2:52 PM
I propose at the debinning to enter into details of Alex Trounev's solution.
Here is the potential (in Volts) over the domain. The lines are the equipotentials :
One sees that the boudary of the objects which are conductors are the equipotentials 1 Volt and -2 Volts.
Obviously the boundaries of the square is not a equipotential. They can't be metallic.
Now let see the electrical field :
By definition the electrical fileds lines are orthogonal to the equipotential.
One sees that they are effectively orthogonal to the boundary of the conductors.
But what's happening at the boundaries of the square ?
That means that the electric field doesn't go across the boundary of the square.
This is a typical case of what is called a Neuman Boundary condition (In general there is a value attached to Neumann Boundary condition, here the value is 0. In short one says a Neumann=0 condition)
One can grasp a very intuitive understanding of the signification of Neumann0 with the thermal equivalent of this electrostatic problem.
(some background : the equations are exactly the same, it suffices just to transform the word "Volts" in "degrees". For example the circle is at temperature +1 degree, the rectangle at temperature -2 degrees)
The thermal equivalent of the electric field is the flux of heat.
Back to the "field that doesn't go across the boundary" :
The thermal equivalent to this expression is "no flux of heat goes across the boundary".
That is to say : no heat goes across the boundaries of the square. This boundaries are made of athermal insulation material. A very common situation in the domain of thermics.
The electrical equivalent situation is not that usual. One must even imagine the following contrieved situation :
Take the thermal situation, if you want to have no flux thru the (say) right boundary of the square, you place another symetrical square at the right side of your square (which i will call the first square in the following).
In this situation the 2 squares gives each other the same quantity of heat in the 2 directions. Total = 0 Watts. We have artificially constructed a isolant wall. That the situation corresponding to Neumann=0 on the right side.
Unfortunetely, after that, the left side is still not Neumann=0.
An approach to the solution could be to add a third square on the left side of the first square.
This would be a solution if each lateral square wouldn't have a small long range effect on the other side of the first square (= the central one).
The solution is to imagine that we have a very long row of square, a périodic pattern, long enough so that the side effect are negligeable (the kind of this that happens in critallography).
We are done, execpt that there is also the top and bottom side of the square.
The story is the same : just imagine a 2D periodic pattern.
Electrostatic
