Edits to “Statement”, “Solution”, “Answer”
en/14.4.6.md
+16 −5
| @@ -1,7 +1,11 @@ | |||
| ### Statement | |||
| − | $14.4.6.$ | ||
| + | $14.4.6.$ | ||
| + | A stationary electron is impacted by a longitudinal electric field of intensity | ||
| + | E at the speed of light. How deep will the electron penetrate into this field if | ||
| + | it acts on the electron in the direction of its motion? | ||
| + | |||
| ### 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) | |||
| @@ -13,17 +17,24 @@Solution | |||
| The force modifies the linear momentum: | |||
| − | $\frac{dp}{dt} = eE, | ||
| + | $\frac{dp}{dt} = eE,\qquad p = \gamma m_e v, \quad \gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$ | ||
| − | The total energy of the electron is | ||
| + | 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 | ||
| + | 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; | |||
| @@ -36,4 +47,4 @@Solution | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | $\boxed{x = \frac{m_e c^2}{eE}}$ | ||
| @@ -1,7 +1,11 @@ | |||
| ### Statement | ### Statement | ||
| $14.4.6.$ |
$14.4.6.$ | ||
| A stationary electron is impacted by a longitudinal electric field of intensity | |||
| E at the speed of light. How deep will the electron penetrate into this field if | |||
| it acts on the electron in the direction of its motion? | |||
| ### Solution | ### Solution | ||
| Physical interpretation | Physical interpretation | ||
| The electric field acts with a constant force $ F = eE $ in the direction of motion of the electron. | 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. | The penetration depth is defined as the distance the electron travels until its kinetic energy equals its rest energy. | ||
| Equations of motion (relativistic dynamics) | Equations of motion (relativistic dynamics) | ||
| @@ -13,17 +17,24 @@Solution | |||
| The force modifies the linear momentum: | The force modifies the linear momentum: | ||
| $\frac{dp}{dt} = eE, |
$\frac{dp}{dt} = eE,\qquad p = \gamma m_e v, \quad \gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$ | ||
| The total energy of the electron is |
The total energy of the electron is | ||
| $\mathcal{E} = \gamma m_e c^2$ | |||
| and its variation satisfies | |||
| $d\mathcal{E} = v\,dp$. | $d\mathcal{E} = v\,dp$. | ||
| Combining: | Combining: | ||
| $d\mathcal{E} = eE\,v\,dt = eE\,dx$ | $d\mathcal{E} = eE\,v\,dt = eE\,dx$ | ||
| Integrating from |
Integrating from | ||
| $ x = 0 (where v = 0, \mathcal{E}_0 = m_e c^2)$ | |||
| $\mathcal{E} - m_e c^2 = eE x$ | $\mathcal{E} - m_e c^2 = eE x$ | ||
| $\Rightarrow \quad \gamma m_e c^2 - m_e c^2 = eE x \quad\Rightarrow\quad x = \frac{m_e c^2}{eE}(\gamma - 1)$ | $\Rightarrow \quad \gamma m_e c^2 - m_e c^2 = eE x \quad\Rightarrow\quad x = \frac{m_e c^2}{eE}(\gamma - 1)$ | ||
| For $\gamma = 2 (kinetic energy = m_e c^2)$ | For $\gamma = 2 (kinetic energy = m_e c^2)$ | ||
| $\boxed{x = \frac{m_e c^2}{eE}}$ | $\boxed{x = \frac{m_e c^2}{eE}}$ | ||
| This is the distance the electron must travel for its kinetic energy to reach a value comparable to its rest energy; | This is the distance the electron must travel for its kinetic energy to reach a value comparable to its rest energy; | ||
| @@ -36,4 +47,4 @@Solution | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | $\boxed{x = \frac{m_e c^2}{eE}}$ | ||