# PHYS 201: Fundamentals of Physics II

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# Fundamentals of Physics II

## PHYS 201 - Lecture 4 - Gauss's Law and Application to Conductors and Insulators

### Chapter 1: Derivation of Gauss’s Law [00:00:00]

Professor Ramamurti Shankar: Let’s start with a brief recall of Gauss’s Law. It is so important I wanted to make sure you guys understand not only how it is used but how it’s derived. I derived it for you last time, but I thought the presentation could be improved a little bit, so I’m going to do it one more time, just this time only hitting the highlights. The content of Gauss’s Law is very easy to visualize, so I’m going to draw you a picture that’ll tell you the whole story, then there are all the equations. The picture is the following. You take one charge q and you draw the lines of force coming from the one charge. It’s a point charge. Therefore, you have no freedom but to draw lines that look like this. Now, how many lines you draw is a matter of convention. You can draw eight or twelve, whatever it is, but you’ve got to agree that the number per coulomb should be fixed. So we’ll fix some number. So I’m going to pick for whatever charge is here these lines. You cannot have a distribution of lines in which it’s suddenly more dense in just this direction, for example. That’s not allowed, because if I take this point charge and I rotate it, it looks the same, but if I took the line distribution with more dense here than other places if I rotate it, it looks different. So if the cause of the field lines looks the same after rotation, the effect, which is the field lines, should not change. You can only have a distribution of lines with respect to symmetry of the source of the cause of the field lines.

You understand that? It cannot be biased one way or the other, whereas this distribution, if you turn it around, looks exactly the same. So we draw those lines, and the content of Gauss’s Law is really the following. We are going to count the lines crossing a sphere, S, and we’re going to count the lines crossing some generic surface S’. You can see at a glance without doing any calculations that the following is true. Lines crossing S = lines crossing S’, right? That’s clear to everybody? You don’t need to do any equation for that. That’s because the lines have nowhere to go but out, and you can count them on this circle, that circle as long as you enclose the whole thing, the solid surface. Enclosing this charge you’re going to get the same lines. That’s the heart of Gauss’s Law, but that’s not the final form, but that’s the part that starts everything. Now I’m going to calculate the left hand side, and I’m going to calculate the right hand side. I’m going to equate them.

From that I will get almost everything I want. Now, how many lines are coming out of S? We know — first the notion of line density. I’ve got to remind you what a line density is. Line density is the lines per area, but it’s not enough to say area. I’m going to put a perpendicular symbol there. That means the lines are going, say, from left to right like this. You want to put an area that intercepts it perpendicularly, or if you like, the area vector which is always normal to the area should be parallel to the lines. So it’s the number of lines crossing unit area perpendicular to the lines. That’s called the line density. If you tilted this area; the same area, but at a bad angle, it’ll intersect fewer lines. That you know.

Okay, so this is the meaning of line density. What’s the line density going to look like in this problem? Line density is going to be proportional to the charge you have because we agreed we’ll put some number of lines per coulomb, and it’s inversely proportional to the radius because the area of a sphere of radius r is 4Πr2. So if you cut it with a bigger sphere of bigger radius the density, lines per area, will go like 1/r2. So it’s proportional to q. It’s proportional to 1/r2, okay? So let me write not equal to but proportional to. You have to stop me if you don’t follow anything I’m saying. There’s no reason why you shouldn’t follow this argument. Then I say how about the electric field? The strength of the electric field, not the direction, the electric field E at radius r is also proportional to q and also proportional to 1/r2. Because from Coulomb’s Law the field is the force on a unit test charge, it is q1q2. q2 is 1 because the test charge is 1 and there’s q/4Πε0r2. Forget the 4Π. It’s proportional to q/r2.

From this we may conclude that the line density is proportional to the electric field because they are both proportional to q/r2. Therefore, we can say line density is equal to some constant c times the electric field in magnitude; direction of the lines is along E, but the magnitude. Do you understand that? This guy is proportional to this. This may have some proportionality constant, this may have another one, but nothing changes the fact that this expression’s proportional to that one. The constant will depend on details. Let me just call it some number c. c is up to you. It really depends on how many lines you want to draw per coulomb. I made a particular choice last time, 1/ε0 lines per coulomb, but you don’t have to make that to prove the law. So this time I’m just assuming the lines is proportional to charge, but you pick your own proportionality constant. So I’ve got this line density that’s proportional to electric field. Now I’m going to go back and I’m going to calculate the left hand side.

Now, let’s give this equation a number, equation number 1. I’m going to calculate the left hand side of 1. Left hand side of 1 is the number of lines crossing this sphere, S. So that’s going to be equal to — let me call it ΦS is equal to number crossing S, crossing this sphere S. Φ is the standard symbol for flux. So what is it? Well, ΦS = line density times area of sphere, because on a sphere the lines are uniform in density so the total number crossing the sphere is the line density times area of the sphere. But the line density is a constant times the electric field at radius r (if you like) times area of the sphere which is 4Πr2. That’s the lines crossing the inner surface which is this sphere of radius r. Now, ΦS’ is lines crossing S’. Now you’ve got to be a little careful when you find the lines crossing S’. You understand why that’s more difficult? What are the two complications in finding the lines crossing S’ compared to this one which I did in one second? Any idea? Yes?

Student: The normals are all parallel to the field lines?

Professor Ramamurti Shankar: Which is not parallel?

Student: The normal to the surface isn’t parallel to the ________.

Professor Ramamurti Shankar: Okay, the surface is not — surface vector at each point is not parallel to the electric field. For example, if you’ve got a crazy surface like this — I’ll try to replicate that surface, but I don’t remember what I drew, some weird object like this. Then the electric field point this way, but the normal to the surface will point that way. See, E points that way. Tiny area there, which I’m calling vector dA, points in a different direction. So each region has a different direction. Secondly, the strength of the field itself is varying on the surface. Unlike on a sphere where everything is at the same distance r some are further, some are closer. But it is still true that if you want the number of lines cross a patch, if you pick a little area here, if you want to know how many lines cross that you take the normal to the area, call it a vector dA, and you multiply by line density.

Therefore, this is equal to the integral of the electric field at some point r dot product with a tiny area which is there. At each region you’ve got to draw a different dA. So you take the surface, you chop it up into tiny little tiles. Each tile is small enough to be considered a plane. Then the area vector is perpendicular to the plane, outward from the surface, and of magnitude equal to the numerical area of the plane. So the lines crossing every tiny little square on the big surface is the number of lines per unit area times the area times the cosine of the angle which you need, because the area vector may not be parallel to the flow. I have explained many times why the flow there, the flow vector and the area vector are not parallel you’ve got to put in a cosθ. That’s what comes into this dot product. And we write the symbol like this to tell you not only is it a double integral, meaning integral over a two-dimensional region, but it’s over a closed two-dimensional region. That symbol is to remind you.

So now, ΦS equal to ΦS’ is what I’m going to write. So I’m going to write — oh, I made one mistake. Can anybody see one thing missing in this equation here? This equation’s not correct. To find the lines crossing S’ I took E times dA. It’s a constant c because the line density is not simply the electric field it’s the constant. So I forgot the constant, but I knew that I would need it because when I now equate ΦS to ΦS’ I get constant times the electric field at r times 4Πr2 is equal to constant times the surface integral of EdA on some surface S’. But E(r) times 4Πr2 is c times q0 because E is q/4Πε0r2. You see that? This much you should know. I’m not going to write the details. So that is equal to c times surface integral of EdA. Now you see the constant does not matter, so whatever the constant is I can make the following statement and that’s going to be the beginning of Gauss’s Law.

So Gauss’s Law tells me that if you took a charge and you sat under it with any surface whatsoever — I’m not going to put a prime on this anymore. S is going to be the surface. Then the surface integral of the electric field due to that charge over a surface is equal to the charge that is sitting inside divided by ε0, and it does not depend on the details of the surface. Okay, everybody should know in principle how to do the surface integral. If I give you a surface, and at the surface at every point I give you a vector E, which is electric field, you should know how to numerically calculate it. You may not know how to do the integral analytically, but operationally you should know the meaning of this, okay? If E were actually the velocity of some stuff running out of the volume you’re counting in each region how many guys are escaping per second and you’re adding it all up. And to find how many guys are leaving a tiny area, you multiply the area by the velocity of flow times the cosine of the angle between the velocity in the area. That’s it.

Okay, now we do the next thing. Let’s call this charge q1 because I’m now going to put a second guy q2 here. If you now think in terms of lines of force you’re going to get into a big mess because lines of force, due to two charges is a complicated thing, you remember drawing some pictures. But now here’s the beauty. Forget the lines of force completely, not interested. I’m interested in the electric field. Focus on the electric field for which there’s a very simple rule. The electric field due to two charges 1 and 2 is the electric field due to 1 plus the electric field due to 2. That’s the great principle of superposition. Therefore, on the left hand side you might say if both charges are inside the surface integral of E1,2meaning when both are present is equal to surface integral of E1dA + surface integral of E2dA, because E1,2 is E1 + E2. But the first guy is q1/ε0, second is q2/ε0. Once you can do two charges you can do any number of charges.

And now we write Gauss’s Law almost in its final form. The final form is that if you took the total electric field, I’m not going to put total, E will stand for the total, on a surface S that is equal to the sum of all the charges, i from 1 to whatever many charges you have, divided by ε0. The condition is these are charges inside the volume bounded by the surface. You do not count the charge here. You only count charges which are inside not charges which are outside. So let’s make a notation that tells you that i is inside the volume V. And this surface is the boundary of the volume V. This partial derivative symbol is the universal convention for the boundary of any region. Okay, if you draw a real circle, this is the circle. The boundary of the circle is this. If you a have a cube, or if you have an orange the boundary of the volume orange is the skin of the orange. That’s the definition of boundary.

You can see this is like the skin of the solid region, and the surface integral on the skin of the electric field is a charge inside the volume whatever be the shape of the volume. And it works for all kinds of surfaces. I mean, here’s one example. Suppose you have a charge here. You take a surface that kind of intersects itself and so on. Then what can happen is if I put a charge here the lines may go like this. It doesn’t matter because over this when it switches back on itself some lines leave, some lines enter, and some lines leave again. So it might as well be that surface because double entering and leaving doesn’t matter. So the rule is if q is inside it contributes to the surface integral, if q is outside it doesn’t. The last variation is what if q is not made up of a discrete set of charges, which you can say one is here one is there, but there’s some continuous blob, some lump of charge where the charge density — so then we say suppose q is continuous.

You’ve got to realize in real life it’s never continuous. Real charges are made up of electrons and protons. They are discrete, but if your eye is not looking at it in great detail the charge can appear to be continuous. For example, if you take a glass of water the water is really made up of point molecules, water molecules, but to your eye it looks like a continuous medium. So for certain purposes we can replace a discrete distribution by a continuous one. For a continuous distribution what you have to tell me is the following. This is space x, y, and. Here’s the point. I take a tiny cube whose sides are dx, dy and dz, and ρ of x, y, and z times dxdydz is equal to charge in that little cube. In other words, like the usual density with this mass per unit volume the ρ is the charge per unit volume. So you multiply by volume you get the charge, but it need not be uniform.

So if you have a solid region then the charge density could be varying from point to point, so. Oh, let’s keep it here. In that case you will write the surface integral of EdA over a surface that binds a volume V would be 1/ε0 times the volume integral of ρ of (x, y, z) dxdydz. This is the final form of Gauss’s Law. Now, how many people have ever done integrals in more than one dimension? Great, very good, made my day because I didn’t have Plan B. But I did — no actually I had a Plan Tiny B which is to give you a quick refresher. If there is time at the end of the class I will remind you of how the multiple integrals are done. But right now I’m going to show you the efficacy of this law and that doesn’t require doing complicated integrals. All the integrals will be so trivial you can just read them off. So when we have time we’ll do questions of how to do these integrals, but I’m assuming this symbol means something to you. In words it means all the charge inside.

### Chapter 2: The Electric Field Due to a Spherical Distribution of Charge [00:21:12]

So, again, by taking the volume, dividing into tiny regions, multiplying each volume, tiny volume, by the density to get the charge there, and summing over all the volumes inside the big volume V. That’s a charge. So Gauss’s Law still says the surface integral of E is the charge inside over epsilon 0. All right, so this is the great law. Now we’re going to get some mileage out of it to solve certain problems which normally would be very difficult. The first problem, I think I probably did that last time but it’s worth going over, is what is the electric field of a uniform density of charge. Suppose there’s a charge Q and this is sphere of radius r, and it is a ball of radius r. You take all these charges and you pile them on top of each other. They don’t like it, but you glue them or do whatever you have to do. You’ve got a ball of charge Q. You want the field due to that.

It’s the same problem Newton had with gravity. The earth is a solid ball. It’s got mass everywhere and you would like to find the force on an apple. And it’s not so evident that the force of the earth on the apple is as if the entire mass of the earth but at the center. You have to prove that. You have to prove that by dividing the earth into little pieces and finding the gravitational force of every little piece on the apple and adding all those vectors up. And when you do all the hard work you will find, in fact, that it’s true, but it took Newton several years to prove that even though he was pretty certain that was true. That’s very important because when you do the force law for apple versus the earth the distance you use for m1m2/r2 is a center-to-center distance as if the entire earth were at the center. A similar law is true for electric forces.

In other words, if you’ve got a spherical distribution of charge the field due to that outside that ball is as if all the charge were at the center. That’s what we want to prove, but we want to prove that without doing any difficult integrals, and the trick is the following. You are going to use Gauss’s Law, and you can use Gauss’s Law on any surface you like. For example, you can take that surface and you can say surface integral of E on the surface is Q/ε0, but that’s not very helpful because what you’re trying to find out is not the electric field integrated over a whole region, but the electric field at each point. You want to know what’s going on right there. So even though it’s a true statement that whatever E is doing its integral on any surface is Q/ε0 that doesn’t tell you what is E at every point, and we’re going to find that by the following trick.

The trick is to first decide. This surface S, by the way, it’s called a Gaussian surface. It’s not a real surface. In real life there are no Gaussian surfaces. You draw them for a calculation, and you can move it around in your mind anyway you like, for each surface you’ll get a different answer. We’ll pick a Gaussian surface equal to a sphere of radius r. So I’m going to pick a Gaussian surface of radius r. I’m going to argue two things and you have to follow the reasoning because all the work is in the argument. You don’t do any integrals, but you do some serious arguing. The arguing says it’s a ball, positive charge. It’s going to repel a test charge. Which way is the repulsion going to be at this point? It has to be radially away from the center. That’s your argument. At each point the repulsion has to be away from the center. In other words the electric field has to have this distribution. We’re not saying how it varies with distance, but it has to be radial.

Again, the argument is the radially outgoing thing like a hedgehog. It’s the only distribution of lines which will look the same if you rotate them because then one line will go to another line. The hedgehog will look exactly the same. And that is a necessary requirement because if you took this ball of charge you all understand that if you rotate the ball the field lines have to rotate due to any calculation you did, but if you rotate the ball it looks the same. So the field lines have to look the same. So the distribution you draw must be invariant under that rotation. This is a very powerful principle of symmetry and invariance. The principle says if something is the cause of something, and if I do something to the cause that leaves it alone, namely it looks the same, the effect has to look the same.

It’s a very reasonable argument. And the cause here is a ball of charge which is completely isotropic and spherical. You turn it, it looks the same, so the field distribution looks the same. That’s all you need to proceed because that distribution then is going to have radial lines of unknown strength, but the strength can vary only with r, but not at a given r. At a given r the direction varies, but the strength is some E of r, and we’re asking what is E of r? So we know that E looks like r, which is the vector of length in the radial direction times E of r. And sometimes I also write this as er times E of r, and we just want to know what E is a function of r. But now we can use Gauss’s Law. So the left hand side is a surface integral of this radial E on a sphere of radius r. That’s going to be E times 4Πr2, because E is a constant on the surface.

The integral of a constant on the surface is the constant times the area of the surface. It’s like saying if you want to integrate a function which is constant from here to here it’s the constant height multiplied by the interval over which you’re integrating. There’s no integral to do. It’s very trivial. It’s a rectangle. Similarly a constant function integrated on a sphere is the area times this, but E is the function of, of course, r. That’s going to be charge enclosed. The charge enclosed is Q and you’ve got to put the ε0, and then you find E(r) is Q/4Πε0r2. And if you want to put all the vectors back you can write here times er. So by this trick one can show that the field of a sphere is as if the charges were at the center, provided you are talking about a point outside the sphere.

As long as this field is outside the real ball no matter what its radius is, the charge enclosed is always Q. The right hand side doesn’t vary with Q. The left hand side looks like r squared times electric field so E goes like 1/r2. But now I’m going to do some variations on this. The first variation is what if I want the electric field at some point here, in here? You’re allowed to ask that. You can take an instrument and put it there and ask what field you have. You have to calculate that too, so let’s calculate it for a sphere. The formula is the same, same formula. Everything is the same, but Q is now the charge enclosed. So I’m going to write, again, 4Πε0, I’m sorry, 4Πr2 times E(r) is the charge enclosed over ε0. What is the charge enclosed in this sphere, now, which is a mathematical sphere with radius as little r, but it’s smaller than big R? Can you make a guess? How much charge do you think that is in a smaller sphere? Yeah?

Student: The charge density times volume.

Professor Ramamurti Shankar: Right. So he said it’s the charge density times the volume. I’m going to write down the answer without going through the intermediate step, but it’ll coincide with what he said. The answer I’m going to write down is the total charge is Q on the whole ball. The charge is proportional to the volume. Then the volume of this little guy is r3 and the volume of the big guy big R3. That’s how much charge will be enclosed. Can you see that? If you take a square, and you take a part of the square that’s inside that’s half as big, the area of it will be ¼, because areas go like 1 over the distance squared. Volumes go like 1 over distance cubed. But we can also do what he just said. You can take the Q divide it by the volume 4/3Πr3. That’s the density. Then you can multiply the density by the volume of the little sphere 4/3Π little r cubed, and all I’m saying is forget these 4 thirds pis. It’s just little r3 over big R3. Therefore, the electric field now, E(r), looks like Q over 4Πε0r divided by r3. This is for r < R, is equal to Q/4Πε0 (1/r2) for r > R.

Student: Isn’t it supposed to be r cubed ________?

Professor Ramamurti Shankar: Sorry, here?

Student: Yeah, down on the second one.

Professor Ramamurti Shankar: Here you mean? This line?

Student: No, I mean like on the line above. You have like r over r cubed.

Professor Ramamurti Shankar: I have little r cubed over big R cubed.

Student: Right, but no. I mean the next line.

Professor Ramamurti Shankar: Here?

Student: Yeah.

Professor Ramamurti Shankar: This is r less than R, r bigger than R.

Student: [inaudible] Oh, gotcha.

Professor Ramamurti Shankar: Oh, you mean this 4Πr squared came downstairs? Yeah, that’s right.

Student: Gotcha.

Professor Ramamurti Shankar: Okay, eternally vigilant, that’s good. Yes?

Student: What if little r equals big R?

Professor Ramamurti Shankar: Let’s ask. What if little r grows up and become big R, right? We’ll find out. What do you expect? You’ve got to get the same answer at the field at a certain point no matter how we approached it, right? You’ll find that’s true in this formula, because if you put little r equals big R and big R over big R3 you’ll get 1 over big R2 and this will also give you 1 over big R2. These formulas will match on the surface of this sphere. Yes?

Student: So can you say again why we can assume that we don’t have to do the integral ________________________?

Professor Ramamurti Shankar: Why did I not do the integral?

Student: Why we ________ 4 pi squared times E of r.

Professor Ramamurti Shankar: Why do I do that you mean?

Student: Yeah.

Professor Ramamurti Shankar: Right. So the correct way to do it is to really do EdA, right? If I did that it’ll be fine with you, correct? This is what you’re saying I should do, right? You asked me why I didn’t do this integral, why I just wrote 4Πr2, is that your question? Right. The assumed form for E is the function that depends on r times unit vector in the radial direction. And since I surrounded it with this sphere the tiny area vector is also er times the value of the tiny area. In other words, dA and E are both parallel if the surface is a sphere, right? So when you take the dot product this will just become E(r) times dA times cosine of 0 which you don’t care about.

That’s the first thing. You still have to integrate over this sphere, but E of r is a constant on this sphere because it depends only on r. So throughout the whole sphere if it’s a constant, it’s like the number 19, you just pull it and then you get E of r times integral dA and that is my 4Πr2. So in general you have to do an integral, but if you’re lucky and the integrand is a constant then the integral is trivial. So for a charge at the center of the spherical ball, and if you went to a sphere then the integrant is a constant. Okay, so this is what you get. If you draw a picture of this it looks like this. Electric field as the function of r it will grow with r then it will decline as 1/r2. And here is little r equals big R. Yep?

Student: What if it were hollow and all the charges were on the outside ________?

Professor Ramamurti Shankar: We’ll come to that. Yes?

Student: Why can we ignore the charge that’s outside the small sphere?

Professor Ramamurti Shankar: Why don’t we count it?

Student: Uh-huh, did you explain that already?

Professor Ramamurti Shankar: Yeah, well it’s a very interesting point, right? Suppose you make a hole in the earth, of if you go deep into the earth. We said if you’re outside the earth the entire ball is pulling you in, but the claim I’m making is if you’re inside the earth, if you’re here, only the ball underneath your feet is pulling you, but not these guys, right? That’s the question, why don’t they exert a force, right? Well, Gauss’s Law says, no, but we can try to verify that explicitly. So we are going to do the hollow sphere, but you followed the solid sphere now? So why is the field getting stronger as you leave the center, why is it falling like 1/r2 once you leave the sphere? You’ve got to think about that. Why is it growing when you leave the origin? Yep?

Student: Because there is more charge _________.

Professor Ramamurti Shankar: Right. The charge you are enclosing grows like r cubed, and the distance from the center of that charge is r squared which comes downstairs. So as you go further out inside the ball you are gaining, but once you reach the surface of the ball and you go further out you’re getting more area for Gauss’s Law, but not more charge which is stuck at capital Q. That’s why once you leave the sphere it starts declining, but if you’re inside the sphere it increases. This is also true for the gravitational force. The gravitational force if you go to the center of the earth and you move away by 1 inch you’ll be pulled back with an extremely weak force that’s proportional to the distance from the center of the earth. Okay, then we come to another question which came up, which is let’s take a hollow sphere. I take a sphere and I scoop a hole in it and I want the field here. Okay, so if you want the field here you take a Gauss’s Law for a circle like that, or a sphere like that and you’ll get E times 4Πr2 is equal to 1 over ε0 at times what?

Student: Nothing.

Professor Ramamurti Shankar: Nothing. So the electric field inside a hollow sphere is 0. In other words, if you took the earth and you scooped a big hole in the middle you can float. It won’t pull you at all. So inside a hollow sphere there is no electric field. Even though there’s charge everywhere there’s no electric field. We can understand that in some cases, and other cases require a little more thought. Let me first show you that for a very, very thin shell why that happens. Once it’s true for a very, very thin shell and it doesn’t do anything you can put one shell inside another and make it as thick as you like because 0 is going to be the contribution from each guy.

So let me take a very thin shell. Here’s a very thin shell. So remember, this is not a ball. This is a shell whose thickness is the thickness of the chalk here. I think you will agree that at the center of this sphere the force has to be 0. You agree with that? Because which way should it go? For every guy pushing, say from here in that direction, there’s another one from there pushing in this direction. They all cancel. It’s very clear that at the center you don’t feel a force. But what is truly amazing is that even when you’re off center you don’t feel a force. So I’m going to give you the argument for that one, argument for that namely I’m once again going to pair off canceling charges, okay? I’m going to cancel one thing against another thing. When you were at the center you canceled a little charge here with a little charge here, right? When you are here here’s what you do. Take a cone like that. It is not two lengths. It is a three dimensional cone, but I’m able to show you only a slice right down the middle, you understand? This is really a cone that’ll cut the sphere there; it’ll cut the sphere there.

Take all these guys. They have a certain charge, which is some charge density which is uniform, times the area of this thing divided by the square of the distance r1. Here, these charges here will push you to the right with a force which is the area of density, the charge density per unit area times the area of the smaller circle divided by r2 squared. But now I claim that I can equate these two. I can equate these two if this area of vector and this area of vector were radiating outwards, and this is actually a true statement. This area is proportional to r1 squared. This area is proportional to r1 squared, therefore they will cancel. But this is a very subtle argument. Sometimes even textbooks get it wrong. And generally if you take two cones like this you can see that the area vector points that way, but the electric field vector points this way. They are not the same. The area vector, area on the sphere, right, will point in one direction, but the r vector is different.

So you have to take the cosine θ on both sides, but it turns out cosine θ is also the same because if you took a line and you cut it this way that angle will equal that angle. So that’s a subtly we don’t need, but that’s how you really prove for every shell. There’s a cancellation. If you know what a solid angle is it’s very easy to say. This is called, with the cosine thetas in it, it’s called a solid angle enclosed by this point, and the two things have the same solid angle and they cancel. But if you don’t want the most complicated proof you can take points like this where it’s clear that this area vector and this area vector are both perpendicular, and to the radial direction, and therefore the electric field and this vector are parallel. Then you don’t need the cosine thetas. They are clearly equal at possibly even 1. Then A/r2 being the same is the reason it works.

In other words, if the force of gravity or the electric force did not fall exactly like 1/r2 the field inside a hollow sphere will not be 0. So this is how people tested the 1/r2 force law in the old days. They went inside a hollow sphere and tried to see if there is any field inside by putting test charges and finding they don’t respond. That’s the most reliable way to prove 0 field. Yes?

Student: Can you explain again why the A over r squareds are equal?

Professor Ramamurti Shankar: Well, you can see that if you took a cone and if you just did similar triangles you can see that this area is growing like that radius square. It’s a matter of how areas grow when you scale anything. Take a cone, okay, and you slice it at some point, and slice it even higher. Here’s a cone. You slice it there and you slice it there, and my claim is this area is to that area that distance squared is to that distance squared. Why is that, right? That’s your question. The area of this cone is something, something radius times something, something height, right? So you can calculate, if you like. The radius of this circle and the radius of that circle are growing linearly as you go out. If you go distance r here, and you get a radius here, if you go distance r there you get another radius. Is it clear to you that the radii are proportional to the distance from the center in this cone?

Let me draw it again. This is a bad picture. Here’s my cone. Take two circles. Well again, I screwed up. This is supposed to be circles. This is the center. That radius is to that height what this radius is to this height. They’re just similar triangles. Therefore, this height is what I’m calling r1 and this height is what I’m calling r2. Therefore, if this radius is proportional to r2 the area of that circle is Πr22. This is Πr12. So that’ll go like that distance squared is to that distance squared. Okay, whenever you scale things for an area by factor of 2 it’ll go like the square of that factor. You can already see that if you draw these cones this’ll intersect the circle, the sphere on the small circle. This will intercept it on a bigger circle. That much is clear, but the extra is not linearly proportional.It is quadratic because the circle has two dimensions both of which are growing linearly with the distance from the apex of the cone. That’s why it’s r squared.

### Chapter 3: Electric Field Due to an Infinitely Long Wire [00:44:51]

All right, another thing one can calculate is the electric field due to an infinitely long wire. Here’s an infinitely long wire with lambda coulombs per meter. So you want to find the electric field. By the way, I should tell you that Gauss’s Law — let me do this one example. I’ll tell you what the restriction of Gauss’s Law is. So we come to this problem and we can again argue by symmetry the field lines have to be radially away from the axis of their line, and if you look at it from the edge, from one edge it’ll look like this. If you look at the wire from the edge the lines should be going out like that. It’s like a hedgehog but it’s cylindrical. It’s not in all three dimensions. It’s radial in this direction.

And the question is and the field has to be constant along the length of the infinite wire at each point at the same distance because if you move the wire horizontally it looks the same so the field distribution cannot vary. If it is a finite wire you cannot make the argument. In a finite wire as you come near the edges lines will start tilting. But a finite wire doesn’t look the same when you move it; an infinite wire does. Therefore, for an infinite wire if you don’t stop here the lines will always look the same, so that if you shift them over they should look the same. So the only unknown question is: I know the field is in this direction radially away from the wire in all directions whose magnitude is fixed at a distance r, but I want to know what the magnitude is. Again, I’m going to take Gauss’s Law and I’m going to apply to the following surface. Surface is a cylinder and the cylinder has some length l, and it’s got some radius r and the flat faces of the cylinder are parallel, so if you really want to give the nice picture here they look like this. So Gauss’s Law can be applied only to a closed surface.

You understand that? You cannot do it for an open surface because only if you trap the charge completely in all directions will you count every line. If you’ve got holes in your surface then stuff can escape and you cannot promise anything. So I need a closed surface and my closed surface is the cylinder. So I’m going to write, once again, the surface integral of the electric field on that cylinder is the charge enclosed. The charge enclosed is the easiest part. I’m going to give you 10 seconds to think in your head. What is a charge enclosed by the cylinder? Okay? It’s charge per unit length times the length of wire trapped inside the cylinder and then, of course, I have this 1/ε0. It still is a statement about the integral of E on the surface, and it could be any surface, but the beauty of this surface is the following. There are contributions to EdA from the curvy side of the cylinder and from the flat side. In the flat side the surface vector, area vector dA for any small portion, or in fact for the entire face, is like that. The electric field runs along that face so the dot product is 0.

So I got 0 from the left side, 0 from the right side, then I got non-zero from the curvy cylinder. On the curved cylinder you can, I hope, see that every area vector is actually parallel to the field lines. So just like on a sphere this integral will be E(r) times the surface of the curvy part of the cylinder which is 2Πr times L. That’s the area of the curved part of the cylinder. So you see L cancels out, and it better cancel out because L is an artificial construction. We made up L. The answer should not depend on the Gaussian surface you picked. I told you it’s a figment of our imagination, so it doesn’t depend on it, and we find E(r) = λ/2Πε0r, which is the result we got earlier on by doing all the brute force integration. Remember we took a point here. We took some segment. We drew the arrows, did the integrals, sine thetas and whatnot, but you can get that in one shot.

So it looks like Gauss’s Law is the easy way to do stuff, and you may wonder why we bother to do any difficult integrals. The reason is that if I change the symmetry of the problem in the slightest way I cannot calculate the field. For example, if on the spherical charge distribution instead of a sphere I made a little blip here, maybe did a little surgery and put that guy here. We’re dead. I mean, there’s a formula for the field, but no one can calculate it in any simple way. You can, again, take a Gaussian surface and it’ll still be true that the integral of the electric field on that surface will be the same Q/ε0, but the problem is the E on the surface is no longer constant. You understand? Not every point is symmetric any more. E may be stronger where there’s a bulge. E may be weaker where there is hole, so and also its direction is changing in a crazy way. So you can make one true statement about the integral of that crazy function over the whole region. That cannot be used to deduce the value of E at every point. So you still need the integral. You may have to do the integral maybe on a computer, but that’s the answer to all problems, but for simple problems with a great deal of symmetry we can use Gauss’s Law to get these things very easily, okay? That’s the Gauss.

### Chapter 4: Electric Conductors and Insulators [00:51:42]

Now, I’m going to introduce you to a second notion which is pretty important to study electricity, and that’s the notion of conductors. So we’re going to divide the world into two things, conductors and insulators. As you know, matter is made up of positive and negative charges and the negative charges circle the positive charges, and they pretty much stay near their parent atoms, near the parent nuclei, and you cannot separate them. But in a metal at least some of the electrons from each atom become communal. In other words, they can run around the whole solid. They don’t belong to any one nucleus. So that’s a conductor. In a conductor the negative charges, if you like, are free to move. So here’s the first result. E is equal to 0 inside a conductor.

That really follows from the meaning of the word conductor. If you took a chunk of perfect conductor, maybe copper is good enough; they’ll be no electric field inside copper. Why? Because if this is the chunk of material, this electric field, then the charges that I said are free to move will respond to the electric field and they’ll be moving. So I should say E equal to 0 inside a conductor in a static situation. In other words, once the charges have stopped moving the electric field will be 0 in a conductor. Let me explain to you a little more what’s going on so it’s not a big mystery.

Suppose there is a uniform electric field going from left to right? In that uniform field I take a chunk of copper like a nice rectangular chunk and I stick it in there. What will happen? In the beginning the electric field will penetrate the copper and the field lines say to the positive charges, “You go to the right,” and to the negative charges, “You go to the left.” Negative charges will race to the left until they cannot go anymore without leaving the solid, and that they’re not allowed to do, leaving behind some deficit on the other side. But look what’s happening now. These guys produce their own electric field which goes from here to here.

Therefore, inside the conductor, the electric field by the superposition principle is the field due to whatever outside agency produced this field, plus the field due to these guys, and they will not stop until the field they produced exactly cancels the external field. Then the migration will stop. So how do they know when to stop? They are not that smart, right? But the point is once they’ve produced the potential, I mean, once they produce the field that cancels the external field there’ll be no field inside the bulk to encourage charges to move any more and that’s when they stop moving. Okay, so a conductor, at least in this case, not so hard to understand what they have to do. Negative charges go to one side. Positive go to the other side until the field they produce is an arrow going to the left of the same magnitude as this one then E is 0 inside.

But what’s amazing about these metals is that if you take a potato shaped metal it’s not so easy to see what charge arrangement will exactly neutralize the field everywhere in sight. Here it’s easy to see. I want a right moving one canceled with a left moving one. I draw a line, a plane of charges here and here, and you can see they will cancel. But even this oddball object, I claim, will eventually acquire some density of charges, they’re not uniform or anything, in some complicated fashion until the field inside the solid becomes 0. So these particles will always figure out a way to make the field 0 inside, because it’s the definition. If it’s not 0 they’ve got more work to do, and the charges have to separate even more until there is no hunger for the separation because they managed to produce a field inside that is 0. Then nobody else will join this flow and it’ll stop.

So that’s a very short period, 10 to the minus something, when charges rearrange when you put a conductor in the field, then quickly it’ll stop. If you put an alternating field going back and forth then, of course, it depends on how rapidly it’s oscillating, and charges may not be able to keep up with that. That’s called a plasma frequency, and beyond that a field would start penetrating because charges cannot keep in step. But a DC field, where you’ve got all the time in the world to settle down, they will very quickly come to this arrangement. They will stop. So remember field in a conductor is 0 by definition.

Okay, now the next thing I will show you is that charge inside a conductor equal to 0. By that I mean the following. If you took a conductor and you threw some charge on it where will it go? Okay, here’s a chunk of copper. Throw 10 million electrons. Well, the 10 million electrons don’t like each other. They will try to run away from each other.

In the end they will all sit somewhere here, but they will sit in such a way that the field inside is 0, because you cannot have an electric field. If the electric field is 0 everywhere then the charge is also 0 everywhere, because you can take any surface, any volume you like with the integral of EdA, you’re going to get 0. If you can get EdA is 0 for every possible surface then the charge enclosed by every possible surface is 0, therefore there is no charge in sight. So where does the charge go? It goes to the boundary of the metal. Once they’re at the boundary of the metal the metal starts calling you back because there is something called a work function. It’s like all the electrons in a swimming pool. If they don’t like each other they can go to the edge of the pool, but to leave the pool they’ve got to climb up over the vertical walls, and they cannot scale the walls. If you rip them hard enough they will. If you put enough charges, charges will start flying from here and maybe land there; that we call lightening. For that they’ll have to break the air and produce a conducting path, but if you don’t put such strong fields the charges will remain on the surface.

Now, it turns out we can actually relate the electric field at the surface to the charge density of the surface by a following trick. Let’s go to the surface here and ask, “What’s the electric field? Which way can it point?” I claim the electric field can only point perpendicular to the surface, because if you had parallel components then the charges can move along the surface. No one says you cannot run it on the edge of the swimming pool, you just cannot leave it. So the electric field lines must point radial, I mean, normal to the surface. And I claim that we can actually calculate the electric field here if you knew the charge density here. So let’s do that. So here’s some surface, and I have — take a tiny region here and I’m going to take a Gaussian surface that looks like this. It’s a cylinder, very tiny cylinder, and the field lines are like that there. There is no field lines here because there’s no field inside. And the field here all of these things are 0, and the field here is parallel to the cylinder so it doesn’t contribute.

You see that? I got a cylinder. I rammed it into this solid. It’s a mathematical Gaussian surface. It’s got following faces, flat face, no E because there’s nothing inside the metal, this part of the curvy-face, no E, nothing inside the metal. Here E is parallel to the sides of the cylinder, so there’s no flux. The top end, if the area’s A or dA, the top end will have a flux which will be the E that I’m trying to calculate times dA. That’s going to be equal to the charge enclosed is σdA/ε0, because σ is the charge density which I assume I’m given. Then you can see the dA cancels. The electric field is σ/ε0. Do you remember ever seeing a formula like this sigma or anything like this? Yes?

Okay. Let me remind you where you saw it. I showed you that if you took an infinite plane of charge density σ the electric field was σ/2ε0, and I did that with a long calculation. I took a point there. I took rings and so on. Now, you don’t have to do that. Let’s use Gauss’s Law again to calculate it. Once again, you argue that if you’re in front of an infinite plane the only distribution of field lines that makes sense is if the field line’s always perpendicular to the surface. They can get weaker as you go away, but at a given distance from the plane they should have the same value, because if you slide the plane up and down it shouldn’t matter. So let’s take the side view of the plane. There are your charges. And I’m going to take a Gaussian cylinder that looks like this. The field lines are going like that. They’re going like that. There is some charge trapped here. And the area of that guy is some dA.

So again, I apply Gauss’s Law here. Of the cylinder I got the curvy sides with no contribution, because the field is parallel to the curvy side. I got the flat faces. So I get E times dA for one of them, and E times dA for the other one, because both are outgoing. That is equal to charge enclosed with the σ times dA divided by ε0. If you cancel the dA you find E is σ/2ε0. So the electric field on either side of the plane is σ/2ε0. You just get it from Gauss’s Law. Again, there’s no reason to do the complicated integral. That’s because we’ve reduced everything to one unknown, namely what was the magnitude of the electric field at a certain distance from the plane? Direction is known to be perpendicular and to be constant on this line. Then the one number came out to be, in fact, independent of how far you are. But now go back to the conductor. If you go back to the conductor you find the field at the surface of a conductor is σ/ε0 where σ is the charge density there, whereas for an infinite plane it’s σ/2ε0 on either side.

And you can ask, what’s going on? Why is it σ/ε0? Can anybody guess what may be happening? In other words, if you take a tiny area on a conductor and you go very, very, very close it should be as if you’re next to an infinite plane, because if you’re very, very close to the surface you don’t know if it’s finite or infinite. So you should get the same answer as for the infinite plane. Therefore, if you took only the charge sitting there it should give you σ/2ε0 going out and σ/2ε0 going in. In other words, let me draw a picture here. That’s my tiny area. I expect the field due to just the charge there to be like this, σ/2ε0, and σ/2ε0. But that is not the whole story because I’ve got the rest of the conductor that’s got its own charge. It is going to produce a field.

That field will have some value, and I claim I know the value of that field. It’ll be precisely enough to cancel it on the inside, but if you cancel it on the inside you’ll double it on the outside because it’s got σ/2ε0 pointing outwards to cancel this guy, but then it’ll aid the one due to the little area by doubling it. That’s why it’s σ/ε0 outside and 0 inside. In other words, here is the full story. If you go to a conductor, if you pick a region here, the field at this point is due to what’s here and what is everywhere else. What is here does that as if it was an infinite plane. The rest of the guys do that and that cancels this here, but adds there, so you can see more ε0 outside.

Okay, then there are other variations to this theme. So here’s a conductor, and I make a hole in it, and I put some charge. You can ask if I throw some charge on it where will it sit. Well, some charges will sit here if you put some coulombs, but what will happen on the inner wall? This is a hole. The claim is that there’ll be no charges here. They’ll all be outside even though you’ve got a hole in the middle. Again, Gauss’s Law tells you why. If I take a surface like this and do Gauss’s Law, since the electric field is 0 inside the metal, Q enclosed is 0. So the Q that’s enclosed, which is 0, means that maybe some positive guys and some negative guys, but there’s nothing from stopping the positive guys from rushing to meet the negative guys and canceling. They will, for there is no reason they just stay on opposite sides of the island. They’ll just come together. Therefore, there is no stable equilibrium in which the charges will be ever inside a conductor. So if you have a conductor and you put charge on it, it goes to the surface even if there are holes inside the conductor.

Now, that’s a very interesting paradox. I showed you that the field inside a conductor even if there’s a hole is 0, but let’s ask the following question. I take a conductor with the hole in it and I put a charge q here. Will you know about it from the outside? You’re not allowed to enter it. Will you know from the outside there’s a charge inside or will it get also shielded? Will there be any flux lines coming out here? Yep?

Student: Yes.

Professor Ramamurti Shankar: Because?

Student: Because there’s a charge.

Professor Ramamurti Shankar: Right. If there are no field lines here, and I do Gauss’s Law on this surface, if the surface integral of E is 0 qenclosed is 0, but I know there is some q inside. So what’s going to happen? I want the field lines to be present here and absent here. So what’ll happen is if you put a +q here the material conductor will separate into -q’s where these lines can terminate, and then out here will be some compensating plus charges that’ll produce the field that’ll produce the lines going out. In other words, this is really like a chunk of copper I showed you where it screens the field by polarizing into a negative part in this wall and positive contribution in the outer wall so that inside there is no field. Okay, so these are different variations of this theme and you should be able to do a whole bunch of problems connected with that.

Okay, I’ve used up my board, so I think I’ll give you guys a five minute tour on how to do these integrals in case it ever comes up. I’m just going to do two very trivial integrals and stop, and I’m going to do them in two dimensions and you can worry about higher dimensions. It’s just generalization. Suppose you’re in the xy plane and there’s a function in the xy plane, f (x, y), and someone says find f (x,y) dx dy. What’s the procedure and what are you supposed to do is the question. What you do in Cartesian coordinates is you take a line y equal to something, y equal to something + delta y, x equal to something, x equal to something + delta x, and that region (bounded) by these contours of constant x at different values and constant y at different values, has an area dxdy.

You multiply that area by this function. This could be the number of people living per square mile and this could be the number of square miles, and you add up all the little squares over the area that was given to you. So whoever tells you to do integrals better tell you over what range you’re going to do that integral. Okay? So let’s find out the area for triangle by this process. Here’s a triangle. Let’s say there’s one here and one here, 0, 1; 1, 0. This is y. This is x. I want to find the area of the triangle. Yep?

Student: Isn’t that point 1, 1?

Professor Ramamurti Shankar: Yes, thank you. That point is 1, 1. You’re supposed to take dxdy you write it like this. There’s no function or integral. I just want 1 times dxdy, so the 1 is implicit. dy, you pick a certain x and y should go from here to here. The range of y is clearly 1 - x because this is a curve x + y = 1. So y goes from 0 to 1 - x, and for every choice of x, x goes from 0 to 1. Yep?

Student: Isn’t the line defined as y = x ________.

Professor Ramamurti Shankar: Pardon me?

Student: Is the line ________?

Professor Ramamurti Shankar: Oh, let me see. No, no. This is not the equation of the line. This height here is equal to 1 - x. Oh, you think that’s wrong?

Student: [inaudible].

Professor Ramamurti Shankar: The height of y, isn’t there 1 - x?

Student: [inaudible]

Professor Ramamurti Shankar: Or it’s just x. I’m sorry. Okay, caught on tape. Very good. All right, so here is x, goes up to x and the y integral you do it as lower limit, upper limit gives you an x and x goes from 0 to 1. That gives you x2/2 from 0 to 1. That gives you 1 over 2. If you had a function of x and y you wanted to integrate, in other words instead of just the number one, if you had a function of x and y the rule is the following. You put the function here, and put x equal to a fixed value, don’t take it as a variable, and treat y as a variable and do dy from this limit to this limit. When you are done you will get some mass that depends only on x that you integrate from 0 to 1.

But the final thing I’m going to do is to do the same integral in polar coordinates, area of a triangle in polar coordinates looks like this. In polar coordinates, as you know, you draw circles of r and r + dr, and lines at θ and θ + , and this shaded region here has got one side equal to rdθ and the other side equal to dr.

So area integral will look like rdrdθ times some function of r and θ over the allowed region, but I want to do now a triangle of height 1 and base 1. Here’s what I do. I’m going to pick theta first and hold it fixed, and theta will go from 0 to Π/4. Can you see that? This is supposed to be a square. Theta goes from 0 to Π/4, and I have rdr. What is the range of r? How far is this thing? You can see that if this angle is theta then rcosθ is 1, and for r is 1/cosθ. So r dr is really d of r squared. d of r squared is equal to d of 1/cos2θ, right? I mean, let me do it this way. d of r squared is r2/2 from 1/cosθ to 0. So that gives me 1/2 integral 0 to Π/4 of 1/cos2θ.

Well, you may not know this, but sec2θ is the derivative of tanθ. So derivative of integral of tanθ will turn out to be simply tanθ calculated between Π/4 and 0. That’ll be just 1. So you’re not supposed to follow at lightening speed this integral, okay? Either you’ve seen it before or if you need some help you can look at my math book I mentioned to you. All I’m trying to tell you here is if when you have multiple integrals you pick one coordinate and you freeze it, for example, x when you integrate to the other coordinate over the allowed region that keeps you in the boundary then integrate over the second variable. You may need that, but for most of the problems I’ve given you won’t have to do that. They’re pretty simple.

[end of transcript]