 Good afternoon, so we will continue from where we left and we expect to finish our discussion of electrostatics today. So what we did just before we went for lunch is to try to look at the Maxwell's equation. So what we said is that the Gauss's law was that the flux through any closed surface is given by the amount of charge which is enclosed inside divided by epsilon 0 and we used the divergence theorem to convert this into a differential form of the Gauss's law. So since the flux is defined as the surface integral of the electric field E.ds and this by divergence theorem is divergence of E the volume integral of the divergence of E and that is equal to 1 over epsilon 0 and the enclosed charge is nothing but the integral of the charge density rho over the volume. So if you now compare these two expressions what we get is since this is going to be valid for any volume we must have the integrants equated. So divergence of E which is also written as del dot of E is equal to rho divided by epsilon 0 this is the first equation of Maxwell that we are doing it in differential form. So let us proceed with it I will not repeat this slide because I prove to you that Gauss's law can be derived from Coulomb's law and I repeat because many of you have been pointing it out the thing is you are alright that Gauss's law though completely is equivalent to Coulomb's law in order that we can use Gauss's law effectively the problem that we are doing must have sufficient amount of symmetry and Coulomb's law on the other hand is a basic law. Now do not think that means Coulomb's law is any more effective than Gauss's law because doing these vector sums using Coulomb's law is not always possible. So the ultimately if a problem has symmetry it is easy if a problem does not have a symmetry you will be using Coulomb's law but whether you can still effectively get a result or not depends on your ability of doing some complicated integrations etc. So I gave you this example of how where I do not have a symmetry obvious symmetry I can make it asymmetrical situation remember the surface or the volume that we have it does not have to be a real volume it could be an imaginary volume. In other words Gaussian surfaces could be imaginary I will not repeat this problem which we did earlier and so let me instead go over to a different problem to illustrate the power and in this case it is not just Gauss's law I am using it along with the principle of superposition that so basically what I want to do is that I have a uniformly charged sphere with a cavity inside it now I will go over to the write up here so let us look at that. So remember this is of course cool physics so I what I am doing is this that I have let us say a uniformly distributed charge q so if I have a uniformly distributed charge q so I will define the charge density to be rho and so this is 4 pi by 3 r cube times rho. Now if I have a uniformly charged distributed charge then of course so far as a point outside is concerned the field is given by the Coulomb's law assuming as if the entire charge is located at its center now on the other hand if I look at a distance let us say which is r a smaller distance from the full radius so I am looking at the electric field here now the amount of charge enclosed is now within this 4 pi by 3 small r cube volume and if you look at now first is symmetry so where does symmetry come from since this is spherical these all these points on the surface of a circle of a radius small r are symmetric. So therefore my flux is simply electric field times 4 pi r square the direction of the electric field is obviously radial by symmetry so I am not writing down the vectors now this quantity is equal to q enclosed divided by epsilon 0 now how much is q enclosed because this is a sphere of a radius smaller so it is equal to 4 pi by 3 r cube small r cube times rho and of course divided by 1 hour epsilon divided by epsilon 0 so if you look at it my electric field at a distance small r less than for r less than r is given by rho times r small r divided by 3 epsilon 0 and of course the direction is a radial direction now this is of course you are familiar I am not writing the corresponding expression for r greater than r this is obviously given by Coulomb's law for a point charge okay now the point is the problem that I have here is the following that I look at a sphere of radius r I assume that there is a cavity here so the center of the cavity is located from the center of the sphere at a position given by vector D so this is a sphere with a cavity this is actually an excellent example of the use of superposition principle so how do I handle this so the job is to find out the electric field at a point let us say p this point is inside the cavity located at the position smaller now in order to do this problem the what I do is this I imagine that my cavity is filled up now how do I fill up a cavity remember cavity is nothing but emptiness so therefore I assume that a cavity is replaced by a superposition of charge density rho and a charge density minus rho because rho minus rho is equal to 0 so if you superpose a charge density rho with a charge density minus rho then I have essentially a cavity with no charge at all now by introducing this charge density rho what I have done is fill up the big big sphere so my problem has now a big sphere of radius r having a charge density rho because now everything is filled up having rho plus a small sphere a charge density minus rho now notice one thing that when I have stated the problems in terms of rho the radius of these sphere is not coming into the picture so what I do is this I now write down I now write down what is the what is the field so let us redraw that picture again this time in a much bigger way so here is your cavity this is the center this is the vector D and this is the point P and this is my r so what actually do I do so the field due to the bigger sphere the field due to the bigger sphere is obviously rho by 3 epsilon 0 times vector r now the field due to the smaller sphere now notice the point P is at a vector distance r minus D from there so therefore due to the smaller sphere is rho by but this is rho is minus rho minus rho by 3 epsilon 0 and the position of that is r minus D so this is the power of superposition principle if you add it up you notice that the field is simply given by rho by 3 epsilon 0 times D vector D notice one thing the vector D is a given distance between the center of the original sphere the big sphere with the small sphere in other words vector D is a constant quantity so therefore the field inside a cavity is constant this is really a very interesting result continuing with my discussion I give you a third example and this is let us talk about intersection of two spheres intersection intersection of two oppositely charged spheres two oppositely charged spheres so there is no cavity in the problem but I have two spheres so this is first sphere having a charged density rho I have another sphere having a charged density minus rho the center of this is rho and the center of this is rho prime now the idea is to find out the field in the to find field in the overlap region once again it is an excellent example of or illustration of the principle of superposition so look at any point here supposing this is the point p and this vector is there you can call it anything so and the so due to the first sphere due to the first sphere my field is given by at the point p is given by rho by 3 epsilon 0 and o p vector due to the second sphere it is also given by a similar expression but because rho is oppositely directed it is minus rho by 3 epsilon 0 o prime p which gives me that this is rho by 3 epsilon 0 o p minus o prime p but what is o p minus o prime p this is nothing but the vector joining the two centers so which is nothing but rho by 3 epsilon 0 into o o prime so once again you notice this is an example in which the field is again constant because the field in the interface region did not depend upon actually where your point p is it did not depend upon where your point p is but it depends only on the vector distance between the two centers of the circle so therefore this is an illustration of the let us go over to the screen so this and if you people have any specific problems or questions you want to be answered you can just send it as a chat we will I will take care of it tomorrow because as I said we have spent a lot of time in the morning today so the next question that I want to take up is this that having done so much about electrostatics it is time now to summarize the electrostatic situation first thing to know about that is we have pointed out several times that coulomb force is central meaning thereby that the strength of the force depends only on the distance as we know in this case it is inverse square of the distance and it is along the line joining the source and the charge so therefore it is a central force second thing is that it is a conservative force like you have done gravitational force a conservative force by definition means that I can define a scalar potential so let us do that because my force expression I have not written down the 1 over 4 pi epsilon 0 etc my 4 force expression is basically vector r by r cube so if you take the curl of this expression you get the curl of now this is a vector multiplied by a scalar 1 over r cube times vector r now when you have a vector multiplied by scalar the del cross of that is gradient of the scalar cross product with the vector plus the scalar times the curl or the curl of that vector so you write that down so it is del that is gradient of 1 over r cube which is my scalar cross r plus 1 over r cube times del cross of r now this is very trivial because it depends upon r only so it is like differentiating with respect to r so 1 over r cube differentiated as you know is minus 3 by r to the power 4 which since it is a vector I write it as minus 3 vector r by r to the power 5 cross r del cross of r you can trivially calculate and show it to be equal to 0 and this is r cross r which is equal to 0 so therefore del cross of f that is the field is equal to 0. Now curl of any field being equal to 0 indicates that it is consummative so we have seen this in case of gravitational field it is true of electrostatic field also. Now I need an expression for electrostatic potential I am not going to do a lot of derivation but there are few things which I must point out so notice I am interested in finding out what is that quantity what is that quantity whose gradient gives me 1 over r cube r vector r by r cube by definition of gradient 1 over r minus r prime is minus r minus r prime by r minus r prime cube so therefore my electric field my electric field which by Coulomb's law is written like this because I have charged density and of course this is my inverse square law and this gives me the distance. So I use the fact the r minus r prime by r minus r prime cube is minus the gradient of this so I write this as this quantity. Now you remember the definition of a potential we said vector E is equal to minus grad phi there is technically an extra minus sign that has come in but let me alert to you on one thing it is not necessary that you define this is physics people always like to define their potential with a minus sign that is we define electric field to be equal to minus the gradient of a potential but mathematics people define the things as gradient of a potential there is no controversy that both of them are right it depends upon a you know I mean a sign on your potential. So therefore I have made a mistake with respect to physics so this minus should go and I have phi of r given by this expression and this is the potential. Morning we had questions which I discussed our friends also talked online the point is why a potential now we have realized there are many advantages of using a potential rather than using an electric field the first and the foremost is that a potential is a scalar quantity. Now if a quantity is scalar you know that when you are using superposition principle you simply add them up a plus b plus c you do not worry about what is the direction of that. Now on the other hand if you wanted to use the electric field electric field being a vector use of superposition principle will get you into much bigger trouble. So supposing I have many sources I want to calculate how much is the electric field using the electric field expression due to each one of them and adding them up vectorially is a very tall order. So what one does is to find out the potential add up the potential because they are numbers like thing I mean they are scalar functions add them up and finally on the final thing just do a del operation. So where do we stand now we have been talking so far only about electrostatic field and these two equations this one and that one both of them we derived that is divergence of the electric field is rho by epsilon 0 I have told you that there is a divergence whenever there is a source in this case the charge source. But electric field is conservative static electric field is conservative I want you to remember this that electric field is conservative is an incorrect statement electrostatic field is conservative because we will see later when electric field arises due to time variation of the magnetic field this statement will not be correct. But at this moment we are only doing electrostatics. So therefore these are the two Maxwell's equations that we deal with ok next question what is this potential I mean is there any relationship between potential and potential energy remember potential energy is something which we have always learnt. So let us try to understand there is a relationship but do not be under the impression that potential is the same thing as potential energy. In fact it is probably a very unfortunate nomenclature calling it by the one could have given it some other name but we cannot change history. So understand the following that supposing you have a charge q at let us say infinite distance or anywhere where you decide that the reference point of the potential is 0 normally for coulomb forces we take the potential to be 0 at infinite distance but we know that that is not really always necessary we will give examples to show that is not necessary in fact there are some problems where you should not be using the potential equal to 0 at infinity only. So wherever the potential is equal to 0 that is your reference point. Now from that suppose there is a charge q at that point and u now u meaning not the coulomb force but u as an external agency you bring that charge and put it at a point b. Now remember the work that you do is negative of the work done by the force because you have to overcome that force. And how much of work do you do? The force is given by q times I suppose I have not written a q there but it does not make any difference let us take unit charge. Move from reference point where the potential is 0 to the point r er dot dr. Now this is very important here electric field being conservative this work that is done does not depend upon the path and that you take. So therefore bring it by the shortest simplest path that is the radial line joining your reference point to wherever you want to put it. If it is not along the radial line it does not make any difference you go transverse come to the radial line in line with the point where you want to bring but on the transverse line the work done is 0 because the electric field is radial whereas you are going transverse to it. So therefore ultimately you simply get this electric field is minus the gradient of the potential dq bar prime and this by fundamental theorem of algebra remember the way gradient is defined is nothing but d phi. So which is gives me minus phi of r. So if this is true this tells me the work done by an external agency in bringing a charge q to p is phi times q now this then is stored as the potential energy of the system. Now the potential therefore is nothing but the potential energy now I am saying I am not saying potential energy off but I am saying the potential energy which is associated with the unit charge. So the two words are used and one has to realize that there is a bit of a confusion sometimes but potential has something to do with potential energy but potential is not the potential energy. So let me give you couple of examples of how to calculate potential I have chosen problems which are slightly different. So let us look at suppose I have a line charge line charge is basically an infinite charge with the linear charge density now remember one thing a line charge contains the infinite amount of charge but by definition the line charge has a finite density. So normally a line charge density will be represented by us with lambda. So let us take a cylinder around it having a radius r and let us say that the length of the cylinder is l. Now notice one thing that since the line charge is taken to be infinite then there is no edge effect when I do that that is because both the ends are equal distance from the ends if it is there. So the question is that I have now established a symmetry by Gauss's law because that then on the surface of this cylinder my direction of the electric field by symmetry has to be radial. So therefore my e dot d s which is the flux. So it is nothing but the electric field strength times the area which is of this cylinder and how much is the area of the cylinder 2 pi r l because to I have said length is l radius is the distance from the this thing is r. So it is e times 2 pi r l which is the by Gauss's law e dot d s is nothing but the charge enclosed how much is the charge enclosed the length is l. Since lambda is the charge density lambda times l is the charge enclosed. So this quantity is lambda times l by epsilon 0 cancel out things and get that electric field magnitude is given by lambda by 2 epsilon 0 whatever r the direction has been missing but by symmetry the direction is outward. Now notice this I was not particularly interested but since in calculating this but since this is an illustration of Gauss's law I let it go there. So the electric field magnitude goes as 1 over r. Now if that is 1 over r what is my phi? Now my phi then is remember that the if I have to integrate electric field that is what is that function whose derivative gives me 1 over r very clearly logarithm of r. So it is phi I need a minus because my del minus del phi is electric field. So therefore it is minus lambda by 2 epsilon 0 logarithm of r and like we have done an integration. So therefore there is a constant of integration there. Now notice here this is the type of example I was telling you that in this case you cannot say that let us choose the reference point that is phi equal to 0 at infinity. You cannot do it because if you do that there is a divergence and logarithm of infinity that is not possible. You can say that phi is equal to 0 at 0 because logarithm of 0 also does not make sense. So you can choose the reference point wherever you want but a very convenient point to choose is to eliminate this constant. Now you then you do one thing you choose r is equal to 1 supposing you say at r is equal to 1 my phi is equal to 0. Now you can see immediately that then this constant because log of 1 is 0. So I will get 0 is equal to 0 plus constant which means that constant is 0. So therefore the potential corresponding to line charge is minus lambda by 2 epsilon 0 times logarithm of r. So this problem illustrates an important point that the reference point for a potential is something which one can choose the way we want it. The next thing that I want to do is to calculate or talk about the electrostatic potential of an electric dipole. But before that I need to define what is the dipole vector this is this is a point which a lot of students make a mistake. So it would be a good idea that when you are teaching it please point it out. See basically an electric dipole is nothing but two small charges q and minus q separated by a short distance. The magnitude of the electric dipole moment if you like is q times that distance. But by convention and it is very important to realize that the direction of the dipole moment vector is from negative charge to the positive charge. This is the direction of the unit vector. So my dipole moment vector p is q d times unit vector p and what is the unit vector? This is a vector along the direction joining minus q to plus q. Let me come back to my screen my paper and try to find out an expression for the potential of in the situation that I have given you. So my picture is something like this that supposing this is my minus q and this is plus q though remember that my actual distances must be much smaller. But I am illustrating it so I am drawing a little bigger. Now I am interested in finding out what is the potential at an arbitrary point p. And suppose this point p from the positive charge is at a vector distance r plus and the from a negative charge it is at r minus capital R minus. Now where is how do I define this point p? We take the center point of the dipole and say that this vector is at a negative charge this is the vector r. That is I am interested in finding out what is the potential due to the dipole, due to dipole at position vector r. I need to remind you of a bit of arithmetic or trigonometry if you like this angle is theta. So you remember your triangle law. So that triangle law told you that r plus this distance is given by this vector distance was d. So therefore this is d by 2 and this is also d by 2. So this is equal to r square plus d by 2 whole square which is d square by 4 minus 2 times r times d by 2 times cos theta. And since it is r plus I need to put a trigonometry. So and this quantity then would be notice that the distance d square by 4 is a very small quantity. So what I do is this that I neglect this pull out a r there so I get 1 plus d by r cos theta raised to the power half. And that then gives me r into 1 plus d by 2 r cos theta dynamical expansion and correspondingly I have an expression for r minus which will be very similar it will be r into 1 minus d by 2 r into cos theta. So therefore if I want to write down the potential at p now remember from here from the positive charge it is at r plus from the negative charge it is at r minus so I get 1 over 4 pi epsilon 0 q by r plus minus q by r minus. So plug these 2 in this is actually approximately equal to because I have done a dynamical expansion. So you put them back and well you notice that your r is there in both of them so they will cancel out because of the minus sign and you will be left with 1 over 4 pi epsilon 0 times well there is a 1 over r there and I need that so therefore I get q d cos theta divided by 4 pi epsilon 0 r square. And we have just now said that q times d that is your dipole moment so therefore the expression for the potential becomes 1 over 4 pi epsilon 0 times p dot there is a cos theta there so p dot unit vector r divided by r cube. So this is the expression this is the expression for your the potential due to a dipole. Now why am I talking about potential due to a dipole and that I will be coming to in a short time. Now notice that because the potential was 1 over r square type potential the remember Coulomb forces potential is 1 over r and this is 1 over r square. So the corresponding electric field due to this dipole is goes as 1 over r cube and that is what we have done there just do a gradient of that potential you would get this thing. And you know this is the way one actually plots you could go to a mathematical or things like that and try to find out how the electric field varies here is a negative charge here is a positive charge and you feed in the expression for the electric field and you can get beautiful pictures which you must have seen there. Having done that let me now come back to another important thing I am afraid that this situation will be a little mathematical but let us see what is the best we can do. Now this is Poisson's and Laplace equation I want remember we talked about del dot of E equal to rho by epsilon 0. We also said that the electric field is negative gradient of the potential. So therefore del dot of del phi or rather minus del phi that is equal to rho by epsilon 0 which means del square phi is equal to minus rho by epsilon 0. This is an important equation this equation is known as the Poisson's equation. Now suppose I have a region of space where there is no source then my rho is equal to 0. So in that region I satisfy del square phi is equal to 0. Why what does it mean this is the equation which we will be talking about today this is called Laplace's equation and this is called Poisson's equation. In fact most of the time you would be interested in finding out the potential in a source free space that is the source could remain somewhere but I am interested in finding out what is the potential at a point. In that region where there is no source the field has been created by something. So this is called Laplace's equation. This equation is an extremely interesting equation both from mathematics point of view and from physics point of view. But before I do that I want to talk about one small thing and that is what is known as electrostatic boundary condition. See boundary conditions are things which are usually neglected in a class but on the other hand they are extremely important. Take for example I have given you an example of an infinite sheet of charge. Now you have seen that if I have a charge density sigma then the electric field above the sheet is oppositely directed to that below the sheet. Now this has very interesting consequences look at it. The question we are asking is what is the difference between the electric field immediately above a charge surface and a point immediately below a charge surface. Now look at this from Gauss's law. So we said that consider a Gaussian surface and the Gaussian surface is an imaginary surface which I have taken to be a rectangular parallelepiped. Rectangular parallelepiped of height epsilon half of is above the field above the surface half of is below the surface and this epsilon will be taken to go to 0. Now and you can take some length l some width b does not matter. So I know that e dot d s is q n closed by epsilon 0 which is how much is q n closed the q n closed is nothing but whatever is the area intercepted by this imaginary Gaussian surface supposing that area is a. Then if charge density in sigma the amount of charge that is enclosed in this Gaussian surface is sigma a by epsilon 0. But look at it now that but how do I calculate this? Notice one thing that as this epsilon go to 0 then the contribution to the electric flux from those which have share this height because that area goes to 0. So therefore the flux goes to 0. So what I am left with are these two surfaces only the top surface and the bottom surface which we have already said the area is a. But then on the top surface my electric field is e perpendicular above it is normal normally directed above. So it is the surface integral will be e perpendicular above multiplied by a and the other one is e perpendicular below multiplied there is a double above written this should be below multiplied by a. And there is a minus sign because we know that oppositely directed where the direction of the normal is opposite. So e perpendicular above the normal component of the electric field above minus the normal component of the electric field below cancelling out a is sigma by epsilon 0. Now I want you to understand this is telling me that if the I look at a charged surface. If I look at a charged surface then the normal component of the electric field has a discontinuity the normal component of the electric field above minus electric field below is given by sigma by epsilon 0 the or is the other words sigma is epsilon 0 times this. Now supposing this surface that I have given you where actually a conductor. Now if it were a conductor then of course inside the conductor the electric field is not there. Now so therefore e perpendicular below is 0 and we only only have one of them. So therefore the I only have electric field above and this term is not there because inside the conductor the electric field is 0 because a conductor material of the conductor cannot tolerate an electric field. So e perpendicular in case of a conductor is simply equal to sigma by epsilon 0 this is something which one uses at a later stage. The other problem is now that I have said that the normal component of the electric field has a discontinuity provided I meet a charged surface. If there is a charged surface my normal component has a discontinuity. What about a tangential component? Now in order to find out the tangential component you do the following. You take a loop now this time not a surface but a loop of height epsilon again and run half of it is above and half of it is below and find out the contour integral of the electric field around. Now notice that the because my integral e dot dl I am computing how much is the integral e dot dl because as my epsilon goes to 0 this nothing is enclosed inside this. So this must be equal to 0 it tells me parallel component above is equal to the parallel component below. So summarizing this it means the normal component of electric field has a discontinuity provided there is a charged surface. If it meets a surface there is a discontinuity tangential component of the electric field is however continuous. Now this is these two boundary conditions should be understood by people return back to Poisson's Laplace's equation. So we have said that I am going to be this is del dot of e equal to rho by epsilon 0 and so if I take realize that e is minus del phi this gives me del square phi is minus rho by epsilon 0 and del square phi is 0 this Laplace's equation. Now in addition I know that electric field is a conservative field so therefore it is an irrotational field del cross of e is equal to 0. So these are my workhorses. Now I will go through this section a little fast because a lot of you have pointed out to me that do not waste time on derivations these are there in the books but I need some information out of there. So I will not derive taking time but I will point out what the derivations are and then you can of course look up the notes which will be put up. So look at I am interested in solving Laplace's equation Poisson's equation difficult I will not do that. Incidentally solutions of Laplace's equations are known as harmonic functions. So let us look at supposing I am in three dimension the Cartesian del square is simply d square by dx square d square by dy square d square by dz square I sometimes work in spherical coordinate this is the expression is spherical this is the expression is cylindrical as I promised you I am not going to waste time on this but let us look at some important points. Remember I derived this equation right I mean we have already proved that del square of phi is minus rho by epsilon 0 this is simply repeating this all right. Now I come to a very interesting thing I am going to give you a few interesting points regarding Laplace's equation. Now this statement I am going to make is the following. Now here I have a region of space some close region of space s represented by a volume v inside this I have 3 I mean I have given 3 you can take any number of surfaces in closing volume I have called them s 1 s 2 s 3 and any number s 1 s 2 s 3 s 4 and the final surface the bigger surface which encloses all of them is s. Now I am saying now that I am interested in finding out what is the solution of the Laplace's equation inside. Now we are saying that look that the suppose I will I will prove the following that I have the following conditions given one is that the potential either the potential is equal to 0 on the surface or the derivative of the potential is given to be 0 on the surface. So this is a boundary condition remember whenever you solve differential equation you need boundary conditions or initial conditions. Now the conditions I have given you is either this that is you have given me the values of phi or the values of the first derivative. Now I am saying that the given these conditions I will show that this solution is unique in other words now this is a very important point usually glossed over the point that I am trying to make is this that if you have a Laplace's equation corresponding to a given boundary condition the first boundary type of boundary condition is called Dirichlet boundary condition which simply says that you are given have given me the values of the potential on the conducting surfaces let us say and the other option is what is known as a Neumann boundary condition which has not given me the values of phi but it has given me the normal derivatives of phi. Now my statement is that the solution is unique now let us see what it means. So remember this is my divergence theorem del dot of a d tau is a dot n d s. Now so what I am now going to do is this. So I will first in this expression I am going to be using what is known as the Green's first identity well this is something which you have done del dot of a whatever is a volume integral is that is divergence the volume integral of a divergence is surface integral I am saying that choose a particular a supposing a is phi delta psi substitute here. So what is del dot the del dot this is a chain rule differentiation so I get del phi dot del psi and phi times del square psi this is what I have written down there. So that is my divergence dq bar that is equal to integral a dot n ds now a is given to be this. So therefore, phi times d psi by dn ds now I am now claiming that let me take psi is equal to phi equal to some capital phi if you do that this becomes phi del square phi plus del phi dot del phi equal to this integral phi d phi by dn now remember I have told you that I have given you on the surface on the surface either the value of phi or the value of d phi by dn. So therefore, this integral which is over a contour is equal to 0. So this tells me that the left hand side is 0 whichever is the boundary condition this is 0. So left hand side is 0 which means phi del square phi plus this is equal to 0 and this tells me that the notice this is equal to del phi absolute square this is the term because del square phi is equal to 0 by Laplace's equation this is equal to 0 which tells me that del phi is equal to 0 which tells me since capital phi is phi 1 minus phi 2. So it tells me that the solution is unique what is the importance of this I have actually glossed over this derivation because I did not want to waste a lot of time now but the details of the solution would be there on this what have I actually said I have said the following I have said that given a boundary condition given a boundary condition either by specifying the potential or by specifying the electric field on the boundaries this solution of the Laplace's equation is unique they there cannot be two solutions. Now let us take couple of examples then I will tell you what is the importance of a uniqueness theorem because uniqueness theorem tells me that a Laplace's equation cannot have different solutions but first let us look at these two parallel plate capacitors the lower one is grounded it has negative charge the upper one is at some distance d now I am asking what is the field in this region now you all have done this calculation in your schools but let us look at how does Laplace's equation give me. So notice in this region there are no sources so my Laplace's equation is valid now I take this capacitor plates to be infinite so therefore the only variation of phi can be there in the z direction so del square phi is nothing but d square phi by dz square which is equal to 0 now this is a trivial second order differential equation whose solution is a z plus b a is a constant b is a constant this is grounded so z equal to 0 the potential is equal to 0 put it put this condition you get b is equal to 0 at z is equal to d the potential is given to be phi 0 so a becomes phi 0 by d so put it back you get the potential to be phi 0 by dz the electric field we have already said is minus d phi by dz which is equal to minus phi 0 by d k now notice this electric field is as we expect constant in this region there is a parallel plate capacitor. So therefore the what is the charge density that I have the charge density which we have just now said is epsilon 0 times the normal component of the electric field and normal component of the electric field is already this thing here so this is simply given by minus epsilon 0 because the direction of the normal is this way here that way there so minus epsilon 0 phi 0 by d on the lower plate plus epsilon 0 phi 0 by d on the upper plate. But what is what general statements I can make about the Laplace equation now you will see I am trying to make a statement that Laplace equation does not have any feature it is not an interesting equation you will say what is this if it is not an interesting equation why are you spending so much of time there is a reason for it. So first let us look at the one dimensional equation remember what is one dimensional equation d square phi by dx square is equal to 0. So solution is mx plus this is nothing suppose I have a boundary condition which says that it is equal to 0 at x equal to 1 and it is equal to 3 at x equal to 2 I can write down phi of x equal to 3x minus 3 what does it mean it tells me that the phi x because it is the linear graph at any point is given by the average of the values at the two points where you have been given this again no great feature let us come to two dimension. Now two dimensional Poisson's equation has interesting feature this is actually a function which I have taken to be said suppose I am trying to solve dou square phi by dou x square plus dou square phi by dou y square equal to a. Now the a typical function which satisfies this equation is a by 4 x square plus y square you can see that this equation is satisfied this is a mathematical plot or genu plot of the function phi which is a by 4 x square plus y square and the height is z of course if you plot it you find that this picture it looks like a cup and it looks like a cup with a minimum at the bottom. So it has a minimum at x equal to y equal to 0 the function rises in all other direction. So this is an interesting equation that is something has a minimum. Now let us look at the corresponding situation in the two dimension of a Laplace's equation. Now I want to in this case solve dou square phi by dou x square plus dou square phi by dou y square equal to 0 not a a typical example is for instance a by 4 x square minus y square. Now this function is very interesting if you plot this you find it is picture is a saddle. Saddle if you remember is the thing that you put on a horse's back. Now what happens there is in one direction it rises in the length direction it rises because the man is the jockey is going to sit down on it. On the other direction it has to fall down along the back of the horse. So therefore the this is what is called as saddle and it rises like this but at that point in the another direction it falls. This is the feature that is common in all dimensions. In other words even this does not have a minimum. This is a saddle point it is there only in a particular direction it is a maximum in one direction minimum in another direction. So once again without much of a feature. Now the fact that the potential cannot have a minimum or a maximum if it is satisfying Laplace's equation. This I came I gave you from general plotting and giving you some examples these can be rigorously proved but I am not going to even attempt it. The there is an interesting theorem which goes by the name of arms of theorem. It says that a system of charge cannot be held in static equilibrium by electrostatic forces alone. Now this is very interesting theorem. It says you cannot put a system of charges in equilibrium purely by static equilibrium purely by electrostatic force. Now and the reason is obvious that since in that region there are no other charges. The charge must be the point where the charge is there should be satisfying the Laplace's equation. Now if it satisfies the Laplace's equation the potential cannot have any minimum. Now the potential cannot have a minimum I cannot have a stable equilibrium. So potential energy also has no minimum. What I will do is this that I want to there are many interesting points about solving Laplace's equation is spherical coordinates. If time permits we will do that in our tutorial session but let me go over before that for the remaining time that I have for to discuss another method which is known as the method of images which you might find into. So what I want to do now is the following. I will tell you something about the method of images. Now why do I want to talk about images? Let me consider an infinite grounded conducting plane. Remember grounding means the potential is 0. Now in front of that I have a charge q which is located. The question is what is the potential at any arbitrary point p due to this combination of charge q and the infinite grounded sheet that we have. Now you will say that what is this problem? Say the point is this that the potential of this thing has to be kept because it is grounded it has to be kept at 0. But on the other hand because of this charge the there would be certain amount of electric field here and I want to keep the potential at 0 potential I mean this surface at 0 potential. So therefore the presence of this charge this you have learnt from school that the presence of this charge on this metal surface will induce charges. The question is what are the charges that is induced? How do I solve this problem? Normally if you try to solve this problem by application of Coulomb's law and things like that it is almost a an impossible problem to work out. But I have just now told you something very interesting. We have said there is something called a uniqueness theorem. And the uniqueness theorem told us that if there is a region of space for example this point p is a point where there are no charges in this region I can. Now then at the point p a Laplace equation must be satisfied. So Laplace equation simply says phi is equal to 0 I mean sorry del square phi is equal to 0. So how does one solve this problem? Now remember what is the what is the reason for doing the method of images? The method of images told me that suppose you have been able to guess a solution. Now if you have guessed a solution uniqueness theorem tells you that is the only solution that is possible. Remember what I am doing? I am saying suppose by hook or by crook you have managed to say this solution is good enough. What should be the characteristic of that solution? Firstly it must satisfy Laplace equation. Secondly it must keep the boundary condition given namely that on the surface the potential is equal to 0. How do I manage it? In order to manage it I do what is called a method of images. I will guess a solution and uniqueness theorem will tell me that if I have been able to guess a solution the solution must be correct and must be the only solution. So what I do is I go back to my knowledge of optics and I say that look supposing when I stand in front of a mirror then I find an image of mine on a mirror at a distance which is equal to the object distance plane mirror. So I if I am in front of a mirror at a distance d the image which is virtual is it is virtual because you cannot get it on a screen at a distance d behind the mirror. Now what I have done here is something this idea is what I have done. I said alright you have a distance d from the top imagine that beneath now remember this is virtual. So I am not really interested in doing anything and it is a infinite sheet. So there is nothing like a below just as there is nothing like the other side of a mirror. So these are virtual world. So I imagine there is a charge q prime located at a distance d prime. What is q prime what is d prime I do not know but the condition is that the potential at p my original problem that is this charge plus this conductor is equivalent to this charge and a charge q prime at a distance d prime in below the conductor in such a way that the potential at p satisfies Laplace's equation and the boundary condition on the surface. If I can do that then the solution that I get must be the only solution and that is the powerful technique. Now so let us look at it I have said forget about this surface imagine you have a charge q and you have a charge q prime at a distance this is I take this as the origin I take the plane the infinite plane as the x y plane and this as the z axis this as the minus z axis. So I said this distance is r 1 this distance is r 2 original problem of charge plus the surface has been replaced by a charge plus another virtual charge and due to these two if I can satisfy my condition that is Laplace's equation is satisfied and the boundary condition is valid then I will get a solution. So look from q now notice I am only interested in obtaining the solution above the surface because below the surface does not mean anything. So do not put the negative values on your solution. So potential due to q is q by 4 pi epsilon 0 r 1. Now notice I can since my surface is the x y plane I can take go over to so what is my distance r 1 my distance r 1 is so I am calculating the distance of this point now this point is at this is at d. So therefore if the coordinate of this is x y z this coordinate of this point is 0 0 0 0 0. So the distance is x square plus y square plus z minus d whole square. So the potential due to this charge q is q by 4 pi epsilon 0 square root x square plus y square plus z minus d whole square potential due to this virtual charge is phi 2 which is q prime by 4 pi epsilon 0 r 2 and that is given by q prime by square root of x square plus y square plus z plus d whole square because this is extra distance here. So the my total potential is simply the sum of these two now what do I want now I still do not know what is q sorry what is q prime what is d. So what I say is that look firstly my net potential is this plus this but I know phi 1 satisfies Laplace's equation phi 2 satisfies Laplace's equation. So phi 1 plus phi 2 of course satisfies Laplace's equation. Now then I do the following I say all right I want the potential to be 0 at the ground. So I want this term if I come to x y z equal to some x y any point here which is at the x y 0 take z is equal to 0. Now then the net potential calculated by this formula should give me 0 which means that by putting that I get q square plus x square plus y square plus d prime square because I put there z is equal to 0 equal to q prime square plus x square plus y square plus d square. You rewrite it and you find that the solution is q prime equal to minus q at d prime equal to d. Now if you take q prime equal to minus q that is meant by what is meant by that is I have a negative charge equal to magnitude to this positive charge situated at a distance d prime equal to d below the curve. If I do that then my Laplace's equation is of course satisfied but my potential on the surface also turns out to be 0. So I have now because of the uniqueness theorem been able to obtain an expression for the potential for this problem. Now once I get an expression for the potential this is mathematics so I am not doing it here but it might be there on your. So this plus this is my expression replace q prime by minus q and d prime by d and you get this plus this there is a minus sign here there is a plus sign there this plus this is the expression for the potential. Now once you have got the potential you can calculate the electric field and you can calculate how much is the normal component of the electric field. Now you notice something very interesting that if you now take the normal component of the electric field to calculate how much is the charge density that is induced on that surface. Now realize that the charge density is negative because they of the fact that you know there is a positive charge there the charge density is negative. Now this is the plot of what the charge density is like I mean I have not given you the expression because we are really running short of time on electrostatics. What do you find is the charge density on the plane is maximum or because it is a negative thing does not matter that is why a dip is being shown at a point at points directly opposite the real charge and decreases as the distance on I mean on the surface from the direct point of the real charge increases. So this is the way the field lines will be looking like see I have just about 5 minutes the I need to still talk about you know dielectrics which I will try to do it tomorrow method of images other examples will be there in tutorial. If there are quick questions clarifications related to what we talk about in this session I will make it now otherwise please send your questions which we will talk about which is this college Don Bosco. Yes Don Bosco go ahead. Sir can you please go to page number 24 the question the clarification that I would like is in this case you have applied the Gauss theorem the Gauss law in this particular problem. So as per the on the left hand side there is an integral of E dot ds and over here left hand side the complete integral of E dot ds is there. So we have one surface above and one surface below because only these two surfaces contribute to the total flux. Now my question is we above it is pointing upwards and ds on the upper surfaces also pointing upwards. I know that but I have pointed out that there is a printing error there twice above is written. While I was talking you are right but I am saying that I have pointed it out that there is an error in that slide. Sorry about that obviously that equation does not make sense it is 0 on the left hand side I had realized it but I also made a statement to that effect when I was talking about yes go ahead next. Thank you yes this is petroleum technology is that. Sir this is regarding a problem by applying Gauss law. Yeah. If I want to calculate electric field at a point. Yeah. Inside a charged non conducting sphere without using Gauss law can I do that. What do you want you want to calculate electric field inside. I want to calculate the electric field. No no just. Inside a uniformly charged non conducting sphere. Yes. Without using Gauss law. Of course you can do it. Because if I apply Gauss law. Yeah. Then for a point inside this charged sphere. Yeah. Which we generally take on a Gaussian surface. Yeah. And the charge enclosed by Gaussian surface will only contribute to the electric field at that point. Yeah. Right. But why the charge which is not enclosed by the Gaussian surface. Right. Does not contribute to electric. But this is this is precisely what I tried to explain in the morning when I was talking about that the charges which are outside if you take a Gaussian surface a typical charge which I when I was talking to you about the solid angle problem. I said that the flux lines if you are putting lines from the outside point to the Gaussian surface it cuts it in two places. It cuts it in two places because it is an external point. Now if it cuts it in two places then irrespective of what is the surface the amount of the solid angle is the same. And one of them is negative the other one is positive. So therefore it cancels out is because of that if you remember right in the morning I talked about that that in fact that question came up yesterday that if there is a charge outside take any surface and you try to any closed surface you try to draw lines from there which goes in at one place comes out at the other place. So there is a flux which is going out there is a one direction of the normal direction and when it is entering it is another direction flux is entering flux is going out. So therefore there is no contribution to the surface integral for charges which are outside. The best way to look at is through that and doing it by Coulomb law is possible but is difficult. But am I supposed to get a different result if I do not use Gauss law? No, no, no. If you do not have to say Gauss's law makes this type of problem with symmetry very easy but Coulomb's law is always valid. So is Gauss's law but if there is no symmetry you will not be able to use it properly. Now but however if you want to use Coulomb's law now what you have to do is take all charge distribution including the charges which are outside then find out what is the strength of the electric field then add it out. In this particular case because of this spherical symmetry doing those integrations would not be too difficult but in general if the symmetry is not there using a Coulomb's law poses a very big challenge. Thank you. Thank you. So we will break now for T and you please assemble for the next lecture and please send your questions on chat. I will as today this morning I did I will take up all your questions as long as it is relevant to whatever I am teaching tomorrow morning the first thing. Thank you.