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I'm not asking for a solution to the problem, I'm confused about what I should set the boundary conditions to, it's obvious that $V=0$ at $z=0$ because of the grounded $xy$ plane, but I don't know what to do with the other boundaries, the problem is that the line charge itself extends to infinity parallel to the x-axis, so I feel like it isn't right to assume that $V$ goes to zero as $s$ goes to zero, not like the case of a point charge where $V$ actually goes to zero away from the charge, so what should I do?

edit: One fix I thought about is to assume that $a >> d$ where $a$ is the reference point of the potential of the line charge since the charge extends to infinity and $d$ is the distance between the line and the $xy$ plane parallel to the x-axis, then set the boundary conditions as $V=V(y.z)$ where $V=\frac{\lambda}{4 \pi \epsilon_0} ln(\frac{s_{-}^{2}}{s_{+}^{2}})$, neglecting the potential produced by the grounded plane, is that reasonable?

  • $\begingroup$ Do you know the solution for the line charge replaced by a point charge ? $\endgroup$ – Kutsit Sep 27 at 10:59
  • $\begingroup$ I know the analogy with a charged sphere that is equivalent to a point charge at the center, but for the line charge, I don't recall. $\endgroup$ – khaled014z Sep 27 at 11:17
  • $\begingroup$ This is the general method : en.wikipedia.org/wiki/Method_of_image_charges $\endgroup$ – Kutsit Sep 28 at 6:54

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