[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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[플라즈마 물리][Plasma Physics]Debye Shielding 디바이 차폐

Example Problem: Debye Shielding
Consider a positive point charge immersed in a plasma as we discussed in the class. Show that the net charge in the Debye shielding cloud exactly cancels the test charge. Assume that the ions are fixed and that \(e\phi << KT_e\). Note that we assumed that ion distribution is similar to that of the electrons in the class.

Answer:

While deriving, we assume potential to be

\[\begin{aligned} \frac{e\phi}{KT_e} << 1 \quad \quad \quad n_i \simeq n_e\\ \lambda_D = \sqrt{\frac{\epsilon KT}{ne^2}}\\ \phi(r) = \frac{e}{4\pi\epsilon_0 r }e^{-\frac{r}{\lambda_D}}\end{aligned}\]

Assume \(\frac{m_i}{m_e} \rightarrow \infty\) so that ions are fixed; \(n_i(r)=n\). Further assume electrons obey the Boltzmann distribution.

\[\begin{aligned} f(u) &= A e^{\frac{\frac{-1}{2}mv^2 + q\phi}{KT_e}}\\ n_e(r) &= ne^{\frac{e\phi(r)}{K_B T}}\end{aligned}\]

\[\begin{aligned} \rho &= -e(n_i - n_e) \\ &=-e \left(n - ne^{\frac{e\phi(r)}{K_BT}} \right)\\ &= ne \left(e^{\frac{e\phi(r)}{K_BT}} - 1 \right)\\\end{aligned}\]

for \(\frac{e\phi}{KT_e} << 1\),

\[\begin{aligned} e^{\frac{e\phi(r)}{K_B T}} = 1 + \frac{1}{2}\frac{e\phi(r)}{K_B T} + ...\end{aligned}\]

\[\begin{aligned} \rho &= ne(1+\frac{e\phi(r)}{K_B T} - 1) \\ &=\frac{ne^2 \phi(r)}{K_B T} \\ &= \frac{\epsilon_0}{\lambda_D^2}\phi(r)\end{aligned}\]

\[\begin{aligned} \int \rho dv &= Q \\ &= \int_V \frac{\epsilon_0}{\lambda_D^2}\frac{e}{4\pi\epsilon_0 r }e^{-\frac{r}{\lambda_D}} r^2 \sin\theta dr d\theta d\phi\\ &= \int \frac{4\pi \epsilon_0}{\lambda_D^2}\frac{e}{4\pi \epsilon_0} r e^{-\frac{r}{\lambda_D}}dr \\ &= \frac{e}{\lambda_D^2} \int_{0}^{\infty} r e^{-\frac{r}{\lambda_D}}dr \\ &= \frac{e}{\lambda_D^2} (\lambda_D^2 e^{-\frac{r}{\lambda_D}}-\lambda_D r e^{-\frac{r}{\lambda_D}}) =0\end{aligned}\]

On the last step, when \(r = \lambda_D\), charge becomes zero. Hence, the Debye sheilding cloud exactly cancels the test charge.

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[플라즈마 물리]F.F. Chen Problem 1.5 Solution Debye Shielding

Problem In a strictly steady state situation both the ions and the electrons will follow the Boltzmann relation

\[n_i = n_0 e^{\frac{-q_i \phi }{KT_i}}\]

For the case of an infinite, transparent grid charged to a potential \(\phi\), show that the shielding distance is then given approximately by

\[\lambda_D^{-2} = \frac{ne^2}{\epsilon_0} \left( \frac{1}{KT_e} + \frac{1}{KT_i} \right)\]

Answer

\[\begin{aligned} \nabla \cdot \vec{E} &=\frac{\rho}{\epsilon_0} = \frac{e}{\epsilon_0}(n_i - n_e) \nonumber \\ &= - \frac{e}{\epsilon_0} n_0 \left( e^{-\frac{e\phi}{KT_i}} - e^{\frac{e\phi}{KT_e}} \right) \nonumber \\ &= - \frac{e}{\epsilon_0} n_0 \left( 1 -\frac{e\phi}{KT_i} - 1 - \frac{e\phi}{KT_e} \right) \nonumber \\ \frac{d^2 \phi }{dx^2 }&=\frac{e^2 n_0}{\epsilon_0}\left( \frac{1}{KT_i} + \frac{1}{KT_e} \right) \phi \nonumber\end{aligned}\]

Hence,

\[\begin{aligned} \frac{1}{\lambda_D^2} = \frac{e^2 n_0}{\epsilon_0}\left( \frac{1}{KT_i} + \frac{1}{KT_e} \right) \nonumber\end{aligned}\]

\[\lambda_D = \sqrt{\left( \frac{\epsilon_0 K T_e T_i}{ne^2 (T_i + T_e)} \right)}\]

• \(T_i << T_e \quad \quad \lambda_D \simeq \sqrt{\left( \frac{\epsilon_0 K T_i}{ne^2} \right)}\)

• \(T_e << T_i \quad \quad \lambda_D \simeq \sqrt{\left( \frac{\epsilon_0 K T_e}{ne^2} \right)}\)

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[플라즈마물리]Single Particle Motion - E cross B drift

Problem Consider a particle of charge \(q\) and mass \(m\), initially at rest at (0,0,0), in the presence of a static magntic field \(\vec{B}=B\hat{z}\) and \(\vec{E}=E\hat{y}\).

(a) Taking \(E, B > 0\), sketch the orbit of the particle when \(q>0\).

(b) Derive an exact expression for the orbit \([x(t),y(t),z(t)]\) or the particle. Express your answer in terms of \(E\), \(B\), and \(\omega_c\)

(c) Find the drift velocity after averaging the motion in time. If there were many particles of various charges and masses present, would there be any net current?

(d) Suppose the electric field were replaced by a force \(F\) in the \(y\)-direction. What would be the drift velocity?

Answer (a)

Answer (b)

\(\vec{B}=B\hat{z}\) and \(\vec{E}=E\hat{y}\). From Lorentz equation,

\[m\frac{dv}{dt}=q(\vec{E} + \vec{v}\times \vec{B})\]

\[\begin{aligned} \dot{v_x}&= \quad \quad \quad \quad \frac{q}{m}v_y B \nonumber \\ \dot{v_y}&= \frac{q}{m}\vec{E_y} - \frac{q}{m}v_xB \nonumber \\ \dot{v_z}&= 0\end{aligned}\]

In z-direction \[z(t) = constant = 0\]

\[\begin{aligned} \frac{dv_x}{dt} &= \frac{qB}{m}v_y = \omega_c v_y \nonumber \\ \frac{dv_y}{dt} &= \frac{qE}{m}-\frac{qB}{m}v_x = \frac{qE_y}{m}-\omega_c v_x \nonumber\end{aligned}\]

\[\begin{aligned} \frac{d^2 v_y}{dt^2} &= -\omega_c^2 v_y \nonumber \\ \frac{d^2 v_x}{dt^2} &= -\omega_c^2 \left(v_x-\frac{E}{B} \right) \nonumber\end{aligned}\]

since \(\frac{E}{B}\) is constant, \[\frac{d^2}{dt^2} \left[ v_x-\frac{E}{B} \right] = \frac{d^2 v_x}{dt^2} = -\omega_c^2 \left(v_x-\frac{E}{B} \right)\]

\[\begin{aligned} v_x &= iv_{\perp}e^{i\omega_c t} + \frac{E}{B}\nonumber \\ v_y &= v_{\perp}e^{i\omega_c t} \nonumber \end{aligned}\]

\[\begin{aligned} x(t) &= x(0) + r_L \cos{\omega_c t} +\frac{E}{B}t \nonumber \\ y(t) &= y(0) + r_L \sin{\omega_c t} \nonumber \end{aligned}\]

Answer (c)

\[<v_d> = \frac{\int_{-\infty}^{\infty}v_d f(v) dv}{\int_{-\infty}^{\infty}f(v) dv} = \frac{E}{B}\]

There will be no current because \(\vec{E} \times \vec{B}\) drifts the ions and electrons in the same direction.

Answer (d)

\[\begin{aligned} \vec{v_f}&= \frac{1}{q} \frac{\vec{F} \times \vec{B}}{B^2} \nonumber \\ &=\frac{FE}{qB}\hat{y}\end{aligned}\]

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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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[대학원-전자기학]문제풀이 델타함수와 발산

Problem Prove that \[\nabla^2 (\frac{1}{r}) = -4\pi \delta(\vec{x})\] where \(r=|\vec{x}|\)
Answer

\[\begin{aligned} \nabla^2 \frac{1}{r} = \nabla \cdot (\nabla \frac{1}{r}) = \nabla \cdot (-\frac{1}{r^2}\hat{r}) = \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \left( \frac{-1}{r^2}\right) \right) \nonumber\end{aligned}\]
Hence, the intergral of \(\nabla^2 \frac{1}{r} = 0\)
\[\int_v \nabla^2 \frac{1}{r} dv = 0\]
However, when we use the divergence theorem, we get different result.
\[\begin{aligned} \int_v \nabla^2 \frac{1}{r} dv &= \oint_s -\frac{1}{r^2}\hat{r} d\vec{a} \nonumber \\ &=\oint_s - \frac{1}{r^2} r^2\sin\theta d\theta d\phi \nonumber \\ &= -4\pi \nonumber\end{aligned}\]
This problem arise in the origin. Hence, we introduce delta function to solve this problem.
\[\nabla^2 (\frac{1}{r}) = -4\pi \delta(\vec{x})\]
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[대학원-전자기학]문제풀이

Problem Prove the symmetry property of the Green’s function satisfying the Dirichlet boundary condition \[G(\vec{x},\vec{x}^{\prime}) = G(\vec{x}^{\prime}, \vec{x})\]

Refer to Jackson Problem (1.14)

Answer

Green function \(G(\vec{x},\vec{x}^{\prime})\) satisfies Dirichlet boundary conditions; bound region \(\Omega\) with boundary \(d\Omega\), \(G(\vec{x},\vec{x}^{\prime})=0 \quad \forall x^{\prime} \in \partial \Omega\)

\[\int_V (\phi \nabla^2 \psi - \psi \nabla^2 \phi)d^3x = \oint_s \left[ \phi \frac{\partial \psi}{\partial n} - \psi \frac{\partial \phi }{\partial n}\right]da\]

Substitute \(\phi = G(\vec{x},\vec{y})\) and \(\psi = G(\vec{x}^{\prime},\vec{y})\).

\[\begin{aligned} \int_{\Omega} \left[ G(\vec{x},\vec{y})\nabla^2 G(\vec{x}^{\prime},\vec{y}) - G(\vec{x}^{\prime},\vec{y}) \nabla^2 G(\vec{x},\vec{y}) \right]d^3y =\nonumber \\ \int_{\partial \Omega}\left[ G(\vec{x}, \vec{y})\frac{\partial }{\partial n}G(\vec{x}^{\prime},\vec{y}) - G(\vec{x}^{\prime},\vec{y}) \frac{\partial}{\partial n} G(\vec{x},\vec{y}) \right]da \nonumber\end{aligned}\]

\(G(\vec{x}, \vec{x}^{\prime})\) satisfies Dirichlet B.C. Hence \(RHS=0\); \(\Phi\) is known on the surface and \(F\) can be chosen to make \(G_D(\vec{x},\vec{x}^{\prime}=0)\) Since \[\nabla^2 G(\vec{x},\vec{y}) = -4\pi \delta^3(\vec{x}-\vec{y})\]

\[\begin{aligned} \int_{\Omega} \left[ G(\vec{x},\vec{y})\nabla^2 G(\vec{x}^{\prime},\vec{y}) - G(\vec{x}^{\prime},\vec{y}) \nabla^2 G(\vec{x},\vec{y}) \right]d^3y = \nonumber \\ \int_{\Omega}\left[ -4\pi G(\vec{x},\vec{y}) \delta^3(\vec{x}^{\prime} - \vec{y})+G(x^{\prime},\vec{y}) 4\pi \delta^3(\vec{x}-\vec{y}) \right]d^3y \nonumber \\\end{aligned}\]

Finally

\[G(\vec{x},\vec{x}^{\prime}) = G(\vec{x}^{\prime}, \vec{x})\]

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