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PHYS 201: Fundamentals of Physics II
Lecture 6
 Capacitors
Overview
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.
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htmlFundamentals of Physics IIPHYS 201  Lecture 6  CapacitorsChapter 1: Review of Electric Potential [00:00:00]Professor Ramamurti Shankar: 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, “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 really a reasonable question. I 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 now 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 0, 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. This is called a potential. So that ½mv_{1}^{2} (this is a little v this is big V) V(r_{1}) = ½mv_{2}^{2} + V(r_{2}). That is our goal. Ah, missed one mistake already. Anybody know what’s missing here? Student: [inaudible] Professor Ramamurti Shankar: Pardon me, yes? What’s missing? Student: The charge. Professor Ramamurti Shankar: 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 1 kilogram mass to that height you do 1 times gh, 5 kilograms 5 times gh. So depending on the mass you want to lift the work done will be different and 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. So this is the old potential energy you want. So the potential energy is q times the potential, okay? So think of V as the electrical height at a point and q is like the mass, and q times V is like mgh. Okay, we’ve got a formula for V. The general formula for V is that V of r_{2} minus V of r_{1} is the line integral of E⋅dr from r_{1}to r_{2}. 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 r_{1}to r_{2} you apply force minus E to compensate that, and the work done by you, this really work done by you to move it from r_{1} to r_{2}. 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, and that’s 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 E. That ⋅dr is the work done by you. That’s what you invest. That’s what you’ll 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 — these should be all vectors — 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’ll 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 1 charge it is q/4Πε_{0} times r for a 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^{2} and there is no vector, okay? The potential due to a charge here at the point here is a number given by this, and if you’ve got lots of charges that’s just fine. V due to many charges is sum over all those charges, the value of each charge (divided) by 4Πε_{0 }times the length from where that charge is to where you want the potential. That’s the principle of superposition, okay? 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 this 1 over 4Πε. 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 ix partial derivative minus jy partial derivative. You should expect two derivatives because the force you are looking for now is a vector. You’re living in 2D so you cannot say the force is a derivative of the potential, in what direction? If you want the force in the x direction take the x derivative. If you want it 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 or the gradient, and you go from the field to the potential by doing the integral. Again, the subtlety is the integral is of a dot product of E with a 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 do some work, E⋅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 a correct statement. It’s 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 it’s the field here times that length, but 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 have 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 and another angular segment. So you approximate the smooth pass by a bunch of radial, angular, radial, angular, and from a long distance it all looks 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, and I want to walk from here to there. So you know I go a distance √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. And these are very tiny differences. You can say it’s the same thing, but this path is of length two. 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 √2. So just because things are approximately the same it 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 when you 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 is 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 it on that segment along some straight line joining the two. What I want to show you is that not only does that approximate this, from a 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. It’s 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 and then this end. Let me call it delta r, delta r is the spacing here between that curve and that curve. We agree the dot product is trivial. E is this way. Displacement is this way, no cause thetas to worry about. That’s the work done. Work done here is 0 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(r)⋅dr . When I say E at r, r is, of course, not 1 point it’s kind of spread out, but this is an infinite decimal region, so you can talk about the E at one point and not worry about the variation. So don’t forget it’s an infinite decimal triangle. E⋅dr will be E at the distance r, the length of dr and the cosine of the angle between them which is this angle. The length of dr times cosine is precisely that distance, Δr and therefore you do the same amount of work. So basically I am saying I should wander around. You may not choose to go radial angle or radial angular. 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⋅dr does for you. That’s why you can deform the fact anyway and get the same answer. That’s the first thing. Second thing is I forgot to tell you anything about units. So the electric field is force on the 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 units 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 12 volts 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 park next to the fire hydrant. I have tenure. I tear up the tickets. I don’t do jury duty. I don’t pay taxes, you don’t have to do anything when you reach — and Alan is going to tell you the joys of being a graduate student. The joys of being tenured prof are 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. As soon as the little guy can crawl he attaches himself to a rock, and he 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, anyway. So you can, if you work hard, you can be like me, okay? 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’re right or wrong, right? For any problem the answer is 19 in some units, right? So if you don’t give the units 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. Chapter 2: Advantages of Electric Potential, V [00:15:53]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, and 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 are going. First thing is conservation of energy, there’s no need to belabor 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 if you can do kinetic plus potential = 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. It 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 every instant 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 the 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. So that’s the first advantage, conservation of energy. 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 you’ll do a dipole q and +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 explain 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 net 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 form this guy and that guy. Then once you’ve got the potential you find the field by taking derivatives. You want E_{x} take the x derivative, the minus sign. If you want E_{y}you take minus dV/dy. So what’s the E here? You remember it’s q/4Πε_{0} times [1/r^{+} (r^{+} is that length) − 1/r^{}]. That’s it. No arrows, nothing, that’s the total V. This is r^{}. Let’s define r to be if you want 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’s working. This is not an approximation. This is exact. So let’s do the following then. We come by the denominators. You get r^{+} times r^{}. Numerators r^{} − 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 2ar cosine theta or something. In the bottom we ignore that difference and we write everything as 4Πε_{0}r^{2}. In other words 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 plus maybe a number a divided by r and also a^{2} divided by r^{2}, but these are all negligible. In the numerator you can again say, “Hey, 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 in 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. Then if that angle is theta, theta is the angle then it’s very easy to show that this distance is simply 2a cosine theta because if theta is this angle then you can say cosine of this angle here will be by whatever complimentary angle is 2a cosine theta. Therefore in the top I have 2a times cosine theta, and that I will write as dipole moment p cosine theta over 4Πε_{0}r^{2}. That’s what I was trying to get. Yep? Student: Sorry, which angle is theta? Professor Ramamurti Shankar: Theta is the angle measured from here. Let me draw this. 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, therefore this difference is 2a cosine 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/4Πε_{0}r^{2}, and cosine theta is x/r. So the whole thing is x/r^{3} which is the same as x/(x^{2} + y^{2})^{3/2}, right? r is the square root and r^{3} is three powers of that. Now, you can easily take the derivative of this guy with respect to x, or with respect to. Derivatives are very easy to take. Integrals are hard to do. And that’ll give you the electric field E_{x} and E_{y} everywhere, okay? So E_{x} is equal to dV/dx, and E_{y} is dV/dy. It’s 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 V and then to find E. For example, if I have a ring of charge and I want to find the V here then let’s say the ring has charge of 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 4Πε_{0}. And what’s the distance? If that is z then simply a root of z squared + a^{2}. That’s due to this little charge, but every unit of charge on this ring is equal distance from this point. You understand? It’s a ring and I’m perpendicular to the ring. So they all contribute the same number and sum of all the dq’s is just the q. So that’s the potential on the axis. Remember, 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, where 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. Let’s say there’s a hollow conducting sphere. It’s got charge Q. So 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’s going to depend only on the radial distance from the center because it has 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 it 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 because it’s spherically 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/4Πε_{0}r^{2} as long as r is outside this sphere. So here’s one problem where it’s easier to get E because E is pretty trivial, it’s radial. 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 this sphere it’s as if Q were sitting at the origin and the potential V(r), which is Q+Q/4Πε_{0}routside that sphere, because remember V(r) minus V of infinity is equal to integral of E⋅dr from infinity to r with a 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 falling like 1/r. Now, what about for r less than r? How do I continue this graph? Student: [inaudible] Professor Ramamurti Shankar: Pardon me? Student: Use Gauss’s Law. Professor Ramamurti Shankar: You can use Gauss’s Law. If you use Gauss’s Law inside what will you get for E? Student: 0. Professor Ramamurti Shankar: E is 0. Integral of E⋅dr is 0, so potential does not change inside this sphere, but don’t think potential is 0. It does not change. It’s stuck at this value. So throughout the hollow sphere there’s only one potential, okay? Electric field is a different story. Electric field is, in fact, 0 inside then falls like 1/r^{2}. The electric potential is a constant. So 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 then integrate that 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 has the advantage that it lets us compute the electric field by first doing a sum of scalar quantities rather than vectors, then taking derivatives. The third advantage has to do with being able to visualize things. It gives you a landscape and that’s intuitively very helpful. So here’s one example. Let’s take two parallel plates and 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 σ for unit area and they’re infinite. I’m just showing you a finite portion. This guy will produce sigma over 2ε_{0} here and here. This one, because of negative charge, will do this; σ/2ε_{0} 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’s coming in. That guy’s going out. And remember, 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 has the strength which is double what you have on either side if you had only one sheet. That’s the electric field E. That’s my electric field. You 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 = 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 this statement. You cannot pick V equal to 0 over an extended region. You have freedom to pick it to be 0 at one point, but 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 in the lower plate, but it won’t be 0 anywhere else, but you can see that may be V = 1, that may be V = 2, V = 3, upper plate maybe V = 4 volts. So lines of E are vertical and lines of constant V are horizontal. This is really like the gravitational 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 the charge. It’s straight down and the potential just increases with height. Yep? Student: Can you just go over real quick, please, how you do the calculus to get the constant voltage _________? Professor Ramamurti Shankar: Oh, okay. Think of it as a plate coming outside the blackboard. Student: For the hollow sphere. Professor Ramamurti Shankar: Oh, for 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 is a point charge at the origin, so I pushed up against that until 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 V that you got up to that point. You don’t drop it. You 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’ve got this 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 4 volts here, if you take a coulomb and you let it drop it’ll pick up 4 joules by the time it hits this lower plate. That’s the meaning. That’s 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 1 coulomb here. If it was 10 coulombs 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. See, if an electron is released here in fact it’ll go like this. You don’t have any analogs in gravity. Gravity everything falls down. Suppose some things had negative mass and some things had 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 q times E, so everything has got a minus sign. So an electron going from here to here will in fact lose energy. We will lose potential energy and it will gain and turn in kinetic energy. So an electron falls through 1 volt. The work done is charge of an electron times 1 volt which is 1.6 times 10^{19} joules. That happens to be called an electron volt, just a convenient thing to use. When the charges that are pushing around are not coulombs but this tiny electron it’s good to know in electron volts. For example, in the hydrogen atom the coulomb potential is negative because the charge at the origin is a proton +q (electron has got some q) divided by 4Πε_{0}r. 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 electrons, 13 point 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^{19} joules of work. So as if this were 13 volts down compared to infinity. It’s a hole in the ground 13 point 4 or 6 volts deep. So if you want to liberate that electron you have to furnish that amount of energy. Okay, now going on with the 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; we’ve done it a million times. Now, what about the electric potential? For a point charge the potential is 1/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/r. You’ll 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 understand if you’re very near one charge forget about the rest of the world. It dominates everything. Potential would look like circle 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 plainer 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 or equipotentials are 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. You remember that the V_{2}  V_{1} is  integral of E⋅dr from 1 to 2. So let’s say V_{2} and V_{1}, 2 and 1 are very close. In that case let’s call it delta V = E⋅dr. There’s no integral to do because I’ve picked two points which are near by an infinite decimal amount dr, so E⋅dr is equal to delta V which is a minus of V_{2}  V_{1}. 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 the electric potential is replaced by height in two dimensions, you know, top of Sleeping Giant 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 onestep, 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?” And cosine theta is 1, right? Because then there’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 in that direction. If you release a marble in the gravitational field it’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 quickest time if you can compute quickly enough, okay? But that’s the process. At every point there’s a 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’s us say you’re on this mountain and you don’t want to go down and 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? Student: You move perpendicular to the gradient. Professor Ramamurti Shankar: You move perpendicular to the field. To move perpendicular to the field the cause theta is 0. That’s why, now going back to this, this is called Mt. Coulomb, 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 of the field, but if 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 equal potential is always — in fact in general the change in any potential is gradient of the potential dot dr. Except for the minus sign it’s the same story that if dr is parallel to grad V or antiparallel 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 mountaintops, and these lines of constant height are like constant V, so it helps you visualize. And then if you’re told you release 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. Chapter 3: Conductors as Equipotentials [00:43:32]Okay, 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. At the boundary of the metal there are forces bringing them back. Inside they are free to move. Therefore, if you take a conductor and you stick it into a field initially the electric field will penetrate the conductor. The charges, because they are free to move, will move. The negatives 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 0. So here’s a generic conductor. If you put it in an external field it will have some charges on its surface, will produce a field this way that will oppose the external field in which you put it until at the end there’s no field inside the metal. Or, if you did not put it in the field and 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 the metal are in the boundary. There is no electric field inside because of Gauss’s Law. If I want to find the charge here I do a tiny sphere and do integral E⋅da. I get 0, and that’s the 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 is no field inside. Q = 0 inside metal. This is a solid metal, and E = 0 inside solid. The other variation you can have is suppose they 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’re not in a static situation. Let the charges move. They’ll keep moving until 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 0, but that does not preclude the possibility that there are maybe some 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 neutralize, 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 that’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 0 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 a field inside the cavity let’s take that path and another path which are two ways to go from here to here. On one path I get a line integral of E to be not 0. Second one I get line integral equal to 0 because there is no E, but 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 an 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 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 the hole will not have charges on the edge either. So now, what else can we say about conductors? Today I used 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. Second thing is every conductor is an equipotential. It’s an equipotential. Why is that? Can you tell me why that is true if it is not equipotential, the whole lump? Yep? Student: If it wasn’t equal potential then there would be an electric field inside there. Professor Ramamurti Shankar: And how do we know that? Student: Because there’s a potential difference ________________. Professor Ramamurti Shankar: That’s right. In other words, if you take the gradient, namely taking derivatives, you’ll get a nonzero 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 it’s got charge on it that’s the same, and what it means if you start at infinity the work done to come there, the work done to come there, the work done to come there are 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, a sort of 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 challenge. 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 are positive charge and negative charge here what’s the force of attraction between this charge and the plane? In general that’s a very difficult problem. If I gave you a potato, aluminum potato, put a charge next to it, it will have a force, but it’s not easy to calculate it. 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. Forget 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 at somewhere there? You cut it right down the middle. So this plane was an equipotential at V = 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 are equipotentials, and solve problems in which a charge is in front of a conductor if you’re lucky enough to find the 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. That you can easily calculate at all points on the plane. And once you’ve got the electric field you remember σ/ε_{0} is the electric field so you can find σ, the charge density. And if you integrate the charge density you will get exactly q, because the number of lines coming from here have to terminate on the plane rather than on this and that’ll be q. And 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. Students: Same as that of the other charge? Professor Ramamurti Shankar: Same as the force that this negative charge would exert on this positive charge, and I’ll tell you why. This guy is looking around, okay, seeing some forces. It’s due to these negative charges on the plane, but they’re identical to the field created by this guy. So it cannot tell the difference and therefore 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 which is q times 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 equipotential. 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 equipotential and it’ll happily sit on equipotentials because there are no forces on the surfaces of the conductor moving the charges. So the moral of the story is you can sneak in a solid object bounded by the surface of an equal potential for any problem. So if you can solve certain problems and the equal 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. So it’s like this charge looks into 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 with energetics.The question is I want to bring a whole bunch of charges, q_{1}, q_{2}, q_{3} 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 is my goal. I’ve got to do some work, bring everybody together. They may 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 q_{2} times the potential here, potential there is q_{1}/4Πε_{0}r_{12}, r_{12 }is the distance between 1 and 2. Now I nail the two charges. Nobody is 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 infinity to here due to these two charges. So that means the work I’ll have to do is equal to q_{1}q_{3}/4Πε_{0}r_{13} + q_{2}q_{3}/4Πε_{0}r_{23}. You follow that? That’s the work that I have to do. The first guy, q_{2}, 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 q_{i}q_{j}/4Πε_{0}r_{ij} where i goes from 1 to — let’s see, from 1 to n, where j goes from 1 to n, but i cannot be equal to j and you divide by 2. So 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 the condition that i should not be equal to j times all these. You can see that’s valid here. Yep? Student: V or U? Professor Ramamurti Shankar: 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/4Πε_{0} is the potential, but I multiply by q_{2}. That is actually an energy, so these are all energies denoted 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 q_{1}q_{3}, then again it counts q_{3}q_{1}, but there’s only one such thing so I divided it by 2 and I have nothing like q_{1}q_{1}, 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 it’ll 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 couple of coulombs come in they don’t run into anything, so I put some charge here. That makes it harder to bring the next set of coulombs and 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/4Πε_{0} 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 capital Q will be a Q^{2}/2 times 1/4Πε_{0}R. Sorry, capital 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, so the work done is the integral of x that becomes x^{2}/2. Similarly to bring in an extra charge you’ve got to worry about how much charge is already there on this sphere that’s opposing you with this potential. You work against that by bringing dq from infinity and you integrate it you get q^{2}/2. So this is a good problem to explain the difference between potential and energy. This is the potential and this is the charge I brought from infinity, and that together is the du and integral of the dU is my total u. So one can ask, “What’s the energy to produce any arrangement 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 can find the work to add a thin layer to it and integrate it. Chapter 4: Capacitors [01:01:48]Okay, so final topic is one of capacitors. So a 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 lake here at the foot of the tank, and I go to the top, and go to the top and empty little buckets until I’ve got some water at the 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, and it may turn a turbine blade and give me back thermo electric power. You want to do similar things electrically. You want to invest with some work and it’ll give you back. So one thing 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 are both electrically neutral, but they’re conductors. 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 are not the same, but this whole conductor has one potential, this has 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 0 by convention. That is called the capacitance. That’s the ability of that system to hold charge. If they’re very, very tiny metal spheres there’s some amount of work. If they’re huge metal spheres you can transfer a lot of charge, okay? That’s the definition. It’s measured in farads. Again, it’s got some units coulomb for volt, but again it’s got a different name, named after Mr. Faraday, measured in farads. Usually you will have microfarads in a lab. Farad is a huge unit. So let’s find the capacitance of certain simple arrangements. Simplest arrangement is the parallel plate capacitor. Here it is. There’s one plate. Here’s 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 transferred. So suppose I transfer some charge Q. My 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⋅dr is just E times d. E, if you remember, is σ/ε_{0} between the plates, and sigma is Q/A. Charge density is the total charge over that. And that is by definition Q over C, therefore the capacitance of this parallel plate capacitor is ε_{0}A/d. So 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 — hey, where is all these blackboards? Ah, bad choice. So you guys took all this down? Yes? Student: Should that A be on the bottom for the A Q over ________________? Professor Ramamurti Shankar: Oh, yes. σ is Q/A. Yeah, this is what comes from knowing the answer. This is what you guys do in the exam. When you turn that page then 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. We’re getting there no matter what, but you’re quite right. I didn’t put the A in the right place. So any 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. Here’s another problem you could do. These are also common. Spherical capacitors are one sphere of radius a surround by second sphere of radius b and you shove charges, maybe plus on this one and equal number of minus on that one. So what’s my goal? Goal is to find the electric field in the region between them, right, and integrate it from here to here, and 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 4Πε_{0}r^{2}, because by Gauss’s Law if I draw a sphere here that encloses charge Q, and it’s isotropic and uniform, so 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 V_{2}  V_{1} will be Q/4Πε_{0} times [1/a  1/b]. Well, we are done because this is what I call V. The capacitance is Q/V and that’s going to be 4Πε_{0} divided by [1/a − 1/b] which is 4Πε_{0}ab/(b − 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. 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. 4Πε_{0}ab/(b − 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 parallel plate capacitor? Yep? Student: Make the a close. Professor Ramamurti Shankar: a what? Student: b  a is close. Professor Ramamurti Shankar: Right, you can take b  a very close, but it still may not look like a parallel plate capacitor. So here’s what I say we do. Make b and a astronomical, size of the galaxy, okay? And b  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 galaxy is just like saying the earth is round which is not obvious from short excursions. In that case we can write 4Πε_{0}. a and b are really equal to some common number R and b  a is the distance between the inner shell and the outer shell. That you can see. Wait. I’m sorry, 4ΠR^{2} is the area times ε_{0} over d, right? So that’s one test. It doesn’t mean it’s right, but if this test failed it means you’re definitely 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, the delta q times the voltage at that stage. Voltage at that stage is equal to q/C, which is the q you have up to that point. Therefore, the total work you do is ½q^{2}/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 it as ½CV^{2} because Q/C is a V. Multiply top and bottom by C. You can write the energy in a capacitor either in terms of the charge stored or in terms of the voltage between the plates. So we’ll resume this next time. [end of transcript] Back to Top 
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