 So, in the first session I had talked about Coulomb's law and Gauss's law and just when I was about to give some examples of applications of Gauss's law, we had the session terminated. So, let us come back to where we left from. So, what I want to do is I do not want to do the traditional problems, but let us take couple of interesting problems, but before I do that a couple of clarifications. When I ended the lecture last time someone had asked me that you put del dot of E is equal to rho over epsilon 0. So, what happens if I am you know what does it mean actually am I in vacuum or what see till now all that you are doing is vacuum electrodynamics ok. The change that occurs if we have material is a lot more complicated I will try to give you a flavor of that that is what happens when you have dielectric, but because of shortage of time I would not have occasion to do it in any great detail. So, the point is this that we had seen that two things another thing is that the superposition principle just now as being asked by somebody superposition principle is another hypothesis with which we work that is if a body in fact this is something which we have been accustomed to in mechanics for a very long time. That is if a body is acted upon by two forces the net result is given by vector addition of the two forces and that is all that we are talking about. In fact the whole of quantum mechanics is based on superposition principle the linear algebra is what decides the validity. So, but even in mechanics accepting where non-linear effects come in the superposition is a pretty good effect. So, we are working with superposition principle and here is an example of superposition principle. So, what I have here is a sphere which is it has a uniformly charged sphere with let us say charge density rule. Now, what we have done is to have a another small sphere of let us say radius a which is becomes a cavity inside this. I have not given you the total charge because it is a it makes the algebra is little messy, but I have said that the charge density is low. Now, the question is how does one calculate the field at any point? Now, the point is this to realize the following that supposing I had a field a sphere having a charge q then I know the expression which we have done from our school of how to calculate the field at any distance within the sphere and as you know the field goes linearly with r and once the you reach the surface edge after that of course, it goes as 1 over r square because the effect is the same as if all the charges were concentrated at the center. Now, this is something which we have done in school. Now the point is this I do not quite have a field sphere. Yeah. Just non-conductive conductors are not yet in there you see I said that the charges are uniformly distributed. So, that statement itself makes it clear that it is not a conducting sphere. Actually you are getting a charge sphere this will be a sphere of charge. What difference is there? The sphere of charge is non-conductive sphere and charge sphere is a conducting sphere. Why? I mean if a sphere has a charge I can call it charge sphere it is the English language what is the problem? The sphere of charge is non-conductive sphere. Why is it? I mean does anybody understand why a sphere of charge and charge sphere are different? It is the same thing, the fact that I am talking about a charge sphere at this moment and I have said uniformly distributed my physics is contained in that. If it is a conductor charges cannot be inside all right. I am studying somewhere in Residic and Holiday. All of us have studied from Residic and Holiday that is not a problem but the point is that the moment I say I have a uniformly charged sphere that statement makes it very clear that what I am talking about is not a conductor because no charge can be uniformly spread inside a sphere if it is a conductor because the charges will then only go to the surface ok. So, uniformly charged sphere automatically implies that it is a non-conducting sphere ok. But you see the point is this that at this moment let us forget about this cavity. I had this whole thing filled up with the charge density rho then I know that the field is linear. Remember that usual picture which you have that it goes linearly and then drops as 1 over r square. The field due to this sphere had it been totally filled up with charge would be given by q by 4 pi epsilon 0 linear factor r by r cube ok. Now so, if I realize that q by the volume volume is 4 pi by 3 r cube right that is my volume then of course, this statement is the same as rho over 3 epsilon 0 I. So, if I am given the charge density then this is the expression for the electric field. Now, let us go back to the smaller sphere. So, in the smaller sphere I have again it goes as 1 over r. Now so, if this distance is whatever is that distance and D is the distance between the two centers then the with respect to this origin this distance is given by r minus d right. If this is r this is d then this o prime p is r minus d. So, therefore, the field due to smaller sphere this minus sign I will explain would have been this, but I have a cavity. So, what I do is I say alright a cavity is same as filling up the cavity with positive charges and filling up it up with negative charges cavity means there are no charges that no charges same as plus charge plus minus charge. So, what I do is I first fill the cavity with positive charge density rho. So, that I have a big sphere of charge density rho and then I am left with the smaller sphere with the charge density minus rho is this clear. So, the original problem because of superposition principle is equivalent to a sphere of radius capital R superposed with a sphere of radius a, but filled with negative charge of equal density. So, therefore, I would have this rho by 3 epsilon 0 r minus d which is the o prime p, but with a minus sign. So, the cavity is being replaced by positive charge density and negative charge density as if superposed with each other. And now you use your superposition principle add up these things you find that the net field is given by rho by 3 epsilon 0 d and notice the interesting thing that the field inside the cavity field inside the cavity does not depend upon where you are calculating field because the only dependence that is there is on d and d is simply the distance between this sphere and that sphere. So, no matter which point inside the cavity you calculate the field the field is uniform ok. This is a consequence of the superposition principle. I will apply the superposition principle for yet another problem and this is that suppose I have two spheres. I have two spheres which are partly overlapping one with a center at o the another with a center at o prime both of them have the same range. Now notice this thing again what has happened is the following that I have a sphere of certain density I have a sphere of another opposite charge density and I am looking at what is the field in the intermediate region. Now once again I use the superposition principle and you can calculate the field at any point p due to this sphere it will be q by 4 epsilon 0 1 over r q o p vector. Due to this sphere it is o prime p vector and when you subtract them you are left with only o prime vector. So, once again you notice that field in the intersecting region is uniform because it depends only on the distance between the two so the centers of the two spheres ok. So, let us summarize again we had made the statement that Coulomb force is central. I did make out a point that look the Coulomb force is also inverse square law all, but 1 over r square is an important force. The force is conservative in fact this statement is true for any central force it does not have to be 1 over r square you could take any dependence of the force on the distance the force can be shown to be conservative. Now if the force is conservative then there exists a potential function remember the definition of conservative force. A conservative force is one where the work done in going from one point to another does not depend upon the path that you take it only depends upon the two points the your beginning point and the end points. So, the this is my f I have written 1 over r square, but since its direction is against along the radius this is the same as vector r by r cube. Now if you calculate the curl of this vector then you find curl of this quantity you are taking. Now this is curl of a scalar times a scalar times a vector. So, which is grad 1 over r cube cross r plus 1 over r cube times del cross r. Now you notice here this of course I can calculate immediately del of 1 over r cube which is minus 3 by r to the power 5 actually r to the power 4 which I have written as vector r by r to the power 5 and cross r and this is del cross r ok. So, this quantity is 0 and I am left with r cross r which is 0 instantly curl of a vector becoming 0 is also a test of the vector being the force field being conservative. So, you can calculate if you are given a force you can calculate its curl and if it is 0 the force is conservative and this follows from the Stokes theorem. Because if del cross r or del cross f is 0 if I take the integral any surface integral del cross f dot ds is the same as integral f dot dl, but del cross f being 0 this tells me that the path integral would depend only on the two end points ok. So, say that because the force is conservative I have a potential function there and this what is the point in having potential I already have field. The greatest advantage of using potentials is the potentials are scalars they are numbers which can be added up. Whereas if you deal with field this is equally satisfactory provided you can add up vectors properly. And obviously, working with large number of vectors is much more difficult than working with scalars which are simply additive and that is the reason why people want to and since the superposition field principle holds good. So, if you have different sources you calculate the potentials due to different sources add them up and then take the gradient of that it is equally good. So, that is what happens. So, notice this that the del gradient of 1 over r minus r prime is this and so therefore, if you calculate from here that electric field works out to situation like this I will not be explaining any mathematics, but they will be on the notes you can find it. So, therefore, an expression for the scalar potential is minus 1 over 4 pi epsilon 0 integral rho r prime by r minus r prime dv prime. Now, you can see what happens sorry minus sign is unnecessarily coming here because there is a minus already there. So, minus should not be there 1 over 4 pi epsilon 0 rho r prime by r minus r prime dv prime look at this expression ok. See in this expression you would say I have similar thing, but remember that here I have to worry about the vector addition which is not being explicitly talk to work here. So, I should put actually this one and then I say that that is the gradient of that that becomes a lot more difficult to work out alright. Now, since the curl of the electric field is 0 this is my second electrostatic equation. So, the first Maxwell's equation was del dot of E equal to rho over epsilon 0 del cross of E is equal to 0 del dot of E equal to rho by epsilon 0 would change if I am looking at dielectric material, but we will be talking about it later. Del cross E is a result of electrostatic force we are discussing only electrostatics when we have time dependent phenomena that equation will also change. So, at this moment we are in electrostatics only. Now, the old potential it always becomes confusing. You see the potential is not potential energy though there is a connection between potential and potential energy to understand what exactly is the connection. Suppose you have a charge q which you want to bring it from a point which you define as the 0 of the potential. Now, we will see later that though more commonly we take the point at infinity to be 0 of the potential this is not always feasible. This is very similar to in gravitational potential we normally take the 0 of the potential to be at infinity, but if I am calculating in this room I would take the 0 of the potential on the floor. So, it is just a matter of reference point. Now, so, if I want to bring the I have to do work when I say I have to do work it means that the external agency has to do work somebody else because what you are doing actually is a it is like you are going up the slope you had a you were at 0 and you are going to a point with higher potential. So, when you go up the slope or push a body up the slope you have to do work and this work is negative of the work done by the force and as you are aware that the work done by the force is f dot dr. So, f is minus del phi dot dr this is nothing, but d phi itself which is what I told you in the morning. So, therefore, it is minus f r. So, therefore, the work done by the external agency in bringing a charge q to the point p is phi times q times phi p. Now, this amount of work that is done by work energy theorem will be stored as the potential energy of the system. Now, see the problem is the what is meant by potential energy of the system you see when you are doing mechanics problem you have a mass connected with a spring you push it. Now, when you push it the work that you have done gets stored as the potential energy of the spring that you have compressed. Now, here the point is slightly different where is this energy stored is there a physical place where the energy stored the answer is no. The this energy is stored in the electric field in the region and you can always extract it when you release that body and let it go back. If it goes back then it will spend that potential energy to get to the becomes the kinetic energy of that one. So, that is. So, therefore, if you look at this expression it tells me potential can be regarded as the potential energy which is associated with a unit charge. So, the names are confusing, but that is the way it was and. So, therefore, let us proceed with that. Let me give you a couple of examples. These examples have been chosen because of certain reasons. The problem is very trivial the problem is what is the potential due to a line charge. Now, once again since it is a infinite line charge it has cylindrical symmetry and once you have a cylindrical symmetry what you do is you enclose this charge line charge with a cylinder of certain radius this is your Gaussian cylinder and you say that I will calculate how much is the flux from this surface. Now, the top and the bottom surface because the electric fields are directed outward from this line the and the normal to the top and the bottom cap are perpendicular to it. So, there is no flux from there. So, only from the side. So, this is and this area I know is 2 pi r time cells. So, therefore, the flux is e times 2 pi r time cells and that is equal to charge contained and that is lambda is the charge density. So, lambda L by epsilon 0 and this is something which you have calculated the electric field is given by this. Now, if you look at the electric field the dependence is 1 over r. If the dependence is 1 over r the potential remember that differentiation of the potential will give you the field. So, the result is going to be 1 over r I must start with the logarithm of r. So, therefore, the potential because minus grad phi is e. So, potential is minus lambda by 2 pi epsilon 0 times log r plus a constant. Now, this constant is the one point of our discussion. See, I cannot any longer choose the potential at infinity to be equal to 0 because if I do that I will say phi at infinity equal to 0 I am left with something quantity called log infinity which is nonsense. But I know that potential is something which I can decide on my own reference of the potential. So, you can choose it anywhere, but not at infinity. You cannot choose at infinity you cannot obviously choose at 0 because both these are not defined. But a very convenient thing would be to say arbitrarily let the point at r is equal to 1 define the point of 0 potential because if I do that then you can see that this constant will become 0. So, in linear problems one chooses the 0 of the potential not at infinity, but at some arbitrary distance you could have chosen it r equal to 2 r equal to 5 whatever it felt like, but not at r equal to infinity. Let us come to another important thing. So, this is a simple dipole just a charge plus q and a minus q separated by a short distance d. Remember by convention the direction of an electric dipole is from the negative charge to the positive charge not the other way round. From negative charge this would be the direction of the dipole moment. So, dipole moment would be essentially the magnitude of the charge q times the distance between them and is along this distance this direction from minus q to plus q. Now as trivial algebra I am not going to do it that phi of r is 1 over 4 pi epsilon 0 q by this distance minus q by that distance and you can immediately write down you see this r plus can be written in terms of using the triangle law it is the r plus is square root of r square plus this a by 2 square minus 2 r a by 2 cos theta. So, you do that and then you can show that the potential expression becomes p cos theta by 4 pi epsilon 0 r square and that is the generic expression for the potential due to a dipole. But ask you know this picture you must have seen on the books and the point is this that how are these pictures drawn? It would be a good idea if you could ask your students to plot this picture not just by looking at a book, but say that look I have a field here. Now you are plotting something in true dimension. So, try to say what they will do is to draw a few points you know what happens to the field here, what happens to the field here, what happens to the field there and then of course, manually I did. These were the morning session people asked about lines of forces and all that that is where it came from ok. Finally, I want to talk about a another thing these always go with electrostatics we want to talk about what are known as electrostatic boundary conditions. See a boundary condition is basically a condition that has to be met whenever you meet a surface. So, for example, I think I have a charged plane here. Now let me assume the infinite plane, but it does not have to be. Now the question is this plane is charged I am asking that there is a field up there and there is a field down there. Now how do I connect these two fields? There is a surface there. Now this discontinuity that you may have whenever you meet a surface between two bodies what you get is what is called as a boundary condition. And they are extremely important in doing certain types of problems. So, let me let me explain how these boundary conditions work. I recall I repeat again we have been talking so far only about vacuum electrodynamics there are no material. So, I have a charged sheet I have I know what is the field above I know what is the field below I am asking what is the connection between them. So, what you do is this that firstly any field is can be it is a vector. So, I can resolve it into a component which is perpendicular to that surface and a component parallel to itself. So, what we do is we take a rectangular parallel pipe small one this surface has negligible thickness. So, what it does is to half way e by 2 above and e by 2 below. Now, let us suppose I am looking for what is the condition on the perpendicular component of the electric field. Now, then I look at the following way I said what is the surface integral of e dotted. Now notice that surface integral of e dotted by Gauss's law is q enclosed by epsilon. Now, if you have a rectangular parallel pipe which is cutting this surface, then whatever is the area of this supposing it is a on the surface that amount of charge is contained let me sigma and and of course, by epsilon. But on the flux side the as epsilon goes to 0 there is no contribution from the edges. On the top I have e perpendicular because I remember that I have to take the normal component e dot ds. So, e perpendicular above times a minus e perpendicular below times b this should have been below a and a will cancel out. So, what it tells me that the normal component of the electric field above the surface if you compare it with the normal component of the electric field below the surface there is a discontinuity. And this discontinuity depends upon the charge density that is there on that surface. If I did not have a surface with charge then there would not be a discontinuity. And if I meet with a charge surface there is a discontinuity and this is incidentally true in many situations. Wherever you meet a surface with a for example, in this case charge density you get a discontinuity. Similarly, you will find in magneto statics if you have a surface with a current sheet you will also find a discontinuity. Now a similar thing I am now talking about what happens to the parallel component tangential component. Well in this case you take a rectangle half of it above half of it below and I know that by conservative of nature of the electric field the line integral is 0 and that tells me that the parallel or the tangential component of the electric field will be continuous. So, this is this is very important that if I have a discontinuity and if that surface contains charges then the parallel component of the electric field will have a discontinuity. The tangential component must be continuous. And this is actually what happens when you look at problems of reflection of electromagnetic waves from a boundary. You have to make sure that these conditions are met all right. So, with this let me go over to a little bit of mathematical work. So, notice that we have so far talked about two equations. One is del dot of E equal to rho by epsilon 0 and del cross of E equal to 0. Both of them are in electrostatics. Now this gives me because E is minus gradient of phi del square of phi equal to minus rho by epsilon 0. Now if rho is equal to 0 this equation incidentally is called Poisson's equation del square of phi equal to something that is called Poisson's equation. And if there is no charge density then I will get del square of phi equal to 0 that is called a Laplace's equation that is valid when we have source free field. And in addition of course I must always have del cross of equation. We are talking about little bit about the solutions of these equations because they turn out to be extremely important. First again I will not go through the detailed derivation anywhere. So, mathematics is something all of you can do as good as I can do, but let me just point out the important point ok. So first these solutions of Laplace's equations are known as harmonic functions. There is a name given to it. And remember what is the Laplacian? Laplacian is d square phi by dx square plus d square phi by dy square plus d square phi by dz square. Now if you look at it in spherical coordinates this is the expression incidentally this expression since all of you are physicists if you have not done it in electrodynamics you would have done it in quantum mechanics. Because this is exactly the form in which you write down for the hydrogen atom problem. You see the physics becomes the same no matter where you go. So, this is an equation. So, let us look at. So, we have already said that this is phi of r ignored. Now I am trying to solve Laplace's equation that is easier to solve sometimes. What are the characteristics of the Laplace's equation? What statements can you make about the Laplace's equation even before you have opened the method of solution? Now here is an example I give supposing I have a volume v some region with a surface s and volume v. And in that region in that region I am looking at source free region so that my Laplace's equation is valid. Inside that region there are some conductors. Remember conductors are surfaces of equipotential ok. Now the I can have either a condition on the potential itself that is supposing I have sort of pointed out that on the bounding surfaces I have phi equal to 0 on possibility or phi equal to phi 1 or I can give you the derivative of that. Remember the derivative of the potential is the electric field. So, what I am doing is this I am saying that inside I have certain conductors and all those conductor surfaces I give you the values of the potentials or I give you the normal component of the potential itself that is normal derivative of the potential that is the normal component of the electric field. This set of boundary condition that is if I give you what is phi 1 and what is phi 2 or supposing this is phi this is phi this is called a Dirichlet boundary condition. The other one is called a Neumann boundary condition where you have given the derivative. Now a statement I am making I am saying that once you have given the conditions of either type there could be of course different types of conditions and I am talking about these two classes of conditions. If you give me the boundary conditions of either type the solutions of the Laplace equation is unique that is you once you have got a solution you do not have to worry about finding another solution. And the way one proves it is the following that supposing what I said is not true supposing I have phi 1 and phi 2 are two solutions. Then let me define a capital phi which is phi 1 minus phi 2. So, you notice that del square phi ok is equal to 0 because del square phi 1 is 0 and del square phi 2 is 0 because both of them satisfy Laplace equation. So, I have del square phi equal to 0 that is phi 1 minus phi 2 satisfies that and on the bounding surface supposing you have given me the values of the potential. So, what we are saying is since the boundary conditions are given. So, phi 1 minus phi 2 is 0 on the surface or d phi 1 by d phi 1 minus phi 2 d n is 0. Now with this this has been fortunately shaded so that even if you miss it it does not matter. There is something known as Green's first identity this is not a great that name is big, but ultimately it is nothing but divergence theorem. So, look at the divergence theorem you know what it says if you have a is a vector field del dot of a d tau is a dot n d s. So, if you choose some arbitrary scalars phi and psi you will get this equation which is known as Green's first identity. Now in this equation if you choose psi equal to phi equal to chi then you start immediately with a trivial algebra you find phi 1 equal to phi. Forget about the mathematics which all of you can look up and things like that what is it that we have said. We have said that given certain types of boundary conditions the Laplace's equations solutions will be unique. So, that now this is something which we are interested in and look at this that they this is a very trivial example you have a parallel plate capacitor. If you have parallel plate capacitor with let us say this plate is at a potential phi 0 this plate is grounded meaning thereby potential is 0. Now remember that I had said that I have given you the directly at boundary condition. Potential on one conductor is given to be constant phi 0 on the other conductor is given to be 0. Now this tells me now it is a 1 dimensional problem because of the symmetry of the problem the potential can only vary in this direction. So, del square phi is nothing but d square phi by dz square if that is equal to 0 the solution is trivial phi is equal to a z plus b. Now put in the 1 2 boundary condition phi equal to 0 at z equal to 0 and phi equal to phi 0 at z equal to b. If you do that you immediately get what is the potential and the electric field like. This is of course some the result which is known to you that these are the parallel plate capacitor. So, this is another problem once again it is in a different it is in a cylindrical reference frame. So, let me let me skip that for the moment. Let me make a few comments about Laplace's equation. Supposing I have a 1 dimensional Laplace's equation which is nothing but d square phi by dx square equals 0 or d square phi by dz square equals 0 does not matter. The solution of that equation is mx plus c. Now suppose I give you a boundary condition let us say phi equal to 0 at x equal to 1 and phi equal to 3 at x equal to 2. Now I can write down the solution uniquely. Now immediately you realize that general phi x is given as half of phi x plus a plus half of phi x minus. That is the solution is absolutely without any features. There is nothing interesting about that solution. It is a linear function and so what we have said is this that if you are calculating at any point now you just go plus a minus a and whatever is the average you get it. Then that solutions of Laplace's equations are featureless is carried over to other dimensions as well. Now this is a mathematical plot of the following function. Look at this. Supposing I have a function which is a by 4 x square plus a square or this is supposing I am looking at a Poisson's equation del square phi by dx square plus d square phi by dy square equal to a. Look at that function. Now this quantity you can immediately solve intuitively you can solve. You can immediately see that if this is the equation I have mind you this is not a Laplace's equation this is a Poisson's equation del square phi by dx square plus del square phi by dy square equal to a. I can immediately see that phi equal to a by 4 x square plus y square is a solution. Now what I have done is I have plotted this. Now if you look at that picture this is what I have drawn in 3 dimensional plot that you have the x axis and the y axis and this is the z axis. Now if you can plot this function in a 3 dimensional plot what you will find it is like a bed sheet which has been pulled below. Imagine a bed sheet being extended from both sides and you are pulling it from below x equal to 0 and y equal to 0 is the minimum of that. This function has a minimum at x equal to 0 y equal to 0 and it looks like a cup. Supposing I did not have this equation but I had Laplace's equation. Now in Laplace's equation I have del square phi by d square phi by dx square plus d square phi by dy square equal to 0 instead of a. Now that equation has this type of a solution. Here a by 4 is not important but basically x square minus y square. This picture if you draw is not like a cup but is given like this. This is like a horses saddle. If you have seen horses saddle you might not have written over it. The horses saddle you see it falls down like this but on the other two sides it rises ok. So, this is a different type of function. This function does not have a maximum or a minimum. It has instead what is known as a saddle point and this fact that Laplace's equation basically has no interesting feature. The pictures may be more interesting than the other one but physics wise there is nothing there. There is no minimum no maximum. Then why am I talking like that? There is a theorem known as Ornstor's theorem. Ornstor's theorem tells you that an electric charge cannot be kept in equilibrium using only electrostatic forces. I am not saying a charge cannot be in equilibrium. Charge can be in equilibrium if you have other types of forces. So, the charge is a mass you can use a part of gravity of gravitational force. You can use electromagnetic the magnetic force you can use anything. But if the only force that you have is electrostatic force a charge cannot be held in equilibrium. The reason I have just now proved. I have proved because if a charge is in a force free field then in that region Laplace's equation is valid. Now, if the Laplace's equation is valid by whatever we just now talked about the potential as calculated for such a problem does not have a minimum. As you know that stability is connected with maximum or minimum of a potential energy function. And if it is a stable equilibrium you want the potential energy to be minimum. If it is an unstable equilibrium it could be a maximum. But it requires its existence of a potential function which is either minimum or maximum and Laplace's equation does not allow us to do it. So, what we are trying to say is if potential has no minimum potential energy also does not have a minimum. And as a result given only electrostatic forces a charge particle cannot be in equilibrium. This is a very important concept of physics. Well let me once again I will skip the mathematics, but since this mathematics also familiar to you while solving the hydrogen atom problem. Let me talk about Laplace's equation solution in spherical coordinates. I am really not interested in the complete solution, but look at it. So, Laplacian this is the expression which you have seen several times. That is the advantage of talking to a teachers because had I been talking to students I would have spent one full lecture in explaining or deriving this. I am solving Laplacian. One of the methods which works very well in such cases is known as a separation of variable. You recall your hydrogen atom problem also you did by separation of variable. What it means is that if you have an equation in r theta phi or x y z or z r phi rho phi what you do is to write down the solution as a product of a function of r alone theta alone and phi phi is the azimuthalic. Put it back into this equation. Now remember because the right hand side is 0 I can bring the phi equation to one side only. Then you say that look the left hand side is a function of theta and r whereas the right hand side is a function of phi only and that is not possible unless each one of them is constant. So, therefore, you first say that this is constant. If this is constant f of phi becomes e to the power i m phi. But if phi changes by 2 phi the result must be the same which tells you that m must be an E t. This is I am not going to repeating it, but this is something which we have repeatedly done while solving hydrogen atom problem. This is the same mathematics I am not interested in that much. So, where does this separation of variable assumption gets violated? Sorry the thing is separation of variable may not always work. Yeah. So, in the in the type of problem that we are doing ok. Classically they the problems connected with Laplacian equation problems connected with waveguides ok it works right. These are certain mathematical techniques which are known to be working. If the equation were little more complicated it will not work ok. It works for even hydrogen atom as well. Hydrogen atom is the same problem. Yeah. So, hydrogen atom is just the same problem there is no difference. See look at what is hydrogen atom problem. No, what I am looking for is that what is there is some general guidelines to let say I am encounter with an equation like this. Yes. And then I want to know from the very beginning. Yeah. Whether I can actually look for a separation. Yeah. So, let me let me look at it the following way. That if you have a differential equation ok. It is a little complicated differential equation. There is no prescribed method for solving it ok. But one good thing about standard differential equation is lots of people before us have worked on it ok. And so, certain techniques for example, when you did your calculus first course people told you that there is something called a integrating factor which works in first order differential equation. It does not work with second order differential equation right. Yeah. So, in other words if you have a particular form of first order differential equation then putting in an integrating factor will work. But I can give you a first order differential equation where it will not work correct. So, solving differential equations the method that will work partly depends upon your experience and have you seen it in some context. Yeah. Seem it yes we do it. Ok the symmetry does not have to do anything. But you see supposing I had the same equation. But on the right hand side I had a complicated function it does not even have to be complicated. Let us say on the right hand side I wrote equal to 1 over r square plus r square sin square by theta. It will not work. The thing is thing is working because you have a 0 on the right hand side you cannot have a simpler function than 0 ok del square of something equal to 0. The point is that there are certain techniques which are known to work for solving differential equations. One of them is separation of variable. So, one of the first thing that everybody does is to look at a separation variable working. Now if it works you do not worry about it again ok. Now having done this I rewrite. So, I now know that the this must be now my equation. So, again I say all right on the left hand side I have a function of r on the right hand side I have a function of theta. So, each must be equal to some constant. See the idea is r is the variable which is an independent variable. Theta is the variable which is an independent variable. Phi is the variable which is an independent variable. So, keeping r the same I can vary theta I can verify. So, if a function of r is equal to a function of theta for all r and all theta the each one of them separately must be constant that is the whole idea. Now this solution for theta equation they are known as associated Legendre function. Now again I will not be telling you what those are, but some of the early associated Legendre function or for these are just Legendre polynomials for m equal to 0 they have been well tabulated. For example, p 0 is always equal to 1, p 1 of cos theta equal to cos theta. This and finally, p 2 of cos theta is half 3 cos square theta minus 1. Normally one remembers these three the rest of them you can find out from a book right they have been tabulated for several. So, these are this is the way they your polynomial looks like Legendre polynomials looks like p 0 is a constant which is equal to 1 and p 1 is cos theta. So, as a function of cos theta it is linear p 2 is here 3 cos square theta minus 1. So, 1 this thing here p 2 p 2 p 2 here this this is p 2. So, this is the way it looks like. Let me let me give you one example of application of Laplace's equation. Let me talk about a conducting sphere in a uniform electric field conducting sphere. So, I put a conducting sphere in a uniform electric field uniform meaning there is a constant electric field going all over the space. First let us look at what statements we can make about it. One is that the field supposing I have a conductor here which is in a uniform electric field. What this conductor will do because it is a conducting sphere it will change the electric field in its proximity only. But far from the conductor if you look at for example, at infinite distance whether you have a conductor here or not into the material. So, one of the conditions that we have is the electric field at very large distance is the same constant electric field that we had and let us take it to be along the z direction. This tells me if the electric field is along the z direction then the potential at large distances must be minus e 0 times z, but z is r cos theta plus a constant ok. Now, we know that conductors are equipotentials. If the conductors are equipotentials the field lines will only strike the surface normally. So, field lines are perpendicular to that surface. Now, so as a result of that what will happen is the conductor will pick up charge. So, on the surface the potential is constant and since there are no sources anywhere the potential satisfies the Laplace's equation. So, you see I wrote down a few things without actually mathematically solving the equation. That is the way to look at a problem that I have a conductor which is presumably small. I am looking at infinite distance. So, I do not expect the conductor to have influence the electric field at large distances. So, the field will be uniform there. So, I look at all those and so the that is the first job that I do. So, what we said is this that my form of the remember that I still have not said what is c. We have said phi r theta is minus e 0 cos theta that is my z plus c ok. So, the thing is this remember that I started writing down the full expansion for the potential function. So, this potential function the angular part is p l cos theta namely the associated Legendre function or actually Legendre polynomials. The radial equation had a solution which is a r to the power l plus b divided by r to the power l plus 1. This is I did not solve them as I said that you had seen such solutions when you had done hydrogen atom problem. So, the point is this that out of this what do we pick up this I am sorry for the red color tomorrow I will see that I do not use that. So, firstly del square of 1 over r is known to be a delta function. So, therefore, l is equal to l cannot be equal to 0 in the second term. Then we have already said that phi at r going to infinity is of this form ok. So, therefore, I must have a 1 equal to minus e 0 r if you do that then your form of the equation becomes this. Only r cos theta is picked up this as r goes to infinity because it goes to 0 it is l not equal to 1 b l by r to the power l plus 1 b l cos theta this term remains. Now, I put the boundary condition that at capital R theta the constant is phi 0. So, therefore, it tells me that b l is equal to 0 if l is not equal to 1. So, if you do that you notice you immediately get an expression for the potential. This is mathematical, but that is the way to look at it and the remaining things are fairly trivial I will put it up in the notes let me not spend more time on it. This is what this is what a spherical conductor does to the electric field near it. So, notice what we have done we said as you go away far away the scale is not the figure is not to scale the field will remain uniform, but when you come close to it the field will get modified and these field lines will approach this sphere in a normal fashion. Now, if that happens you see on the one part the field lines are getting in from the other parts the field lines are getting out. This tells me that one phase will be positively charged and the diametrically opposite side will be negatively charged. You can calculate how much of what is the charge density and what it does. So, the sphere modifies the potential of the uniform field by a term of this type E 0 r square by r square cos theta and if you recall the potential of a dipole was given by this. So, the sphere in a uniform field is equivalent to a dipole having this dipole. Let me go to another technique this technique is based on what we talked about a little while back this given a boundary condition if you are solving a Laplacian equation this solution is unique. We proved it this theorem is known as uniqueness theorem in electrodynamics. What is the advantage of a uniqueness theorem? Uniqueness theorem tells me that if I am given a differential equation with specified boundary condition there cannot be two different types of solution. Whatever solution I find that has to be the only solution. The point to be noted is since it is going to be the only solution I can get the solution any way I like ok. Now, once I have as phrase would be if I get a solution by hook or by crook then I do not have to worry about whether I have missed out a solution that has to be the only solution. The question was asked that does the separation of variables always work the answer is no, but supposing it works you try it out and you have got a solution then no matter whichever other way you try it that has to be the only solution that is the uniqueness of the solution. If the solution is unique then the solution of a Laplacian equation subject to a given boundary condition has to be the same irrespective of the method that I adopted to get this and this is the basic principle behind what is known as method of images. As you recall image is something which you do in optics. So, here I have given you I have a conducting plate which as you see is grounded that is the plate is at 0 potential. I have a charge cube the question that I am asking is what is the electric field like? Does this charge induce any charges on this conducting plate. Now, so what is my problem mathematically? My mathematical problem is I have a Poisson's equation to solve because there is a charge I have been given a boundary condition. Boundary condition is that there is a conductor here which is at 0 potential. Therefore, I make a statement that this solution of this problem I will not solve by actually trying to tackle differences. I will do the following I will say look suppose I think of a charge cube prime which is it is a virtual which is below this. So, cube there at a distance d cube prime there at a distance d prime below the plane. I look at the potential at any point b. But look at it this way that due to this charge which is the real charge my potential is q by 4 pi epsilon 0 r 1 and that is if I have this as the x axis y axis z axis this is simply q by 4 pi epsilon 0 x square plus y square plus z minus d whole square. And due to this artificial charge q prime which I mentally thought of something the potential is given by q prime by 4 pi epsilon 0 r 2 and that is given by this. And the net potential due to the real charge plus the virtual charge is nothing by superposition principle is nothing but 5 1 plus 5 2. But then the net potential must satisfy the boundary condition that I have given that at z equal to 0 the potential must be 0. So, look at what happens. So, I add 5 1 and 5 2 and I say what is the value of 5 1 plus 5 2 at x y 0 and I put it equal to 0 what happens. Now, if I put it to be equal to 0 you find that this equation is satisfied if you take q prime to be equal to minus q and d prime to be equal to d. In all the words that the problem is uniquely solved. Remember again the uniqueness principle. Uniqueness principle said that if you have a solution which satisfies the given boundary condition then that has to be one and the only one solution. I got my knowledge from optics. I said that look if you had a mirror then I can think of a charge which is at a distance d away from the mirror and imagine that it has an image at a distance d below it. And by some reason that just as you have left right symmetry in a mirror the positive charge has an image which is a negative charge ok. These are all you know dream top, but this dream is good because if you did that then the boundary condition is satisfied. Remember the virtual image was virtual charge does not exist, but we said that the original problem is equivalent to the problem of a charge q at a distance d at a given and a charge minus q at a distance d below it. Replace one problem by the other solution of this problem is the same as the solution of the other this is the principle of method of images. This is where the name was taken. Look at this picture you can find out how much is the charge density on the plane because everything you know now ok. Do I use a mathematical plot? What you will find is that all the plate there would be charges induced and the charge density would be have a maximum magnitude in the point directly opposite the original charge. This is shown below because the charge is negative. So, this is a plot which anyone can do and this is once you have got the problem you can easily draw the field lines and find out what it does. The method of images is a very powerful technique. The name comes from the image of an object in front of a mirror, but the method of images has been also done to do different type. For example, in this case I have a sphere which is grounded means the sphere is at 0 potential. There is a charge q which is located at a distance a from here. Now, the point is find the potential at all points. So, this mirror it modifies the potential due to the charge q. So, once again I make a statement. So, let us see whether it is doable by method of images or not. I repeat again method of images may not work, but if it works and you have got a solution then you do not have to look any further because of uniqueness theorem. So, let us look at what is happening. So, I say all right that imagine there is an image charge which is q prime at a distance b from here. Take an arbitrary point p, find out what is this distance r 1, this distance r 2 and let that point p be at r. Now, you write down what is phi arbitrary. If you do that and then this is standard triangle relations r 1 is a square plus r square minus 2 a r cos theta and this is b square plus r square minus 2 r b cos theta do that. And you say the net potential is 0 at r is equal to capital R for all theta. Now, when is this equation valid? That is the question that we are asking. That turns out that if you choose q prime to be given by minus r by a times q and b to be given by r square by a, the b is called the inverse and distance. Then the conditions of the problem is satisfied. So, the method of images is a particular method can be used to solve many physical problems in electromagnetics and if it works then you have a lot of work saved. I will end this session with some comments on dielectric because of shortage of time I will not find time to discuss in detail what happens when we introduce dielectric into this. See so far the discussions that we are having are all with respect to vacuum. Vacuum air whatever you call it, electrodynamics of vacuum if you like. The charges that we have talked about they are located in vacuum or in air. But I know that physical world is not that. So, what happens? There are two types of things. One is class is conductor. What are conductors? Conductors are those where there are free charges. The charges can are mobile and when you apply an electric field the charges can move around and that is why there is current. Now, other type of material is dielectric for example, take this piece of wood which will not conduct electricity. Now, these non-conductors in these non-conductors the electrons are tightly bound to the atoms ok, atoms are molecules. So, the charges are what are known as bound charges depending upon the type of molecules you have. There are two types of think one is called non-polar molecule or another class is called polar molecule. In some of them even in the absence of an electric field there is a small separation between the charges. In some others the positive and the negative charges their centers coincide, but when you apply an electric field they get separated by a small amount. Now, so what we are talking about are things which are electrically neutral piece of wood, piece of rubber or whatever you have. So, insulators is another name for dielectric. Now, notice that even though they are neutral since the subject of our discussion is what happens to the behavior of substances in the presence of electric field whether this material is a polar molecule or a non-polar molecule whether their charge centers coincide or do not coincide. When you apply an electric field there is a separation of charges and if there is a separation of charges these dielectric would act as a source of electric field and they will be in a position to modify the electric effect am I clear. So, look at it this way. So, this is what happens say you have a situation where positive and the negative charge centers coincide. So, net dipole moment of that is you the if you apply an electric field there is a slight separation between the positive and the negative charges. So, what happens is that when the electric field is 0 the molecular dipoles are randomly oriented. Now, if they are randomly oriented then the net dipole moment is 0. When the dipoles are placed in an electric field the medium will be polarized that is the word used meaning thereby that each molecule has a dipole moment and in the presence of the electric field they are trying to align they may not be completely aligned does not matter, but there is still a vector sum which is not equal to 0. So, look at what I am doing here I am saying that here is a dielectric and I am looking at what is the field at a point p due to let us say a you know charge which occurs here. So, this is standard Coulomb's law expression 1 over r minus r prime which is given like this r square plus r prime square minus 2 r prime cos theta. Now, if you this distance at which you calculate the field is usually much larger than the dimensions of this. So, as a result you can do an expansion in powers of r prime by r this expansion is known as the multiple expansion of the Coulomb field ok. I am not going to the multiple expansion because I will stop at one point. So, look at it this way supposing you did an expansion of this straight forward there will be many terms, but let us stop it here 1 over r r prime by r cos theta plus r prime square by 2 r cube 3 cos square theta minus 1. The first term is nothing, but the usual Coulomb term 1 over r potential is at 1 over r. Now, so this is what we have said that our expansion is like this. So, potential due to a dielectric is how much. So, potential due to a dielectric is given by the same expression, but now I have done that expansion. Can I do that expansion? First term as I told you is nothing, but the Coulomb term. The second term is the dipole moment term r prime rho r prime that is the dipole moment and the corresponding potential is there. One defines the polarization vector of the medium as dipole moment per unit volume. So, if you look at the potential, the potential term is written like this and after a bit of mathematics which is fairly standard which I will not be working on you find that this can be split into a surface term and a volume term. Now, mathematics is fairly straightforward I will not do it we do not have time. Now, look at what we have said. We have said the modification of the Coulomb potential due to the presence of the dipole moment. Net dipole moment per unit volume is what we call as the polarization that is the definition of polarization. So, it turns out that the potential due to this has two parts. One part is a surface integral over the material which is written as sigma b b x I will explain ds by r minus r prime and the second term is a volume integral. So, you notice that if you compared it with the Coulomb field it is as if the dipoles the dielectric is equivalent to a charge on its surface and a charge in the volume. So, you replace the dielectric with a charge density on the surface and a charge density in the volume. These charges to distinguish them from free charges are known as bound charges that is because in dielectric the charges are actually bound they are not moving around. If we went back you will immediately see what that sigma b is. It turns out that sigma b is nothing, but the polarization vector dotted with n normal component of the polarization. The bound charge is minus the divergence of the polarization. Now, let us come back to our Maxwell's equation. So, what we said is this we had said del dot of E equal to rho by epsilon 0. Now, let me repeat E is always the actual electric field. So, the electric field E is not only due to the external charges that you may have, but it also takes into account the effect of the dielectric. Whatever is the actual if you put a test charge whatever is the actual field it will see that is always represented by E electric field. Now, the bound charges have two components. One is on the surface of the dielectric another is inside the dielectric as a volume density. I am not putting anything I am just saying that you have a dielectric. The effect of dielectric is taken care of if you replace the original problem by a surface charge density and a volume charge. You started considering a kind of journal space filled up with charges probably, but to start with you don't assume. Space filled up with dielectric. No, but this what I am saying is that you can do this multiple expansion thing even if you don't consider whether you have a dielectric or not. You just consider a collection of charges in space and you try to calculate the electric field at some arbitrary position are you still get the same. Terminating it at dipole. I am not going to any what nuclear phases do I am not doing. I am saying just consider I have real charges and dipoles. So, my multiple expansion is a big name. All that I did is actually to terminate at second part. No, probably I got confused. What I am saying is that you still can do the same kind of multiple expansion. Yeah. Not considering it to be a dielectric. Yeah. But how does that because but if you don't consider it to be a dielectric then the concept of these bound charges and everything don't appear. Agree. So, how do you reconcile the same formalism and try to. I understand. See the reason why that is done is if you have a bare 1 over r minus r prime term then you need to add up all the multiples. Okay. Okay. Now, so therefore the truncation procedure that I did that doesn't remain valid. Okay. Thank you. So, what I am saying don't truncate. I truncate it at dipole. Okay. So, when I truncated at dipole, my collection of dipole was equivalent to a surface charge and a volume charge. This is artificial. And this surface charge is given by p dot n, normal component of the polarization. And volume charge is minus the divergence of the volume charge. Yeah. And but when you write this equation del dot e equal to rho by epsilon 0. Yes. You are thinking of the electric field now inside the dielectric field. No, no, no. Or anywhere, anywhere in this space. Del dot e equal to rho by epsilon 0. Okay. Rho by epsilon 0 is the electric field as experienced by an external test charge. Okay. Okay. Which is due to free charges as well as due to direct. Okay. So, e is the real electric field. Yes. Then the bound charge density 0. Yeah. No. See, that is what I have tried to tell you that the bound charge density is not 0 because you have put in external electric field which has made a charge separation. But usually the surface and volume bound charge. That is a total charge. Don't confuse it with the charge density. Total charges will be 0 because after all there is the whole system is neutral. No, no, no. There is a surface charge density and there is a bound charge density. You are talking about whether the system is neutral or not. A system neutral does not mean that the density is 0. Total charge is 0. Bound charge is 0. Surface bound and the volume bound, they come up in such a way that the total charge enclosed in the dielectric will become 0. Yeah. But not the densities are not individualism. Now, so please take this that electric field e is the real electric field. If you put a test charge, that test charge you will experience the field which is called e. So, therefore, del dot of e equal to rho by epsilon 0. This rho has two components. One is the free charge and another is the bound charge by epsilon 0. One defines a quantity which is known as, in fact, it is always called vector D. It does not have a name. It is called vector D always. I mean people said it is displacement vector, but it is not displacement vector. Sorry. It is flux density. No, no, no, flux density is not. The flux density language is used more with respect to magnetic field, right? Not so much in electric field. It is not flux density. But D is in some sense the flux density corresponding to say what is D? We have real charge and bound charge. Imagine somehow or other I could switch off the bound charges. It may not be possible, but mentally I can always do anything. Now, if I could switch it off, then the effect of the real charge if I could separate. Remember there is no way of separating because my external charge will always get all the contribution. That field would be given by D. So, we define D as a quantity having the dimension of the polarization vector which is epsilon 0 e plus 2. If you substitute it there you find the divergence of the vector D simply is the free charges, but there is no way of determining this other than mathematically. So, in some sense the vector D is some artifact in our mind that as if we can switch off the bound charges and if I had real charges only what will happen? D and E in the absence of the polarization is very easy. It is just epsilon 0 times. Any other question? When you try to go inside the material and try to measure the field. Yes. What do you expect to see? The question is how do I measure the field? The only way of measuring the field experimentally is to put a test charge and find out what is the force it is exactly. And that field is the E field all completely dressed up field. So, E field is the real physical field ok. That has the effect of external charges and whatever that external charger field has done to my direct. D field is an artificial construct that supposing I could say that this consists of two parts then the part that can be ascribed to the real charges is this. So, at the end of it my electrostatic equations are del dot of D equal to rho free incidentally this equation physicists rarely use. The electrical engineers use it. D is lot more common by use commonly used by electrical engineers ok. But because E is the real field you will find most physics books will almost ignore D other than as a footnote. Any other question? I want to ask something which relates to something which you discussed slightly longer back. Right. It is regarding this Ernst's theorem. Yeah. So, you actually stated it for electrostatic force. Yes. There is an equivalent statement for this magnetostatic means it is just. See the point is that there cannot be an equivalent statement for magnetostatic because there are nothing like a magnetic monopoles. We are talking about whether electric charge can be kept in equilibrium under the action of electrostatic forces. If there were a magnetic equivalent how would you state it? That can you since I cannot have only north poles the question does not arise. Had it been there I can obviously give the same statement because source of the field is unimportant. But magnetism unfortunately the north poles and the south poles cannot be separated. So, the Ernst's theorem said that pure electrostatic forces cannot put a charge in equilibrium. Okay. There is something else I want to ask if questions and physics are over. Yes. See, Mr. I want to ask regarding Coulomb's law. For example, say I have two charts Q1 and Q2. Right. But both are in different dielectric medium. Different dielectric. Yes. Then how I will write Coulomb's law. No, no. The thing is this that look the point is this that the Coulomb's law is always the manner in which I wrote it namely Q over 4 pi epsilon 0. It has nothing to do with the medium in which you put it. But the moment you say I have put it in a medium what is it that that charge is doing to that medium? The role of the medium is that charge will polarize that medium. Okay. Now once the charge polarize that medium the electric field due to the bare charge will be changed by this discussion that I had. I had taken another simplified discussion, but this is what will happen. Supposing you had a linear medium. A linear medium is one where I can talk about a constant dielectric constant. That is instead of epsilon 0 I have a k times epsilon 0. Then I can say that all right in which case probably instead of 1 over 4 pi epsilon 0 I will have 1 over 4 pi epsilon, but it is a very dangerous statement. The Coulomb's law is not modified in the presence of a material. What the Coulomb's law what the charge does is to modify that medium and create a large number of dipoles in our approximation. Actually speaking it creates that was his original question that what happens if you take the quadrupole the octopole and things like that we have we have ignored that because they are in macroscopic physics those are negligible. Yes, any other question? Okay if not thank you.