# Solving for potential outside a square with outgoing flux condition?

## Context

As a follow up/ simplification of this question I would like to compute the potential created by a charged square outside of the square. I would like to impose that far enough the outgoing flux behaves as though the cube was a point mass.

Let me define a region between a square and a disk as

reg =
RegionDifference[Disk[{0, 0}, 2], Rectangle[{-1, -1}, {1, 1}]]


and extract a mesh out of it

Needs["NDSolveFEM"];
mesh = ToElementMesh[reg, MaxCellMeasure -> 0.05,
"MaxBoundaryCellMeasure" -> 0.01];
Show[mesh["Wireframe"], Frame -> True, PlotRange -> All] Let me try and solve of the Laplacian, aiming to fix the potential on the square and the outgoing flux on the disk

  edge = DirichletCondition[1, {x, y} \[Element] Rectangle[{-1, -1}, {1, 1}]]


so that

  sol = NDSolveValue[{-Laplacian[u[x, y], {x, y}] ==
NeumannValue[1, {x, y} \[Element] Disk[{0, 0}, 2]], edge}, u, {x, y} \[Element] mesh]


I get this error message

NDSolveValue::femibcnd: No DirichletCondition or Robin-type NeumannValue
was specified for {u}; the result may not be unique.


and the plot looks wrong (it does not have the symmetry of the problem and the values are very large).

ContourPlot[sol[x, y], Element[{x, y}, mesh], AspectRatio -> 1,Contours -> 20] ## Question

What am I doing wrong with the boundary condition?

I am guessing this is some trivial issue but...? I would be great if the preprocessing of mathematica could guide (more) the user in this context?

sol = NDSolveValue[{-Laplacian[u[x, y], {x, y}] == • @chris, fixed the ^2 issue. I did look at the question - but I do not understand it. – user21 Jan 28 '18 at 16:47