| ### Statement | | ### Statement |
| | | |
| $14.3.23.$ | | $14.3.23.$ |
| Solve problem 14.3.22 in the case of a capacitor moving at a velocity βc | | Solve problem 14.3.22 in the case of a capacitor moving at a velocity βc |
| pendicular to the plates | | pendicular to the plates |
| | | |
| ### Solution | | ### Solution |
| | | |
| 1. Rest frame S' | | 1. Rest frame S' |
| | | |
| Plates parallel to the x'y' plane, separated in z' by a distance h. | | Plates parallel to the x'y' plane, separated in z' by a distance h. |
| Uniform electric field: | | Uniform electric field: |
| $\displaystyle \mathbf{E}' = (0,\,0,\,E_0), with E_0 = \dfrac{\sigma}{\varepsilon_0} (SI)$. | | $\displaystyle \mathbf{E}' = (0,\,0,\,E_0), with E_0 = \dfrac{\sigma}{\varepsilon_0} (SI)$. |
| Uniform leakage current: $\mathbf{j}' = (0,\,0,\,j)$. | | Uniform leakage current: $\mathbf{j}' = (0,\,0,\,j)$. |
| There is no magnetic field $(\mathbf{B}' = 0) $by symmetry. | | 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. | | The capacitor moves with velocity $\mathbf{v} = v\,\hat{\mathbf{z}} (v = \beta c)$, parallel to the electric field and the leakage current. |
| $\gamma = 1/\sqrt{1-\beta^2}$ | | $\gamma = 1/\sqrt{1-\beta^2}$ |
| | | |
| Electromagnetic fields (boost in z): | | Electromagnetic fields (boost in z): |
| | | |
| $\begin{aligned} | | $\begin{aligned} |
| E_z &= E'_z = E_0, \quad \mathbf{E}_\perp = \gamma(\mathbf{E}'_\perp + \mathbf{v}\times\mathbf{B}') = 0,\\ | | E_z &= E'_z = E_0, \quad \mathbf{E}_\perp = \gamma(\mathbf{E}'_\perp + \mathbf{v}\times\mathbf{B}') = 0,\\ |
| B_z &= B'_z = 0, \quad \mathbf{B}_\perp = \gamma\!\left(\mathbf{B}'_\perp - \frac{\mathbf{v}}{c^2}\times\mathbf{E}'\right) = 0. | | B_z &= B'_z = 0, \quad \mathbf{B}_\perp = \gamma\!\left(\mathbf{B}'_\perp - \frac{\mathbf{v}}{c^2}\times\mathbf{E}'\right) = 0. |
| \end{aligned}$ | | \end{aligned}$ |
| | | |
| $\boxed{\mathbf{E} =(0,\,0,\,E_0),\qquad \mathbf{B} = 0}$ | | $\boxed{\mathbf{E} =(0,\,0,\,E_0),\qquad \mathbf{B} = 0}$ |
| @@ -28,7 +28,9 @@Solution |
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| The electric field does not change relative to the rest system. |
| The electric field does not change relative to the rest system. |
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| |
| The four-current $(c\rho, \mathbf{j}) transforms as a four-vector. In S': \rho' = 0, \mathbf{j}' = (0,0,j)$ |
| The four-current $(c\rho, \mathbf{j})$ |
| |
| transforms as a four-vector. In S': |
| |
| $\rho' = 0, \mathbf{j}' = (0,0,j)$ |
| |
| |
| $\begin{aligned} |
| $\begin{aligned} |
| \rho &= \gamma\!\left(\rho' + \frac{v}{c^2}j'_z\right) = \gamma\frac{v}{c^2}j,\\ |
| \rho &= \gamma\!\left(\rho' + \frac{v}{c^2}j'_z\right) = \gamma\frac{v}{c^2}j,\\ |
| j_z &= \gamma\,(j'_z + v\rho') = \gamma\,j,\\ | | j_z &= \gamma\,(j'_z + v\rho') = \gamma\,j,\\ |
| j_x &= j'_x = 0,\quad j_y = 0. | | j_x &= j'_x = 0,\quad j_y = 0. |
| \end{aligned}$ | | \end{aligned}$ |
| | | |
| The volumetric current density increases: | | The volumetric current density increases: |
| $ j_z = \gamma\,j$. | | $ j_z = \gamma\,j$. |
| | | |
| A net charge density appears in the dielectric: | | A net charge density appears in the dielectric: |
| $\rho = \gamma v j / c^2$ | | $\rho = \gamma v j / c^2$ |
| | | |
| However, the total current I through a cross-section remains invariant because the area of the plates (perpendicular to the motion) contracts: | | 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$. | | $ A = A'/\gamma$. |
| | | |
| $I = \int j_z\,dA = (\gamma j)\left(\frac{A'}{\gamma}\right) = j A' = I'$ | | $I = \int j_z\,dA = (\gamma j)\left(\frac{A'}{\gamma}\right) = j A' = I'$ |
