本文主要參考 Griffiths, David J. Introduction to Electrodynamics (4th edition), Pearson, ISBN 13: 978-0-321-85656-2 Section 10.2。
程度：中等／大學部二年級電磁學
複習一下，在位勢與場（上）的最後，我們選用勞倫茲規範，使馬克士威方程式的位勢表述有了較簡潔的形式：
上式也是一個非齊次波動方程式。對於靜態情況，上式會簡化成四個泊松方程式：
這些位勢的解是
![\begin{array}{l}\displaystyle V(\mathbf{r})=\dfrac{1}{4\pi\epsilon_0}\int\frac{\rho(\mathbf{r}')}{\mathscr{R}}\text{d}\tau',\\[15pt]\displaystyle \mathbf{A}(\mathbf{r})=\frac{\mu_0}{4\pi}\int\frac{\mathbf{J(\mathbf{r}')}}{\mathscr{R}}\text{d}\tau',\end{array}](https://s0.wp.com/latex.php?latex=%5Cbegin%7Barray%7D%7Bl%7D%5Cdisplaystyle+V%28%5Cmathbf%7Br%7D%29%3D%5Cdfrac%7B1%7D%7B4%5Cpi%5Cepsilon_0%7D%5Cint%5Cfrac%7B%5Crho%28%5Cmathbf%7Br%7D%27%29%7D%7B%5Cmathscr%7BR%7D%7D%5Ctext%7Bd%7D%5Ctau%27%2C%5C%5C%5B15pt%5D%5Cdisplaystyle+%5Cmathbf%7BA%7D%28%5Cmathbf%7Br%7D%29%3D%5Cfrac%7B%5Cmu_0%7D%7B4%5Cpi%7D%5Cint%5Cfrac%7B%5Cmathbf%7BJ%28%5Cmathbf%7Br%7D%27%29%7D%7D%7B%5Cmathscr%7BR%7D%7D%5Ctext%7Bd%7D%5Ctau%27%2C%5Cend%7Barray%7D&bg=ffffff&fg=2b2b2b&s=0&c=20201002)
其中 
Some English Idioms
In this article 13 common idioms in English is introduced and defined.
For better understanding of the idioms, Some of renowned dictionaries are cited, with the detailed copyright information presented at the end of the article.
Some idioms are accompanied with Chinese translation similar in meaning.
繼續閱讀 Some English Idioms位勢與場（上）
本文主要參考 Griffiths, David J. Introduction to Electrodynamics (4th edition), Pearson, ISBN 13: 978-0-321-85656-2 Section 10.1。
程度：中等／大學部二年級電磁學
在得到馬克士威方程組
![\begin{array}{lllll} \text{(i)}&\boldsymbol\nabla\boldsymbol\cdot\mathbf{E}=\dfrac{1}{\epsilon_0}\rho, & & \text{(iii)}&\boldsymbol\nabla\boldsymbol\times \mathbf{E}=-\dfrac{\partial \mathbf{B}}{\partial t}, \\[10pt]\text{(ii)}&\boldsymbol\nabla\boldsymbol\cdot\mathbf{B}=0, & &\text{(iv)}&\boldsymbol\nabla\boldsymbol\times \mathbf{B}=\mu_0\mathbf{J}+\mu_0\epsilon_0\dfrac{\partial \mathbf{E}}{\partial t}\end{array}](https://s0.wp.com/latex.php?latex=%5Cbegin%7Barray%7D%7Blllll%7D+%5Ctext%7B%28i%29%7D%26%5Cboldsymbol%5Cnabla%5Cboldsymbol%5Ccdot%5Cmathbf%7BE%7D%3D%5Cdfrac%7B1%7D%7B%5Cepsilon_0%7D%5Crho%2C+%26+%26+%5Ctext%7B%28iii%29%7D%26%5Cboldsymbol%5Cnabla%5Cboldsymbol%5Ctimes++%5Cmathbf%7BE%7D%3D-%5Cdfrac%7B%5Cpartial+%5Cmathbf%7BB%7D%7D%7B%5Cpartial+t%7D%2C+%5C%5C%5B10pt%5D%5Ctext%7B%28ii%29%7D%26%5Cboldsymbol%5Cnabla%5Cboldsymbol%5Ccdot%5Cmathbf%7BB%7D%3D0%2C+%26+%26%5Ctext%7B%28iv%29%7D%26%5Cboldsymbol%5Cnabla%5Cboldsymbol%5Ctimes+%5Cmathbf%7BB%7D%3D%5Cmu_0%5Cmathbf%7BJ%7D%2B%5Cmu_0%5Cepsilon_0%5Cdfrac%7B%5Cpartial+%5Cmathbf%7BE%7D%7D%7B%5Cpartial+t%7D%5Cend%7Barray%7D&bg=ffffff&fg=2b2b2b&s=0&c=20201002)
之後，我們想要求它的通解。這就是說，給定任一時刻、任一點的電荷組態（包含電荷密度 
A blue-sky wanderer 