 Good morning. We have had two lectures on electrostatics, but primarily the problems that many of you have seem to come from the vector calculus part. And what I will do is I will begin this session first clarifying the questions that have been raised by you which could not be taken online. So, the I will fortunately most of the questions that have been raised are all on one thing, but let me come to that a little later. But the first there was this question from SDM college 1047 center the person who asked this question did not understand the way I defined the divergence by means of a limiting value. So, let me explain what do I mean by that see the point is this that you are all accustomed to defining divergence in terms of its differential expression. What I did yesterday is to define divergence of a vector field f as a limiting value when I take a very small volume let us say delta v going to 0. And I said it is the ratio of the surface integral of the vector field normal f dot n d s divided by this delta v in the limit of delta v going to 0. Now couple of comments I will make to make this definition clear and I will also tell you how this definition is the same as what you have learnt as a differential relationship. One of the things is that I told you that the fields are characterized by having a point definition. In other words the a field is defined at every point in certain region of space and a vector field again is defined at points at all points, but having not only a magnitude, but also as a direction. So, what I told you is this that supposing you have a point and you are looking at you have a vector field at that point of course and you are looking at how do I define a divergence in that. So, the way to do it is this according to my definition now is to consider a small volume it is not material what type of shape of the volume that you take about that point around that point and calculate how much is the surface integral of the vector field f from all faces of whatever shape you have taken and divide it by the volume of this quantity. Now this is the definition I gave and from here getting into the what we called as the divergence theorem was very easy. So, we said I will come back to this definition again. So, we said this is the one of the most important things that we need to understand and this is divergence theorem which tells me that the normal component when it is integrated that is the surface integral of the vector field is given by the volume integral of the divergence of the vector field. Remember that divergence is a scalar. So, this is at this side is a scalar because of this dot product and this side of course by definition of scalar. So, this is my you know I mean simplest way of deriving the divergence. Now, let us see whether this divergence that I told you is the same as what you have learnt in your calculus. So, what I do is the following. So, let me first draw a I will take for simplicity a cube and let us suppose this is my x axis this is my y axis and this is my z axis. Let me orient a cube here. So, supposing I draw a cube like this and let us say that this is oriented along the axis. So, that it is intercept when the x axis is delta x on the y axis is delta y and on the z axis is delta z. So, the point is this that to realize what I am trying to say is this. So, notice first let me let me look at any two surfaces opposite surfaces. So, here is one surface which is parallel to the y axis, but the normal direction is along minus y direction and here is the opposite phase which is along the plus y direction. So, the direction of the normal are different. So, let us look at how much is the contribution of flux due to these two phases. Now, what I do is this I consider the corresponding points the corresponding points where the x and the y coordinates remain the same. So, what I said is that the left hand sides normal is along minus j direction the right hand sides normal is along plus j direction that is the y direction. Now, how much is the flux from the left side let us look at the flux of the field from left. So, if you look at the flux of the field from the left which is which has the normal as minus j. So, since I am considering only points opposite points whose values of x and y are the same. So, I do not have to worry about the x and the y component I simply look at the y dependence of the vector field only y of course, there is x and z, but they are the same on both sides. So, f y of y is the combination here and since minus direction is there and it is mine. So, I have a minus and this times the surface area which is delta x times delta z. Now, if you look at the flux from the right which is of course, a outward flux. So, I have a plus this time again the y component and, but this is y plus delta y because the x and the z have been taken to be the same and I have a y plus delta y, but the surface area multiplying is delta x into delta y. Now, look at what is the contribution to the flux from these two opposite side. So, in order to do that I need I notice that delta x and delta z are the same delta x delta z I am sorry here are the same. So, therefore, I simply have this minus this. So, my net outward flux is given by f y y plus delta y minus f y of y. Now, this is net outward flux from the two opposite faces whose area is delta x into delta z. Now, since delta y is small I can do a Taylor expansion of this and I can give get you then delta f y by y times delta y and I have delta x delta z. Now, I can do the similar calculation from the other pairs of faces and if I do the for example, the x face I would get f x x plus delta x minus f x of x delta y delta z and so, from the six faces my contribution will be now notice this is going to be delta x delta y delta z which is nothing but delta v. So, from six faces the flux the surface integral is given by delta f x by delta x plus delta f y by delta y plus delta f z by delta z multiplied by the volume delta v. Now, remember this is your traditional definition of the divergence with the Cartesian expression for the delta f v. So, what I have on this side is delta v times the sorry delta where divergence of f times delta v. My definition was of the divergence was divide this expression by delta v and take the limit. Now, limit actually in some sense I have already taken when I said that when you subtract only keep the first term of the Taylor series. So, this limit has already been taken. So, therefore, the definition I gave you yesterday is the same as the definition of the divergence that you are familiar with namely the divergence of f is given by d f x by d x plus d f y by d y plus d f z by d z, but that is the original if you like the geometrical definition of the divergence and so, therefore, that is what I thought was a good idea to define the divergence here. Then done that let me go to other questions. There was a question from that I will simply answer in one line. Center number 1339 College of Engineering Bhuvaneshwar. There are a couple of questions, but let me first go to the simpler of the question. The question was is a potential a scalar field? Obviously, yes you notice how did I define my field? I said a field is a quantity which has a value at every point and it would be a scalar field if that quantity only has magnitude, but no direction. So, therefore, since potential is a you can define potential at every point supposing I have an electric field you define potential due to the electric charges at every point in space. So, it is a point function. Secondly potential is a scalar. So, therefore, there is no question that potential is an example of a scalar field. However, the derivative of the potential the gradient of the potential which we know is essentially equal to electric field that is a vector field because again it is a point function, but which is defined from point to point having both a magnitude and direction. I have a set of 7 or 8 questions after that. And in fact since so many of you have asked this question and I guess a very large number of you have a confusion on a statement made by me that I said that every vector field can be split into two parts. The one part is one for which the curl is equal to 0 we call it as irrotational and another part for which the divergence is equal to 0 we call it solenoidal. Now, most of you all these questions that I said want to know two things one is what is the nomenclature you see the irrotational is very standard I mean there is no curl. So, therefore, obviously the field does not have a rotational motion there. So, this splitting that I talked to you about it is called Helmholtz decomposition. And they so let me let me try to explain that and also along with that take questions like why. So, basically the Helmholtz decomposition says incidentally I did not mention the mathematical restrictions that are there on some of those like the fields must fall faster than 1 over r square. So, let me let me not be try to be mathematically very rigorous, but the statement is that a vector field can be decomposed into one part which has 0 curl which is irrotational this nomenclature is obvious. And another part whose divergence is 0 which is known I am sorry it is written wrongly there it should be called solenoidal that second bracket irrotational is wrong it should be called solenoidal. And so therefore, I have stated there that f is equal to minus gradient of phi which is and we know that a curl of a gradient is equal to 0. The second one is this is of course very trivial because notice that the curl of a gradient makes sense that is because we are talking about the curl must be of a vector. So, anyway curl of a gradient that is equal to 0 and the second one is divergence of a curl is always equal to 0. So, therefore, now why is it called solenoidal in order to explain why it is called solenoidal because some of these are very old nomenclature and you have to try to understand the history a little bit while irrotational is very trivial to understand. You remember the physical meaning of the divergence as I told you I told you divergence essentially tells me how much a field is spreading out. So, in other words the supposing you are looking at a point if the if the field spreads apart at that point then I called it as positive divergence if the the the fields are coming in towards that we call it negative. Now, so when I say divergence is 0 it automatically implies that the there is the the the fields are not spreading out. Now, imagine lines which are not spreading out if the lines are not spreading out and you are trying to draw the field lines basically you will have to draw them as parallel lines which remain close. Now, the another way of saying that is that it is as if those field lines can be put inside a tube because they are not spreading out. So, therefore, they will not go out of the tube and you will recall from your magnetism what is a solenoid a solenoid is actually a tube like thing. So, therefore, it is as if the whole thing can be contained inside a solenoid and therefore, the name came that the the there is no divergence there. So, it can be put inside tube the field lines can be put inside tubes and therefore, there those with 0 divergence. So, that is about nomenclature, but that is not the main question most of you wanted to know you know what is the you know wanted me to explain this theorem a little better. So, let me let me go through I need a few more I did not plan on proving this theorem, but since very large number of you have asked that the theorem should be explained better I have no option, but to do it. So, let me I need a little more mathematics, but most of you would know it. So, therefore, I would hurry through those mathematics first comment that I want to make is remember grad curl and divergence they are all first order differential operators grad is gradient of course, curl up and divergence we have been using it. Now, so what we do is we define an operator which we call as the del operator which is written like this i d by dx plus j d by dy plus k d by dx. Now, the when it acts on a scalar function we call it grad. So, grad of a function f is written like either written like this or written like that that is i d f by dx j d f by dy plus k d f by dx. So, that is your grad. Now, divergence which is written by the del operator del dot there is a dot missing because of math mismatch this is an operator which acts on a vector function because del dot is a divergence. So, del dot of a vector function is the same operator which I wrote to you acting on the vector which is i f x plus j f y plus k f z which means d f x by dx plus d f y by dy plus d f z by d z. The curl similarly is also written like this I am not going to derive these expressions these you are all very familiar with all right. Now, let me define using this operator a thing called a Laplacian. Now, Laplacian is written as del square which is nothing, but a short form of del dot del. Now, notice that the second del second del being a gradient operator has to act on a scalar function and then it will give me a vector it will give me a vector function on which the del dot or the divergence will operate. So, therefore, the meaning of del dot del which is also known as the Laplacian operator is this that is i d by dx plus j d by dy plus k d by dz dotted with this and that gives you del square d square f by dx square dou square f by dou x square dou square f by dou y square plus dou square f by dou z square. So, I define my Laplacian operator as d square by dx square plus dou square by dou x square plus dou square by dou y square plus dou square by dou z square this is called a Laplacian operator. Now, let us look at this Laplacian operator a little better, but in order to do something like this and particularly because we work with electric charges which have a 1 over r potential I need another bit of a mathematics and I will spend a bit of a time on it because most of you might have done it as a course, but it requires a bit of a an explanation. I need to define and explain what is meant by a Dirac delta function Dirac as you know was a well known physicist. Now, if you look at it strictly it is not a function in fact, many people give it different name, but physicists like to call it a function. So, let me stick with that that it is called Dirac delta function the definition of the Dirac delta function is this when the value of its argument is not equal to 0. In this example, if delta of x is if x is not equal to 0 then the function becomes equal to 0 on the other hand if x is equal to 0 I do not define that function at all I do not define that function, but I define something crazy I say if the function is 0 everywhere accepting at a point what it is at that point I do not explain I say simply that if you integrate that function integrate any function multiplying it with that delta function and the range of your integration includes this point of argument the point at which this argument becomes 0. So, any interval from minus epsilon to plus epsilon will do f x delta x dx that is equal to simply the value of the function at the point 0 and this is we require that f should be continuous at x equal to 0. Now, you would immediately ask me why is it not a function in the proper sense. Now, in order to understand it go back to your definition of an integral. Now, remember that the integrals that we do they are usually called Riemann integrals and in the Riemann integral there is a theorem which you must have learnt that supposing I have a function which is 0 everywhere accepting at one point. Now, if I have a function like this then the area under that curve becomes equal to 0 because if a function irrespective of whatever its value at one point if the width of a rectangle which it encloses is 0 then your integral has to be equal to 0. But what I am saying here the delta function is such a function for which the integral of f x delta x dx equal to f 0 or integral delta x dx is equal to 1. Now, what does it actually mean if it is not a function what is it? If it is not a function you can look at it in many ways. So, for example, you can look at delta function as a limit of sequence of functions. So, let me explain by giving couple of example. For example, look at a sequence of quantities like delta n of x which are all 0 as long as x is less than either less than minus 1 over 2 n or greater than 1 over 2 n. But if it is between these 2 values then its value is n. Now, that is these are basically rectangle within this range of n height which is equal to n. So, look at what the picture is like this is the type of picture that you take various values of n and you get rectangles of different value. Now, what happens when you increase the value of n? Now, if you increase the value of n what is happening is that your this width is decreasing, but the height is increasing because height is proportional to n. But in such a way that the area of that rectangle becomes equal to 1. So, therefore, when you take this idea and take the limit of n going to infinity. So, what has happened is the width here is shrunk to 0, but the height has essentially gone to infinity if you like, but in such a way that the area which is the product of the height and the base has remained equal to 1. So, this is one of the ways of understanding it. There are many other ways this can be looked as a limit of Gaussian functions which you are familiar with I will just because we are going a little slow I will skip it, but this is what it is the picture is, but there is just an axis here. So, I take one Gaussian function which is in the green. Now, what I do is I keep on reducing its width, but increasing its height which is given by the blue and you could keep on doing this. You keep on doing this the Gaussians have been normalized in such a way that the integral is equal to 1. So, integral will remain 1 when this width keeps on decreasing and the height keeps on increasing, but so ultimately it will be a very spiked thing and therefore, this is also another example of a delta function. Delta function has many interesting properties, but I have just listed a few of them since I am not going to be using it right now I will just leave it in my notes. Now, one of the reasons why I introduced the concept of a delta function is to tell you what is the gradient what is the Laplacian operating on 1 over r I told you remember that 1 over r is my Coulomb potential. So, it is a very important potential that I have. So, let us look at that firstly supposing r is not equal to 0 remember 1 over r is not defined at r is equal to 0. So, supposing r is not equal to 0 the gradient because this is just a function of r. So, gradient is nothing but just d by d r. So, therefore this gives you minus vector r by r cube right 1 over r square unit vector r the del square is del dot del of this and so that is del dot of this quantity. Now, this is the divergence of a vector multiplied with a scalar and this is a very standard formula you can calculate it trivially and you get that del square of 1 over r is equal to 0, but provided r is not equal to 0, but what happens if r equal to 0 is included. Now, for that I am claiming that del square of 1 over r is essentially a delta function accepting for a factor of minus 4 pi which comes with it. So, del square so look at this definition del square d cube r I said it is equal to del dot del 1 over r d cube r. Now, d cube r is essentially the volume integral my integrand has nothing to do with angles, but d cube r has. So, d cube r as you know is r square sin theta d theta d phi. So, put all this now if you do that integration this integration does not vanish it becomes equal to minus 4 pi. So, I have worked it out here and so therefore, del square of 1 over r is a delta function, but for a factor of minus 4 pi that we need and that is because if r is not equal to 0 it is equal to 0 when the r is equal to 0 is included the volume integral turns up to equal to minus 4 pi. Now, this is a very important you know mathematical relationship which we will be requiring there. Now, why am I doing this I need it for proving the Helmholtz decomposition in the best possible shortest possible way there are many derivations, but most of them are long. So, therefore, this is the fastest. So, what I do is this supposing f is a vector field I define a vector v which is a function of vector r as an integral of f of r prime. Now, notice that this is a definite function depending upon the integration variable by r minus r prime d tau prime. Now, what happens if you take del square of v i i is the i th component. So, since f of r prime depends upon integration variable. So, it comes out because the del square is being taken with respect to r. So, it is del square 1 over r minus r prime see del square is being taken with respect to r because the left hand side v is a function of r only the vector r only. So, this del square of 1 over r minus r prime as we have just now seen is minus 4 pi delta r minus r prime. So, this gives me f of r. So, notice one thing the we have a given vector field f using this by applying the Laplacian operator. I would proved that they I can find a v defined through f corresponding to f I can find a v and this is my definition of v. So, let us let us proceed with that I need a an algebraic identity with which you are all very familiar which is del cross del cross of a vector field del cross del cross of v is given by gradient of divergence of v minus the Laplacian of v del square of v. Now, define remember I have defined v taking the expression for f. Now, what I do is using that v I define a scalar phi as minus del dot v and again using the same vector field I define a another vector a which is given by minus del cross of v. Now, that tells me del square v from this identity which is del of del dot of v minus del cross del cross v. Now, you can see immediately my del dot of v was minus phi. So, I have got it that minus del phi my del cross of v was minus a. So, therefore, this is del square v equal to minus gradient of phi plus del cross of v. Now, this is this is what we have, but del square v I just now proved is equal to f. So, you notice what we have done we have a vector field v f using that f I have constructed a v and showed that del square of v is equal to f. And now I have shown that del square of v is can be written as minus gradient of phi where phi and a are defined with respect to v itself plus del cross, but this is precisely what you want to prove that vector field f can be written as gradient of a scalar plus a curl of a vector and that is your Helmholtz decomposition. The other many people ask the question give us an example of vector decomposition. There are actually empty number of example, but let me try to give you some from our own subject namely electricity magnetism. So, you are all familiar with that the mean Maxwell's equation del dot of p is rho by epsilon 0 del cross of p is equal to 0. Now, notice that I can write e as equal to minus gradient of phi that is it does not have that it has only one component it is written as a quantity whose curl is equal to 0. And this plus 0 the the the solidarity part is 0. And this expression for phi is given by this the way we defined our v. So, this is an example of Helmholtz decomposition this is a very trivial decomposition because the second part is 0. Likewise, I can define corresponding to the magnetic field. So, this is I am going back to the same example I said phi is this 1 over 4 pi epsilon 0 rho by r minus r prime you can check that if you do del dot of p on this using the property of the delta function that we talked about it gives me rho by epsilon 0. So, that is there. Magneto statics provides you another example because del dot of b is equal to 0 del cross of b is equal to mu 0 j. And b is equal to del cross a once again this is an example where I have a solenoidal field, but I do not have a corresponding irrotational field. So, the expression for a is given by this. So, these are two examples from electrostatics and magnetostatics, but look at a little more general situation. I I claim that that I could define E the electric field for instance as equal to gradient of something V plus curl of some other field F where V is given by this expression what I do I have been given remember what I am trying to do I have been given an E. So, using E I construct this I know E. So, I say all right I have a scalar V of r which is del prime dot e r prime by 4 pi r minus r prime d tau prime. I have a vector F which is again constructing using E with this expression. Now, how do I know that this is correct? The way to know it is correct is the following I will give just one explain one I I am trying to say that E is given like this. So, what is supposing I calculated del dot of E? Now, this part is of course, identically 0. So, this is minus del square V is minus del square of this integration and once again the integration is with respect to r prime variable. So, therefore, it will go inside it will go inside giving me del square of 1 over r minus r prime which as you know is a delta function. So, therefore, this is gives me an identity and. So, therefore, this expression for V or similarly F that I have written down is correct. So, this is the way to understand Helmholtz decomposition. So, as I said I have spent more than an hour now in explaining the questions that came up and since I do not have really that much of time to start a new lecture what I will do is let me try to see what the new questions are coming in can you. Meghna the institute Meghna Chah Kolkata. Yeah, good morning Professor Ghavash. Good morning. Yeah, yesterday we talked about gradient of a scalar field and then we talked about curl of a vector field and we associated those things with some physical interpretation. And now we are talking about curl of a gradient of a scalar field is 0. And then we know divergence of curl of a vector field is also 0. So, what are the physical significance of those two mathematical forms? Number one. Number two, yesterday we talked about electrostatics Coulomb's law. So, which is essentially applied for charges at rest. So, can we apply Coulomb's law for charges in motion? If yes, why? If no, then why? So, let me let me talk about the first question first. You wanted to know to understand physically. You see the physically it should be obvious from the way you define the curl and the divergence because see we said that the meaning of the curl is how much the you know thing is rotating you know I mean it has a sense of rotation changing its direction. So, therefore, the if you take that physical interpretation of the curl right. So, the point is this that it is then understandable that such a field because it is curling out it is not diverging actually. So, I am just giving you a very hand waving argument on that. So, the and the same thing that comes out that see everything comes out from like for example, why it is called solenoidal that was the question which we tried to answer in the morning. The this is historical, the name does not have anything to do with solenoid, but it has got something to do with a tube because a solenoid has a tube like character. Now, if a vector field does not spread, if it does not spread the field lines can be confined in a tube and or in other words the field lines can be put inside a solenoid. So, the thing is this some of these mathematical things you can always prove rigorously mathematically, but at every stage of mathematics it may not always be possible to bring in a physical intuition, but but roughly this is what happens one of them is the field is diverging ok. The another one is the field is curling around and. So, therefore, there is it is a it is a divergence less field. So, they they I mean if you want a hand waving argument it is good enough, but it is good to sort of say that at every stage of mathematics you do not look for a hand waving argument. I mean I think it is best best to prove it rigorously, yes. Actually I am wondering I mean whenever we are taking gradient of a scalar quantity and it always gives us a curl 0. So, what really makes it so special that a vector which comes out from a scalar field always gives us curl 0, that was my. So, you know I mean I cannot probably answer it much better than what I did because what you are asking me essentially is this that take an arbitrary vector and then take a gradient and look at what the gradient I mean sorry take an arbitrary scalar then calculate a gradient and see what does that gradient look like pictorially and and then try to show that why such the curl of such a thing should be 0. I think it would the mapping of all this physically is going to be troublesome I mean because you know I can always think of one scalar, but then you will think of another scalar it is probably not never a good idea to try to understand every step of mathematics physically I mean I do not personally think. Now, the next question was connected with the Coulomb's law the Coulomb's law is valid whenever there are charges ok whether the I will be trying to tell you if not today maybe tomorrow whether the charges are at rest or at motion the thing that we will try to point out is this the electric field and the magnetic field are different manifestation of the same situation. So, we call say for instance supposing you have a charge at rest in your lab now you would say ok Coulomb's law is valid it is electrostatic from Coulomb's law is valid now supposing the charge is still the same, but you are moving now with respect to you the charge would appear to be in motion then you would say there is there is effectively a current if there is a current you would say there is also a magnetic field right. So, your friend who is sitting down there in the lab will say there is an electrostatic field and no magnetic field because the charges are at rest whereas you will say that I have a charge. So, therefore, there is an electric field I agree, but my charge is in motion. So, therefore, there is a magnetic field also the point is that I mean you can rigorously establish this if you go over to relativity, but I will try to give you some example of how they this thing that electric field and the magnetic field are also related to your frame of reference. Thank you. Yes, Silliguri go ahead. Good morning. Good morning. I just was just wondering then Helmholtz theorem you stated that it is not valid for just any vector field. I guess there are two conditions first one is it has to be twice differentiable and the second one I think it has goes faster than 1 by r I was just wondering why is the second condition needed. Okay, I mean I knew that somebody will come up with that question see my problem is that I have to mathematically justify it and it is sort of taking a lot of time. So, what I could do is if you mail me that question I can send an answer because I really cannot spend that much of time on mathematical niceties, but yes the second condition that the field must fall faster than 1 over r square is correct you are absolutely right and we need that condition in fact then you might ask me the question in which case how is it being used for electrostatics where the Coulomb's law is does not go faster than 1 over r square, but then that is a very limiting case. So, I will not be able to spend that much of time on mathematics here you know I mean we have already spent almost three two and a half three hours on mathematics today. So, if you send me a mail my mail address must be there on the model I will try to answer that So, yesterday when we concluded we had said that the there are two things which are going to be useful for us one is the Coulomb's law and of course we know what Coulomb's law is and we said that the if there is a test charge q at a point e then the it experiences a force q times e and the second session point is that if you have large number of charges the effect due to them add up that I called as vectorially add up and I call that as the principle of superposition. Now it is important that your students understand the that principle of superposition is a very important concept because it is not clear that when there are two effects which are acting at the same place simultaneously right it is a matter of simply adding them up sometimes you know the effects multiply, but so superposition principle says that if there are different forces the forces simply add up. So, that is a very important point so the we our entire basis of electrostatics electromagnetic theory is based on the you know validity of the superposition principle. So, that if I have a charge q i at the position r i then the field at the point p which is has a position vector r is given by 1 over 4 pi epsilon 0 sum over i q i r minus r i by r minus r i cube. So, this is my superposition principle and Coulomb's law. Now having talked about Coulomb's law let me jump to we discussed yesterday the definition of electric flux and Gauss's law. So, let me make one statement very clear Gauss's law is completely equivalent to the Coulomb's law, but the reason why you use Gauss's law is it is much simpler to deal with in some cases and particularly in cases which exhibits symmetry the Gauss's law is much better used over Coulomb's law. In fact, later on when you do magneto statics you will realize the role of Coulomb's law is taken by Barrett Savart's law there and there also we have a situation corresponding to the magnetic Gauss's law and once again or even there are things like Ampere's law even there such laws which depend upon symmetry are much easier to deal with. So, basically what it says is this remember how we define the flux as the surface integral of the electric field. Now here is for example a surface and there is this electric field coming in and so the there are incoming from one side outgoing from the other side. Now let us try to understand this Gauss's law a little bit because yesterday actually there was a question also which says why is it that if I have a charge inside a volume I get a flux, but if the charge is outside I do not get a flux. Now in order to understand that we need another mathematical concept and that that is the concept of a solid angle. Now what I told you yesterday is the solid angle is a straightforward generalization from the way we define an ordinary angle. Now remember in an ordinary angle what we do is to say that supposing I have two radial lines from a point at equal distances I draw an arc. Now this arc is of course perpendicular to the two radial lines and so therefore my definition of the angle that it makes is given by the length of the arc divided by the radial line and as a result the angle is a dimensionless quantity it is a different matter that we made some units like degrees radians and things like that, but it does not have any dimension. Now the situation with respect to solid angle is identical. So here what we do is instead of an arc we take a surface supposing I have this surface ds and at any point imagine that from that point we start drawing tangents all over on to that surface. So this will make a cone at the point b. Now this quantity which it makes that is obviously feels like an angle, but you notice that it is not a linear angle it has a conic angle if you like. Now how does one define remember our definition of angle theta was the length of the arc divided by the radial. Here I define exactly the same way I say it is the area the perpendicular area like there also we said it is the perpendicular one. Perpendicular area divided by r square because this has a dimension of r square this also has a dimension of r square. So therefore, this becomes and since if the radial line is vector r and the normal to the this surface remember that these are all valid for infinitesimal surfaces is n and if this angle makes an angle alpha then this is given by ds cos alpha by r square. So, now let us look at why did we suddenly talk about Gauss's law and solid angle. Now look at this point here supposing there is a volume there if I have a volume there then I am looking at what is the flux through this surface. Now so I have a q here and this is a enclosing thing remember the my definition of Coulomb's law. So, it will be 1 over 4 pi epsilon 0 q divided by r square is my thing there. So, therefore, since the way we defined our solid angle. So, solid angle was essentially the normal component of this thing divided by r square. So, therefore, what I did is to say the flux through this infinitesimal area is given by q times d omega by 4 pi epsilon 0 because remember that how d omega was defined it was this area the perpendicular area divided by r square. So, r square is there in the Coulomb's law and ds which will come in here perpendicular that is there in the definition of flux. So, therefore, the flux from this area element is written as d phi which is equal to q times d omega divided by 4 pi epsilon 0. Now if this is the situation then if you are looking at the total flux from all over then you notice that my solid angle is going to be have to be integrated and the total solid angle around is 4 pi. So, therefore, if I take how much is the flux. So, I will simply get this this will give me 4 pi I will get q by epsilon 0. So, now let us look at the same situation when this charge is outside because this was the question why do not you get a why do you say in Gauss's law that the flux from a surface is given by a closed surface is given by the charge enclosed divided by epsilon 0. So, look at this here now if you want to look at this point P and draw a tangential line this is going to intersect this closed surface not at one point, but at two points simultaneously. So, this is one area which it intersects this is another area which it intersects. Now the point is this that there is a slight difference between the two areas. So, far as this area is concerned the angle between the radial line and the normal is acute angle. So, far as this is concerned the angle between this area and the radial line is an octagon I mean actually it is the normal direction is outside above the. So, there is a opposite direction. So, therefore, since the both of them irrespective of the size of the surface that they enclose since they will give me the same solid angle. So, therefore, one of them gives me a positive angle another of them gives me a negative one. So, therefore, the total solid angle that is substituted is equal to 0 and so therefore, if the charge is outside it does not subtend any angle sorry about the any flux. So, that was of course, my the statement of the Gauss's law. So, we have our first Maxwell's equation here that is saying that the flux is equal to enclosed charge divided by epsilon. This is of course, what we learnt in school, but let us try to cast it in differential language. So, what we do is this remember we are talking about closed surfaces. So, flux is integral e dot d s integrated over the closed surface that is the definition of the this side. On the right hand side let me come back to this next, but on the right hand side which is there by the last term I have a 1 over epsilon 0 times enclosed charge. Now, how much is the enclosed charge now obviously, it is given by the integral of charge density inside that volume integrated over the volume. So, now, but we did what is called a divergence theorem this is a surface integral e dot d s. So, this must be the same as divergence of e volume integral. So, therefore, the Gauss's law also says that the volume integral of the divergence is 1 over epsilon 0 times the integral of rho d v which is nothing, but the charge. Now, if you look at this relationship, if you look at this relationship then you realize that the this is this must be true of any volume. Now, a relationship like this which is true of irrespective of the size of the volume mean that the two integrants must be the same because you know you can always satisfied for a particular volume, but if you say this is true of arbitrary volumes then the two integrants must be the same which tells me that the divergence of the electric field is equal to rho by epsilon 0. So, this is the first of the Maxwell's equation that we have derived that is basically equal to the Gauss's law electrostatic Gauss's law. So, I made a statement that Gauss's law is completely equivalent to the Coulomb's law. So, let me let me try to get that. So, to do that let me let me try to get this del dot p from Coulomb's law. So, what we said is this that del dot of p is given by 1 over 4 pi epsilon 0 del dot integral rho of r prime r minus r prime divided by r minus r prime cube d cube r prime. So, this is the this is equal to 1 over 4 pi epsilon 0. Now, what I do is this I take this gradient take this divergence del dot inside, but remember that the integration variables are r prime whereas, I am differentiating or this divergence is being with respect to the r variable. So, I put it like this this thing del dot of this quantity which is r minus r prime by r minus r prime cube d cube r prime. And I need a minus sign now because what I now do is this I say all right this is this was divergence with respect to r you notice this quantity is entirely a function of r minus r prime. So, what I could do is I could make this divergence with respect to the prime variable and since this is a function of r minus r prime I put a minus sign there. So, I get minus 1 over r minus 4 pi epsilon 0 rho r prime I have got already del prime cube dot now this quantity here r minus r prime by that I write it as del prime. So, I write it as del prime of 1 over r minus r prime. Now, notice what I have done I am just doing some mathematical tricks the gradient of 1 over r minus r prime if it is with respect to minus r prime variable is plus 1 over r minus r prime cube into 1 minus r prime because the variable there is a minus sign that will come in successive differentiation. But with this there is a del prime dot del prime. So, which gives me minus 1 over 4 pi epsilon 0 integral rho r prime, but this is nothing but del prime square that is the Laplacian of 1 over r minus r prime. But this we have shown is given by minus 4 pi epsilon sorry minus 4 pi times a delta function. So, if you use the delta function to do this integration you find this is equal to rho r divided by epsilon 0. So, this is the mathematical derivation look at that here is Don Boskoff. Hello. Yes. What I want to say is in spite of the ease of applying the Goss law I believe it has got limited scope compared to the Coulomb's law here is because the Goss law can be applied only in case there is some kind of a symmetry. In case there is no symmetry obviously we have to fall back on the Coulomb's law only because the Goss law in the most general case it will only give you the total flux not the electric field. You are absolutely right I made that statement earlier also I said that Goss's law or for that matter Ampere's law later when we do it they have the limitation that they can be effectively used only in situations where there are sufficient symmetry. But remember that Coulomb's law is not very easy to use in general cases. So, in general cases both of them are equally bad. But in some high synthesis the Goss's law gives you an advantage over Coulomb's law they are completely identical in some situations you should use Goss's law because of the symmetry. I have I completely agree with you that the Goss's law can be used in very limited situation but mathematically it is as rigorous as Coulomb's law. So, do not be under the impression that see all that the difference that is there is I may not be able to calculate it using Goss's law whereas Coulomb's law at the worst case you can go to a computer and do a work. I agree with you you are right thank you which is that. Maulana Azad NIT yes. Hello very good morning sir. My question is related to the concept of potential that what is the... Related to concept now? ... concept of potential whether it is just a mathematical concept which is used for the sake of mathematical ease. Potential are you talking about? Yeah. Is that a potential you are talking about? Yes sir. Okay, so the point is this that is it just mathematical ease? Answer is only partly yes. Firstly you realize that it is much more difficult to deal with vector fields than scalar ones okay the potential is a scalar field the advantage of a scalar field is you can since superposition principle is valid the you can go on adding things up without having to worry about what supposing you are working with electric field you will need to worry about the directions of the electric field how to add up vectorially but what I can do is when there are multiple situations I can just add the scalar fields just like that okay and then after I have added all the effects I take a final gradient to find out the net electric field. Now I have avoided the entire problem of having to do a vector addition so it is not just a triviality now the question is there also a meaning to that the answer is yes in fact it is similar to you know many of these concepts even the name potential came from for example the fluid dynamics the point is this that you see when we are talking about fluid flow the level at or the height at which a fluid is located is also important because you make a statement that the fluid has the natural tendency to go flow from a higher potential or a higher level to a lower level now I actually borrow the same thing here also that the potential has a meaning because it is essentially telling me what is the level what is its electric level and supposing you feel put a charge there will it simply which direction will it go okay will it climb up will water climb up the answer is no. So these are also extremely intuitive concepts the the third thing is the potential though it is not equal to potential energy there is a intimate relationship between potential and the potential so therefore 1 it is a scalar function 2 it sort of tells me what is the natural tendency of the charged particles in an electric field which way they will move so I it is much easier instead of calculating the force to tell me the look if I know the potential I know it goes from a higher potential to a lower potential right and the third thing is its direct association with potential energy thank you. So let me give you a few examples of as I told you that we are not you know I mean strictly following a tutorial but we will be having some look at this interesting problem it is a question of I have a small charge charge q and a and a cube and this charge q is located at one of the corners my question is that what is the flux through one of these surfaces remember some of our friends pointed out the Gauss's law is useful only when there is sufficient symmetry at this moment it does not look like there is a symmetry because it is in one corner of a cube but you see if you are smart enough you can create a symmetry how do you create a symmetry what you do is you pack up this I have shown you here that put one more cube another cube below another cube here. So make it into four different cubes such that this charge q which was at one corner of the cube comes to the center of this four cube situation. Now I asked the following question supposing you had a charge which was at the center of a cube now then what is the flux that is going out from any one of the surface now you would say that look this is so symmetrical that from this bigger size cube supposing there is a charge at the center of a cube since a cube has six faces and with respect to the central position there is no special position of any of the faces there are the six faces. So q by epsilon 0 was the is the total flux through the whole surface. So q by 6 epsilon 0 should be the flux through any one of these surfaces fine but then this surface that I have got is four times as big as my original surface. So therefore my net through the shaded side here is just q by 24 epsilon 0 you notice a simple idea you try to use Coulomb's law you will be nowhere but beautiful symmetry of this problem which we have constructed. In the second session I will continue with application of Gauss's law and since we seem to be running out of time for the current session and you all we you all as well as myself we need to go out for tea we have just couple of minutes maybe one more quick question if there is a clarification you can take it up RBS engineering college can you quickly if there is a quick comment or a question you can tell me because otherwise we will be answering them all together yes go ahead good morning sir good morning sir my question is that apart from magnet magnet is there is any example where net divergence is zero well yesterday I told you velocity field for the there are lots of situations where the things could be zero magnetic field I talk about because this is subject of our course here but there are velocity fields which you can have where the divergence can be zero because the question is that if in through a particular volume the amount of fluid coming in is equal to amount of filling going down then of course the divergence could be zero because the fluid is incompressible so the row the density does not change there there I mean if you want to look up other examples from various fields we I am sure we can look it up because it is a very standard field and but we stick to because our magnetic field because of absence of magnetic monopole is that but remember even with electric situation if I have a dipole for example right I can create a situation where my situation would be zero but the you I will remember this question particularly if I mean fluid flow is one of them which I often remember if there are other examples which I can find out I will answer it next time thank you very close okay we have a 15 minutes break now for T and we'll come back and continue with quantum mechanics session