Plasma Sheath

Source lecture(s): pc368_lec06_shealth

Intuition

A plasma cannot touch a solid wall directly. Because electrons are lighter and faster, they would otherwise rush to the wall and charge it negatively, repelling further electron flow while pulling in ions. The result is a sheath—a non-neutral boundary layer that sustains a potential drop and enforces charge neutrality across the plasma–wall interface.

Formal Definition

A sheath is a space-charge region adjacent to a material surface where quasi-neutrality is violated to satisfy the Bohm sheath criterion:

\[c_s = \sqrt{\frac{\gamma_e T_e + \gamma_i T_i}{m_i}}\]

Ions must enter the sheath with at least the sound speed $c_s$.

Mathematical Formulation

In a stationary, collisionless sheath with isothermal electrons:

Ion continuity: $n_i v_i = \text{const}$
Ion momentum: $m_i v_i \frac{dv_i}{dx} = -e \frac{d\phi}{dx}$
Electron Boltzmann: $n_e(x) = n_{e0} \exp\Bigl(\frac{e\phi(x)}{k_B T_e}\Bigr)$
Poisson: $\frac{d^2\phi}{dx^2} = -\frac{e}{\varepsilon_0}\bigl[n_i(x) - n_e(x)\bigr]$

Derivation

Assume ions enter the sheath at $x=0$ with $v_i(0) \ge c_s$.
Integration of ion momentum gives the Bohm–Chen relation:

$$\frac{1}{2}m_i v_i^2 + e\phi(x) = \frac{1}{2}m_i c_s^2$$

Combine continuity ($n_i = J_i / v_i$) with Boltzmann electrons.
Substitute into Poisson. The resulting integral describes the sheath potential.
The only self-consistent solution requires $v_i(0) = c_s$ exactly at the sheath edge ($\phi=0$).

Worked Example

Langmuir probe: A cylindrical probe biased negative relative to the plasma draws ion current $I_i = en_i A c_s$. The electron current is exponentially suppressed by the sheath drop. If $T_e = 2$ eV and $m_i = 40$ amu (argon), compute $c_s$. $c_s = \sqrt{(\gamma_e T_e)/m_i} \approx \sqrt{1\times 2\,\text{eV} / (40\times 1836 m_e)} \approx 1.6\times 10^3\,\text{m/s}$.

Common Mistakes

Sheath is a material boundary condition. It is not a collisionless plasma wave.
Ignoring electron inertia. Electrons are Boltzmann-distributed; ions are the dynamical carriers.
Ion supersonic entry. Sub-sonic ions are reflected; the sheath requires $v_i \ge c_s$ at its edge.

Quiz Questions

Conceptual: Why does a plasma “see” a solid surface as a prospectively negative plane?
Computational: Derive the Bohm criterion from Poisson’s equation for a planar, collisionless, isothermal sheath.
MCQ: The entrance velocity into a sheath must be:
A) $v_i < c_s$
B) $v_i = c_s$
C) $v_i > c_s$
D) $v_i = 0$