Problem 14.3.22

Solution

Rest frame S'

In S', the capacitor is at rest:

· Plates perpendicular to the z-axis; separation h.
Surface charge density$\pm\sigma$
Volumetric leakage current j flowing in the z-direction (from the positive plate to the negative plate).

By symmetry (infinite plates), the magnetic field in S' is zero:
$\mathbf{B}' = 0$
The electric field is uniform and perpendicular to the plates (according to Gauss's law):

$\mathbf{E}' = (0,\,0,\,E_0), \qquad E_0 = \frac{\sigma}{\varepsilon_0}$

The leakage current is j in the z-direction; it does not create a magnetic field because it is uniform and infinite in the plane.

Transformation to the laboratory system S

The capacitor moves relative to S with velocity
$\mathbf{v} = \beta c\,\hat{\mathbf{x}}$ (parallel to the plates).

For a boost in the x-direction, the fields transform according to:

\begin{aligned} E_x &= E'_x, \quad & E_y &= \gamma(E'_y + v B'_z), \quad & E_z &= \gamma(E'_z - v B'_y) & B_x = B'_x, \quad B_y = \gamma\left(B'_y - \frac{v}{c^2}E'_z\right), \quad B_z = \gamma\left(B'_z + \frac{v}{c^2}E'_y\right)\end{aligned}

with

$\gamma = \dfrac{1}{\sqrt{1-\beta^2}}$

Substituting $\mathbf{E}' = (0,0,E_0)$
and $\mathbf{B}' = 0$:

$\mathbf{E} = (0,\,0,\,\gamma E_0), \qquad \mathbf{B} = \left(0,\,-\gamma\frac{v}{c^2}E_0,\,0\right)$

The electric field remains perpendicular to the plates, but its magnitude increases by a factor of$\gamma$:

$\boxed{E_z = \gamma E_0 = \frac{E_0}{\sqrt{1-\beta^2}}}$

Consistency with the sources

The total charge on each plate is invariant.
Moving parallel to the plates, the length of the plates in the direction of motion contracts:$L_x = L'_x/\gamma$
The area of each plate decreases: $A = A'/\gamma$
Thus, the surface charge density in S is $\sigma_S = Q/A = \gamma\sigma$.