 In the last lecture, we had talked about Gauss's law of electrostatics. Basically, what we said is the amount of flux that comes out of any surface, a closed surface is equal to essentially the amount of charge that is contained within that surface and divided by epsilon 0. This surface which we take need not be a physical surface, but one could think of it as an artificial boundary which a geometrical surface that we have in mind and the Gauss's law is true irrespective of whether the surface is real or is an imaginary surface. As we said that the physical content of Gauss's law is identical to that of Coulomb's law. However, it is lot more difficult to work directly with Coulomb's law whereas, in certain situations particularly those where there are enough symmetry of the problem, one finds application or using Gauss's law is lot more convenient. Today, we will continue our application of Gauss's law to couple of problems and after that we will be discussing the concept of potential. .. So, now let us look at this interesting problem of two charged spheres which are oppositely charged, charge density is uniform in each one of them and we are looking at the field in the region of intersection of the two spheres. Now, we had already seen that when we look at the electric field within a material, within a surface inside a surface, the Gauss's law that we apply that says we have to consider the amount of charge that is contained therein. So, we had also seen that if I have a sphere of radius r containing a charge q, then if the charge density is uniform that is if the charge is uniformly distributed over the surface, the electric field at a point r which is less than the radius capital R of the sphere is actually linear whereas, linear with the distance from the center whereas, if the distance at which we are trying to get an expression for the electric field is outside the sphere then of course, like Coulomb's law it sort of goes as 1 over r square. So, let us look at this problem in which I have two spheres whose centers are at o and o prime and I am looking at the electric field at the point p here. So, what I do is this that I know according to superposition principle the electric fields due to distribution of charges different charge distributions simply add up and the point p is inside this sphere 1 as well as it is inside this sphere 2. So, therefore, according to what we have learned from application of Gauss's law the field at the point p due to this sphere is given by q by 4 pi epsilon 0 1 over r cube the capital R is the radius times o p I told you this is linear and the other one is o prime p. Now, notice that the center of the second one is at o prime and the minus sign is because the charge density is opposite. Now, so therefore, what happens is the following that we if you look at how much is o p minus o p prime o prime p you notice this is just by vector addition and it turns out that this is along the vector o o prime. So, in other words the y component or the perpendicular component of the electric field will become 0 and the x component is q by 4 pi epsilon 0 divide into 1 by r cube times the vector o o prime. So, now notice that the this only depends on the distance between the two centers in other words the field in the intersecting region is uniform it does not matter where inside the intersection you want to evaluate the field. So, this is the method of producing uniform field. Let us take a slightly different example which is an application of the principle of superposition this is a sphere of radius r and inside that there is a sphere of smaller radius which I have called it as a, but this sphere is not concentric. So, there is a cavity there is a cavity of radius a at located at the point o prime and we are interested in finding out how much is the field inside the cavity. Now, look at this problem. So, what we do to begin with is this supposing the charge density of the bigger sphere is rho and I have a cavity. Now, I can make that sphere the bigger sphere into a uniform sphere without cavity by filling up the cavity with a charge density rho, but however remember that cavity does not have any charge. So, what I do is mentally I assume that the cavity is equal to a charge density rho plus rho and a charge density minus rho one over the other. So, that the charge density is 0. Now, so therefore, my original problem with the cavity is now split into two different charge distribution one charge distribution which is this sphere of radius capital R with center at o containing a uniform charge distribution rho and we had seen that in such a case the field at an arbitrary point r will be given by q by 4 pi epsilon 0 vector r by capital R q and you recall the density of charge rho is the total charge q divided by 4 pi by 3 capital R cube. If you put that in the electric field due to a uniformly distributed charge density rho at position vector capital R small r is given by rho by 3 epsilon 0 vector r. So, that is due to the bigger sphere not having a cavity, but containing a charge density rho. Now, I need to superpose this with the field due to a sphere of radius a, but having a charge density minus rho. So, notice this that since I am interested in finding the electric field at p which is at a vector position vector r with respect to o then what I am interested is in the vector o prime p. Now, this o prime p is nothing, but vector r minus vector d where vector d is the displacement vector from the center of the bigger sphere to the center of the smaller sphere. And so therefore, the field due to the smaller sphere e 2 is minus rho which is the charge density there by 3 epsilon 0 vector r minus d. Using the superposition principle the field at the point p is simply the sum of these two and you notice that vector the term proportional to the vector r cancels out and I am left with rho by 3 epsilon 0 vector d which is which only again depends on the distance between the point o and the point o prime the center of the cavity and the field at any point within the cavity then is constant. So, these are two applications of the Gauss's law that we find will be of interest. Let me now come back come to a slightly different issue that is the nature of the electric field. We had seen that the electric field which is a inverse square law, but a central force is essentially a conservative force. If you recall when we discussed vector calculus we had said that the conservative force is characterized by the curl of the vector field becoming 0. Now, one can easily calculate what is the curl of this vector field. So, del cross f now this is the ratio of a vector or multiplication of a vector with a scalar. So, del cross of this quantity is first term is gradient of 1 over r cube cross r vector plus 1 over r cube comes out and del cross r. Now, you can easily calculate firstly the gradient of 1 over r cube is along the vector r. So, therefore, r cross r will give me 0 and a direct calculation will tell you that del cross of the position vector r is also equal to 0. So, del cross of f is equal to 0 which means that the force is conservative. Now, if you recall that if the force is conservative that is if the curl of a vector field is 0 we have also said that it means that the field can be expressed as a gradient of a scalar function. And the conventionally one defines the electric electrostatic field f instead of just the gradient minus the gradient of a scalar field which is the potential field that is called a potential. First we will be talking about what is potential at great length in this lecture, but let us look at first how does one express it for the case of Coulomb's field. . So, basically what I need is remember that in the Coulomb field the force I have vector r by r cube or if I am looking at a field at the point r prime due to a charge placed at the point r I need vector r minus r prime by r minus r prime cube. Now, you notice that this quantity is nothing, but the gradient of 1 over the distance between the charge and the position r prime where I am interested in calculating the electric field. Now, the minus sign comes in because it is gradient of 1 over r minus r prime. Now, if you look at the expression for the electric field for a continuous charge distribution we had seen that this is given by 1 over 4 pi epsilon 0 integral over the volume of rho r prime by r minus r prime cube into d v this is straight forward extension from the Coulomb's law and we have just now seen that this quantity incidentally this should have had a r minus r prime vector here. So, this quantity r minus r prime vector by r minus r prime cube is nothing, but negative gradient of this quantity. So, therefore, I can write this as minus the gradient of this and the gradient is taken with respect to the first variable r. So, therefore, it comes out and so therefore, I define potential phi which is by this expression here that my potential phi is given by 1 over 4 pi epsilon 0 integral of v rho r prime r minus r prime d v. So, that is that is what the expression for the potential electrostatic potential for a continuous charge distribution is fair enough. So, we have now let us look at some consequences there of firstly if I calculate the divergence of the electric field the divergence of the electric field is given by del dot of this quantity here which is del square of this del dot del is del square. Now, what I do is this that the what I require now is to take this del square because this is a derivative with respect to the first variable r inside. So, because the integral is with respect to d v prime. So, I get del square of 1 over r minus r prime and if you recall we are during our module on calculus we had shown that del square of 1 over r is minus 4 pi times delta of r. So, therefore, del square of 1 over r minus r prime is minus 4 pi times delta of r minus r prime where this delta actually it is a delta cube because it is a 3 dimensional delta function is the Dirac delta function. Now, once I have a Dirac delta function inside an integral the integral is easy to do the integral then is simply the value. So, 4 pi 4 pi will cancel out and I will have 1 over epsilon 0 and this only one point contributes namely r prime equal to r. So, I get rho r by epsilon 0. So, del dot of E is rho by epsilon 0 and del cross of E is equal to 0 are the 2 major equations of electrostatics that we have learnt so far. .. Now, let us look at the fact that the electric field can be written as a negative gradient of a potential function. So, now let me start with E dot d r now what is E dot E at r prime dot d r prime. Now, suppose you are bringing in a unit charge now you remember that by definition of electric field if I have a charge q the force that is exerted on the charge is q times the electric field strength at a point. So, therefore, what I do is this that first purely mathematically. So, suppose I integrate it formally from some reference point to the point r E r prime dot d r prime E is written as minus gradient of a scalar function phi dot d r prime. Now, we had seen that this quantity this quantity is nothing but d phi this is by definition of the gradient. So, therefore, this is minus the integral from the reference point to r of d phi which is minus phi at r plus phi at the reference point. But it is at the reference point let me define the value of the potential phi to be equal to 0. So, that it simply becomes minus phi of r. Now, supposing you looked at electric field due to a charge point charge at the origin then I know what is q E of vector r prime which is simply q by 4 pi epsilon 0 r prime square. And in that case this integral can be very trivially done and by comparing the result of this integral with the potential I find that phi of r for a point charge at the origin is given by q by 4 pi epsilon 0 r. Now, so the question is this there are two issues to be discussed. As you realize the name has a lot of similarity with another term with which we are familiar namely potential energy. Now, it is somewhat unfortunate that this quantity which is whose negative gradient is the electric field is called a potential because though there are there is some relationship between the potential and the potential energy these are actually two different things. And but however, this name has stuck and so we need to also we will be talking about what is this relationship and we need to be clear in our mind the difference between the term potential and the potential energy. So, first thing is that suppose I want to bring in a test charge q and let us say my reference point is some point A and I want to bring it to some point P whose position with respect to some arbitrarily chosen origin is vector r. Now, if the electric field is E this test charge q at any point r experiences a force which is q times E the electric field at the point. Now, if the electric field is the electric field exerts a force now if I want to bring the charge from a point let us say r A which is the position vector of the reference point A to the point r I have to do work against this force. And hence the work done by external agency which is me is minus the work done by the charge. So, which is W that is the work done by the external agency is minus f dot d l from r A which is the reference point to the point r. So, which is same as minus q E dot d l from r A to r, but we had just now seen that E is negative gradient of the potential. So, it becomes q integral grad phi dot d l which is integral simply of d phi. So, what you get is that it is q times the potential at the point p minus the potential at the reference point which will take it to be 0. So, it becomes q times the potential at the point p. So, let us look at what are the consequence thereof. So, what if you look at this expression you find that the work done in bringing a charge from a reference point where we define the potential to be 0 is q times the potential at the point to which it is being brought in. So, this amount of work that we have done the external agency has done obviously becomes the potential energy of the system or we can say it is the potential energy associated with the charge q. Now, remember the potential energy of is of the system. So, what it means is there is a source charge it could be a distribution of electric charges which is producing that electric field and this term this work that has been done is the contribution of the to the potential energy of this source charges and this charge. So, in other words this is the potential energy of the system which can be associated with the charge that is being brought in. Now, if I take the charge to be a unit charge that test charge to be unit charge then you notice that the work done is simply the potential at the point p. In other words what it means is that potential can be interpreted as the potential energy associated with a unit charge at a point. So, that is the relationship between the potential and the potential energy. . So, let us look at what we the first equation that we wrote down was divergence of the electric field. There are two equations which we talked about one was curl of the electric field was 0 and divergence of the electric field is rho over epsilon 0. Now, notice we have said the electric field can be written as negative gradient of the potential. So, this is del dot minus grad phi is rho by epsilon 0 which means del square phi. Remember del square is the Laplacian operator is minus of the charge density divided by epsilon 0. This equation is known as the Poisson's equation. We will have occasion to discuss this Poisson's equations in detail. Now, supposing I am looking at the solutions of the Poisson's equation that is I am trying to evaluate the potential at some point and supposing in that region there are no charges. If there are no charges then this equation becomes del square phi equal to 0. This is in the source free region source free region and this equation which we will be discussing in detail later is known as the Laplace's equation. So, Poisson's equation is to determine the potential where there are sources available and Laplace's equation is when the source it is in a source free region. Let me look at a few examples of calculation of the potential. I take a simplest possible example namely that of a line charge an infinite line charge. So, what I do is this that I have a line charge I which I have taken as the positive charge and it is infinite. Now I know how to evaluate the field and this is simply done we had done it earlier by explicit calculation from the Coulomb's law, but what we do is we apply Gauss's law here. So, let me take a cylinder of radius small r and length l and this is the Gaussian surface and I enclose the charge within it. So, notice since the Gaussian Gauss's law says that the flux integral E dot d s is equal to the amount of charge that is enclosed by the surface. This is an imaginary surface in this case. The amount of charge that is enclosed obviously is if lambda is the charge density is lambda times the length l of this cylinder which is symmetrically placed with respect to this long wire line charge. So, the flux E now notice the flux is due to electric field on the surface. So, by symmetry it is very clear that the electric fields must be radially outward from the line charge. So, therefore, the contribution to the flux from the top and the bottom surfaces will be 0 because the normal directions are perpendicular to the direction of the electric field. So, the only thing that we have to consider will be the charges will be the contribution to the flux from the curved surface of the cylinder and the magnitude of the electric field obviously is the same everywhere. So, therefore, it is magnitude E times the curved surface area which is 2 pi r times the length of this cylinder. So, that tells me the magnitude of the electric field is lambda by 2 epsilon 0 1 over r and of course, in the radial direction. Now, this is if you want to write it as negative gradient of the potential. Now, electric field is simply a function of r. So, therefore, gradient boils down to essentially a derivative with respect to r. So, as a result the potential phi is given by minus lambda by 2 pi epsilon 0, longer than up r and of course, a constant of integration. Now, notice unlike that of a point charge where we could take the reference point at infinity that is the point at infinity would be taken as the point where the potential will become 0 because potential expression was 1 over r. I cannot do that for a line charge because the logarithm r equal to 0 the logarithm is not defined. So, one has to take this reference point in case of a line charge little more carefully and if you want this constant to vanish then you can sort of artificially take some unit r is equal to 1 and say that the 0 of the potential is at r is equal to 1. What is your r is equal to 1? You have to sort of course, decide. So, this is the potential due to a line charge. As a second example, I would consider a rather well known case this arises this is called screened coulomb potential. This arises in many cases for example, when you have a charge put in inside a semiconductor medium then what happens is because of the fact that the medium itself is a dielectric we will be discussing this in detail. The interaction between charges in the medium is not bare coulomb, but the charges are screened charges are screened by an exponential factor. And the potential instead of being just 1 over r it is given by q by 4 pi epsilon 0 e to the power minus r by lambda. Lambda is some quantity some length parameter at which the strength of the potential reduces by a factor of 1 over e. So, instead of just 1 over r potential I have a e to the power minus r by lambda by r. This also is sometimes known as yukawa potential because of its importance in some other problems of nuclear physics. So, let us look at how does one calculate how does one calculate the electric field due to such a screened coulomb potential. So, let me work it out. So, electric field the first thing to notice is the potential phi is q by 4 pi epsilon 0 e to the power minus r by lambda divided by r. Now, electric field is negative gradient of that. So, minus grad phi, but remember grad is because this function depends only on r. So, grad is nothing but d by d r and the direction is along the unit vector r. So, this is minus q by 4 pi epsilon 0 the direction r and d by d r of this quantity here namely e to the power minus r by lambda divided by r. So, another trivial differentiation to be done. So, I take d by d r. So, 1 over r gives me minus 1 over r square and exponential differentiated chain rule differentiation. So, straight forward it gives me q by 4 pi epsilon 0 unit vector r 1 over r square I take out as common and I will get as 1 plus r by lambda into e to the power minus r by lambda. So, this is my electric field. Now, let us look at the what is the charge distribution? What is the we are interested in finding out? What is the charge distribution that gives rise to this electric field? And for that we need to calculate the divergence of this del dot of e and equate it with rho by epsilon 0. So, let us look at this little detailed calculation, but fairly straight forward. Once again electric field only has radial component. So, calculating divergence would be easy. So, let us look at first this is q by 4 pi epsilon 0. So, I have got del dot of this quantity in the here which is r by r square there are 3 terms there 1 plus r by lambda into e to the power minus r by lambda. So, we need to do this differentiation. Remember del dot of a vector times a scalar is scalar multiplied by del dot of that vector plus gradient of the scalar dotted with the vector itself. So, therefore, what I get is the following the this is equal to q by 4 pi epsilon 0. So, let me take this as let me take this as my one term. So, which is first term is del dot of r by r square which I could write as vector r by r cube into 1 plus r by lambda e to the power minus r by lambda plus the vector r by r square times gradient of this quantity namely 1 plus r by lambda times e to the power minus r by lambda. So, you note this thing coming back to this here this term here this is what I have written down. Now, the one vector r by r square is nothing but the minus the gradient of 1 over r. So, therefore, this term can be written as minus del square of 1 over r there is no dot there into this term plus whatever we have written down here namely r by r square. So, notice that this is r by r square dot gradient of this and since the gradient is also along the unit vector r. So, I get a 1 here 1 by r square d by d r of this thing and these are straight forward differentiation you can do. So, what do you get out of it the once again you recall that del square of 1 over r del square of 1 over r is minus 4 pi times a delta function. So, if you do this calculation the straight forward differentiation first term which is del square of 1 over r gives you 4 pi delta cube r into 1 plus r by lambda e to the power minus r by lambda and this term simply gives you minus 1 over r lambda square e to the power minus r by lambda. But remember delta function only contributes at r is equal to 0. So, but r times delta cube of r is 0. So, this term actually does not exist. So, this is what I am left with this is what I am left with that there is a delta function at the origin multiplied by this quantity and times this thing. So, you notice this and of course, I put r is equal to 0 here also this will also become 1 and this has to be equated to rho by epsilon 0. So, the charge density rho is given by q times delta cube r that is a delta function at the origin minus q by 4 pi lambda square r e to the power minus r by lambda. The reason for this long exercise is to tell you that one can from the knowledge of electric field get the charge distribution and vice versa. The electric field can be obtained from a knowledge of for example, potential. So, why talk about a potential? Notice that a potential is a scalar quantity. So, therefore, if I am looking at calculating electric field due to a charge distribution if I am directly dealing with vectors I have to do vector superposition vector additions that is of all the sources at a whatever is the electric field they give at a point I have to make vector addition and that is a little more difficult problem. On the other hand if I deal with potential these are scalars. So, they can be just added up nicely and having added them up I can take its negative gradient and that will be then that will then give the electric field at that point. So, dealing with scalars is one of the greater advantages of introducing the concept of a potential. So, let me illustrate now by calculating potential for a couple of cases. So, one very important problem which you will be dealing with lot more detail later is that of a dipole. A dipole is basically two charges q and minus q they are separated by a distance d which for an ideal dipole is infinitesimally small. So, by convention the direction of the dipole dipole is a vector it is from the negative charge to the positive charge. This has to be sort of understood that dipole moment p is the magnitude of the charge q times the distance between them these are two equal magnitude charges one positive one negative. So, q times d along this vector from minus charge to the plus charge which is what I call as unit vector p. Now, let us calculate we are interested in calculating the field and the potential due to such a dipole. So, I will calculate it by the potential method that we talked about. So, let us talk about a point p which let me orient this dipole along the z direction and the point p is at a distance r from the origin making an angle theta with the z axis. So, therefore, this point p is at a distance r plus from the positive charge and r minus from the negative charge. Now, you can just drop perpendicular from this charges to this o p and. So, r plus is actually less than r the. So, r plus is this distance I have given a positive because it is a positive charge I have written plus the way I have given it this is not really greater it is less than r plus is less than the vector r and r minus is greater than the distance o p. So, as a result r plus minus r plus is r minus d by 2 cos theta and r plus is this r minus is this distance which is r plus d by 2 cos theta. Now, what I need is 1 over r plus minus. So, I take the divide 1 by this and. So, therefore, what do I get? So, what I get is phi of r is 1 over 4 pi epsilon 0 q by r minus because it is opposite charge by r minus. So, as a result I can put this approximation there 1 and 1 will cancel out and I will be left with I will be left with q d cos theta q d cos theta divided by 4 pi epsilon 0 r square there is a r there and there is a r there. So, this is equal to p cos theta by 4 pi epsilon 0 r square because p is nothing but q times d, but recall that the angle between p and vector p and r is theta. So, as a result p cos theta is vector p dot r by 4 pi epsilon 0 r square. So, we have obtained an expression for the potential phi due to a dipole at a point r that is given by vector p dotted with r by 4 pi epsilon 0 r square. Now, I can calculate the electric field at the point r because this is nothing but negative gradient of the potential. Now, notice the gradient because the potential depends now on not only r, but it also depends upon the angle theta there is no phi dependence. So, as a result I need to know what does the gradient operator look like in spherical polar coordinates, but so this is the expression for the gradient in the spherical polar coordinate r d by d r plus unit vector theta 1 over r d by d theta. This term is of not much interest to us here because the potential expression does not have any phi dependence, but this is a term which would be important if there is an azimuthal dependence. So, this is 0. So, I am interested in calculating this and this is cos theta. So, this is very fairly straight forward to calculate. Now, notice this. So, this quantity that we are interested let us rewrite it is minus p by 4 pi epsilon 0 gradient of cos theta divided by r square which is equal to minus 4 p by 4 pi epsilon 0. We had seen that the gradient is vector r unit vector r times d by d r the partial with respect to r of 1 over r square. So, 1 over r square gives me minus 2 by r cube times cos theta. Then I have a theta 1 over r and now d by d theta, but d by d theta of cos theta is simply sin theta. So, I got 1 over r square which is already there times sin theta, but with the minus sign because it is a cosine being. So, let me put this minus here. Now, look at this. This gives you the geometry. This gives you the geometry. I have oriented my dipole along the z axis. This is the radial direction. So, this is the direction of p and you recall that the tangential unit vector theta is always defined in the direction of increasing angle. So, as a result since this is the theta angle theta. So, the direction of increasing theta is this which means if I want if I have a vector p the unit vector p, I can resolve it along the radial direction along the radial direction and the tangential direction by saying vector p is vector unit vector r cos theta minus unit vector theta times sin theta. This is an expression which will require. So, let us look at what we have we got. So, notice that there are the minus signs go away. So, I am left with p by 4 pi epsilon 0 unit vector r by r cube. Now, notice this what is cos theta? Cos theta is p dot unit vector. So, let me write this 2 times unit vector p dotted with sorry vector p let me take this away. So, vector p dotted with unit vector r divided by r cube plus because this minus has been already absorbed there 1 over r cube 1 over r cube the p that has to come inside and I have got theta sin theta fair enough. So, this quantity is equal to 1 over 4 pi epsilon 0. So, let us look at what do I get out of this. Remember we said that the unit vector p is r cos theta minus theta sin theta. So, what I do to the second term to this term here is write my theta sin theta as unit vector p or rather vector r cos theta minus unit vector p and if you just add them up properly what you get is 3 times vector p dot unit vector r along the vector r minus the vector p divided by r cube. This is an expression which is independent of your coordinate system because it is all written in terms of the vector. This is a rather important expression for the field of a dipole. If I now try to plot what is the electric field I could take this expression and plot the electric field like we have been doing for any vector field. This is the way it looks like it should not come as a surprise to you because the dipole is basically one positive charge and one negative charge. So, the field lines must leave the positive charge and enter the negative charge. So, this is the way the electric field due to a dipole looks like. . So, let us quickly summarize as to what we have done today. We have introduced the concept of a potential. We have realized that electric force the Coulomb force being a conservative field of force the curl of the electric field is 0 and hence the field itself can be written as a gradient of a quantity which we call as a potential. In definition we take a negative gradient, but that is convention. Now, potential is related to potential energy, but is not the potential energy. Potential is measured in joules per Coulomb which is called a volt. So, essentially potential at a point is the amount of potential energy associated with a charge test charge unit test charge at that point. Potential is something like pressure when a liquid is flowing. So, for example, if you have a pipe through which one end of the pipe has a higher pressure than the other end the liquid will flow from the high pressure region to the low pressure region. So, potential is a very similar concept. If a positive charge is at a higher potential than a negative charge it will have a tendency to go into region of lower potential. Potential is a scalar quantity. Superposition principle can be applied much easier and it is easy to deal with because once I have added up the scalar potentials I can then take the gradient and calculate the electric field.