| @@ -1,40 +1,38 @@ |
| | − | ## I confused the task number with 7.3.11. I'll post this solution tomorrow. |
| | + | ### Condition |
| | + | $7.3.13^*$. A rarefied plasma in a high‑frequency electric field of strength $E = E_0 \sin \omega t$ acquires a positive potential. Determine this potential, given that the ion mass $M \gg m_e$. |
| | | |
| | − | ### 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. |
| | − | |
| | | ### Solution |
| | − | The initial velocity is zero, therefore |
| | + | |
| | + | Since the ion mass is much larger than the electron mass, the ions practically do not have time to shift during the period of the high‑frequency field. |
| | + | |
| | + | The electrons, on the contrary, are easily driven by the field and perform forced oscillations about the ions. However, at the boundaries of the plasma, these oscillations cause the accelerated electrons to leave the volume. As a result, an excess positive charge (the ions remain, some electrons have left) appears inside the plasma. |
| | + | |
| | + | This positive charge creates a retarding electric field. The plasma charges itself until this potential barrier stops the fastest electrons from escaping. |
| | + | |
| | + | From the solution of problem [7.3.11](https://savchenkosolutions.com/en/7.3.11) one can obtain the expression for the maximum electron velocity: |
| | | $$ |
| | − | v=\int_0^t a \, dt \tag{1} |
| | + | v_{\max} = \frac{2eE_0}{m_e\omega} \tag{1} |
| | | $$ |
| | − | Newton's second law: |
| | + | The condition for stopping the electron leakage is: |
| | | $$ |
| | − | F=-Ee=am_e \tag{2} |
| | + | W = e\varphi = \frac{1}{2}m_e v_{\max}^2 \tag{2} |
| | | $$ |
| | − | Let $a_0=\frac{eE_0}{m_e}$, then from (2) into (1): |
| | | $$ |
| | − | v(t)=-a_0\int_0^t \sin (\omega t + \varphi) \, dt=\frac{eE_0}{m_e\omega} \left(\cos (\omega t + \varphi)-\cos \varphi\right) \tag{3} |
| | + | \varphi = \frac{2e}{m_e}\left(\frac{E_0}{\omega}\right)^2 \tag{3} |
| | | $$ |
| | − | 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 |
| | + | |
| | + | Important: here it is assumed, as in [7.3.11](https://savchenkosolutions.com/en/7.3.11), that initially all electrons are at rest and the field is switched on instantaneously. If one considers a steady‑state regime, the maximum velocity will be determined simply by the oscillation amplitude: |
| | | $$ |
| | − | v_{max}=\frac{eE_0}{m_e\omega}(1+|\cos\varphi|). |
| | + | v_{\max} = \frac{eE_0}{m_e\omega} \tag{4} |
| | | $$ |
| | − | It is clear that the velocity changes periodically with period $T=\frac{2\pi}{\omega}$. Then the average value is |
| | + | and the answer will be |
| | | $$ |
| | − | \bar{v}=\frac{1}{T}\int_0^T v(t) \, dt \tag{4} |
| | + | \varphi = \frac{e}{2m_e}\left(\frac{E_0}{\omega}\right)^2 \tag{5} |
| | | $$ |
| | − | 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: |
| | − | $$ |
| | − | \bar{v}=\frac{2eE_0}{m_e\omega}\cos\varphi. |
| | − | $$ |
| | | |
| | − | There is a typo in the author's answer. |
| | + | In the author's answer there is a misprint in any case, but it is closer to (3). |
| | | |
| | | #### Answer |
| | | $$ |
| | − | v_{max}=\frac{eE_0}{m_e\omega}(1+|\cos\varphi|) |
| | − | $$ |
| | − | $$ |
| | − | \bar{v}=\frac{2eE_0}{m_e\omega}\cos\varphi |
| | + | \varphi = \frac{2e}{m_e}\left(\frac{E_0}{\omega}\right)^2 \quad \text{or} \quad \varphi = \frac{e}{2m_e}\left(\frac{E_0}{\omega}\right)^2 \tag{5} |
| | | $$ |
