Motion of charged particles in an electric field > Motion in an alternating electric field > Problem 7.3.13

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 $M \gg m_e$.

Solution

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 one can obtain the expression for the maximum electron velocity:
$$
v_{\max} = \frac{2eE_0}{m_e\omega} \tag{1}
$$
The condition for stopping the electron leakage is:
$$
W = e\varphi = \frac{1}{2}m_e v_{\max}^2 \tag{2}
$$
$$
\varphi = \frac{2e}{m_e}\left(\frac{E_0}{\omega}\right)^2 \tag{3}
$$

Important: here it is assumed, as in 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:
$$
v_{\max} = \frac{eE_0}{m_e\omega} \tag{4}
$$
and the answer will be
$$
\varphi = \frac{e}{2m_e}\left(\frac{E_0}{\omega}\right)^2 \tag{5}
$$

In the author's answer there is a misprint in any case, but it is closer to (3).

Answer

$$
\varphi = \frac{2e}{m_e}\left(\frac{E_0}{\omega}\right)^2 \quad \text{or} \quad \varphi = \frac{e}{2m_e}\left(\frac{E_0}{\omega}\right)^2 \tag{5}
$$