The moving capacitor To what extent can a charged capacitor mounted on a moving platform (e.g. a rotating wheel) be considered an electric current generator? Electric current, after all, is nothing more than the transport of charge. Some extra detail:
Because a capacitor is a charge separation device, do the charges on the two plates effectively cancel to produce no net current?
If we ground one side of the capacitor, does the situation change?
If we rotate one plate and counter-rotate the other, do we succeed in producing net current?
 A: The rotating of charge does produce what's essentially a current, but there's some subtlety to it.
A simple definition of current is the charge through an area per unit of time. This is usually taken in the limit as time approaches 0, making it a derivative.
$I=\frac{\mathrm{d} q}{\mathrm{d} t}$
Typically, the area the charge is moving through is a cross-section of wire, though that isn't necessary. Even though current isn't a vector quantity, it has kind of a direction associated with it, depending on whether the charge is flowing in one direction through the area, or the other way. Since the choice of positive direction is arbitrary, you just choose a positive direction, and then negative current is just current in the opposite direction.
This definition doesn't distinguish between the kind of charge carrier, nor the medium through which the charge moves. For example: electrons moving through a wire, ions diffusing in a battery, beams of protons, as well as rotating capacitor plates would all be considered electric current.

It's important to note though that you have to take the sign of the charge into account. Negative charge (e.g electrons) moving in the positive direction is actually a negative current.
If you want to draw the areas small enough, your rotating capacitor actually produces two currents of equal magnitude in opposite directions, one for each plate, separated by the distance of the plates. Practically though, the distance is small enough to be considered a net zero current.
If you ground one of the plates, nothing should change. Charge won't flow out of the capacitor unless you ground both plates (due to the attraction between the opposite charges). Same net zero charge rotating, same zero current.
The last case though, where you rotate the plates in opposite directions, does create a measurable current! The average current would be twice the charge on one of the plates, divided by the period of rotation. Rotating the plates faster would produce more current.
I've simplified the calculus in this explanation a bit. For more detail, look up the concepts of current density, flux, and surface integral. Usually, one defines current in terms of a surface integral of current density (which is a kind of flux).
Source: Detailed electromagnetic/electrical textbooks, or some materials textbooks should cover the definition of current, which describes this situation.
A: Most capacitors' plates are metals, where charge is free to flow, rather than dielectrics with charge embedded, I believe. Were the plates embedded with charge, rotating them would definitely produce an effective current.
However, I suspect that rotating metal plates with charge on them would initially produce a current due to the moving charge, but that current would set up a magnetic field which would quickly move the electrons to oppose their movement on the plate. A quick force diagram sketch tells me that if $v$ is in the $\hat \theta$ direction and $B$ is in the $\hat z$ direction (using cylindrical coordinates), then the force F on an electron is either in the $\hat r$ or the the $-\hat r$ direction, though I'd have to do the math to figure out which. 
A: Rotating a net-charged thing like a metallic plate around would indeed create an effective current, which would achieve an stable steady state (assuming no friction, etc.) if we have the plate insulated from the axis in the middle to avoid the charge fleeing away by Hall Effect (described by the other answer).
However, when you start turning it, you do feel transient effects. The magnetic filed will be increasing in the (imaginary) loop and so a force would oppose your accelerating turning, but if you overcome that force and stop, it'll stay there.
A: This is really a suggested answer to the following question, which I think is a little different to the one it is linked to: You can make some estimates: Consider one plate of radius $R$ with an amount $+Q$ of charge uniformly distributed over the plate, giving a charge density $\sigma = \frac{Q}{\pi R^2}$, which is rotating at an angular velocity $\omega$. Now consider an annulus of radius $r$ to $r + dr$. The amount of charge in this annulus is $d q = \sigma \times 2 \pi \times r dr$. This charge is rotating with the angular speed $\omega = \frac{2 \pi}{T}$ where $T$ is the period of the rotation, and leads to an effective current of $I = \frac{dq}{dt} = \frac{\sigma 2 \pi r dr}{2 \pi/\omega} = \omega \, \sigma \, r dr.$ The magnetic field due to this current loop a distance $l$ along the $z$ axis (from symmetry other components vanish) is $$d B(l) = \omega \, \sigma \, r dr \times \frac{\mu_o}{4 \pi} \frac{1}{r^2 + l^2} \cos \theta = \frac{\omega \, \sigma \, \mu_0}{4 \pi} \frac{r^2 dr}{(r^2+l^2)^{3/2}}$$ and adding up all the contributions gives $$B(l) = \frac{\omega \, \sigma \, \mu_0}{4 \pi} \int_0^R \frac{r^2 dr}{(r^2+l^2)^{3/2}}$$ According to wolfram we have $\int_0^R \frac{r^2 dr}{(r^2+l^2)^{3/2}} = \ln \frac{\sqrt{l^2+R^2} + R}{l} - \frac{R}{\sqrt{l^2+R^2}}$. You can similarly determine the magnetic field due to the plate carrying the charge $-Q$ and then add the fields (remembering the relative directions).
