Built on Facts

Potential II: Electric Boogaloo

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:

i-7b7142f1efa09b47c3b8242f89bcb9f4-graph.png

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.

Comments

  1. #1 Braxton Thomason
    February 18, 2009

    So, in my answer yesterday, I had assumed that the “box” was solid and conductive – a “cube” instead of a “cubical box” (clearly I need to read better). I assumed that current would flow through the box producing the voltage gradient – but if the interior of the box is a vacuum, it wouldn’t necessarily create a current, right? Depending on the size of the box, it may or may not arc – but you would still measure a potential inside the box with the above results, even with no current flow, right?

  2. #2 rob
    February 18, 2009

    i like the answer that arne gave. very elegant use of the superposition principle. i tried to do too much math and made a bad assumption about the effect of the sides. they do have an effect–they cause the potential to drop faster than linearly.

  3. #3 rob
    February 18, 2009

    Braxton: yes, you would still measure a potential whether or not the box were solid. the case where the box was a solid conductor and a hollow box with vacuum would give different answers for the potential, but there would still be a potential.

  4. #4 Braxton Thomason
    February 18, 2009

    Rob: Why would the two cases yield different voltage results? (assuming the cube was of uniform resistance)

  5. #5 rob
    February 18, 2009

    Braxton: the two cases are different because in one case there is nothing in the cube and the other case there is matter in the cube.

    electric fields interact with matter. these interactions can lead to polarization, redistribution of charge, currents etc.

    any rearrangement of charge will change the electric potential from what it would have been if the matter was not there.

  6. #6 Braxton Thomason
    February 18, 2009

    Rob: The cases are definitely different in terms of current flow. But look at it this way:

    Say you have a resistor (some arbitrarily high resistance) and apply 0V at one end, and 100V at the other. There is still some potential in the middle of the resistor (50V) even though no current is flowing.

    Replace the resistor with something with lower resistance, and you get a current, but still the same voltage gradient through the resistor.

    I think. I technically have an EE degree, but I haven’t done this stuff in 15 years. Matt, can you clarify?

  7. #7 Matt Springer
    February 18, 2009

    All other things being equal, the current is going to follow the electric field. The field is the gradient of the potential, so if you imagine that an electron is a little BB that you set down on that potential surface it will follow the curve of steepest descent.

    Which is one reason this particular configuration can’t be exactly recreated in the lab – the potential gradient (and thus the field) is infinite where the 100 V face meets the 0 V face.

  8. #8 bbbeard
    February 19, 2009

    Okay, now suppose you have a solid cube made out of some isotropic material. Five sides of the cube area maintained at zero degrees C and the top is maintained at 100 degrees C. What is the temperature at the center of the cube? ;-)

    For extra credit, repeat with the two sides heated… does it matter which two sides?

    BBB

  9. #9 rob
    February 19, 2009

    there are two categories that most materials can be divided into: conductor and dielectric (insulator)

    a resistor is a conductor (well, semi-conductor) so there will be a current because there are free carriers. you can show that the electric field is uniform across a resistor (or a wire) which implies the potential is linear. in this case you will find that the voltage at the center should be 50 V for 100 V applied, no matter what the resistance is.

    if the material is a dielectric, there are no free carriers, so no current will flow. there will be a potential that is a solution to laplace’s equation. the dielectric typically reduces the voltage from that you would find in vacuum by a function of the dielectric constant. so, if the metal box for this problem were filled with a dielectric, you won’t generally find that the voltage will be the same as when there is vacuum.