The final solution should be something like $$V\left (x, y, z \right) = \frac{\sigma_0}{2\kappa \epsilon_0}\sin\left(\kappa x \right)e^{-\kappa |z|}$$.
The main idea is to solve the Poisson equation outside the plane, $$\nabla ^2 V = 0$$. and to apply boundary conditions only after having found the general form for the potential.
Since the solution is unique, we can guess a form and if it solves the equation we have solved the problem.
To do this, we will study the symmetries of the problem and write down the consequences of these:
* Translation invariance along $y$
This means that the potential is function of $x$,$z$ only.
$$V\left (x,y,z\right) = V\left(x,z\right)$$
* Translation invariance along $x$ by $\frac{2\pi}{\kappa}$
$$V\left(x, z \right) = V\left(x+\frac{2\pi}{\kappa}, z\right)$$
* Symmetry under spatial reflection with respect to the $x-y$ plane.
$$V\left(x, z \right) = V \left(x, -z \right)$$
Notice that the latter condition can be used to study the problem in the half space $z>0$. 418
Indeed once we find $V$ in the superior half space, we get it also in the $z<0$ half space by making the substitution $z \rightarrow |z|$ in $V\left( x,z\right)$.
Thanks to this consideration, we will focus on the $z>0$ half space from now on we.
We will look for a separate variable solution (we are in the vacuum outside the plate!) $$V\left( x,z\right) = A\left(x \right)B\left( z\right)$$. Plugging in <PERSON> equation and dividing by $V$ we get $$\frac{A''(x)}{A(x)} + \frac{B''(z)}{B(z)} = 0 $$
A possible solution which respect the symmetry we want the solution to have is given by $$V\left(x,z\right)=V_0\sin(\kappa x) e ^{-\kappa z}$$
We only need to check if the electric field (calculated by taking the derivative along the normal direction on the plate) is equal to $\frac{\sigma}{2\epsilon_0}$ (the electric field near a charged sheet) This can be done by choosing $V_0 = \frac{\sigma_0}{2\kappa\epsilon_0}$. This is the potential in the upper half space ($z>0$).
In order to find $V$ in the whole space, we can do the substitution $z\rightarrow |z|$ as we discussed earlier. 418