A charged disk or sphere will create currents around its turning axis if a rotation is added. The total current can then be calculated by adding all concentric currents together. Every current (I take the example of a thin charged disk, as it is the most easiest casus) is as big as:

$$ dI = dQ / dt = f*\sigma*2 \pi *rdr $$

So the total charge would be something like:

$$ I = \omega\int_0^R \sigma r dr $$

With $R$ the radius of the disk,$f$ the frequency, $\omega$ the angular speed (so we can leave the $2\pi$ out) and $\sigma$ the charge density.

In my physics book they consider $\sigma$ as a constant. But I was wondering: as electrons (who determine the charge) have mass, they also have a centrifugal force, which would imply that the charge density is NOT the same at every distance from the centre, so we would have to define a $\sigma$ in function of $r$ (so $\sigma$ is not just total charge on total surface).

Would this centrifugal force give considerable different results and if so, could anyone show me a calculation of the difference in total current? Also if the difference is not considerable for a small disk, would it be for a very large disk, because the centrifugal force is bigger at a bigger radius (for example a disk with radius 1000km)?


1 Answer 1


you are right, rotation would cause charge density to become more negative at the edges than at the centre due to the centrifugal force on free electrons (if present). but the effect is practically negligible because all these discs and rings are made of non conducting materials which means there are practically no free electrons.

But yes surface charge density however can be present in both conductors and non conductors. (unlike volume charge density which is only for non conductor) in such cases u can presume that the effect is negligible for the sake of solving the problem :P

  • $\begingroup$ You only answered part one of my question, not the part about a fast spinning BIG object, therefore I will not accept this response, as it is incomplete $\endgroup$ Commented Jun 14, 2014 at 8:15

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