Subfield: Electromagnetism

Problem3. Simplest model of gas discharge

An electric current flowing through a gas is called a gas discharge. There are many types of gas discharges including glow discharge in lighting lamps, arc discharge inwelding and the well known spark discharge that occurs between the clouds and the earth in the form of lightning.

Part A. Non-self-sustained gas discharge 

In this part of the problem the so-called non-self-sustained gas discharge is studied. To maintain it a permanent operationan external ionizer is needed, which creates $Z_{\text {ext }}$ pairs of singly ionized ions and free electrons per unit volume and per unit time uniformly in the volume.

When an external ionizer is switched on, the number of electrons and ions starts to grow. Unlimited increase in the number densities of electrons and ions in the gas is prevented by the recombination process in which a free electron recombineswith an ion to form a neutral atom. The number of recombining events $Z_{\text {rec }}$ that occurs in the gas per unit volume and per unit time is given by

$$
Z_{\mathrm{rec}}=r n_{e} n_{i},
$$

where $r$ is a constant called the recombination coefficient, and $n_{e}, n_{i}$ denotethe electron and ion number densities, respectively.

Suppose that at time $t=0$ the external ionizer is switched on and the initial number densities of electrons and ions in the gas are both equal to zero. Then, the electron number density $n_{e}(t)$ depends on timetas follows:

$$
n_{e}(t)=n_{0}+a \tanh b t
$$

where $n_{0}, a$ and $b$ are some constants, and $\tanh x$ stands for the hyperbolic tangent.
Context question:
A1 Find $_{0}, a, b$ and express them in terms of $Z_{\text {ext }}$ and $r$.
Context answer:
\boxed{$n_{0}=0$ , $a=\sqrt{\frac{Z_{e x t}}{r}}$ , $b=\sqrt{r Z_{e x t}}$}


Extra Supplementary Reading Materials:

Assume that there are two external ionizers available. When the first one is switched on, the electron number density in the gas reaches its equilibrium value of be $n_{e 1}=12 \cdot 10^{10} \mathrm{~cm}^{-3}$. When the second external ionizer is switched on, the electron number density reaches its equilibrium value of $n_{e 2}=16$. $10^{10} \mathrm{~cm}^{-3}$.
Context question:
A2 Find the electron number density $n_{e}$ at equilibrium when both external ionizers are switched on simultaneously.
Context answer:
\boxed{$20.0 \cdot 10^{10}$}


Extra Supplementary Reading Materials:

Attention!In what follows it is assumed that the external ionizer is switched on for quite long period of time such that all processes have become stationary and do not depend on time. Completely neglect the electric field due the charge carriers.

Assume that the gas fills in the tube between the two parallel conductive plates of area $S$ separated by the distance $L \ll \sqrt{S}$ from each other. The voltage $U$ is applied across the plates to create an electric field between them. Assume that the number densities of both kinds of charge carriers remain almost constant along the tube.

Assumethat both the electrons (denoted by thesubscripte) and the ions (denoted by the subscripti) acquire the same ordered speed $v$ due to the electric field strength $E$ found as

$$
v=\beta E
$$

where bisa constant called charge mobility.

Question: A3 Express the electric current $I$ in the tube in terms of $U, \beta, L, S, Z_{\mathrm{ext}}, r$ and $e$ which is the elementary charge.

Solution:
In the steady state, the balance equations of electrons and ions in the tube volume take the form

$$
Z_{\text {ext }} S L=r n_{e} n_{i} S L+\frac{I_{e}}{e}
\tag{A3.1}
$$
$$
Z_{\text {ext }} S L=r n_{e} n_{i} S L+\frac{I_{i}}{e}
\tag{A3.2}
$$

It follows from equations (A3.1) and (A3.2) that the ion and electron currents are equal, i.e.

$$
I_{e}=I_{i}
\tag{A3.3}
$$

At the same time the total current in each tube section is the sum of the electron and ion currents

$$
I=I_{e}+I_{i}
\tag{A3.4}
$$

By definition ofthe current density the following relations hold

$$
I_{e}=\frac{I}{2}=e n_{e} v S=e \beta n_{e} E S
\tag{A3.5}
$$
$$
I_{i}=\frac{I}{2}=e n_{i} v S=e \beta n_{i} E S
\tag{A3.6}
$$

Substituting (A3.5) and (A3.6) into (A3.1) and (A3.2), the following quadratic equation for the current is derived

$$
Z_{e x t} S L=r S L\left(\frac{I}{2 e \beta E S}\right)^{2}+\frac{I}{2 e}
\tag{A3.7}
$$

The electric field strength in the gas is equal to

$$
E=\frac{U}{L}
\tag{A3.8}
$$

and solution to the quadratic equation (A3.7) takes the form

$$
I=\frac{e \beta^{2} U^{2} S}{r L^{3}}\left(-1 \pm \sqrt{1+\frac{4 r Z_{e x t} L^{4}}{\beta^{2} U^{2}}}\right)
\tag{A3.9}
$$



It is obvious that only positive root does make sense, i.e.

$$
I=\frac{e \beta^{2} U^{2} S}{r L^{3}}\left(\sqrt{1+\frac{4 r Z_{e x t} L^{4}}{\beta^{2} U^{2}}}-1\right)
\tag{A3.10}
$$

Final answer: $I=\frac{e \beta^{2} U^{2} S}{r L^{3}}(\sqrt{1+\frac{4 r Z_{e x t} L^{4}}{\beta^{2} U^{2}}}-1)$