Mind if I give the readers who've taken an electrostatics class something to noodle over?
Consider a cubic box consisting of six sides of which five are held at a uniform potential of 0 volts. The top side is held at a uniform 100 volts. What is the potential at the center of the cube?
Here's a hint: The hard way is pretty hard. There is an easy way, and it's very easy.
It's a classic example of one of the most important principles in physics - laziness. Ok, that's a joke. What it's actually an example of is the idea that sometimes the formally general mathematics obscures the important parts of the problem, which in this case happens to be the fact that the symmetry of this problem itself contains enough information to give us the potential at the center. Knowing how to see those fundamentals can both save you time and allow you to understand the entire problem more cleanly.
- Log in to post comments
More like this
This reminds me of a similar problem I presented to a math class I taught a few years ago:
Consider a woman in a kayak wearing a hat. She is paddling at 4 mph on the river that is moving at 7 mph (so that total speed relative to the land is 11 mph). Her hat falls off in to the river, but she doesn't realize it for 2.7 hours. Once she realizes it she does an instantaneous turn around and paddles at the same rate upstream. How long until she finds her hat?
(of course, assume that the hat and boat/person flow down the river at the same rate, etc.)
Hrm. Is it 16.6... ?
It seems like the center would have to be at the average voltage of the 6 sides, since it's equally far away.
I'm not sure how much sense this makes electrostatically, but that's the only "easy" way I can think of.
An interesting question is, what is the voltage at a spot 1 inch into the cube from one of the corners where the 100 V side meets.
Daniel,
She'd get it quicker if she just sat on the bank and waited (it'll come to her at 7 mph instead of 4--which means 1.54 hours instead of 2.7).
Roger -
I agree. It's too bad she didn't pick that solution.
0, no? Charge stays on the outside, as I recall (if it's placed on the surface of a closed conductive container)
Dreikin,
He's not applying a charge, he's applying a voltage to the outside of the box. If the cube is conductive, current will flow, producing some voltage gradient across the interior of the cube.
1 billion.
I'd probably treat this cube as a very thick and very short wire with a voltage difference of 100 Volt from one end to the other and apply Ohm's law (OK, I'm a chemist, I like my physics simple).
What do you think?
the first thing to come to mind to solve this is the uniqueness theorem. then image charges set up to give potential of zero and 100 V at the appropriate spots.
now i need to think.
Mmm.... I'd say something that may be silly, but sounds good to me:
Because it's at the center of the cube, and because of the "side" symmetry, it doesn't matter what is the voltage of the sides. So we only have to take into account the upper and bottom sides. So... I think it'll be 50V.
Solving the Laplace equation in a cube isn't THAT bad, the problem you've given even has Dirichlet boundary conditions...
because of the symmetry, all you have to do is find the solution to laplace's equation along a line from the center of the 100 V edge to the center of the opposite edge. the solution will be linear with 100 V and 0 V for the boundary conditions, which means at the center it is 50 V, like xhaju said.
There is not enough information to give a definitive answer. Therefore I must ask: what sort of cube is it? Is it contiguous, conductive and hollow? Solid conductive? Or am I over-thinking it?
Well, electrostatics follow the principle of superposition.
We can see that the problem is symmetrical under the choise 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 useed superpositon and symmetry)
Thus with only ome loaded side we will have one sixth of the one hundred volts; 16.6666...
"Therefore I must ask: what sort of cube is it?"
Hollow, with a vacuum inside, bounded by the six simply-connected square sides at their respective potentials. Really that pretty much requires in practice that the six sides are nonconductive, but it's not strictly necessary.
The correct solution has been given, but I'm going to defer actually giving the solution myself until tomorrow's post.
Braxton Thomason @6:
D'Oh - yeah, I was thinking of the wrong problem...
Arne @14:
Under those conditions, voltage would be 0 at the center - all the sides would cancel at the middle rather than contribute to a non-zero voltage.
It occurs to me, from a simplistic I'm-just-about-to-go-to-bed standpoint that it would depend on the size of the box, due to charge drop-off over distance - but I also assume you're ignoring those types of effects, so, I'd concur with xhaju and rob
It's actually pretty cool to look at in my head, if I've got it right. The fields, that is. Mind including an image of the potentials so I can check it?
(also, what is the laplace equation? Haven't got that far yet)
Consider a box with 50V applied to all six sides - in the middle, the voltage is also 50V.
Consider another box, with -50V applied to the top, +50V applied to the bottom, and +50V applied to each of the sides - in the middle, the voltage is 0V.
Superimpose both these solutions and you get the box with 100V applied everywhere but the top (which has 0V applied) and the solution for the center is 50V.
Or I've got it completely confused...
Consider a box with 50V applied to all six sides - in the middle, the voltage is also 50V.
Consider another box, with -50V applied to the top, +50V applied to the bottom, and +50V applied to each of the sides - in the middle, the voltage is 0V.
Superimpose both these solutions and you get the box with 100V applied everywhere but the top (which has 0V applied) and the solution for the center is 50V.
Or I've got it completely confused...
Andrew@18. Your second box is just the original box with all the sides offset by 50V. This is exactly the same problem and you haven't justified why it's 0V at the centre.
You can do it by superposition, though. The solution is the sum of the solutions for (top & bottom) + (left & right) + (front & back) = 50V + 0V + 0V.
I also think Matt is wrong to suggest it doesn't matter that the plates are non-conductive - the charges would get "moved around" on conductive plates giving a completely different solution. You also have to remember the angles of the field lines are different on the two sorts of plates - viz., at right angles to a conductor.