 In the last lecture, we had brought in the concept of a dielectric. We said that in while in conductors, the charges are free in the sense that the carriers of electricity they do not remain tightly bound to an atom or a molecule. There are situations in insulators where what you find is that the electrical charges retain their identity to the molecule or the atom to which they belong. The we had talked about two different types of materials, one where even in the absence of an electric field, there is a separation of the positive and the negative charges, thereby the molecule getting a what is we can say as a permanent dipole moment. On the other hand, there are molecules or systems where in the absence of an electric field, the charge centers that is the positive and the negative charge centers they coincide. However, if you apply an electric field, these charge centers are pulled apart and as a result the system or the atoms and the molecules they develop what is what can be called as an induced type of moments. So, basically these are the class of materials which go by a common name of dielectric as opposed to the conductors where the charge carriers are said to be free. So, what we did last time is to look at the concept of what happens in a situation like this and what we said is that the system because in the presence of an electric field for instance, the there is a separation of the charges. So, the molecules they aligned in the way it is shown here, for instance you notice the negative there end to end negative positive negative positive etcetera. Now, what you notice is in this area in the central region there is a negative charge for instance which sort of acts as a source for the electric field. So, notice that here I have a source which has come because of the way the molecules of the material have become aligned and this is what we have been calling as the bound volume charge. . The other thing that happens is that if you look at the boundary of the thing then you notice again that there is a separation of charge that is the edges seems to get charged. For instance the here I have shown on a particular type of alignment in which this edge becomes negatively charged and this edge becomes positively charged. Now, the second thing that we did last time is to look up a what we called as a multi pole expansion. The basically we are interested in finding out the potential due to a an arbitrary charge distribution. And so as you know that the arbitrary charge distribution is I have 1 over 4 pi epsilon 0 supposing rho of r is my density. So, the potential at the point r is given by the usual expression 1 over 4 pi epsilon 0 integration over the whole volume rho r prime 1 over r minus r prime d cube r prime r prime is my integration variable. Now, what we do here is this that since r is a fixed position at which we are interested in finding out the potential this quantity we expand. There is a 1 over r minus r prime which is being expanded and I sort of in principle this is an infinite series, but what has been generally found is that for most physical situation if you retain up to the first the first term as you realize is just I am expanding 1 over r minus r prime by r and 1 over r I have taken out. So, in the first term I simply get rho r prime d cube r prime as you can see this is just the total charge q. So, therefore, this term is the total charge q divided by r which is exactly the expression that we have when a point charge located at the origin its potential is calculated at the point r. Now, notice the reason why it is true I have made this expansion for r very large that is further away from the charge distribution. Now, if r is very large if I am looking at a potential at a large distance from a charge distribution as a very crude first approximation it turns out that that charge distribution seems like a point charge. Now, since it looks like a point charge to the leading order in the expansion I just get the potential due to a point charge which is q by r which is exactly what is shown here. Now, when you go to the next term. So, I have got 1 minus r prime by r to the power minus 1. So, therefore, my leading term now becomes this the other than 1 over r square which comes out. So, I get r by r cube dotted with this integral integral r prime rho r prime d cube r prime. Now, let us look at this quantity here. So, we said integral r prime rho r prime d cube r prime. Now, this is the quantity which we define to be the dipole moment. They as you can see that this is a straight forward generalization from the way we define a point dipole where we had simply 2 charges which were separated by a distance d and we said that the dipole moment has a magnitude q times d and its direction is from the negative charge to the positive charge. Now, what I have done here is to essentially extend the same concept to a charge distribution and here again I get a distance times the charge density and the direction will still be from negative charge to the positive charge. Now, this is what we call as the dipole and we are going to be looking at this term in lot more detail as we go along. The next term in the expansion is what we call as a quadrupole term and I had shown you last time that the quadrupole term is given by I am basically looking at this expression here which is 3 x i prime x j prime minus r prime square delta i j rho r prime d cube r prime. And you notice that this x i and x j these are the components of the position r prime. So, for example, 1 2 is x y etcetera and so this quantity this integral is what is known as a quadrupole moment tensor. You can see there are 9 quantities there because i and j can each run from 1 to 3 that is x y and z. So, and for most practical situations this essentially is the distance to which we have to do the expansion and further expansion which would be octa pole and things like that they are normally not of much use. Just a comment on the quadrupole the notice that q i j is d cube r 3 x i x j minus delta i j r square rho r. Now, this quantity has certain properties which you can very easily show. First is that quadrupole moment tensor is traceless. Now, you can see that you remember that trace is simply the sum of the diagonal components of a matrix. So, if you just look at how much is the diagonal component you can trivially prove that this quantity is 0. The second thing that you can prove which I will talk about it in my lecture that typically the quadrupole field goes as 1 over r to the power 4. Now, like point dipole one can think of quadruples. Now, the quadruples there can be various types. For instance we could think in terms of a n to n quadrupole meaning there by I have got let us say minus q then plus q then plus q and minus q that is that is a charge plus 2 q at one end at the center and a minus q minus q there. This is what I will call as a linear quadrupole. You could think of the 4 charges distributed in a slightly different fashion. For example, they could be distributed at 4 corners of a square. So, I get q here minus q there plus q there and minus q there. So, this also is an orientation or a configuration for a quadrupole. We will be looking at quadrupole as we go along with the lecture slate. The point about dipole which is going to be the primary subject of discussion in today's lecture is that if you look at the dipole expression dipole moment expression vector p equal to integral rho r vector r d q r. You notice this could in general depend upon a choice of origin. For instance if you define a vector r prime equal to r plus a then the my dipole moment becomes rho r prime this is just a variable change and r is r prime minus a and d q r of course is d q r prime. So, look at the first term you notice that this is nothing but the dipole moment with respect to the new variable that I have chosen minus since a is a constant I have got rho r prime d q r prime. This quantity is nothing but the total charge. So, the expression for the dipole moment will not depend upon the origin if the total charge is equal to 0. If q equal to 0 p is independent of origin. So, this is about some general comments about dipole moment and let us proceed with what we were talking about. So, I defined the polarization vector as the dipole moment per unit volume. So, I take a unit volume I take a volume v make a vector sum of all the dipoles there in and divide it by the volume that is being considered. Now, the we derived the expression for the potential of this collection of dipoles potential due to this collection of dipoles and what we found last time was that the general term general expression for the potential due to a collection of dipole is given like this 1 over 4 pi epsilon 0 of course and I have got a this is actually consists of two terms. The first term is a surface term and the surface term is p dot n d s prime p is the polarization vector that we talked about over r minus r prime and the second term which is a volume integral term which is 1 over 4 pi epsilon 0 integral 1 over r minus r prime minus del prime dot p r prime d q r prime. Now, if you look at these expressions you notice that this is the type of expression we would get for the potential due to a charge distribution for example, in this case it is a surface charge distribution. So, if I had a surface charge density sigma in place of this p r prime dot n which is nothing but the normal component of the polarization vector if I replace this with sigma b then this would be the potential expression due to such a surface charge distribution and similarly if minus del prime dot p prime is replaced by a volume charge density rho b then this should be identical to the potential that I get due to a volume charge distribution rho b within this volume. So, this tells me that the polarization vector is related to the bound charges by this expression that rho bound let me give the symbol rho b this is nothing but the minus del dot p where p is the polarization vector and sigma b which is the surface charge distribution is given by the normal component of the polarization vector. .. The effect of the dielectric is to essentially given by a sum of a volume charge density rho b and a surface charge density sigma b and how does it modify the Maxwell's equation that we are familiar with remember del dot b was equal to rho by epsilon 0. Now this is the actual electric field, but now my rho can have two components first is of course could be the free charges that are there in the system. So, let me call it rho free and the bound charges that have been developed because of the polarization. So, it is rho f plus rho b divided by epsilon 0. Now notice this makes it more difficult to compute the electric field which is the actual electric field that a test charge at the location where the electric field is calculated will feel. Now in order to make some sense out of it what one does is to define an auxiliary vector which we call the displacement vector or is simply called the vector d. Now this is defined to be epsilon 0 times the electric field plus the polarization vector p. Now note that the vector d and vector e do not have the same dimensions because there is a permittivity constant coming in there. Now using this I can write down the Maxwell's equations like this what is del dot d. So, notice del dot d is epsilon 0 del dot e plus del dot of p and we have just now seen that del dot of e is rho free plus rho bound divided by epsilon 0 and minus del dot of p is just rho bound. So, therefore, this is minus rho bound. So, after canceling the epsilon 0's what I am left with is simply rho free. First thing there are two interesting points to note the displacement vector satisfies a Maxwell's equation that its divergence is given in terms of the free charge density. Now remember I it is much easier than to calculate d because I would know how I would have a better information about free charge density than I have about the bound charge density because bound charge density is an internal mechanism. Mind you that the bound charges are as much of a real charge as the free charges are there is nothing they are not fictitious charges they are actual charges accepting that they still remain attached to their parent molecules. So, del dot of d is rho free. Now one can then write down the Maxwell's equation in the integral form recall that integral of del dot of d d q r is then equal to integral rho f d q r and I use the divergence theorem to convert this divergence of d over a volume to a surface integral of d over a closed surface containing the volume. So, it will be d dot d s and that is equal to integral rho f d q r which is nothing but the free charge total free charge. So, this is the integral form of the Maxwell's equation and once again note that the vector d is related to the free charge density whereas the vector e is related to both free charge density and the bound charge density and the realistic field is actually the vector e though we might occasionally find it easier to work with vector d which is determined in terms of the free charge density about which we may have little more information. Having done this let us look at the boundary conditions that we satisfy. Now notice this is very similar to what we did in case of a conducting boundary accepting that you have to remember that in when we took a Gaussian pill box half outside a conductor and half inside a conductor the electric field inside was 0 that is no longer true here. So, let me first just take assume there are no free charge makes life simple assume there are only bound charges. Now if I now have a Gaussian pill box with of a very small width height half into the polarized medium namely the dielectric and half outside namely to the vacuum then I can write del dot of e d cube r because remember that I have no free charges. So, e we have already seen that this would generally then mean this is minus integral del dot p divided by epsilon 0 d cube r we only assume bound charges that is integral e dot d s is 1 over epsilon 0 if I convert this to a surface integral this becomes 1 over epsilon 0 p dot d s. Now, now what we do is this that since these have now been converted to surface integrals I look at the two surfaces remember that the thickness is so small that from the curved surfaces there is no contribution to this integral. So, what we do is this and we also realize that the direction of the normal for the outside surface is opposite to the direction of the normal for the inside surface. Now as a result what we get is this that e out that is the field outside minus e n dot n d s the minus sign comes because the direction of n is opposite and that is equal to p dot n d s and the reason is that there is no polarization outside. So, therefore, I have only contribution from inside and we already know that this p dot n d s is nothing p dot n is nothing but my bound charge density the surface bound charge density there is delta is missing here. So, if you look at that what I get is that the discontinuity in the electric field e out minus e n as was before given by the surface charge density divided by epsilon 0 and this is precisely what we had earlier as well accepting that in this particular case the surface charge density is purely due to the bound surface charge density. This is one of the boundary conditions that we have. The second boundary condition is actually on the tangential component and this is done exactly the same way as before that is take a circuit and since the electric field is a conservative field just go round the circuit and you get if this is the medium 2 this is the medium 1 I get e 2 t is equal to e 1 t. They look at this we had said that d is equal to epsilon 0 e plus p and so, therefore, you can now convert this boundary condition that we did for e to a boundary condition on d and that gives me d out normal component out minus d normal component in is freeze free charge density surface charge density of the free carriers. So, at all stages the vector d is related to free charge density whether they are on the surface as it happens in case of a conductor or inside the material. So, d n normal component of d is discontinuous only if there is a charge surface across which there is a discontinuity. This is exactly the same thing as we had talked about for the normal component of the electric field when we were talking about conductors. The difference is that because of the difference in their dimensionality the d n out minus d n in is simply equal to sigma b no epsilon 0 etcetera and the tangential component of the full electric field e is continuous. So, these are the things that we need to learn. Now, there is a class of dielectric known as the linear dielectric. Now, in a linear dielectric the polarization vector that is the dipole moment per unit volume is proportional to the electric field e itself. Now, so therefore, what one does is to write the polarization vector as some constant which we will write as epsilon 0 times some number chi e and chi e is normally defined to be a susceptibility and so epsilon 0 chi times e. Now, notice that d is epsilon 0 e plus p by definition. So, therefore, if you replace p by epsilon 0 chi this gives you epsilon 0 into 1 plus chi times e. This quantity 1 plus chi is called the dielectric constant that is a number kappa. So, epsilon 0 times kappa this kappa is the dielectric constant and the epsilon which is the product of epsilon 0 times kappa is usually called the permittivity of the medium that is the polarized medium just as epsilon 0 was called as the permittivity of the free space. So, epsilon and epsilon 0 are related by a dielectric constant. Now, let me now go over and try to illustrate some specific methods of doing problems with dielectric. This problem is where we consider two semi infinite dielectrics. Well, one we will take to be epsilon 2 which could be epsilon 0 for example and epsilon 1 and I assume that the epsilon 1 fills the entire half space z greater than 0 and the interface is z equal to 0 which is perpendicular to the plane of the figure and epsilon 2 you could take it as vacuum, but once again it is in general another dielectric. Now, so what we want is this in the dielectric when I use the word dielectric I mean epsilon 1 here and I will occasionally call epsilon 2 as epsilon 0 which is my vacuum. So, epsilon 2 exists in the space z less than 0 and therefore, there is since there are no charges in the region z less than 0 I know that epsilon 2 times del dot e must be equal to 0. The in the region z greater than 0 epsilon 1 del dot e I know is nothing, but the charge density rho. There is a charge at least here because we have put a charge at the position d the how does not solve this problem. The basic principle of solving this problem is to observe that the normal component of the vector d and the tangential component of the vector e must be continuous at the surface which is at z is equal to 0. And since my surface is z is equal to 0 my normal direction is the z direction. So, since the normal component let me assume a linear dielectric. So, if I have a linear dielectric the relationship between the electric field and the field d is given by d is equal to epsilon that is the permittivity of the medium times the electric field. So, therefore, the normal component of the electric field in this medium is epsilon 1 times e z and the normal component of the electric field in the other medium is epsilon 2 times e z well I have said in out. So, these are essentially the z components I know tangential component of the electric field e is continuous. And since z is equal to 0 is nothing, but the x y plane you could take any two directions there let us call it x and the y direction. So, that I have x component of the electric field e is the same as the x component inside and the same is true for the y component. So, basically the tangential components are continuous. Now, you recall that when we did the problem of a charge in front of a conducting grounded conducting medium or conducting plane we had seen that the problem can be solved by what we called as a method of images. And what we did there is to put an equal and opposite charge at the image position. This was done so that the boundary condition at the surface of the conductor namely the electric field becoming equal to 0 can be satisfied. So, let us look at what happens in this case. Now, what we do is this that look at this picture once again before I describe the method I recall that whenever we are doing solutions of problems like this we depend on the famous uniqueness theorem namely that this solution of the Laplace's or the Poisson's equation that we are getting they are unique. And so therefore, it is immaterial how you get it once you have got it and it satisfies this equation you are perfectly. So, let us look at that situation. Now, in the z greater than 0 that is in the medium which I have been calling as the dielectric. Now, what we could do is that the solution in this medium is the same as what one would obtain if one were to fill up the entire space with a dielectric medium having a dielectric permittivity epsilon 1. And replacing and putting a charge q prime at a distance d from the origin in the left hand side. So, notice that what I have done is to this q is my original charge and this is at a distance d into the medium. Now, I am saying that this imagine that the entire medium is uniform medium and you put a charge q prime at a distance d here. So, that I have got a charge q and its image q prime. Now, since these are just two point charges in a dielectric medium epsilon 1 I can write down at any arbitrary point p the potential due to such a charge distribution. And I notice that I can write phi 1 is equal to 1 over 4 pi epsilon 1 q by z minus d whole square plus rho square rho is nothing but x square plus y square and q prime by z plus d whole square plus rho square to the power half. Now, notice that if you took the Laplacian of this equation this gives you the singularity at the point where the charge q is situated. So, therefore, this indeed is the solution of the Laplace's equation that we set out or Poisson's equation that we set out to solve. Now, what happens to the field in the left hand side that is z less than 0. Once again I assume that the entire space can be considered to be filled up by a charge by a dielectric epsilon 2 all over. And in this case I substitute or I put an image charge which I have called as q double prime. So, in all words q double prime gives me the solutions of the Laplace's equation in this problem. And this is again written down 1 over 4 pi epsilon 2 because I have filled up the entire medium with a dielectric of permittivity epsilon 2 q double prime by z minus d whole square plus rho square to the power half. Once again if you did the Laplacian of this you will find that this the Laplace's equation is satisfied there are no singularities in this problem. So, what do I do now? So, what we do is this I have got two expressions. I have got an expression for phi 1 which is the field in the right hand side and I have got an expression for phi 2 which is an expression for the potential in the region to the left. And I need to satisfy the boundary conditions and these boundary conditions are things which we have discussed a little while back namely the normal component of d is continuous. So, let us look at what is normal component of d. So, the normal component of d let us just do calculation for one of them. So, what is d 1 m? I know that d 1 m remember that this is a linear dielectric. So, all that I need is epsilon 1 times e 1 m. So, that will cancel out this epsilon 1 that is here and I will be left with 1 over 4 pi this is I know that negative gradient of the potential is what I need to calculate. So, it is minus d phi 1 by d r because these are normal derivative. So, since well d actually d n and since the normal direction is just the z direction. So, d 1 n is same as 1 over 4 pi times minus d by d z of this big expression that we wrote down. This is a very easy differentiation to do because I have got 1 over z minus d whole square. So, you notice this is 1 over 4 pi minus sign I have got q of course, let me keep take the q outside. I have got z minus d whole square plus rho square to the power 3 by 2 and a minus half and d by d z of this thing. So, which is simply 2 times z minus d and a very similar calculation for this one. So, let me put the q inside here and a q prime there. So, this will be simply replacing with 2 z plus d q prime and by z plus d whole square plus rho square to the power 3 by 2 and of course, minus a half again. You can simplify this and notice that you have to take z equal to 0. So, that both these terms will have the same denominator and as a result these 2 terms can be added up very simply. So, this is this will give you one equation. I do a similar job for the second term I am not working it out and you notice that this will give me an equation of this type q minus q prime by d square plus rho square to the power 3 by 2 equal to q double prime divided by d square plus rho square to the power 3 by 2. The second condition that I must satisfy is that the tangential component of E must be continuous and what is tangential component? Tangential component is differentiation with respect to either x or y. Remember rho square in my denominator is nothing but x square plus y square. So, once again the it is a fairly straight forward derivation and with this you get these 2 equations. They are fairly easy to solve and what you find is that the this gives you the amount of 2 image charges that you have. Turns out q prime is epsilon 1 minus epsilon 2 by epsilon 1 plus epsilon 2 times q and q double prime is 2 times epsilon 2 by epsilon 1 plus epsilon 2 by q. So, this is a method of solving the problem at with image charges. So, notice that what we have done is this. We have assumed that the dielectric is linear and there are no free charges accepting probably at z is equal to d and you notice this implies since it is a linear dielectric I am a polarization vector is proportional to E. So, that the dielectric vector d, d is proportional to E. So, that del dot of p becomes equal to 0 and since del dot of p is equal to 0 there are no volume charge densities. However, there is a bound surface charge and the bound surface charge you notice can be easily calculated because we had seen that the surface charge density is the normal component of the polarization vector. So, that it is p 1 dot n 1 plus p 2 dot n 2 and I know that the direction of the normal is the z direction which is towards the positive z for the medium to the right hand side and towards the negative z for the medium on the left hand side. So, therefore, one can write this as minus p 1 dot z plus p 2 dot z. We have just now written down that p is proportional to E and so I just rewritten it here and you notice that since the normal component of electric field the d field is continuous that will cancel two of the terms there and I will be left with simply sigma b equal to epsilon 0 e 1 z minus e 2 z. So, this is the surface charge density. So, this is the way one does use uses image charges for these problems. Let me look at a problem which once again we did it for the case of a conducting sphere. We had seen how the field lines are modified near a conducting sphere if it is put in a uniform electric field along the z direction. So, let me look at what happens to a dielectric sphere. Let me say that there is an electric field E along the z direction. So, let me write it down by electric field E is E 0 z. The corresponding potential which should be applicable at far distances external potential at large distances should be minus E 0 z and since I am going to be using spherical symmetry. Let me write it as minus E 0 z cos theta. So, this should be the form of the potential at large distances from the sphere. Now, we had seen that one can have a an expression for the potential in general using associated Legendre functions which is nothing but an expansion in various cosine theta cos theta cos square theta etcetera. Now, since I know that whatever is the form of the potential it must reduce to minus E 0 cos theta at large distances. So, what we will do is this that the field in vacuum has to be given by. So, let me call this phi 1 of r theta that is equal to a 1 r cos theta plus b 1 by r square cos theta. So, this is the field outside the dielectric. Notice that all that I can have is a cos theta term. So, that at large distances I recover this back and there is no 1 over r term because there is no point source of charge. So, b 1 by r square cos theta is what I have which is of course, a term which would vanish at infinite distance leaving me only with a 1 r cos theta. The field in the dielectric I should write in a very similar way accepting that the constant let it be different. Let it be a 2 r cos theta I do not write down b 2 by r square cos theta because the point r is equal to 0 is a part of my dielectric. So, as a result the I cannot have the potential diverging at that point. Now, first thing is to notice that at large distances this must boil down to minus E 0 r cos theta that tells me that the constant a 1 is nothing but minus E 0. The other thing to notice is that the potential itself is a continuous function because if it were not then the field at the surface of the dielectric would become infinite. So, since the potential is continuous for all theta. So, we put r is equal to a in these two expressions and equate them. So, that tells me that minus E 0 a plus b 1 by a square is equal to a 2 a cos theta has been cancelled because this relationship must be valid for all angles theta. The third condition is there are no free charges on the surface. So, as a result the normal component of D must be continuous that gives me a another relationship that I am looking for the if you solve these expressions you can write down what are a 1 a 2 and b 1 and get an expression for the potential phi 2 for instance and get what is electric field. What we will do is that take this result next time and find out what is the effect of the polarization of the medium on the electric field lines that are there due to the external uniform electric field.