Правка разделов «Statement», «Solution», «Answer»
en/14.4.12.md
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| @@ -1,7 +1,11 @@ | |||
| ### Statement | |||
| − | $14.4.12.$ | ||
| + | $14.4.12.$ | ||
| + | An electron flying at velocity v into a uniform electric field flying towards it | ||
| + | at velocity u penetrates the field to a depth of l. Determine the field strength | ||
| + | if it is directed along the electron velocity. | ||
| + | |||
| ### Solution | |||
| We work in the system S' where the electric field is at rest. In S', the electron moves with speed v' and is decelerated by the constant field E. The penetration depth in S' is l', and the work–energy relation gives: | |||
| $eE\,l' = (\gamma' - 1)m_e c^2, \qquad \gamma' = \frac{1}{\sqrt{1 - v'^2/c^2}}$ | |||
| In the laboratory, the electron travels with speed v and the field with speed u (in opposite directions). In S': | |||
| $v' = \frac{v + u}{1 + \dfrac{vu}{c^2}}$ | |||
| The penetration distance measured in the laboratory l is contracted with respect to l': | |||
| $l = \frac{l'}{\gamma_u}, \qquad \gamma_u = \frac{1}{\sqrt{1 - u^2/c^2}}$ | |||
| From the velocity composition law, it follows: | |||
| @@ -20,7 +24,7 @@Solution | |||
| $\gamma' = \gamma\gamma_u\left(1 + \frac{uv}{c^2}\right), \qquad \gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$ | |||
| − | Substituting$ l' = \gamma_u l$ and \gamma' $into the energy equation and dividing by $\gamma_u$ | ||
| + | Substituting$ l' = \gamma_u l$ and $\gamma' $into the energy equation and dividing by $\gamma_u$ | ||
| $E = \frac{m_e c^2}{e l} \left[ \gamma\left(1 + \frac{uv}{c^2}\right) - \frac{1}{\gamma_u} \right]$ | |||
| Replacing $ \gamma = 1/\sqrt{1 - v^2/c^2} and 1/\gamma_u = \sqrt{1 - u^2/c^2} $ | |||
| $\boxed{E = \frac{m_e c^2}{e l} \left( \frac{1 + \dfrac{uv}{c^2}}{\sqrt{1 - \dfrac{v^2}{c^2}}} - \sqrt{1 - \dfrac{u^2}{c^2}} \right)}$ | |||
| @@ -30,4 +34,4 @@Solution | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | $\boxed{E = \frac{m_e c^2}{e l} \left( \frac{1 + \dfrac{uv}{c^2}}{\sqrt{1 - \dfrac{v^2}{c^2}}} - \sqrt{1 - \dfrac{u^2}{c^2}} \right)}$ | ||
| @@ -1,7 +1,11 @@ | |||
| ### Statement | ### Statement | ||
| $14.4.12.$ |
$14.4.12.$ | ||
| An electron flying at velocity v into a uniform electric field flying towards it | |||
| at velocity u penetrates the field to a depth of l. Determine the field strength | |||
| if it is directed along the electron velocity. | |||
| ### Solution | ### Solution | ||
| We work in the system S' where the electric field is at rest. In S', the electron moves with speed v' and is decelerated by the constant field E. The penetration depth in S' is l', and the work–energy relation gives: | We work in the system S' where the electric field is at rest. In S', the electron moves with speed v' and is decelerated by the constant field E. The penetration depth in S' is l', and the work–energy relation gives: | ||
| $eE\,l' = (\gamma' - 1)m_e c^2, \qquad \gamma' = \frac{1}{\sqrt{1 - v'^2/c^2}}$ | $eE\,l' = (\gamma' - 1)m_e c^2, \qquad \gamma' = \frac{1}{\sqrt{1 - v'^2/c^2}}$ | ||
| In the laboratory, the electron travels with speed v and the field with speed u (in opposite directions). In S': | In the laboratory, the electron travels with speed v and the field with speed u (in opposite directions). In S': | ||
| $v' = \frac{v + u}{1 + \dfrac{vu}{c^2}}$ | $v' = \frac{v + u}{1 + \dfrac{vu}{c^2}}$ | ||
| The penetration distance measured in the laboratory l is contracted with respect to l': | The penetration distance measured in the laboratory l is contracted with respect to l': | ||
| $l = \frac{l'}{\gamma_u}, \qquad \gamma_u = \frac{1}{\sqrt{1 - u^2/c^2}}$ | $l = \frac{l'}{\gamma_u}, \qquad \gamma_u = \frac{1}{\sqrt{1 - u^2/c^2}}$ | ||
| From the velocity composition law, it follows: | From the velocity composition law, it follows: | ||
| @@ -20,7 +24,7 @@Solution | |||
| $\gamma' = \gamma\gamma_u\left(1 + \frac{uv}{c^2}\right), \qquad \gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$ | $\gamma' = \gamma\gamma_u\left(1 + \frac{uv}{c^2}\right), \qquad \gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$ | ||
| Substituting$ l' = \gamma_u l$ and \gamma' $into the energy equation and dividing by $\gamma_u$ | Substituting$ l' = \gamma_u l$ and $\gamma' $into the energy equation and dividing by $\gamma_u$ | ||
| $E = \frac{m_e c^2}{e l} \left[ \gamma\left(1 + \frac{uv}{c^2}\right) - \frac{1}{\gamma_u} \right]$ | $E = \frac{m_e c^2}{e l} \left[ \gamma\left(1 + \frac{uv}{c^2}\right) - \frac{1}{\gamma_u} \right]$ | ||
| Replacing $ \gamma = 1/\sqrt{1 - v^2/c^2} and 1/\gamma_u = \sqrt{1 - u^2/c^2} $ | Replacing $ \gamma = 1/\sqrt{1 - v^2/c^2} and 1/\gamma_u = \sqrt{1 - u^2/c^2} $ | ||
| $\boxed{E = \frac{m_e c^2}{e l} \left( \frac{1 + \dfrac{uv}{c^2}}{\sqrt{1 - \dfrac{v^2}{c^2}}} - \sqrt{1 - \dfrac{u^2}{c^2}} \right)}$ | $\boxed{E = \frac{m_e c^2}{e l} \left( \frac{1 + \dfrac{uv}{c^2}}{\sqrt{1 - \dfrac{v^2}{c^2}}} - \sqrt{1 - \dfrac{u^2}{c^2}} \right)}$ | ||
| @@ -30,4 +34,4 @@Solution | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | $\boxed{E = \frac{m_e c^2}{e l} \left( \frac{1 + \dfrac{uv}{c^2}}{\sqrt{1 - \dfrac{v^2}{c^2}}} - \sqrt{1 - \dfrac{u^2}{c^2}} \right)}$ | ||