Suppose that the potential Function is given by $V(x,y,z)$ then the boundary conditions would be

1. $lim_{x \rightarrow \pm \infty} V(x,y,z) = 0$ and the same will be true for $y$ and $z$ going to infinities
2. $V(x,y,z) = 0, z \leq 0$

regarding the Electric fields you will have the following

1. $E(x,y,\mu) = 0\hat{i} + 0\hat{j} + \frac{2qd}{4\pi\epsilon_0(x^2 + y^2 + d^2) }\hat{k} $ upto first order as $\mu$ goes to 0+

2. and below that plane E is 0 vector

Suppose that the potential Function is given by $V(x,y,z)$ then the boundary conditions would be

1. $\lim_{x \rightarrow \pm \infty} V(x,y,z) = 0$ and the same will be true for $y$ and $z$ going to infinities
2. $V(x,y,z) = 0, z \leq 0$

regarding the Electric fields you will have the following

1. $E(x,y,\mu) = 0\hat{i} + 0\hat{j} + \frac{2qd}{4\pi\epsilon_0(x^2 + y^2 + d^2) }\hat{k} $ upto first order as $\mu$ goes to $0+$

2. and below that plane $E$ is $0$ vector