If we put a voltage on a big square piece of aluminum sheet how will the distribution of the current look like?

Question

Picture an enormous square piece of aluminum sheet, say of $100\times100(m^2)$, on which' two opposite corners we put a potential difference. The ensuing current is low enough to do no damage to the sheet or the voltage supplier.

How will the current distribution between the two points of the square, throughout the sheet, look like? Will the current be the highest on the diagonal, getting less approaching the edges of the sheet (and maybe be zero somewhere, which I can't imagine)?

I can remember from high school that according to Ohm's law $V=IR$ ($V$ being the potential difference, $I$ the current, and $R$ the resistance), and that by putting a piece of metal between the anode and cathode of a battery we can take $R=0$, so we short-circuit the battery, but because of the internal resistance of the battery, the current won't be infinite. In reality, this is, of course, not true, especially for such a big piece of metal as stated above. And in a structure as the big sheet of aluminum $I$ isn't a scalar but a vector $\vec I$ (it has a magnitude as well as a direction at every point of the sheet), which means $V$ becomes also a vector somehow.

Maybe we can compare the situation with a sheet of water between two closed, square pieces of glass with two little holes on two opposite sides of a diagonal, which we use (by connecting little water hoses to them) to create a pressure difference. By putting little, colored balls (with the same mass density as the water) in the water we can see the distribution of the water current between the two plates.

Is there a way to make the current in the aluminum visible? If so we can test the theoretical prediction of the current distribution, which is the core of my question.

Answer 1

One should always think of any electrical phenomenon as organizing itself in such a way that it has minimum energy. And minimum energy implies a minimum total potential distributed through space. The more potential throughout space, the more energy the system has, and it will never organize itself in such a way than it has more energy than it absolutely must.

Imagine a system made out of infinite parallel plates. I can already tell you that, if we take V(inf)=0 on both extremes, the field outside the system will be zero. And I know this because the way the system reduces the energy is to create an equipotential null area in its largest extension of uniform field-space, which is outside the system towards both extremes of infinity. The charge in the conductors will distribute itself in such a manner to produce this configuration, because that's how it will have the minimum energy it can have.