Special theory of relativity > The transformation of electric and magnetic fields > Problem 14.3.13

Statement

$14.3.13.$
Between the stationary plates of the capacitor, a plate made of a substance
with a dielectric constant ε moves at a speed βc. Electric field strength be-
tween the dielectric and the plates E. What is the electric field strength and
magnetic field induction inside a dielectric?

Solution

Data:

Plane capacitor fixed in the laboratory, electric field in vacuum (between plates) E.
Dielectric plate of constant $\varepsilon$ moves parallel to the plates with velocity $\mathbf{v} = \beta c\, \hat{\mathbf{x}}$

the field in vacuum is uniform.

Charge density on the plates

In the laboratory system S, the plates are at rest. The relation between field and surface density in CGS is$ E = 4\pi\sigma$, hence:

$\sigma = \frac{E}{4\pi}$

Rest frame of the dielectric (S')

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

The capacitor plates and their charges move with velocity $-\beta c$

The length in the direction of motion contracts; since the total charge is invariant, the surface density increases by$ \gamma = 1/\sqrt{1-\beta^2}$

$\sigma' = \gamma\sigma = \frac{\gamma E}{4\pi}$

· The electric field in vacuum (outside the dielectric) in S' is:

$E' = 4\pi\sigma' = \gamma E$

In S' the dielectric plate is at rest and there is no magnetic field inside it $(\mathbf{B}'_{\text{diel}} = 0)$

Electric field inside the dielectric in S'

The vacuum–dielectric boundary is perpendicular to the field. The boundary condition for the electric displacement $(\mathbf{D}) $in the absence of free surface charges is $D'_{\text{vac}} = D'_{\text{diel}}$. In CGS$ (\varepsilon_0 = 1)$

$E' = \varepsilon,E'_{\text{diel}} \quad\Longrightarrow\quad
E'_{\text{diel}} = \frac{E'}{\varepsilon} = \frac{\gamma E}{\varepsilon}$

Transformation back to the laboratory (S)

Now we return to S by applying a boost of velocity +$\beta c$ to the internal fields of the dielectric$ (S' \to S)$.

Since $\mathbf{B}'_{\text{diel}} = 0$

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

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

$\boxed{\mathbf{E}_{\text{diel}} = \frac{E}{\varepsilon(1-\beta^2)},\hat{\mathbf{y}}},\qquad
\boxed{\mathbf{B}_{\text{diel}} = \frac{\beta E}{\varepsilon(1-\beta^2)},\hat{\mathbf{z}}} \quad (\text{CGS system})$

Answer

$\boxed{\mathbf{E}_{\text{diel}} = \frac{E}{\varepsilon(1-\beta^2)},\hat{\mathbf{y}}},\qquad
\boxed{\mathbf{B}_{\text{diel}} = \frac{\beta E}{\varepsilon(1-\beta^2)},\hat{\mathbf{z}}} \quad (\text{CGS system})$