Новое решение

Alexphysics правка от 19 июн. 2026 г., 01:58

правка #19020 позже →

en/14.4.6.md +39 −0

		@@ -0,0 +1,39 @@
		+	### Statement
		+
		+	$14.4.6.$ [Insert the problem statement]
		+
		+	### Solution
		+
		+	Physical interpretation
		+
		+	The electric field acts with a constant force $ F = eE $ in the direction of motion of the electron.
		+	The penetration depth is defined as the distance the electron travels until its kinetic energy equals its rest energy.
		+
		+	Equations of motion (relativistic dynamics)
		+
		+	The force modifies the linear momentum:
		+
		+	$\frac{dp}{dt} = eE, $\qquad p = \gamma m_e v, \quad \gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$
		+
		+	The total energy of the electron is $\mathcal{E} = \gamma m_e c^2$ and its variation satisfies
		+	$d\mathcal{E} = v\,dp$.
		+
		+	Combining:
		+
		+	$d\mathcal{E} = eE\,v\,dt = eE\,dx$
		+
		+	Integrating from$ x = 0 (where v = 0, \mathcal{E}_0 = m_e c^2)$
		+
		+	$\mathcal{E} - m_e c^2 = eE x$
		+
		+	$\Rightarrow \quad \gamma m_e c^2 - m_e c^2 = eE x \quad\Rightarrow\quad x = \frac{m_e c^2}{eE}(\gamma - 1)$
		+
		+	For $\gamma = 2 (kinetic energy = m_e c^2)$
		+
		+	$\boxed{x = \frac{m_e c^2}{eE}}$
		+
		+	This is the distance the electron must travel for its kinetic energy to reach a value comparable to its rest energy;
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

	@@ -0,0 +1,39 @@
			### Statement

			$14.4.6.$ [Insert the problem statement]

			### Solution

			Physical interpretation

			The electric field acts with a constant force $ F = eE $ in the direction of motion of the electron.
			The penetration depth is defined as the distance the electron travels until its kinetic energy equals its rest energy.

			Equations of motion (relativistic dynamics)

			The force modifies the linear momentum:

			$\frac{dp}{dt} = eE, $\qquad p = \gamma m_e v, \quad \gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$

			The total energy of the electron is $\mathcal{E} = \gamma m_e c^2$ and its variation satisfies
			$d\mathcal{E} = v\,dp$.

			Combining:

			$d\mathcal{E} = eE\,v\,dt = eE\,dx$

			Integrating from$ x = 0 (where v = 0, \mathcal{E}_0 = m_e c^2)$

			$\mathcal{E} - m_e c^2 = eE x$

			$\Rightarrow \quad \gamma m_e c^2 - m_e c^2 = eE x \quad\Rightarrow\quad x = \frac{m_e c^2}{eE}(\gamma - 1)$

			For $\gamma = 2 (kinetic energy = m_e c^2)$

			$\boxed{x = \frac{m_e c^2}{eE}}$

			This is the distance the electron must travel for its kinetic energy to reach a value comparable to its rest energy;

			#### Answer

			[Insert a concise answer or boxed result]