Is electrical capacitance different at relativistic velocity? A 2 parallel plate capacitor has a capacitance proportional to the area of the plates and inversely proportional the distance between the plates. If the plates are perpendicular to the direction of travel at very high speed, relativity tells us that the distance between the plates is reduced relative to the observer. Does this mean the observer sees a higher capacitance?
If the capacitor is rotated 90 degrees so that the plates are parallel to the direction of travel, the area of the plates is now smaller but the distance between them is the same as if it were stationary , does this mean the observer sees a lower capacitance now?
 A: This is an interesting question. First of all, we should compare the electromagnetic field seen by an observer in the rest frame S of the capacitor with the one seen by an observer in a reference frame S' moving with a certain constant velocity $V$ with respect to the capacitor.
I denote the space-time coordinates of the reference frame S by $t,x,y,z$. In order to avoid unnecessary complications, I assume an infinitely thin plate filling the whole $y$-$z$ plane with a homogeneous surface charge density $\sigma$. The second plate with surface charge density $-\sigma$ is positioned parallel to the first plate at a distance $d$. The corresponding charge density is then given by $\rho(t,x,y,z) = \sigma [\delta(x)-\delta(x-d)]$, where $\delta(x)$ is the Dirac delta function. As we are in the rest frame of the charge distribution, the current density $\vec{j}(t,x,y,z)$ vanishes. The resulting electromagnetic field is immediately obtained as
$E_x(t,x,y,z)= E_0\Theta(x) \Theta(d-x)$, $E_y=E_z=0$, $\vec{B}=\vec{0}$,
where $\Theta(x)$ denotes the Heaviside step function and $E_0=4 \pi \sigma$ (I am using the Gauss system).
We now consider a reference frame S' with coordinates $t^\prime, x^\prime, y^\prime,z^\prime$ related to the coordinates of S  by the Lorentz transformation
$t^\prime = \gamma ( t + \frac{V}{c^2} x)$, $x^\prime =\gamma (x+Vt)$, $y^\prime = y$, $z^\prime = z$, $\gamma=(1-V^2/c^2)^{-1/2}$,
i.e. the system S' is moving with respect to the system S with speed $V$ in the negative $x$-direction. Note that the inverse transformation is obtained by interchanging the primed and the unprimed quantities and replacing $V$ by $-V$.
As the 4-current density $(c \rho, \vec{j})$ transforms as a 4-vector field, the charge density and the current density in S' are related to the charge density in S by
$\rho^\prime(t^\prime, x^\prime, y^\prime, z^\prime)=\gamma \rho(t,x,y,z)$, $j_x^\prime(t^\prime,x^\prime, y^\prime, z^\prime) = V \gamma \rho(t,x,y,z)$, $j_y^\prime = j_z^\prime =0$.
Explicitly, one obtains
$\rho(t^\prime, x^\prime, y^\prime, z^\prime)= \gamma \sigma [\delta(x)-\delta(x-d)]= \gamma \sigma [\delta(\gamma(x^\prime-V t))-\delta(d-\gamma(x^\prime-V t))=\sigma[\delta(x^\prime-Vt)-\delta(d/\gamma-x^\prime +V t^\prime))]$
and $j_x^\prime(t^\prime, x^\prime, y^\prime, z^\prime) = V \rho(t^\prime, x^\prime, y^\prime, z^\prime)$.
This result was to be expected. The two charged plates are moving with velocity $V$ in the positive $x^\prime$-direction, their distance is reduced by a factor $\gamma^{-1}=\sqrt{1-V^2/c^2}$ because of Lorentz contraction and the surface charge density remains unchanged as there is no Lorentz contraction orthogonal to the relative velocity of S and S' and the fact that electromagnetic charge is a Lorentz scalar. As the charge distribution in S' is moving with velocity $V$, we consistently find the current density $j_x^\prime = V \rho^\prime$.
Our next task is the determination of the electromagnetic field generated by the charge/current distribution in S'. In principle, we could do this by solving Maxwell's equations for $\rho^\prime$ and $\vec{j}^\prime$, but this would simply be an exercise in self torture. We take advantage of the behaviour of the electromagnetic field strength tensor $F^{\mu \nu}$ under our Lorentz transformation. Expressed in terms of $\vec{E}$ and $\vec{B}$, the electromagnetic field transforms as
$E_x^\prime = E_x$, $E_y^\prime =\gamma (E_y+\frac{V}{c}B_z)$, $E_z^\prime =\gamma(E_z-\frac{V}{c} B_y)$,
$B_x^\prime=B_x$, $B_y^\prime = \gamma(B_y-\frac{V}{c}E_z)$, $B_z^\prime=\gamma(B_z+\frac{V}{c}E_y)$.
Inserting the explicit form of the electromagnetic field observed in the reference frame S, we obtain
$E_x^\prime (t^\prime, x^\prime, y^\prime, z^\prime) = E_0 \Theta(x^\prime-V t^\prime) \Theta(d \gamma^{-1} -x^\prime +V t^\prime)$, $E_y^\prime=E_z^\prime =0$, $\vec{B}^\prime =\vec{0}$.
We see that the electric field stays confined between the two plates with its strength remaining unchanged. At first glance, a vanishing magnetic field seems surprising in view of the nonvanishing current density in S'. However, a moment's thought reveals that this must indeed be the case: Imagine one of the charged plates consisting of uniformly distributed point charges. The current generated by a single point particle moving with velocity $V$ will indeed generate a magnetic field, but the superposition of the magnetic fields of all point particles will cancel exactly. This may also be checked formally by observing that the right-hand-side of the Maxwell equation
$\vec{\nabla}^\prime \times \vec{B}^\prime = \frac{4 \pi}{c} \vec{j}^\prime + \frac{1}{c} \frac{\partial \vec{E}^\prime}{\partial t^\prime}$
vanishes.
Our result for the strength of the electric field between the two moving plates can be obtained even without any detailed calculation. Remember that $\vec{E} \cdot \vec{E}-\vec{B} \cdot \vec{B}$ is a Lorentz invariant quantity. In the system S, we have $\vec{E} \cdot \vec{E}-\vec{B} \cdot \vec{B} =E_0^2$. As $\vec{B}^\prime =\vec{0}$, we find $\vec{E}^\prime\cdot \vec{E}^\prime = E_0^2$.
In conclusion, for an observer moving perpendicular to the plates of the capacitor, the charge density on the plates and the strength of the electric field between the plates remain unchanged. There is no magnetic field and the distance between the plates is Lorentz-contracted.
Let us now rotate the capacitor by $90^0$ such that the plate with surface charge density $\sigma$ coincides with the $x$-$z$ plane. The charge density in the rest frame of the capacitor becomes $\rho(t,x,y,z)= \sigma[\delta(y)-\delta(y-d)]$ and the electromagnetic field in S is given by
$E_x = E_z =0$, $E_y(t,x,y,z)=E_0 \Theta(y) \Theta(d-y)$, $\vec{B}=\vec{0}$.
Transforming to the reference frame S', we find
$\rho^\prime(t^\prime, x^\prime, y^\prime, z^\prime)= \gamma \sigma [\delta(y)-\delta(y-d)]$, $j_x^\prime = V \rho^\prime$, $j_y^\prime = j_z^\prime =0$.
The surface charge density becomes larger by the factor $\gamma$ because of the Lorentz contraction in $x$-direction, the distance between the plates remains unchanged (no Lorentz contraction in $y$-direction).
The electromagnetic field measured in S' is now given by
$E_x^\prime =E_z^\prime=0$, $E_y^\prime(t^\prime, x^\prime, y^\prime, z^\prime)=\gamma E_0 \Theta(y^\prime)\Theta(d-y^\prime)$,
$B_x^\prime = B_y^\prime =0$, $B_z^\prime = \frac{V}{c} E_y^\prime$.
The electric field in S' is larger than in S by a factor $\gamma$, in addition we have a magnetic field between the two plates being orthogonal to the electric field.
We check our result by employing the Lorentz invariants $\vec{E} \cdot \vec{E} -\vec{B} \cdot \vec{B}$ and $\vec{E} \cdot \vec{B}$:
$\vec{E} \cdot \vec{E} -\vec{B}\cdot \vec{B} = E_0^2$, $\quad$ $\vec{E}^\prime \cdot \vec{E}^\prime-\vec{B}^\prime \cdot \vec{B}^\prime = \gamma^2 E_0^2- (V/c)^2 \gamma^2 E_0^2=\gamma^2 (1-V^2/c^2) E_0^2= E_0^2$,
$\vec{B}=\vec{0} \Rightarrow \vec{E} \cdot \vec{B} = 0$, $\quad$ $\vec{E}^\prime \perp \vec{B}^\prime \Rightarrow \vec{E}^\prime \cdot \vec{B}^\prime = 0$.
Summarizing the results for an observer moving parallel to the plates of the capacitor, the surface charge densitity and the electric field become larger by the $\gamma$ factor, in addition we encounter a magnetic field orthogonal to the electric field. The distance between the plates remains unchanged.
Turning now to the question of the capacitance: As the presence of the capacitor singles out a specific reference frame (namly its rest frame), there is no harm in defining the capacitance with respect to a capactor at rest. However, one might ask the question if one could construct a Lorentz covariant and gauge invariant object reducing to the capacitance in the rest frame of the capacitor. (I am thinking of some analogue to the spin 4-vector.) If I find some time, I will think about it.
A: Here is a partial answer to this question, which has been in my mind for several years now in a slightly different form.
We start by imagining a lab inside a rocket traveling at relativistic velocities through the airless vacuum of outer space. On a lab bench inside the rocket is a parallel-plate capacitor situated near a porthole so an observer on the outside can see the plates and the spacing between them as the rocket zooms by.
Connected to the plates is a capacitance meter and since we know the formula for capacitance has the plate spacing in the denominator and the plate area in the numerator, we'll perform the following experiment:
If the plate spacing is subject to contraction when the normal to the plane occupied by the plates is aligned with the direction of travel, one might suspect that the capacitance meter will show a capacitance increase that is proportional to the velocity of the rocket.
Furthermore, if we rotate this capacitor arrangement by 90 degrees, then the foreshortening will reduce the apparent area of the plates and their capacitance would then decrease in proportion to the rocket's speed.
By this reasoning, the capacitance experiment could be used to determine the rocket's velocity relative to some absolute rest frame and thereby determine the rocket's absolute velocity without reference to any external observations.
Now note that an observer outside the rocket will see through the porthole that the capacitor gets squished in a way that depends on which way the setup is rotated.
However, if we are inside that rocket, we get foreshortened along with all our tape measures and clocks and to us, none of the dimensions of the capacitor change at all no matter which way we rotate the setup relative to the nose of the rocket, and no change in its capacitance is detectable with any change in its speed.
And if the digital display of the capacitance meter is visible to the external observer through the porthole, they will see that the capacitance measurement doesn't depend on the speed of the rocket nor the orientation of the setup relative to the rocket's direction, even though the capacitor is visibly squished in appearance.
A: Let's say we have oppositely charged plate-shaped spaceships very close to each other. They are so close that electric field energy is approximately zero.
These spaceships accelerate to the same direction (perpendicular to plates) at the same acceleration. Spaceship crew says spaceships separate, capacitance decreases, work is done on the charges so that electric field energy E is created.
Launchpad crew says spaceships stay at the same distance, capacitance stays the same, fields of the charges deform and separate so that electric field energy $E/\gamma$ is created.

Let's say we have oppositely charged plate-shaped spaceships very close to each other. They are so close that electric field energy is approximately zero.
These spaceships accelerate to the same direction (perpendicular to plates) keeping a constant proper distance. Spaceship crew says nothing changes.
Launchpad crew says spaceships get closer, capacitance increases by $\gamma$, fields of the charges contract but still overlap more and do work, and electric field energy decreases by $1/\gamma$. We know the fields do work, because work is done when opposite charges get closer to each other, and from that we know the fields must overlap more although they contract.
A: Hyperon's abstract, mathematical partial answer yields the following important result:

In conclusion, for an observer moving perpendicular to the plates of
the capacitor, the charge density on the plates and the strength of
the electric field between the plates remain unchanged. There is no
magnetic field and the distance between the plates is
Lorentz-contracted.

I will now attempt to complete a sensible model of moving capacitance by switching to a more concrete, physical point of view.
If the field remains unchanged, the Lorentz contraction of the distance indicates a similar Lorentz contraction of the voltage on the capacitor(same field, smaller distance). Capacitance is defined as $C=Q/V$, so, in this case, the capacitance of the moving capacitor, $C^\prime$ is:
$$C^\prime=\gamma C_0$$
where $C_0$ is the capacitance of the capacitor in its own rest frame.
Now, consider making an LC oscillator using the capacitor. In the rest frame of the oscillator, its period is simply $2 \pi \sqrt{L_0 C_0}$. The relative orientations of the capacitor and inductor don't matter. Then, since time dilation doesn't depend on the orientation of the clock, we may conclude that $C^\prime$ is independent of orientation. From time dilation we get:
$$\sqrt{L^\prime C^\prime}=\gamma \sqrt{L_0 C_0}$$
Which leads to:
$$L^\prime=\gamma L_0$$
In conclusion, a sensible model for moving capacitors and inductors has capacitance and inductance increasing by the Lorentz factor $\gamma$. The orientation of the devices doesn't matter.
