电场
库伦定律
$$ \mathbf{E}(\mathbf{r}) = \sum_{i = 1}^{n}\frac{q_{i}}{\mathfrak{s}^{2}}\hat{\mathfrak{s}}\\ \Downarrow\\ \mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\epsilon}\int\frac{1}{\mathfrak{s}^{2}}\hat{\mathfrak{s}}\mathrm{d}q $$
不同类型分布的表达式:
$$ \mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\epsilon_{0}}\int_{\mathcal{P}}\frac{\lambda(\mathbf{r}’)}{\mathfrak{s}^{2}}\hat{\mathfrak{s}}\mathrm{d}\mathbf{l}’\\ \mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\epsilon_{0}}\int_{\mathcal{S}}\frac{\sigma(\mathbf{r}’)}{\mathfrak{s}^{2}}\hat{\mathfrak{s}}\mathrm{d}\mathbf{a}’\\ \mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\epsilon_{0}}\int_{\mathcal{V}}\frac{\rho(\mathbf{r}’)}{\mathfrak{s}^{2}}\hat{\mathfrak{s}}\mathrm{d}\mathbf{\tau}' $$
通量定理(高斯定理)
$$ \oint\mathbf{E}\cdot\mathrm{d}\mathbf{a} = \frac{1}{\epsilon_{0}}Q_{\text{ene}} = \frac{1}{\epsilon_{0}}\int_{\mathcal{V}}\rho\mathrm{d}\tau $$
散度定理: $$ \oint_{\mathcal{S}}\mathbf{E}\cdot\mathrm{d}\mathbf{a} = \int_{\mathcal{V}}(\nabla\cdot\mathbf{E})\mathrm{d}\tau $$
所以有
$$ \begin{aligned} \int_{\mathcal{V}}(\nabla\cdot\mathbf{E})\mathrm{d}\tau &= \int_{\mathcal{V}}\frac{\rho}{\epsilon_{0}}\mathrm{d}\tau\\ \Downarrow\\ (\nabla\cdot\mathbf{E}) &= \frac{\rho}{\epsilon_{0}} \end{aligned} $$
电场散度
$$ \mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\epsilon_{0}}\int_{\text{All Space}}\frac{\hat{\mathfrak{s}}}{\mathfrak{s}^{2}}\rho(\mathbf{r}’)\mathrm{d}\tau' $$
$$ \nabla\cdot\mathbf{E} = \frac{1}{4\pi\epsilon_{0}}\int 4\pi\delta^{3}(\mathbf{r} - \mathbf{r}’)\rho(\mathbf{r}’)\mathrm{d}\tau’ = \frac{1}{\epsilon_{0}}\rho(\mathbf{r}) $$
以上为微分形式, 积分即为
$$ \int_{\mathcal{V}}(\nabla\cdot\mathbf{E})\mathrm{d}\tau = \oint_{\mathcal{S}}\mathbf{E}\cdot\mathrm{d}\mathbf{a} = \frac{1}{\epsilon_{0}}\int_{\mathcal{V}}\rho(\mathbf{r})\mathrm{d}\tau’ = \frac{1}{\epsilon_{0}}Q_{\text{ene}} $$
电场旋度
球坐标中 $$ \mathrm{d}\mathbf{l} = \hat{r}\mathrm{d}r + \hat{\theta}r\mathrm{d}\theta + \hat{\phi}r\sin{\theta}\mathrm{d}\phi $$
所以
$$ \begin{aligned} \mathbf{E}\cdot\mathrm{d}\mathbf{l} &= \frac{1}{4\pi\epsilon_{0}}\frac{q}{r^{2}}\hat{r}\\ \int_{a}^{b}\mathbf{E}\cdot\mathrm{d}\mathbf{l} &= \frac{1}{4\pi\epsilon_{0}}\int_{a}^{b}\frac{q}{r^{2}}\mathrm{d}r = \frac{1}{4\pi\epsilon_{0}}\left(\frac{q}{r_{a}} - \frac{q}{r_{b}}\right) \end{aligned} $$
闭合路径下即有
$$ \oint\mathbf{E}\cdot\mathrm{d}\mathbf{l} = 0 $$
使用 Stokes 定理即有
$$ \nabla\times\mathbf{E} = 0 $$
电势
$$ \mathbf{E} = -\nabla V $$
Possion 方程 & Laplace 方程
导出
既然 $$ \mathbf{E} = -\nabla V,\quad\nabla\cdot\mathbf{E} = \frac{\rho}{\epsilon_{0}},\quad\nabla\times\mathbf{E} = 0 $$
得到 Possion 方程:
$$ \nabla^{2}V = -\frac{\rho}{\epsilon_{0}} $$
当 $\rho=0$, 即考虑区域内无电荷, 即有 Laplace 方程: $$ \nabla^{2}V = 0 $$
界面条件(边界条件)
$$ \begin{aligned} E^{\perp}_{\text{up}} - E^{\perp}_{\text{down}} &= \frac{\sigma}{\epsilon_{0}}\\ E^{\parallel}_{\text{up}} &= E^{\parallel}_{\text{down}} \end{aligned} $$
将两者合并即有
$$ \mathbf{E}_{\text{up}} - \mathbf{E}_{\text{down}} = \frac{\sigma}{\epsilon_{0}}\hat{\mathbf{n}} $$
使用电势 $V$ 表示:
$$ \frac{\partial V_{\text{up}}}{\partial n} - \frac{\partial V_{\text{down}}}{\partial n} = -\frac{1}{\epsilon_{0}}\sigma\\ \Downarrow\\ \frac{\partial V}{\partial n} = \nabla V\cdot\hat{\mathbf{n}} $$
电场能量
离散:
$$ W = \frac{1}{2}\sum_{i}^{n}q_{i}\left(\sum_{j=1,j\neq i}\frac{1}{4\pi\epsilon_{0}}\frac{q_{j}}{\mathfrak{s}_{ij}}\right) = \frac{1}{2}\sum_{i = 1}^{n}q_{i}V(\mathbf{r}_{i}) $$
连续:
$$ \begin{aligned} W &= \frac{1}{2}\int\rho V\mathrm{d}\tau = \frac{\epsilon_{0}}{2}\int(\nabla\cdot\mathbf{E})V\mathrm{d}\tau = \frac{\epsilon_{0}}{2}\oint V\mathbf{E}\cdot\mathrm{d}\mathbf{a} - \frac{\epsilon_{0}}{2}\int\mathbf{E}\cdot(\nabla V)\mathrm{d}\tau\\ &= \frac{\epsilon_{0}}{2}\left(\int_{\mathcal{V}}E^{2}\mathrm{d}\tau + \oint V\mathbf{E}\cdot\mathrm{d}\mathbf{a}\right)\Rightarrow W = \frac{\epsilon_{0}}{2}\int_{\text{All Space}}E^{2}\mathrm{d}\tau \end{aligned} $$
导体
性质
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导体内部电场为零;
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导体内部 $\rho = 0$;
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净电荷分布在导体表面;
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导体是等势体;
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电场垂直于导体表面.
界面条件
根据上述性质, 有
$$ \mathbf{E} = \frac{\sigma}{\epsilon_{0}}\hat{\mathbf{n}} $$
电场在导体带电面处不连续, 受力应当是上下电场的平均:
$$ \mathbf{f} = \sigma\langle E\rangle = \frac{1}{2}(\mathbf{E}_{\text{up}} + \mathbf{E}_{\text{down}}) = \frac{1}{2\epsilon_{0}}\sigma^{2}\hat{\mathbf{n}} $$
所以静电压力:
$$ P = \frac{\epsilon_{0}}{2}E^{2} $$
电容
$$ C = \frac{Q}{V} $$
平行板电容器
$$ \sigma = \frac{Q}{S},\quad E = \frac{1}{\epsilon_{0}}\frac{Q}{S},\quad V = \frac{Q}{\epsilon_{0}S}d\\ \Downarrow\\ C = \frac{\epsilon_{0}S}{d} $$