Excellent job on yesterday’s physics problem. Several people got the right answer, and in lieu of answering it again myself I’m going to let commenter arne fill us in:
Well, electrostatics follow the principle of superposition. We can see that the problem is symmetrical under the choice of loaded side. We know that if we apply the same voltage to all the sides we will have uniform potential inside the cube.
Therefor each loaded side contributes one sixth of the voltage in the middle. (Here we used superposition and symmetry)
Thus with only one loaded side we will have one sixth of the one hundred volts; 16.6666…
Precisely so. If we take a 2D cross-section through the center of the cube going from the 100 V end to the 0 V end, the potential looks something like this:
Ignore the wiggles near the 100 V end, that’s a computational artifact of the series being truncated at a reasonable number of terms. Ninety, though lots of them are 0. I’m being a slightly lazy programmer and making the computer evaluate them anyway. For the value of the central point computed with 100 terms I get V = 16.66666736 volts, very close to our exact value of 16.666… volts. Actually in this case just the first term by itself is not bad: V = 17.37729 volts.
So here we have a purely theoretical prediction for what the real-life results would be from this charge configuration, and a computational prediction as well. Fortunately both match. In real-life physics research we’d want to do an experiment as well to make sure the theory wasn’t wrong to start with. And in fact this sort of experiment has been done to extreme precision on too many occasions to count, fitting the predictions of Maxwell’s equations very well. Perfectly in fact, if the corrections from quantum and relativistic effects are too small to matter – as they usually are in the macroscopic world.