9 October 2002 Physics 217, Fall 2002 1Today in Physics 217: the method of images Solving the Laplace and Poisson equations by sleight of hand Introduction to the method of images Caveats Example: a point charge and a grounded conducting sphere Multiple imagesyxbbbbaaqq-q-qaa9 October 2002 Physics 217, Fall 2002 2Solving the Laplace and Poisson equations by sleight of handThe guaranteed uniqueness of solutions has spawned several creative ways to solve the Laplace and Poisson equations for the electric potential. We will treat three of them in this class:Method of images(today).Very powerful technique for solving electrostatics problems involving charges and conductors.Separation of variablesPerhaps the most useful technique for solving partial differential equations. You’ll be using it frequently in quantum mechanics too. Multipole expansionFermi used to say, “When in doubt, expand in a power series.” This provides another fruitful way to approach problems not immediately accessible by other means.9 October 2002 Physics 217, Fall 2002 3Introduction to the method of imagesA point charge lies a distance d above a infinite, conducting, grounded plane. Calculate the potential Veverywhere above the plane. This looks like a Laplace-equation problem, and we know some boundary conditions at the plane: But there’s charge induced on the grounded plane.The electrostatic potential can not be calculated directly without knowing the induced charge distribution.zqd0, 0.xyVEE===9 October 2002 Physics 217, Fall 2002 4Introduction to the method of images (continued)Consider alternatively the situation of two point charges q and –q, separated by 2d.The potential can be calculated directly and is equal toNote that is automatically satisfied. This also gives V = 0 on the plane z = 0, just as it would need to be for the grounded plane.qzd-qd()()()2222,, qVs zszdqszdφ=+−−+++24Vπρ∇=−9 October 2002 Physics 217, Fall 2002 5Introduction to the method of images (continued)The potential yields the electric field, as usual: ()() ()()()()()()()()()()32 32222232 3222223222, above32221ˆˆˆ,,ˆˆ ˆ2Note , ,0 ,,,0whence , .42VVVsz Vss zqqszd szdqz d qz dszd szdqdzssdEsqdssdφφφφσφππ⊥∂∂∂=− = + +∂∂∂=−+− ++−++−+− ++=−+==−+Esφ zsszzE—No φdependenceE perpendicular to z = 0 plane, at z = 0, just as it would need to be with the grounded plane.9 October 2002 Physics 217, Fall 2002 6Introduction to the method of images (continued)Thus, for, the two-charge potential satisfies the Poisson equation and the boundary conditions for the single charge – grounded plane problem: it is a solution to this problem. But there is no “a” solution, only “the” solution, because solutions of electrostatics problems are unique.qzd-qdzqd≥z0=9 October 2002 Physics 217, Fall 2002 7Introduction to the method of images (continued)How did we know this would work? Is there a method by which we could guess these solutions in general?Yes. The auxiliary charge is the imageof the original charge in the “mirror” that comprises the grounded conducting plane. Other configurations of charges and grounded conductors can be treated similarly: as if they were objects, images and mirrors in geometrical optics. qz-qMirror9 October 2002 Physics 217, Fall 2002 8Caveats for the method of images The solution for the images is only the same as that for the conductor, in the region outside the conductor! In particular, the field is still zero, and the potential constant,inside the conductor. Remember that the image charge doesn’t really exist.  In particular, the potential energy of the charge+conductor arrangement is quite different from the charge-image charge combination, because the field is finite for the latter in locations where the field of the former is zero. (Got it?) Remind yourself of these facts by noting, in every image solution, that you can calculate the induced charge on the surface of the grounded conductor.9 October 2002 Physics 217, Fall 2002 9Potential for a point charge and a grounded sphere (Example 3.2 + Problem 3.7 in Griffiths)A point charge qis situated a distance Z from the center of a grounded conducting sphere of radius R. Find the potential everywhere. Find the induced surface charge on the sphere, as function of θ. Integrate this to get the total induced charge. Calculate the potential energy of the system.RZqV = 09 October 2002 Physics 217, Fall 2002 10Potential for a point charge and a grounded sphere (continued)We need to find a position to put charge q’ such that the boundary conditions on the sphere are satisfied. Start by determining where this charge should lie, using the points along the z axis:RZqV = 0RZqV = 0q'z'd'd()()00PQqqVZR RzqqRzZRqqVZR RzqqRzZR′== +′−−′′⇒=− −−′== +′++′′⇒=− ++PQ9 October 2002 Physics 217, Fall 2002 11Potential for a point charge and a grounded sphere (continued)Now consider an arbitrary point on the sphere, P’:RZqV = 0q'z'd'dPQ() ()()()()()()()()()()222So Rearrange:22,qqRz RzZR ZRZRRz ZRRzzZ R zZ R ZzRZ R RZ R RqqRRRzq Rz R qZZR ZRZZ′′−= +−+′′+−=−+′′ ′−+ +==+−−=′′ ′⇒ = =− + =− + =−++P’224222 222cos2cos 2 cosdRZ RZRRdRz Rz R RZZθθθ=+−′′′=+− =+−θ9 October 2002 Physics 217, Fall 2002 12Potential for a point charge and a grounded sphere (continued)The potential should come out to be zero there, and sure enough,Thus the potential outside the grounded sphere is given by the superposition of the potential of the charge q and the image charge q’.22 4 22222 4 22222 222cos2cos2cos2cos02cos 2cosPRqqq qZVddRZ RZ R RRRZZRqqZRZ RZ R RRRZZqqRZ RZ RZ RZθθθθθθ′−′=+ = +′+−+−−=++−+−=−=+− +−9 October 2002 Physics 217, Fall 2002 13Potential for a point charge and a grounded sphere (continued)So the potential at some point outside the sphere is given by Now for the induced charge density:Differentiate the formula above for the potential, and evaluate it at r = R:22 4 22222 222cos2cos2cos2cosRqqZVrZ rZ R RrrZZqqrZ rZrZRrZRθθθθ−=++−+−=−+−+−(),,rθφ1144rVErσππ∂==−∂9 October 2002 Physics 217, Fall 2002 14Potential for a point charge and a grounded sphere (continued)()()()()223/2 3/2222223/2 3/222