"배워서 남주자"라는 가치를 가진 산동일크무크를 응원해주세요!
Debye Shielding
Consider a test charge \(e\) and electron cloud around it in a singly charged plasma. Assume that
\(\bullet\) The ions are fixed; \(\frac{m_i}{m_e} \rightarrow \infty\)
\[\begin{aligned} \label{eq_6} n_i(r)=n \end{aligned}\]
\(\bullet\) The electrons obey Boltzman relation. In the presence of a potential energy \(q\phi\), the electron distribution function is
\[f(u)=A e^{-\frac{\frac{1}{2}mu^2+q\phi}{KT_e}}\]
Integrate this over \(du\)
\[\begin{aligned} \label{eq_7} n_e(r)=ne^{\frac{e\phi(r)}{K_BT}} \end{aligned}\]
Poisson’s equation in one dimension is
\[\begin{aligned} \epsilon_0 \nabla^2 \phi = \epsilon_0 \frac{d^2 \phi}{dx^2}=-e(n_i - n_e) \end{aligned}\]
substitution in to the Poisson’s equation,
\[\begin{aligned} \nabla^2 \phi(r) = \frac{e}{\epsilon_0}n(e^{\frac{e\phi(r)}{KT}}-1)\end{aligned}\]
assume that \(|\frac{e\phi}{KT}|<<1\), then by using taylor expansion; \(e^{\frac{e\phi(r)}{KT}}=1+\frac{e\phi(r)}{KT} +...\)
\[\begin{aligned} \nabla^2 \phi(r) &= \frac{e^2n}{\epsilon_0 KT}\phi(r)\\ &=\frac{\phi(r)}{\lambda_D^2}\end{aligned}\]
here we define Debye length \(\lambda_D\) as
\[\lambda_D = \sqrt{\frac{\epsilon_0 K T }{ne^2}}\]
In spherical coordinate
\[\begin{aligned} \frac{1}{r^2}\frac{d}{dr}\left( r^2 \frac{d\phi}{dr}\right) -\frac{\phi}{\lambda_D^2}&=0\\ \phi^{\prime \prime} + \frac{2}{r}\phi^{\prime} - \frac{1}{\lambda_D^2}\phi&=0\\ r\phi^{\prime \prime} + 2\phi^{\prime} - \frac{1}{\lambda_D^2}r\phi&=0\end{aligned}\]
Let \(\psi(r)=r\phi\), then \(\psi^{\prime}=\phi+r\phi^{\prime}\) and \(\psi^{\prime \prime}=2\phi^{\prime}+r\phi^{\prime \prime}\) so that
\[\psi^{\prime \prime}-\frac{1}{\lambda_D^2}\psi =0\]
\[\begin{aligned} \psi(r) &= C_1 e^{-\frac{r}{\lambda_D}}+C_2 e^{\frac{r}{\lambda_D}}\\ \phi(r) &= \frac{C_1}{r} e^{-\frac{r}{\lambda_D}}+\frac{C_2}{r} e^{\frac{r}{\lambda_D}}\end{aligned}\]
Applying the boundary conditions,
\[\begin{aligned} \phi &\rightarrow 0 \quad \quad \quad \quad ,\quad \quad r \rightarrow \infty \\ \phi &\rightarrow \frac{e}{4\pi\epsilon_0 r} \quad \quad , \quad \quad r \rightarrow 0\end{aligned}\]
we can find the constants.
\[C_1 = \frac{e}{4\pi \epsilon_0} \quad \quad \quad C_2 = 0\]
Hence, the solution is given by
\[\begin{aligned} \label{eq_debyelength} \boxed{\phi(r) = \frac{e}{4\pi\epsilon_0 r}e^{-\frac{r}{\lambda_D}}}\end{aligned}\]
The quantity \(\lambda_D\), called the Debye length, is a measure of the shielding distance or thickness of the sheath over which the influence of an individual charged particle is dominant.
\(\bullet\) \(\lambda_D\) = how long the shielding is effective.
\(\bullet\) Notice that electron temperature is used to define Debye length because it is more mobile; most of time this is true.
\(\bullet\) Useful forms of Eq.([eq_debyelength]) are
\(\lambda_D = 69\sqrt{\frac{T_e}{n}}[m]\) \(T_e\) in \(^{\circ}K\)
\(\lambda_D = 7430\sqrt{\frac{KT_e}{n}}[m]\) \(KT_e\) in \(eV\)
\(\bullet\) Trend
effective shielding \(\quad n \uparrow\) \(\lambda_D \downarrow\)
poor shielding \(\quad T \uparrow\) \(\lambda_D \uparrow\)
4차 산업혁명에 걸맞는 인터넷 기반 고등교육기관
이메일: ilkmooc@ilkmooc.kr
홈페이지주소: http://ilkmooc.kr
산동일크무크란? https://goo.gl/FnvqXd
댓글 없음:
댓글 쓰기