 We continue our discussion of solutions of Laplace equations and last time we had shown that the Laplace equation in spherical polar can be written by this expression which you see on your screen namely 1 over r square d by d r of r square d phi by d r plus 1 over r square sin theta d over d theta of sin theta d phi by d theta plus 1 over r square sin square theta d square phi over d phi square equal to 0. Now, what we did is to show that this equation can be solved by separation of variable technique and the general solution to this equation is given in terms of what are known as spherical harmonics. The radial part of the solution is has two parts one part which goes as r to the power l and the second part which goes as 1 over r to the power l plus 1 where l of course, goes from 0 to infinity this is a general solution. What we tried to do last time is to specialize to the case of where there is azimuthal symmetry. Azimuthal symmetry simply implies that the solutions do not have phi dependence azimuthal angle dependence. In such a case the potential phi which only depends upon the radial distance r and the polar angle theta is given by sum over l is equal to 0 to infinity. You have this a l r to the power l plus b l over r to the power l plus 1 times Legendre polynomial p l cos theta and we had also seen that p l cos theta are polynomials in cos theta of degree l with p 0 being defined to be equal to 1. So, let us return back to a special application of this. We have a conducting sphere which has been put in an electric field which is uniform and the electric field is directed along the z direction. Therefore, very far from the sphere e r theta when the distance r goes to infinity is a uniform field in the z direction. So, e 0 times the unit vector k. Now, what we are interested in knowing is that because you have now put in a conductor in this electric field what is the corresponding potential outside the sphere at both distances which are close to the sphere and far from it. So, we start with a few observation. The first observation what is what we just now talked about that is the electric field at far distances from the sphere is not expected to be very much different from the uniform field with which I started with. So, as a result e r theta as r goes to infinity is e 0 times k. Correspondingly if you look at the potential at large distances can be written as minus e 0 r cos theta which is nothing but z because the electric field is along the z direction. So, the potential must be minus e 0 times z plus of course, a constant which we write as c. The there are few physical observations which you can make. Since the conductor has to be an equipotential the lines of forces have to strike the conductor normally. So, this is what it is this is the direction of the electric field. So, close to the sphere the lines of forces must strike it normally and of course, exit from the other end and of course, the electric field cannot penetrate the sphere. So, therefore, there are some bending effect as you close look at close distances. Now, this edge or this side of the sphere obviously, picks up a positive charge induced charge positive charges will be there and the rear end will pick up negative charges. Now, this is required because we have put the sphere in an electric field and we need to make the electric field inside the sphere 0. So, as a result there must be a an internal electric field which will be responsible for cancelling the field e 0 that we have started with. So, let us look at a now we need to now try to find out what these constants are. So, let us start with a condition that on the surface of the sphere that is at r is equal to r the potential phi is equal to phi 0. Let us say the second thing that we realize is this that outside the sphere in the empty space there are no sources. So, I need to solve a Laplace equation and the general solution to the Laplace equation is what we have written down. Now, let us look at this thing first thing that we notice is this that this contains this contains terms which go as r to the power l and terms which go as r to the power l plus 1. Now, when l is equal to 0 this second term of the potential is a 1 over r potential. Now, you recall that we are aware that del square of 1 over r is minus 4 pi delta 3 dimensional delta function at the origin. So, this since I do not have a charge at the origin and I have actually the origin surrounded by a sphere. So, I cannot have a term in the potential which goes as 1 over r what this implies is that in this summation l is equal to 0 term will not be there and as a result b 0 must be identically equal to 0. Now, the second thing that we do is to compare this expression that is the asymptotic form of the potential and try to compare it with this at large distances. Now, you notice that this has a term other than a constant it has a term which is proportional to cos theta. Now, we are aware that P l of cos theta is a polynomial of degree l in cos theta. So, if you just have a simple cos theta the corresponding l value must be equal to 1. So, we start with a form which is a 1 r plus b 1 by r to the power 1 plus 1 that is b square b by r square times cos theta and we have said already l is equal to 0 term is not there. Now, so therefore, let us rewrite this equation phi of r theta this is equal to well c l c minus e 0 cos theta is what I have e 0 times cos theta actually e 0 times r cos theta plus sum over l is equal to 1 to infinity b l by r to the power l plus 1 p l cos theta of course, as we have said just now the only term that is should be there in this is p 1 cos theta term. So, if you compare these two expressions and remember that the potential on the surface the potential on the surface that is at small r is equal to capital R must be constant equal to phi 0. So, therefore, I require looking at this expression what I require is e 0 times r cos theta must be equal to b 0 divided by r times cos theta. Now, this is required because at r is equal to capital R the cos theta dependence from these two terms must cancel out. Now, this immediately tells you that b 0 must be equal to e 0 times r square and of course, r to the power l for l is equal to 0 give me a term which is equal to a 0 and that must be identified with the constant c. So, therefore, combining all these things what I get is the phi at r theta is given by phi 0 minus e 0 times 1 minus r cube divided by r cube times r cos theta. So, this is actually there is a e 0 r cos theta term and as expected l is equal to 1 term which gives me 1 over r square term. So, this is my expression for the potential once I have the expression for the potential I can find out what the electric field is by simply taking a gradient. Now, remember that I do not have any azimuthal dependence. So, the gradient is simply r unit vector r d by d r plus unit vector theta 1 over r d by d theta acting on this potential. So, fairly straightforward integrations to be done and you can immediately see that the electric field is given by unit vector r times a term proportional to cos theta and an unit vector theta times a term proportional to sin theta. Now, I know that the charges induced on the surface is equal to the normal component of the electric field at r is equal to r. So, therefore, take this expression and the normal component is obviously the radial direction and put small r is equal to capital R. What you get is the normal component the induced charge which is equal to the normal component of the electric field at r is equal to r is given by 3 times e 0 times cos theta. And as you can see in the northern hemisphere the induced charge is positive and in the southern hemisphere the induced charge is negative. You can actually calculate by integrating over this function over the angles and show that the total induced charge is equal to 0. This is very straightforward integration this sin theta comes from the integration surface integration term and this works out to be 0. . So, notice what we did what we said is that in the presence of the sphere the potential changes or modifies the potential due to a uniform electric field by a term which is equal to e 0 r cube by r square times cos theta 1 over r square term. Now, if you want to look at an equivalent problem you recall that the potential of a dipole which if it is fixed at the origin is given by p cos theta divided by 4 pi epsilon 0 r square. Now, if you compare these two it tells you that the sphere is equivalent to or the sphere can be equivalently replaced by putting a dipole of dipole moment 4 pi epsilon 0 r cube times e 0 at the origin. Now, what happens this was an uncharged sphere now what happens if this sphere had a charge well this is not a very difficult problem because all that we need to do is that at far distances we are aware that if the sphere has a charge it looks like a point charge at the origin. So, with the potential that we have derived in the just a little while back I must add the potential due to a potential point charge located at the origin that would be the potential if the sphere and a charge n is put in an electric field of strength e 0 along these z axis. Let me give you another example in this case I have two conducting heavy spheres I have just pictorially separated them and the northern hemisphere is connected to a potential phi 0 while the southern hemisphere is connected to a potential minus phi 0. Now, what we are interested in is what is the potential inside the sphere. Now, look at this I have got phi 0 the general expression for phi r theta is what I started with last time and what we have is this phi of r theta that is equal to sum over l is equal to 0 to infinity this problem also has azimuthal symmetry. So, therefore, I have got a l r to the power l plus b l by r to the power l plus 1 times the Legendre polynomial p l of cos theta. The boundary conditions on the potential r phi at radius r theta that is equal to phi 0 in the northern hemisphere namely 0 less than theta less than phi by 2 and is equal to minus phi 0 for pi by 2 less than theta pi. Now, notice that inside the potential must be finite and therefore, I cannot have a term which is proportional to either 1 over r or any of its power. So, as a result I do not have actually the b l term at all all b l are 0. Now, how do I handle this problem in other words I need to determine what are various a l's. Now, in order to do that you recall the Legendre polynomials are orthogonal polynomials and the. So, if you take for example, p m mu integral unless m is equal to n it has to be equal to 0 and that is the reason I put this delta chronicle delta m n and the normalization happens to give me 2 by 2 m plus 1. Now, if you use this you can find out a formal expression for what a l is by simply multiplying both sides of this expression which you recall I do not have b l anymore by p m of mu and integrate. Now, if you do that then by orthogonality of the Legendre polynomial this expression on the right hand side gives me delta l m into 2 by 2 m plus 1 and of course, since l must be equal to m I get an a m and r to the power m. So, as a result the coefficient a m that is there is given by 2 m plus 1 by 2 capital R to the power m and an integral from minus 1 to plus 1 of phi with p m mu. Now, this is the integral that I have to evaluate from minus 1 to plus 1 and you remember I know the value of phi in the northern hemisphere and in the southern hemisphere. Now, I could as an illustration let me sort of tell you how does one how does it work. So, a m is given by this expression and from minus 1 to 0 the potential is minus phi 0. So, therefore, phi has been replaced with minus phi 0 from 0 to 1 the potential is just plus phi 0. So, this integral is split into two terms and you can trivially find out what do you get. Now, once you look at reorganize this and convert both the integrals to 0 to 1 that can be trivially done by taking mu going to minus mu in this expression what you find is all even a l's must be equal to 0. The odd a l's survive which is given by this expression here. Now, we can as an illustration calculate a few values of a l for instance if you take a 1 I know that p 1 of cos theta is just cos theta which is mu. So, therefore, this is just doing an integration over mu and this gives me a 1 equal to 3 pi 0 over 2 r. If you take m is equal to 2 is absent now if you take m is equal to 3 you have essentially a cubic 5 mu cube minus 3 mu by 2. So, substitute that and once again you can show that a 3 is minus 7 by 8 phi 0 etcetera. So, we get this as a series of terms with only polynomials or Legendre polynomials of odd degrees being there. Now, that brings us to an end the our discussion on these solutions of Laplace equation in spherical coordinates. Now, what we wish to do now is to look at the solutions of the Laplace equation in cylindrical coordinates. Now, recall the cylindrical coordinate what we have are the distance it is basically you recall that cylindrical coordinate is nothing but a polar coordinate in two dimension r theta which we write as rho theta here and a height which is same as the Cartesian z direction. So, this illustrates that where is my r theta phi. So, I have a point p here and the x y plane is here and this point p from p I drop a perpendicular on to the x y plane and the foot of the perpendicular is what makes an angle theta which is the same as this angle theta and the height of that is z and the distance from the origin is given by rho. Now, in the cylindrical coordinates the Laplacian is written as del square equal to 1 over rho d by d rho of rho d by d rho plus 1 over rho square d square over d theta square and of course, the usual d square over d z square which comes from the z direction. So, if you look at the Laplace equation here I get del square phi is equal to this quantity equal to 0. Now, like the case of the spherical polar I can solve this problem by first talking about a separation of variables. Remember the variables here are rho theta and z. So, I write this as a function r of rho, a function q of polar angle theta and a functional capital Z of the z variable. Now, substitute this and if you substitute this you can write down 1 over r. Remember what we do? We substitute this into this equation then divide the entire equation by r q and z and this gives me this form here and this form is 1 over r d by d rho of rho d by d rho. There is one more 1 over r there, 1 over rho there plus 1 over q 1 by rho square d square q over d theta square plus 1 over z d square z over d z square equal to 0. Our principle of solving this is identical. What you do is you first take the z part to the right hand side and you get minus 1 over z d square z over d z square and since the left hand side is a function of rho and theta only whereas, the right hand side is a function of z for all rho theta and rho theta and z in order that such an equation may be satisfied I must have each one of the terms equal to a constant and let me put that constant to be equal to minus k square. Now, if you look at now the z equation the z equation then tells me that d square z over d z square minus k square z is equal to 0 which tells me that the solutions of the z equation is simply exponential of plus or minus k z. What is k? We have not even said whether k is real or imaginary if it were to be imaginary this would of course be the sine cosine functions. Let us rewrite this now putting this k square there write this equation for the rho theta. Now, by I what I have done is to multiply all over by rho square and this has given me an equation of this type. You notice that what I have done is to also open up this d by d rho of rho d r by d rho and if you do that you get an equation of this type. Now, once again remember that this equation can be split into and a term which depends only on rho and a term which depends only on theta. So, bring this to that form. So, I get this as my left hand side rho square by r d square r by d rho square plus rho by r d r by d rho plus k square rho square equal to this quantity. Once again I demand that each one of these terms must be equal to a constant and this time I put that constant to be equal to some nu square. The equation theta equation which is d square q by d theta square plus nu square q equal to 0 must give me q equal to plus or minus i nu theta. Now, remember that the this I have done intentionally. So, that I get sinusoidal functions or sine cosine functions. The reason is that my problem must have must be single valued as theta changes from 0 to 2 pi I must return back to the same solution. I could not do it if I had taken nu instead of nu square there if I had taken a minus nu square then I will not have trigonometric functions, but I would have had hyperbolic function. But in order that the potential is single valued function for theta. So, that potential value at theta plus 2 pi is the same as its value at theta equal to 0. I must choose these nu's to be real. So, that I get plus or minus i nu theta. So, that is that is my solution for q. Now, once I have substituted that into the r equation my r equation is becomes like this d square r over d rho square plus 1 over rho d r by d rho plus k square minus nu square by rho square equal to 0. Now, this equation I am not going to be actually deriving its solution, but let me first bring it to a more familiar form. Suppose I check take the variable to be x equal x which is equal to k times rho. Now, this equation becomes d square r over d x square plus 1 over x d r by d x plus 1 minus nu square by x square r equal to 0. So, this equation is known as the Bessel equation. So, let me rewrite this equation which is d square r by d x square plus 1 over x d r by d x plus 1 minus nu square by x square times r equal to 0. Recall that the single valuedness of the q requires that my nu must be an integer. Now, this equation is well known in mathematics and you will probably come across it in many other areas as well. This is known as the Bessel equation and the solutions of Bessel's equation are Bessel functions. Now, let me I will not be able to spend time on these detailed nature of the solution, but let me sort of make some remarks on what the solutions look like. Firstly, there are this is a second order differential equation. So, as a result there are two solutions. Now, it turns out that when nu is an integer the solutions are not really independent. The solutions are Bessel's functions. Now, the Bessel function for x this is a general expansion for the Bessel function. It is a power series in x and j nu of x behaves like this. Now, for x much less than 1 that is very close to the origin j nu of x goes as x to the power nu that is other than for nu equal to 0 where x to the power 0 is equal to 1. All others Bessel functions they at origin they become 0. Now, for large distances they are oscillating function cosine functions and as you can see it all these functions oscillate. Now, the second function which is also a the second solution of the Bessel function is what is known as N of nu it is given a name Neumann function also called Bessel function of the same second kind. Now, it turns out that these functions they at the origin they are divergent, but as at long distances again they are sinusoidal functions. One sometimes expresses the solutions in terms of linear combinations of Bessel functions of first kind and second kind or that is Bessel function and Neumann function and these are then called the Henkel functions of first kind and second kind. Other than knowing the nature of the solution I am not really going to be talking about the any detailed properties of those solutions. Now, let us look at the what is the general solution of the cylindrical wave equation. Now, I had shown you that the Bessel functions oscillate. In other words Bessel functions have several 0s just look at this picture. So, you notice that this is take any degree for example, this is the Bessel function of degree 1 you notice it becomes 0 here then 0 here 0 there etcetera. Now, these are 0s of Bessel functions which are very well documented and you can see that if the n is equal to 0 or nu is equal to 0 the Bessel functions roots of the Bessel function are at 2.4, 5.5 to 8.65 etcetera and this table gives you the locations of the 0s of the Bessel functions and for larger values they sort of go like this. Bessel functions also satisfy some orthonormality property orthonormality property that if you take the product of j m with j n and integrate over rho rho d rho then you get 0 unless the degree of this Bessel functions are the same and when they are equal then we get this expression here. Now, any piecewise continuous function which is defined in the range 0 to a such that f of a equal to 0 can be expanded in terms of Bessel functions and like the way we did with the Legendre polynomials I can use this orthonormality property and this form of the function f of r to determine what the a m ns are in terms of an integral over an f r with and the product of a Bessel function. As a result the general solution to the cylindrical wave equation has other than e to the power plus minus i n theta that is the solution for q and e to the power plus minus k m z which is the solution for the z part the other part the the radial part or the radial part is given by linear combination of Bessel function of the first kind and the second kind. Once again in doing applications I will take a rather simple case I will assume that my problem has symmetry its solution has no z dependence. Now, if that happens my del square phi becomes like this. So, let me rewrite this del square phi since there is no more z dependence this is equal to 1 over rho d by d rho of rho d phi by d rho plus 1 over rho square d square phi over d theta square equal to 0. This is rather simple to solve by doing a separation of variable you find out that this is equal to that and as we talked about earlier the solution for q has to be sin cosine function with n being an integer and you substitute it here and this is the equation that you get. Remember that when z was there I had obtained a Bessel equation, but this equation is rather simple because this equation has a power series solution. You can check that if I take a combination of rho to the power n and 1 over rho to the power n this equation is satisfied. The only exception that exists if the value of n is equal to 0 if the value of n is equal to 0 then I get rho square d square r by d rho square plus rho d r by d rho equal to 0 which is a first order differential equation in d r by d rho and it can be very trivially solved and the solution in that case is r 0 of rho is c 0 plus a logarithm of rho. Therefore, the expression for phi is r n of rho which in general is a power series in rho and 1 over rho times q n of theta which is a linear combination of cos theta and cos n theta and sin n theta. Now, to look at a specific example let me take a problem very similar to what we did for the spherical problem. There I had put in a sphere in an electric field which is otherwise uniform that is uniform electric field at large distances. Now, this time what I am going to do is to put a cylinder in a the source cylinder is here perpendicular to the plane of the screen and this is the electric field that I have got. So, I put in a cylinder there and let us look at the modifications that I have. Now, physical picture is very much the same because I have a conductor the lines of forces must arrive at or leave from the conductor surface normally and at large distances I must have the potential corresponding to an electric field. So, let us the principle is very similar. So, once I say now this time I have taken the electric field at large distances along the x direction which means at large distances my potential rho going to infinity is minus E 0 times x which is nothing but minus E 0 times rho times cosine of theta. Now, once again I must cancel this potential on the surface of the sphere I can take the potential on the surface of the cylinder to be a constant which I take I have taken a 0. So, a charge distribution must now arise on the surface of the cylinder which will cancel the potential in which I have put in this charge distribution. So, let us look at how does it go. So, I have the expression for phi function of rho theta and we have said this is n is equal to 1 to infinity of C n rho to the power n plus d n by rho to the power n times a n cos n theta plus b n sin n theta. Now, remember that only if I had a line charge I would have expected a potential with a logarithmic term and since there is no line charge here I have not taken n is equal to 0 term which we had shown that gives me a logarithmic solution. So, that is why this sum goes from 1 to infinity. Now, what is my asymptotic limit? My asymptotic limit is that the potential goes as minus E 0 rho cos theta. Now, if the potential goes as minus E 0 rho cos theta it tells me that that all the b n's must be 0 because I do not have a sin theta anywhere. So, in principle I can have a n cos n theta, but however I do not have any cos 2 theta 3 theta terms. So, as a result the only term that I can have is a 1 cos theta. Now, this tells me that since this is just the cos theta term and so I take n is equal to 1. So, I get c rho a 1 cos theta and for convenience I have said c 1 times a 1 is equal to it must be equal to E 0. So, this is what I get there and similarly I have said c times a times d must be equal to another constant a and so therefore, the potential form is like this. Now, I now need to determine the constant a which is the only unknown quantity there. Now, this is done by realizing that the potential on the surface of the cylinder is equal to 0. So, if I take small r is rho is equal to capital R then I get phi r theta is equal to 0 which tells me that a is equal to E 0 r square. So, phi of rho theta is equal to minus E 0 rho plus this constant E 0 r square by rho times cos theta. So, I have got an expression for the potential and now I need to find out the potential on the electric field that is there. So, let us look at the electric field. The electric field is negative gradient of this potential. The negative gradient gradient is rho d by d rho plus 1 over rho d by d theta along the unit vector theta and this is a fairly straight forward differentiation to be done. Now, when you do this differentiation you find a term which is proportional to cos theta which is E 0 plus E 0 r square by rho square cos theta times rho and a term which is obtained by the differentiation of this cos theta which gives me a sin theta term which is minus E 0 E 0 r square by rho square. Now, notice what is happening once I know the electric field I can find out what is the normal component of the electric field and that will give me the charge distribution. So, notice that the field lines crowd near theta equal to 0 and theta equal to 2 pi and if you take theta is equal to 0 this term of course goes and you will be left with cos theta which is equal to 1 and if you put rho is equal to r that is on the surface of the sphere you notice that the field strength becomes twice E 0. So, that is what we get on the surface of the cylinder the charge density on the surface of the cylinder is given by minus E 0 normal component namely d phi by d rho at rho is equal to r it gives me 2 times epsilon 0 E 0 cos theta. Now, once again I can find out what is the total amount of charge induced on the surface q is the theta integral from 0 to 2 pi of sigma r d theta and the sigma is proportional to cos theta. So, therefore, this is equal to 0. Now, notice one interesting thing that like the case of the sphere I have an potential expression where I have minus E 0 rho cos theta which was already there and this is modified by a term which is proportional to 1 over r and so what we do is this that we are now asking the question what is the equivalent dipole moment that can be put at the origin so that the potential will become 0 on the surface. Now, this is simply done by equating that extra term that we had with the expression for the potential due to an electric dipole on the surface of the cylinder. So, put rho is equal to r and that gives you the equivalent dipole moment is 4 pi epsilon 0 E 0 r q. So, this is the solutions of Laplace's equation in the cylindrical coordinates. There are more formal ways of solving Laplace's equations some of them we will be talking as we go along, but what I would do next time is to tell you that instead of just taking up mathematical ways of solving as we have been doing there are some beautifully intuitive ways of solving these problems. And one of them is what is known as a method of images where you replace the original problem by an equivalent problem which you intuitively get and then using the fact that if I have obtained a solution of the Laplace equation corresponding to a given boundary condition, then the solution that I obtain must be unique this is because of the uniqueness of the solution.