Motion of charged particles in an electric field > Motion in an alternating electric field > Problem 7.3.11

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
$$
v=\int_0^t a \, dt \tag{1}
$$
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:
$$
\bar{v}=\frac{eE_0}{m_e\omega}\cos\varphi.
$$

There is a typo in the author's answer.

Answer

$$
v_{max}=\frac{eE_0}{m_e\omega}(1+|\cos\varphi|)
$$
$$
\bar{v}=\frac{eE_0}{m_e\omega}\cos\varphi
$$