[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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