 Good morning to all of you. As we start our third day's lectures, today the plan is to complete the electrostatics in the morning and do a lecture on magnetostatics in the afternoon. The reason why we spend so much of time on electrostatics, though planning to spend only one lecture on magnetostatics is because the electrostatics essentially establishes all the principles that we want. The magnetostatics almost goes parallel, so therefore there is minor cosmetic changes that we have to make as we go along. Like yesterday, we start our session taking up the questions which came on the chat session which you could not take up. We still seem to be getting a lot of questions on the meaning of curl, divergence, solenoid or irrotational. See the point is that I have to also proceed with physics and one thing I would like to point out, please understand these are mathematical techniques. I have been asked questions like given example of where it is used in communication engineering. The point is this that this is really not a very relevant question because tools of vector calculus are used in any subject which uses mathematical. It does not have to be specific to a particular area. Since I am not a communication engineer, I will not be able to tell you, okay, look at this theorem where it is useful. But supposing I ask this question of thermodynamics or other areas of physics, electrical engineering etc., maybe I would have been able to give it. But basically the mathematical tools, they should be taken up as tools and even this question which keeps on coming that what is the meaning of the curl? Why it is so called? See you have to realize that definitions or nomenclatures like curl, solenoidal, irrotational etc. came from the topic where the vector calculus was used in detail and that is actually the subject of fluid dynamics. And so therefore the word curl as you know that when you curl your finger like this which means something which is having a rotatory motion. So in case of vector what we say is the direction of the vector is changing. So that is how the name curl came from and it is irrotational because the idea is that if curl is equal to 0, it means it does not rotate, it does not curl. So that is the whole idea. But as I said these are historical introductions and there is not much physics in nomenclatures and things like that. Same goes about solenoidal. So I will not be really spending much of a time on re-explaining or re-visiting the questions on vector calculus. What I take up today? First a question which is asked by Thiagraju College of Engineering Madurai Center 1173. The question was asked that what is the difference between potential and potential energy? In fact somebody said supposing instead of taking a unit charge, I take a charge Q, will it become potential energy? I think because it is a thing which confuses a lot of students, let me explain to you so that you have no doubt that we are talking about two different things. Historically the name potential is unfortunate but once the name was given we have to stick to it. There is a relationship with potential energy but they are two distinct things. So let me try to explain the difference between potential and potential energy in a little clear way. The best way to explain this is to look at what happens in gravitational field. As you know that the gravitational field is very much like electrostatic field, it is a conservative field. Of course the gravitational forces between mass is always attractive whereas in case of charges they could be attractive or repulsive depending upon the charges. Let us look at the difference between gravitational potential energy and gravitational potential and if you try to understand what I am trying to talk about here then you will have no problem even in understanding mass. First let us assume that I am in this room and where let us say gravity is constant. So gravity G is constant, you usually take it to be 9.81 meter per second square and the direction of gravity is downwards. This is your gravity. Now I am just saying that suppose I have a particle which is at a height h. These things are very well known to you or also to your students and let us say the mass of the particle is m. Now we say the potential energy of this mass because the force is constant is given by mg times h. This is the potential energy. Now notice that the force of gravity is mg however it is downwards. So let me I can write this as the magnitude of the force of gravity multiplied by h. Now how do I understand this? See the idea is this that supposing I started with the floor of the house where I take the reference of the potential energy to be 0. Supposing I say that this is potential energy is equal to 0 then if I, I meaning an external agency, if I want to take this mass to from ground to the height h I have to provide a force in a direction opposite to that of gravity. And let us suppose I just give it a slightly you know bigger force and so that the mass moves to a height h with a you know constant velocity, constant speed. Now then this particle comes here. Now what does it mean that it has a potential energy mg h? What it means is supposing you leave it, leave this mass at this point then the mass will of course fall down converting continuously the potential energy into kinetic energy and when it falls on to the floor all its potential energy has been converted into a kinetic energy. Now so in other words the potential energy is the amount of energy that is stored in this mass by virtue of the fact that a work has been done by an external agency against gravity which is the prevailing force. Now let us look at what would happen suppose I talk about the corresponding electrostatic phenomena the how do I produce a constant electric field I mean it is not difficult I at least not difficult to imagine. So let me take an infinite charged plane let us say positively charged and look at above it I know that the field is uniform. So supposing I have a positive charge I which I take from here to this point now notice that since this charged plane is positive the charge that I have is also positive the there is a repulsive force between them. So therefore in principle they should actually just go up but let us look at supposing the charge is already here at a high touch and then I want to bring it down. Now if I want to bring it down I have to do work against the repulsive force of this charged plane and exactly when I bring it down supposing this was my reference point of 0 potential energy then I would get the same principle as there. Now identical statement you can make for let us say the electrostatic potential energy due to a charge q supposing there is a particle at a point a. Now let me also for convenience take both of them to be charged similarly positive or negative and so let this distance along the radial distance be r a and suppose I want to move this charge q to a position r b. Now the way to do it is this remember the gravity always acts along the line joining the two particles. So but what I can do first is to okay let me draw a radial line here now what I do first is move from here in a transverse direction like this and since at every point on its path the electrostatic force is perpendicular to the direction of motion the work done is 0 on this path. On the other hand I can come from here to there I will do work because I will be working against the direction of gravity and you can sort of work out that the amount of work done by the external agency that would be qq the integration is trivial by 4 pi epsilon 0 1 over r b minus 1 over r a and this would then be the potential energy that is stored work has been done by an external agency the energy has increased. So that is stored as the potential. Now the question is in what is the meaning of the word potential I told you that potential is a point function now suppose potential is a point function which is actually a scalar field. So supposing I have a charge q here capital Q now all around it is a seat of electric field and every point has a potential whether there is a charge small q there or not this is the basic question. Now what do I mean by that what I mean by that is the following suppose you define your 0 of the potential somewhere conveniently in this example I will take the 0 of the potential phi is equal to 0 at infinite distances. Now in that case the potential energy the potential energy now this is very important that a unit charge would have if it is located there. Now what is the difference the difference is potential energy is an energy potential is its basic capability that is if you put a charge 1 unit of charge there the potential and the potential energy expression wise will be the same. If you put a charge q there it would be the potential multiplied by q if you put a charge 2 q there it will be a potential multiplied by 2 q in other words the potential has nothing to do with how much charge is there what is the charge unit charge 10 units of charge 20 units of charge but potential is a property of that point. In the presence of q the potential is the property of that point you put a charge q there it will have that much of potential energy. So in other words the one is the capability that if you go somewhere it would happen the. So let me give you a crude example supposing I say that there is a there is a fireplace in my room. Now if I go to that fireplace I will feel it out but property of that point that there is a fire point there the temperature is high there that has nothing to do with whether I am there or not it is a property of the space it is a property of a point in the space. So one is a capability the other one is an actual value supposing you have a wallet or a purse now I can say wallet or that purse is the place where you can keep money but on the other hand the money does not have to be there. So please understand there is a difference between the word potential which is a point function which is a property of a point and it is related to potential energy because if you were to put a charge q there then the potential energy that charge would have is q times the potential. So that is that is about the potential and the potential energy. The next question is more in the nature of clarity this question asked by 1092 KIT college Kolhapur question was that the discussion of boundary conditions was not clear enough. So can you explain the boundary conditions little better I will come back to boundary conditions later also but since the question has been asked that let me let me first give you an example. You all know because light is something which you have been learning from the beginning that suppose I have a medium I have an interface between two medium for example there is glass here and there is a block of glass here and there is light there is air here. So this is medium number one is air and medium number two is let us say glass. Now what happens to a ray of light which falls here now the point is that when a ray of light falls at the interface between two medium certain continuity condition has to be satisfied this is maybe I will come to it when I discuss the optics the you know that of course ray optics wise the incident angle must be equal to the angle of reflection and the there would be a refraction which will depend upon the Snell's law. But what happens here is that there is a question of continuity of the electric field associated with light. So boundary conditions are conditions that must be met when we talk about the what happens at the interface between two medium in this example that I gave you is the interface between glass and air. Now what we were trying to say yesterday is this that suppose I consider the electric field let us say generated by a charged plane supposing this is a charged plane let us assume it is also positively charged. Now what we said is this that if there is a charged surface notice this is a good example of a boundary function. You know that the electric field here of course the above the plane and below the plane they are constant but for example this is directed that way. So let us call this E above the electric field below the plane is directed that way let us call it E below. So you notice immediately that there is a discontinuity in this case the there is a electric field which is directed above there is a electric field which is directed below. So you can subtract one from the other and say that there is a discontinuity in the electric field the electric field is not continuous. Now the way to look at it is the following suppose I define E perpendicular this is a perpendicular component of the electric E perpendicular above and E perpendicular below I take both of them with the same sense in other words supposing this has been taken this way this will be taken that way you will say how well that is not very important you just put in a minus sign there that is not very important. So now what we said is this that the fact that there is a charged surface there it tells me that there is a difference between E perpendicular above E perpendicular below and that is equal to the surface charge density sigma divided by epsilon 0. Now this is a discontinuity this is a discontinuity with this sense the reason was of course the your direction of N is above here direction of N is below on that surface and so this is your discontinuity. So whenever I am talking about an electric field and if the electric field is looked at the interface between two medium later on when we talk about dielectric we will point this out little more carefully that then the perpendicular component of the electric field is continuous. However if you compare the parallel fields component E per parallel above it turns out to be continuous which is equal to E parallel below. Now look at it this way you know that the since electric field is a conservative field your E dot dl is equal to 0. Now this tells me that suppose I am to look at a the potential now you see potential difference between the point B and point A that is nothing but minus integral from A to B E dot dl. Now since the point A and B are very close one just above the charge and the other below the charge this integral goes to 0 when the distance between A and B becomes negligible. So therefore it tells me though the normal component of the electric field is discontinuous the potential remains continuous okay there are quick questions on what is this earth connected and earth bounded see or grounded. Say you have to realize that the earth is a huge reservoir it can absorb any amount of charge and it can release any amount of charge. So basically a proper earth connection because it is an infinite reservoir is because the earth is at 0 potential. So therefore we use in electrostatics the grounding is the same as talking about that for example yesterday I gave an example in method of images that the lower plate was grounded and I use the fact that the potential of the lower plate is 0 this is used synonymously. However in real grounding when you want to do real earth connection if you like now if there is a significant resistance then the 0 potential may not actually be true. But you notice that grounding is not necessarily connected that language is not necessarily connected with earth even in for example your mobile phone or other handheld device. Now you would find there is a large conductor which is attached to one side of your power supply. So that is the place to which the all return parts of various currents coming from different components come. This is usually come by the called by electrical engineers as common. So basically it serves the same thing that it provides the presence of a reservoir with fairly large you know capability so that it can be regarded as a reference potential. The last question that I take up is it actually came I think by an email from somebody and this was a question connected with Laplace's equation and I said something about the Laplace's equation which is del square phi is equal to 0 and I said that it shows saddle point. The question was raised that when my second derivative is 0 I normally talk about it as a local or a global maximum depending upon what the condition is. Now the you know what in what connection I am calling about a saddle point. Now I have to sort of tell you that derivative equal to 0, second derivative greater than 0 is a minimum, second derivative less than 0 is a maximum and these are strictly one dimensional observation. But supposing I go to higher dimension supposing I go to even two dimension. Now in that case I can define my local minimum. Now a point AB is a local minimum provided if I am looking at some region around the point AB. If around point AB if I look at an arbitrary point XY then the value of the function at such a point XY remains greater than F of AB and this is in a region around the point AB. Now this does not have to be a global maximum or a minimum. For example, you look at this type of a thing say this is a minimum this is a local minimum. But this is not obviously a global minimum because there are other points which are lower than that. Now what happens in two dimension is you look at a different quantity which is D which is defined as Fxx, Foy. Minus Fxy square where Fxx means D square F over Dx square and Fxy means D square F over Dxdy. Now depending upon the value of this F I have various types of things. For example if I say that D is greater than 0 then my Fxx being greater than 0 gives me one of the local minima D less than 0 and Fxx less than 0 gives me a local maximum. But D is less than 0 sorry D greater than 0 in both cases D less than 0 gives me a saddle. Now if D is equal to 0 then I need still higher order derivatives. But you know this is multivariable calculus if I continue with it it will take a very long time. So please look up some standard calculus books. With this we complete the questions that I received yesterday. If your question has not been answered either you will receive an email with an answer or it means that this question has already been answered you are repeating that question and I will in that case please look up the any standard textbook. With this let me start today's balance lecture on electrostatics. End of the session today if I am left with time I will come back to Laplace's equation which yesterday I could not do because of shortage of time. So I was talking to you about method of images. Now first let me tell you that where did this method of images come from. The method of images came from a general statement that I proved yesterday which is called the uniqueness theorem. Now this is a very important theorem in electrostatics. The uniqueness theorem tells me that if I am trying to solve Laplace's equation in a region of space and suppose in that region there are boundary conditions given to me. For example, supposing I am solving a general potential problem in this room and then there is a conductor in this room. I know that whatever is my solution the solution must be such that when I come to the conductor my potential must remain constant that is a boundary. No matter how you solve it you have to you could be solving an actual differential equation corresponding to a real problem in electrostatics that you have but your solution must satisfy the boundary conditions. Now the uniqueness theorem was in some sense unique because it told us that once you have given Laplace's equation. In fact the statement can be made even for Poisson's equation given Poisson's equation for a given charge density. In Laplace's equation of course you do not provide a charge density del square phi equal to 0. Now if you have del square phi equal to 0 and you have been given conditions. Now I talked about two types of conditions there. One I call as the Dirichlet condition. The Dirichlet condition is like what happens on a conductor. For example, it tells me that at certain point the potential values are specified. The other possible condition is called a Neumann condition is that instead of specifying the potential we specify the normal derivative of the potential which is nothing but basically the electric field component. Now if this is given the uniqueness theorem tells me that solution is unique. What do I mean by solution is unique? It means that if you take two solutions phi 1 and phi 2 or psi 1 and psi then these two solutions can at best differ by a constant. A constant different does not make any difference to a potential. So this is uniqueness theorem. Coming out of that uniqueness theorem one of the answers is something like this that if you have by some means you have a method of guessing a solution. If you have guessed a solution then since the solution can only be unique your solution that you have guessed must be the same solution. Now you see this is not a very unusual statement. Let me first tell you in connection with for instance the you know for those of you who are mathematically oriented you know that if you are given a very large number and if you are asked to find out factors. Now if the factors are very big also there is not a very easy job to find out a factor because one does not have very efficient algorithms meaning thereby as computer scientist will tell you there are no polynomial time algorithms standard algorithm there are theories which will enable you to factorize a number. But suppose you are a bright person and you say okay I have seen this number and I believe that this is a factor. Now how you got it I do not know maybe you dreamt it but supposing you dreamt it now all that you need to do is now do a division. Take the original number take the number you dreamt and divide it and if you find there is no remainder then of course your guess was right. Now if your guess is right and that solution then has to be correct it is a unique solution it is not true that if another person dreams of the same number and he divides it he will not he will find a remainder. So in method of images we actually use our intuition. So what I did is this I said supposing I have a conductor which is grounded that it is in that connection that the grounding question came up I meant supposing I have a body which is kept at zero potential this is one of the smartest way of providing a zero potential very large conductor which is grounded. Now so I have a charge q and a conductor. So I need to find out what is the potential due to this charge anywhere in this space but my solutions must have the property that when I come take this expression come to the surface of the conductor then the value of my potential must become zero. Now I will now say that look let me try to guess this solution. So what I did is to say that imagine now this is this language is borrowed from optics that imagine there is a charge q prime behind the plane what is meant by behind the plane it is a virtual world remember when you are standing in front of a mirror the image is not is there you say that it is at a distance deep behind the mirror but you cannot take a tape and measure it because it is in a virtual world there. So in that sense I have said that let there be a charge q prime located at a distance d prime behind the plane. Now what I now do is this I say alright now let us calculate the potential due to q and q prime at a point p okay. Now these are very standard expression I have taken the z axis along the line joining q and q prime and this distance is d along the z direction this distance is d prime along the z direction minus d prime. So my potential at the point p by superposition principle is q1 by 4 pi epsilon 0 r1 which is in the denominator square root of x square plus y square okay plus z minus d whole square okay and phi 2 is q prime by this this equal to this. So what is my total potential my total potential by superposition principle is the sum of these two. So this is my total potential and now I say that supposing I take the grounded plane itself okay supposing I look at only the point z is equal to 0 that is the grounded plane if I put z is equal to 0 in this expression this becomes q by 4 pi epsilon 0 x square plus y square plus d square and this becomes q prime by x square plus y square plus d prime square. So I have this plus this that must be equal to 0. Now I need to satisfy that now what is the best way I can satisfy this I will say that look this has been done by squaring it up but does not matter. So what I do is this that look one of the ways to satisfy this is to let q prime be equal to minus q d prime be equal to my d. Now so therefore my image charge has the same magnitude as my real charge the image charge must be located at a distance d behind the conductor identical to what happens in case of an image and that is why the name came up. Now if you do that then this combination of two charges satisfies Laplace's equation because phi 1 satisfies Laplace's equation phi 2 satisfies Laplace's equation. So therefore remember the solution is only in the region above the plane. So since phi 1 and phi 2 both satisfy Laplace's equation phi 1 plus phi 2 will also satisfy Laplace's equation. Now we have also shown that if you choose q prime equal to minus q and d prime is equal to minus d then the potential is 0 on the surface. In other words Laplace's equation this new problem that I have got it satisfies Laplace's equation this region satisfies the condition that there is on the conductor the potential is equal to 0 by uniqueness theorem the solution is unique. So that must be the only solution let me continue with by giving another example. So this I sort of told you that you can now use this expression for potential to calculate the charge density and this is just a plot of the charge density on Mathematica and you find that charge density that is induced obviously negative is very large magnitude just points directly opposite to the real charge and sort of spreads out around but these are problems which if you ask your students they are all more proficient in computers than many of us are they will be able to draw it without any problem. Now another thing which I am not really sort of showing that if you wanted to calculate the force between the charge q and remember that there is a induced charge on the plane and so you could since the induced charge is negative if this is a positive charge there is a force of attraction between the plane and the positive charge. Now this one can show is exactly equal to the force of attraction between this charge and the image charge which is much easier to calculate. Let me take another example where again method method of images is used the solution is slightly more difficult so let me explain that. So I have taken this time a spherical conductor in fact this was one of the question that was asked that if the conductor is a sphere will the charge be on the surface the answer is yes but I did not take up that question for the simple reason that I was going to do it here in detail. So the point is this that I have a spherical conductor which is grounded again by grounded I mean that it is kept at zero potential. Incidentally you can use this method of images it does not have to be always grounded you can maintain it at constant potential so these are all possibilities and but I am of course able to illustrate only those which are reasonably simple. So suppose I have a charge q at a distance a from the center of this sphere. Now I am interested in calculating what is the field at a point p which is the region I consider is outside the sphere. So let us suppose this distance is R 1. Now without doing anything I say that alright let me put a image charge q prime. Now some are inside the sphere because it has to be in the virtual world. So essentially what I have done is to take that along the line joining the center and this and suppose the distance from the point p is R 2 the coordinates of point p itself with respect to the center of the sphere is R theta phi and so therefore let this distance be equal to b. Now let us look at this problem my total potential is 1 over 4 pi epsilon 0 q by R 1 remember phi 1 satisfies Laplace's equation phi 2 is q prime by R 2 R 2 which I have written as b should have been little more the R 2 is this distance from q prime to p R 1 is q to p. So this is the net potential phi 1 and phi 2 satisfy Laplace's equation. So phi also satisfies Laplace's equation. Now only we have to satisfy the boundary condition of the surface of the sphere but let us look at what is R 1. My R 1 by triangle equality is a square plus R square minus 2 a R cos theta. So this is what it is. Now what about R 2? R 2 is this. So this is equal to this square plus this square that is b square plus R square minus 2 b R cos theta. So this is my phi. Now I want that this plus this when small r becomes equal to capital R for any angle because I am on the surface of the sphere put small r equal to capital R for any angle this plus this must be equal to 0. Now what is the condition I get? So I have written down this plus this is equal to 0 square then this very trivial algebra and try to see what should be the relationship between q. Remember this relationship so it has to be true for all theta all theta this is very important. So therefore the first thing that you do is because this relationship has to be true for all theta I must have that b by a must be equal to q prime by q and that is equal to minus square root of b by a. So you do that and what you find is that your charge ratio of the charge q prime must be equal to minus r by a multiplied by q. It is a negative charge but unlike the plane conducting plane case it is not equal and opposite to that charge that we have to get. There is a there is a ratio there. Now you would say then why call it mirror? Well you see the point is that object distance equal to image distance as you know is true for plane mirrors that is not true for spherical mirrors for instance. So there is no 1 to 1 correspondence but what we are saying is that if at a distance b given by r square by a this point is actually called the point of inversion. If you put a charge q prime equal to minus r by a times q then again the potential on the sphere at any point would be equal to 0 which means that once again I have solved the Laplace equation subject to a given boundary condition. So this is method of images I repeat again method of images draws its power from the uniqueness theorem which says if you want to solve either Laplace's equation or Poisson's equation then for a given charge density of Poisson's equation and for Laplace's equation if you are given boundary conditions in the nature of potential and in the nature of normal component of the electric field then the solution is unique. There are some quick questions if I can take up 3 questions coming up if there are clarifications we will first is from Motilal Nehru College Institute. Yes sir my question is what is the specialty of using method of images to find the potential we can also find out it by using conventional method what is the specialty. So let me explain the point actually is that you know that what we are trying to do is basically to solve differential equations multivariable differential equation because it is a del square of phi that is given subject to boundary conditions. Now if you have solved differential equations you are given an ordinary you know either ordinary or partial differential equation subject to boundary conditions you must have realized how much of labour it involves most of the time that you do not unless you have differential equations of very special type you do not have special prescription for solving differential equation most of the time today you will find people use computer packages for solving differential equation and of course lot of effort has gone into this. I am not saying that method of images gives you a way in which for example have an integrating factor multiply and do it that is not what we are talking about we are saying in some problems where your intuition can play a role like supposing you are given a differential equation now if instead of trying to solve it by multiplying a integrating factor using a package and solving it supposing you have a great intuition that look I can look at what the solution could be looking like then obviously you will not be spending time on solving that differential equation completely. So method of images provides you a very convenient and a quick way of getting into and checking that there is a solution now you check that there is a solution then you do not have to go any further because we have proved that any solution that you have guessed if you put it back into the equation and say that it is satisfying the boundary condition that must be unique solution. So therefore when you said that what is the way let me let me explain what is it what it means supposing I asked you that what are the factors of 105 now you could sit down start using an Euler program for factorizing but many of your friends will say 105 last digit is 5 so 5 must be a factor. Now you understand what I am trying to say what did you go do here 5 is a factor indeed you went by intuition now is it any less rigorous but how much of time did you save on that. So what I am trying to tell you is the method of images is not a method which can be applied in all cases but there are certain cases where a particularly physicists would love to use their intuition rather than trying to say that ok let me spend next one hour and doing the factorize right see of course 105 is trivial example but I could give you a much bigger number for instance I give you a huge number now but a school child will say alright let me add up all the digits ok now I add up all the digits and find that whatever sum I got is divisible by 9 now what do you what can you conclude out of it if some of the digits is divisible by 9 you take a 20 digit fact number I am sure somebody knows the answer there what does it mean that number is divisible by 9 its intuition are you going to do the factor of that number the answer is no I know I will immediately test thank you next MGM college yes sir for the conductor the distance is b is equal to r square by b you according to you and q dash is equal to minus q r into a whatever and for the plane q is equal to minus q and d is equal to d dash so if I apply this law for the non-conducting there is a difference or same ok good question actually they if you took a non-conducting surface then of course the potential does not have to be kept constant you see a conductor maintains the surface at the constant potential so all that I am saying is can you give me a surface where the potential is specified now if you give me a non-conductor then of course you cannot say potential is constant you will then have to give me the complete potential distribution on the surface of that non-conductor now if you can do that then technically the solution by method of images is the same but it will not be an easy problem to solve in which case you will probably not try to solve it by method of images they are usually done for conductors because I can easily keep the potential constant thank you yeah anymore finding the non-conducting like this any formula yeah not by method of images of course there are say the basically you can always find a potential you know I mean we have you have you have done such calculations see analytical calculations are easy only if the problem have sufficient symmetry the Coulomb's law or the expression for the potential is usually for multi particle system is usually so complicated that if you want to find out how much is the potential you usually have to take recourse to a computer program but analytical methods are very difficult so method of images provides you an analytical method and of course it is applicable only for very specific and simple cases I will stop this question here because the questions are piling up and since this is math questions are on method of images please send the questions tomorrow first thing in the morning I will take it out ok so so far we have been talking about conductors ok what are conductors conductors are basically material where there are free charge carriers and which can flow when an electric field is applied but a large class of material they are actually non-conductors in a non-conductor the electrons are tightly bound to the atoms of the molecules to which they belong and and because of that that because they are bound to they do they do not have much of a mobility that is why they are called bound charges now suppose in a non-conductor we apply an electric field then you see a typical charge neutral nonconductor there the charge center for the positive charge and the negative charge they coincide I mean this is many many classes of materials are like that they are electrically neutral and sometimes if you apply an electric field it would lead to a visible difference or separation between the elect positive charge center and the negative charge center now when such a thing happens it creates a essentially a dipole moment now let us look at what the situation is supposing this is the situation you have the positive charge center negative charge center they are the same the the the electron cloud overlaps in the atom with the positive charges and they they they are centers coincide so that the net thing as a 0 dipole moment now if you apply an electric field it leads to a small separation between the two but the electrons do not actually leave this atom so therefore they remain bound to this atom but if you look at their charge densities there is a small separation between them so what actually happens is that this creates in this atom or molecule a dipole moment this is this is an an induced dipole moment because it arises only when an electric field is applied there are examples of dialectics or nonconductors where even in the absence of electric field there are charge separation but let us look at that suppose I am looking at such a situation where I have applied no electric field but if you look at the net dipole moment net dipole moment that is total dipole moment of a combination that is equal to 0 so let me explain this with a much easily understandable comparison you take a piece of iron now you know if you pick up a piece of iron from the road that piece of iron is not a magnet though iron is a magnetic material that piece of iron is not a magnet now the way one explains it is to say that the iron actually consists of a large number of small domains and these domains are in the absence of a magnetic field are randomly oriented so that the net vector sum is equal to 0 and I make the same state material if you look at this picture you notice that I have given a collection of large number of dipoles now if I do not have an electric field then these dipoles even if they are permanent dipoles these dipoles are not are randomly oriented giving me that the total electric dipole moment of this assembly is equal to 0 now what happens when you apply an electric field now if I apply an electric field like this the these dipoles will be persuaded to align fully or partly depending upon the strength of the electric field now since remember I told you yesterday the direction of the dipole moment is from negative charge to positive charge so therefore this is the way they would be oriented and in the presence of an electric field therefore I will develop a net dipole moment that of this assembly okay so this is a this is what is called a dielectric now what I am now going to do is this that basically I have I have taken a collection here this this is a random piece of material an insulator but you see I told you already that in the insulator if you look microscopically in small regions there might be dipoles so let us look at how does one handle this problem the way to handle this problem is the following that suppose I have a point p now this and I choose some origin it does not matter where you choose your origin now the vector r is from this chosen origin to the point p and let me take a small volume here located at a distance r prime now I am trying to find out now that how does one calculate the effect of the potential due to this volume element which may have charges or the dipole moments or whatever now let us look at first remember that we are dealing primarily with Coulomb's law so it is very important to find out how does one handle one of our r minus r prime in mathematics now first let us look at that so one over r minus r prime this is a very important thing particularly those who do nuclear physics they are very much involved into it so what I do is this this is r prime and this is r so what is r minus r prime this is this distance so this is given by by triangle law r square plus r prime square minus 2 rr prime cos theta to the power half okay now what I do is this that I am going to be going to assume that this distance r is much greater than this distance r prime now if you do that then look at I do a binomial expansion of this so I have got 1 over r square plus r prime square minus 2 rr prime cos theta raised to the power half and so this is nothing but same thing and I do a binomial expansion assuming that r prime by r is small if you do that trivial algebra I am not repeating it you get 1 over r that is the first r prime by r cos theta plus r prime square by r cube 3 cos square theta minus 1 at dot dot because I am not really interested in going further now look at each term this is what I wrote down and what I now do is I sort of put it in a slightly more compact form for instance I supposing I define r dot r prime I know it is rr prime cos theta so therefore I multiply and r there in the numerator rr dot cos theta I have got so therefore I get this relationship there should be square and then this relationship etc. Now I look at the various terms that is there I look at the various terms that is there in the expression for potential since I am only dealing with dipole the next term is incidentally called quadruple I am not really at the moment interested in it but let us just look at the first two terms. Now if I look at the first two terms now this is my potential because my potential is rho r prime 1 over r minus r prime d cube r integrated over the volume remember my first term in the expansion was 1 over r so since I take and take out 1 over r this integration is over r cube first term in the expansion was 1 over r so I have just taken it out now if I have taken it out I am left with integral rho r prime d cube r prime with this you recognize as the charge so the first term is simply q by r now this is the potential that a point charge you would have at a distance r how does it come it simply says remember my definition was my r is much greater than r prime so what we are saying is if your distances are very large even if you are looking at distributed charge distribution at large distances it seems like a point of course we know it any finite body if you go far enough it looks like a point so because of that I have a coulomb correction here a coulomb term here the second term is this you notice again in the in the top I had written r r prime r dot r prime so again everything connected with r I take out I am left with r prime rho r prime d cube r prime you recognize this as nothing but the dipole moment of the assembly. So the dipole moment term is rho r prime this is charge multiplied with distance and so therefore the corresponding dipole factor is 1 over 4 pi epsilon 0 r dot p by r so this is the way one does the dipole moment expansion. So let us come back now and concentrate only on the dipole moment and that is phi of r there is an algebra but I will go through it because of certain reason. So phi of r is 1 over 4 pi epsilon 0 now p r prime is defined as dipole moment per unit volume it is called the polarization so this is the expression that I have here. So this is the expression that I have and so it is p r prime dot r minus r prime by r minus r prime cube and this we have done several times that whenever you have got r by r cube you can write it as the derivative in this case if I took the derivative with respect to r I will pick up a minus sign but I am doing derivative with respect to the integration variable namely r prime so there is no minus sign there. So I got p r prime del dot gradient with respect to r prime d cube r prime now this term I combine by using the chain rule differentiation. So I said this is del prime dot this into this minus the scalar term out and del prime dot this is nothing but rearranging this term as del prime dot this into this minus this times del prime dot p which is simply saying f differentiation of two quantities is a differentiation b plus b differentiation that is all that I use. So once I have used that notice my first time was first term is del prime dot p r prime one over r minus r prime now this is divergence of a quantity integrated over a volume. So therefore this term by standard divergence theorem which we have been talking about several times can be written as the surface integral of this quantity p r prime dot n r minus r prime this is surface integral and I of course have a volume integral remaining there which is del prime dot p etc. Now look at these two terms for the interpretation purpose. So in the first term notice I have a one over r minus r prime and I have p dot n ds this is that since it is surface integral this is that type of term that you would get if you had a surface charge in density sigma b. So how do I define sigma b? Sigma b is nothing but the normal component of the polarization vector and the polarization vector I have defined as the net dipole moment per unit volume. The second term here has a minus del prime dot p but it is a volume integral. So therefore and again one over r minus r prime is there. So therefore this would be what I would get if there is a volume charge density here. So the original potential to the approximation that we have done has two terms. One is a surface term I put a subscript b to indicate that these are bound systems. So these are called bound charges. So I have a bound surface charge density sigma b. What is sigma b? Sigma b is nothing but normal component of the polarization. Then I have a volume charge density which is bound which is rho b and what is rho b? My rho b is minus del prime dot b. So this is the bound volume charge density and bound surface charges. So these are the two definitions. Sigma bound I am talking about bound charges because these charges are not flowing around. p dot m rho bound that is volume charges is minus del dot b. p is the polarization vector. Now let us look at what happens to our Maxwell's equation. Remember we proved that del dot of p is equal to rho by epsilon but then my rho could have two components. I could have one of the parts for due to the free charges which are moving in. So rho free and another part due to the bound. So I can write del dot of p equal to rho free plus rho bound by epsilon. Now notice that I could define a new vector which is I could define epsilon 0 del dot of p. I could define epsilon 0 e plus p. So I suppose I define a vector d. Incidentally though again do not ask me the nomenclature because the nomenclature may be historical but it does not mean anything. This vector d in fact physicists always call it vector d. The vector d is defined as epsilon 0 e plus p. Now remember from here this rho bound is minus del dot p. So substitute minus del dot p here take that to the other side. Multiply with epsilon 0 delta del dot e take this term to the other side. You will be left with the new version of our Maxwell's equation which is del dot of d equal to rho free. What is this rho free? Now the thing is this that we are assuming that there are bound charges in the system. Now the and there are two charges in the system which are the rho free. Now if you are only concentrating on the two charges not the induced bound charges then del dot of p does not give you the rho free by epsilon. But if you define a different vector then del dot of d gives you that rho free. So in other words now mind you it is not very easy to make this separation. Not easy to make this separation because this is intuitive. This is you have to imagine how that if I subtracted out the bound charge if I could then use the mechanism to calculate my d vector then my del dot of d would be equal to rho free. This is the way now our new the Maxwell's equation would look like my del dot of d is equal to rho free. And along with that I need my subsidiary definition what is d epsilon 0 e plus p. Remember e is the actual electric field which a test charge would actually experience. Now let me see if there are quick questions which I can take up and this is silly good even yes. I want to ask you question on Laplace's equation. Yes. You taught me taught us earlier. Yeah. First thing del square phi equals to 0 that is the Laplace's equation. Right. So applicability of Laplace's equation. Suppose you have shown two conductors aborted by distance placed in air medium. Yeah. So that is a charge free. Yes. So we can apply Laplace's equation to find out potential. Sure. But if we if we place dielectric material in this parallel plate capacitor. Yeah. Between two conductors sorry two conductors. Right. So what is the situation del square phi equals to 0. Can we use this Laplace's equation to find out potential or we use a Poisson's equation. See since your solution you are trying to solve the problem not in the dielectric but in the charge free medium. See the point is you are saying that along with my conducting plates I also have a dielectric there. Okay. Now if you are trying to solve in a region where there are no charges then of course the Laplace's equation is valid. So you can solve it but the problem is this with Laplace's equation that you know other than in you know I mean nowadays of course there are many techniques of solving Laplace's equation particularly if you are interested in going to a computer. These are packages which are available but so the otherwise solving a multivariate second order differential equation is not an easy task. I mean you do not have analytical methods of doing it and that is the reason why we are trying to talk about subjects like method of image. You know relatively simple problems which give us intuition about how to find it out. But the thing is this that first thing is is the Laplace's equation valid. If the Laplace's equation is valid then of course all that you need now that you have to change the thing depending on the dielectric. If the dielectric is uniform and linear then you can again make some approximations and do it. But if you have a general dielectric right without any simplifying assumptions then it is not going to be easy. See there is a there is a statement on in principle is it possible the answer is yes. In practice it is possible the answer is no I mean you have to then go to a computer. There are no methods that I know which will be able to solve it. But I was planning actually to come to a solution of Laplace's equation using some methods like you know we have some ways of doing it. So let me take that up because you have raised this question. Let me take that up will maybe in the next session you come back to your question again. Since I some of you have shown interest in solving Poisson's Laplace's equation let me repeat a few things that I said and try to solve it for some problems. You realize that I always my starting point is this del square phi equal to rho by epsilon 0 that is Poisson's equation and del square phi equal to rho 0 that is my Laplace's equation in source field. So in addition of course I am looking at the electrostatic field. So I have del cross of E is equal to 0. Let me try to tell you how to solve Laplace's equation in some specific cases. But first you need the expression for Laplace's n. Now Laplace's n in Cartesian of course all of you know del square phi by dx square dou square phi by dou square dou square phi by dou z square. This is the expression in spherical coordinates. This is what I will be doing today and this is the expression in cylindrical coordinates one can do solutions there also. Now there are certain methods of solving these equations. So let us look at that first. I have talked about uniqueness theorem. Basically the uniqueness theorem told me that if I have two solutions of Laplace's equation let us say phi 1 and phi 2. This is a region in which I am looking at solutions in regions let us say outside as and I have potential or the normal derivative of electric field given on various surfaces that one there. Then my statement was uniqueness theorem statement was that the solutions are unique. I am not going to prove it again we have done that last time. So I made a statement that one dimension Laplace's equations basically are featureless equations. So I gave an example for example supposing you had one dimensional equation. So what is one dimensional equation? d square phi by dx square equal to 0. The solution of that is mx plus c which is a straight line. So therefore you can find out supposing phi is given to be equal to 0 at x equal to 1, phi is given to be equal to 3 at x equal to 2. You can immediately write down that phi at any point x is nothing but the averages. I go to next dimension. This is of course Poisson's equation and I told you that there is a feature in this equation. The Laplace's equation I told you does not have that interesting feature that is you generally do not come up with a potential minimum or a potential maximum. For example in this example it is a saddle point. Let us come back to in the next few minutes that I have a technique of solving Laplace's equation. So I told you that this is incidentally a technique which is useful even in Cartesian coordinates and that is del square phi is given by this expression. This is the expansion of the Laplacian in spherical coordinates. Now the standard technique of solving these equations is to do what is known as separation of variables. That is write phi which is a function of r theta and phi as a function capital R of r P of theta and f of phi. Now substitute it here. If you substitute it here you realize that I can write it, substitute it here then divide things properly. This is well I have used r r P theta f phi. If you do that since d by dr takes only on capital R, d by d theta is only on capital P, d by d phi is only on capital F. Now this is the equation that you get. Now when you get this equation then you say by left hand say the equation that you get is this plus 1 over f del dou square phi by dou phi square equal to 0. Then you say this is going to be equal to this. Now this depends upon r theta, this depends upon phi. Now since theta phi and r they can vary independently. How can these two things be constant same? They can be same if the left hand side is constant and right hand side is constant. You put that equal to constant. Now if you put that equal to constant you immediately get a solution for the azimuthal which is e to the power i m phi. Put it back there get back to the equation. Now remember e to the power i m phi I can find out what is m because if phi goes around by 2 pi then I must return back to f of phi. In the same way now I go to the polar equation namely the theta equation. Now when I go to the theta equation once again I say by separation I have left hand side I have a function of r. Second the middle term is a function of theta and they can be equal only if each one of them is constant. For convenience I put that constant not as some constant l but as l into l plus 1. There is a reason but do not worry about it. So my theta equation now looks like this. Now this equation which you have got is a very well studied equation and what you do to study this equation is to substitute cos theta equal to mu. You do that this is your equation and you can mathematicians will tell you we have known this equation for years and even physicists use this equation. This is the solution of this equation are known as Legendre polynomials. The Legendre polynomials see the point is that these are already in standard mathematics texts. Now so if I am given a standard equation I do not try to solve it again because if other people have solved it I take borrow it from them. The solution of this equation are known as associated Legendre polynomial P and it properties that we have is P0 cos theta equal to 1 P0 P1 cos theta is cos theta P2 cos theta is half cos 3 cos square theta minus 1 etcetera and this is the way the picture looks like. So this is your P1 P0 is of course 1. So therefore that is here P1 is this is plotting against cos theta. So therefore it is a straight line P2 is this with a single minimum P3 P4 this is the way the functions look like. I would in one of the sessions where we take up problems try to solve a very interesting case of conducting sphere in a uniform electric field using the method of Legendre polynomials but at this moment I am running out of time so therefore I would stop it here. Thank you.