Solution

Reference frame and geometry

S': system at rest with respect to the capacitor.
S: laboratory system, in which the capacitor moves with velocity v along the z-axis (axis of the cylinder).

The capacitor consists of two coaxial cylinders of radii a and b (b > a) and proper length L' (very large compared to the radii). In S' the electric field is purely radial and there is no magnetic field.

For a boost along z, the components of the fields transform according to:

\begin{aligned} E_\parallel' &= E_\parallel, \quad B_\parallel' = B_\parallel,\ \mathbf{E}_\perp' &= \gamma (\mathbf{E}_\perp + \mathbf{v} \times \mathbf{B}),\ \mathbf{B}_\perp' &= \gamma \left(\mathbf{B}_\perp - \frac{\mathbf{v}}{c^2} \times \mathbf{E}\right). \end{aligned}

In our case, in S' (at rest):

$\mathbf{E}'$
is radial (perpendicular to z), therefore $\mathbf{E}_\perp' = \mathbf{E}', E_\parallel' = 0$
$\mathbf{B}' = 0$

In S
using the inverse transformation (changing v \to -v) :

$\mathbf{E}_\perp = \gamma (\mathbf{E}_\perp' - \mathbf{v} \times \mathbf{B}') = \gamma \mathbf{E}_\perp', \quad \mathbf{E}_\parallel = 0$

$\mathbf{B}_\perp = \gamma \left(\mathbf{B}_\perp' + \frac{\mathbf{v}}{c^2} \times \mathbf{E}'\right) = \gamma \frac{\mathbf{v}}{c^2} \times \mathbf{E}'$

The radial electric field in S is:

$\boxed{E_r = \gamma E_r'}$

Potential difference between the cylinders

The potential difference in S' is $V' = \int_a^b E_r' \, dr$
In S:

$V = \int_a^b E_r \, dr = \int_a^b \gamma E_r' \, dr = \gamma V'$

The transverse distances (a and b) do not change because the motion is parallel to the axis.

$\boxed{V = \gamma V' = \frac{V'}{\sqrt{1-\beta^2}}}$.

The potential difference increases by a factor of $\gamma$

Capacitance

The total charge Q on the plates is a relativistic invariant (scalar). In S' the capacitance is $C' = Q/V'$
In S:

$C = \frac{Q}{V} = \frac{Q}{\gamma V'} = \frac{C'}{\gamma}$

$\boxed{C = C' \sqrt{1-\beta^2}}$

The capacitance decreases by a factor of \gamma (it is multiplied by
$\sqrt{1-\beta^2})$