Problem 14.3.14^∗

Solution

Rest frame of the dielectric (S')

We go to the system S' that moves with the dielectric (velocity $+\beta c$relative to the laboratory). In S':

The external magnetic field transforms as
$B' = \gamma B, \qquad \gamma = \frac{1}{\sqrt{1-\beta^2}}$
An electric field induced by the motion appears:
$E' = \gamma \beta B \quad (\text{in CGS}), \qquad E' = \gamma v B \quad (\text{in SI})$.
This field is perpendicular to the plates.

In S' the dielectric plate is at rest. The magnetic field B' does not act on it (there are no magnetic properties). The electric field E' is reduced inside the dielectric according to the boundary condition for the electric displacement $(D'_\perp continuous)$:

$E'_{\text{diel}} = \frac{E'}{\varepsilon} = \frac{\gamma\beta B}{\varepsilon}$

Transformation back to the laboratory (S)

We return to S by applying a boost of velocity $+\beta c$to the fields inside the dielectric. Since in S' there is no magnetic field inside the dielectric $(B'_{\text{diel}} = 0)$

\begin{aligned} E_{\text{diel}} &= \gamma\left(E'_{\text{diel}} + \beta B'_{\text{diel}}\right) = \gamma E'_{\text{diel}} = \frac{\gamma^2\beta B}{\varepsilon}, \[4pt] B_{\text{diel}} &= \gamma\left(B'_{\text{diel}} + \beta E'_{\text{diel}}\right) = \gamma\beta E'_{\text{diel}} = \frac{\beta^2\gamma^2 B}{\varepsilon}. \end{aligned}

Using $\gamma^2 = 1/(1-\beta^2)$

$\boxed{E_{\text{diel}} = \frac{\beta B}{\varepsilon(1-\beta^2)}} \quad (\text{CGS}), \qquad \boxed{E_{\text{diel}} = \frac{v B}{\varepsilon(1-\beta^2)}} \quad (\text{SI})$

Potential difference between the plates

The electric field inside the dielectric is uniform and perpendicular to the plates. The potential difference between the open plates is simply:

$\Delta V = E_{\text{diel}}\,h$

In CGS:

$\boxed{\Delta V = \frac{\beta B h}{\varepsilon(1-\beta^2)}}$

In SI:

$\boxed{\Delta V = \frac{v B h}{\varepsilon(1-\beta^2)} = \frac{\beta c B h}{\varepsilon(1-\beta^2)}}$