Hello, I'm trying to understand the application of Green's function to find the potential better. I apologize in advance if I start mixing things up a little. From what I understood and seen, we use this method (Green and method of images) in known symmetries (cylindrical/spherical/planar) and the whole catch is to find the function F in G= 1/|x-x'| + F where F has to fulfill Laplace's equation. This F represents the potential of the charge distribution or point charges outside the volume V that is bounded by the surface S whilst fulfilling the boundary condition (Dirichlet/Neumann). Also, an important note, Green's function is dependent only on the geometrical shape of the surface we have. Assuming what I said is fully correct, I'd like to ask the following: 1. Most of the examples I've seen always solved by Green's function and the method of images for the case where the potential on the surface (the boundary condition) is zero (grounded surface). Well what about the case where the potential on the surface is nonzero? I know that the method of images is good for replacing some problem with point charges distributed where, at some point given, they all add up to have the potential zero there so I'm not sure how it can be applied here. I tried answering it by saying, well, I know Green's function for an infinite grounded plane (for example), why not simply add up the constant to it. It fulfills the boundary conditions so it seems okay. I plug it into the solution for a Dirichlet boundary condition and I'll have that extra element in the final answer (the integral of the charge distribution in the volume -rho- multiplied by the constant potential). 2. Green's function is dependent only on the shape of the surface I have. Well does that mean that for each of the symmetries (that I know their Green function) it is pretty much the same Green function and the only difference from one problem to another is the boundary conditions or how many image charges I have to add? 3. I forgot what I wanted to ask more, I'll update later when I remember. Thanks in advance!