Question 3.5: A problem that is found in the study of a plasma is to compu...

Due to the spherical symmetry of the system, we can neglect any variation of potential in two of the coordinates (\theta, \phi)  and write Poisson’s equation in spherical coordinates with the assistance of (3.9), using only the dependence on the radial coordinate  ρ , where the charge distribution is given by \rho_{ v }(\rho)=\left( n _{ e }(\rho)- n _{0}\right) q  :

\nabla^{2} V=\frac{1}{\rho^{2}} \frac{\partial}{\partial \rho}\left(\rho^{2} \frac{\partial V}{\partial \rho}\right)+\frac{1}{\rho^{2} \sin \theta} \frac{\partial}{\partial \theta}\left(\sin \theta \frac{\partial V}{\partial \theta}\right)+\frac{1}{\rho^{2} \sin ^{2} \theta} \frac{\partial^{2} V}{\partial \phi^{2}}         (3.9)

\frac{1}{\rho^{2}} \frac{d}{d \rho}\left(\rho^{2} \frac{d V}{d \rho}\right)=-\frac{\rho_{v}}{\varepsilon_{0}} \equiv-\frac{n_{0} q\left(\exp \left(-\frac{q V(\rho)}{k_{B} T_{e}}\right)-1\right)}{\varepsilon_{0}}

This differential equation is non-linear one, which has to be solved numerically in general – this is considered in the next section. The charge density has to be linearized in order to be solved analytically

-\frac{\rho}{\varepsilon_{0}}=-\frac{ n _{0} q }{\varepsilon_{0}}\left\{\left[1-\frac{ qV }{ k _{ B } T _{ e }}+\ldots\right]-1\right\}=\frac{ n _{0} q ^{2}}{\varepsilon_{0} k _{ B } T _{ e }} V

or

  \frac{1}{\rho^{2}} \frac{d}{d \rho}\left(\rho^{2} \frac{d V}{d \rho}\right)=\frac{V}{\lambda_{D}^{2}}

where \lambda_{ D }=\sqrt{\varepsilon_{0} k _{ B } T_{ e } /\left( n _{0} q ^{2}\right)}  is the Debye length. We have used a small potential expansion for the exponential term: \exp (\pm x) \approx 1 \pm x . The solution of this equation is facilitated if we define a new variable W = ρV . Substitute this variable into this equation to obtain

\frac{d}{d\rho}\left(\rho \frac{d W}{d \rho}-W\right)=\frac{\rho W} {\lambda_{D}^{2}}

or

\frac{d^{2} W}{d \rho^{2}}=\frac{W}{\lambda_{D}^{2}}

The solution of this equation which is finite when \rho \rightarrow \infty   is

W=\exp \left(-\frac{\rho}{\lambda_{D}}\right) \text { or } V=\frac{1}{\rho} \exp \left(-\frac{\rho}{\lambda_{D}}\right) \text {. }

This states that the effect of the additional charge will be “screened” away in a few Debye lengths.

We can think of the Debye length as being analogous to a “time constant” in an electric circuit that has a potential that decays to zero in a few time constants.