Новое решение

Alexphysics правка от 17 июн. 2026 г., 14:23

правка #18985 позже →

en/14.3.22.md +58 −0

		@@ -0,0 +1,58 @@
		+	### Statement
		+
		+	$14.3.22.$ [Insert the problem statement]
		+
		+	### Solution
		+
		+	Rest frame S'
		+
		+	In S', the capacitor is at rest:
		+
		+	· Plates perpendicular to the z-axis; separation h.
		+	Surface charge density$ \pm\sigma$
		+	Volumetric leakage current j flowing in the z-direction (from the positive plate to the negative plate).
		+
		+	By symmetry (infinite plates), the magnetic field in S' is zero:
		+	$\mathbf{B}' = 0$
		+	The electric field is uniform and perpendicular to the plates (according to Gauss's law):
		+
		+	$\mathbf{E}' = (0,\,0,\,E_0), \qquad E_0 = \frac{\sigma}{\varepsilon_0}$
		+
		+	The leakage current is j in the z-direction; it does not create a magnetic field because it is uniform and infinite in the plane.
		+
		+	Transformation to the laboratory system S
		+
		+	The capacitor moves relative to S with velocity
		+	$ \mathbf{v} = \beta c\,\hat{\mathbf{x}}$ (parallel to the plates).
		+
		+	For a boost in the x-direction, the fields transform according to:
		+
		+	$\begin{aligned}
		+	E_x &= E'_x, \quad & E_y &= \gamma(E'_y + v B'_z), \quad & E_z &= \gamma(E'_z - v B'_y),\\[4pt]
		+	B_x &= B'_x, \quad & B_y &= \gamma\!\left(B'_y - \frac{v}{c^2}E'_z\right), \quad & B_z &= \gamma\!\left(B'_z + \frac{v}{c^2}E'_y\right),
		+	\end{aligned}$
		+
		+	with
		+
		+	$\gamma = \dfrac{1}{\sqrt{1-\beta^2}}$
		+
		+	Substituting $\mathbf{E}' = (0,0,E_0)$
		+	and $ \mathbf{B}' = 0$:
		+
		+	$\mathbf{E} = (0,\,0,\,\gamma E_0), \qquad
		+	\mathbf{B} = \left(0,\,-\gamma\frac{v}{c^2}E_0,\,0\right)$
		+
		+	The electric field remains perpendicular to the plates, but its magnitude increases by a factor of$ \gamma$:
		+
		+	$\boxed{E_z = \gamma E_0 = \frac{E_0}{\sqrt{1-\beta^2}}}$
		+
		+	Consistency with the sources
		+
		+	The total charge on each plate is invariant.
		+	Moving parallel to the plates, the length of the plates in the direction of motion contracts:$ L_x = L'_x/\gamma$
		+	The area of each plate decreases: $A = A'/\gamma$
		+	Thus, the surface charge density in S is $\sigma_S = Q/A = \gamma\sigma$.
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

	@@ -0,0 +1,58 @@
			### Statement

			$14.3.22.$ [Insert the problem statement]

			### Solution

			Rest frame S'

			In S', the capacitor is at rest:

			· Plates perpendicular to the z-axis; separation h.
			Surface charge density$ \pm\sigma$
			Volumetric leakage current j flowing in the z-direction (from the positive plate to the negative plate).

			By symmetry (infinite plates), the magnetic field in S' is zero:
			$\mathbf{B}' = 0$
			The electric field is uniform and perpendicular to the plates (according to Gauss's law):

			$\mathbf{E}' = (0,\,0,\,E_0), \qquad E_0 = \frac{\sigma}{\varepsilon_0}$

			The leakage current is j in the z-direction; it does not create a magnetic field because it is uniform and infinite in the plane.

			Transformation to the laboratory system S

			The capacitor moves relative to S with velocity
			$ \mathbf{v} = \beta c\,\hat{\mathbf{x}}$ (parallel to the plates).

			For a boost in the x-direction, the fields transform according to:

			$\begin{aligned}
			E_x &= E'_x, \quad & E_y &= \gamma(E'_y + v B'_z), \quad & E_z &= \gamma(E'_z - v B'_y),\\[4pt]
			B_x &= B'_x, \quad & B_y &= \gamma\!\left(B'_y - \frac{v}{c^2}E'_z\right), \quad & B_z &= \gamma\!\left(B'_z + \frac{v}{c^2}E'_y\right),
			\end{aligned}$

			with

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

			Substituting $\mathbf{E}' = (0,0,E_0)$
			and $ \mathbf{B}' = 0$:

			$\mathbf{E} = (0,\,0,\,\gamma E_0), \qquad
			\mathbf{B} = \left(0,\,-\gamma\frac{v}{c^2}E_0,\,0\right)$

			The electric field remains perpendicular to the plates, but its magnitude increases by a factor of$ \gamma$:

			$\boxed{E_z = \gamma E_0 = \frac{E_0}{\sqrt{1-\beta^2}}}$

			Consistency with the sources

			The total charge on each plate is invariant.
			Moving parallel to the plates, the length of the plates in the direction of motion contracts:$ L_x = L'_x/\gamma$
			The area of each plate decreases: $A = A'/\gamma$
			Thus, the surface charge density in S is $\sigma_S = Q/A = \gamma\sigma$.

			#### Answer

			[Insert a concise answer or boxed result]