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If we ignore the inner section, we have a box with 3 sides held at V = 0 and the top edge at V = V1. I'm pretty sure this is easily solvable by separation of variables using an oscillatory solution in x with a decaying solution along y. Using superposition we can then treat the inner box as a separate problem of similar geometry/boundary conditions. The ...


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The method of images is useful when you have a point charge near a $V=0$, a.k.a. grounded, surface; usually a plane or a sphere. In this configuration you can substitute the surface by an additional point charge (of opposite charge) and the problem becomes finding the potential due to two point charges. In these other problems, you don't have this ...


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If you look at the two particles in the centre of mass frame you'll see something like: Your task is to find $b$. In the COM frame both particles move in an effective central potential, and the two trajectories are symmetric. I'm going to ignore the green one and just show how to calculate $b$ for the red trajectory. The closest approach is then $2b$. In ...


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OK, I will try to say something a bit more useful than my other response. I am still on the skeptical side of getting a closed-form, simple solution to this problem, specially using separation of variables. I think that the problem is that the potential outside of a square with that border conditions cannot be attacked using separation of variables. My ...



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