[플라즈마 물리][Plasma Physics]CH1 Introduction - Plasma Frequency

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Plasma Frequency

Consider a hypothetical slab of plasma, where we assume that the ions have infinite mass(immobile) and the electrons can move freely through the ions.

1. Suppose the electron slab is displaced a distance \(x\) to the right of the ion slab and then allowed to move freely.

2. An electric field will be set up, causing the electron slab to be pulled back toward the ions.

3. When the electrons exactly overlap the ions (when \(x=0\)), the net force is zero, but the electron slab overshoots.

4. The net result is harmonic oscillation. The frequency of the oscillation is called the electron plasma frequency.

Derivation for Plasma Frequency

From Gauss’s law,

\[\oint \vec{D} \cdot d\vec{a} = Q\]

we have

\[\epsilon_0 EA = neAx\]

or

\[E=\frac{nex}{\epsilon_0}\]

Since the force is given by

\[F=QE=(-neAx)(\frac{nex}{\epsilon_0})=nAxm_e \frac{d^2x}{dt^2}\]

we obtain the equation of motion

\[\begin{aligned} \frac{d^2x}{dt^2} = - \left( \frac{ne^2}{m_e \epsilon_0} \right) x\end{aligned}\]

or

\[\begin{aligned} \frac{d^2x}{dt^2} +\omega_{pe}^2x=0\end{aligned}\]

where

\[\begin{aligned} \boxed{\omega_{pe}=\sqrt{\frac{ne^2}{m_e \epsilon_0}}}\end{aligned}\]

This is electron plasma frequency.
\(\bullet\) If there are no collisions, the disturbance will oscillate indefinitely.
\(\bullet\) Debye length, thermal velocity, and plasma frequency are inter-related.

\[\frac{v_{th}}{\omega_{pe}}=\frac{\sqrt{\frac{K_BT}{m}}}{\sqrt{\frac{ne^2}{m\epsilon_0}}}=\sqrt{\frac{\epsilon_0 KT}{ne^2}}=\lambda_D\]

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[플라즈마 물리][Plasma Physics]CH1 Introduction - Plasma Parameter

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Plasma Parameter

\(\bullet\) The plasma parameter is defined as

\[\begin{aligned} \boxed{ N_D =n\frac{4}{3}\pi \lambda_D^3 }=1.38 \times 10^6 \frac{T^{\frac{3}{2}}}{n^{\frac{1}{2}}} [T](in^{\circ}K)\end{aligned}\]

which is the number of plasma particles in a Debye sphere.

\(\bullet\) For Debye shielding to occur, and for the description of a plasma to be statistically meaningful, the number of particles in a Debye sphere must be large; that is \(N_D>>1\)

\(\bullet\) Plasma parameter is a measure of the ratio of the mean plasma kinetic energy to potential energy.

\[\frac{K.E.}{P.E.} \simeq \frac{\frac{3}{2}K_BT}{\frac{e^2}{4\pi\epsilon_0 \lambda_D}} \simeq \frac{9}{2}N_D >> 1\]

\(\bullet\) Thus \(N_D>>1\) means that the potential energy of a particle due to its nearest neighbor is much smaller than its kinetic energy. If this were not the case, there would be a strong tendency for electrons and ions to bind together into atoms, thus destroying the plasma.

\(\bullet\) An ideal gas corresponds to zero potential energy between the particles. Since the plasma parameter is large, the plasma may be treated as an ideal gas of charged particles, that is, a gas that can have a charge density and electric field but in which no two discrete particles interact.

\(\bullet\) In deriving the Debye potential, we assumed that the electrostatic energy was small compared to the thermal energy. The largeness of the plasma parameter guarantees the validity of the Debye potential.

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[플라즈마 물리][Plasma Physics]CH1 Introduction - Debye Shielding 디바이 차폐

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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 \(\quad T \uparrow\) \(\lambda_D \uparrow\)

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[전자기학][Fundamentals of Engineering Electromagnetics]Electrostatic potential energy, Example 3-17 Cheng (P122)

[Fundamentals of Engineering Electromagnetics]

Electrostatic potential energy, Example 3-17 Cheng (P122)

전자공학 학생이 아래 질문을 가지고 질문한 사항에 대해 답변을 하고자 작성하였습니다. 틀렸으면 알려주세요. 제 자신도 공부하기 위함입니다. (I just upload the image from Cheng p122; if there is problem with the copyright please let me know. I am trying to discuss the question that student has)

Problem: Find the energy required to assemble a uniform sphere of charge of radius \(b\) and volume charge density \(\rho_v\)

In terms of total charge

\[\begin{aligned} Q= \frac{4\pi}{3}\rho_v b^3 \end{aligned}\]

Hence we have

\[\begin{aligned} W_e = \frac{3Q^2}{20 \pi \epsilon_0 b}\end{aligned}\]

By looking at the equation below, a lot of people asks why does the half goes away?

\[\begin{aligned} W&= \frac{1}{2}\int_v \rho_v V dv\end{aligned}\]

And the answer is it is not actually going away. Those who might have twice differece might solved problem as below ( used \(V =\frac{q}{4\pi \epsilon_0 r}\))

\[\begin{aligned} W&= \frac{1}{2}\int_v \rho_v V dv \\ &= \frac{1}{2}\int_v \frac{\rho_v^2 \frac{4}{3} \pi r^2}{4 \pi \epsilon_0} r^2 \sin \theta dr d\theta d\phi \\ &=\frac{2}{15}\frac{\rho_v^2 r^5}{\epsilon_0}\\ &=\frac{3}{40}\frac{Q^2}{\epsilon_0 r \pi}\end{aligned}\]

Notice that \(Q = \rho_v \frac{4}{3}\pi r^3\). So there is twice difference. The reason for this is that potential is different inside and outside the effective sphere.
Answer: Think we have sphere with radius R; charge exist in this region. For \(r<R\)

\[\begin{aligned} \oint_s \vec{E} \cdot d\vec{a} &= \frac{\rho}{\epsilon_0}\frac{4}{3}\pi r^3 \\ \vec{E} &= \frac{\rho}{3\epsilon_0} r \\ &=\frac{\rho r}{4\pi \epsilon_0 R^3}\end{aligned}\]

On the last step, we used the fact that charge is inside the \(\frac{4}{3}\pi R^3\). For \(r>R\),

\[\begin{aligned} \vec{E} = \frac{Q}{4\pi \epsilon_0 r^2}\end{aligned}\]

Finally \(E\) field and potential becomes

\[\begin{aligned} \vec{E}(r) = \frac{Q}{4\pi \epsilon_0} \times \begin{cases} \frac{r}{R^3} \quad (r<R) \\ \frac{1}{r^2} \quad (r>R) \end{cases}\end{aligned}\]

\[\begin{aligned} \vec{V}(r) = \frac{Q}{4\pi \epsilon_0} \times \begin{cases} -\frac{1}{2} \frac{r^2}{R^3} + \frac{3}{2R} \\ \frac{1}{r} \end{cases}\end{aligned}\]

Notice that constant term from the potential comes from the Boundary Condition; \(at r=R\), \(V = \frac{Q}{4\pi \epsilon_0 R}\). Finally, let us calculate the energy.

\[\begin{aligned} W &= \frac{1}{2} \int_{0}^{R} \frac{Q}{4\pi \epsilon_0} \rho \left( -\frac{r^2}{2R^3} + \frac{3}{2R} \right) 4\pi r^2 dr \\ &=\frac{1}{2}\int_{0}^{R} \frac{Q \rho}{\epsilon_0}\left( -\frac{r^4}{2R^3} + \frac{3r^2}{2R} \right) dr \\ &= \frac{4}{20}\frac{Q \rho R^2}{\epsilon_0}\\ &= \frac{3}{20}\frac{Q^2}{\pi \epsilon_0 R}\end{aligned}\]

Hence, we get the same answer as David K. Cheng.

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프로젝트 팀: 논리설계 카드뉴스

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[대학원-전자기학]Vector Calculus Problem 2

Problem 2: Prove \(-\nabla \frac{1}{|\vec{r}-\vec{r}\prime|}=\frac{\vec{r}-\vec{r}\prime}{|\vec{r}-\vec{r}\prime |^3}\)

Let \[\vec{r}=x_1\hat{x} + y_1\hat{y} + z_1\hat{z}\] \[\vec{r}\prime=x_2\hat{x} + y_2\hat{y} + z_2\hat{z}\] \[\zeta =(\vec{x_1}-\vec{x_2})^2+(\vec{y_1}-\vec{y_2})^2+(\vec{z_1}-\vec{z_2})^2\]

\[\nabla \frac{1}{|\vec{r}-\vec{r}\prime|} = \frac{\partial}{\partial x}\hat{x} + \frac{\partial}{\partial y}\hat{y} + \frac{\partial}{\partial z}\hat{z}\frac{1}{|\vec{r}-\vec{r}\prime|}\]

Let us consider only \(\hat{x}\) components. \[\begin{aligned} \frac{\partial}{\partial x}\hat{x}\frac{1}{\zeta^{\frac{1}{2}}} &= \frac{1}{2}\zeta^{-\frac{3}{2}}\zeta\prime \nonumber \\ &=\frac{x_1 - x_2}{\zeta^{\frac{3}{2}}}\hat{x} \nonumber \\ &=\frac{x_1 - x_2}{|\vec{r}-\vec{r}\prime|^3}\hat{x} \nonumber\end{aligned}\]

Same procedure occur for the other two component. When we add all, final result become \[-\nabla \frac{1}{|\vec{r}-\vec{r}\prime|}=\frac{\vec{r}-\vec{r}\prime}{|\vec{r}-\vec{r}\prime |^3}\]

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[대학원-전자기학]Vector Calculus Problem 1

Problem 1: Prove \(\nabla \times \frac{\vec{r}-\vec{r}\prime}{|\vec{r}-\vec{r}\prime |^3}=0\)

Answer:
\[\nabla \times \frac{\vec{r}-\vec{r}\prime}{|\vec{r}-\vec{r}\prime |^3}= \begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \frac{x_1-x_2}{|\vec{\eta}^3|} & \frac{y_1-y_2}{|\vec{\eta}^3|} & \frac{z_1-z_2}{|\vec{\eta}^3|} \end{vmatrix}\] Where \[\vec{r}=x_1\hat{x} + y_1\hat{y} + z_1\hat{z}\] \[\vec{r}\prime=x_2\hat{x} + y_2\hat{y} + z_2\hat{z}\] \[\vec{\eta} = \vec{r}-\vec{r}\prime\], \[|\vec{\eta}|=\sqrt{(\vec{x_1}-\vec{x_2})^2+(\vec{y_1}-\vec{y_2})^2+(\vec{z_1}-\vec{z_2})^2}\] \[\zeta =(\vec{x_1}-\vec{x_2})^2+(\vec{y_1}-\vec{y_2})^2+(\vec{z_1}-\vec{z_2})^2\] \[\begin{aligned} \nabla \times \frac{\vec{r}-\vec{r}\prime}{|\vec{r}-\vec{r}\prime |^3} &= (\frac{\partial}{\partial y}(\frac{z_1-z_2}{|\vec{\eta}^3|})-\frac{\partial}{\partial z}(\frac{y_1-y_2}{|\vec{\eta}^3|}))\vec{x} \nonumber \\ & -(\frac{\partial}{\partial x}(\frac{z_1-z_2}{|\vec{\eta}^3|}) - \frac{\partial}{\partial z}(\frac{x_1-x_2}{|\vec{\eta}^3|}))\vec{y} \nonumber \\ & +(\frac{\partial}{\partial x}(\frac{y_1-y_2}{|\vec{\eta}^3|})-\frac{\partial}{\partial y}(\frac{x_1-x_2}{|\vec{\eta}^3|}))\vec{z} \nonumber\end{aligned}\]

Since \[\frac{\partial}{\partial y}(\frac{z_1-z_2}{|\vec{\zeta}^{\frac{3}{2}}|})=-3\frac{(z_1 - z_2)(y_1 - y_2)}{|\vec{\zeta}^{\frac{5}{2}}|}\]

\[\frac{\partial}{\partial z}(\frac{y_1-y_2}{|\vec{\zeta}^{\frac{3}{2}}|})=-3\frac{(y_1 - y_2)(z_1 - z_2)}{|\vec{\zeta}^{\frac{5}{2}}|}\]

They cancel out each others. Same procedure occur for the other two components. Hence, \(\nabla \times \frac{\vec{r}-\vec{r}\prime}{|\vec{r}-\vec{r}\prime |^3}=0\)

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프로젝트 팀: 논리설계 카드뉴스

산동일크무크 프로젝트 팀: 논리설계 카드뉴스

산동일크무크 그룹 프로젝트 팀장이 이끄는 팀에서 (학부생 그룹) 아래와 같이 카드뉴스를 만들어 보았습니다.  앞으로도 다양한 컨텐츠를 만들어 배포할 수 있었으면 좋겠습니다.

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