# Boundary conditions for an infinite line charge and grounded conducing plane

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?

• Do you know the solution for the line charge replaced by a point charge ? – Kutsit Sep 27 at 10:59
• 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. – khaled014z Sep 27 at 11:17
• This is the general method : en.wikipedia.org/wiki/Method_of_image_charges – Kutsit Sep 28 at 6:54