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# PHYS 201: Fundamentals of Physics II

## Lecture 4

## - Gauss's Law and Application to Conductors and Insulators

### Overview

Lecture begins with a recap of Gauss’s Law, its derivation, its limitation and its applications in deriving the electric field of several symmetric geometries—like the infinitely long wire. The electrical properties of conductors and insulators are discussed. Multiple integrals are briefly reviewed.

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html## 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]
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. So it’s proportional to ^{2}q. It’s proportional to 1/r, 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 ^{2}E at radius r is also proportional to q and also proportional to 1/r. Because from Coulomb’s Law the field is the force on a unit test charge, it is ^{2}q. _{1}q_{2}q is 1 because the test charge is 1 and there’s _{2}q/4Πε. Forget the 4Π. It’s proportional to _{0}r^{2}q/r.^{2}From this we may conclude that the line density is proportional to the electric field because they are both proportional to 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/ε 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._{0}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 Φ r. Now, Φ_{S’} 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?
Therefore, this is equal to the integral of the electric field at some point So now, Φ E⋅dA on some surface S’. But E(r) times 4Πris ^{2} c times q/ε_{0} because E is q/4Πε_{0}r. You see that? This much you should know. I’m not going to write the details. So that is equal to ^{2}c times surface integral of E⋅dA. 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 ε Okay, now we do the next thing. Let’s call this charge q 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 _{2}Emeaning when both are present is equal to surface integral of _{1,2}E⋅_{1}dA + surface integral of E⋅_{2}dA, because E is _{1,2}E + _{1}E. But the first guy is _{2}q/_{1}ε, second is _{0}q/_{2}ε. Once you can do two charges you can do any number of charges._{0}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, 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 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 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 ## 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 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 r is a center-to-center distance as if the entire earth were at the center. A similar law is true for electric forces.^{2}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 The trick is to first decide. This surface 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 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 4Πr, because ^{2}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. 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._{r}As long as this field is outside the real ball no matter what its radius is, the charge enclosed is always Q is now the charge enclosed. So I’m going to write, again, 4Πε, I’m sorry, 4_{0}Πr times ^{2}E(r) is the charge enclosed over ε_{0}. 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?
R. 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 ^{3}Q divide it by the volume 4/3Πr. That’s the density. Then you can multiply the density by the volume of the little sphere 4/3^{3}Π little r cubed, and all I’m saying is forget these 4 thirds pis. It’s just little r over big ^{3}R. Therefore, the electric field now, ^{3}E(r), looks like Q over 4Πε divided by _{0}rr. This is for ^{3}r < R, is equal to Q/4Πε(1/_{0 }r) for ^{2}r > R.
R and this will also give you 1 over big ^{2}R These formulas will match on the surface of this sphere. Yes?^{2}.
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 e times the value of the tiny area. In other words, _{r}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 r it will grow with r then it will decline as 1/r. And here is little ^{2}r equals big R. Yep?
_{0} at times what?
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 r 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 _{2}r squared. This area is proportional to _{1}r 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 _{1}r vector is different.So you have to take the cosine In other words, if the force of gravity or the electric force did not fall exactly like 1/ r 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?^{2}
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 r. Therefore, if this radius is proportional to _{2}rthe area of that circle is _{2} Πr. This is _{2}^{2}Πr. 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 _{1}^{2}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 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/ E on the surface, and it could be any surface, but the beauty of this surface is the following. There are contributions to E⋅dA 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 2 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 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. 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 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 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 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 dA cancels. The electric field is σ/ε_{0}. 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/2ε 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 And you can ask, what’s going on? Why is it σ/ε 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/2ε 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, 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 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 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
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 So area integral will look like Well, you may not know this, but sec [end of transcript] Back to Top |
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