 In the last lecture, we had introduced the concept of electric field and talked about Coulomb's law as well as the principle of superposition, that is when there are multiple charges which produce electric field at a point, we had said that the effect is essentially a linear sum of the effects due to each individual charge. And we had also talked about what happens in a continuous charge distribution where we wrote down an expression for the electric field in terms of integrals, which involve the charge density at a point. In this lecture, we will be trying to talk about essentially two ideas. The first one is known as Gauss's law, which is essentially a law very similar in content to the Coulomb's law. And we will see that in spite of the fact that Coulomb's law and Gauss's law are essentially equivalent statements, but under certain conditions it is much easier to use the Gauss's law. But before we do that, let us introduce the concept of a flux. The flux essentially comes in the same way as when for example, you have a flowing water. Now, when you have flowing water passes through a surface, basically the amount of flow that goes out of that surface, it depends upon not only the velocity of the flow, but it also depends upon both the area and the orientation of the area to the direction of the velocity. Now, exactly the same way we define the flux of an electric field. So, you recall that we had talked about the area being considered as a vector. So, if you look at an arbitrary area, then you could consider an infinitesimal part of it. The outgoing normal to such an infinitesimal area is the direction associated with such an area. It of course has a magnitude. So, in that sense an area though it is two dimensional, one can consider it as a vector. It is a vector, however there is a convention associated with it, namely that of the outgoing normal. Now, in this particular example, what we are showing for instance is a surface and there is an electric field which is coming and so the direction of that surface is perpendicular to this s and as a result the electric field makes an angle with the direction of that surface. And one defines the flux of the electric field as surface integral of the dot product of the electric field with the area element integrated all over. So, in other words if you look at a particular point where the area element the vector the normal vector is making an angle theta with the electric field, then the definition of the flux would be integral of s e d s cos theta. So, this is what we are going to be talking about, but before we continue with it we would like to introduce what is known as a solid angle. You are all familiar with the concept of an ordinary angle, you recall how does one define an angle in two dimension. Now, in angle in two dimension is defined by so it is something like this. So, supposing you have this angle. So, what is this angle theta, this is you draw an arc there and this angle theta is related to this length of the arc and this radius. So, if this is radius r then r times theta is this length of the arc. In other words you define this angle theta measured in radians as the ratio of the length of the arc to the radius. The only thing that you recall is that this arc is a part of a circle. So, for instance if you did not have this as a part of the circle, but you had a curve like this, then I could still define this angle theta by saying that I take the projection of this arc along the normal. Now, recall that in spite of the fact that an angle being ratio of two lengths r theta by r is dimensionless. We still conventionally measure it by using if you like an unit called degrees or radians. So, the this angle theta in radians is given by the length of the arc divided by the radius. Now, the concept of a solid angle is essentially very similar. So, what we do is this that suppose I am looking at what is the angle what is the solid angle that a an area makes at a particular point B. Now, what I do is this I join I draw tangents from that point on to this area and remember if I take this angle to be small then the direction. So, I have the following thing that the direction of the length which I take to be infinitesimal is perpendicular to the direction of perpendicular to the surface. And this is the direction r which is essentially joining P to some to that element and extending it and this angle this angle is alpha. Now, one defines the solid angle d omega made by this area element which is infinitesimally small area element as the projection the perpendicular component of this surface the perpendicular component of this surface divided by r square which means that since I am taking the projection perpendicular to this r this is d s cosine of alpha divided by r square. So, once again you notice that d omega and hence omega is dimensionless in spite of that we measure it in terms of an unit known as steridians just to look at this a little more in detail. Let us look at let us suppose I have a cone I have a right circular cone having a vertical angle half semi vertical angle equal to theta. Now, let me make a statement this red things that you see these are the direction of the electric field which is threading through this cone the from the base of the cone it enters perpendicularly and leaves through the slanted surface of the cone. Now, I am interested in calculating how much is the flux coming out of that cone. Now, before you do that please note that this cone has essentially two surfaces. So, let me draw a cone here. So, there is a base there and since we define n as the outward normal that would be the outward normal of the base. Now, suppose I am interested in finding out how much is the flux through this base. So, this we have seen is integral of E instead of dot d s let me write a dot n d s since the electric field is constant and has a magnitude E it will come out of this sign. So, basically and the second thing that you notice is the direction of n and the direction of E are opposite to each other because electric field is entering the base. So, as a result this E dot n is nothing but minus the modulus of E and I am left with only the integral over d s and assuming that r is the radius of the base of the cone. So, this is nothing but minus E times pi times r square the reason why I have a negative sign is because in my definition I had talked about outward flux, but as you can see from the picture the electric fields are threading into the cone as a result the net outward flux through the base is negative. However, let us look at what happens if I look at the slanted area now actually speaking you could we could supposing it were a water flowing through you could have got an answer to that without even working it out. The reason is whatever water is entering through the base must leave through the slanted surface of the base and as a result if the flux coming in is E times pi r square the flux going out through the slanted surface must also be equal to E times pi r square. Let us look at whether that is correct or not now what we do is this let us consider the following. .What is take a strip now this is this is a strip of radius smaller this is a strip of radius smaller at and width d l length and width will be d l at a distance at a height h below. Now, so if I do that look at this the how much so I have a length l and radius smaller here now. So, let us look at this what is E dot d s the E dot d s is of course, the magnitude of E the area of that strip area of that strip which is equal to 2 pi r because r is the radius there times d l which is the slanted width and the second thing is that the angle which the normal to the slanted surface is making with the electric field is obviously, pi by 2 minus theta. So, I get 2 pi r d l cosine of pi by 2 minus theta. So, let us write it again. So, E dot d s is equal to magnitude of E 2 pi r d l times sin theta. Now, let us look at this. So, let me redraw this picture on this. So, I have taken this strip here this height here is h this angle is theta and this is of course, r I can immediately look up a relationship between them you can see that the what is r what is the relationship between l h and r. So, r is obviously, equal to h tan theta and l which is this length which is this length is equal to h by cos theta because this is h. So, l cos theta is h. So, it is h by cos theta. So, if you plug them in here what I get is E dot d s is equal to magnitude of E times 2 pi and of course, I have got r d l sin theta. So, therefore, you notice there is already a sin theta by cos theta here and I have got this thing. So, I get h square times tan theta. Now, the tan theta is actually constant tan theta is actually constant and. So, therefore, what I get here is this. So, what is my flux phi I have to only integrate this quantity there. So, therefore, what I get is after integration I get E magnitude of E times if you do it properly and put in the relationship as we have done in the slide times pi r square. This is the flux through a cone, but then I already knew the answer that if the amount of flux that is getting in is E times pi r square the amount of flux which is going out should also be E times pi r square. Let us look at some results that is coming out of this. Let me take a sphere of radius capital R and let me put a charge q at the center of this sphere. Now, I already know that the lines of forces which are the directions of the electric field due to this charge q are all symmetric and if I have since I have got a sphere these are all perpendicular to the surface of this sphere and I am only interested in the electric field on the surface of the sphere. Now, since the direction of the electric field and the direction of normal to the surface are parallel the flux E dot d s is nothing but the E times d s and when I integrate. Now, remember that electric field strength is the same all over at all points on the surface of the sphere namely it is equal to 1 over 4 pi epsilon 0 q by r square. So, that is my strength of the electric field and since the electric field strength is the same everywhere the integral is only over the surface which gives me 4 pi r square that is the area of that surface and if I now calculate it it becomes q by epsilon 0. Notice that the amount of flux I put in a charge at the center of the sphere single charge q the flux coming out of the surface is simply given by q by epsilon 0 which is independent of which is independent of the radius of the sphere. Let us proceed with little more we already introduced the concept of a solid angle. Now, suppose I had a surface of arbitrary safe and there is a charge q there is a charge q that I have kept somewhere inside that surface. Now, notice supposing I am interested in calculating the a the flux through the surface. So, what I do is this I need to calculate electric field at every point on this surface. So, let me take an infinitesimal surface area d s at a distance r along the vector r from the point q. Now, what I do is this using this surface area I try to I connect them to all q and find out the solid angle d omega that this surface element d s makes at the point q. Now, let us look at what do I get out of it. Now, the so far as the electric field is concerned the electric field at the distance r is q by 4 pi epsilon 0 1 over r square along the radial direction along this direction. I have to now take the dot product with d s. So, I multiply with d s times cos theta whatever is the angle that makes it makes here, but you remember that d s cos theta is nothing but the projection of this area element d s in a direction perpendicular to the r. So, that is nothing but in our previous notation d s perpendicular. So, as a result what I get is that d s perpendicular divided by r square that gives me d omega. So, the d phi is q times d omega divided by 4 pi epsilon 0. So, notice that the flux due to the area element d s depends only on the solid angle that it subtends at the point q. Now, when I do the integration all that I need to do is to take the area element all over the surface and you notice that every place the direction of the radial direction is outward and the normal to the surface will make an acute angle with it. And as a result I get the total solid angle that is subtended at the point q which obviously is 4 pi. So, that I get the total flux equal to q times 4 pi divided by 4 pi epsilon 0 namely q by epsilon 0. Now, let us suppose this charge q is not inside that surface, but is outside that surface at this point. Now, I do exactly the same thing, but notice that the radial if you draw rays from this point q it will cut the surface in two places. One is here let us call it d s 1 another is here that is called d s 2. The normal the normal direction to this is that way whereas, the normal direction to this one is in the opposite direction. Now, so as a result as you go around this surface for every positive solid angle there would be a negative solid angle and the net effect would be 0. So, what have we actually got? What we got is that integral of e dot d s over a closed surface is equal to q by epsilon 0 when the charge q is inside the surface and is equal to 0 if q is outside. This is in general true if I have multiple charges as well because the principle is the same it is a superposition principle. So, what you actually find is this that the flux phi electric flux phi which is the same as the integral e dot d s is given by the amount of charge that is enclosed by the surface divided by epsilon 0. This is the statement of Gauss's law if there are no charges enclosed inside it is equal to 0. If there are some then of course, you have to simply talk about the amount of charge that is enclosed. So, that is my Gauss's law stated in the form of flux phi which is integral e dot d s. Now, let me now recall for you that suppose this surface s enclosed as a volume v. So, this integral e dot d s by divergence theorem is same as the integral over volume v and the surface integral will become a volume integral and I will be left with del dot e that is divergence of e d v and that is equal to the amount of charge that is enclosed divided by epsilon 0. The amount of charge that is enclosed can be written as integral over the volume the integration over the density. Now, since this is true for an arbitrary volume I get a corresponding differential statement of the Gauss's law which says del dot of e is equal to rho by epsilon 0. So, this is the differential form of the Gauss's law. So, let us do a little bit of detailed arithmetic and try to see whether one can directly derive it from Coulomb's law. So, for that I need the expression of Coulomb's law for a continuous charge distribution. So, now notice that that we had earlier written down the electric field e is given by 1 over 4 pi epsilon 0 this electric field at a point r integral over the volume rho at an arbitrary point inside that volume r minus r prime vector by r minus r prime cube and the integration is over the primed variable. So, if I want now del dot of this if I need the del dot of e this remember this del operator is with respect to the variable r which is the point at which I am trying to find my electric field. So, this can be written as 1 over 4 pi epsilon 0 integral over the volume rho r prime does not have any dependence on r it is a quantity dependent on r prime and then I write down del dot of r minus r prime divided by r minus r prime cube. So, this is this is del dot of this thing. Now, I can do a bit of a slight of hand and I notice that this quantity here depends on r minus r prime. So, instead of doing the derivative over the variable r supposing I am to take the derivative with respect to r prime then all that I need is just a minus sign. So, therefore, I put a minus sign here and put a prime there. So, the difference now is that this quantity now is del prime dot r minus r prime divided by this. Now, this is equal to minus 1 over 4 pi epsilon 0 integral over the volume rho r prime del prime dot. Now, notice this is essentially a vector like vector r by r cube. So, I can write it as a gradient I can write it as a gradient, but because I am writing it as a gradient over the primed variable I do not pick up that minus sign. So, this is del prime dot del prime of 1 over r minus r prime d cube r prime, but this is nothing but del prime square. This is nothing but del prime square. So, this is equal to minus 1 over 4 pi epsilon 0 integral over the volume rho r prime del prime square of 1 over r minus r prime d cube r prime. If you recall when we talked about the Laplacian operator we said del square of 1 over r this is a magnitude is equal to minus 4 pi times the delta function there is actually 3 dimensional delta function. So, therefore, this quantity here is minus 4 pi times delta function. Now, since the delta function is a 3 dimensional delta function I indicate it by del delta cube r minus r prime, but once I have a delta function this integral is easy because at all points other than at r is equal to r prime the delta function vanishes. So, I only have contribution from r is equal to r prime there is already a minus sign 4 pi cancels. So, I will be left with 1 over epsilon 0 times rho of r which is exactly what we derived using the definition of the flux, but you notice this is a hard work. Let me illustrate the calculation of flux taking a few examples. The I have here a cube now I have a cube and at the corner of the cube at the corner of the cube I have put a charge cube, but before I illustrate this notice a couple of things. The physical content of the Gauss's law is the same as that of the Coulomb's law there is nothing more nothing less. However, we can sometimes use Gauss's law to our advantage particularly where the symmetry makes it easy for us to write the left hand side namely integral E dot d s as a function explicit function of the electric field. Now, then if I can calculate that I can get an expression for the electric field this obviously cannot be done excepting in cases where there is significant amount of symmetry and that is exactly what we are going to talk about. Now, for instance let us take first a cube now supposing I put a charge cube at the center of the cube and I ask the question how much is the flux that is going out of any of the sides for instance this side. Now, first you invoke symmetry there is no reason why since the charge cube is at the center one side of the cube should be different from any other side. So, therefore, whatever is the flux through all the six surfaces of the cube should be divided equally by each surface and what is the amount of flux. So, the amount of flux that I have according to Gauss's law is nothing but charge enclosed cube divided by epsilon 0 cube enclosed which is equal to small cube by epsilon 0 this is the amount of flux that would go out through all six surfaces. So, if you want to talk about one surface when the charge cube is at the center I get flux through one surface or one face of the cube has to be equal to one sixth of cube by epsilon 0 that is good. Now, let us ask what if this charge was not at the center of the cube, but at one of the corners the symmetry is obviously lost. The symmetry is lost because that charge is at the corner intersection of three surfaces so these three surfaces are a different from the other surfaces, but we can do something better. So, notice that I am interested in calculating the outward flux through the shaded surface. Now, what I do is this a charge is at the corner I already know how to solve the problem if the charge happened to be at the center of a cube. So, why not artificially make this charge at the center of a cube, but please note that these surfaces that we are talking about on which we apply Gauss's law they are known as Gaussian surfaces. A Gaussian surface does not have to be a physical surface it could be an imaginary surface as well because the physical property of a surface never came into our discussion. So, I could be talking about artificial volumes artificial surfaces of course, the charge is real. So, let us look at this so what I now do is this that I put in eight such cubes I put in cubes like this. So, over this cube I put in another cube and put two on the side and things like that. So, therefore, this cube the charge which is shown in red is now at the center of a cube artificial cube, but this cube has twice the side of the original cube. Now, as we have just now seen that if a charge is at the center of a cube the flux going through any one of the faces is q by 6 epsilon 0 this expression did not depend upon the size of the side the length of the side. So, whether it is a or it is 2 a if the charge is at the center of a cube the flux through any face is q by 6 epsilon 0, but I am not interested in calculating flux out of the entire 2 a by 2 a face I am only interested in my original face. So, this original face now you notice is one fourth of one of the faces here and there is perfect symmetry there as well. So, you had q by 6 epsilon 0 through the 2 by 2 a by 2 a face through my original face of the size a by a it is one fourth of that namely q by 24 epsilon 0. .. Let me come to yet another example. So, let me take a uniformly charged sphere. So, there is a uniformly charged sphere containing charged cube now if you are looking at. So, let me illustrate this here supposing this is my sphere containing a charged cube and I am interested in finding out a field at a distance r from the center. Now, if this r is greater than the radius capital R of the sphere what I do is I enclose this in what I call as a Gaussian surface which is a concentric sphere of radius r the electric field of course is outward from here. Now, notice that at a distance r I am interested in calculating the electric field the magnitude of the electric field is everywhere the same because it is a sphere this is the Gaussian sphere imaginary sphere. So, that my flux is e times 4 pi r square the direction of electric field is the radial direction which I will put in by hand at a later stage and that is equal to q divided by epsilon 0. So, therefore, the magnitude of the electric field is q by 4 pi epsilon 0 r square and if you want to write down the electric field itself then of course you put q by 4 pi epsilon 0 r square times the unit vector r. Notice this expression is identical to what you would get if all the charges of the sphere instead of being uniformly distributed where to be concentrated at the origin of the sphere. Suppose instead suppose instead we have the sphere where the charge is uniformly distributed radius capital R, but I am interested in finding out the field at a distance r less than r. Now, once again I take a concentric Gaussian sphere of radius smaller the field directions are perpendicular to the surface of this imaginary sphere. So, the left hand side still calculates to be the same namely e times 4 pi r square, but remember in our right hand side I had q enclosed the amount of charge that is enclosed in this Gaussian sphere is not the full charge q because the volume is less by an amount small r cube by capital R cube. So, amount of charge enclosed is small q capital Q times r cube by r cube this is the q enclosed this divided by epsilon 0. So, electric field magnitude is q by 4 pi epsilon 0 notice it is now linear r by capital R cube the electric field direction is still the same sorry this is. So, let me put this as a vector field. So, therefore, what happens in this case is this that up to the radius r it goes linear and then it falls off quadratically following Coulomb's law. One more exercise let us look at a spherical shell spherical shell of inner radius A and the outer radius B. Now, what is the field at three different distances first if you look at totally outside the shell somewhere in this region my calculation is identical because the amount of charge that is contained is all that we have to calculate I have given that this is a spherical shell with charge density k by r square for a less than r less than b. So, rho is some constant k by r square. So, how much is the total charge the total charge then is integral rho d cube r. But notice this only depends upon r. So, therefore, the theta and phi integration do not come into picture and that gives me 4 pi angle integration gives me 4 pi and I have rho which is k by r square and d cube r which is r square sin theta d theta d phi the sin theta d theta d phi has been taken care of already in my 4 pi. So, this times r square d r from a to b. So, this is nothing but 4 pi k into b minus a that is my charge. So, if I am looking at radius r greater than b the directions are all radian. So, the magnitude of the electric field will be 1 over epsilon 0 the amount of charge enclosed namely the whole charge 4 pi k b minus a divided by 4 pi r square that came from that e dot d s. So, this also goes as the 4 pi will cancel out and I will be left with this expression k by epsilon 0 into b minus a by r square. So, once again it is as if the entire amount of charge is concentrated at the origin. Now, let us look at the region first let me look at the region r less than a. Now, if r is less than a I take a Gaussian sphere of radius r and inside that there are no charge because it is within the hollow region of the sphere. So, my electric field will be equal to 0 no field in the hollow space. Now, what about a less than r less than b. So, if a is less than r less than b the calculation is identical accepting now you have to worry about what fraction of charge is included there in. So, this is the amount of charge that is contained between this sphere of radius small a and this Gaussian surface marked with red and I can calculate how much is the volume of this simply the difference between the volume of the radius r sphere of radius r and the volume of sphere of radius a and repeat the same calculation. You will find the electric field is given by k times r minus a divided by epsilon 0 r square times r. So, this is this is something which we this is an application of the spherical shell. Now, let me take another yet another example. Let me take the example of a charged sheet an infinite sheet containing charge which I will take to be the positive charge. So, this is a sheet which is positively charged I am interested in calculating how much what is the field first let us look at what we what statements we can make without actually working out anything it is an infinite sheet. So, therefore, the field can only depend upon the distance from the sheet. So, field depends upon the distance from the sheet only the other thing is if this charge is taken to be positive. Then in the upper half of the plane above that plane the direction of the electric field will be outward perpendicular to the field below the thing it will be in the reverse direction, because the direction of what is outward changes if it is above it is upward if it is below it is downwards. So, the electric field is oppositely directed we will be discussing about it later, but you notice that there is obviously a discontinuity of the electric field at the charged surface. Now, what I do is this supposing I am interested in calculating the electric field at a distance r what I do is I take a cylinder I take a cylinder this is a Gaussian cylinder and let it intersect let it intersect the sheet. Now, look at the various directions the electric field on this face here on the top face is upward on the bottom face is downward on the side faces it is like this perpendicular to the this. Now, so when I calculate E dot d s there is no contribution from the slanted face and since the direction of n and the direction of e are parallel on the top and the bottom face the net outward flux that I get there are two faces pi r square each. So, I get 2 pi r square times e and that is equal to the charge enclosed which is pi r square times sigma, sigma is the surface charge density by epsilon 0. So, that the magnitude of the electric field is given by sigma by 2 epsilon 0. We will continue with more applications of Gauss's law and introduce the concept of potential in our next lecture.