that any image charges lie on the line joining the center of the sphere to the charge outside. Let the radius of the sphere be a,and choose coordinates with origin at the center of the sphere and the charge outside on the positive y-axis at (0,r,0).A single image charge of g'=-ga/r'and location (0,a2/r',0)does the trick. 「1 or)=imorr-a 0 (63) To see that o vanishes on the sphere note that putting r=ar gives: |r-r=a2+rP-2r'r… (64) r-/r=心+an-2arr=re+2-2rr (65) This potential is proportional to the Dirichlet Green function for the exterior of a sphere.We just have to put the exterior charge g=eo at a general point r'and adjust the normalization: Gn(r,)= 1「1 4mr-r可r-a2r/网] (66) This can be plugged into Eq.(46)to solve the arbitrary Dirichlet problem for a sphere.For this we need the normal derivative of Gp on the surface of the sphere: n.VGD OGD r-a Or a-r2/a 4π(a2+r2-2ar.r'3/2 (67) Then the solution for the boundary condition (r=a,,)=V(0,)is oir,g,p0=-av0,pl+-2nrr0 a(a2-2) (68) Here ds=sin dod,andf.r'=r'(cos cos'+sin sin'cos()).While the method of images is very powerful for the simple geometry of a single sphere,it becomes less so with more complicated geometries.In the homework you will see this for the case of two conducting spheres,which require an infinite number of image charges. Even the case of the region between two concentric spheres requires an infinite number of image charges,in spite of the spherical symmetry.This motivates the development of other approaches to electrostatic problems. 3.2 Method of Separation of Variables Partial differential equations put conditions on functions of several variables.The separa- tion of variables method first tries to find solutions that are products of functions of single variables.For instance o(x,y,2)=X(x)Y(y)Z(z) (69) 15 ©2010 by Charles Thornthat any image charges lie on the line joining the center of the sphere to the charge outside. Let the radius of the sphere be a, and choose coordinates with origin at the center of the sphere and the charge outside on the positive y-axis at (0, r 0 , 0). A single image charge of q 0 = −qa/r0 and location (0, a 2/r0 , 0) does the trick. φ(r) = q 4π0  1 |r − r 0yˆ| − a r 0 |r − a 2yˆ/r0 |  (63) To see that φ vanishes on the sphere note that putting r = arˆ gives: |r − r 0 yˆ| 2 = a 2 + r 02 − 2r 0 r · yˆ (64) |r − a 2 yˆ/r0 | 2 = a 2 + a 4 /r02 − 2a 2 r · yˆ/r0 = a 2 r 02 (r 02 + a 2 − 2r 0 r · yˆ) (65) This potential is proportional to the Dirichlet Green function for the exterior of a sphere. We just have to put the exterior charge q = 0 at a general point r 0 and adjust the normalization: GD(r, r 0 ) = 1 4π  1 |r − r 0 | − a r 0 |r − a 2r 0/r02 |  (66) This can be plugged into Eq.(46) to solve the arbitrary Dirichlet problem for a sphere. For this we need the normal derivative of GD on the surface of the sphere: nˆ · ∇GD    r=a = − ∂GD ∂r = a − r 02/a 4π(a 2 + r 02 − 2arˆ · r 0 ) 3/2 (67) Then the solution for the boundary condition φ(r = a, θ, ϕ) = V (θ, ϕ) is φ(r 0 , θ 0 , ϕ 0 ) = − Z dΩV (θ, ϕ) a(a 2 − r 02 ) 4π(a 2 + r 02 − 2arˆ · r 0 ) 3/2 (68) Here dΩ = sin θdθdϕ, and rˆ·r 0 = r 0 (cos θ cos θ 0 +sin θ sin θ 0 cos(ϕ−ϕ 0 )). While the method of images is very powerful for the simple geometry of a single sphere, it becomes less so with more complicated geometries. In the homework you will see this for the case of two conducting spheres, which require an infinite number of image charges. Even the case of the region between two concentric spheres requires an infinite number of image charges, in spite of the spherical symmetry. This motivates the development of other approaches to electrostatic problems. 3.2 Method of Separation of Variables Partial differential equations put conditions on functions of several variables. The separa￾tion of variables method first tries to find solutions that are products of functions of single variables. For instance φ(x, y, z) = X(x)Y (y)Z(z) (69) 15 c 2010 by Charles Thorn