 We have been talking about solutions of Laplace's equation and Poisson's equations subject to a given boundary condition. One of the things that we learnt is what is known as the uniqueness theorem. Uniqueness theorem implies that when we obtain a solution to Laplace's or Poisson's equation which satisfies the given boundary condition, that solution is one and the only one solution. Now, this has some very interesting consequences. One of the consequences is that if by some means we are able to let us say guess a solution. We have seen there are ways of getting at the solutions analytically. These are not always feasible and they also require certain use of high level mathematics. But supposing we appeal to our intuition and for a given problem, we are able to guess that this could have been the solution. Now, if we can show that that solution that we have guessed satisfies the Laplace's or the Poisson's equations as the case may be and also it satisfies the boundary conditions for the given problem. Then because of the uniqueness theorem, we are guaranteed that the solution that we have guessed is the only solution. Today, what we are going to do is to appeal to this sense of intuition that we have and we will be talking about some a method which goes by the name of method of images. The name as you can see is borrowed from optics and we will see why such a name comes out. So, let me start with an example. Supposing I have a plane a conducting infinite conducting plane I have taken to be in the x y plane. Now, this plane I will take it to be grounded. This is for convenience we could also make it at a given constant potential, but let us say it is at 0 potential. Now, what we have is we have a charge q located at a distance d above this conducting plane. So, that is for convenience I will take this direction as the direction of my z axis. So, I have an x y plane which is at 0 potential and a charge q located at a distance d along the z axis. Now, notice everywhere in space other than at the location of charge q I have Laplace's equation because there are no sources there. So, I need to solve Laplace's equation everywhere in space subject everywhere actually in the half space above the infinite conductor subject to the condition that the potential on the conducting plane is equal to 0. Now, let us see what is the best way in which I can do it, but notice that supposing I am interested in calculating the potential at a distance r 1 from the charge q. I imagine that there is a charge which I will name as the image charge which is located at a distance d prime, but below the conductor. Now, please try to understand what is meant by below the conductor in this context. Just as when we stand in front of a mirror we know what is the distance of let us say the object the us or the person who is standing in front of a mirror from the mirror. Now, that is a real distance from the mirror a real distance. Now, image of this person is also visible to us and if it is a plane mirror we know that the image is located at a distance which is equal to the object distance, but behind the mirror. Now, behind the mirror also implies that the whole process is a fictitious the idea of a of the image is fictitious because you cannot actually put a screen and catch that image there. So, in that sense what I am talking about is a fictitious charge q prime which is located at a distance d prime below the mirror that is along the z axis. And let us look at that as a result of the real charge q which is shown here by a red dot and the image charge q prime which is shown by the green dot what is the potential at the point r 1 from the point q and let us also assume that the distance happens to be r 2 from the point q prime that is from the image charge. Now, it is easy to write down the potential equation we of course, see that the potential at the point p due to the charge q is q by 4 pi epsilon 0 r 1. Well, notice that the plane is the x y plane. So, therefore, the point p's coordinates can be written as x y and so the distance is what I am looking at. So, it is x square plus y square and plus z minus d whole square. So, this is this is the distance from the charge q to the point p. Now, I can write down the expression for the potential phi 2 due to the image charge q prime from which the point p is located at a distance r 2 and that is then given by 1 over 4 pi epsilon 0 q prime divided by square root of x square plus y square plus z plus d prime whole square. So, this is the two potentials I have got. Now, my total potential at the point p is thus given by phi 1 plus phi 2. Now, suppose I now put in the condition now this is a general expression for the point the potential at an arbitrary point p which is x y z and so now suppose I now impose the condition the potential at on the plane must be equal to 0. Then my requirement is just to go back remember the plane is at z equal to 0. So, the potential which is phi 1 plus phi 2 I must just add them up and put z is equal to 0 in those and that gives me q square into x square plus y square plus d prime square is equal to q prime square into x square plus y square plus d square. Now, notice that this equation I need to solve, but one thing I observe that in order that phi 1 plus phi 2 should be equal to 0. Then it is required that the potential phi 1 and the potential phi 2 must be of opposite sign. In other words q and q prime must have opposite sign. So, as a result though I have written here as q square the two denominators I have equated and so therefore, you notice that this gives me q minus q prime whole square into x square plus y square plus q square d prime square minus q prime square d square is equal to 0. Of course, I know q prime must have a sign which is opposite to that of q. Now, this equation should be satisfied at all points x y on the plane. So, therefore, if you refer to this equation again you notice q square must be equal to q prime square because this is I require at all points x square plus y square at all points x and y this equation must be satisfied. So, therefore, these two terms must be separately be equal to 0. Now, which implies q square must be equal to q prime square, but we have already observed that q and q prime must be of opposite sign. So, that tells me q prime should be equal to minus q and you substitute it into this equation then you find that I must have d prime equal to d. Now, notice what it actually means the image charge has the same magnitude, but has an opposite sign to that of the object charge and the image distance that is the distance at which the image charge is located is equal to the object distance, but below the conducting plane. So, now notice the electric field only exists in the upper half plane because there is this infinite plane which is conducting plane which is of course, cleaning out the field below the plane. .. Now, for convenience let me write this total potential phi as equal to q by 4 pi epsilon 0. So, 1 over instead of carrying on that square root of x square plus y square let me denote it by rho square x square plus y square is equal to rho square. So, I have got rho square plus z minus d whole square minus because of the opposite sign and I have got root of rho square plus z plus d whole square. Now, if this is the potential then I know how to calculate the electric field the electric field as we all know is minus the gradient of the potential phi and you must have observed that this is basically a cylindrical geometry the I have the x y plane which is my polar plane the usual r theta plane which rho theta plane in this case and of course, this z is the cylindrical z direction. So, that my gradient is basically given by rho cap d by d rho plus well there is a minus sign in front plus minus k d by d z. So, this of phi. So, therefore, if I look at the electric field I have a component in the plane or component parallel to the plane and a component perpendicular to the plane and these differentiations are extremely easy to work out. So, this is equal to q over 4 pi epsilon 0. So, notice that I have a rho the minus sign will go because I am differentiating 1 over square root and. So, therefore, this gives me rho square plus z minus d whole square to the power 3 by 2 the half of the differentiation will go with the 2 that I will get from rho square differentiation. So, therefore, I get a rho there and minus rho this from the second term and I get from here the same type of thing namely rho square plus z plus d whole square to the power 3 by 2. Now, that is the rho component and then of course, I have a k component. So, the k component will be differentiation with respect to z and which is again very similar. So, I have once again rho square plus z minus d whole square to the power 3 by 2 and this term gives me z minus d and the other term gives me z plus d fairly straight forward differentiation. Now, I know that the charge density the charge density is induced on the plane and the charge density is nothing but epsilon 0 times the normal component of the electric field. Since it is a conductor the tangential component of the electric field must be 0 and the normal component of the electric field is nothing, but the z component. So, therefore, the sigma which is my charge density which is epsilon 0 times E z. So, therefore, I have to simply worry about this component, but notice that the plane is at z is equal to 0. So, as a result both these denominators happen to be the same and I am left with z minus d by rho square plus d square to the power 3 by 2 and here z plus d. So, z goes away and I will be left with epsilon 0 times. So, there is a minus d minus d. So, 2 minus d is come in and I will be left with minus q is already there. I will write down a d and that 2 d and 4 pi epsilon 0 gives me 2 pi times rho square plus d square to the power 3 by 2. So, this is my charge density that is induced on the plane. Now, let us look at what is the total charge that I get. Now, I know that in order to get the total charge that is induced on the plane I must integrate the charge density sigma over the entire plane. So, sigma d x d y if you like, but we have written it in r theta coordinate. So, therefore, this is equal to minus I had q d over 2 pi which is there and the charge density is given by 1 over rho square plus d square to the power 3 by 2 and we know that the angle integration gives me 2 pi because there is no angle dependence of my integral and of course, rho d rho and rho integral is from 0 to infinity. This is a fairly straight forward integral because rho d rho is the differentiation of rho square and what I am left with then is simply q d divided by rho square plus d square to the power half and this limits are from 0 to infinity which simply gives me minus q. So, notice that the total charge induced on the surface is exactly equal to minus the charge that is there in front of the plane. So, that is the amount of charge that is induced. Now, here what I have done is to sort of give you a picture of what does the charge density look like. Now, notice that charge density is negative. So, close to the directly opposite to the point where the object charge is located, the charge density is large and of course, it sort of tapers off as we go to longer distances along the plane. You could physically look at what do the lines of forces look like. Let me assume here of course, I have drawn the conducting plane on the side, but that should not matter. So, here is the charge positive charge with a red dot and the lines are emanating from it shown by this green arrows and I know that on the conductor they must approach normally. So, this is the way the lines of forces are and these circles which are actually spheres they are my equipotential surfaces. So, this is the way the geometry looks like. Now, notice that so what we have done so far is to say that my original problem is replaced by taking a charge of magnitude equal to, but sign opposite to that of the object charge located at a distance d below the conducting plane. Now, let us look at what is the field that is generated at the position of the charge. Now, where does it come from? You see there is an induced charge on the conductor. So, this charge density will be giving a field at the location, but remember we have calculated the electric field everywhere due to the problem. And so therefore, what we are interested in is to look at what is the electric field? The electric field everywhere it has been done near the position of the charge and remember the position of the charge is rho is equal to 0 and the distance is equal to d. So, if you recall the original expression which I give back here. So, I am looking at rho is equal to 0. Now, once you put rho is equal to 0 and z is equal to d. So, I am an electric field which since rho is equal to 0 both these numerators vanish and only the k component will remain and you put here rho is equal to 0. You notice that this term this is we put rho is equal to 0 and also z is equal to 0 both these terms become the same. And I am left with this term here which is q by 4 pi epsilon 0 k and this we have written down minus z plus d divided by rho square plus z plus d whole square. The first term which is z minus d in the numerator having vanished because I must I am having it at the location which means at z is equal to d. So, therefore, only one term remains and this term is as you can see from here I have to put z is equal to d. So, that is a 2 d there and rho is equal to 0. So, I am left with a 4 d square there. So, e 0 d is along minus k direction and q by 4 pi epsilon 0 times 1 over 4 d square. Now, the electric field is this at the location of the charge q then the force that the charge q experiences due to the induced charges on the surface is simply obtained by multiplying the electric field at that point with the charge q. And as a result the electric field the force between the charge the force on the charge due to the induced charges is given by this expression. And this is nothing, but if I replace that conducting surface and I just have a charge and its image which are now separated by a distance 2 d. So, this is nothing, but the force as dictated by Coulomb's law. Let me look at a slightly different thing what is the field on the surface due to the charge q. Now, remember I have calculated the total electric field, but I must only consider half of this electric field. If I am looking at how much is the force that a charge q on the surface will experience because the image charge is actually a fictitious charge. So, therefore, of the field that I have calculated I must look at my conducting plane is at z is equal to 0. So, I must look at half the field that we have calculated. And this is again the E z prime if you look back the expressions that we had once again this is the expression that survives. And I must put here z is equal to 0, but remember rho can be anything now. And as a result the expression that I get is q by 8 pi epsilon 0 minus 2 d by rho square plus d square to the power 3 by 2. So, I have got the electric field on the surface and I know that the surface has a charge density sigma. So, therefore, force exerted on the surface that is f that is nothing, but sigma times the electric field at the surface times d rho that is my actually d square rho because it is integration is over the surface. And this integration is fairly straight forward to do look at this that I already had an electric field which is rho square plus d square to the power 3 by 2. The charge density also has a very similar variation. So, if you plug in charge density there you will find this is equal to q square d square divided by 8 pi square epsilon 0. And rho square plus d square to the power 3 by 2 being the expression for both E z and sigma. So, I have a cube 2 pi is my angle integration and of course, rho d rho and this is from 0 to infinity. Calculate this and you will get this to be equal to q square over 16 pi epsilon 0 d square. So, which is of course, the same expression as what we got earlier that is the force that is exerted by the plane on the charge. Obviously, by Newton's third law these must be equal and opposite, but notice one thing there is a bit of a hand waving came in because it is important to realize that the field that is there on the surface is not the total field due to the charge and its image, but we must take half of that. Now, I can look at the electrostatic energy of the problem. How do I calculate the electrostatic energy? I simply try to bring in supposing initially I just had that plane and then a charge q is to be brought from infinity to the point where it is to be located which is at a distance d from the conducting plane. I know that electric forces are conservative and as a result I can bring it the charge from infinity anyway I like and it is very convenient because I know the force to bring it from infinity along the z axis because just a little while back I have calculated what is the force that is exerted on the charge when it is at a distance d and I can use the same expression and say that when it is at a distance z what is the force that is exerted and that force as we have seen when it was a distance d it was q square by 16 pi epsilon 0 d square. So, when it is at a distance z it is obviously q square by 16 pi epsilon 0 z square and the force is along the minus k direction. So, since the force is minus q square by 16 pi epsilon 0 z square along the negative z direction the total energy the e is not to be confused with the electric field, but the total energy is simply this quantity integrated from infinity to d and that is equal to minus q square by 16 pi epsilon 0 d. So, that is my total energy of the problem. This method of images can also be applied for many other problems. Let me give you an example of when I have got two conducting surfaces making a right angle with each other. So, look at this situation I have one conducting plane semi infinite conducting plane at y is equal to 0 and another semi infinite conducting plane at x equal to 0 and the I have got a charge here which is located. In other words this is in x z plane this is in y z plane z axis is out of the screen and I have a charge plus q which is located here. Now, let us look at what would be the image problem corresponding to this. The as I know that if I had two plane mirrors at right angle to each other I will first of course, have an image of this at a distance along this mirror at a distance this which is minus q and another image on the other screen which is at a distance again if this is at a distance b it is at minus b which is also minus q. But these images for example, this image which is due to this will have its image here and similarly this will have an image here and the whole process will be completed if I have a plus q here minus q there plus q there and a minus q there. Now, this of course I did by hand waving. So, what you do is to write down and as an exercise the potential due to these charge distribution and show that that by simple symmetry if you want the total potential on the two plane surfaces to be equal to 0 then these must be the image charges that is plus 1 plus q at a diagonally opposite corner and a minus q there and a minus q on the other side. Now, let us look at I will not repeat the potential expression they all work out fairly straight forward, but let us look at what is the force that is exerted on this charge due to the charges that are induced on these two planes. The charges that are induced on these two planes we have seen equivalently we can consider the force between this charge and the image charges. Now, notice that this image is between a charge plus q and a minus q located at a distance 2 a along the x axis. So, therefore this gives me a force along the positive x axis this of course does not give me any force along the x axis because it is along the y axis. So, this force is minus q by 4 pi epsilon 0 minus 1 over 4 a square minus because it is along the minus x axis. Now, let us look at these two points these are similar charges the distance between them is 2 times square root of a square plus b square and since this is along the diagonal what I need is to take the x component of it which is a divided by the square root again. So, as a result my net force is q square by 4 pi epsilon 0 1 over minus 1 over 4 pi 4 a square plus because this is between two positive charges a by square root of a square plus b square to the power 3 by 2 and f y similarly is obtained by replacing a with a b by symmetry. So, these are the forces that are exerted. .. Let me take a little more complicated problem and this problem is the problem of again I have the same two conductors intersecting at a right angle one in the x z plane the other in the y z plane, but instead of a single charge what I have is at the location a b I have a line charge line charge of charge density lambda. Now, clearly now remember that this line charge is parallel to z axis it is an infinite line charge. Now, we know the electric field and the potential due to infinite line charge by simple intuition it follows that I must have a line charge object charge plus lambda here a image line charge minus lambda there image line charge minus lambda there and an image line charge plus lambda here. If you combine these four then the potential condition on both these surfaces will be satisfied. We all know how to write down the potential at an arbitrary point x y due to a line charge and that was the lambda by 4 epsilon 0 logarithm of a constant which I have written as c square there divided by x minus a square plus y minus b square. Now, this is the potential this term is the potential due to your original real line charge the I have now let us look at the pictures again I have a line charge here at a distance a that is along the x axis at a location minus a at a location minus b along the y axis and this at a minus a minus b. So, look at that. So, I have got at a b that is my original line charge at minus a minus b which is of the same sign as this because that is the one along the diagonal corner then minus because the line charge has changed from lambda to minus lambda and this is at a minus b and this is at this is at a minus b this is at minus a plus b. Now, you can check that the this expression if you take the tangential component of the electric field remember that the two operators are in the x z plane and the y z plane. So, the tangential component on the y equal to 0 plane is d by d x of the electric field and tangential component on the x equal to 0 plane is the d by d y of their potential. So, these automatically are satisfied if you take y equal to 0 and x equal to 0 respectively. Now, let us look at what happens to the normal component. So, the normal component on the this plane is along the y direction. So, therefore, look at this expression. So, let me calculate what is the normal component namely e z that is equal to minus d phi by d y, but I must take this at y equal to 0. Now, this is this will have two terms. So, I need d phi by d y and at y equal to 0 and this all these logarithms I can I know log a by log b log of a by b is log a minus log b. So, open these up and logarithm of x has a differentiation 1 over x. So, straight forward differentiation substitute y is equal to 0 and you find that this is given by 4 lambda a b x by pi epsilon 0 x minus a whole square plus b square into x plus a whole square plus b square and this quantity is the expression for charge density on the y equal to 0 plane. You do not actually have to work out the electric normal component of the electric field on the x equal to 0 plane. The symmetry is obvious you simply put you know a going to b and then you get the result. So, therefore, I got this and this is the result for the expression and of course, x going to y. So, I get these two what I have done here is to plot the charge density for different locations of the original image the original line charge. These are only x y coordinates .and. So, notice this that for example, this graph is the charge density that I have got and this is only along the this distance is along x. Remember that one of the planes is at y is equal to 0 x is equal to 0. So, obviously, I am looking at the other plane now. So, if I have the x distance equal to 2 and y distance equal to 1 this is just arbitrary location x distance equal to 2 and y distance equal to 1 how much is the charge density that is generated on that plane. So, let us suppose it is given by this expression you can plot it using any standard plotting. . Now, let us reduce the x distance. So, notice as x distance is reduced the charge density spreads out much more and here for example, I am increasing the for a correspondingly I am increasing the y distance and this is of course, become much more shallow and now remember in all cases the total charge that is there would still be the same. But obviously, if I am closer to one of the conducting plane then the charge density directly opposite that point should be much more. So, this is what you see here because here the line charge is at 2 1. So, therefore, the peak is at the point 2 here it is at 1 1. So, the peak is at the point 1. Now, let me just pick up y is equal to 0 plane and calculate how much is the total charge induced on that. We had seen that the all that we need to do is to calculate the electric field the normal component of the electric field on the surface and multiply it with epsilon 0 and that gives me the total charge density. So, the charge density is obtained by this expression on the y is equal to 0 plane. Now, obviously, on the x equal to 0 plane all that we need to do is to replace a with b and x with y. So, I am not repeating that argument. Now, how much is the total charge induced on the y is equal to 0 plane. So, this is simply an integration over x remember these are semi infinite plane. So, it goes from 0 to infinity. Now, these are standard 1 over x square plus a square type of integral giving me tan inverse function. So, this works out to minus 2 lambda by pi tan inverse of a by b by symmetry the total charge on the x equal to 0 plane is given by 2 lambda by pi minus 2 lambda by pi times tan inverse of b by a. Now, what is the total charge remember tan inverse x plus tan inverse 1 over x is equal to pi by 2. So, if you just add up these 2 expressions you get that the total charge that is induced on these 2 surfaces is just equal to minus lambda. So, once again the net charge that is induced on all these plane problems happen to be equal and opposite to they the opposite to equal in magnitude, but opposite to they charge or the charge density that you have put in. What I am going to do is this that now is it restricted to only the planes planar conductor it is much easier the name method of images came from there. However, by suitable adaptation of this technique you can also use the method of image for for instance a spherical conductor. Now, let us look at how this works out. . .So, what you see here is a spherical conductor which is grounded I have a charge q which I have taken at a distance a from the center of the spherical conductor. Now, I do not know where is its image charge, but let me state that let the image charge be located at a distance b let it is the charge b q prime from the center again and let us calculate the potential at the point p due to the charge q and I still call it is its image charge q prime at an arbitrary point. Let the distance of that arbitrary point p be r 1 from q and r 2 from the its image the net potential is phi r theta is equal to 1 over 4 pi epsilon 0 q by r 1 plus q prime by r 2 remember q prime is my image charge I have not quite said how much is q prime I have not even stated where is q prime. However, if you look at the picture again you notice that I can use that triangle law if this angle that is the point p is at a location distance r from o if this angle between the line joining the center and the location of the charge q is theta then normal triangle law tells me that this r 1 is r square plus a square minus 2 a r cosine theta. So, let us write it down 1 over 4 pi epsilon 0 q divided by square root of a square plus r square minus 2 a r cos theta then plus q prime divided by well suppose the image is at a distance b from the center I will write it as b square plus r square minus 2 b r cos theta. So, this is the general expression for the potential now remember on the surface of the sphere that is at r is equal to r my potential must vanish. So, first thing that you observe is these two must have opposite sign otherwise I cannot make the vanish. So, q prime must have a sign opposite to that of q, but let us calculate its magnitude the way to calculate the magnitude is to say this plus this is equal to 0. So, I square them and I find that q square into b square plus r square minus 2 b r cos theta is equal to q prime square into a square plus r square minus 2 a r cos theta. Now this condition is to be satisfied at all values of theta. Now if you want to satisfy it for all values of theta this means these two terms must be equal. So, for any theta this has to be true. So, therefore, I write down that q square by q prime square is equal to q square divided by q prime square is nothing but a by b that is q prime is equal to now I put in a minus sign by hand because I know it has to be opposite square root of b by a into q. Now this is really not a determination because I still do not know a is a distance given to me, but I need to know where is the image located in other words I need to calculate the distance b. Now that is done by equating the remaining two terms that I have. So, therefore, what I require is b square plus r square is equal to q prime square by q square into a square plus r square and we have already seen that q prime square by q square is b by a. So, this times a square plus r square. So, this tells me that a times b must be equal to r square and this r is just the radius of I have to put r is equal to r because it is on the surface of this way. Now notice what I have done I have said q prime is equal to minus square root of b by a q and a b is equal to r square. So, that tells me that the image charge magnitude. So, the image charge magnitude which is q prime which is equal to minus square root of b by a into q and b is root of r square by a. So, that gives me an a square into q. So, which is minus r by a into q and b is since a b is equal to r square b is equal to r square by a. So, this is the net solution that is I have an image charge which is located at a distance equal to r square by a. Now notice a b product is r square. So, therefore, if the object charge is outside the sphere the image must be inside the sphere at a distance b equal to r square by a and the image charge magnitude is simply r by a into q. Now I can continue with this exactly the way we did that is write down an expression. Now that I know exactly what is the image charge location as well as the magnitude I can write down an expression for the potential and then calculate the charge density on the surface of the sphere by calculating how much is the normal component of the electric field. We will take up this problem and solve the complete sphere problem as well as some other application of the image problems next time.