[amelius]

Thanks but I was thinking more about how fields drop off in 2D space versus 3D space. Simple electrostatic example: consider a 1D string of identical resistors. Voltage drops linearly as you go along this string. Now consider a 2D grid of resistors: voltage does not change linearly anymore if you move between two points (current will move in a more complicated spread-out pattern). So the dimensionality changes how fields behave.

[Steuard]

Ah, I see what you're getting at. My instinct here is that (exactly as you've pointed out) fields like E and B will fall off like 1/r instead of 1/r^2, but that all of the qualitative behavior will be basically the same. So I wouldn't trust this simulation to predict the precise behavior of a real circuit (even one whose shape was basically planar), but I suspect that it will behave more or less right.

Looking at the examples, it seems like you can make 1D and 2D strings/grids of resistors here in much the same way you would in a 3D model; you just can't make a 3D grid (or non-planar circuits). My general experience working with and teaching basic circuits is that it's rare that we consider current flow in a genuinely 3D medium: the vast majority of problem-solving examples approximate wires as simple 1D paths for charge to follow, and more careful treatments that talk about where charges accumulate to guide current flow around corners, etc. still almost always illustrate their points in 2D diagrams/examples.

So my impression is that this simulation is likely to give a pretty solid qualitative sense of how these systems work, despite its 2D framing.

[ajb]

That's true, but it's actually a property of the circuit. Any circuit that fits into a 2d space will work the same if simulated in 3d: voltage will still drop off linearly along a 1d resistor.

This is because it's actually an emergent property already in 2d space.

Consider a resistor shaped like a capital letter Z in 2d space, with ground at one end and 1V the other. (Assume also that the Z has a square aspect ratio). The potential along the bar in the middle will initially be equal, because all points on the bar are equidistant from our voltage sources (AKA charges) . But the potential will drop along the arms of the Z. So charge will move along the arms and accumulate at the corners, until there is also a voltage drop along the bar, and ohms law holds.