[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]Normalization Constant & Average Kinetic Energy

Problem The 3D Maxwellian distribution is given by
\[f(\vec{v}) = A_3 e^{\left( -\frac{m(v_x^2 + v_y^2 + v_z^2)}{2KT} \right)}\]
with \[n = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(\vec{v})dv_x dv_y dv_z\]
(a) Show that the normalization constant is given by \[A_3 = n \left( \frac{m}{2\pi KT} \right)^{\frac{3}{2}}\]
(b) Show that the average kinetic energy is \[E_{av} = \frac{3}{2}KT\]
Answer (a)

Notice that
\[\begin{aligned} \label{eq_gauss} \int_{-\infty}^{\infty} e^{-ax^2}dx = \sqrt{\frac{\pi}{a}}\end{aligned}\]
\[\begin{aligned} n &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(\vec{v})dv_x dv_y dv_z \nonumber \\ &= A_3 \int_{-\infty}^{\infty} e^{ -\frac{m(v_x^2)}{2KT}}dv_x \int_{-\infty}^{\infty} e^{ -\frac{m(v_y^2)}{2KT}}dv_y \int_{-\infty}^{\infty} e^{ -\frac{m(v_z^2)}{2KT}}dv_z \nonumber \\ &= A_3 \left( \sqrt{\frac{ 2 K T \pi}{m}} \right)^{3}\end{aligned}\]
Hence,
\[A_3 = n \left( \frac{m}{2\pi KT} \right)^{\frac{3}{2}}\]
Answer (b)

Average energy can be calculated as
\[\begin{aligned} E_{av} &= \frac{\int_{-\infty}^{\infty} \frac{1}{2}mv^2 f(\vec{v})dv}{\int_{-\infty}^{\infty}f(\vec{v})dv} \nonumber \\ &=\frac{m}{2n}\int_{-\infty}^{\infty} (v_x^2 + v_y^2 + v_z^2) A_3 e^{ -\frac{m(v_x^2 + v_y^2 + v_z^2)}{2KT} }dv\end{aligned}\]
The denominator is \(n\) as we calculated in previous problem. Notice that, \[\int x^2 e^{-ax^2}dx = \frac{1}{2}\sqrt{\frac{\pi}{a^3}}\]
If we calculate the first term of equation,
\[\begin{aligned} E_{av_x} &= \frac{m}{2n} \int_{-\infty}^{\infty} A_3(v_x^2 ) e^{ -\frac{m(v_x^2 + v_y^2 + v_z^2)}{2KT} }dv_x \nonumber \\ &=\frac{m}{2n}n \left( \frac{m}{2\pi KT} \right)^{\frac{3}{2}} \frac{1}{2} \left( \frac{2 KT}{m} \right)^{\frac{3}{2}}\sqrt{\pi}\frac{2KT \pi}{m} \nonumber \\ &=\frac{1}{2}KT\end{aligned}\]
Hence, if we calculate \(v_y\) and \(v_z\) components, final result becomes \[E_{av} = \frac{3}{2}KT\]
4차 산업혁명에 걸맞는 인터넷 기반 고등교육기관
이메일: ilkmooc@ilkmooc.kr
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