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en/14.3.13.md
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| + | ### Statement | ||
| + | |||
| + | $14.3.13.$ [Insert the problem statement] | ||
| + | |||
| + | ### 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 | ||
| + | |||
| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $14.3.13.$ [Insert the problem statement] | |||
| ### 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 | |||
| [Insert a concise answer or boxed result] | |||