$7.3.11.$ 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
$$
v=\int_0^t a \, dt \tag{1}
$$
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
$$
\bar{v}=\frac{1}{T}\int_0^T v(t) \, dt \tag{4}
@@ -24,7 +24,7 @@Solution
$$
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:
$$
−
\bar{v}=\frac{2eE_0}{m_e\omega}\cos\varphi.
+
\bar{v}=\frac{eE_0}{m_e\omega}\cos\varphi.
$$
There is a typo in the author's answer.
#### Answer
$$
@@ -34,5 +34,5 @@Answer
v_{max}=\frac{eE_0}{m_e\omega}(1+|\cos\varphi|)
$$
$$
−
\bar{v}=\frac{2eE_0}{m_e\omega}\cos\varphi
+
\bar{v}=\frac{eE_0}{m_e\omega}\cos\varphi
$$
### 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.
$7.3.11.$ 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
### Solution
The initial velocity is zero, therefore
The initial velocity is zero, therefore
$$
$$
v=\int_0^t a \, dt \tag{1}
v=\int_0^t a \, dt \tag{1}
$$
$$
Newton's second law:
Newton's second law:
$$
$$
F=-Ee=am_e \tag{2}
F=-Ee=am_e \tag{2}
$$
$$
Let $a_0=\frac{eE_0}{m_e}$, then from (2) into (1):
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}
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
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|).
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
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}
\bar{v}=\frac{1}{T}\int_0^T v(t) \, dt \tag{4}
@@ -24,7 +24,7 @@Solution
$$
$$
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:
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:
