Electric field given flux through a plane Suppose you have a hidden, arbitrary, static charge distribution below the plane $z=0$ and you know the electric flux through the plane at every point on that plane. There is no charge above the plane. Is it possible to determine the electric field at every point above that plane? I would think so. Heuristically, if you know the flux, then don't you have full knowledge of the field lines? I'd think that would imply some Neumann boundary conditions which then uniquely determines the potential and so the field, but I've only managed to get so far.
The differential flux element is $d\Phi=E_zrdrd\theta$. The flux and area elements are known, so $E_z$ is known, consequently the $\theta$ and $r$ derivatives of $E_z$ are known.
Since a static electric field is irrotational, the $z$ derivative of $E_r$ and $E_\theta$ are also known on the plane.
Can any more be deduced? Not sure where to go from here.
 A: I believe this should be possible.  Is the region above the plane a charge-free region?  If so, then you should be able to write down a general solution to Laplace's equation using separation of variables.  Then apply boundary conditions at $z=0$, and $r\rightarrow \infty$, basically at the boundary of the upper hemisphere.  The BC at the surface will involve $E_\perp=-\partial V/\partial n$, and likely you would choose $V(r\rightarrow \infty)=0$.
With azimuthal symmetry, you should be able to write
$$V(r,\theta) = \sum_l{\left(A_l r^l + \frac{B_l}{r^{l+1}}\right)P_l(\cos{\theta})}$$
Note that $r$ is the distance from the origin and $\theta$ is the polar angle measured from the z-axis.
Imposing the boundary conditions in this coordinate system might prove tricky, because you need to evaluate $(\partial V/\partial z)_{z=0}$.  However, the boundary condition that $V\rightarrow 0$ at $r\rightarrow \infty$ immediately gives $A_l=0$, so you only have the $B_l$ to determine from the boundary condition at $z=0$.
