Fundamentals of Physics II

PHYS 201 - Lecture 7 - Resistance

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

Professor Ramamurti Shankar: When I look at you guys, I realize that I don’t know what most of you look like. That’s the problem in a big class like this. Usually it would be nice if you know who your students are. If they say hello, you can say hello back, but I don’t know what I can do about it, because if I focus on you, I forget what I’m going to talk about. But you know what I look like, and it’s kind of an unnerving asymmetry, because you might become my doctor one day. You’ll be wearing a mask and carrying a knife, and you’ll be thinking of problem set number three, right? And I won’t even know who did it, so it’s very asymmetric. All right.

Okay, so anyway, maybe I’ll get to know you if you ask a lot of questions, or your favorite pastime, finding something wrong on the blackboard. Look at that. I didn’t do that. I cannot even reach that far. So let me now go back a little bit, because I think I rushed some of the topics near the end and I thought about it some more, so I think it will be helpful to revisit this question about image charges. I want you to think about what the point was.

So here’s the main thing: you all know how to think about the potential or the field, right?

I give you a bunch of charges and you calculate the potential and you take derivatives, you get the field. Sometimes that’s not the only kind of problem you have to solve. You may have to solve the following problem. There is some chunk, maybe a potato, and you bring some electric charge, what does it do? That’s a meaningful question, because it does something, right? And nature calculates right away and does something. But if you want to figure out what happens, it’s quite complicated, so a somewhat easier problem is the case of the perfectly conducting potato, which is kind of solid metal, shaped like whatever you like. You bring a charge next to it, what happens?

That question occurs a lot, because we do electrical experiments in the presence of many conductors. And the simple thing about a conductor is, the whole conductor is one potential. So I want to take a simple problem and see if we can answer it. Now most problems, you cannot answer. By “cannot answer” I don’t mean in principle. In principle, we know the equations, we can solve them, including the real potato. You can actually figure out everything, but it’s too hard. You cannot write it in simple form. There are a few simple textbook examples where you can ask the question and you can answer.

And the question you want to ask is, here is an infinite conducting plane and it is grounded. Grounded means what? You take a wire and you connect it to the earth. Now the point of connecting to the earth is that the earth is pretty close to being an object at 0 potential.

In other words, if you bring a charge from infinity to the earth, the amount of work you do you can take to be 0. And on a daily basis, people are dumping charge, people are taking charge out of it, but it’s so huge, it doesn’t matter. It’s like one of the reservoirs you studied in thermodynamics. You can take some heat, you can put some heat. It doesn’t change its temperature. So the earth is a huge repository where you can take a few charges now and put a few back. It will always remain at 0. It will force you also to go to 0 potential.

So that is that plane, and you have another charge, q, you slowly bring towards that plane. You may like to know what’s the answer to this problem. What does the field look like everywhere? Now this plane, for convenience, I’m going to be taking to be infinite. If you want a side view of that plane, it looks like this. It divides the whole universe into two parts, where you are with your charge and the rest on the other side.

So what will this plane do? You can see that if it did nothing, it’s in trouble. If it did nothing, this charge at that distance will produce a positive potential. At that distance, it will produce an even bigger positive potential, because it’s closer. So different parts of the sheet will be at different potentials. That’s not allowed. It’s an equipotential. It’s a metal. It’s got to arrange its potential to be constant, and because it’s connected to the ground, that constant has to be 0. So the way it will lower its potential from your attempts to raise it is to suck up some negative charges from the ground. So negative charge will leave the ground and somehow come and stand in front of this positive charge, maybe something like this, over some region roughly proportional to the distance between you and the plane. Then what will happen is, the electric field lines, which normally go through the metal if it wasn’t there, will now of course have to come and terminate on that metal, and they have to terminate perpendicularly because the electric field is always perpendicular to the conductor.

You know why, right? Because if you move along the conductor, the potential cannot change, so the line integral of E must be 0, but basically, you should not feel any force as you move along the plane. So E is perpendicular. So some bunch of charges will be drawn from the ground, as necessary, and this is what will happen. The question is, can you say anything more in detail? Can you actually calculate the charge distribution on this infinite plane? Can you calculate the force of attraction between the charge you have and the infinite plane? You know there’s going to be attraction because your charge I take to be positive, and this has got negative. They’ll attract.

What is the attraction? These are all well-defined questions and they exist no matter what this is made of. But for this infinite plane, we can actually answer it

So we answer by the following device, which is what I was trying to tell you last time. If you take that infinite plane, your conditions are that it should be at 0 potential and to the left of it should be the charge q that I put in. These are the requirements. So you dream up another problem. That’s a charge -q, same distance d from this one. Take out the plane. Forget the plane now.

Think of a new problem, + and -q, and you’ve done that many, many times. The lines go like this. And you know that on the perpendicular bisector, the lines will be perpendicular, because the potential everywhere is 0, right? Potential is a scaler and whatever this guy does, that will do - of that, because it’s at the same distance, but has -q. So V is 0, so E will be perpendicular. Now if you take that arrangement and look what’s happening to the left of the plane, it looks like exactly what we want in our problem. What we demand in our problem is in this part of the universe where I live, there should be charge q, there should be infinite plane at 0 potential. That condition is satisfied by this field configuration in the following sense. Start with this configuration and stick in the infinite plane, but give it the charge it needs to terminate all these field lines. How much charge will it need?

You can tell right away. What’s the total charge on this plane? Because all the lines leaving this have to terminate on this. You can see it’ll be -q. It won’t be point charge; it’ll be spread out, but you need -q on this. So what we say is, if you brought the plane in and either you give it -q in advance, or you connected it to the ground, in which case it will itself suck up -q. It’ll form this configuration here, so the infinite plane is able to remain where it is without disturbing the solution to the problem. Therefore the answer to the original question of what happens when you put a q in front of an infinite plane is that as far as you’re on the left of this, you can compute anything you want by taking q and the image charge, -q. The field the two of them will produce will be exactly the field this guy and all these induced charges will produce.

So this is a way of taking a solution for a simple problem with an equipotential, sticking a conductor into the equipotential, and giving it the right charge. Then the problem of a charge in front of a conductor is thereby solved. Now there are some mathematical theorems — I don’t have time to prove but they’re not too hard in fact. I thought of lecturing on that, but there isn’t enough time — that show that if you can find one answer that has the right behavior at the boundary of the region, that has got the charges in the right place, is the only answer. Yes.

Student: Is the -q evenly distributed across the plate?

Professor Ramamurti Shankar: Let us ask. Where is the -q, right? We know that it’s got to be -q because these lines are all ending here. But what’s your intuition? Where do you think it will be? You think it’ll be evenly distributed? Pardon me?

Student: I don’t think so.

Professor Ramamurti Shankar: Where will it be? What’s your feeling?

Student: Probably more towards the middle.

Professor Ramamurti Shankar: Yeah, more towards in front of this guy, and less when you go up. Now we can actually calculate that quantity. We can tell you exactly how much it will be by the following trick: we agree that to the region to the left, this charge is able to fake the effect of all these charges. That’s why it’s called a mirror charge. When you stand in front of a mirror, you are here and there is another person in the mirror. And the light bouncing off and coming here looks like it’s coming from this image. So to the left of the mirror, either you can talk about you and the mirror and the reflected light, or forget the reflected light. It’s as if there’s another person behind the mirror. Of course, you know there is no other person behind the mirror, right? I don’t want to be the one to break that news. There isn’t. But as far as this side is concerned, it’s a lot easier to draw a line from the mirror, because that will simulate the effect of what’s happening through the mirror. Similarly, as long as you are interested only in the left of this region, you cannot tell the difference between the infinite plane with its induced charge, and this guy, versus this guy and just one other charge.

Another question was, where is the charge and how is it distributed? You agree that if you knew the electric field here, you can find the charge density, because if you draw a little Gaussian surface like that, then we know the σ/ε₀ is the electric field. So if you know the electric field, you can find sigma, and I find the electric field very simply by saying that’s the repulsion from this q, there’s an attraction from that q. If you add the two, you get the sum, which I can calculate, a very simple sum. It’s the contribution from two point charges. You can calculate it as a function of this angle here or as the function of the distance here, and that will tell you what σ is. And it will have the property you expect. It’ll be largely in front of this guy and will vanish very quickly when you move away from him. If you integrate that surface charge density over the plane, you will in fact get -q. So you’ve answered some of the questions, namely, how will the charge distribute itself on this infinite plane?

Next thing you can ask is, what’s the force of attraction between this one and this one? Now, is it really the same as this one and this one? The answer is yes, and the reason is the following: in this region, for example, the field is due to this guy, and due to all these, but that’s the same as this guy and this guy. This one of course doesn’t respond to its own field. It responds to the field of all of these. But all of these precisely have the same field to the left as this one charge, therefore this will be pulled towards the one charge. It thinks it’s falling towards the other charge, but it’s not actually falling towards the plane. And the force will be q times -q over 4Πε₀ times 2d, right? That’s the force of attraction. It will be −q²/4Πε₀, the distance between them is 2d, so you can calculate that too. Normally it’s a difficult problem, because the image charge reduces the problem of a very complex conductor to that of a point charge. Yes?

Student: Shouldn’t it be 2d squared?

Professor Ramamurti Shankar: Yes, thank you. Then you can also ask, what is the energy it takes to assemble this distribution, starting from this one at infinity and this one here, and if you integrate this force, you can do the integral and you’ll find −q²/4Πε₀ divided by 4d. That will be the energy. Naïvely, you may have thought it should be (q²/4Πε₀)2d, because 2d is the distance between them, but it’s only half as much. It’s half as much because if in real life you had a + and a -q and you brought them together, you will be exerting forces on both of them and calculating the work done. But in this actual example of an infinite plane, you only do work on this guy, therefore the answer will be half. So not everything is the same. For example, the energy in the electromagnetic field, which I’ll tell you about later today, will be half as much as before, because in the fake problem, the energy is here and here. In the real problem, the field is non-zero only here, but 0 here. Yes?

Student: For the potential energy, is it 1 over 4d or 1 over…?

Professor Ramamurti Shankar: You would think it’s 1 over 2d, right?

Student: Yes.

Professor Ramamurti Shankar: But I’m telling you it’s 1 over 4d.

Student: Because when you take the integral, you have 2 on… okay.

Professor Ramamurti Shankar: All right. Now what if this plane was not grounded, it’s a standalone plane? So let’s not take it to be infinite in this problem. It’s 1 million miles in radius, huge disc. You bring the charge in. What’s it supposed to do? It cannot suddenly acquire a negative charge. Charge conservation says you only have the charge you have, which was 0. So what do you think you will do if you have that infinite plane? You have to remain at some potential. Yes?

Student: Push positive charges out?

Professor Ramamurti Shankar: It will split into positive and negative charges. Let’s see what to do with the positive charges in a minute. The negative charges, it will arrange exactly the way it wants to in this problem, okay? That takes care of this being at a constant potential of 0. But it’s got an equal number of positive charges, which you’ve got to dump on this infinite plane, without ruining the constant potential. The way to do that is to put it uniformly on this plane. If you spread it uniformly on the plane, you don’t destroy the equal potential nature. But it turns out that field will be very, very, very small. It will be very small, because you’re taking a charge q and you’re putting it on a huge disc, like the radius of the galaxy. That’s the charge density and that over ε₀ will be the electric field. You can see that that can be made vanishingly small.

So the infinite plane, basically you borrow from somewhere a positive charge and negative charge, and you put the negative in this particular arrangement. Positive is smeared over the whole thing uniformly. It is so dilute, because of the size of it, it doesn’t matter. But it principle, that amount of positive charge will be a background constant. That way the infinite plane will be a little strange, because you don’t see the richness of the phenomenon of the infinite plane because it can basically create negative charge and not fully account for the positive charge, because it’s smeared over a huge distance.

So a problem that’s more interesting would be this one. I take a sphere of radius a, it’s a conducting sphere. And after distance b, I put a charge q. What will happen now?

Again, now let me say this is a grounded sphere, meaning it’s connected to the earth. It’s maintained at 0 potential. So it’s not very different from the plane, except you’ve got a huge sphere now instead of an infinite plane, but it’s finite. You come around from infinity, you can see that some negative charges will be induced on this, and some lines will come here. And some lines may go past like this. Because this is a finite sphere, not every line has to terminate here. Now the question is, what will be the charge distribution? How will it arrange itself, and what will be the surface electric field and so on? That’s what you want to calculate? So what kind of problem do you think we should try to solve? If it had an easy solution, what form do you think it will take? Yes?

Student: You could put a charge on the other side of the sphere?

Professor Ramamurti Shankar: Okay, that’s not a bad guess. If you put a charge on the other side of the sphere, then in the universe in which you’re living, namely outside the conductor, things have changed, right? In the image charge problem, remember, the fake charge was never in your universe. It was outside. The problem you defined yourself should remain the same where you are, but you can do stuff inside the region that’s excluded from you. So you’re almost right. We want to put a charge inside here. And where do you think you should put it, roughly? Can you make a guess where in the sphere it will be?

Student: Center?

Professor Ramamurti Shankar: Center is a good guess, but it turns out, if you put it in the center, this doesn’t turn out to be a sphere. But you agree that it should be on this line, just from symmetry. You don’t want to put it anywhere else. So I will give you the answer. It’s not very hard to calculate. If you put a charge q’ at a distance s, which is a²/b, and the size of q’ is -qa/b, a/b is less than 1. So q’ is less than q, and is at a distance s from the center, which is a²/b. This is something you can do for fun. You can go and calculate now the potential of this negative and smaller charge together with this positive charge q and you will find, if you put q’/r for that, and q/r for this guy, it will be exactly 0 on this sphere, centered here.

So it’s another fortunate thing that you can get a spherical equipotential by taking two unequal charges — not two equal, two unequal charges — and putting it slightly off center by this amount. Let’s not do the algebra now. You can all imagine doing q’/r’ and you’ll find it’s equal to q/r, up to a sign, and therefore it will be in fact — you can always make two things cancel at one point, but they cancel over an entire sphere. It’s very amazing, but true. Yes?

Student: Are the field lines always perpendicular to the surface of the sphere?

Professor Ramamurti Shankar: They must end up perpendicular. Tell me why. Can they end up anything but perpendicular to a conductor?

Student: Oh, right, no.

Professor Ramamurti Shankar: No, because if it’s got a tangential part, it will move the charges. So charges, once they’ve come to equilibrium, the only force you can apply to a conductor is normal to the surface. So the lines will terminate in the normal way. So the problem you want to solve, the fake problem, the image problem, is a −q’ and a +q so located you get a sphere.

Now you go and say, let me take another problem of a conducting sphere. There’s nothing in it, grounded. What it will do is suck from the ground this charge and spread it exactly the way you want in this problem. And then it will come to 0 potential. So ground is here, 0 potential. If you want to know, what is the electrostatic potential there in this problem, we can do that now.

What’s the electrostatic potential here? It is due to this guy at that distance and the fake charge at that distance. You should add them with the proper sign. So the problem of the field due to very complicated distribution of charges on a conducting sphere and a point charge is reduced to the problem of 2 point charges. Very easy to add their potentials, take the derivative and find the field. If you want to know the surface charge density, you find the electric field at the surface. It will come out to be perpendicular. Then σ/ε₀ = E, therefore if you knew E in magnitude, you can find sigma and you can find the charge distribution on the whole sphere. If you integrate that guy, what will you get for the charge distribution on the surface? What should it give for the total charge? Pardon me?

Use Gauss’s law for the real problem and the fake problem. In the real problem, surface integral of E is the charge on the sphere. But the same Eis produced in this combination with this guy here. The surface integral of E is a charge enclosed, which is q’, after some epsilons. Therefore the charge on the sphere will turn out to be exactly the same as the q’. Just like in the plane, there’ll be some negative charge, but it’s spread out in a particular way and you can verify that.

Now a more interesting question is, what if this is not grounded? In other words, it’s an isolated sphere and you’re bringing a charge near it. It wants a certain negative charge to spread on the surface to produce a zero potential here, but you don’t have negative charge. You have zero charge.

So what will you do if you are that sphere? You will say 0 = q’ + -q’. You would split into q’ and −q’. The q’ will arrange itself on the sphere to exactly terminate these field lines. The only difference is, now you’ve got a charge, -q’, left over. You have to spread it around and you don’t want to screw up the equal potential nature of the sphere by doing that. You can all guess what you should do. You should spread it uniformly on that sphere. So it will contain negative charge distribution, which is biased in this direction, and a positive charge distribution that’s uniform in it. Then the three of them together, this one, the positive charge, I put here of −q’ and this one, three of them together will keep the sphere at a constant potential.

The potential it ends up with will however not be 0, because q’ and q made it 0, but you have a −q’ at the center, therefore you must find the potential due to −q’ on the surface of the sphere, which is that. That’s the potential to which the sphere will come.

So we have answered many questions. We have said if you bring a charge q near the sphere, what potential will it acquire? It will acquire exactly this potential. You can predict the charge on the sphere. You can predict the force of attraction between this and that sphere. How?

Because the field that this one feels generally, the field at any point is due to this one, this one and this one. But if you want the force on this one, if you’d only find the field due to the other two guys. But the other two guys, namely the sphere with all the charges on it, is simply equal to 2 point charges. So you must find the force between these 2, add to that the force between these two and you’ll get something. That will be the force. It will be a force of attraction, but you can actually calculate it. This is the trick by which you can solve a variety of problems by finding equal potentials of nice shapes. Maybe one will be a nice ellipsoid, because if it is an ellipsoid, you can stick an ellipsoid there. And inside the ellipsoid will be the image charge. And if the ellipsoid is grounded, you’ll get whatever charge is needed to maintain that at 0. If it’s not grounded, it will split into positive and negative charges, where the negative will distribute this way, and the positive will distribute itself in some way, so that the potential is a constant. For a sphere, we know what that some way is, which is uniform.

So the reason I took some time to describe this is that problems are not always finding the potential due to a bunch of charges. That’s the easiest problem, but more generally you are asked, if you’re given a set of conductors and a bunch of charges, what happens? If the conductors have a nice shape, like a sphere or a plane, or certain solids of revolution, we can appeal to a different problem with image charges and solve it. Now there’s a wonderful theorem that says if you get a solution this way, it’s the only solution. In other words, there’s a theorem, which I’m not going to write down, which says if in all of space that you’re interested in, or at least in a region bounded by something, if you know the potential of the boundary and you’ve got a whole bunch of conductors, each one at a known potential, and you’ve got a whole bunch of charges, there can be only one potential function, V(r), in this whole region. You cannot have two answers.

There’s only one answer to that question. In other words, the potential is completely determined by knowing the charge distribution, and the potential on the various conductors you stick into that. They don’t all have to be to 0. This can be at 5 volts, this can be at 9 volts, this can be at 13 volts. This can be 1 coulomb, 2 coulombs, 3 coulombs, -10 coulombs. This is say this sphere at infinity of 0 potential, there’s only one answer. It’s called the uniqueness theorem, which is why, if you can find some way to fudge the answer in a given region, which is completely specified, that is the answer. So that is the interlude on metallic objects called conductors.

Chapter 2: The Parellel Plate Capacitor [00:28:45]

Now I’m going to go to the other problem that I did towards the end, which is the notion of a capacitor. So if you take two blobs of metal, they are both neutral. Then you grab maybe a coulomb from this and stick it there, so that becomes positive and this becomes negative. Then you want to take more, you take more stuff, you can see that you’re going to run into resistance, because these guys are getting positively charged. They don’t want more positive charges. Meanwhile the negative charge you leave behind wants it to come back, so you’re working against that. But you do some work and you start pumping charge into this. So in the end, suppose you have a charge Q there and charge -Q there. Then there will be a potential energy difference between the two, because there’s a certain amount of work needed to go from here to there. The potential difference is a unique number, because the whole solid has one potential, other solid has another potential.

No matter where you start and where you end, if you find the work done, that potential difference we like to call V. Therefore V is always going to be proportional to Q and the constants of proportionality we like to write downstairs and call it capacitance. So that’s the ability of the system to hold charge. If you’ve got any two metal containers, you can store energy by drawing the charges from one to the other, putting them there. I did calculate the capacitance for a very simple system and I’m going to stick to the simple system because we don’t want to get lost in the details. It’s the parallel plate capacitor in which I put some charge +Q on the upper plate and -Q on the lower plate. Someone says, what’s the capacitance of the system? For that, you must find the voltage difference between the two plates, and you want to take Q over V. Now there’s going to be an electric field here, E, which is σ/ε₀ pointing down. So the voltages difference will be the electric field times the distance, because that’s what the line integral will be. That’s the distance between the plates. And σ is Q/A.

Now in a real capacitor, near the ends there’ll be some funny business. It’s not going to be simply like this, but we take it to be so large in this extent that the edge effects are neglected. Then this formula is valid. Then you can compare it to Q/C and you can see C is ε₀a/d. So this is what I have done towards the end of class. That’s the capacitance of that. Once again, you can talk about capacitance, but you cannot always calculate it. You can take two irregular metallic objects and in principle for a transfer of charge q. There will be a voltage v, and the ratio is the capacitance, but you cannot compute it. But for a parallel plate geometry, or another one I did in class near the end, you take two concentric spheres and put some charge on this one, maybe + and - on that one, you all know how to find the field in the region in between. You can integrate it and you can find the potential difference. And then you can see it is proportional to Q, and you take the ratio, you’ll get the capacitance of that.

Now the point, one thing I want to calculate, which also I think I started doing is, if I take a capacitor and I charge it, from 0 to some charge Q₀, what’s the total amount of energy stored in the capacitor? So the way I think about it is, I take some intermediate situation, when there’s a charge Q on the capacitor, not the final amount. Then I want to take a little amount of charge dQ and I want to move it from the negative to the positive. Since at that stage the potential difference is Q/C that potential difference times dQ is the work you do. That’s the definition of potential difference, how much work it takes to move a coulomb from one plate to the other. If you’re moving dQ coulombs, that’s the work you do. You integrate that from 0 to some maximum value Q₀, you see you get Q₀²/2C. So the energy in a capacitor is q²/2C. Forget the 0, subscript 0. Usually Q stands for the charge of a capacitor.

But you can also write it in another way. Since Q = CV, you can write it as ½CV² .

There are two ways to write the energy. So the energy in the capacitor is ½CV², but I’m going to play with that expression and get a very interesting result, very profound. Capacitance is ε₀A/d, what about the voltage? It is E/d. Sorry, voltage E times d. E squared, d squared. So what do you get? You get ε₀E²/2 times a times d. What is A times d? It’s a volume of the region that contains the electric field. Before you charge a capacitor, there is no energy and there is no field. Once you charge the capacitor, you’ve got a field. And if you wish, you can say I’ve got to ascribe that energy to the fact that it’s a non-zero field, in which case, this is the energy per unit volume, due to an electric field. That’s a very interesting notion. Little u is the energy per unit volume and that happens to be ε₀E²/2.

In other words, it takes energy to establish the electric field. So in this room, if you’ve go to a tiny region where E is essentially constant over the tiny region. ε₀E²/2 times the tiny volume is the energy in that tiny region. So once you’ve got an electric field, it just cannot just disappear. Law of conservation of energy will require that you account for it, and this is the energy per unit volume. It turns out to work even if you’ve got radio waves, electromagnetic waves, going anywhere, to take that energy, electric field squared times epsilon over 2, that’s the energy density.

All right, so this is really the end of what I wanted to finish last time, but I wanted to go back to the study of equipotentials and conductors. But now I am setting the stage for electrical circuits. Now this is the kind of thing some of you probably did somewhere in high school. So how many people have done basic circuits in high school? So I’m going to assume you’ve done some of it, so I won’t do it in that detail, but I will mention all the essential facts. For the few of you that didn’t do it, you have a chance to keep up with the class. The first thing in electrical circuits is, you’ve got some wire and you’ve got an electric current flowing in it. We need a description of the current. The current is defined as follows. Imagine this is the perfect cylinder, cross section A. You cut it somewhere and you watch all the charges go by, and you see the number of coulombs that go by per second. That’s called the electric current and is measured in amperes. So 1 coulomb per second is 1 amp.

All right, now let us ask, what’s the connection between the electric current and what’s going on microscopically? So let n = number of carriers per unit volume. Let e be the charge of a carrier. Now here is one of the biggest nuisances in life. As you know, in a wire, the current is carried by electrons. Because the charge is negative, when you draw a picture like this, the current to the right, electrons are actually moving to the left. What we will instead do is to just keep an eye on the direction of the current. You imagine there are positively charged objects carrying the current in the direction of the current, whereas in reality, it’s negatively charged objects moving in the opposite direction that produce the same current. So you can ask yourself, if you wait 1 second, how many coulombs will go past this checkpoint? You can see that it’s a cylinder whose length is the velocity, because in 1 second, it will have gone v times 1 second. All those guys have crossed the finish line.

Therefore the current will be A times v (is the volume of stuff that’s gone), that’s the number of carriers in the volume, that’s the charge of each carrier. That’s the total current. So we like to define a quantity called the current density, which is the current per unit area. That will be equal to nev. And actually, current density is a vector so if you want, because velocity is a vector, you can make it a vector like that. In this problem, the current density is uniform. In fact, in any wire, the current here and the current there and the current there do not change, because if it changed, charges will pile up in some places like a traffic jam. Then it will resist it till this thing evens out. Current is constant on a wire. Need not be uniform across a cross section of the wire, but I’m going to take it to be uniform. But if it’s not constant, then you can have a J that’s varying with space, so the current crossing that surface will be the surface integral of J⋅dA. In other words, divide that area into tiny little patches, dA, then as we have seen many times, whenever something is flowing, it’s the dot product of the flow rate with the area that measures the actual flow. You add it over the surface, that’s the current crossing a surface. We won’t use this formula very much. I’m just mentioning it for completeness.

All right, so the next question is, I’ve told you there is no electric field inside a conductor. But if you take a resistor, even a nice thing like copper, it’s not a perfect conductor. Actually, there’s an electric field inside a conductor. That’s what makes a current flow. If you want any wire to carry current, except for ideal wires that have no resistance, any realistic conductor needs a field to drive it. The reason is that if you look microscopically at these carriers, they’re all going like crazy in all directions. That’s precisely why they don’t carry a current. They don’t have a common direction. They’re going very fast but they’re going nowhere. One of these guys is going to the right. For every one going to the right, there’s one to the left, one going northeast and one going southwest. The velocity all averages out to 0. But if you apply an electric field to the right, you can imagine somehow in the middle of all this chaos, there’ll be an overall tendency to drift to the right. So let’s see how much there is.

I found that the current density was nev. Forget about the vector sign now. Just everything is along x. I want to find here an average velocity. What’s going to be the average velocity over all the particles at a given time? Now here’s the picture of conduction. You should in fact find it very paradoxical that when you have an electric field, the electric field produces a force e times E, has an acceleration a, so the charges should accelerate. If they accelerate more and more, the current should keep on growing because velocity is growing. But you get a steady current, in spite of a force acting to the right. Do you know why that happens? Why doesn’t everything accelerate and pick up more and more and more speed?

Student: Resistance.

Chapter 3: The Concept of Resistance [00:42:47]

Professor Ramamurti Shankar: It is resistance, but microscopically, why don’t these particles pick up speed forever? They collide. They collide with basically the impurities in the solid. And every time they hit their head on one of the impurities, they don’t know what happened, and they bounce off the collision in a totally random direction. So they may go in this direction, hit an impurity. There’s no telling which way they’ll come out. They can come out in any direction with equal probability. So they’ll lose all memory of what they were doing after each collision. Now if I look at my clock, and I look at the entire set of electrons and I say, what’s the average velocity? The average velocity, which I denote with a bar, is obtained by averaging individual velocity. Individual velocity is the velocity at time 0 since the last collision, plus eE/m times t_i.

In other words, let me explain to you slowly what I mean. Take particle number i. Let us say it has been t_i seconds since it had its last collision. So just after the collision, it has got initial velocity which is completely random, but since that thing, because the electric field it has been accelerating, it has not collided with anything yet, so its velocity will be eE/m times t_i. Therefore the average velocity, which you find by the averaging of everything, has the following property: the first one will give you 0. If you average over all possible initial velocities, they’ll be 0, because they’re pointing in random directions. So what you really need is eE/m times some τ, where τ is the name for the average time since the last collision. That time will have a range of values. Some guys will have just collided; some won’t have collided for a long time.

So you have to find that average, and we don’t know how to compute it right now, but that is some average t_i for a material. The larger the t_i is, the better the conductor it is, because it can go for a long time on average without colliding. So the whole idea is, the minute you collide, you lose everything, because you scatter off, forgetting your memory. So any coherent motion you have in the direction of the field is there only because you have not yet collided. So the net current is a function of how many seconds have elapsed since the last collision for each guy. On average, it gives you this. So if you use the symbol tau for that, therefore the current density will be ne times the average v. The average v is eE/mτ. So if you write it as ne²τ/m times the electric field. So this is the very, very important result. So I’m going to go back and study this result. It tells you that the current density in a wire is there thanks to the electric field and the number in front of it happens to be the number of carriers e²τ/m. Does that formula make sense?

Take a minute to look at that formula. You apply a field. That’s why they even know which way to flow. If you don’t have a field in a wire, the current doesn’t know which way to flow. It’s just random motion, not going anywhere, like molecules in this room. They’re not going anywhere in particular, even though they’re moving. You are saying that the current that you’re going to get for unit area is bigger if you have a bigger density of carriers. That makes sense, because they’re the ones carrying the charge. This is inversely proportional to the mass. You can understand why. The electric field, whatever force it produces, the acceleration is inversely proportional to the mass. The bigger the τ, the bigger the response, because they can go for a longer time on average before colliding, therefore they have more time to pick up speed in the direction of the field. e² is interesting. One e comes because the force on the carrier is little e times big E. The second e comes because the current it carries is itself proportional to e. You understand?

The charge of the carrier affects it in two ways. Any time it moves, it carries a charge e. How much it moves depends on the force the electric field exerts on it. That’s another e. So this is our expression. And we write it as σ times E, where sigma is called the conductivity. You notice that this argument doesn’t care what material it is. It could be copper, it could be aluminum, it could be some alloy. For all of them, the answer depends on ne²τ/m. What depends on the actual carrier, the mass is just the mass of the electron. τ is what varies from problem to problem. Some materials have very large τ, some materials are very small τ. That’s what decides how good or bad a conductor it is

So this is the formula. Now sometimes you can write it in another way. Let’s see now. Suppose you have a wire. What is the total current in the wire? The total current in the wire is the current density times area, which is σAE.

Now let us say this E was obtained by applying the voltage difference V between the two end points of a wire of length L. There is σAV/. Do you understand? The electric field times the length of the wire is the voltage difference between the two end points. So rather than saying the current is driven by the electric field, let’s say the current is driven by the voltage difference between the two ends of the wire. That’s why I wrote E as V/L. But now you see it looks like V/R that R is by definition is called the resistance of that wire. It’s measured in ohms, as denoted by this symbol. This is how you get ohm’s law, because just by going through the microscopic equation, and applying it for a wire of length L, I’m able to find that the current is proportional to V, divided by some number. That number looks like L/Aσ, also written as Lρ/A, where ρ = 1/σ is called the resistivity. It’s just the inverse of the conductivity.

So the bigger the resistivity, the bigger the resistance, but notice, if the wire is twice as long, the resistance of the wire will be twice as big. Does that make sense to you? For a given material, if you double the length of the wire, resistance is double. If you double the area, resistance is turned into half its value, because if you’ve got a big wire, you can think of it as two wires that are carrying the current together, therefore the resistance is half as much. So this is the relation between resistivity and the resistance. As far as we are concerned, the main thing for us is just ohm’s law, which we are going to use: V = IR, and this is telling you roughly where you get it.

So what’s the summary of all this talk? In a wire, unlike in a perfect conductor, there is an electric field. And it’s the electric field that keeps the charges moving. But whereas an electron, for example, in the vacuum electric field that will accelerate indefinitely. The carriers in the wire do not accelerate indefinitely because they keep bumping into stuff, and every time they bump into stuff, they lose the gain they had. They start all over again.

So at any given time, the activity I have or the motion I have depends on how many guys are still around since the last collision. They’re the ones who have been picking up speed, and that’s how you get conductivity proportional to the applied field. So it’s hard to get a velocity proportional to force. You always get acceleration proportional to force, but when you’ve got random motion, the velocity is proportional to the applied field. By the way, I should tell you, in a real solid, if you ask what do they collide into, can you imagine? You have a mental picture of a solid where the atoms form a nice array. The nuclei form a nice periodic array. The electrons in a metal are free to travel the length and width of the solid. But who do you think I’m talking about when I say collisions? Pardon me?

Student: Other electrons.

Professor Ramamurti Shankar: Other electrons is possible. How about with the nuclei? If you’re an electron, you’re going through, you see a nuclei every whatever, 10 to the -8 centimeters, there’s another nucleus, but that’s not what matters. This is more advanced theory, when you try to find the conductivity of materials. A perfectly periodic lattice of nuclei, electrons have a way to travel through them without ever colliding. It’s like, if you cannot see and the furniture is all in a certain place, you can navigate freely around them. But if somebody moves something and you’re not expecting it, that’s when you have a collision. And that motion comes about when you heat the solid. When you heat the solid, the nuclei start vibrating, so you don’t know quite where somebody will be.

That gives a small probability for the electron to collide with them. Therefore the conductivity will quite often depend on the temperature, but it can also depend on the force of interactions between electrons. But even if electrons don’t interact with each other in a serious way, the collision with the nucleus is controlled by lattice vibrations. But no matter how sophisticated the calculation is, you can go to my office and ask, what are people doing for conductivity?

Everyone is trying to compute this quantity τ and ne²τ/m. The n, the e², they’re all what you think they mean. τ is more sophisticated, and you have to calculate it in a quantum theory. But in the end, after all the work, you get a number τ, you put it in the same formula, ne²τ/m to get the conductivity of a material.

So now I want to do a little electric circuits for you. So here’s one simple circuit I’m going to do. I’m going to take a capacitor, charge it up to some amount Q, then I’m going to put my resistor here like this and ask what happens when I close that switch. So when I close the switch, we can all imagine what will happen. These positive guys are dying to get over to the negative side, but they cannot jump the gap here, because it’s a vacuum. But if you give them a path, they will go through that and come back to the other side.

So charges will start going round if only I close the switch. So when I close the switch, let me write down an equation for what will happen. There’s a resistance here. This is the current I flowing here. The fundamental equation you write down in any circuit is if you start at any point, you take any closed path and find all the changes in potential, the change has to be 0. Because it’s coming from a conservative field, the integral of the electric field on a loop is 0. That means the total change in potential from anywhere back to the same place is 0, because it’s like a height. So when I start here and I go through the capacitor, I go up in electrical height by an amount Q/C. This conductor is assumed to be a perfect conductor, so there is no electric field inside here, there’s no change in potential until I come to this resistor. A resistor will not carry current unless there’s a voltage applied to it. You’ve seen there, and this is the higher end of the voltage, this is the lower end of the voltage, because it’s flowing downhill. So you have a drop in voltage by an amount RI, then you come back to the starting point. All the changes you had better add up to 0. That’s the fundamental statement that in any electrical circuit, if you start anywhere and go through any path in the circuit and come back to a starting point, the net change in voltage has to be 0, because every point has an electrical height. So this is our equation. Now we can write it as RI = Q/C. But what’s the relation between I and Q? Can you find a relation between I and Q? Suppose a current flows for a short time. Who’s paying for it? Where is the charge coming from? Yes?

Student: Isn’t I Q divided by time?

Professor Ramamurti Shankar: It’s not necessarily divided by time, but it’s a derivative. I is dQ /dt, because in a small time dt, when the current I flows, Idt coulombs flow through here, but they’ve got to come from there. But the only thing missing is the - sign. Because if I is defined this way and is positive, a positive I depletes the charge, so I is −dQ/dt. So I have the equation here that looks like — so go back to RI = Q/C and write it as RdQ/dt. This is my equation. That can be written as dQ/Q = -dt/RC.

Now if you integrate that from start to finish and start to finish, this will give you log of Q over the initial Q will be - the time divided by RC. Or Q of time t will be Q₀e^-t/RC. That means after you close the switch, if you measure the charge and the capacitor as a function of time, it starts with some Q₀ and it decays exponentially. How long does it take for it to completely discharge? The answer is infinite amount of time. Why is the capacitor not able to discharge it completely? Can you think about that? Why doesn’t it just get it over with, right? Just dump all the charge to the other plate? Why is it taking forever?

Student: There’s always resistance.

Professor Ramamurti Shankar: There is always resistance. There has always been resistance, so what’s happening as time goes by?

Student: [inaudible]

Professor Ramamurti Shankar: The voltage on the capacitor plate is decreasing with time, right? That’s the voltage driving the current through the resistor. So as it drives current through the resistor and it begins to become empty, it’s able to drive less current through the resistor. So it’s trying to work against itself, but it doesn’t have enough Q on it to drive more Q down. That’s why. That’s the meaning of the equation. The rate of flow of Q is proportional to Q itself. So as long as there is some Q left, it will be decaying, but it will never come to 0, because the driving force for the decay is Q itself. And if you say how long should I wait? There’s a certain time called 1/RC, which is called the time constant. But any exponential function, e^-t/t₀, if t is much bigger than t₀, then you’ve got e to the minus huge number and that’s a negligible number. So whenever you say something’s falling exponentially, it doesn’t mean that it may take forever, or It may happen very quickly.

It depends what’s up in the exponent. All exponential functions will look like e to the minus time over some other number with dimension of time. That’s the unit in which you measure time. If little t is many times big t₀, then it’s e to the minus big number, which is negligible. So you’ve got to wait many time constants before a capacitor will discharge.

So capacitors are pretty dangerous. In fact, your computer, if you open it, there can be capacitors inside which are charged, even though it’s not plugged into the mains. So don’t think it’s safe to open an electrical device, just because it’s not plugged into the mains. Because people tell you, “Hey, pull the plug, then you can do what you want.” Not quite, because if you do that, that’s a big fat capacitor, that R stands for you. That current is going to go right through you. That’s why they always tell you, “Do not take this computer to your bathtub” for example.

So these warnings, even though they sound ridiculous, part of it is true. Capacitors are very dangerous. What you want to do is discharge all your capacitors first. Even in flash bulbs, that’s what happens. You charge up a lot of charge in a capacitor, and the resistance there is the bulb itself, the little coil in the bulb. When we close the circuit, the capacitor dumps all its charge and then in the brief moment, the coil heats up, namely R heats up and glows and you have a flash. But there you want time constant to be very small, because how long are you going to tell people to keep smiling? So you’ve got to get on with it, so you put in a very quick, very rapid time constant, so R is going to be very small. Sometimes you want the decay to be very, very slow. People are even thinking of getting rid of batteries and just buying capacitors. With a capacitor, you can use it to drive something, but slowly the voltage will go down. So if your device can operate over a range of voltage, the voltage is just Q(t)/C. That’s the voltage at a given time. If it can operate up to that voltage, between that and that voltage, you can run it for that time.

Okay, so this is the fate of the current. Now what is Q₀? At t = 0, what is the charge in the capacitor? Well, that’s the initial charge in the capacitor. What is the current in this problem, I? It’s just dQ/dt. It’s −dQ/dt, and if you do −dQ/dt, you’ll find there’s Q₀/RC. Just take the derivative of this function times e^−t/RC. At t = 0, Q₀/C is the voltage in the capacitor, divided by R as a current. That’s a current it starts with, but the current also decays exponentially, so with another prefactor.

But there’s one calculation I wanted to do, which is the energy calculation. If I had a capacitor that was charged, I had an energy ½CV², or if you want, Q²/2C. At the end of the day, my capacitor’s completely discharged. You want to ask yourself, what happened to the energy I had in the capacitor? And we all know the answer is that it went through the resistor and heated it up, and the heat energy or light energy is where you got your rewards. But we’ve got to make sure that the energy deposited in the resistor over all time is equal to the energy you had in the capacitor.

That’s the last one thing I wanted you to check. So when current goes through a resistor — here’s a current going through a resistor — what is the rate at which energy is being consumed? So you must think, let me draw the resistor this way. Its’ very suggestive. You’re flowing downhill electrically through voltage V. Every coulomb that falls down loses an energy Q times V as it falls. And that’s the energy that’s given to the wire by colliding with the stuff in it and heating it up. Therefore in 1 second, the number of coulombs falling down is VI and that’s the power. So power in the resistor is VI. That’s the rate of energy consumption.

So in this problem, V is Q/C and I, I got somewhere here — I’m too close to the board to see where anything is. I want to first find the current. Here is my current. I’m sorry, there’s an easier way to do this. Instead of writing VI, let me write it as I²R, because 1V is IR. So I want to integrate from 0 to infinity the quantity I²R. Now I’m not going to waste your time, guys. First of all, you know how to do this integral. Square this, multiply by R and do the integral, and believe me, you will just get Q₀²/2C. I don’t want to spend time doing that. So that’s where the energy goes. You follow that? The capacitor discharges through the resistor and the energy you pumped into it comes out in the form of heat, or light if it’s glowing. And we have seen the balance of energy. But now the trouble with this circuit is that it doesn’t last very long.

You’ve done it once, you’re finished. The capacitor discharges. If you want an experiment where the current can keep on running, then you need what’s called a cell or a battery. I’ve got to explain to you a little bit about the cell. It’s something I did not fully understand the first time around, so I want to share with you whatever understanding I have, how the cell works. People generally tell you a cell provides a certain voltage difference between the end points, maybe 1.2 volts between this and the terminal. That’s certainly true, but in light of what we have learned, here is what is the correct way to think about it. The analogy is with the ski resort. So here is the ski resort. All these guys are coming down. Let’s follow one person coming down the ski. Gravity’s acting down here, then you sort of loop around and you come here for free. Then there is a lift that takes you to the top.

So let’s call it the force of the lift. The force of the lift has nothing to do with gravity, driven by some other engines and so on. And we define force of the lift on the closed loop to be something called curly E. This is the mechanical analog of what’s called electromotive force. Now this line integral of this force is not 0 on a closed loop, because on the way up, it is doing some work, but it doesn’t do anything the rest of the circuit. It’s always acting up, moving a distance h. Maybe that force times h is the electromotor force. So you go to the top, you come down, and the electromotive force comes and pushes you back to the top, and again, gravity brings you down.

So in a battery, what you have is the following: that’s where you have to follow this very closely. Here is an electrical cell. In an electrical cell, there are a lot of positive charges and there are a lot of negative charges, and the field in fact looks like this. So when a current flows, it goes like that, goes through resistor, comes down. But the current, you realize, has to flow up here. Inside the cell, if the current flows this way, the current has to go like this. The electric field is actually pointing down between the plates. That’s not what makes it move, because there’s an extra force. This is your electric field.

There’s an extra force called E’, which is of chemical origin, which pushes your charges against their will from lower potential to higher potential. That is like the ski lift. Electric field everywhere is like gravity. Its line integral that way and line integral that way are all the same. But this E’ has a line integral, E’⋅dr on a closed loop, which is not equal to 0 and in fact is called the emf. You can see it’s not 0, because it’s non-zero here. Everywhere else it is 0. And in the region where it’s non-zero, there’s no cancellation. The displacement and the force are all in the same direction. So it’s a non-conservative force sitting inside a battery. It’s a chemical force.

The usual electric field that we all like, E⋅dr, in fact has line integral equal to 0. But when you’re inside this battery, if you ask how big is this E’ and how big is this E, what happens inside the battery is, once you start piling up charges there, this electric field opposed to the charge is coming, the chemical force will exactly balance the electrical force inside. So inside this region, the electric field will be - that chemical force. E’ is the chemical force per charge. Inside the battery, there is something that takes the charge against its natural instinct, against gravity if you like, and takes it up. You need something like that to make this thing run, otherwise charges will just sit in the bottom. They won’t go up. So you need an external agency. Sometimes you can have a belt. In a Van der Graaf generator, you have a belt that runs up and down and carries positive charge to the top, even though the top is already positively charged.

Left to itself, a positive charge would never like to go to the dome which is positively charged, but you drag it on that belt, kicking and screaming. You’ve got to do that to charge it up against its will. That’s what provides the force there. You can connect that guy to the resistor and the current will keep flowing. Now here is the interesting thing. Since inside this loop, the electric field and the chemical force are exactly balanced. Now let’s look at E⋅dr between here and here, between the positive and negative terminal. That’s the same as E⋅dr inside. So E⋅dr from the positive to the negative terminal is −E’⋅dr from the positive to the negative terminal is V_- - V₊. Therefore V₊ - V_- is the emf.

So if the chemical force has an emf, which is the line integral of the chemical force around a loop, which is basically inside a battery, that will be also equal to the voltage difference for a person living outside the battery. If a person outside the battery went on a loop like this from here to here, the line integral done by the electric field, which is the voltage dropped, if you want, the voltage gain between this one and this one, will be exactly equal to the emf. So V₊ and this is V_-. Let me just make sure this part is correct. The line integral is 0, or if you like, the integral of the electric field from here to here, doesn’t matter which path you take, but the electric field is conservative. That’s equal to the integral of -E from here to here, but that’s the same as being the polarities reversed in the integral, and that will be the emf. So that = integral E⋅dr from - to +. And that will = V₊ - V_-, if you write it that way.

Okay, so the point is, when you have a battery, if you don’t want to get into any of the details, this is what you guys have to know. If you don’t look under the hood, just look at those two terminals. There’ll be a voltage difference between them equal to what’s called the emf of the battery. The subtle thing to remember is, there are non conservative forces at work inside the battery, this is what you may not realize, whose line integral is in fact not 0, but you don’t have to know all that. You just have to know the integral, you have to know what happens if I go from here to here in an electrical circuit. I go up in voltage equal to V . That’s what I want you to remember. For those of you who want to know the details, this is the reason that happens. From now on, we won’t look inside the terminal. We’ll take every battery to have an emf E, meaning if you jump from a negative to positive, you gain a voltage.

[end of transcript]

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