Debye Shielding

Plasmas generally do not contain strong electric fields in their rest frames. The shielding of an external electric field from the interior of a plasma can be viewed as a result of high plasma conductivity. According to this explanation, electrical current can generally flow freely enough through a plasma to short out any interior electric fields. However, it is more useful to consider the shielding as a dielectric phenomenon. According to this explanation, it is the polarization of the plasma medium, and the associated redistribution of space charge, that prevents penetration by an external electric field. Not surprisingly, the lengthscale associated with such shielding is the Debye length.

Let us consider the simplest possible example. Suppose that a quasi-neutral plasma is sufficiently close to thermal equilibrium that the number densities of its two species are distributed according to the Maxwell-Boltzmann law (Reif 1965),

$$n_s = n_0\,\exp\left(-e_s\,{\mit\Phi}/T\right),$$ (1.11)

where ${\mit\Phi}({\bf r})$ is the electrostatic potential, and $n_0$ and $T$ are constant. From $e_i = - e_e = e$ , it is clear that quasi-neutrality requires the equilibrium potential to be zero. Suppose that the equilibrium potential is perturbed, by an amount $\delta{\mit\Phi}({\bf r})$ , as a consequence of a small, localized, perturbing charge density, $\delta\rho_{\rm ext}$ . The total perturbed charge density is written

$$\delta\rho = \delta\rho_{\rm ext} + e\,(\delta n_i - \delta n_e) \simeq \delta\rho_{\rm ext} -2 \,e^2\, n_0 \,\delta{\mit\Phi}/T.$$ (1.12)

Thus, Poisson's equation yields

$$\nabla^2 \delta {\mit\Phi} = - \frac{\delta\rho}{\epsilon_0} = -\... ...{\delta\rho_{\rm ext} -2 \,e^2\, n_0\, \delta{\mit\Phi}/T}{\epsilon_0} \right),$$ (1.13)

which reduces to

$$\left(\nabla^2 - \frac{2}{\lambda_D^{\,2}}\right) \delta {\mit\Phi} = - \frac{\delta\rho_{\rm ext}}{\epsilon_0}.$$ (1.14)

If the perturbing charge density actually consists of a point charge $q$ , located at the origin, so that $\delta\rho_{\rm ext} = q\,\delta({\bf r})$ , then the solution to the previous equation is written

$$\delta{\mit\Phi}(r) = \frac{q}{4\pi\,\epsilon_0\,r}\,\exp\left(-\frac{\sqrt{2}\,r}{\lambda_D}\right).$$ (1.15)

This expression implies that the Coulomb potential of the perturbing point charge $q$ is shielded over distances longer than the Debye length by a shielding cloud of approximate radius $\lambda_D$ that consists of charge of the opposite sign.

By treating $n$ as a continuous function, the previous analysis implicitly assumes that there are many particles in the shielding cloud. Actually, Debye shielding remains statistically significant, and physical, in the opposite limit in which the cloud is barely populated. In the latter case, it is the probability of observing charged particles within a Debye length of the perturbing charge that is modified (Hazeltine and Waelbroeck 2004).