What does a relativistically moving capacitor "look" like? Suppose I have a parallel plate capacitor with a vacuum between the plates, a voltage $V$ across them and a capacitance of $C$.
What will this arrangement look like to an observer in a uniformly moving frame of reference with velocity $v$ at right angles to the capacitor plates? Obviously the apparent plate separation is changed, so presumably the capacitance changes, but what about the electric field between the plates and the stored energy?
(Not a homework question - though I'm exploring whether it could be...!)
 A: One expects the energy stored in the capacitor to transform like the zeroth component of the four-vector $(U,\vec p)$. In its rest frame the field configuration around the capacitor has $$(U,\vec p)_\text{rest}=(U_0,\vec 0),$$ and by the Lorentz transformation the moving observer will see $$(U,\vec p)_\text{moving}=(\gamma U_0, \gamma\vec\beta U_0),$$ where $\gamma=(1-\beta^2)^{-1/2}$ and $\vec\beta=\vec v/c$ as usual.
If the moving observer calculates the capacitance 
$$
C_\text{rest} = \frac{\epsilon_0 A}{d}
$$
he'll see a length-contracted separation $d_\text{moving}=d_\text{rest}/\gamma$ between the plates but the same area $A$ and the same vacuum permittivity $\epsilon_0$. That gives us $C_\text{moving}=\gamma C_\text{rest}$, which translates the way we expect for the stored energy $U=\frac12 C V^2$ — assuming that the potential difference $V$ is the same in both frames.
However, also asking about the fields brings out the interesting parts of the problem. By the symmetry argument using Gauss's Law, the electric field inside the volume of the capacitor has the same magnitude $|\vec E| = \sigma/{\epsilon_0}$, where $\sigma=Q/A$ is the surface charge density on the capacitor plates and is the same in both reference frames. The moving observer sees this field occupying a smaller volume than the stationary observer, and so the part of the energy stored in the electric field,
$$
U^\text{electric} = \int d^3x \frac{\epsilon_0}2 E^2,
$$
is actually smaller by $1/\gamma$ for the moving observer, even though we've already decided that the total energy for the moving observer increases.
The way out, of course, is to remember that each of the moving charged plates is a strangely-shaped current distribution which produces closed loops of magnetic field.  Imagine that the capacitor is approaching you with the positive plate first. As it gets near you'll feel counterclockwise-going magnetic fields; if you happened to pass through the uniform field region the magnetic field would vanish; there'd be a clockwise-going magnetic field as the negative plate receded from you. 
The magnetic field will be strongest in regions where the electric field changes rapidly, as the moving observer passes through the fringe field of the capacitor.
The stored energy in this magnetic field $U^\text{magnetic}$ apparently grows faster than $U^\text{electric}$ shrinks.
I think this argument suggests that a moving observer also sees a smaller potential difference $V$ across a capacitor than an observer in the capacitor's rest frame, which surprises me a little.
