The question is as follows:
A cross section of a uniform conducting sheet of charge density $\sigma$ lies along the $x$-axis in the $z$-plane. Take $\phi(x,0)=V$ on the charged sheet, and let $k=\sigma/\epsilon_0$. (Note: the electric field satisfies $E=-\nabla\phi$, and has components $(0,k)$ for this case.)
Find the $\phi(x,y)$ above the sheet that satisfies Laplace’s equation $\nabla^2\phi(x,y)=0$ in the upper half plane $y > 0$. Also identify the harmonic conjugate $\eta(x,y)$ and the complex potential $\Phi(z)=\phi(x,y)+i\eta(x,y)$.
I am having a lot of trouble starting this problem. I am stuck on how to get a general phi from $\nabla^2\phi(x,y)=0$ and $E=-\nabla\phi$ so I can apply the Dirichlet boundary condition.
The first answer should be $\phi(x,y)=V-ky$, which makes intuitive sense.
If i understand correctly, they want the potential over the plate, which is lying on the $xz$ plane., that is the potential for $y \gt 0$ Given that the plate is at the potential $V$. Since the electric field due to an infinite charged plane must always be perpendicular to its surface, the potential is simply the integral:
$-\int \vec E \cdot dy = -\int k.dy$
So, the potential $\phi(x,y) for y\gt 0$, is given by : $\phi(x,y) = -ky + C$ . Clearly, since for $y =0, \phi(x,0) = V$, the answer is as you mentioned. This is the only solution for the given boundary conditions, as stated by the uniqueness theorem.
$$ANOTHER APPROACH$$
Just guess a function $\phi(x,y)$ satisfying the boundary conditions:
$\phi(x,0) = V, \nabla^2 \phi(x,y) = 0$. Clearly, it cannot be a constant, as it does attract/repel a point test charge. So to keep the second derivative zero, we choose a linear function. So $\phi(x,y) = Ky + C$. Make this satisfy the boundary conditions, and you will get the previous answer. Uniqueness theorem gurantees, that this is the only solution satisfying the laplace eqn. And the given boundary conditions.