电荷

  • 分为+，-两种属性
  • 量子化，元电荷e≈1.6×10−19Ce \approx 1.6\times 10^{-19}C
  • 电荷量守恒

电荷的形态

点电荷：电荷大小qq
线电荷：电荷线密度ρl=lim⁡Δl→0ΔqΔl\rho_l = \lim_{\Delta l \to 0}\frac{\Delta q}{\Delta l}
面电荷：电荷面密度ρs=lim⁡ΔS→0ΔqΔS\rho_s = \lim_{\Delta S \to 0}\frac{\Delta q}{\Delta S}
体电荷：电荷体密度ρv=lim⁡ΔV→0ΔqΔV\rho_v = \lim_{\Delta V \to 0}\frac{\Delta q}{\Delta V}

ΔV\Delta V：“宏观小，微观大”

  • 相对于问题求解区是宏观小
  • 相对于微观带电粒子很大

点电荷的密度

ρ(r⃗)=qδ(r⃗−r⃗0)={∞r⃗=r⃗00r⃗≠r⃗0\rho(\vec{r}) = q\delta(\vec{r}-\vec{r}_0) = \left\{ \begin{array}{lr} \infin & \vec{r} = \vec{r}_0 \\ 0 & \vec{r} \neq \vec{r}_0 \end{array} \right.

考虑均匀的带电小球

ρ(r⃗)={q4/3⋅πR3r⃗∈V0r⃗∉V\rho(\vec{r}) = \left\{ \begin{array}{lr} \frac{q}{4/3 \cdot \pi R^3} & \vec{r} \in V \\ 0 & \vec{r} \notin V \end{array} \right.

那么点电荷密度就等于

ρ(r⃗)=lim⁡R→0ρR(x⃗)=qδ(x⃗−x⃗0)\rho(\vec{r}) = \lim_{R \rightarrow 0} \rho_R(\vec{x}) = q\delta(\vec{x}-\vec{x}_0)

δ\delta函数具有下面的性质

∫Vδ(x⃗−x0⃗)dx={1x0⃗∈V0x0⃗∉V∫Vf(x⃗)δ(x⃗−x0⃗)dx=f(x0⃗)    x0⃗∈V\int_V\delta(\vec{x} - \vec{x_0})dx = \left\{ \begin{array}{lr} 1 & \vec{x_0} \in V \\ 0 & \vec{x_0} \notin V \end{array} \right.\\ \int_Vf(\vec{x})\delta(\vec{x} - \vec{x_0})dx = f(\vec{x_0}) \space \space \space \space \vec{x_0} \in V

电流

线电流：

∣I⃗∣=lim⁡Δt→0ΔQΔt|\vec{I}| = \lim_{\Delta t \to 0}\frac{\Delta Q}{\Delta t}

（经过某一点的电流）
面电流：

∣JS⃗∣=lim⁡Δl→01ΔLlim⁡Δt→0ΔQΔt|\vec{J_S}| = \lim_{\Delta l \to 0}\frac{1}{\Delta L}\lim_{\Delta t \to 0}\frac{\Delta Q}{\Delta t}

（面的电流密度）
体电流：

∣JV⃗∣=lim⁡ΔS→01ΔSlim⁡Δt→0ΔQΔt|\vec{J_V}| = \lim_{\Delta S \to 0}\frac{1}{\Delta S}\lim_{\Delta t \to 0}\frac{\Delta Q}{\Delta t}

电荷与电流的关系

  • J⃗=ρv⃗\vec{J} = \rho \vec{v}（v⃗\vec{v}是电荷的运动速度）
  • ∇⋅J⃗+∂ρ∂t=0\nabla \cdot \vec{J} + \frac{\partial \rho}{\partial t} = 0（电荷守恒）（电流连续性方程）

静电场Coulomb定律

静电力：

F⃗=14πϵ0q1q2r3r⃗\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^3}\vec{r}

Q在x⃗\vec{x}处产生的电场：

E⃗(x⃗)=14πϵ0qr3r⃗\vec{E}(\vec{x}) = \frac{1}{4\pi\epsilon_0}\frac{q}{r^3}\vec{r}

叠加原理：

E⃗=∑iE⃗i\vec{E} = \sum_i \vec{E}_i

推广到连续分布电荷：

E⃗(r⃗)=14πϵ0∫Vρ(r′⃗)∣r⃗−r′⃗∣3(r⃗−r′⃗)dV′\vec{E}(\vec{r}) = \frac{1}{4\pi\epsilon_0}\int_V \frac{\rho(\vec{r'})}{|\vec{r}-\vec{r'}|^3}(\vec{r}-\vec{r'})dV'

Biot-Savart定律

稳恒电流场：

∂J⃗∂t=0,∇⋅J⃗=0\frac{\partial \vec{J}}{\partial t} = 0, \nabla \cdot \vec{J} = 0

磁感应强度：

B⃗(r⃗)=μ04π∫J⃗(r′⃗)×(r⃗−r′⃗)∣r⃗−r′⃗∣3dV′\vec{B}(\vec{r}) = \frac{\mu_0}{4\pi}\int \frac{\vec{J}(\vec{r'})\times(\vec{r}-\vec{r'})}{|\vec{r}-\vec{r'}|^3}dV'

磁场力：

  • 运动电荷：

F⃗=qv⃗×B⃗\vec{F} = q\vec{v}\times\vec{B}

线电流源：

dF⃗=I⃗×B⃗dld\vec{F} = \vec{I}\times\vec{B} dl

体电流源：

dF⃗=J⃗×B⃗dVd\vec{F} = \vec{J}\times\vec{B}dV

Lorentz力：电场力+磁场力

点电荷

F⃗=qE⃗+qv⃗×B⃗\vec{F} = q\vec{E} + q\vec{v}\times\vec{B}

体元：

dF⃗=ρE⃗dV+J⃗×B⃗dVd\vec{F} = \rho\vec{E}dV + \vec{J}\times\vec{B}dV

Maxwell方程组

静电场的散度

∇⋅E⃗=14πϵ0∇⋅∫Vρ(r′⃗)∣r⃗∣3r⃗dV′=14πϵ0∫V∇⋅ρ(r′⃗)∣r⃗∣3r⃗dV′=14πϵ0∫Vρ(r′⃗)∇⋅r⃗∣r⃗∣3dV′\begin{aligned} \nabla \cdot \vec{E} &= \frac{1}{4\pi \epsilon_0}\nabla \cdot \int_V \frac{\rho(\vec{r'})}{|\vec{r}|^3}\vec{r}dV' \\ &= \frac{1}{4\pi \epsilon_0}\int_V \nabla \cdot \frac{\rho(\vec{r'})}{|\vec{r}|^3}\vec{r}dV' \\ &= \frac{1}{4\pi \epsilon_0}\int_V \rho(\vec{r'})\nabla \cdot \frac{\vec{r}}{|\vec{r}|^3}dV' \\ \end{aligned}

根据 Gauss 定理

∫V∇⋅dV⃗=∮SE⃗⋅dS⃗=14πϵ0∫S[∫Vρ(r′⃗)r⃗∣r⃗∣3dV′]dS⃗\begin{aligned} \int_V \nabla \cdot d\vec{V} &= \oint_S \vec{E} \cdot d\vec{S} \\ &= \frac{1}{4\pi \epsilon_0}\int_S [\int_V \rho(\vec{r'}) \frac{\vec{r}}{|\vec{r}|^3}dV']d\vec{S} \\ \end{aligned}

考虑一个点电荷的情况

E⃗=14πϵ0qr3r⃗\vec{E} = \frac{1}{4\pi \epsilon_0}\frac{q}{r^3}\vec{r}

∮SE⃗⋅dS⃗=q4πϵ0∮Sr⃗r3⋅dS⃗=q4πϵ0∮Sr^r2dS=q4πϵ0∮sdΩ=q4πϵ04π=qϵ0\begin{aligned} \oint_S \vec{E} \cdot d\vec{S} &= \frac{q}{4\pi\epsilon_0}\oint_S \frac{\vec{r}}{r^3} \cdot d\vec{S} \\ &= \frac{q}{4\pi\epsilon_0}\oint_S \frac{\hat{r}}{r^2} dS \\ &= \frac{q}{4\pi\epsilon_0}\oint_s d\Omega \\ &= \frac{q}{4\pi\epsilon_0}4\pi \\ &= \frac{q}{\epsilon_0} \end{aligned}

q位于S外的时候

∮SE⃗⋅dS⃗=0\oint_S \vec{E} \cdot d\vec{S} = 0

多个电荷的情况

∮SE⃗⋅dS⃗=∑iqiϵ0\oint_S \vec{E} \cdot d\vec{S} = \sum_i \frac{q_i}{\epsilon_0}

这里要求qiq_i在积分面内

对于连续分布电荷

∮SE⃗⋅dS⃗=1ϵ0∫Vρ(r′⃗)dV′∫V∇⋅E⃗dV=1ϵ0∫Vρ(r′⃗)dV′∇⋅E⃗=ρϵ0\oint_S \vec{E} \cdot d\vec{S} = \frac{1}{\epsilon_0}\int_V \rho(\vec{r'})dV'\\ \int_V \nabla \cdot \vec{E}dV = \frac{1}{\epsilon_0}\int_V \rho(\vec{r'})dV'\\ \nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}

对于点电荷：

E⃗=14πϵ0qr3r⃗,ρ(x⃗)=qδ(x⃗−x′⃗)q4πϵ0∇⋅r⃗r3=qδ(x⃗−x′⃗)∇⋅r⃗r3=4πδ(r⃗)\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{q}{r^3}\vec{r}, \rho(\vec{x}) = q\delta(\vec{x} - \vec{x'})\\ \frac{q}{4\pi \epsilon_0}\nabla \cdot \frac{\vec{r}}{r^3} = q\delta(\vec{x} - \vec{x'})\\ \nabla \cdot \frac{\vec{r}}{r^3} = 4\pi \delta(\vec{r} )\\

∇⋅E⃗=14πϵ0∇⋅∫Vρ(r′⃗)∣r⃗∣3r⃗dV′=14πϵ0∫V∇⋅ρ(r′⃗)∣r⃗∣3r⃗dV′=4π4πϵ0∫Vρ(r′⃗)δ(x⃗−x′⃗)dV′=ρ(x⃗)ϵ0\begin{aligned} \nabla \cdot \vec{E} &= \frac{1}{4\pi \epsilon_0}\nabla \cdot \int_V \frac{\rho(\vec{r'})}{|\vec{r}|^3}\vec{r}dV' \\ &= \frac{1}{4\pi \epsilon_0}\int_V \nabla \cdot \frac{\rho(\vec{r'})}{|\vec{r}|^3}\vec{r}dV' \\ &= \frac{4 \pi}{4\pi \epsilon_0}\int_V \rho(\vec{r'})\delta(\vec{x}-\vec{x'}) dV' \\ &= \frac{\rho(\vec{x})}{\epsilon_0} \end{aligned}

静电场的旋度

考虑点电荷

E⃗=14πϵ0qr3r⃗\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{q}{r^3}\vec{r}

∮lE⃗⋅dl⃗=q4πϵ0∮lr⃗r3⋅dl⃗=q4πϵ0∮lr^r2dl=q4πϵ0∮ldrr2=0\begin{aligned} \oint_l \vec{E} \cdot d\vec{l} &= \frac{q}{4\pi\epsilon_0}\oint_l \frac{\vec{r}}{r^3} \cdot d\vec{l} \\ &= \frac{q}{4\pi\epsilon_0}\oint_l \frac{\hat{r}}{r^2} dl \\ &= \frac{q}{4\pi\epsilon_0}\oint_l \frac{dr}{r^2} \\ &= 0 \end{aligned}

由Stokes定理

∮lE⃗⋅dl⃗=∫S(∇×E⃗)⋅dS⃗∇×E⃗=0\oint_l \vec{E} \cdot d\vec{l} = \int_S (\nabla \times \vec{E}) \cdot d\vec{S}\\ \nabla \times \vec{E} = 0

静电场是无旋场，所以存在电势（静电场是势场的梯度场）

∇×E⃗=0⇒E⃗=−∇φ\nabla \times \vec{E} = 0 \Rightarrow \vec{E} = -\nabla \varphi

静电势 φ\varphi

定义：把单位电荷从规定的零电势点移到某一点克服电场力所做的功

φ(x⃗)=−∫x0⃗x⃗E⃗⋅dl⃗\varphi(\vec{x}) = -\int_{\vec{x_0}}^{\vec{x}} \vec{E} \cdot d\vec{l}

计算φ\varphi沿着l^\hat{l}方向的导数

lim⁡Δl→0=−E^⋅l^=∇φ⋅l^\lim_{\Delta_l \to 0} = - \hat{E} \cdot \hat{l} = \nabla \varphi \cdot \hat{l}

E⃗=−∇φ∇⋅E⃗=−∇2φ=ρϵ0\vec{E} = -\nabla \varphi\\ \nabla \cdot \vec{E} = -\nabla^2 \varphi = \frac{\rho}{\epsilon_0}

∇2φ=−ρϵ0\nabla^2 \varphi = -\frac{\rho}{\epsilon_0}

此为Poisson方程
点电荷电势

φ=14πϵ0qr\varphi = \frac{1}{4\pi\epsilon_0}\frac{q}{r}

连续分布电荷电势

φ(r⃗)=14πϵ0∫Vρ(r′⃗)∣r⃗−r′⃗∣dV′\varphi(\vec{r}) = \frac{1}{4\pi\epsilon_0}\int_V \frac{\rho(\vec{r'})}{|\vec{r}-\vec{r'}|}dV'

∇1r=−r⃗r3∇21r=0\nabla \frac{1}{r} = -\frac{\vec{r}}{r^3}\\ \nabla^2 \frac{1}{r} = 0

静磁场的散度

B⃗(x⃗)=μ04π∫J⃗(x′⃗)×r⃗r3dV′=−μ04π∫J⃗(x′⃗)×∇1rdV′=μ04π∫∇×J⃗(x′⃗)rdV′=μ04π∇×∫J⃗(x′⃗)rdV′=μ04π∇×A⃗\begin{aligned} \vec{B}(\vec{x}) &= \frac{\mu_0}{4\pi}\int \frac{\vec{J}(\vec{x'})\times \vec{r}}{r^3}dV'\\ &= -\frac{\mu_0}{4\pi}\int \vec{J}(\vec{x'})\times \nabla \frac{1}{r} dV'\\ &= \frac{\mu_0}{4\pi}\int \nabla \times \frac{\vec{J}(\vec{x'})}{r}dV'\\ &= \frac{\mu_0}{4\pi} \nabla \times \int \frac{\vec{J}(\vec{x'})}{r}dV'\\ &= \frac{\mu_0}{4\pi} \nabla \times \vec{A} \end{aligned}

其中A⃗\vec{A}是磁矢势

A⃗=μ04π∫J⃗(x′⃗)rdV′\vec{A} = \frac{\mu_0}{4\pi}\int \frac{\vec{J}(\vec{x'})}{r}dV'

∇⋅B⃗=∇⋅(∇×A⃗)=0\nabla \cdot \vec{B} = \nabla \cdot (\nabla \times \vec{A}) = 0

磁场是无源场

静磁场的旋度

∇×B⃗=∇×(∇×A⃗)=∇(∇⋅A⃗)−∇2A⃗\nabla \times \vec{B} = \nabla \times (\nabla \times \vec{A}) = \nabla(\nabla \cdot \vec{A}) - \nabla^2 \vec{A}

先考虑分解式的第一项

∇⋅A⃗=μ04π∇⋅∫J⃗(x′⃗)rdV′=μ04π∫J⃗(x′⃗)⋅∇1rdV′=−μ04π∫∇′1r⋅J⃗(x′⃗)dV′=μ04π∫1r∇′⋅J⃗(x′⃗)dV′−μ04π∫∇′⋅J⃗(x′⃗)rdV′=−μ04π∫∇′⋅J⃗(x′⃗)rdV′=−μ04π∮J⃗(x′⃗)r⋅dS⃗=0\begin{aligned} \nabla \cdot \vec{A} &= \frac{\mu_0}{4\pi}\nabla \cdot \int \frac{\vec{J}(\vec{x'})}{r}dV' = \frac{\mu_0}{4\pi}\int \vec{J}(\vec{x'})\cdot \nabla \frac{1}{r}dV'\\ &= -\frac{\mu_0}{4\pi}\int \nabla' \frac{1}{r} \cdot \vec{J}(\vec{x'})dV'\\ &= \frac{\mu_0}{4\pi}\int \frac{1}{r}\nabla' \cdot \vec{J}(\vec{x'})dV' - \frac{\mu_0}{4\pi}\int \nabla' \cdot \frac{\vec{J}(\vec{x'})}{r}dV'\\ &= -\frac{\mu_0}{4\pi}\int \nabla' \cdot \frac{\vec{J}(\vec{x'})}{r}dV'\\ &= -\frac{\mu_0}{4\pi}\oint \frac{\vec{J}(\vec{x'})}{r} \cdot d\vec{S}=0\\ \end{aligned}

最后一步考虑到S内部是包含所有的电流的，所以积分为0
接下来考虑分解式的第二项

∇2A⃗=μ04π∇2∫J⃗(x′⃗)rdV′=μ04π∫∇2J⃗(x′⃗)rdV′=μ04π∫J⃗(x′⃗)⋅∇21rdV′=−4πμ04π∫J⃗(x′⃗)⋅δ(x⃗−x′⃗)dV′=−μ0J⃗\begin{aligned} \nabla^2 \vec{A} &=\frac{\mu_0}{4\pi}\nabla^2 \int \frac{\vec{J}(\vec{x'})}{r}dV' = \frac{\mu_0}{4\pi}\int \nabla^2 \frac{\vec{J}(\vec{x'})}{r}dV'\\ &= \frac{\mu_0}{4\pi}\int \vec{J}(\vec{x'})\cdot \nabla^2 \frac{1}{r}dV'\\ &= -\frac{4\pi\mu_0}{4\pi}\int \vec{J}(\vec{x'})\cdot \delta(\vec{x}-\vec{x'}) dV'\\ &= -\mu_0 \vec{J} \end{aligned}

所以

∇×B⃗=μ0J⃗\nabla \times \vec{B} = \mu_0 \vec{J}

并且有

∮lB⃗⋅dl⃗=∫S(∇×B⃗)⋅dS⃗=μ0∫SJ⃗⋅dS⃗\oint_l \vec{B} \cdot d\vec{l} = \int_S (\nabla \times \vec{B}) \cdot d\vec{S} = \mu_0 \int_S \vec{J} \cdot d\vec{S}

这是经典的Ampere环路定理

磁矢势有

∇2A⃗=−μ0J⃗\nabla^2 \vec{A} = -\mu_0 \vec{J}

这是矢量Poisson方程