the smooth path by a bunch of radial  angular, radial angular, and from a long distance it all  goes the same, so it should be the same amount  of work.  That's wrong.  I'll give you an example of why such arguments can be wrong.  Let's take a triangle.  This side is 1 and this side is 1.  I want to walk from here to there.  So you know I go a distance square root of 2.  But then you say, I'm going to take a different  path, I'm going to go horizontal, vertical,  horizontal, vertical, equal amounts like that.  So these are very tiny differences.  You can say it's the same thing.  But you know this path is of length 2.  Understand why?  All the horizontal parts add up to 1,  all the vertical parts add up to 1.  So going zigzag is longer by a factor root 2.  So just because things are approximately the same,  doesn't mean the answer will be the same,  because if you make an infinite number of infinitesimal  mistakes, you can add up to something finite.  Likewise here, you've got to make sure that  you can deform your path from what was really given to you to  what you did, which is radial and angular.  The work done should be the same.  The path length is clearly not the same in this example,  and it can all add up, but the work will be the same.  That's what I did not show to you.  So let me take a segment here where that's the radial part and  an angular part, but the person really wanted  to take a segment along some straight line joining the two.  What I want to show you is that not only does that approximate  this from long distance, the work done on that plus  that is the work done here.  Okay?  So for your convenience, I'm going to blow up that picture  here so you can all see what I'm trying to do.  This should be the radial direction.  This is the radial direction.  This is E.  This is some perpendicular direction,  and this is the direction in which the move dr was made.  That's the actual path which is going like that,  and you manage to just deform it.  So what's the line integral on the radial path?  Well, it's the magnitude of the electric field at that  distance times the difference in radius between this end and  this end.  Let me call it delta r.  Delta r is the spacing here between that curve and that curve.  You agree?  The dot product is trivial.  E is this way.  Displacement is this way.  No cos theta is to worry about.  That's the work done.  Work done here is zero because perpendicular to the field.  So that's the total work done going like that.  How about going along this line?  It's going to be E at r dot dr.  When I say E at r, r is of course not one point.  It's going to spread out, but this is an infinitesimal  region, so you can talk about the E at one point and not  worry about the variation.  So don't forget it's an infinitesimal triangle.  E dot dr will be E at that distance r,  the length of dr, and the cosine of the angle  between them, which is this angle.  But length of dr times cosine is precisely that distance.  Delta r, and therefore you do the same amount of work.  So basically I'm saying as you wander around,  you may not choose to go radial angular,  radial angular.  But if you go at a slant to the radial,  the work you do really depends on how much radial distance you  cover.  That's what E dot dr does for you.  That's why you can deform the path anyway and get the same  answer.  That's the first thing.  The second thing is I forgot to tell you anything about units.  The electric field is forced on a unit charge,  so the unit for that is Newton's per coulomb.  The electric potential looks like the work done on a unit  charge, so the unit for that is Joules per coulomb.  And the name for Joules per coulomb is a volt.  It means if there's a voltage difference between two points  of the two poles, it's going to take you 12 Joules of work to  lug a coulomb from the lower point to the higher point.  So you guys have to get your units right, okay?  Now you might say, why don't you use units?  Right?  You know why?  Because I have tenure.  When you have tenure, you don't have to use units.  I have tenure, I have a car next to the fire hydrant,  I got tenure, I tear up the tickets,  I don't do duty, duty, I don't pay taxes.  You don't have to do anything when you reach.  You know, Allen was going to tell you the joys of being a  graduate student.  The joys of being tenure prof, hard to describe, okay?  Priceless.  In fact, one of my sons, when he saw my lifestyle,  said it reminded him of a jellyfish deep in the ocean.  He saw a rock, he attaches himself to a rock,  and then never moves and eats his own brain for food,  because he has no further use for it.  So apparently that's what I reminded him of.  So you can, if you work hard, you can be like me.  You can do that.  Anyway, look, so you guys have to get the units because  seriously speaking, if you don't write any units,  we don't know if you write the rock, right?  The problem answered is 19, in some units, right?  So if you don't give the unit, the number is not worth  anything.  Another advantage of units is you can keep track of your  calculation.  You multiply one quantity by another quantity.  Everything has units.  You cross out the lengths and the meters.  You've got to make sure the stuff you get has the right units  on the two sides.  That helps you also keep track of calculations.  Okay, so let's go back now.  This is all stuff left over from before.  So now for the new stuff.  I'm going to tell you the first part of this lecture is going  to be what are the advantages of V?  I will just mention them, then I will elaborate on them.  The first advantage is the conservation of energy.  The second advantage is the computation of E.  The third advantage is a lot of good visual pictures,  which are very helpful.  So you won't know what these mean, but this is what we're  going to do, so you know where we're going.  First thing is conservation of energy, there's no need to  belabor at this point.  I mean, that's why we did the whole thing.  The whole conservative forces, gradient, this,  that, all that was to make sure in the end you can get a  potential energy out of it.  And as you know, from the roller coaster problem,  you can do kinetic plus potential equals kinetic plus  potential, it saves you a lot of trouble.  In other words, in this room right now there's a lot of  electric fields.  A lot of charges everywhere applying a field here,  and I want to carry a charge from here, and I want to drag it  over here.  Or maybe the charge was shot by a cyclotron or something came  flying out at some speed here, and it comes here a little  later, and I want to know how fast it's moving.  One way to do that is to follow the charge as it moves at  the same time.  You can, for instance, find the force on it,  the acceleration from that, change in velocity, add up all  the changes, you'll get the final velocity.  But as you know, with the law of conservation of energy,  you can skip all the intermediate stuff.  You just have to know the initial kinetic energy,  initial potential energy, final kinetic energy,  final potential energy.  They are equal if you sum them.  So you can find one number from it, one of the numbers you can  find.  In particular, if you didn't know the final kinetic  energy, you can find it.  That's the energy.  The second thing is that it makes life easy when you want to  compute the electric field.  So this is an example I did again near the end of the class,  and I think it was needlessly complicated.  So I will try to repeat the derivation a little more  quickly, because it's a lot easier than I gave the impression.  So we are trying to find, let us say, the electric field  is a simple minus q and plus q separated by a distance 2a.  If you did it with forces, with forces I didn't even try  to do it everywhere.  I remember I did it here and I did it there.  The force here was like this, the force there was like that.  Then I said if you do a lot of work, you'll get all these  lines.  What makes life difficult is that if you go to this point  here, this I explained to you last time,  this guy will repel it with some force,  this guy will attract it, and you've got to add the two  arrows and get a new arrow.  That's the electric field at that point.  So you've got to add arrows, and that's more difficult than  adding the potential, which is just a number from  this guy and that guy, then once you've got the  potential, you find the field by taking derivatives.  You want to take the derivative of ex,  take the derivative of the minus sign,  you want ey, you take minus dv dy.  So what's the e here?  You remember there's q over 4πε0 times 1 over r+,  r+, at that length, minus 1 over r-,  that's it.  No arrow is nothing.  That's the total v.  This is r-, let's define r to be  the distance from the center.  And I want you to imagine this is long way off.  This picture is not drawn to scale.  You should imagine these two points are like this and I'm  somewhere there.  That's when the dipole approximation is working.  This is not an approximation, this is exact.  So let's do the following then.  We combine the denominators, you get r+, times r-,  and the denominator is r-, minus r+.  Now you can see that r+, and r-, differ from r because of  this little guy, a.  If you use the law of cosines, you know,  it's that squared plus that squared minus 2a r cosine theta  or something.  In the bottom, we ignore that difference and we  write everything as 4πε0r².  Because r+, I approximate with r, r-, I approximate with r.  So this is not the exact answer.  There are corrections here that look like 1+, maybe a number  a divided by r and also a² divided by r².  But these are all negligible.  In the numerator, you can again say,  why don't you replace this by r and this by r,  then you get 0.  It is true.  If you replace everything by r, then you do get 0,  and that's because in the difference the big guy cancels  and what's left over.  And the difference is what's going to keep this from  vanishing.  So we have to find the difference in that distance and  that distance.  So for that what you do, you draw a line here.  Now you cannot draw a real perpendicular here,  but if these things are very far and these lines are almost  parallel, you can draw a perpendicular there.  And if that angle is theta, theta is the angle,  then it's very easy to show that this distance is simply  2a cos theta.  Because if theta is this angle, then you can say cosine of  this angle here will be, by whatever complementary angle,  is 2a cos theta.  So if I draw a line here, I have 2a times cosine theta,  and that I will write as dipole moment P,  cosine theta over 4πε0 r².  That's what I was trying to get.  Yep?  Student is asked which angle is theta?  Theta is the angle measured from here, let me draw this.  You can, one can worry about, is it this angle or that angle?  But I'm saying draw the angle from the middle, okay?  That angle is theta.  And roughly speaking, that angle is also theta.  Now drop a perpendicular here, and this side is 2a,  and this side is 2a, and this side is 2a.  So this difference is 2a cos theta.  So now I have v in terms of r and theta,  but I can write it in terms of x and y very easily.  It is P over 4πε0 r², and cosine theta is x over r.  So the whole thing is x over r cubed,  and it's the same as x over x² plus y² to the 3 over 2.  Right?  r is the square root and r cubed is three powers of that.  Now you can easily take the derivative of this guy with  respect to x, or with respect to y.  That's very, derivative is very easy to take.  Integrals are hard to do, and that'll give you the electric  energy, and you'll get the ex and ey everywhere.  So ex is equal to minus dv dx, and ey is minus dv dy.  So one of the homework problems is to really calculate this.  And if you calculate it and you plot it, you will see this  thing emerging.  Okay.  In other problems, most of the time it's easy to find  b and then to find e.  For example, if I have a ring of charge,  and I want to find the b here, then let's say the ring has  charge q and some radius a.  Then a little guy here produces a little potential dv,  which is the charge dq.  dq is not any distance, it's like the differential,  divided by 4πε0.  And what's the distance?  If that is z, then it's simply root of z²  plus a².  So that's the potential of the x²,  due to this little charge.  But every unit of charge on this ring is equidistant from  this point.  You understand it's a ring and I'm perpendicular to the ring.  So they all contribute the same number and some of all the  dq is just the q.  So that's the potential on the axis.  Now you can take a z derivative to get the field.  Okay?  Generally, it is easier to do the v.  Either it's a sum or integral over the charges producing the v,  then take derivatives.  But I want to give you one example in which the opposite  is true.  But in fact, it's easier to get e than to get v.  And that is, there are many such examples,  I'm just giving you one.  So I say that's a hollow conducting sphere.  It's got charge q.  This guy's hollow and conducting.  I want to find the potential everywhere due to this.  And the charge q is uniformly spread on the surface.  So what's going to be the potential?  First of all, by symmetry, you can see that if you pick any  point, anywhere, the answer is going to depend  on the radial distance from the center,  because of spherical symmetry.  But you still have a lot of hard work to do,  because you can divide the sphere into little rings like  this, slice it at different latitudes,  and find the potential due to each ring using this formula I  just did somewhere, then integrate over all the  rings, and you will get an answer.  But that's a decent amount of calculus.  But you don't have to do that, because in this problem,  you can't do that.  Because it's vertically symmetric, and there's a charge  q here, we know from Gauss's law the electric field is going to  be purely radial, and will look like q over 4  point epsilon 0 r square, as long as r is outside the sphere.  So here's one problem, but it's easier to get E,  because E is pretty trivial.  You just need to know the magnitude.  Gauss's law tells you the magnitude.  You put the unit vector back and you get this.  So as long as you're outside the sphere,  it's as if q were sitting at the origin,  and the potential V of r, which is minus q  plus q over 4 pi epsilon 0 r, outside that sphere.  So that's because remember V of r minus V of infinity is  equal to integral of E dot dr from infinity to r,  the minus sign.  And if you just take the field of a point charge and do the  integral, you will get the potential of a point  charge, which is this.  So if I draw the potential as a function of radius,  it will look exactly like that of a point charge,  and I'm going to draw the potential of radius over r.  Now what about for r less than r?  How do I continue this graph?  Pardon me?  You can use Gauss's law.  If you use Gauss's law, inside what will you get for E?  E is 0.  Integral of E dot 0 is dr is 0, so potential does not change  inside the sphere.  But don't think potential is 0.  If it does not change, it's stuck at this value.  So throughout the hollow sphere, there's only one potential.  Electric field is a different story.  Electric field is in fact 0 inside, then falls like 1 over r  squared.  Electric potential is a constant.  So these are, there are certain problems like a  cylinder of charge surrounded by another cylinder of charge,  where again it's easier to get E from symmetry arguments than  E to get the potential.  Okay, so this is my second point, namely,  barring a few exceptions like this one with high symmetry,  the potential as the advantage that it lets us compute the  electric field by first doing a sum of scalar quantities rather  than vectors, then taking derivatives.  Now third advantage has to do with being able to visualize  the potential.  It gives you a landscape, and that's intuitively very  helpful.  So here's one example.  Let's take two parallel plates, put a lot of plus charge  here, and equal amount of minus charge here.  What will the electric field look like in this region?  Let's say both have the same density sigma for unit area and  that infinite.  I'm just showing you a finite portion.  So this guy will produce sigma over 2ε0 here and here.  This one, because it's negative charge, will do this,  sigma over 2ε0 on this side.  In the region between them, you can see the two arrows are  additive.  So the field will look like this.  In the region outside, the two arrows are opposite.  This guy is coming and that guy is going out.  So the field doesn't diminish with distance due to either plate  so they can completely cancel each other.  So the electric field is trapped between the two plates.  And as the strength, which is double what you would have  on either side if you had only one sheet,  that's the electric field E.  That's my electric field.  We can ask what's the potential difference?  Or what's the potential at various points?  This whole thing being metallic is at one potential.  I say v is equal to 0.  Throughout this line, let's say v is 0.  In fact, it's very clear v is 0 because if I move horizontally  on this line, I don't do any work because the electric field  is perpendicular to me.  So if v was 0 at one point, it is 0 throughout this plane.  Yes, I should qualify the statement.  You cannot pick v equal to 0 over an extended region.  You have freedom to pick it to be 0 at one point.  I'm telling you, if you pick the one point in this plane,  you can navigate throughout the plane without changing v,  because you move perpendicular to E.  So v is 0 on the lower plate, but it won't be 0 anywhere  else, but you can see that may be v equal to 1,  that may be v equal to 2, v equal to 3,  upper plate may be v equal to 4 volts.  So lines of E are vertical and lines of constant v are  vertical.  This is really like the gravitation problem.  An electric field raining down on you is just like the force  of gravity.  This might as well be gh.  This might as well be g.  You multiply a mass, you get mg.  You multiply by the q, you get the force on a charge.  It's straight down and the potential just increases with  height.  Yep?  Student is asking how do you do the cutters to get the constant  voltage?  Oh, okay.  Think of it as a plate coming outside the blackboard.  Student is asking for the hollow sphere.  Oh, the hollow sphere.  The potential is always the work done to go from infinity to  wherever you are, right?  So when I come from here, what do I think is at the  origin?  I think there's a point charge at the origin.  So I push up against that.  I come here.  So I do that amount of work.  Once I'm in here, there is no electric field inside.  So e.dr is 0 from then on.  But don't forget the e, the v that you got up to that point.  You don't drop it.  You add, keep on adding 0 to it.  So it is 0 everywhere.  You add 0 to all points here because you can go from here to  here, no electric field.  So the whole sphere has the same potential as the surface.  So now, here I got these potentials, and it gives you a  picture that you're running uphill, and these are the steps  you've got to climb to go uphill.  And if you've got four volts here, if you take a cool arm and  you let it drop, it will pick up four joules by the  time it hits the slower plate.  That's the meaning.  So the work done by the electric field, just like gravity,  is the potential difference, which is the work done on unit  charge times the charge, which is one cool arm here.  If it is ten cool arms, you do 40 joules of work.  If an electron is released, you've got to be a little  careful.  By definition, the charge of an electron is negative.  So if an electron is released here, in fact, it will go like  this.  You don't have any analogs in gravity.  Gravity, everything falls down.  So what was something said negative mass and something said  positive mass?  All guys with negative mass, you've got to nail them to the  ground, otherwise they'll just float up.  That happens in electricity, so any electron you leave here  will go up.  So the force on the electron is minus q times e, so everything's  got a minus sign.  So an electron going from here to here will in fact lose  energy.  It will lose potential energy, and it will turn into kinetic  energy.  So if an electron falls through 1 volt, the work done is  charge of an electron times 1 volt, which is 1.6 times 10 to  the minus 19 joules, that happens to be called an  electron volt, just a convenient thing to use.  When the charges you're pushing around are not cool arms,  but this tiny electron is good to know in electron volts.  For example, in the hydrogen atom,  the cool arm potential is negative because the charge at  the origin is a proton plus q.  Electron has got some minus q divided by 4πε  0r.  That's the potential.  And in quantum mechanics, only at certain radius you can  have electrons.  So the lowest electron is somewhere here,  and they say it is minus whatever some number of  electron, 13 points something electron volts.  That means if you want to drag an electron from this hole and  liberate it from the atom, you've got to do 13 times 1.6  times 10 to the minus 19 joules of work.  So as if this were 13 volts down compared to infinity.  It's a hole in the ground, 13 points,  and it's got a 4 or 6 volts deep.  So if you want to liberate that electron,  you have to furnace that amount of energy.  Okay, now going on with visualization,  let me say let's take another case we can all do,  which is a single point charge.  Here's a point charge.  The electric field due to that looks like this,  and it is million times.  Now what about the electric potential?  For a point charge, the potential is 1 over r.  That means it's a constant on a circle of radius r.  That means if you pick any circle or a sphere, I'm sorry,  v is one number here, v is a different number there,  a different number there.  So that's the whole surface surrounding the charge on which  v is some fixed number.  It depends on that radius.  You go further out, it's a different number depending on  that radius, and it falls like 1 over r.  You notice once again that the lines of constant potential are  perpendicular to the lines of the field.  That was true there, that is true here.  In fact, let me do one more example.  If you take a dipole, negative and positive,  and the field lines go like this,  if you compute the potential, if you go very,  very close to this guy, I think you all seem to  be in the same position.  If you're very near one charge, forget about the rest of the  world.  It dominates everything.  Potentially, it will look like circles centered here,  but as you go further out, the shape will change and it will  look like this.  A similar thing on the other side, then there will be this  huge infinite sphere that becomes planar,  that goes all the way to infinity.  These are contours of constant v.  If you really plot them, you'll again find contours of  constant v.  If you plot them, you'll find that they're  perpendicular to the electric field lines,  perpendicular to e.  So I want to explain to you that this is a very general  result.  This is not something true in these isolated cases of nice  symmetry.  Why is that?  So let's understand why that is true.  The v2 minus v1 is minus integral of e.dr from 1 to 2.  So let's say v2 and v1 are 2 and 1 are very close.  In that case, let's call it delta v is equal to  minus e.dr.  There's no integral to do because I've picked two points  which are nearby an infinitesimal amount dr,  and e.dr is equal to delta v,  which is minus of v2 minus v1.  What does that mean?  That means if you start at some point and you move a distance  dr, you take a step dr and you find the change in potential,  the answer depends not only on the length of your step,  but on which direction you move.  Because of this dot product, the change in potential will be  the magnitude of the electric field at that point,  the magnitude of the step length, and the cosine of the angle  between them.  So if electric field, if the electric potential is  replaced by height in two dimensions,  you know, top of sleeping giants and the hill is going up and  down, up and down, you're at some point.  There's a height function h of x and y instead of v of x and y,  and I want to take one step and I can ask what are the  consequences to my height?  The answer is if the electric field is pointing in some  direction and you take a step dr in a generally different  direction, this is the change you get.  So you can ask, how do I get the biggest drop in height,  or the biggest drop in potential, when cosine theta is 1?  Right?  Because then it's this, that times 1 with a minus sign.  So the electric field points in the direction of the greatest  rate of change of potential.  It drops that direction.  If you release a marble in a gravitational field,  you'll roll along the gravitational field,  which is obtained the same way from the gravitational  potential.  If you want to climb up the mountain,  suppose there's a tsunami coming,  and you want to get to the top as quickly as possible,  don't panic, compute the gradient,  and move along the gradient.  Okay?  Then stop, think again,  recompute the gradient, and in that manner you will get to  the top in the quickest time if you can compute quickly enough.  Okay?  But that's the process.  At every point, that's the vector pointing in the  direction of greatest increase, the electric field happens to  be the direction of greatest decrease because of this minus  sign in our convention.  Now, let us say you are in this mountain,  and you don't want to go down, you don't want to go up,  but you still want to take a walk.  Can you do that?  Yes?  Or are you stuck at one place?  What do you have to do, guys?  Yes?  You move perpendicular to the field.  To move perpendicular to the field, the cos theta is 0.  That's why, now going back to this,  this is called Mount Kula.  It's coming out of the blackboard, really tall,  and you are here.  If you want to go downhill, you go along the field,  you run away from the hill, all the way to the valley,  to the plains.  If you want to go to the top, you go opposite to the field,  and you go along this equipotential at every point,  you do not do any extra work, so your potential does not  change.  So it's clear from this that the equipotential is always,  in fact, in general, the change in any potential  is gradient of the potential.dr,  except for the minus sign.  It's the same story that if dr is parallel to grad v or  v parallel to e, you get the biggest change in voltage.  All right.  So this is what you should understand in general.  What is good about the potential is you can take a whole bunch  of charges, put them here, there, there,  there, and you can draw little contours,  maybe of constant v.  It's just like the graph you have of the weather,  of the temperature, for example,  or if you have a topographic map of the mountain,  there are some valleys and there are plains and there are some  mountain tops, and these lines of constant  height are like constant v.  So it helps you visualize.  And then if you're told you really something here,  and it rolled down and came here, you can find the speed here,  given the difference of these two numbers.  You don't have to go through the intermediate step.  So these are the advantages.  Okay.  So now I'm going to return to the conductors briefly to tell you  something, one other thing that is new compared to what we did  before.  Remember what I told you about a conductor?  A conductor is one in which the electrons do not belong to  any one parent.  They run around to the whole sample,  but they cannot leave the metal,  or the boundary of the metal that a force is bringing them back.  Inside, they're free to move.  Therefore, if you take a conductor and you stick it into a  metal, initially an electric field will penetrate the  conductor.  The charges, because they're free to move,  will move.  Negative charges will move.  One way positive charges will move.  The other way, until they set up a back reaction to this,  that blocks the field inside the metal,  until the field inside is zero.  So here's a generic conductor.  If you put it in an external field, it will have some charges  on its surface.  It will produce a field this way that will oppose the  external field in which you put it,  until at the end there is no field inside the metal.  Or if you did not put it in the field,  it just took this material and dumped some charge on it,  threw some charge on it, and asked, where does it go?  Well, the charges are all plus, and they don't like each other.  They want to move away from each other,  and they move as far as they can until they come to the  boundary, and they cannot go any further.  That's where they stop.  That we saw before.  So we know all the charges in a metal are in the boundary.  There's no electric field inside because of Gauss's law.  If I want to find the charge here, I do a tiny sphere,  do integral e.da, I get zero, and that's a charge inside.  And I can move this little microscope all over the inside,  I keep getting zero.  That's why I know there's no charge inside.  There's no charge inside, there's no field inside.  Q equal to zero inside metal.  This is a solid metal, and e equal to zero inside solid.  The other variation you can have is, suppose I make a hole in  this.  There is no electric field anywhere here because it's a  metal.  The whole logic is, if you've got a field,  you've got to find a static situation.  Let the charges move.  They'll keep moving till they have no reason to move.  That reason will stop when the field vanishes.  It's quite amazing.  They can always find an arrangement in which they  kill the external field.  That's the perfect metal.  So there's no field here.  So if you do a Gaussian surface like this,  the charge inside that is zero.  But that does not preclude the possibility that there are maybe  two plus charges there and some minus charges there.  Last time I told you that if that happens,  then the pluses and minuses will run around and meet and  destroy each other and utilize.  But that's a little subtle because maybe if they were the  only two things in the world, I can understand that,  but there's also these charges outside.  Maybe they'll impede their motion.  But here's one way to prove that there can be really no  charge even on the inner surface.  If there were some plus charges here and minus charges there,  remember q is zero inside, so you can only have equal number,  then electric field lines will have to leave the plus charge  and terminate on the minus charge.  So there'll be an electric field inside the cavity.  And I'm saying that's not allowed because of the following.  If there is field inside the cavity,  let's take that path and another path,  which are two ways to go from here to here.  One path, I get line integral of e to be not 0.  Second one, I get line integral to be 0 because there is no e.  The line integral cannot vary with the path in an electrostatic  situation.  Therefore, there cannot be any electric field inside the cavity  due to anything you put outside.  You can have electric field inside the cavity if you go into  the cavity and put by hand some extra charge there.  Then what will happen is that guy will produce a field and the  electric field lines will come to the boundary and terminate on  some negative charges.  Again, there'll be no field here.  But a hollow cavity with no charges inside cannot have charges,  no charges inside the hole, will not have charges on the edge  either.  So now what else can we say about conductors?  Today I use the fact that the line integral of e is 0 to tell  you that in a hollow cavity you cannot have any charges in the  wall.  The second thing is every conductor is in  equipotential.  It's in equipotential.  Why is that?  Can you tell me why that is true?  If it is not equipotential, the whole lump.  Yep?  Student is asking, can there be an equipotential  in a hollow cavity?  And how do we know that?  That's right.  In other words, if you take the gradient,  namely taking derivatives, you'll get a non-zero derivative  because I'm telling you v is not constant.  In some direction it's got to change.  There's an electric field in that direction,  at least.  But that's not allowed.  So v is a constant.  So take any metal shaped like anything.  You can associate a single potential to all of it.  Okay?  If you've got charge on it, that's the same.  And what it means, if you start at infinity,  the work doesn't come there, the work doesn't come there,  the work doesn't come there, all the same because they're all  measuring the potential.  The whole metal is at one potential.  That allows you to do certain tricks.  So here's one trick you do.  There are more tricks, but this is the only one we can  discuss in this class.  So here's an infinite conducting plane,  and you've got a ground attached to the ground.  And over this guy I put a charge Q.  And I ask you, find the electric field everywhere now.  That's your talents.  Without this guy, fields are just radial.  So radial, radial.  But the fields come near this metal.  They cannot penetrate.  They've got to stop at the surface of the metal.  So the field lines will do that.  They will all terminate on the metal.  And to terminate, you need some negative charge,  which the metal will suck out of the ground.  That's the reason for grounding it.  So the question is, what kind of charge distribution do you  get on the metal?  How much charge is there on the metal?  And because there's positive charge and negative charge here,  there's the force of attraction between this charge and the  plane.  In general, that's a very difficult problem.  If I give you a potato, aluminum potato,  put a charge next to it, it will have a force,  but it's not easy to calculate.  But this infinite sheet turns out very easy to calculate.  The reason is that we can solve this problem by borrowing  from another problem.  The problem I borrow from is the following.  Take this plane, continue these lines,  and think of the problem of a dipole with an equal negative  charge here.  So the lines from them go like this, right?  And if you slice it in the perpendicular bisector of this  distance between them, you will hit every line  perpendicularly.  You guys remember that part about the dipole field,  it's somewhere there.  You cut it right down the middle.  So this plane was an equipotential at v equal to 0.  If it's an equipotential, there is nothing there,  it was just a mathematical surface on which v was constant  and equal to 0.  If v is constant, you can slide in a conductor there and  nobody will know, because the field is 0 along the surface  of the conductor, nothing happens.  So you can reconcile the fact that conductors at equipotentials  and solve problems in which a charge is in front of a  conductor, if you're lucky enough to find a situation  where that's an equipotential surface of the same shape as  your conductor, then you can slip your conductor there.  So in this example, what happens is that the electric  field here, throughout the plane,  is exactly what you would get due to a dipole minus and plus,  you can easily calculate at all points on the plane.  Once you've got the electric field, you remember,  σ over ε0 is the electric field, so you can find σ,  the charge density.  And if you integrate the charge density,  you will get exactly minus Q, because the number of lines  coming from here have to terminate on the plane rather  than on this, and that'll be minus Q.  So finally, what's the force of attraction between the plane  and this charge?  Can you make a guess?  What's the force with which the plane attracts the charge?  I'm asking you to guess, so it cannot be too difficult.  Same as the force that this negative charge would exert on  this positive charge.  And I'll tell you why.  So this guy is looking around, okay, seeing some forces.  It's due to these negative charges on the plane,  but they are identical to the field created by this guy.  So it cannot tell the difference, and therefore it will move,  if you're releasing it, it'll move the same way in  this problem as in the other problem.  So it'll be drawn to the plate with an attractive force,  and Q times minus Q divided by square of that distance.  So the summary of what I did now is that if you solve any  problem, let's say the dipole problem,  and you get some shape, it's an equi-potential,  you're allowed to then replace that with a chunk of aluminum  and keep the field lines the way they are,  because that aluminum will be an equi-potential and it'll  be a sit-on equi-potential, because there are no forces on  the surface of the conductor moving the charges.  So the model of the story is you can sneak in a solid object  bounded by the surface of an equi-potential for any problem.  So if you can solve certain problems and the equi-potentials  have nice shapes, you can put conductors of that  shape in that precise location.  That'll answer a different problem in which the second  charge is absent.  You only have the first charge.  It's like a phantom charge, an image charge.  It's like this charge looks at the mirror and sees the image.  The image is minus of itself and it's drawn to it,  because the field produced by this guy and this guy is the  same as the field produced by all of these and this.  All right, so now we've got a final problem of energetics.  Let's see.  The question is, I want to bring a whole bunch of  charges, Q1, Q2, Q3, from infinity and bring them to this  arrangement.  So when they're infinitely far, they don't even know about  each other.  They don't feel any force.  So here's my goal.  I've got to do some work, bring everybody together.  They all be positive.  They may not want to be in the same region,  but I'm going to force them.  The question is, what work do I have to do?  We're going to calculate that.  We do that by saying, first let's take charge 1,  let it be wherever it has to be.  Then I bring charge 2 from infinity and put it here.  The work done for that is by definition Q2 times the  potential here.  The potential there is Q1 over 4πε0 R12.  R12 is the distance between 1 and 2.  Now I nailed the two charges.  Nobody's allowed to move, so I don't do further work.  Then I bring a third charge from infinity to its location  here.  How much work should I do?  The work that I have to do is the potential energy from  the two charges I've calculated here due to these two  charges.  That means the work I'll have to do is equal to Q1Q3 over  4πε0 R13 plus Q2 Q3 divided by 4πε0 R23.  You follow that?  That's the work I have to do to bring the first guy,  and Q2 had to fight only this one.  The third guy has to fight these two,  and the fourth guy will have to fight these three,  and so on.  I'm just going up to three charges.  From this, it should not be too hard to guess.  In general, if you've got many, many charges,  you do Qi Qj over 4πε0 Rij,  then Qi goes from 1 to N,  where j goes from 1 to N, but i cannot be equal to j,  and you divide by 2.  If you're not able to read this, I'll tell you the summation  is 1 over 2, i from 1 to N,  j from 1 to N, but with a condition that i should not  be equal to j times all these.  You can see that's valid here.  Yes?  This is the, oh, I'm sorry.  You're absolutely right.  This is not the potential, it's the potential energy,  it should be called U.  Did I say, yeah, that's right.  Because Q over 4πε0 is the potential,  but when I multiply by Q2, that is actually an energy.  These are all energies I denote by U.  Now if you have a continuous set of charges,  you must replace a sum by the integral.  Do you see what's going on here?  Why there's a half?  Because this one counts Q1 Q3,  then again it counts Q3 Q1,  but there's only one such thing, so I divide by 2,  and I have nothing like Q1 Q1, so you don't allow that.  That's the energy to assemble a whole bunch of charges together.  So that's the work you do.  If you do that work and you take your hands off,  you can fly apart and give you back that energy in the form  of kinetic energy.  Now let me do another simple example.  I want to take a sphere of radius r,  and I want to put Q coulombs on it,  and I want to know how much work I have to do for that one.  What work do I have to do to charge this sphere with Q  coulombs?  Well, first a couple of coulombs come in,  they don't run into anything.  I want to put some charge here.  That makes it harder to bring the next set of coulombs,  the next set of coulombs.  At some intermediate stage, when the charge on this sphere  is Q, and I want to bring in a charge dQ,  how much work do I have to do?  When the charge is Q, the potential of the surface  of the sphere is Q over 4π ε0 over r.  The whole sphere is at that potential,  and I'm trying to bring in a tiny more dQ from all sides  and smear it on this.  The work for that will be that times dQ.  Then the total work I do, when I go all the way from  0 to Q, will be Q squared over 2 times 1 over 4π ε0 r.  Sorry, R.  Again, you see why you get the half,  because the last coulomb is much harder to bring than the  first one.  The first one is free.  It's like a spring.  When you pull it, the first inches are easier,  the other ones are harder, because the force is Kx.  The work done is integral of x that becomes x squared over 2.  Similarly, to bring in an extra charge,  you've got to worry about how much charge is already there on  the sphere.  That's opposing you with this potential.  You work against that by bringing dQ from infinity,  and when you integrate it, you get Q squared over 2.  So this is a good problem to explain the difference between  potential and energy.  This is the potential.  This is the charge I brought from infinity,  and that together is du, and integral of du is my total  u.  So one can ask, what's the energy to produce any  changement of charges, and you should be able to  either do the sum or do the integral.  One homework problem will be how to assemble a solid ball of  charge, not a sphere solid ball.  It's got no radius, and slowly it grows up to some  size, and you can do similar integrals where you slap on  more and more spherical shells on this onion to build it up.  Again, if you knew the potential at any stage,  you could find the work to add a thin layer to it and integrate.  Okay, so final topic is one of capacitors.  So capacitor is like a way of storing charge and storing  energy.  In other words, if I want to store gravitational  energy, I can do the following.  I build this tank, and maybe there's some water in  the tank, and I make a bake here at the foot of the tank,  and I go to the top and go to the top and empty little buckets  until I got some water at a big height.  Then I've done some work, and it'll pay me back when I  open the tap and let it all rush down.  Then it may turn a turbine blade and give me back a  thermoelectric power.  You want to do similar things electrically.  You want to invest with some work, and it'll give you back.  What I can do is I can take a plus and minus charge,  which love to be next to each other and separate them.  Now I've done some work, but I cannot walk around like  this.  I've got other things to do.  I've got to find a better way to store energy.  How do you shake hands with people?  So you have to find another way.  So here's what you do.  You take one conductor and you take another conductor,  any shape you like.  It doesn't matter.  They're both electrically neutral.  They're both electrically neutral.  So you take a conductor, so you take a tiny charge from  this guy and put it there.  Then take more and more charge and start putting it there.  At any stage, when you stop, you will have a lot of  positive charges on this conductor and a lot of negative  charges on this conductor.  This conductor will be at one potential because it's a  conductor.  That conductor will be at another potential because it's  a conductor.  But both have unique potentials.  They're not the same, but this whole conductor has  one potential.  This is another potential v'.  And because it's harder and harder to pump the coulombs,  you can imagine that there is a potential difference between  them, because you've got to go uphill to pump even more charge  into this already charged object.  So we define the capacitance as the charge that you transferred  divided by the voltage difference you got.  So let me call this zero by convention.  Let's call it the capacitance.  That's the ability of that system to hold charge.  If there are very, very tiny metal spheres,  if someone want to work, if there are huge metal spheres,  you can transfer a lot of charge.  Okay?  That's the definition.  It's measured in Farad's.  Again, it's got some units, coulomb for volt,  but again it's got a different name,  named after Mr. Faraday, measured in Farad's.  So usually you'll have micro-Farad's in a lab.  Farad's a huge unit.  So let's find the capacitance of certain simple arrangements.  Simplest arrangement is the parallel plate capacitor.  Here it is.  One plate is a second plate.  The area of the plate is A.  The distance between them is D.  And I put plus charges on this one,  and I got them from this one.  Do you understand that?  You don't have to bring your own charge.  You remove charge from one conductor,  you therefore leave it negatively charged,  and you dump it on the other conductor,  thereby leaving it positively charged.  So you don't need to bring charge from the outside world.  You just redistribute the neutrality so that one is now  negative and one is positive.  So q stands for the magnitude of charge that you transfer.  So suppose I transfer some charge q.  The only job is to find q divided by v,  and that's my capacitance.  Okay, so how am I going to find the voltage difference  between these two?  I think you guys should know by now voltage difference is the  electric field times the distance between the plates.  There's no line integral.  E dot dr is just E times D.  E, if you remember, is sigma over epsilon 0  between the plates.  So sigma is q over A.  Charge density is a total charge over that.  That is by definition q over C.  Therefore, the capacitance of this parallel plate capacitor  is epsilon 0A over D.  That's one of the easier calculations to do.  So anytime someone gives you two objects and says,  find for me the capacitance, you will take charge from one,  you will put it on the other one, and then you will,  and you will get the same voltage.  Where is all these black holes?  Bad choice.  So you guys took all this down?  Yes?  Oh, yes.  Sigma is q over A.  This is what comes from knowing the answer.  This is what you guys do in the exam.  You turn that page and all kinds of stuff happens.  Signs change, numbers change, and the right answer appears  on the next page.  That's really what I did, because I knew this formula.  You're getting there no matter what, but you're quite right.  I didn't put the A in the right place.  So in a way, it's not good to know the answer every time,  because you're not that alert doing this.  Anyway, this is the correct answer.  So here's the capacitance of this parallel plate.  So another problem you could do.  These are also common.  Spherical capacitors are one sphere of radius A surrounded  by a second sphere of radius B, and your shove charges may be  plus on this one and equal number minus on that one.  So what's my goal?  My goal is to find the electric field in the region between  them and integrate it from here to here to give the potential  difference and divide by the charge.  So if I put a charge Q here, you can see that the electric  field in this region is just Q over 4.0r2,  because by Gauss's law, if I draw a sphere here,  that encloses charge Q, and it's isotropic and uniform,  and it's like that of a point charge.  Therefore, the potential difference between these two  points will be like the potential difference between  these two points with a point charge of the origin.  So V2 minus V1 will be Q over 4p epsilon 0 times 1 over  A minus 1 over B.  So this is what we have done, because this is what I call V.  So capacitance is Q over B, and that's going to be 4p epsilon  0 divided by 1 over A minus 1 over B,  which is 4p epsilon 0AB divided by B minus A.  So in every geometry, if they want capacitance,  move charge from one guy to the other guy and find the field  between them, then integrate it to find the  potential, then the potential will always be  proportional to the charge, take the ratio,  and that's capacitance.  I'm going to put this guy to a test.  Let me write down clearly the answer here.  4p epsilon 0AB divided by B minus A.  What test can I do if I want to compare it to a parallel plate  capacitor?  Can you see how I can relate it to a parallel plate capacitor?  Yep?  Student is asking, make the A.  A what?  Student is asking, make the B minus A.  Right.  You can take B minus A very close,  but it still may not look like a parallel plate capacitor.  So here's what I say we do.  We take B and A astronomical, size of the galaxy,  okay, and B minus A is just 1 inch.  Then in that case, it will really look like two  infinite planes.  The fact that it gets wrapped around over the length of the  galaxies, just like saying the Earth is round,  which is not obvious from short excursions.  In that case, we can write 4p epsilon 0,  A and B are really equal to some common number r,  and B minus A is the distance between the inner shell and the  outer shell.  That you can see.  Wait.  I'm sorry, 4pi r squared is the area times epsilon 0 over  d.  Right?  So that's one test.  It doesn't mean it's right, but if this test failed,  it would definitely be wrong.  Okay.  Now, how much energy does it take to charge up a capacitor?  It's what I did earlier, but let me do it again.  If you take two plates or any two surfaces,  and there's a certain amount of charge q on it,  and I transfer delta q, the work I do is q,  the first a q times the voltage at that stage,  voltage at that stage is equal to q over C,  which is the q you have at that point.  Therefore, the total work you do is 1 over 2 q  square over 2, q square over C.  So, when you want to charge a capacitor with a charge q on it,  either you can write it that way, or you can also do the  following.  You can write as 1 over 2 C v square,  because q over C is a v multiplied top and bottom by C.  You can write the energy in a capacitor,  in terms of the charge stored, or in terms of the voltage  between the plates.  We'll resume this next time.
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Okay, let's turn to the subject proper. I will start by doing the usual stuff of just refreshing your memory on what happened last time. I mean, last time was pretty heavy, a lot of mathematical machinery which I like to work with. Now, you might say, you know, I'm going to be a psychologist or art historian and I don't need all those details. Just tell me what I need to get through this course. So, that's a very reasonable question. Probably would have the same attitude if I took a course in your field. So, here it is. What I've told you so far is that the electric field, which we understand as basically Coulomb's forces, allows you to define a potential because it's a conservative force. That means the integral of the electric field on any closed loop is zero, and remember that's what it takes to define a potential. And the electric force, which is Q times E, will also satisfy the condition. That means we can define for this problem a potential v, which is called the potential, so that 1⁄2mv1 squared. This is a little v. This is big v. Capital V at the point r1 is equal to 1⁄2mv2 squared plus capital V at the point r2. That is our goal. Ah, missed one mistake already. Did you know what's missing here? Pardon me, yes? What's missing? Yes, the charge. So, this should really be Q times v, and this should be Q times v. So, let me remind you why it is. The force on any charge is really not just the electric field. It's Q times the electric field, and therefore the potential difference is associated with the work done by the electric field times Q. So, you've got to remember Q times v is the real energy. So, v is like an electrical height. For example, if this is the ground, there's a mountain like that, you give me the height. The potential energy is not just the height, it's mg times h. So, let me say you give me g times h. That's like the potential. That tells you the whole story, because if you want to lift a kilogram mass to that height, you do 1 times g h, 5 kilograms, 5 times g h. So, depending on the mass you want to lift, the work done will be different, or the potential energy will be different. But this doesn't depend on the mass you're planning to carry around. So, you take out the mass, you can always put it back. Similarly, the potential concerns itself with the unit charge. If Q were 1, nothing I wrote is wrong. But in general, Q is not 1. So, this is the actual energy. This is the old potential energy, U1. So, the potential energy is Q times the potential. So, think of V as the electrical height at a point, and Q is like the mass, and Q times V is like mg h. Okay, we got a formula for V. The general formula for V is that V of R2 minus V of R1 is the line integral of E dot dr from R1 to R2. This minus sign has the following meaning. The electric field exerts the force E on the unit charge, but if you want to drag the charge against the field from R1 to R2, you apply force minus E to compensate that. So, this really worked done by you, this really worked done by you, to move it from R1 to R2. Let me give you the gravity example. Gravitational force acts this way. But if you want to lift something from here to here, you oppose it, namely it's doing minus mg, your force mg, you move it to height h. That's the work you do, and that you know is the potential at the top minus potential at the bottom. So, that's the origin of the minus sign. The electric field wants to go one way, and if you want to take it against that force and raise it electrically speaking, the force you apply is precisely minus E. That dot dr is the work done by you. That's what you invest. That's what you will get back if it rolls downhill again. Now, if you want V at some point R, minus V at infinity, that integral will be minus, starting from infinity to the point, and then we all vector to the point R of E.dr. We will now take the convention that V of infinity is equal to 0. Remember, potentials are defined only by the differences. So, you can put the 0 anywhere you like. We will choose the 0 to be at infinity. Then in this integral, it's going to be 0, and if you do this integral from infinity to R, you will find for one charge, it is q over 4 pi epsilon 0 times R for single charge, and R is the distance from that charge. It looks like Coulomb's Law, but a couple of differences. There's no 1 over R squared and there is no vector. The potential due to a charge here at the point here is a number given by this. So, if you look at lots of charges, let's just find V due to many charges is sum over all those charges, the value of each charge before pi epsilon 0 times the length from where that charge is to where you want the potential. That's the principle of superposition. The Coulomb potential due to one charge is 1 over R due to many charges is each q times its own 1 over R times its own 1 over 4 pi epsilon. And finally, if you want the potential difference due to the field, you integrate it, and if you want the field from the potential, you take the derivative called the gradient, which is identical to minus i x partial derivative minus j y partial derivative. So, you expect two derivatives because the force you're looking for now is a vector. You're living in 2D. So, you cannot say the force is the derivative of the potential in what direction? If you want the force in the x direction, take the x derivative. If you want in the y direction, take the y derivative. And you put them together into a single vector, and that's the electric field. So, you go from potential to the field by taking the derivative of the gradient, and you go from the field to the potential by doing the integral. So, the subtleties, the integral is of a dot product of E with the displacement because everything is a vector now. It used to be fx dx in the old days, but f has become a vector, dx has become a vector dr, so you need the dot product. Okay, so this is a summary, but I want to make two course corrections from last time. First one was, when I tried to prove to you that the electric field is conservative, I did one thing which I realized was a swindle. I didn't realize it myself, and I wasn't waiting for anyone to correct me. I did not think about it hard enough. So, here is the mistake I made. Remember what I said? I said that's a potential due to point charge. I'm trying to look at it. These are different lines of the electric field. Let's say you want to go from here to here. You want to do some work, E dot dr along this line. The claim was that's the same work you do if I take another path that looks like this. Go this way, go that way, and go that way. Same work, and that's still correct statement. The same work because here dr is angular, E is radial, the dot product is 0. Likewise, 0. Here at every segment you have here, E is radial, dr is radial, the dot product is the field E times that length. Here is the field, here times that length, the field is constant at a given radius, so they're also equal. Still, no wrong statements. Then I said, what if I do this? A little radial, a little angular, and so on, and that's also okay. Then I finally said, what if I had some nice smooth path which is not made up of radial and angular, and my claim was, well, you take for this segment, for example, you take a radial segment, then you take an angular segment, then you may take another radial segment, another angular segment. So you approximate the smooth path by a bunch of radial angular, radial angular, and from a long distance it all goes the same, so it should be the same amount of work. That's wrong. I'll give you an example of why such arguments can be wrong. Let's take a triangle. This side is 1 and this side is 1. I want to walk from here to there. So you know I go a distance square root of 2. But then you say, I'm going to take a different path, I'm going to go horizontal, vertical, horizontal, vertical, equal amounts like that. So these are very tiny differences. You can say it's the same thing. But you know this path is of length 2. Understand why? All the horizontal parts add up to 1, all the vertical parts add up to 1. So going zigzag is longer by a factor root 2. So just because things are approximately the same, doesn't mean the answer will be the same, because if you make an infinite number of infinitesimal mistakes, you can add up to something finite. Likewise here, you've got to make sure that you can deform your path from what was really given to you to what you did, which is radial and angular. The work done should be the same. The path length is clearly not the same in this example, and it can all add up, but the work will be the same. That's what I did not show to you. So let me take a segment here where that's the radial part and an angular part, but the person really wanted to take a segment along some straight line joining the two. What I want to show you is that not only does that approximate this from long distance, the work done on that plus that is the work done here. Okay? So for your convenience, I'm going to blow up that picture here so you can all see what I'm trying to do. This should be the radial direction. This is the radial direction. This is E. This is some perpendicular direction, and this is the direction in which the move dr was made. That's the actual path which is going like that, and you manage to just deform it. So what's the line integral on the radial path? Well, it's the magnitude of the electric field at that distance times the difference in radius between this end and this end. Let me call it delta r. Delta r is the spacing here between that curve and that curve. You agree? The dot product is trivial. E is this way. Displacement is this way. No cos theta is to worry about. That's the work done. Work done here is zero because perpendicular to the field. So that's the total work done going like that. How about going along this line? It's going to be E at r dot dr. When I say E at r, r is of course not one point. It's going to spread out, but this is an infinitesimal region, so you can talk about the E at one point and not worry about the variation. So don't forget it's an infinitesimal triangle. E dot dr will be E at that distance r, the length of dr, and the cosine of the angle between them, which is this angle. But length of dr times cosine is precisely that distance. Delta r, and therefore you do the same amount of work. So basically I'm saying as you wander around, you may not choose to go radial angular, radial angular. But if you go at a slant to the radial, the work you do really depends on how much radial distance you cover. That's what E dot dr does for you. That's why you can deform the path anyway and get the same answer. That's the first thing. The second thing is I forgot to tell you anything about units. The electric field is forced on a unit charge, so the unit for that is Newton's per coulomb. The electric potential looks like the work done on a unit charge, so the unit for that is Joules per coulomb. And the name for Joules per coulomb is a volt. It means if there's a voltage difference between two points of the two poles, it's going to take you 12 Joules of work to lug a coulomb from the lower point to the higher point. So you guys have to get your units right, okay? Now you might say, why don't you use units? Right? You know why? Because I have tenure. When you have tenure, you don't have to use units. I have tenure, I have a car next to the fire hydrant, I got tenure, I tear up the tickets, I don't do duty, duty, I don't pay taxes. You don't have to do anything when you reach. You know, Allen was going to tell you the joys of being a graduate student. The joys of being tenure prof, hard to describe, okay? Priceless. In fact, one of my sons, when he saw my lifestyle, said it reminded him of a jellyfish deep in the ocean. He saw a rock, he attaches himself to a rock, and then never moves and eats his own brain for food, because he has no further use for it. So apparently that's what I reminded him of. So you can, if you work hard, you can be like me. You can do that. Anyway, look, so you guys have to get the units because seriously speaking, if you don't write any units, we don't know if you write the rock, right? The problem answered is 19, in some units, right? So if you don't give the unit, the number is not worth anything. Another advantage of units is you can keep track of your calculation. You multiply one quantity by another quantity. Everything has units. You cross out the lengths and the meters. You've got to make sure the stuff you get has the right units on the two sides. That helps you also keep track of calculations. Okay, so let's go back now. This is all stuff left over from before. So now for the new stuff. I'm going to tell you the first part of this lecture is going to be what are the advantages of V? I will just mention them, then I will elaborate on them. The first advantage is the conservation of energy. The second advantage is the computation of E. The third advantage is a lot of good visual pictures, which are very helpful. So you won't know what these mean, but this is what we're going to do, so you know where we're going. First thing is conservation of energy, there's no need to belabor at this point. I mean, that's why we did the whole thing. The whole conservative forces, gradient, this, that, all that was to make sure in the end you can get a potential energy out of it. And as you know, from the roller coaster problem, you can do kinetic plus potential equals kinetic plus potential, it saves you a lot of trouble. In other words, in this room right now there's a lot of electric fields. A lot of charges everywhere applying a field here, and I want to carry a charge from here, and I want to drag it over here. Or maybe the charge was shot by a cyclotron or something came flying out at some speed here, and it comes here a little later, and I want to know how fast it's moving. One way to do that is to follow the charge as it moves at the same time. You can, for instance, find the force on it, the acceleration from that, change in velocity, add up all the changes, you'll get the final velocity. But as you know, with the law of conservation of energy, you can skip all the intermediate stuff. You just have to know the initial kinetic energy, initial potential energy, final kinetic energy, final potential energy. They are equal if you sum them. So you can find one number from it, one of the numbers you can find. In particular, if you didn't know the final kinetic energy, you can find it. That's the energy. The second thing is that it makes life easy when you want to compute the electric field. So this is an example I did again near the end of the class, and I think it was needlessly complicated. So I will try to repeat the derivation a little more quickly, because it's a lot easier than I gave the impression. So we are trying to find, let us say, the electric field is a simple minus q and plus q separated by a distance 2a. If you did it with forces, with forces I didn't even try to do it everywhere. I remember I did it here and I did it there. The force here was like this, the force there was like that. Then I said if you do a lot of work, you'll get all these lines. What makes life difficult is that if you go to this point here, this I explained to you last time, this guy will repel it with some force, this guy will attract it, and you've got to add the two arrows and get a new arrow. That's the electric field at that point. So you've got to add arrows, and that's more difficult than adding the potential, which is just a number from this guy and that guy, then once you've got the potential, you find the field by taking derivatives. You want to take the derivative of ex, take the derivative of the minus sign, you want ey, you take minus dv dy. So what's the e here? You remember there's q over 4πε0 times 1 over r+, r+, at that length, minus 1 over r-, that's it. No arrow is nothing. That's the total v. This is r-, let's define r to be the distance from the center. And I want you to imagine this is long way off. This picture is not drawn to scale. You should imagine these two points are like this and I'm somewhere there. That's when the dipole approximation is working. This is not an approximation, this is exact. So let's do the following then. We combine the denominators, you get r+, times r-, and the denominator is r-, minus r+. Now you can see that r+, and r-, differ from r because of this little guy, a. If you use the law of cosines, you know, it's that squared plus that squared minus 2a r cosine theta or something. In the bottom, we ignore that difference and we write everything as 4πε0r². Because r+, I approximate with r, r-, I approximate with r. So this is not the exact answer. There are corrections here that look like 1+, maybe a number a divided by r and also a² divided by r². But these are all negligible. In the numerator, you can again say, why don't you replace this by r and this by r, then you get 0. It is true. If you replace everything by r, then you do get 0, and that's because in the difference the big guy cancels and what's left over. And the difference is what's going to keep this from vanishing. So we have to find the difference in that distance and that distance. So for that what you do, you draw a line here. Now you cannot draw a real perpendicular here, but if these things are very far and these lines are almost parallel, you can draw a perpendicular there. And if that angle is theta, theta is the angle, then it's very easy to show that this distance is simply 2a cos theta. Because if theta is this angle, then you can say cosine of this angle here will be, by whatever complementary angle, is 2a cos theta. So if I draw a line here, I have 2a times cosine theta, and that I will write as dipole moment P, cosine theta over 4πε0 r². That's what I was trying to get. Yep? Student is asked which angle is theta? Theta is the angle measured from here, let me draw this. You can, one can worry about, is it this angle or that angle? But I'm saying draw the angle from the middle, okay? That angle is theta. And roughly speaking, that angle is also theta. Now drop a perpendicular here, and this side is 2a, and this side is 2a, and this side is 2a. So this difference is 2a cos theta. So now I have v in terms of r and theta, but I can write it in terms of x and y very easily. It is P over 4πε0 r², and cosine theta is x over r. So the whole thing is x over r cubed, and it's the same as x over x² plus y² to the 3 over 2. Right? r is the square root and r cubed is three powers of that. Now you can easily take the derivative of this guy with respect to x, or with respect to y. That's very, derivative is very easy to take. Integrals are hard to do, and that'll give you the electric energy, and you'll get the ex and ey everywhere. So ex is equal to minus dv dx, and ey is minus dv dy. So one of the homework problems is to really calculate this. And if you calculate it and you plot it, you will see this thing emerging. Okay. In other problems, most of the time it's easy to find b and then to find e. For example, if I have a ring of charge, and I want to find the b here, then let's say the ring has charge q and some radius a. Then a little guy here produces a little potential dv, which is the charge dq. dq is not any distance, it's like the differential, divided by 4πε0. And what's the distance? If that is z, then it's simply root of z² plus a². So that's the potential of the x², due to this little charge. But every unit of charge on this ring is equidistant from this point. You understand it's a ring and I'm perpendicular to the ring. So they all contribute the same number and some of all the dq is just the q. So that's the potential on the axis. Now you can take a z derivative to get the field. Okay? Generally, it is easier to do the v. Either it's a sum or integral over the charges producing the v, then take derivatives. But I want to give you one example in which the opposite is true. But in fact, it's easier to get e than to get v. And that is, there are many such examples, I'm just giving you one. So I say that's a hollow conducting sphere. It's got charge q. This guy's hollow and conducting. I want to find the potential everywhere due to this. And the charge q is uniformly spread on the surface. So what's going to be the potential? First of all, by symmetry, you can see that if you pick any point, anywhere, the answer is going to depend on the radial distance from the center, because of spherical symmetry. But you still have a lot of hard work to do, because you can divide the sphere into little rings like this, slice it at different latitudes, and find the potential due to each ring using this formula I just did somewhere, then integrate over all the rings, and you will get an answer. But that's a decent amount of calculus. But you don't have to do that, because in this problem, you can't do that. Because it's vertically symmetric, and there's a charge q here, we know from Gauss's law the electric field is going to be purely radial, and will look like q over 4 point epsilon 0 r square, as long as r is outside the sphere. So here's one problem, but it's easier to get E, because E is pretty trivial. You just need to know the magnitude. Gauss's law tells you the magnitude. You put the unit vector back and you get this. So as long as you're outside the sphere, it's as if q were sitting at the origin, and the potential V of r, which is minus q plus q over 4 pi epsilon 0 r, outside that sphere. So that's because remember V of r minus V of infinity is equal to integral of E dot dr from infinity to r, the minus sign. And if you just take the field of a point charge and do the integral, you will get the potential of a point charge, which is this. So if I draw the potential as a function of radius, it will look exactly like that of a point charge, and I'm going to draw the potential of radius over r. Now what about for r less than r? How do I continue this graph? Pardon me? You can use Gauss's law. If you use Gauss's law, inside what will you get for E? E is 0. Integral of E dot 0 is dr is 0, so potential does not change inside the sphere. But don't think potential is 0. If it does not change, it's stuck at this value. So throughout the hollow sphere, there's only one potential. Electric field is a different story. Electric field is in fact 0 inside, then falls like 1 over r squared. Electric potential is a constant. So these are, there are certain problems like a cylinder of charge surrounded by another cylinder of charge, where again it's easier to get E from symmetry arguments than E to get the potential. Okay, so this is my second point, namely, barring a few exceptions like this one with high symmetry, the potential as the advantage that it lets us compute the electric field by first doing a sum of scalar quantities rather than vectors, then taking derivatives. Now third advantage has to do with being able to visualize the potential. It gives you a landscape, and that's intuitively very helpful. So here's one example. Let's take two parallel plates, put a lot of plus charge here, and equal amount of minus charge here. What will the electric field look like in this region? Let's say both have the same density sigma for unit area and that infinite. I'm just showing you a finite portion. So this guy will produce sigma over 2ε0 here and here. This one, because it's negative charge, will do this, sigma over 2ε0 on this side. In the region between them, you can see the two arrows are additive. So the field will look like this. In the region outside, the two arrows are opposite. This guy is coming and that guy is going out. So the field doesn't diminish with distance due to either plate so they can completely cancel each other. So the electric field is trapped between the two plates. And as the strength, which is double what you would have on either side if you had only one sheet, that's the electric field E. That's my electric field. We can ask what's the potential difference? Or what's the potential at various points? This whole thing being metallic is at one potential. I say v is equal to 0. Throughout this line, let's say v is 0. In fact, it's very clear v is 0 because if I move horizontally on this line, I don't do any work because the electric field is perpendicular to me. So if v was 0 at one point, it is 0 throughout this plane. Yes, I should qualify the statement. You cannot pick v equal to 0 over an extended region. You have freedom to pick it to be 0 at one point. I'm telling you, if you pick the one point in this plane, you can navigate throughout the plane without changing v, because you move perpendicular to E. So v is 0 on the lower plate, but it won't be 0 anywhere else, but you can see that may be v equal to 1, that may be v equal to 2, v equal to 3, upper plate may be v equal to 4 volts. So lines of E are vertical and lines of constant v are vertical. This is really like the gravitation problem. An electric field raining down on you is just like the force of gravity. This might as well be gh. This might as well be g. You multiply a mass, you get mg. You multiply by the q, you get the force on a charge. It's straight down and the potential just increases with height. Yep? Student is asking how do you do the cutters to get the constant voltage? Oh, okay. Think of it as a plate coming outside the blackboard. Student is asking for the hollow sphere. Oh, the hollow sphere. The potential is always the work done to go from infinity to wherever you are, right? So when I come from here, what do I think is at the origin? I think there's a point charge at the origin. So I push up against that. I come here. So I do that amount of work. Once I'm in here, there is no electric field inside. So e.dr is 0 from then on. But don't forget the e, the v that you got up to that point. You don't drop it. You add, keep on adding 0 to it. So it is 0 everywhere. You add 0 to all points here because you can go from here to here, no electric field. So the whole sphere has the same potential as the surface. So now, here I got these potentials, and it gives you a picture that you're running uphill, and these are the steps you've got to climb to go uphill. And if you've got four volts here, if you take a cool arm and you let it drop, it will pick up four joules by the time it hits the slower plate. That's the meaning. So the work done by the electric field, just like gravity, is the potential difference, which is the work done on unit charge times the charge, which is one cool arm here. If it is ten cool arms, you do 40 joules of work. If an electron is released, you've got to be a little careful. By definition, the charge of an electron is negative. So if an electron is released here, in fact, it will go like this. You don't have any analogs in gravity. Gravity, everything falls down. So what was something said negative mass and something said positive mass? All guys with negative mass, you've got to nail them to the ground, otherwise they'll just float up. That happens in electricity, so any electron you leave here will go up. So the force on the electron is minus q times e, so everything's got a minus sign. So an electron going from here to here will in fact lose energy. It will lose potential energy, and it will turn into kinetic energy. So if an electron falls through 1 volt, the work done is charge of an electron times 1 volt, which is 1.6 times 10 to the minus 19 joules, that happens to be called an electron volt, just a convenient thing to use. When the charges you're pushing around are not cool arms, but this tiny electron is good to know in electron volts. For example, in the hydrogen atom, the cool arm potential is negative because the charge at the origin is a proton plus q. Electron has got some minus q divided by 4πε 0r. That's the potential. And in quantum mechanics, only at certain radius you can have electrons. So the lowest electron is somewhere here, and they say it is minus whatever some number of electron, 13 points something electron volts. That means if you want to drag an electron from this hole and liberate it from the atom, you've got to do 13 times 1.6 times 10 to the minus 19 joules of work. So as if this were 13 volts down compared to infinity. It's a hole in the ground, 13 points, and it's got a 4 or 6 volts deep. So if you want to liberate that electron, you have to furnace that amount of energy. Okay, now going on with visualization, let me say let's take another case we can all do, which is a single point charge. Here's a point charge. The electric field due to that looks like this, and it is million times. Now what about the electric potential? For a point charge, the potential is 1 over r. That means it's a constant on a circle of radius r. That means if you pick any circle or a sphere, I'm sorry, v is one number here, v is a different number there, a different number there. So that's the whole surface surrounding the charge on which v is some fixed number. It depends on that radius. You go further out, it's a different number depending on that radius, and it falls like 1 over r. You notice once again that the lines of constant potential are perpendicular to the lines of the field. That was true there, that is true here. In fact, let me do one more example. If you take a dipole, negative and positive, and the field lines go like this, if you compute the potential, if you go very, very close to this guy, I think you all seem to be in the same position. If you're very near one charge, forget about the rest of the world. It dominates everything. Potentially, it will look like circles centered here, but as you go further out, the shape will change and it will look like this. A similar thing on the other side, then there will be this huge infinite sphere that becomes planar, that goes all the way to infinity. These are contours of constant v. If you really plot them, you'll again find contours of constant v. If you plot them, you'll find that they're perpendicular to the electric field lines, perpendicular to e. So I want to explain to you that this is a very general result. This is not something true in these isolated cases of nice symmetry. Why is that? So let's understand why that is true. The v2 minus v1 is minus integral of e.dr from 1 to 2. So let's say v2 and v1 are 2 and 1 are very close. In that case, let's call it delta v is equal to minus e.dr. There's no integral to do because I've picked two points which are nearby an infinitesimal amount dr, and e.dr is equal to delta v, which is minus of v2 minus v1. What does that mean? That means if you start at some point and you move a distance dr, you take a step dr and you find the change in potential, the answer depends not only on the length of your step, but on which direction you move. Because of this dot product, the change in potential will be the magnitude of the electric field at that point, the magnitude of the step length, and the cosine of the angle between them. So if electric field, if the electric potential is replaced by height in two dimensions, you know, top of sleeping giants and the hill is going up and down, up and down, you're at some point. There's a height function h of x and y instead of v of x and y, and I want to take one step and I can ask what are the consequences to my height? The answer is if the electric field is pointing in some direction and you take a step dr in a generally different direction, this is the change you get. So you can ask, how do I get the biggest drop in height, or the biggest drop in potential, when cosine theta is 1? Right? Because then it's this, that times 1 with a minus sign. So the electric field points in the direction of the greatest rate of change of potential. It drops that direction. If you release a marble in a gravitational field, you'll roll along the gravitational field, which is obtained the same way from the gravitational potential. If you want to climb up the mountain, suppose there's a tsunami coming, and you want to get to the top as quickly as possible, don't panic, compute the gradient, and move along the gradient. Okay? Then stop, think again, recompute the gradient, and in that manner you will get to the top in the quickest time if you can compute quickly enough. Okay? But that's the process. At every point, that's the vector pointing in the direction of greatest increase, the electric field happens to be the direction of greatest decrease because of this minus sign in our convention. Now, let us say you are in this mountain, and you don't want to go down, you don't want to go up, but you still want to take a walk. Can you do that? Yes? Or are you stuck at one place? What do you have to do, guys? Yes? You move perpendicular to the field. To move perpendicular to the field, the cos theta is 0. That's why, now going back to this, this is called Mount Kula. It's coming out of the blackboard, really tall, and you are here. If you want to go downhill, you go along the field, you run away from the hill, all the way to the valley, to the plains. If you want to go to the top, you go opposite to the field, and you go along this equipotential at every point, you do not do any extra work, so your potential does not change. So it's clear from this that the equipotential is always, in fact, in general, the change in any potential is gradient of the potential.dr, except for the minus sign. It's the same story that if dr is parallel to grad v or v parallel to e, you get the biggest change in voltage. All right. So this is what you should understand in general. What is good about the potential is you can take a whole bunch of charges, put them here, there, there, there, and you can draw little contours, maybe of constant v. It's just like the graph you have of the weather, of the temperature, for example, or if you have a topographic map of the mountain, there are some valleys and there are plains and there are some mountain tops, and these lines of constant height are like constant v. So it helps you visualize. And then if you're told you really something here, and it rolled down and came here, you can find the speed here, given the difference of these two numbers. You don't have to go through the intermediate step. So these are the advantages. Okay. So now I'm going to return to the conductors briefly to tell you something, one other thing that is new compared to what we did before. Remember what I told you about a conductor? A conductor is one in which the electrons do not belong to any one parent. They run around to the whole sample, but they cannot leave the metal, or the boundary of the metal that a force is bringing them back. Inside, they're free to move. Therefore, if you take a conductor and you stick it into a metal, initially an electric field will penetrate the conductor. The charges, because they're free to move, will move. Negative charges will move. One way positive charges will move. The other way, until they set up a back reaction to this, that blocks the field inside the metal, until the field inside is zero. So here's a generic conductor. If you put it in an external field, it will have some charges on its surface. It will produce a field this way that will oppose the external field in which you put it, until at the end there is no field inside the metal. Or if you did not put it in the field, it just took this material and dumped some charge on it, threw some charge on it, and asked, where does it go? Well, the charges are all plus, and they don't like each other. They want to move away from each other, and they move as far as they can until they come to the boundary, and they cannot go any further. That's where they stop. That we saw before. So we know all the charges in a metal are in the boundary. There's no electric field inside because of Gauss's law. If I want to find the charge here, I do a tiny sphere, do integral e.da, I get zero, and that's a charge inside. And I can move this little microscope all over the inside, I keep getting zero. That's why I know there's no charge inside. There's no charge inside, there's no field inside. Q equal to zero inside metal. This is a solid metal, and e equal to zero inside solid. The other variation you can have is, suppose I make a hole in this. There is no electric field anywhere here because it's a metal. The whole logic is, if you've got a field, you've got to find a static situation. Let the charges move. They'll keep moving till they have no reason to move. That reason will stop when the field vanishes. It's quite amazing. They can always find an arrangement in which they kill the external field. That's the perfect metal. So there's no field here. So if you do a Gaussian surface like this, the charge inside that is zero. But that does not preclude the possibility that there are maybe two plus charges there and some minus charges there. Last time I told you that if that happens, then the pluses and minuses will run around and meet and destroy each other and utilize. But that's a little subtle because maybe if they were the only two things in the world, I can understand that, but there's also these charges outside. Maybe they'll impede their motion. But here's one way to prove that there can be really no charge even on the inner surface. If there were some plus charges here and minus charges there, remember q is zero inside, so you can only have equal number, then electric field lines will have to leave the plus charge and terminate on the minus charge. So there'll be an electric field inside the cavity. And I'm saying that's not allowed because of the following. If there is field inside the cavity, let's take that path and another path, which are two ways to go from here to here. One path, I get line integral of e to be not 0. Second one, I get line integral to be 0 because there is no e. The line integral cannot vary with the path in an electrostatic situation. Therefore, there cannot be any electric field inside the cavity due to anything you put outside. You can have electric field inside the cavity if you go into the cavity and put by hand some extra charge there. Then what will happen is that guy will produce a field and the electric field lines will come to the boundary and terminate on some negative charges. Again, there'll be no field here. But a hollow cavity with no charges inside cannot have charges, no charges inside the hole, will not have charges on the edge either. So now what else can we say about conductors? Today I use the fact that the line integral of e is 0 to tell you that in a hollow cavity you cannot have any charges in the wall. The second thing is every conductor is in equipotential. It's in equipotential. Why is that? Can you tell me why that is true? If it is not equipotential, the whole lump. Yep? Student is asking, can there be an equipotential in a hollow cavity? And how do we know that? That's right. In other words, if you take the gradient, namely taking derivatives, you'll get a non-zero derivative because I'm telling you v is not constant. In some direction it's got to change. There's an electric field in that direction, at least. But that's not allowed. So v is a constant. So take any metal shaped like anything. You can associate a single potential to all of it. Okay? If you've got charge on it, that's the same. And what it means, if you start at infinity, the work doesn't come there, the work doesn't come there, the work doesn't come there, all the same because they're all measuring the potential. The whole metal is at one potential. That allows you to do certain tricks. So here's one trick you do. There are more tricks, but this is the only one we can discuss in this class. So here's an infinite conducting plane, and you've got a ground attached to the ground. And over this guy I put a charge Q. And I ask you, find the electric field everywhere now. That's your talents. Without this guy, fields are just radial. So radial, radial. But the fields come near this metal. They cannot penetrate. They've got to stop at the surface of the metal. So the field lines will do that. They will all terminate on the metal. And to terminate, you need some negative charge, which the metal will suck out of the ground. That's the reason for grounding it. So the question is, what kind of charge distribution do you get on the metal? How much charge is there on the metal? And because there's positive charge and negative charge here, there's the force of attraction between this charge and the plane. In general, that's a very difficult problem. If I give you a potato, aluminum potato, put a charge next to it, it will have a force, but it's not easy to calculate. But this infinite sheet turns out very easy to calculate. The reason is that we can solve this problem by borrowing from another problem. The problem I borrow from is the following. Take this plane, continue these lines, and think of the problem of a dipole with an equal negative charge here. So the lines from them go like this, right? And if you slice it in the perpendicular bisector of this distance between them, you will hit every line perpendicularly. You guys remember that part about the dipole field, it's somewhere there. You cut it right down the middle. So this plane was an equipotential at v equal to 0. If it's an equipotential, there is nothing there, it was just a mathematical surface on which v was constant and equal to 0. If v is constant, you can slide in a conductor there and nobody will know, because the field is 0 along the surface of the conductor, nothing happens. So you can reconcile the fact that conductors at equipotentials and solve problems in which a charge is in front of a conductor, if you're lucky enough to find a situation where that's an equipotential surface of the same shape as your conductor, then you can slip your conductor there. So in this example, what happens is that the electric field here, throughout the plane, is exactly what you would get due to a dipole minus and plus, you can easily calculate at all points on the plane. Once you've got the electric field, you remember, σ over ε0 is the electric field, so you can find σ, the charge density. And if you integrate the charge density, you will get exactly minus Q, because the number of lines coming from here have to terminate on the plane rather than on this, and that'll be minus Q. So finally, what's the force of attraction between the plane and this charge? Can you make a guess? What's the force with which the plane attracts the charge? I'm asking you to guess, so it cannot be too difficult. Same as the force that this negative charge would exert on this positive charge. And I'll tell you why. So this guy is looking around, okay, seeing some forces. It's due to these negative charges on the plane, but they are identical to the field created by this guy. So it cannot tell the difference, and therefore it will move, if you're releasing it, it'll move the same way in this problem as in the other problem. So it'll be drawn to the plate with an attractive force, and Q times minus Q divided by square of that distance. So the summary of what I did now is that if you solve any problem, let's say the dipole problem, and you get some shape, it's an equi-potential, you're allowed to then replace that with a chunk of aluminum and keep the field lines the way they are, because that aluminum will be an equi-potential and it'll be a sit-on equi-potential, because there are no forces on the surface of the conductor moving the charges. So the model of the story is you can sneak in a solid object bounded by the surface of an equi-potential for any problem. So if you can solve certain problems and the equi-potentials have nice shapes, you can put conductors of that shape in that precise location. That'll answer a different problem in which the second charge is absent. You only have the first charge. It's like a phantom charge, an image charge. It's like this charge looks at the mirror and sees the image. The image is minus of itself and it's drawn to it, because the field produced by this guy and this guy is the same as the field produced by all of these and this. All right, so now we've got a final problem of energetics. Let's see. The question is, I want to bring a whole bunch of charges, Q1, Q2, Q3, from infinity and bring them to this arrangement. So when they're infinitely far, they don't even know about each other. They don't feel any force. So here's my goal. I've got to do some work, bring everybody together. They all be positive. They may not want to be in the same region, but I'm going to force them. The question is, what work do I have to do? We're going to calculate that. We do that by saying, first let's take charge 1, let it be wherever it has to be. Then I bring charge 2 from infinity and put it here. The work done for that is by definition Q2 times the potential here. The potential there is Q1 over 4πε0 R12. R12 is the distance between 1 and 2. Now I nailed the two charges. Nobody's allowed to move, so I don't do further work. Then I bring a third charge from infinity to its location here. How much work should I do? The work that I have to do is the potential energy from the two charges I've calculated here due to these two charges. That means the work I'll have to do is equal to Q1Q3 over 4πε0 R13 plus Q2 Q3 divided by 4πε0 R23. You follow that? That's the work I have to do to bring the first guy, and Q2 had to fight only this one. The third guy has to fight these two, and the fourth guy will have to fight these three, and so on. I'm just going up to three charges. From this, it should not be too hard to guess. In general, if you've got many, many charges, you do Qi Qj over 4πε0 Rij, then Qi goes from 1 to N, where j goes from 1 to N, but i cannot be equal to j, and you divide by 2. If you're not able to read this, I'll tell you the summation is 1 over 2, i from 1 to N, j from 1 to N, but with a condition that i should not be equal to j times all these. You can see that's valid here. Yes? This is the, oh, I'm sorry. You're absolutely right. This is not the potential, it's the potential energy, it should be called U. Did I say, yeah, that's right. Because Q over 4πε0 is the potential, but when I multiply by Q2, that is actually an energy. These are all energies I denote by U. Now if you have a continuous set of charges, you must replace a sum by the integral. Do you see what's going on here? Why there's a half? Because this one counts Q1 Q3, then again it counts Q3 Q1, but there's only one such thing, so I divide by 2, and I have nothing like Q1 Q1, so you don't allow that. That's the energy to assemble a whole bunch of charges together. So that's the work you do. If you do that work and you take your hands off, you can fly apart and give you back that energy in the form of kinetic energy. Now let me do another simple example. I want to take a sphere of radius r, and I want to put Q coulombs on it, and I want to know how much work I have to do for that one. What work do I have to do to charge this sphere with Q coulombs? Well, first a couple of coulombs come in, they don't run into anything. I want to put some charge here. That makes it harder to bring the next set of coulombs, the next set of coulombs. At some intermediate stage, when the charge on this sphere is Q, and I want to bring in a charge dQ, how much work do I have to do? When the charge is Q, the potential of the surface of the sphere is Q over 4π ε0 over r. The whole sphere is at that potential, and I'm trying to bring in a tiny more dQ from all sides and smear it on this. The work for that will be that times dQ. Then the total work I do, when I go all the way from 0 to Q, will be Q squared over 2 times 1 over 4π ε0 r. Sorry, R. Again, you see why you get the half, because the last coulomb is much harder to bring than the first one. The first one is free. It's like a spring. When you pull it, the first inches are easier, the other ones are harder, because the force is Kx. The work done is integral of x that becomes x squared over 2. Similarly, to bring in an extra charge, you've got to worry about how much charge is already there on the sphere. That's opposing you with this potential. You work against that by bringing dQ from infinity, and when you integrate it, you get Q squared over 2. So this is a good problem to explain the difference between potential and energy. This is the potential. This is the charge I brought from infinity, and that together is du, and integral of du is my total u. So one can ask, what's the energy to produce any changement of charges, and you should be able to either do the sum or do the integral. One homework problem will be how to assemble a solid ball of charge, not a sphere solid ball. It's got no radius, and slowly it grows up to some size, and you can do similar integrals where you slap on more and more spherical shells on this onion to build it up. Again, if you knew the potential at any stage, you could find the work to add a thin layer to it and integrate. Okay, so final topic is one of capacitors. So capacitor is like a way of storing charge and storing energy. In other words, if I want to store gravitational energy, I can do the following. I build this tank, and maybe there's some water in the tank, and I make a bake here at the foot of the tank, and I go to the top and go to the top and empty little buckets until I got some water at a big height. Then I've done some work, and it'll pay me back when I open the tap and let it all rush down. Then it may turn a turbine blade and give me back a thermoelectric power. You want to do similar things electrically. You want to invest with some work, and it'll give you back. What I can do is I can take a plus and minus charge, which love to be next to each other and separate them. Now I've done some work, but I cannot walk around like this. I've got other things to do. I've got to find a better way to store energy. How do you shake hands with people? So you have to find another way. So here's what you do. You take one conductor and you take another conductor, any shape you like. It doesn't matter. They're both electrically neutral. They're both electrically neutral. So you take a conductor, so you take a tiny charge from this guy and put it there. Then take more and more charge and start putting it there. At any stage, when you stop, you will have a lot of positive charges on this conductor and a lot of negative charges on this conductor. This conductor will be at one potential because it's a conductor. That conductor will be at another potential because it's a conductor. But both have unique potentials. They're not the same, but this whole conductor has one potential. This is another potential v'. And because it's harder and harder to pump the coulombs, you can imagine that there is a potential difference between them, because you've got to go uphill to pump even more charge into this already charged object. So we define the capacitance as the charge that you transferred divided by the voltage difference you got. So let me call this zero by convention. Let's call it the capacitance. That's the ability of that system to hold charge. If there are very, very tiny metal spheres, if someone want to work, if there are huge metal spheres, you can transfer a lot of charge. Okay? That's the definition. It's measured in Farad's. Again, it's got some units, coulomb for volt, but again it's got a different name, named after Mr. Faraday, measured in Farad's. So usually you'll have micro-Farad's in a lab. Farad's a huge unit. So let's find the capacitance of certain simple arrangements. Simplest arrangement is the parallel plate capacitor. Here it is. One plate is a second plate. The area of the plate is A. The distance between them is D. And I put plus charges on this one, and I got them from this one. Do you understand that? You don't have to bring your own charge. You remove charge from one conductor, you therefore leave it negatively charged, and you dump it on the other conductor, thereby leaving it positively charged. So you don't need to bring charge from the outside world. You just redistribute the neutrality so that one is now negative and one is positive. So q stands for the magnitude of charge that you transfer. So suppose I transfer some charge q. The only job is to find q divided by v, and that's my capacitance. Okay, so how am I going to find the voltage difference between these two? I think you guys should know by now voltage difference is the electric field times the distance between the plates. There's no line integral. E dot dr is just E times D. E, if you remember, is sigma over epsilon 0 between the plates. So sigma is q over A. Charge density is a total charge over that. That is by definition q over C. Therefore, the capacitance of this parallel plate capacitor is epsilon 0A over D. That's one of the easier calculations to do. So anytime someone gives you two objects and says, find for me the capacitance, you will take charge from one, you will put it on the other one, and then you will, and you will get the same voltage. Where is all these black holes? Bad choice. So you guys took all this down? Yes? Oh, yes. Sigma is q over A. This is what comes from knowing the answer. This is what you guys do in the exam. You turn that page and all kinds of stuff happens. Signs change, numbers change, and the right answer appears on the next page. That's really what I did, because I knew this formula. You're getting there no matter what, but you're quite right. I didn't put the A in the right place. So in a way, it's not good to know the answer every time, because you're not that alert doing this. Anyway, this is the correct answer. So here's the capacitance of this parallel plate. So another problem you could do. These are also common. Spherical capacitors are one sphere of radius A surrounded by a second sphere of radius B, and your shove charges may be plus on this one and equal number minus on that one. So what's my goal? My goal is to find the electric field in the region between them and integrate it from here to here to give the potential difference and divide by the charge. So if I put a charge Q here, you can see that the electric field in this region is just Q over 4.0r2, because by Gauss's law, if I draw a sphere here, that encloses charge Q, and it's isotropic and uniform, and it's like that of a point charge. Therefore, the potential difference between these two points will be like the potential difference between these two points with a point charge of the origin. So V2 minus V1 will be Q over 4p epsilon 0 times 1 over A minus 1 over B. So this is what we have done, because this is what I call V. So capacitance is Q over B, and that's going to be 4p epsilon 0 divided by 1 over A minus 1 over B, which is 4p epsilon 0AB divided by B minus A. So in every geometry, if they want capacitance, move charge from one guy to the other guy and find the field between them, then integrate it to find the potential, then the potential will always be proportional to the charge, take the ratio, and that's capacitance. I'm going to put this guy to a test. Let me write down clearly the answer here. 4p epsilon 0AB divided by B minus A. What test can I do if I want to compare it to a parallel plate capacitor? Can you see how I can relate it to a parallel plate capacitor? Yep? Student is asking, make the A. A what? Student is asking, make the B minus A. Right. You can take B minus A very close, but it still may not look like a parallel plate capacitor. So here's what I say we do. We take B and A astronomical, size of the galaxy, okay, and B minus A is just 1 inch. Then in that case, it will really look like two infinite planes. The fact that it gets wrapped around over the length of the galaxies, just like saying the Earth is round, which is not obvious from short excursions. In that case, we can write 4p epsilon 0, A and B are really equal to some common number r, and B minus A is the distance between the inner shell and the outer shell. That you can see. Wait. I'm sorry, 4pi r squared is the area times epsilon 0 over d. Right? So that's one test. It doesn't mean it's right, but if this test failed, it would definitely be wrong. Okay. Now, how much energy does it take to charge up a capacitor? It's what I did earlier, but let me do it again. If you take two plates or any two surfaces, and there's a certain amount of charge q on it, and I transfer delta q, the work I do is q, the first a q times the voltage at that stage, voltage at that stage is equal to q over C, which is the q you have at that point. Therefore, the total work you do is 1 over 2 q square over 2, q square over C. So, when you want to charge a capacitor with a charge q on it, either you can write it that way, or you can also do the following. You can write as 1 over 2 C v square, because q over C is a v multiplied top and bottom by C. You can write the energy in a capacitor, in terms of the charge stored, or in terms of the voltage between the plates. We'll resume this next time.
means the integral of the electric field on any is Newton's per coulomb. The electric potential the derivative of ex, take the derivative of the change in any potential is gradient of the potential and energy. This is the potential. This a way of storing charge and storing energy. In the capacitance as the charge that you