Правка разделов «Statement», «Solution», «Answer»
en/7.3.13.md
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| @@ -1,11 +1,38 @@ | |||
| ### Statement | |||
| + | $7.3.11.$ A free electron is subjected to an electric field of strength $E = E_0 \sin (\omega t + \varphi)$ starting from time $t = 0$. Find the maximum and average velocity of the electron. | ||
| − | $7.3.13.$ [Insert the problem statement] | ||
| − | |||
| ### Solution | |||
| + | The initial velocity is zero, therefore | ||
| + | $$ | ||
| + | v=\int_0^t a \, dt \tag{1} | ||
| + | $$ | ||
| + | Newton's second law: | ||
| + | $$ | ||
| + | F=-Ee=am_e \tag{2} | ||
| + | $$ | ||
| + | Let $a_0=\frac{eE_0}{m_e}$, then from (2) into (1): | ||
| + | $$ | ||
| + | v(t)=-a_0\int_0^t \sin (\omega t + \varphi) \, dt=\frac{eE_0}{m_e\omega} \left(\cos (\omega t + \varphi)-\cos \varphi\right) \tag{3} | ||
| + | $$ | ||
| + | Taking into account the range of the cosine, the expression in parentheses can take values $[-1-\cos \varphi;\, 1-\cos \varphi]$. One must be careful here: remember that we need the maximum magnitude of the velocity, and that $\cos \varphi$ can be either positive or negative. Finally, analysing (3), we obtain | ||
| + | $$ | ||
| + | v_{max}=\frac{eE_0}{m_e\omega}(1+|\cos\varphi|). | ||
| + | $$ | ||
| + | It is clear that the velocity changes periodically with period $T=\frac{2\pi}{\omega}$. Then the average value is | ||
| + | $$ | ||
| + | \bar{v}=\frac{1}{T}\int_0^T v(t) \, dt \tag{4} | ||
| + | $$ | ||
| + | I will allow myself to skip the calculation of the integral – remembering that the integral of cosine over its period is zero, this can be done very quickly. Eventually, taking the modulus, we get: | ||
| + | $$ | ||
| + | \bar{v}=\frac{2eE_0}{m_e\omega}\cos\varphi. | ||
| + | $$ | ||
| − | 1 | ||
| + | There is a typo in the author's answer. | ||
| #### Answer | |||
| − | |||
| − | [Insert a concise answer or boxed result] | ||
| + | $$ | ||
| + | v_{max}=\frac{eE_0}{m_e\omega}(1+|\cos\varphi|) | ||
| + | $$ | ||
| + | $$ | ||
| + | \bar{v}=\frac{2eE_0}{m_e\omega}\cos\varphi | ||
| + | $$ | ||
| @@ -1,11 +1,38 @@ | |||
| ### Statement | ### Statement | ||
| $7.3.11.$ A free electron is subjected to an electric field of strength $E = E_0 \sin (\omega t + \varphi)$ starting from time $t = 0$. Find the maximum and average velocity of the electron. | |||
| $7.3.13.$ [Insert the problem statement] | |||
| ### Solution | ### Solution | ||
| The initial velocity is zero, therefore | |||
| $$ | |||
| v=\int_0^t a \, dt \tag{1} | |||
| $$ | |||
| Newton's second law: | |||
| $$ | |||
| F=-Ee=am_e \tag{2} | |||
| $$ | |||
| Let $a_0=\frac{eE_0}{m_e}$, then from (2) into (1): | |||
| $$ | |||
| v(t)=-a_0\int_0^t \sin (\omega t + \varphi) \, dt=\frac{eE_0}{m_e\omega} \left(\cos (\omega t + \varphi)-\cos \varphi\right) \tag{3} | |||
| $$ | |||
| Taking into account the range of the cosine, the expression in parentheses can take values $[-1-\cos \varphi;\, 1-\cos \varphi]$. One must be careful here: remember that we need the maximum magnitude of the velocity, and that $\cos \varphi$ can be either positive or negative. Finally, analysing (3), we obtain | |||
| $$ | |||
| v_{max}=\frac{eE_0}{m_e\omega}(1+|\cos\varphi|). | |||
| $$ | |||
| It is clear that the velocity changes periodically with period $T=\frac{2\pi}{\omega}$. Then the average value is | |||
| $$ | |||
| \bar{v}=\frac{1}{T}\int_0^T v(t) \, dt \tag{4} | |||
| $$ | |||
| I will allow myself to skip the calculation of the integral – remembering that the integral of cosine over its period is zero, this can be done very quickly. Eventually, taking the modulus, we get: | |||
| $$ | |||
| \bar{v}=\frac{2eE_0}{m_e\omega}\cos\varphi. | |||
| $$ | |||
| 1 | There is a typo in the author's answer. | ||
| #### Answer | #### Answer | ||
| $$ | |||
| [Insert a concise answer or boxed result] | v_{max}=\frac{eE_0}{m_e\omega}(1+|\cos\varphi|) | ||
| $$ | |||
| $$ | |||
| \bar{v}=\frac{2eE_0}{m_e\omega}\cos\varphi | |||
| $$ | |||