### Solution

A1  ?? If we assume that the velocity of the electrons at a certain point is $\vec{v}$, what is the current density $\vec{j}_t$ at that point?

Answer: $\vec{j}_t = -ne\vec{v}$

A2  ?? How to express the total current $\vec{j}_t$ from the conduction current $\vec{j}$ and the time derivative of the polarization density $\partial \vec{P}/\partial t$.

Answer: $\vec{j}_t = \vec{j} + \frac{\partial \vec{P}}{\partial t}$

A3  ?? Show that for the harmonic external field $\tilde{E} e^{-i\omega t}$ with the angular frequency $\omega$, the redefinition of the $\varepsilon$ is
$\tilde{\varepsilon} = \varepsilon + \frac{i \sigma}{\omega \varepsilon_0}.$

Answer: $\vec{j} = \sigma \vec{E}, \quad \frac{\partial \vec{P}}{\partial t} = - i \omega \varepsilon_0 (\varepsilon - 1) \vec{E}$
Then
$- i \omega \varepsilon_0 (\tilde{\varepsilon} - 1) = \sigma - i \omega \varepsilon_0 (\varepsilon - 1)$

A4  ?? Using that for a spherically symmetric function, the Lapplacian could be written as
$\Delta \Phi = \frac{1}{r^2} \frac{d}{dr} \left( r^2 \frac{d \Phi}{dr} \right),$
calculate the $\lambda_D$. Express your answer in terms of $\langle n \rangle$ and $T$. Estimate the value of $\lambda_D$ in the gas discharge, where$\langle n \rangle = 10^{16}~\text{м}^{-3}$, $T=10^4~\text{K}$.

Let's calculate the Lapplacian of
$\Phi(r) = \frac{q}{4\pi \varepsilon_0 r} e^{-r/\lambda_D}$
$\begin{split} \Delta \Phi &= \frac{q}{4\pi \varepsilon_0 r^2} \frac{d}{dr} \left( r^2 \left[ -\frac{e^{-r/\lambda_D}}{r\lambda_D} - \frac{e^{-r/\lambda_D}}{r^2} \right] \right) = -\frac{q}{4\pi \varepsilon_0 r^2} \frac{d}{dr} \left( \left[ \frac{r}{\lambda_D} + 1 \right] e^{-r/\lambda_D}\right) =\\&= -\frac{q}{4\pi \varepsilon_0 r^2} \left( \frac{1}{\lambda_D} e^{-r/\lambda_D} - \left[ 1 + \frac{r}{\lambda_D} \right] \frac{1}{\lambda_D} e^{-r/\lambda_D} \right) = \frac{q}{4\pi \varepsilon_0 r \lambda_D^2} e^{-r/\lambda_D} = \frac{\Phi(r)}{\lambda_D^2} \end{split}$

Substitution into the equation, introduced in the problem, gives us
$\frac{\Phi(r)}{\lambda_D^2} \simeq \frac{e \langle n \rangle}{\varepsilon_0} \left[ 1 + \frac{e \Phi}{k_\text{B} T} - 1 + \frac{e\Phi}{k_\text{B} T}\right] = \frac{2e^2 \langle n \rangle \Phi(r)}{\varepsilon_0 k_\text{B} T}.$

Finally,
$\lambda_D = \sqrt{\frac{\varepsilon_0 k_\text{B} T}{2e^2\langle n \rangle}}$

Answer: $\lambda_D = \sqrt{\frac{\varepsilon_0 k_\text{B} T}{2e^2\langle n \rangle}} = 0.05~\text{mm}.$

A5  ?? Estimate the average number of particles $N_\text{loc}$ whose field is not shielded at the given point in the plasma.

The number $N_\text{loc}$ could be estimated as the number of particles in the sphere with radius equal to the Debye length:
$N_\text{loc} = 2 \langle n \rangle \cdot \frac{4}{3} \pi \lambda_D^3 = \frac{8\pi}{3 \langle n \rangle^{1/2}} \left( \frac{\varepsilon_0 k_\text{B}T}{2e^2} \right)^{3/2}$

Answer: $N_\text{loc} = \frac{8\pi}{3 \langle n \rangle^{1/2}} \left( \frac{\varepsilon_0 k_\text{B}T}{2e^2} \right)^{3/2}$

A6  ?? Analyze the motion of the electron in the plasma in the external field $\tilde{E}e^{-i\omega t}$ and find the relative permittivity $\tilde{\varepsilon}$ of the plasma.

We have the second Newton's law $m\vec{a} = -e\tilde{E}e^{-i \omega t}$. A straight integration over time gives us
$\vec{v} = \vec{v}_0 + \frac{e\tilde{E}}{i m \omega} e^{-i\omega t}, \quad \Rightarrow \quad \vec{j}_t = \frac{ie^2n}{m\omega} \vec{E}_0,$
where we use that $\vec{v}_0=0$ i.e. it the absence of the $\tilde{E}$ electrons don't move. In the result we have:
$\tilde{\varepsilon}=1-\frac{e^2n}{m\varepsilon_0 \omega^2}$

Answer: $\tilde{\varepsilon}=1-\frac{e^2n}{m\varepsilon_0 \omega^2}$

A7  ?? Find the circular frequency $\omega_p$ of the plasma self-oscillations.

The relative permittivity $\varepsilon$ connects $P$ and $E$:
$P = \varepsilon_0 (\varepsilon-1) E$
and if $\varepsilon=0$ it means that the electric field $E$ is completely generated by the local dipoles! We can also note that the relative perimittivity $\tilde{\varepsilon}$ doesn't have an imaginary part, so $\tilde{\varepsilon}=\varepsilon$. Finally, the circular frequency corresponding to zero relative permittivity is:
$\omega= \sqrt{\frac{m \varepsilon_0}{e^2 n}}.$

Answer: $\omega= \sqrt{\frac{m \varepsilon_0}{e^2 n}}.$