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This stems from thinking about the question If a perfect conductor were to move, what happens to the electrons?.

Suppose we have a rotating disk with no external magnetic field, so this is not a homopolar generator/Faraday disk experiment. The rotation creates a difference in potential energy between the centre and the rim, so does this mean that if we connect a wire (with suitable brushes) between the centre and the rim electrons will flow from the centre through the disk to the rim then back through the wire? That is, does the rotation create an electrical potential difference between the centre and the rim?

It seems obvious to me that the answer is yes, however I have never seen the calculation done. Attempts to Google it fail because the results are swamped by articles on Faraday disks and/or homopolar generators.

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    $\begingroup$ Interesting question. I haven't attempted at even a schematic calculation for this scenario, but I was pondering on just two facts: a) metal surfaces are equipotential; b) any stationary metal object on Earth is actually accelerated (because it is not free falling) by the equivalence principle. $\endgroup$ – Phoenix87 Jun 9 '15 at 12:02
  • $\begingroup$ Can you please define the electrical properties of the wire used ? $\endgroup$ – Gaurav Jun 9 '15 at 13:59
  • $\begingroup$ @Gaurav: does the wire really matter? As long as it conducts electricity the exact properties shouln't make much difference. If you want a precise description replace the wire by a voltmeter with infinite resistance. $\endgroup$ – John Rennie Jun 9 '15 at 14:19
  • $\begingroup$ @JohnRennie I wanted to know whether the wire too is a perfect conducter because if it is, then current wouldn't flow. Let me elaborate in my answer. $\endgroup$ – Gaurav Jun 9 '15 at 14:26
  • $\begingroup$ @Gaurav: I will be interested to read your answer, though it seems to me that Cag has nailed it and I'm sure how you'd improve on his answer. $\endgroup$ – John Rennie Jun 9 '15 at 14:36
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Electrons in a conducting disk in order to maintain equilibrium will have to have a centripetal force on them equal to the local change in potential energy with respect to a change in radius, that is

$$ m_e\omega^2 r = -e{d\phi\over dr} $$

After integrating, we get a potential difference between the center and a point R out

$$ \Delta\phi = -{m_e\omega^2 R^2\over 2e} $$

A conducting disk spinning at a rate of six million radians per second should generate about one volt of potential ten centimeters out from the center. I hope this was helpful. ;)

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    $\begingroup$ Any real uncharged conductor has nuclei as well. They are fixed on a large scale, but not on the scale we are talking about. You would need to consider what dipole moments are created. There are many mechanisms that outweigh this one. Denser charged particles could move outward. There might be a distortion of polar molecules. If you solve the Schrodinger equation for atomic bonds with an extra centripetal potential, you might find that electrons move along the bond, or that bond lengths change. $\endgroup$ – mmesser314 Jun 9 '15 at 14:24
  • $\begingroup$ @mmesser314: true, but it's such a small effect that it seems reasonable to consider movement of the conduction electrons as the primary effect. While the other possibilities you mention are plausible I'd guess they are secondary. $\endgroup$ – John Rennie Jun 9 '15 at 14:26
  • $\begingroup$ @JiK: the model is that of an electron sea, the thing we are measuring is a potential difference, and the assumption is that when the force on the particle from the field produces the centripetal force on the particle, the system is in equilibrium. The force on a particle is the gradient of the potential energy, and the dominant contribution comes from the electric scalar potential in this problem. The nucleus of the conduction material may produce atomic distortions in the field, but the overall shape of the potential curve is what we measure here. $\endgroup$ – Alan Jun 10 '15 at 17:44
  • $\begingroup$ @JiK: The nuclei are bound by metallic bonds, which frees the electrons to belong to the whole metallic complex instead of particular atoms. The metallic bonds let the nuclei stay relatively fixed. The local distortions I was thinking of come from the protons in the nuclei, and since they stay relatively fixed, their overall contribution to the potential shape is to make it a little bumpy on the atomic level. $\endgroup$ – Alan Jun 10 '15 at 20:35
  • $\begingroup$ @JiK: I imagine at some point, the metal around the axis gets torn apart and squishes into a ring or something, but until then, the lattice bonds for the nuclei are deep and local enough that it would only take a little bit of displacement for the metal nucleus to experience the centripetal force. $\endgroup$ – Alan Jun 10 '15 at 21:21
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This answer is valid under the assumption that the wire also rotates with the disk.

@cag's answer reveals two things :

(1) that the electric field is independent of the distribution of charge in the disk. We know that this distribution will vary for materials with different conductivities.

(2) that the field is independent of the shape of the conducter under rotation.

Since the wire is under rotation, a centripetal force exists in the wire as well. And thus, from @cag's formula we get to know that the potential difference in the wire is exactly same as that of the disk and in the same direction.

Now, since you have specified the conductivity of the wire, being zero, we know that the distribution of charge is markedly different from that of the disk.

For a wire having same conductivity as that of the disk, we know that no current would flow as soon as you connect the wire, since the net emf of the two cancel each other.

For a wire of different conductivity than that of the disk, the distribution difference between the two causes a kind of osmosis current which flows until it reaches a state where the distribution of charge along the radius of the disk is same for both.

Forgive me for my simplistic approach at solution, I'm not particularly adept at mechanisms on the microscopic scale.

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  • $\begingroup$ Isn't it possible to have a stationary wire where the two ends are in contact with the disk at all times? I assumed that's what he meant by "with suitable brushes" but maybe I misunderstood. But if there is no rotation of the wire, is there anything incorrect with @cag's answer? $\endgroup$ – tpg2114♦ Jun 9 '15 at 16:11
  • $\begingroup$ @tpg2114 Very true, it seems I have misinterpreted the aforesaid phrase. I find nothing incorrect in cag's answer given the correct definition of the system. $\endgroup$ – Gaurav Jun 9 '15 at 16:26
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    $\begingroup$ Just for confirmation, yes I did mean a stationary wire connected to the disk with brushes of the type typically found in electric motors. Apologies if this was not clear. $\endgroup$ – John Rennie Jun 9 '15 at 17:26

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