jdstokes
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I'm sure I used to have a really simple answer to this. But I've long since forgotten.
BTW This is not my homework.
James
BTW This is not my homework.
James
maverick280857 said:The local charge density varies inversely as the square of the radius of curvature.
So that the potential may satisfy the Laplacian outside the conductor.jdstokes said:Yes, of course. But why??
robphy said:Here's a simple model.
Consider two conducting spheres, with unequal radii A and B, connected by a long thin conducting wire... so the spheres are at the same potential. How do free charges distribute themselves on the two spheres?