Special theory of relativity > The transformation of electric and magnetic fields > Problem 14.3.23

Statement

$14.3.23.$
Solve problem 14.3.22 in the case of a capacitor moving at a velocity βc
pendicular to the plates

Solution

1. Rest frame S'

Plates parallel to the x'y' plane, separated in z' by a distance h.
Uniform electric field:
$\displaystyle \mathbf{E}' = (0,\, 0,\, E_0), with E_0 = \dfrac{\sigma}{\varepsilon_0} (SI)$.
Uniform leakage current: $\mathbf{j}' = (0,\, 0,\, j)$.
There is no magnetic field $(\mathbf{B}' = 0) $by symmetry.

The capacitor moves with velocity $\mathbf{v} = v\, \hat{\mathbf{z}} (v = \beta c)$, parallel to the electric field and the leakage current.
$\gamma = 1/\sqrt{1-\beta^2}$

Electromagnetic fields (boost in z):

$\begin{aligned}
E_z &= E'_z = E_0, \quad \mathbf{E}_\perp = \gamma(\mathbf{E}'_\perp + \mathbf{v}\times\mathbf{B}') = 0,\
B_z &= B'_z = 0, \quad \mathbf{B}_\perp = \gamma!\left(\mathbf{B}'_\perp - \frac{\mathbf{v}}{c^2}\times\mathbf{E}'\right) = 0.
\end{aligned}$

$\boxed{\mathbf{E} =(0,\, 0,\, E_0),\qquad \mathbf{B} = 0}$

The electric field does not change relative to the rest system.

The four-current $(c\rho, \mathbf{j})$
transforms as a four-vector. In S':
$\rho' = 0, \mathbf{j}' = (0,0,j)$

$\begin{aligned}
\rho &= \gamma!\left(\rho' + \frac{v}{c^2}j'_z\right) = \gamma\frac{v}{c^2}j,\
j_z &= \gamma,(j'_z + v\rho') = \gamma,j,\
j_x &= j'_x = 0,\quad j_y = 0.
\end{aligned}$

The volumetric current density increases:
$ j_z = \gamma\, j$.

A net charge density appears in the dielectric:
$\rho = \gamma v j / c^2$

However, the total current I through a cross-section remains invariant because the area of the plates (perpendicular to the motion) contracts:
$ A = A'/\gamma$.

$I = \int j_z\, dA = (\gamma j)\left(\frac{A'}{\gamma}\right) = j A' = I'$