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Added explanation and derivation clarifying the relation between electron density and conductance by relating to Ohm's Law.
iwantmyphd
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In answer to your first question, the electron density is how many free (not bound to an atom) electrons there are in a given volume. Since all plasmas have some degree of ionization, this means that there are electrons that have been stripped from atoms, and are moving around, while the atoms are converted into ions.

As far as units, in plasma physics, it can depend somewhat on the size of the plasma, or simply preference. Typically scientists and engineers use electrons per $\mathrm{cm}^{3}$ or electrons per $\mathrm m^{3}$. In the literature, the electron density is written as a number followed by $\mathrm{cm}^{-3}$ or $\mathrm m^{-3}$. For instance "the electron density of the argon DC glow discharge was $3.5\times10^{11} \mathrm{cm^{-3}}$". The choice of $\mathrm{cm^{-3}}$ or $\mathrm m^{-3}$ stems from using either CGS units or MKS units. It often depends on the length and volume scales that apply to the situation. So, for laboratory-sized plasmas, $\mathrm{cm^{-3}}$ are often used, while for things like inter-stellar or inter-galactic medium, $\mathrm m^{-3}$ makes more sense.

With regard to your follow up question on how electron density is related to electrical conductivity, the electrical conductivity is equal to the electron density multiplied by the elementary charge (the electron charge), multiplied by the electron mobility.

$$ \sigma = n \times e \times \mu $$

The electron mobility, $\mu$, relates the velocity of charged particles in a uniform electric field ($\mu=v/E)$. When these equations are combined, we find:

$$ \sigma = n \times e \times v/E $$

The numerator is in units of charge per area per unit time (or current density, J), and the denominator is electric field, E. Re-arranging and solving for J:

$$ J = \sigma \times E $$

which is Ohm's Law, confirming that the first equation relates electron density to electrical conductivity.

iwantmyphd
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