Topic 33: Green's Functions I – Solution to Poisson's Equation with ...

Topic 33: Green’s Functions I – Solution to Poisson’s Equationwith Specified Boundary ConditionsThis is the first of five topics that deal with the solution of electromagnetism problemsthrough the use of Green’s functions. We will begin with the presentation of a procedurethat permits the solution of electrostatic problems with specified charge distributionswithin a volume V surrounded by a surface S that has specified boundary conditions foreither the potential or the normal component of electric field. To set the stage, considerthe application of Green’s theorem for such a geometry:V22 r r rd ad . rrr(33-1) nnNote that we have chosen a prime to denote the variable of integration. We choose the r be the electrostatic potential in V. Thereforetwo functions as follows. Let where rrS , (33-2)2 ris the charge density. For the other function we choose:0r1 rr (33-3)where r is a fixed point in V. Equation (33-1) becomes: r 1 r1 1 r d 4 0rr 4 rr nVS14r ad Sn1 ad . (33-4)rr As S recedes to infinity the surface integrals vanish, and we recover Coulomb’s Law,which is not particularly interesting or useful. Equation (33-4) does suggest thepossibility of a useful expression for evaluating the potential at points in V if Scorresponds to an actual physical boundary. However, as it stands, this will not work.This is due to the fact that one would need to specify both the potential and its derivativeon the boundary in order to evaluate the integrals. But such a case would correspond toan over-specification since either type of boundary condition suffices to uniquelydetermine the potential inside V.Dirichlet Green’s FunctionElectromagnetism, Topic 33, S.P. Ahlen; 09/26/05; 10:18 PM 1

If we modify our choice for rthe situation improves. Suppose that instead ofEquation (33-3) we use the following function: G where the Green’s Function G is subject to the conditions:r D r,r , (33-5)2 , (33-6)D rr4r , rGandD 0r on S. , rG (33-7)Then Equation (33-4) becomes:14 r r0 VD 1 Ddr,rGr4nSr,r G ad . (33-8)Equation (33-8) can be used to solve Dirichlet type problems for which the potential isspecified on the surface.Neumann Green’s FunctionSuppose that instead of equation (33-5) we use the following function:where G is subject to the conditions: G r N r,r(33-9)2 , (33-10)N rr4r , rGandNnr,r G 4 on S. (33-11)SWe have used the symbol S in equation (33-11) to denote the surface area of the surfaceS. Then Equation (33-8) becomes:14 r G 0 VN 1 dr,rGr N4nSr1ad SS adr . (33-12) Electromagnetism, Topic 33, S.P. Ahlen; 09/26/05; 10:18 PM 2

In this equation, is the angle between the vectors r and r . The Green’s function forthe exterior problem is easily seen to be exactly the same (except that r and r areoutside the sphere in this case).************************************************************************Exercise 33-1Consider an isolated metal sphere of radius a whose center is located at the origin. Thesphere has no charge on it. A positive charge Q is located at z = R, and a charge –Q islocated at z = R. Let’s find the potential outside the sphere.Solution 33-1The potential on the sphere is zero, which can be seen by calculating the work done inmoving a charge from infinity to the surface of the sphere along a line in the xy plane.Equation (33-8) becomes:14 r 0 VD dr,rGr , (33-15)where V is the volume exterior to the sphere. This equation yields the potential at allpoints outside the sphere. By using the specified charge distribution, we get: 1 r 40 VˆRrQˆRrQk dr,rGk . (33-16)DSo we have: Q ˆR,rGk rDD kˆ R,rG . (33-17)40Using equation (33-14):D1 R,rG kˆ 2r2 r1R cos2R R2a2ra4a1R cos rR, (33-18)22rRwhere is the usual polar angle. With equation (33-18) is easy to see that:D1 kˆR,rG 2r2 r1R cos 2R R2a2ra4a1R cos rR. (33-19)22rRThe answer is thus specified in terms of equations (33-17), (33-18), and (33-19).Electromagnetism, Topic 33, S.P. Ahlen; 09/26/05; 10:18 PM 4

An interesting check of the results can be obtained by letting R approach infinity. In thiscase:andCombining terms we have:DD 1 r2a aR,rG kˆ 1 cos 1cosR R rR rR 1 r2a akˆR,rG 1 cos 1cosR R rR rR 0 r E0r cos 2.3E a cos, (33-20)rQ2where E0 is the uniform field produced by the external charges. This is the20R4familiar expression for the electric field outside of an uncharged conduction sphere in auniform field.The charge density on the sphere is easily found to be: 3E00cos. (33-21)Electromagnetism, Topic 33, S.P. Ahlen; 09/26/05; 10:18 PM 5