- #1
Erwin Derek
Consider the scenario where there are two parallel conducting pipes of radius [itex]R[/itex] separated by a distance [itex]d[/itex], with pipe 1 at a potential of [itex]-V[/itex] and pipe 2 at a potential [itex]+V[/itex]. I have seen from many sources that there is a very easy method of images solution to the potential outside the pipes, given that the potential goes to [itex]0[/itex] at [itex]\infty[/itex].
Now I thought of an alternate scenario where the only difference is that instead of [itex]-V[/itex] and [itex]+V[/itex] it is now [itex]0[/itex] and [itex]+V[/itex]. Can this still be solved with method of images?
I have tried using two line charges of unequal charge and the equipotential surfaces end up being slightly distorted circles, so that doesn't work.
Otherwise, how can we do this with separation of variables?
Now I thought of an alternate scenario where the only difference is that instead of [itex]-V[/itex] and [itex]+V[/itex] it is now [itex]0[/itex] and [itex]+V[/itex]. Can this still be solved with method of images?
I have tried using two line charges of unequal charge and the equipotential surfaces end up being slightly distorted circles, so that doesn't work.
Otherwise, how can we do this with separation of variables?