# Is equal charge distribution impossible in most cases?

Let us consider a scenario where one have a cubical stack of 64 similar cubes,like this one.

As it is a single conductor, let us put $ne$ amount of charge in this cube such that $n$ is not a multiple of 64.

And then separate each of the cubes while keeping them connected to adjacent cubes using neutral wires.

Since we take all the cubes to be similar and charge is supposed to be distributed equally.

Charge in each cube=$ne/64=(n/64)e$

The problem is (n/64) is a fraction.

And wikipedia says:

Charge quantization is the principle that the charge of any object is an integer multiple of the elementary charge. Thus, an object's charge can be exactly 0 e, or exactly 1 e, −1 e, 2 e, etc., but not, say, 1⁄2 e, or −3.8 e, etc.

How will be charge distributed? Will the current flow for infinite amount of time through the wires connecting cubes?

• Like you said you can't have an equal number of quanta on every cube. However, quantum mechanics has this way of making the discontinuous continuous. A wavefunction can be a superposition of (one quantum on cube 1 and zero on cube 2)+(one quantum on cube 2 and zero on cube 1). As such the wavefunction can be perfectly symmetric while any observed charge distribution is asymmetric. – user12029 Sep 17 '16 at 22:52
• @NeuroFuzzy That's a great summary, on a point I really think should be more widely known about how QM can't be caught out by apparent inconsistent concepts. If you developed it , or even c & p the above into an answer, I would up vote. No mathjax needed :) Thanks – user108787 Sep 17 '16 at 23:29
• @CountTo10 Sounds good! I wrote an answer and expanded things a bunch. – user12029 Sep 17 '16 at 23:56

The answer is simpler than you might think. To demonstrate my point, imagine you let $n=1$, so there is one elementary charge in your cube. This charge will situate itself in one particular location in the cube (probably in the center, but it doesn't really matter). Whatever mini-cube the charge ends up in will have $1$ elementary charge in it, and all the other mini-cubes will have $0$. If you add another elementary charge, they will distribute themselves in some new way. But still, most of the cubes will have $0$ charges in them, and those two lucky cubes will have $1$ charge each. If we now scale up from $2$ to $n$, you should be able to conclude that even if the charges are distributed "equally," there is no problem with having different numbers of charges in each mini-cube. This answer does not depend at all on the specifics of charges, electrostatics, etc. And contrary to the other answer on here from @NeuroFuzzy, quantum mechanics is definitely not required to solve this problem. There is a perfectly valid classical solution (as I gave in this paragraph).

If you are worried about which particular cubes get more and which ones get less, the answer depends mostly on where in particular you add the charges to the cube as well as the atomic or molecular structure of the cube itself.

There are some incorrect assumptions you have made in your problem which do not affect the answer. I still think it would be useful to provide some additional information so that you have a better idea of how the physics in this problem works. The assumptions all stem from your statement that "charges distribute equally."

In fact, that statement is simply not true in general. Let me tell you what I think you mean by this, and then tell you what the reality is. When you say distributed equally, I think you are saying that they should be equally spaced throughout the volume of the cube. In reality, the situation is a bit more complex for two reasons:

1. You have stipulated that this cube is a conductor. On a conductor, charges always move to the surface since they are repelled by all other charges. So, there will not be any of the charges you add to the cube in the interior—they will all be on the surface.

2. On your conductor cube, the charges on the surface would not distribute themselves uniformly (equally spaced), even if you did use the correct number of elementary charges to match the number of grid regions. The reason is that charges feel larger forces from nearby charges than from distant charges, and the surface of a cube is not uniform. (Some parts of the cube are close to corners and some are close to the center, for example.) To get a true spatially uniform distribution, you would need to use a sphere, which does have a truly uniform surface.

• True! There are definitely different contexts in which to answer the question. – user12029 Sep 18 '16 at 0:24

A conducting cube will have its charge distribute on the surface (of $64-8=56$ cubes), so lets work on a simpler example of two cubes stuck together with one quantum of charge in total.

You can imagine these physically as two tiny cubes of iron or something, like the oil droplets with extra quanta of charge in Millikan's oil drop experiment. All of the electrons and protons balance out to a completely electrically neutral system, except for one pesky extra electron.

If you put the cubes together, separate them, and measure the charge on them (maybe through a Millikan-like experiment), you will indeed have to find that one has a charge on it and the other does not.

In a classical mechanical view of the system, you might imagine the electron as a point particle bouncing around through the two iron cubes. Just due to thermal motion, once you separate the two cubes, it may be on one or the other, with no preference. But at any given time, the electron is on one cube or the other. This is your idea of an electron flowing back and forth. There is a current associated with the moving electron, and the electron is always moving back and forth.

This is not how things have to behave in actuality! Quantum mechanically, you cannot tell even in principle whether the electron is on one cube or the other until you actually make a measurement. That is one moral of the double slit experiment - electrons can be truly delocalized before you make a measurement.

Mathematically you can write down a wavefunction. The system is in a superposition of the two asymmetric states: (one quantum on the pair of cubes)=(one quantum on cube 1 and zero on cube 2)+(one quantum on cube 2 and zero on cube 1). Dirac notation, that would look like: $|\psi\rangle=\frac{1}{\sqrt{2}}(|1\rangle+|2\rangle)$. The probability of measuring the particle on cube 1 is 50% ($|\langle 1|\psi\rangle|^2=0.5$), and the probability of measuring the particle on cube 2 is 50% ($|\langle 2|\psi\rangle|^2=0.5$). However, before you make these measurements, the wavefunction is totally symmetric.

This is similar to how electron orbitals change around a hydrogen atom. When an electron decays from a 2p to a 1s orbital, it emits a photon. Classically, the only explanation is that at some point in time the electron jumped instantaneously. Quantum mechanically, the state can transition smoothly, say like $|\psi(t)\rangle=\cos(t)|2p\rangle+\sin(t)|1s\rangle$. At $t=0$, the electron is in state $|2p\rangle$ with 100% probability. At $t=\pi/2$ the electron is in state $|1s\rangle$ with 100% probability. Inbetween, the electron was in a superposition of both states. It was not even in principle in either state. It made the discrete change continuously. Same with the charges in this example. One quantum of charge can be distributed across two cubes, with no preference for either cube and with no time dependence (so the electron isn't wiggling around classically and travelling from one cube to the other all the time), while every measurement you make of the cubes individually shows that the electron is on one cube or the other (with equal probability).

• Thank you for all your work and time. I self study, so I grab everybody I can on this site as potential teachers. Physics is far more subtle than I assumed before finding this site, so my smugness at solving a problem by myself is always misplaced. Sincerely appreciated. – user108787 Sep 18 '16 at 0:09