介质中的Maxwell方程组

∇×E⃗=−∂B⃗∂t∇×B⃗=μ0(J⃗f+J⃗μ+J⃗p)+μ0ε0∂E⃗∂t∇⋅E⃗=ρf+ρpε0∇⋅B⃗=0\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}\\ \nabla \times \vec{B} = \mu_0(\vec{J}_f + \vec{J}_\mu + \vec{J}_p) + \mu_0\varepsilon_0\frac{\partial \vec{E}}{\partial t}\\ \nabla \cdot \vec{E} = \frac{\rho_f + \rho_p}{\varepsilon_0}\\ \nabla \cdot \vec{B} = 0\\

新定义

电位移矢量

D⃗=ε0E⃗+P⃗\vec{D} = \varepsilon_0\vec{E} + \vec{P}

磁场强度

H⃗=1μ0B⃗−M⃗\vec{H} = \frac{1}{\mu_0}\vec{B} - \vec{M}

maxwell方程组的简洁形式

∇×E⃗=−∂B⃗∂t∇×H⃗=J⃗f+∂D⃗∂t∇⋅D⃗=ρf∇⋅B⃗=0\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}\\ \nabla \times \vec{H} = \vec{J}_f + \frac{\partial \vec{D}}{\partial t}\\ \nabla \cdot \vec{D} = \rho_f\\ \nabla \cdot \vec{B} = 0\\

简单介质本构关系

D⃗=ε0P⃗=εE⃗+ε0XeE⃗=ε0(1+Xe)E⃗=ε0εrE⃗=εE⃗\vec{D} = \varepsilon_0\vec{P} = \varepsilon\vec{E}+\varepsilon_0\mathcal{X}_e\vec{E} = \varepsilon_0(1+\mathcal{X}_e)\vec{E} = \varepsilon_0\varepsilon_r\vec{E} = \varepsilon\vec{E}\\

这里的εr\varepsilon_r是相对介电常数，ε=ε0εr\varepsilon = \varepsilon_0\varepsilon_r是介电常数

H⃗=1μ0B⃗−M⃗=1μ0B⃗−XmH⃗B⃗=μ0H⃗+μ0XmH⃗=μ0(1+Xm)H⃗=μ0μrH⃗=μH⃗\vec{H} = \frac{1}{\mu_0}\vec{B} - \vec{M} = \frac{1}{\mu_0}\vec{B} - \mathcal{X}_m\vec{H}\\ \vec{B} = \mu_0\vec{H} + \mu_0\mathcal{X}_m\vec{H} = \mu_0(1+\mathcal{X}_m)\vec{H} = \mu_0\mu_r\vec{H} = \mu\vec{H}

这里的μr\mu_r是相对磁导率，μ=μ0μr\mu = \mu_0\mu_r是磁导率

边值关系

n^×(E⃗2−E⃗1)=0n^×(H⃗2−H⃗1)=JS⃗n^⋅(D⃗2−D⃗1)=ρsn^⋅(B⃗2−B⃗1)=0\begin{aligned} \hat{n} \times (\vec{E}_2 - \vec{E}_1) &= 0\\ \hat{n} \times (\vec{H}_2 - \vec{H}_1) &= \vec{J_S}\\ \hat{n} \cdot (\vec{D}_2 - \vec{D}_1) &= \rho_s\\ \hat{n} \cdot (\vec{B}_2 - \vec{B}_1) &= 0\\ \end{aligned}

静电场中的导体

在电场中，导体达到经典平衡态后：

  • 电场：
    内部电场为0，E⃗内=0\vec{E}_内 = 0
    外部电场垂直于表面，n^⋅E⃗外=0\hat{n} \cdot \vec{E}_外 = 0
  • 电荷：
    内部无净电荷：ρ内=0\rho_{内} = 0
    电荷集中在表面：ρs=ε0n^⋅E⃗外\rho_s = \varepsilon_0\hat{n} \cdot \vec{E}_外
  • 电势：
    内部为等势体，表面为等势面

    φ∣S=const\varphi |_S = const

    ρ,E⃗,φ\rho,\vec{E},\varphi关系：

    ε0E∣S=ρS=−ε0∂φ∂n∣S\varepsilon_0 E |_S = \rho_S = -\varepsilon_0 \frac{\partial \varphi}{\partial n} |_S

  • 表面单位面积受到的电场力

F⃗=ρSE⃗∣S/2=ρS22ε0n^\vec{F} = \rho_S \vec{E} |_S / 2 = \frac{\rho_S^2}{2\varepsilon_0} \hat{n}

点电荷的势能

w2=q1q24πε0rw3=q3q14πε0r+q3q24πε0rwi=14πε0∑j=1i−1qiqjrijw_2 = \frac{q_1q_2}{4\pi\varepsilon_0r}\\ w_3 = \frac{q_3q_1}{4\pi\varepsilon_0r} + \frac{q_3q_2}{4\pi\varepsilon_0r}\\ w_i = \frac{1}{4\pi \varepsilon_0} \sum_{j=1}^{i-1} \frac{q_iq_j}{r_{ij}}

假设一共有N个点电荷，那么总功为：

W=∑wi=14πε0∑i=2N∑j=1i−1qiqjrij=12∑i=1Nqiφi\begin{aligned} W = \sum w_i &= \frac{1}{4\pi\varepsilon_0} \sum_{i=2}^{N} \sum_{j=1}^{i-1} \frac{q_iq_j}{r_{ij}}\\ &= \frac{1}{2} \sum_{i=1}^{N} q_i \varphi_i \end{aligned}

这里φi\varphi_i是除去第i个电荷外，其余所有电荷在x⃗i\vec{x}_i处所产生的电势

推广到连续分布电势：

W=12∫ρ(r⃗)φ(r⃗)dVW = \frac{1}{2} \int \rho(\vec{r})\varphi(\vec{r})dV

导体系的能量：

W=12∑i=1NQiφiW = \frac{1}{2} \sum^N_{i=1} Q_i \varphi_i

电场的能量

W=12∫ρφdV=−ε02∫φ∇2φdV\begin{aligned} W &= \frac{1}{2} \int \rho \varphi dV\\ &= -\frac{\varepsilon_0}{2} \int \varphi \nabla^2 \varphi dV \end{aligned}

有Green恒等式

∫Vφ∇2ψdV=∫Sφ∂ψ∂ndS−∫V∇φ⋅∇ψdV\int_V \varphi \nabla^2 \psi dV = \int_S \varphi \frac{\partial \psi}{\partial n} dS - \int_V \nabla \varphi \cdot \nabla \psi dV

取ψ=φ\psi = \varphi

∫Vφ∇2φdV=∫Sφ∂φ∂ndS−∫V∣∇φ∣2dV\int_V \varphi \nabla^2 \varphi dV = \int_S \varphi \frac{\partial \varphi}{\partial n} dS - \int_V |\nabla \varphi|^2 dV

当r→∞r \rightarrow \infty时，φ∝1r\varphi \propto \frac{1}{r}，∂φ∂n∝1r2\frac{\partial \varphi}{\partial n} \propto \frac{1}{r^2}，S∝r2S \propto r^2，∮Sφ∂φ∂ndS→0\oint_S \varphi \frac{\partial \varphi}{\partial n} dS \rightarrow 0
则

W=ε02∫∣∇φ∣2dV=∫12ε0E2dVW = \frac{\varepsilon_0}{2} \int |\nabla \varphi|^2 dV = \int \frac{1}{2} \varepsilon_0 E^2 dV

定义电场的能量密度

We=12ε0E2W_e = \frac{1}{2} \varepsilon_0 E^2

在介质中

We=12εE2=12D⃗⋅E⃗=12(ε0E2+P⃗⋅E⃗)W_e = \frac{1}{2} \varepsilon E^2 = \frac{1}{2} \vec{D} \cdot \vec{E} = \frac{1}{2} (\varepsilon_0E^2 + \vec{P} \cdot \vec{E})

其中12P⃗⋅E⃗\frac{1}{2}\vec{P} \cdot \vec{E}是介质的极化储能

静电系统的总能量

W=12∫ρfφdV=12∫εE2dVW=12ε0∫∣E1⃗+E2⃗∣2dV=12ε0∫∣E1⃗∣2dV+12ε0∫∣E2⃗∣2dV+ε0∫E1⃗⋅E2⃗dV\begin{aligned} W &= \frac{1}{2} \int \rho_f \varphi dV = \frac{1}{2} \int \varepsilon E^2 dV\\ W &= \frac{1}{2} \varepsilon_0 \int |\vec{E_1}+\vec{E_2}|^2 dV\\ &= \frac{1}{2} \varepsilon_0 \int |\vec{E_1}|^2 dV + \frac{1}{2} \varepsilon_0 \int |\vec{E_2}|^2 dV + \varepsilon_0 \int \vec{E_1} \cdot \vec{E_2} dV \end{aligned}

于是可以将总能量分为自能和相互作用能

静电势多级展开

已知全空间中区域分布的电荷ρ\rho，求空间电势

φ(x⃗)=14πε0∫ρ(x′⃗)∣r⃗∣dV′\varphi(\vec{x}) = \frac{1}{4\pi \varepsilon_0} \int \frac{\rho(\vec{x'})}{|\vec{r}|}dV'

在远离电荷分布区域的位置，即∣x⃗∣≫∣x′⃗∣|\vec{x}| \gg |\vec{x'}|

1∣x⃗−x′⃗∣=1∣x⃗∣−∇⋅x′⃗∣x⃗∣+12(x⃗′⋅∇)21∣x⃗∣+⋯\frac{1}{|\vec{x}-\vec{x'}|} = \frac{1}{|\vec{x}|} - \nabla \cdot \frac{\vec{x'}}{|\vec{x}|} + \frac{1}{2} (\vec{x}' \cdot \nabla)^2 \frac{1}{|\vec{x}|} + \cdots

记∣x⃗∣=R|\vec{x}| = R

零阶项

φ(0)=14πε0∫ρ(x′⃗)RdV′=14πε0R∫ρ(x′⃗)dV′=14πε0QR\begin{aligned} \varphi^{(0)}&= \frac{1}{4\pi \varepsilon_0} \int \frac{\rho(\vec{x'})}{R}dV'\\ &=\frac{1}{4\pi \varepsilon_0R} \int \rho(\vec{x'})dV'\\ &= \frac{1}{4\pi\varepsilon_0} \frac{Q}{R} \end{aligned}

一阶项

φ(1)=−14πε0∫ρ(x′⃗)x′⃗⋅∇1RdV′=14πε0∫ρ(x′⃗)x′⃗⋅R⃗R3dV′=14πε0[∫ρ(x′⃗)x′⃗dV′]⋅R⃗R3=14πε0p⃗⋅R⃗R3\begin{aligned} \varphi^{(1)} &= -\frac{1}{4\pi \varepsilon_0} \int \rho(\vec{x'}) \vec{x'} \cdot \nabla \frac{1}{R} dV'\\ &= \frac{1}{4\pi \varepsilon_0} \int \rho(\vec{x'}) \frac{\vec{x'} \cdot \vec{R}}{R^3} dV'\\ &= \frac{1}{4\pi \varepsilon_0} [\int \rho(\vec{x'}) \vec{x'} dV'] \cdot \frac{\vec{R}}{R^3}\\ &= \frac{1}{4\pi \varepsilon_0} \frac{\vec{p}\cdot \vec{R}}{R^3} \end{aligned}

其中p⃗=∫ρ(x′⃗)x′⃗dV′\vec{p} = \int \rho(\vec{x'}) \vec{x'} dV'是系统电偶极矩

p⃗=−qx⃗−+qx⃗+=qr⃗\vec{p} = -q \vec{x}_- + q \vec{x}_+ = q \vec{r}

当系统总电荷量为0的时候（Q=0），p⃗\vec{p}与坐标原点的位置选取无关

p1⃗=∫ρ(x′⃗)x′⃗dV′=∫ρ(x′⃗)(x′⃗−o⃗1)dV′p2⃗=∫ρ(x′⃗)x′⃗dV′=∫ρ(x′⃗)(x′⃗−o⃗2)dV′p⃗1−p⃗2=∫ρ(x′⃗)(o⃗2−o⃗1)dV′=(o⃗2−o⃗1)Q=0\vec{p_1} = \int \rho(\vec{x'}) \vec{x'} dV' = \int \rho(\vec{x'}) (\vec{x'}-\vec{o}_1) dV'\\ \vec{p_2} = \int \rho(\vec{x'}) \vec{x'} dV' = \int \rho(\vec{x'}) (\vec{x'}-\vec{o}_2) dV'\\ \vec{p}_1 - \vec{p}_2 = \int \rho(\vec{x'}) (\vec{o}_2 - \vec{o}_1) dV' = (\vec{o}_2 - \vec{o}_1)Q = 0

二阶项

φ(2)=14πε012∫ρ(x′⃗)(x′⃗⋅∇)21RdV′=14πε012∫ρ(x′⃗)xi′xj′∂2∂xi∂xj1RdV′=14πε016∑i,j[∫3ρ(x′⃗)xi′xj′dV′]∂2∂xi∂xj1R=14πε016∑i,jDij∂2∂xi∂xj1R\begin{aligned} \varphi^{(2)} &= \frac{1}{4\pi \varepsilon_0} \frac{1}{2}\int \rho(\vec{x'}) (\vec{x'} \cdot \nabla)^2 \frac{1}{R} dV'\\ &= \frac{1}{4\pi \varepsilon_0} \frac{1}{2}\int \rho(\vec{x'}) x'_i x'_j \frac{\partial^2}{\partial x_i \partial x_j} \frac{1}{R} dV'\\ &= \frac{1}{4\pi \varepsilon_0} \frac{1}{6}\sum_{i,j}[\int 3\rho(\vec{x'}) x'_i x'_j dV']\frac{\partial^2}{\partial x_i \partial x_j} \frac{1}{R}\\ &= \frac{1}{4\pi \varepsilon_0} \frac{1}{6} \sum_{i,j} D_{ij}\frac{\partial^2}{\partial x_i \partial x_j} \frac{1}{R} \end{aligned}

其中Dij=∫3ρ(x′⃗)xi′xj′dV′D_{ij} = \int 3\rho(\vec{x'}) x'_i x'_j dV'是系统的电四极矩

电四极子

Dij=DjiD_{ij}= D_{ji}，其中i代表的是正负电荷分开方向，j代表的是电偶极子分开方向

例：
计算D33D_{33}的电势

D33=∫3ρ(x′⃗)x3′x3′dV′=3qb2−3qa2−3qa2+3qb2=6qb2−6qa2=6q(b2−a2)=6q(b−a)(b+a)=6pl\begin{aligned} D_{33} &= \int 3\rho(\vec{x'}) x'_3 x'_3 dV'\\ &= 3 q b^2 - 3 q a^2 - 3 q a^2 + 3 q b^2\\ &= 6 q b^2 - 6 q a^2\\ &= 6 q (b^2 - a^2)\\ &= 6 q (b-a) (b+a)\\ &= 6 pl \end{aligned}

其中l=(b+a)l = (b+a)是两个电偶极子之间的距离，p=q(b−a)p = q(b-a)是电偶极子的电偶极矩
电势：

φ(x⃗)=14πε0p⃗⋅r⃗+r+3−14πε0p⃗⋅r⃗−r−3=14πε0p⃗⋅∇1r++14πε0p⃗⋅∇1r−=14πε0p⃗⋅(∇1r++∇1r−)=14πε0p⃗⋅∂∂z(1r−−1r−)=14πε0p⃗∂∂zr+−r−r+r−=−14πε0p⃗∂∂zlcosθR2=−pl4πε0∂∂zRcosθR3=−pl4πε0∂∂zzR2=6pl4πε0∂2∂2z1R\begin{aligned} \varphi(\vec{x}) &= \frac{1}{4\pi \varepsilon_0} \frac{\vec{p} \cdot \vec{r}_+}{r_+^3} - \frac{1}{4\pi \varepsilon_0} \frac{\vec{p} \cdot \vec{r}_-}{r_-^3}\\ &= \frac{1}{4\pi \varepsilon_0} \vec{p}\cdot \nabla \frac{1}{r_+} + \frac{1}{4\pi \varepsilon_0} \vec{p}\cdot \nabla \frac{1}{r_-}\\ &= \frac{1}{4\pi \varepsilon_0} \vec{p} \cdot (\nabla \frac{1}{r_+} + \nabla \frac{1}{r_-})\\ &= \frac{1}{4\pi \varepsilon_0} \vec{p} \cdot \frac{\partial}{\partial z}(\frac{1}{r_-} - \frac{1}{r_-})\\ &= \frac{1}{4\pi \varepsilon_0} \vec{p} \frac{\partial}{\partial z} \frac{r_+ - r_-}{r_+ r_-}\\ &= -\frac{1}{4\pi \varepsilon_0} \vec{p} \frac{\partial}{\partial z} \frac{lcos\theta}{R^2}\\ &= -\frac{pl}{4\pi \varepsilon_0} \frac{\partial}{\partial z} \frac{Rcos\theta}{R^3}\\ &= -\frac{pl}{4\pi \varepsilon_0} \frac{\partial}{\partial z} \frac{z}{R^2}\\ &= \frac{6pl}{4\pi \varepsilon_0} \frac{\partial^2}{\partial^2 z} \frac{1}{R} \end{aligned}