Expression for Conductivity of ionized gas (plasma) at low pressure

Friday, December 28, 2012

Expression for Conductivity of ionized gas (plasma) at low pressure

Expression for Conductivity of ionized gas (plasma) at low pressure

A plasma is an ionized gas. As a rule, plasmas contain free electrons and positive ions. Since the ions are more massive than the electrons, the current is carried almost exclusively by the electrons.

At low pressure, we can ignore the collisions between the electrons and the gas molecules, and hence energy losses.

Considering a plane electromagnetic wave (EM wave) propagating, the motion of the charge depends almost entirely on the electric field (E-field).

The conductivity of the gas is given by

Expression for Conductivity of ionized gas (plasma) at low pressure equation 1

where Jt is the electron current density.

Actually, Jt is the sum if all ions in the gas.

Expression for Conductivity of ionized gas (plasma) at low pressure equation 2

where Ni is the number of electrons or ions per metre cube, Qi is the charge of electrons or ions and dxi/dt (= vi) is the speed of the electron or ion along the x-axis.

Now, we require the velocity v of a free electron or ion of mass m subjected to an alternating electric field.

Note that the time derivative of velocity v here can be written as dv/dt = jwv ; assuming for example that v = v0 exp(jwt) where j represents the imaginary component, w the frequency and t the time.

Expression for Conductivity of ionized gas (plasma) at low pressure equation 3

Taking the expressions for the force, we obtain

$\center\sigma E_0 e^{j\omega t} = j\sum_{i} \frac{N_i Q_i ^2}{\omega m_i}E_0 e^{j\omega t - \frac{\pi}{2}} \\ \\ \\ \sigma = -\frac{j}{\omega}\sum_{i} \frac{N_i Q_i ^2}{m_i}$

v leads the field by 90⁰

Then, replacing this expression of v in Jt , taking the pure imaginary part (the "-90⁰") and equating the 2 equations for Jt, we get

Expression for Conductivity of ionized gas (plasma) at low pressure equation 5

As the mass of the ions is much larger than that of the electrons while the charge is of the same order of magnitude as that of the electrons, the contribution of the ions can be neglected.

Thus, the conductivity can be written in terms of the charge and mass of electrons only.

$\sigma = -j \frac{N_e Q_e ^2}{\omega m_e} = -j 4.47\times 10^{-9} \frac{N_e}{f}\; \; (\Omega m)^{-1}$