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If we have a sphere which has $σ$ surface charge density and rotate it in axis z will this create current ? Is it possible without any potential difference ?

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    $\begingroup$ 8.02 by any chance? $\endgroup$ Apr 26, 2013 at 21:59

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Yes. Let's assume that the charge density is fixed relative to the surface of a sphere of radius $R$, then spinning the sphere with angular velocity $\vec\omega = \omega \hat{z}$ would create a surface charge density $\vec K$ given by $$ \vec K(\theta, \phi) = \omega R\sin\theta\sigma(\theta, \phi)\hat\phi(\theta, \phi) $$ where $\theta$ and $\phi$ are the polar and azimuthal spherical coordinates.

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  • $\begingroup$ can we deduce the total charge of the sphere then? $\endgroup$ Apr 26, 2013 at 21:52
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    $\begingroup$ The total charge on the sphere is still just $\int \sigma dA$ even though the sphere is spinning; the motion does not change the amount of charge on the sphere. $\endgroup$ Apr 26, 2013 at 21:54
  • $\begingroup$ I'm sorry, I meant the total current. Is it possible to find that? $\endgroup$ Apr 26, 2013 at 22:03
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    $\begingroup$ By that do you mean the total charge per unit time passing through an arc on the sphere consisting of the portion of a great circle between $\theta = 0$ and $\theta = \pi$? If so, then yes, you would simply compute the integral $\int_0^\pi d\theta \,R \vec K\cdot\hat\phi$. $\endgroup$ Apr 26, 2013 at 22:03
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    $\begingroup$ The spinning sphere constitutes a surface current density. I'm assuming in my response that the charges are fixed relative to the sphere so that the surface charge density relative to the sphere remains the same (imagine somehow "gluing" the charges to an insulator), and I'm assuming that there is some external agent, like a person, who gets the sphere spinning. $\endgroup$ Apr 27, 2013 at 19:04

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