电磁场公式大全
代码及其markdown源代码的百度云链接如下，请注意：里面有些公式可能由于笔者的疏漏有错误，请不要全部采信。本内容仅供学习交流使用，完全免费，因此也请不要用本内容进行任何收费活动，如转载请不要对本内容进行修改。全文共5w+字符，公式全手打，作者也挺辛苦的。以上君子协议，望大家遵守。时间仓促，内容如有错误，还望多多海涵。
链接：https://pan.baidu.com/s/1Wz4_KqqFDf1AppcATzBj1g
提取码：kj7c
超链接目录如下：
文章目录
- 第零章
-
- 常用公式定义
- 第一章——坐标变换与哈密顿算子
-
- 1、坐标变换公式
- 2、哈密顿算子性质
- 3、各坐标系下哈密顿算子性质
- 第二章——位函数与矢量方程
-
- 1、位函数
- 2、矢量方程
- 第三章——边界条件方程
-
- 1、边界条件方程
- 第四章——静电场
-
- 1、对称/非对称场结论
- 2、电偶极子
- 3、电极化
- 4、静电场的哈密顿算符/边界方程
- 5、电场能量/电容
- 6、n个导体平面相对
- 第五章——恒定电流场
-
- 1、电导、电阻、电功率
- 2、JDεσ对偶/恒定电流场边界条件
- 第六章——恒定磁场
-
- 1、对称/非对称场结论
- 2、磁偶极子
- 3、磁位
- 4、磁化
- 5、恒定磁场的哈密顿算符/边界方程
- 6、磁场能量/磁阻
- 7、自感
- 第七章——静态场边值问题
-
- 1、平面电场镜像
- 2、平面磁场镜像
- 3、实心导体球镜像
- 4、圆柱面镜像
- 5、两圆柱电容
- 第八章——正弦电磁场
-
- 1、正弦场量
- 2、色散/坡印廷矢量
- 3、k、E、H转换
- 4、极化
- 5、波的衰减
- 6、极化波/驻波/折射反射
- 第九章——导行波
-
- 1、TE/TM波
- 2、波导中的能量损耗
- 3、TEM波
- 第十章——电磁辐射
-
- 1、赫芝偶极子
- 2、磁偶极子天线辐射
- 3、天线阵
- 最终章——公式推导合集
第零章
常用公式定义
本章用于定义一些常用的方程及其简称，方程简称见右端的括号。
∇ ⋅ D ⃗ = ρ (D) \nabla \cdot \boldsymbol{\vec{D}}=\rho\tag{D} ∇⋅D=ρ(D)
∇ ⋅ B ⃗ = 0 (B) \nabla \cdot \boldsymbol{\vec{B}}=0\tag{B} ∇⋅B=0(B)
∇ × E ⃗ = − ∂ B ⃗ ∂ t (E) \nabla \times \boldsymbol{\vec{E}}=-\frac{\partial \boldsymbol{\vec{B}}}{\partial t}\tag{E} ∇×E=−∂t∂B(E)
∇ × H ⃗ = J ⃗ + ∂ D ⃗ ∂ t (H) \nabla \times \boldsymbol{\vec{H}}=\boldsymbol{\vec{J}}+\frac{\partial \boldsymbol{\vec{D}}}{\partial t}\tag{H} ∇×H=J+∂t∂D(H)
∮ S D ⃗ ⋅ d S ⃗ = ∫ V ρ ⋅ d V (ID) \oint_S{\boldsymbol{\vec{D}}\cdot \text{d}\boldsymbol{\vec{S}}=\int_V{\rho \cdot \text{d}V}}\tag{ID} ∮SD⋅dS=∫Vρ⋅dV(ID)
∮ S B ⃗ ⋅ d S ⃗ = 0 (IB) \oint_S{\boldsymbol{\vec{B}}\cdot \text{d}\boldsymbol{\vec{S}}=0}\tag{IB} ∮SB⋅dS=0(IB)
∮ l E ⃗ ⋅ d l ⃗ = − ∫ S ∂ B ⃗ ∂ t ⋅ d S ⃗ + ∮ l ( v ⃗ × B ⃗ ) ⋅ d l ⃗ (IE) \oint_l{\boldsymbol{\vec{E}}\cdot \text{d}\boldsymbol{\vec{l}}}=-\int_S{\frac{\partial \boldsymbol{\vec{B}}}{\partial t}\cdot \text{d}\boldsymbol{\vec{S}}}+\oint_l{\begin{array}{c} \left( \boldsymbol{\vec{v}}\times \boldsymbol{\vec{B}} \right) \cdot \text{d}\boldsymbol{\vec{l}}\\\end{array}}\tag{IE} ∮lE⋅dl=−∫S∂t∂B⋅dS+∮l(v×B)⋅dl(IE)
∮ l H ⃗ ⋅ d l ⃗ = ∫ S J ⃗ ⋅ d S ⃗ + ∫ S ∂ D ⃗ ∂ t ⋅ d S ⃗ (IH) \oint_l{\boldsymbol{\vec{H}}\cdot \text{d}\boldsymbol{\vec{l}}}=\int_S{\boldsymbol{\vec{J}}\cdot \text{d}\boldsymbol{\vec{S}}}+\int_S{\frac{\partial \boldsymbol{\vec{D}}}{\partial t}\cdot \text{d}\boldsymbol{\vec{S}}}\tag{IH} ∮lH⋅dl=∫SJ⋅dS+∫S∂t∂D⋅dS(IH)
∇ ⋅ D ⃗ = ρ (WD) \nabla \cdot \boldsymbol{\vec{D}}=\rho\tag{WD} ∇⋅D=ρ(WD)
∇ ⋅ B ⃗ = 0 (WB) \nabla \cdot \boldsymbol{\vec{B}}=0\tag{WB} ∇⋅B=0(WB)
∇ × E ⃗ = − j ω B ⃗ (WE) \nabla \times \boldsymbol{\vec{E}}=-j\omega \boldsymbol{\vec{B}}\tag{WE} ∇×E=−jωB(WE)
∇ × H ⃗ = J ⃗ + j ω D ⃗ (WH) \nabla \times \boldsymbol{\vec{H}}=\boldsymbol{\vec{J}}+j\omega \boldsymbol{\vec{D}}\tag{WH} ∇×H=J+jωD(WH)
∇ × ∇ × E ⃗ = ∇ ( ∇ ⋅ E ⃗ ) − ∇ 2 E ⃗ (NXX) \nabla \times \nabla \times \boldsymbol{\vec{E}}=\nabla \left( \nabla \cdot \boldsymbol{\vec{E}} \right) -\nabla ^2\boldsymbol{\vec{E}}\tag{NXX} ∇×∇×E=∇(∇⋅E)−∇2E(NXX)
A ⃗ × ( B ⃗ × C ⃗ ) = B ⃗ ( A ⃗ ⋅ C ⃗ ) − C ⃗ ( A ⃗ ⋅ B ⃗ ) (VXX) \boldsymbol{\vec{A}}\times \left( \boldsymbol{\vec{B}}\times \boldsymbol{\vec{C}} \right) =\boldsymbol{\vec{B}}\left( \boldsymbol{\vec{A}}\cdot \boldsymbol{\vec{C}} \right) -\boldsymbol{\vec{C}}\left( \boldsymbol{\vec{A}}\cdot \boldsymbol{\vec{B}} \right)\tag{VXX} A×(B×C)=B(A⋅C)−C(A⋅B)(VXX)
∇ ⋅ ( A ⃗ × B ⃗ ) = B ⃗ ( ∇ × A ⃗ ) − A ⃗ ( ∇ × B ⃗ ) (NDX) \nabla \cdot \left( \boldsymbol{\vec{A}}\times \boldsymbol{\vec{B}} \right) =\boldsymbol{\vec{B}}\left( \nabla \times \boldsymbol{\vec{A}} \right) -\boldsymbol{\vec{A}}\left( \nabla \times \boldsymbol{\vec{B}} \right)\tag{NDX} ∇⋅(A×B)=B(∇×A)−A(∇×B)(NDX)
本构方程不必言明即可直接应用：
D ⃗ = ε E ⃗ \boldsymbol{\vec{D}}=\varepsilon \boldsymbol{\vec{E}} D=εE
B ⃗ = μ H ⃗ \boldsymbol{\vec{B}}=\mu \boldsymbol{\vec{H}} B=μH
J ⃗ = σ E ⃗ \boldsymbol{\vec{J}}=\sigma \boldsymbol{\vec{E}} J=σE
第一章——坐标变换与哈密顿算子
1、坐标变换公式
{ r ⃗ = { x , y , z } r ⃗ = { ρ cos φ , ρ sin φ , z } r ⃗ = { r sin θ cos φ , r sin θ sin φ , r cos θ } (1.1.1) \begin{cases} \boldsymbol{\vec{r}}=\left\{ x,y,z \right\}\\ \boldsymbol{\vec{r}}=\left\{ \rho \cos \varphi ,\rho \sin \varphi ,z \right\}\\ \boldsymbol{\vec{r}}=\left\{ r\sin \theta \cos \varphi ,r\sin \theta \sin \varphi ,r\cos \theta \right\}\\\end{cases}\tag{1.1.1} ⎩⎪⎨⎪⎧r={x,y,z}r={ρcosφ,ρsinφ,z}r={rsinθcosφ,rsinθsinφ,rcosθ}(1.1.1)
式==(1.1.1)==对应元素相等，例如：
{ x = r sin θ cos φ y = r sin θ sin φ z = r cos θ ⇒ { r = x 2 + y 2 + z 2 sin θ = ρ r , cos θ = z r sin φ = y ρ , cos φ = x ρ (1.1.2) \begin{cases} x=r\sin \!\:\theta \cos \!\:\varphi\\ y=r\sin \theta \sin \varphi\\ z=r\cos \theta\\\end{cases}\Rightarrow \begin{cases} r=\sqrt{x^2+y^2+z^2}\\ \sin \theta =\frac{\rho}{r},\cos \theta =\frac{z}{r}\\ \sin \varphi =\frac{y}{\rho},\cos \varphi =\frac{x}{\rho}\\\end{cases}\tag{1.1.2} ⎩⎪⎨⎪⎧x=rsinθcosφy=rsinθsinφz=rcosθ⇒⎩⎪⎨⎪⎧r=x2+y2+z2sinθ=rρ,cosθ=rzsinφ=ρy,cosφ=ρx(1.1.2)
{ x = ρ cos φ y = ρ sin φ z = z ⇒ { ρ = x 2 + y 2 cos φ = x ρ , sin φ = y ρ z = z (1.1.3) \begin{cases} x=\rho \cos \!\:\varphi\\ y=\rho \sin \varphi\\ z=z\\\end{cases}\Rightarrow \begin{cases} \rho =\sqrt{x^2+y^2}\\ \cos \!\:\varphi =\frac{x}{\rho},\sin \varphi =\frac{y}{\rho}\\ z=z\\\end{cases}\tag{1.1.3} ⎩⎪⎨⎪⎧x=ρcosφy=ρsinφz=z⇒⎩⎪⎨⎪⎧ρ=x2+y2cosφ=ρx,sinφ=ρyz=z(1.1.3)
{ ρ = r sin θ φ = φ z = r cos θ ⇒ { r = ρ 2 + z 2 sin θ = ρ r , cos θ = z r φ = φ (1.1.4) \begin{cases} \rho =r\sin \theta\\ \varphi =\varphi\\ z=r\cos \theta\\\end{cases}\Rightarrow \begin{cases} r=\sqrt{\rho ^2+z^2}\\ \sin \theta =\frac{\rho}{r},\cos \theta =\frac{z}{r}\\ \varphi =\varphi\\\end{cases}\tag{1.1.4} ⎩⎪⎨⎪⎧ρ=rsinθφ=φz=rcosθ⇒⎩⎪⎨⎪⎧r=ρ2+z2sinθ=rρ,cosθ=rzφ=φ(1.1.4)
并且给出坐标间的变换公式：
[ x ^ y ^ z ^ ] = [ sin θ cos φ cos θ cos φ − sin φ sin θ sin φ cos θ sin φ cos φ cos θ − sin θ 0 ] [ r ^ θ ^ φ ^ ] (1.1.5) \left[ \begin{array}{c} \boldsymbol{\hat{x}}\\ \boldsymbol{\hat{y}}\\ \boldsymbol{\hat{z}}\\\end{array} \right] =\left[ \begin{matrix} \sin \!\:\theta \cos \!\:\varphi& \cos \!\:\theta \cos \!\:\varphi& -\sin \!\:\varphi\\ \sin \!\:\theta \sin \!\:\varphi& \cos \!\:\theta \sin \!\:\varphi& \cos \!\:\varphi\\ \cos \!\:\theta& -\sin \!\:\theta& 0\\\end{matrix} \right] \left[ \begin{array}{c} \boldsymbol{\hat{r}}\\ \boldsymbol{\hat{\theta}}\\ \boldsymbol{\hat{\varphi}}\\\end{array} \right]\tag{1.1.5} ⎣⎡x^y^z^⎦⎤=⎣⎡sinθcosφsinθsinφcosθcosθcosφcosθsinφ−sinθ−sinφcosφ0⎦⎤⎣⎡r^θ^φ^⎦⎤(1.1.5)
[ x ^ y ^ z ^ ] = [ cos φ − sin φ 0 sin φ cos φ 0 0 0 1 ] [ ρ ^ φ ^ z ^ ] (1.1.6) \left[ \begin{array}{c} \boldsymbol{\hat{x}}\\ \boldsymbol{\hat{y}}\\ \boldsymbol{\hat{z}}\\\end{array} \right] =\left[ \begin{matrix} \cos \!\:\varphi& -\sin \!\:\varphi& 0\\ \sin \!\:\varphi& \cos \!\:\varphi& 0\\ 0& 0& 1\\\end{matrix} \right] \left[ \begin{array}{c} \boldsymbol{\hat{\rho}}\\ \boldsymbol{\hat{\varphi}}\\ \boldsymbol{\hat{z}}\\\end{array} \right]\tag{1.1.6} ⎣⎡x^y^z^⎦⎤=⎣⎡cosφsinφ0−sinφcosφ0001⎦⎤⎣⎡ρ^φ^z^⎦⎤(1.1.6)
[ ρ ^ φ ^ z ^ ] = [ sin θ cos θ 0 0 0 1 cos θ − sin θ 0 ] [ r ^ θ ^ φ ^ ] (1.1.7) \left[ \begin{array}{c} \boldsymbol{\hat{\rho}}\\ \boldsymbol{\hat{\varphi}}\\ \boldsymbol{\hat{z}}\\\end{array} \right] =\left[ \begin{matrix} \sin \theta& \cos \theta& 0\\ 0& 0& 1\\ \cos \theta& -\sin \theta& 0\\\end{matrix} \right] \left[ \begin{array}{c} \boldsymbol{\hat{r}}\\ \boldsymbol{\hat{\theta}}\\ \boldsymbol{\hat{\varphi}}\\\end{array} \right]\tag{1.1.7} ⎣⎡ρ^φ^z^⎦⎤=⎣⎡sinθ0cosθcosθ0−sinθ010⎦⎤⎣⎡r^θ^φ^⎦⎤(1.1.7)
考虑到此处雅可比矩阵是正交矩阵，将它们转置过来，即可得到逆变换，考虑到我们希望矩阵里由新的基的元素表示，因此如此化简：
[ r ^ θ ^ φ ^ ] = 1 r ρ [ x ρ y ρ z ρ z x z y − ρ 2 − r y r x 0 ] [ x ^ y ^ z ^ ] (1.1.8) \left[ \begin{array}{c} \boldsymbol{\hat{r}}\\ \boldsymbol{\hat{\theta}}\\ \boldsymbol{\hat{\varphi}}\\\end{array} \right] =\frac{1}{r\rho}\left[ \begin{matrix} x\rho& y\rho& z\rho\\ zx& zy& -\rho ^2\\ -ry& rx& 0\\\end{matrix} \right] \left[ \begin{array}{c} \boldsymbol{\hat{x}}\\ \boldsymbol{\hat{y}}\\ \boldsymbol{\hat{z}}\\\end{array} \right]\tag{1.1.8} ⎣⎡r^θ^φ^⎦⎤=rρ1⎣⎡xρzx−ryyρzyrxzρ−ρ20⎦⎤⎣⎡x^y^z^⎦⎤(1.1.8)
[ ρ ^ φ ^ z ^ ] = 1 ρ [ x y 0 − y x 0 0 0 ρ ] [ x ^ y ^ z ^ ] (1.1.9) \left[ \begin{array}{c} \boldsymbol{\hat{\rho}}\\ \boldsymbol{\hat{\varphi}}\\ \boldsymbol{\hat{z}}\\\end{array} \right] =\frac{1}{\rho}\left[ \begin{matrix} x& y& 0\\ -y& x& 0\\ 0& 0& \rho\\\end{matrix} \right] \left[ \begin{array}{c} \boldsymbol{\hat{x}}\\ \boldsymbol{\hat{y}}\\ \boldsymbol{\hat{z}}\\\end{array} \right]\tag{1.1.9} ⎣⎡ρ^φ^z^⎦⎤=ρ1⎣⎡x−y0yx000ρ⎦⎤⎣⎡x^y^z^⎦⎤(1.1.9)
[ r ^ θ ^ φ ^ ] = 1 r [ ρ 0 z z 0 − ρ 0 r 0 ] [ ρ ^ φ ^ z ^ ] (1.1.10) \left[ \begin{array}{c} \boldsymbol{\hat{r}}\\ \boldsymbol{\hat{\theta}}\\ \boldsymbol{\hat{\varphi}}\\\end{array} \right] =\frac{1}{r}\left[ \begin{matrix} \rho& 0& z\\ z& 0& -\rho\\ 0& r& 0\\\end{matrix} \right] \left[ \begin{array}{c} \boldsymbol{\hat{\rho}}\\ \boldsymbol{\hat{\varphi}}\\ \boldsymbol{\hat{z}}\\\end{array} \right]\tag{1.1.10} ⎣⎡r^θ^φ^⎦⎤=r1⎣⎡ρz000rz−ρ0⎦⎤⎣⎡ρ^φ^z^⎦⎤(1.1.10)
2、哈密顿算子性质
∇ f = 1 h a ∂ f ∂ a a ^ + 1 h b ∂ f ∂ b b ^ + 1 h c ∂ f ∂ c c ^ (1.2.1) \nabla f=\frac{1}{h_a}\frac{\partial f}{\partial a}\boldsymbol{\hat{a}}+\frac{1}{h_b}\frac{\partial f}{\partial b}\boldsymbol{\hat{b}}+\frac{1}{h_c}\frac{\partial f}{\partial c}\boldsymbol{\hat{c}}\tag{1.2.1} ∇f=ha1∂a∂fa^+hb1∂b∂fb^+hc1∂c∂fc^(1.2.1)
∇ ⋅ G = 1 h a h b h c ( ∂ h b h c G 1 ∂ a + ∂ h c h a G 2 ∂ b + ∂ h a h b G 3 ∂ c ) (1.2.2) \nabla \cdot \boldsymbol{G}=\frac{1}{h_ah_bh_c}\left( \frac{\partial h_bh_cG_1}{\partial a}+\frac{\partial h_ch_aG_2}{\partial b}+\frac{\partial h_ah_bG_3}{\partial c} \right) \tag{1.2.2} ∇⋅G=hahbhc1(∂a∂hbhcG1+∂b∂hchaG2+∂c∂hahbG3)(1.2.2)
∇ × G = 1 h a h b h c ∣ h a a ^ h b b ^ h c c ^ ∂ ∂ a ∂ ∂ b ∂ ∂ c ( h a G 1 ) ( h b G 2 ) ( h c G 3 ) ∣ (1.2.3) \nabla \times \boldsymbol{G}=\frac{1}{h_ah_bh_c}\left| \begin{matrix} h_a\boldsymbol{\hat{a}}& h_b\boldsymbol{\hat{b}}& h_c\boldsymbol{\hat{c}}\\ \frac{\partial}{\partial a}& \frac{\partial}{\partial b}& \frac{\partial}{\partial c}\\ \left( h_aG_1 \right)& \left( h_bG_2 \right)& \left( h_cG_3 \right)\\\end{matrix} \right|\tag{1.2.3} ∇×G=hahbhc1∣∣∣∣∣∣haa^∂a∂(haG1)hbb^∂b∂(hbG2)hcc^∂c∂(hcG3)∣∣∣∣∣∣(1.2.3)
关于==(1.2.4)==式，详见参考文献1：
∇ 2 = 1 h a h b h c [ ∂ ∂ a ( h b h c h a ∂ ∂ a ) + ∂ ∂ b ( h c h a h b ∂ ∂ b ) + ∂ ∂ c ( h a h b h c ∂ ∂ c ) ] （无论标量矢量） (1.2.4) \nabla ^2=\frac{1}{h_ah_{\begin{array}{c} b\\ \end{array}}h_c}\left[ \frac{\partial}{\partial a}\left( \frac{h_{\begin{array}{c} b\\ \end{array}}h_c}{h_a}\frac{\partial}{\partial a} \right) +\frac{\partial}{\partial b}\left( \frac{h_{\begin{array}{c} c\\ \end{array}}h_a}{h_b}\frac{\partial}{\partial b} \right) +\frac{\partial}{\partial c}\left( \frac{h_{\begin{array}{c} a\\ \end{array}}h_b}{h_c}\frac{\partial}{\partial c} \right) \right] \,\,\text{（无论标量矢量）}\tag{1.2.4} ∇2=hahbhc1[∂a∂(hahbhc∂a∂)+∂b∂(hbhcha∂b∂)+∂c∂(hchahb∂c∂)]（无论标量矢量）(1.2.4)
∇ f = ∂ f ∂ n ^ e ^ n (1.2.5) \nabla f=\frac{\partial f}{\partial \hat{n}}\boldsymbol{\hat{e}}_n\tag{1.2.5} ∇f=∂n^∂fe^n(1.2.5)
∇ × ∇ × E ⃗ = ∇ ( ∇ ⋅ E ⃗ ) − ∇ 2 E ⃗ (NXX) \colorbox{cyan}{$\nabla \times \nabla \times \boldsymbol{\vec{E}}=\nabla \left( \nabla \cdot \boldsymbol{\vec{E}} \right) -\nabla ^2\boldsymbol{\vec{E}}$}\tag{NXX} ∇×∇×E=∇(∇⋅E)−∇2E(NXX)
3、各坐标系下哈密顿算子性质
{ ∇ f = ∂ f ∂ x x ^ + ∂ f ∂ y y ^ + ∂ f ∂ z z ^ ∇ f = ∂ f ∂ ρ ρ ^ + 1 ρ ∂ f ∂ φ φ ^ + ∂ f ∂ z z ^ ∇ f = ∂ f ∂ r r ^ + 1 r ∂ f ∂ θ θ ^ + 1 r sin θ ∂ f ∂ φ φ ^ (1.3.1) \begin{cases} \nabla f =\frac{\partial f}{\partial x}\boldsymbol{\hat{x}}+\frac{\partial f}{\partial y}\boldsymbol{\hat{y}}+\frac{\partial f}{\partial z}\boldsymbol{\hat{z}}\\ \nabla f=\frac{\partial f}{\partial \rho}\boldsymbol{\hat{\rho}}+\frac{1}{\rho}\frac{\partial f}{\partial \varphi}\boldsymbol{\hat{\varphi}}+\frac{\partial f}{\partial z}\boldsymbol{\hat{z}}\\ \nabla f =\frac{\partial f}{\partial r}\boldsymbol{\hat{r}}+\frac{1}{r}\frac{\partial f}{\partial \theta}\boldsymbol{\hat{\theta}}+\frac{1}{r\sin \theta}\frac{\partial f}{\partial \varphi}\boldsymbol{\hat{\varphi}}\\\end{cases}\tag{1.3.1} ⎩⎪⎨⎪⎧∇f=∂x∂fx^+∂y∂fy^+∂z∂fz^∇f=∂ρ∂fρ^+ρ1∂φ∂fφ^+∂z∂fz^∇f=∂r∂fr^+r1∂θ∂fθ^+rsinθ1∂φ∂fφ^(1.3.1)
{ ∇ ⋅ G = ∂ G 1 ∂ x + ∂ G 2 ∂ y + ∂ G 3 ∂ z ∇ ⋅ G = 1 ρ ∂ ρ G 1 ∂ ρ + 1 ρ ∂ G 2 ∂ φ + ∂ G 3 ∂ z ∇ ⋅ G = 1 r 2 ∂ r 2 G 1 ∂ r + 1 r sin θ ∂ sin θ G 2 ∂ θ + 1 r sin θ ∂ G 3 ∂ φ (1.3.2) \begin{cases} \nabla \cdot \boldsymbol{G} =\frac{\partial G_1}{\partial x}&+\frac{\partial G_2}{\partial y}&+\frac{\partial G_3}{\partial z}\\ \nabla \cdot \boldsymbol{G} =\frac{1}{\rho}\frac{\partial \rho G_1}{\partial \rho}&+\frac{1}{\rho}\frac{\partial G_2}{\partial \varphi}&+\frac{\partial G_3}{\partial z}\\ \nabla \cdot \boldsymbol{G}=\frac{1}{r^2}\frac{\partial r^2G_1}{\partial r}&+\frac{1}{r\sin \theta}\frac{\partial \sin \theta G_2}{\partial \theta}&+\frac{1}{r\sin \theta}\frac{\partial G_3}{\partial \varphi}\\\end{cases}\tag{1.3.2} ⎩⎪⎨⎪⎧∇⋅G=∂x∂G1∇⋅G=ρ1∂ρ∂ρG1∇⋅G=r21∂r∂r2G1+∂y∂G2+ρ1∂φ∂G2+rsinθ1∂θ∂sinθG2+∂z∂G3+∂z∂G3+rsinθ1∂φ∂G3(1.3.2)
{ ∇ × G = ∣ x ^ y ^ z ^ ∂ ∂ x ∂ ∂ y ∂ ∂ z ( G 1 ) ( G 2 ) ( G 3 ) ∣ ∇ × G = ∣ ρ ^ ρ φ ^ z ^ ∂ ∂ ρ ∂ ∂ φ ∂ ∂ z ( G 1 ) ( ρ G 2 ) ( G 3 ) ∣ ∇ × G = ∣ r ^ r θ ^ r sin θ φ ^ ∂ ∂ r ∂ ∂ θ ∂ ∂ φ ( G 1 ) ( r G 2 ) ( r sin θ G 3 ) ∣ (1.3.3) \begin{cases} \nabla \times \boldsymbol{G} =\left| \begin{matrix} \boldsymbol{\hat{x}}& \boldsymbol{\hat{y}}& \boldsymbol{\hat{z}}\\ \frac{\partial}{\partial x}& \frac{\partial}{\partial y}& \frac{\partial}{\partial z}\\ \left( G_1 \right)& \left( G_2 \right)& \left( G_3 \right)\\\end{matrix} \right|\\ \nabla \times \boldsymbol{G}=\left| \begin{matrix} \boldsymbol{\hat{\rho}}& \rho \boldsymbol{\hat{\varphi}}& \boldsymbol{\hat{z}}\\ \frac{\partial}{\partial \rho}& \frac{\partial}{\partial \varphi}& \frac{\partial}{\partial z}\\ \left( G_1 \right)& \left( \rho G_2 \right)& \left( G_3 \right)\\\end{matrix} \right|\\ \nabla \times \boldsymbol{G}=\left| \begin{matrix} \boldsymbol{\hat{r}}& r\boldsymbol{\hat{\theta}}& r\sin \theta \boldsymbol{\hat{\varphi}}\\ \frac{\partial}{\partial r}& \frac{\partial}{\partial \theta}& \frac{\partial}{\partial \varphi}\\ \left( G_1 \right)& \left( rG_2 \right)& \left( r\sin \theta G_3 \right)\\\end{matrix} \right|\\\end{cases}\tag{1.3.3} ⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧∇×G=∣∣∣∣∣∣x^∂x∂(G1)y^∂y∂(G2)z^∂z∂(G3)∣∣∣∣∣∣∇×G=∣∣∣∣∣∣ρ^∂ρ∂(G1)ρφ^∂φ∂(ρG2)z^∂z∂(G3)∣∣∣∣∣∣∇×G=∣∣∣∣∣∣r^∂r∂(G1)rθ^∂θ∂(rG2)rsinθφ^∂φ∂(rsinθG3)∣∣∣∣∣∣(1.3.3)
{ ∇ 2 f = ∂ 2 f ∂ x 2 + ∂ 2 f ∂ y 2 + ∂ 2 f ∂ z 2 ∇ 2 f = 1 ρ ∂ ∂ ρ ( ρ ∂ f ∂ ρ ) + 1 ρ 2 ∂ 2 f ∂ φ 2 + ∂ 2 f ∂ z 2 ∇ 2 f = 1 r 2 ∂ ∂ r ( r 2 ∂ f ∂ r ) + 1 r 2 sin θ ∂ ∂ θ ( sin θ ∂ f ∂ θ ) + 1 r 2 sin 2 θ ∂ 2 f ∂ φ 2 (1.3.4) \begin{cases} \nabla ^2f=\frac{\partial ^2f}{\partial x^2}+\frac{\partial ^2f}{\partial y^2}+\frac{\partial ^2f}{\partial z^2}\\ \nabla ^2f=\frac{1}{\rho}\frac{\partial}{\partial \rho}\left( \rho \frac{\partial f}{\partial \rho} \right) +\frac{1}{\rho ^2}\frac{\partial ^2f}{\partial \varphi ^2}+\frac{\partial ^2f}{\partial z^2}\\ \nabla ^2f=\frac{1}{r^2}\frac{\partial}{\partial r}\left( r^2\frac{\partial f}{\partial r} \right) +\frac{1}{r^2\sin \theta}\frac{\partial}{\partial \theta}\left( \sin \theta \frac{\partial f}{\partial \theta} \right) +\frac{1}{r^2\sin ^2\theta}\frac{\partial ^2f}{\partial \varphi ^2}\\ \end{cases}\tag{1.3.4} ⎩⎪⎪⎨⎪⎪⎧∇2f=∂x2∂2f+∂y2∂2f+∂z2∂2f∇2f=ρ1∂ρ∂(ρ∂ρ∂f)+ρ21∂φ2∂2f+∂z2∂2f∇2f=r21∂r∂(r2∂r∂f)+r2sinθ1∂θ∂(sinθ∂θ∂f)+r2sin2θ1∂φ2∂2f(1.3.4)
关于==(1.3.4)==式，矢量拉普拉斯算子与标量算子具有类似的特性，其中柱坐标采用文献公式(21)展开，但球坐标过长，所以保留了原矢量格式，详见参考文献1：
{ ∇ 2 G ⃗ = ∂ 2 G ⃗ ∂ x 2 + ∂ 2 G ⃗ ∂ y 2 + ∂ 2 G ⃗ ∂ z 2 ∇ 2 G ⃗ = ρ ^ [ ∇ 2 G ρ − 1 ρ 2 G ρ − 2 ρ 2 ∂ G φ ∂ φ ] + φ ^ [ ∇ 2 G φ − 1 ρ 2 G φ + 2 ρ 2 ∂ G ρ ∂ φ ] + z ^ ∇ 2 G z ∇ 2 G ⃗ = 1 r 2 ∂ ∂ r ( r 2 ∂ G ⃗ ∂ r ) + 1 r 2 sin θ ∂ ∂ θ ( sin θ ∂ G ⃗ ∂ θ ) + 1 r 2 sin 2 θ ∂ 2 G ⃗ ∂ φ 2 (1.3.5) \begin{cases} \nabla ^2\boldsymbol{\vec{G}}=\frac{\partial ^2\boldsymbol{\vec{G}}}{\partial x^2}+\frac{\partial ^2\boldsymbol{\vec{G}}}{\partial y^2}+\frac{\partial ^2\boldsymbol{\vec{G}}}{\partial z^2}\\ \nabla ^2\boldsymbol{\vec{G}}=\boldsymbol{\hat{\rho}}\left[ \nabla ^2G_{\rho}-\frac{1}{\rho ^2}G_{\rho}-\frac{2}{\rho ^2}\frac{\partial G_{\varphi}}{\partial \varphi} \right] +\boldsymbol{\hat{\varphi}}\left[ \nabla ^2G_{\varphi}-\frac{1}{\rho ^2}G_{\varphi}+\frac{2}{\rho ^2}\frac{\partial G_{\rho}}{\partial \varphi} \right] +\boldsymbol{\hat{z}}\nabla ^2G_z\\ \nabla ^2\boldsymbol{\vec{G}}=\frac{1}{r^2}\frac{\partial}{\partial r}\left( r^2\frac{\partial \boldsymbol{\vec{G}}}{\partial r} \right) +\frac{1}{r^2\sin \theta}\frac{\partial}{\partial \theta}\left( \sin \theta \frac{\partial \boldsymbol{\vec{G}}}{\partial \theta} \right) +\frac{1}{r^2\sin ^2\theta}\frac{\partial ^2\boldsymbol{\vec{G}}}{\partial \varphi ^2}\\ \end{cases}\tag{1.3.5} ⎩⎪⎪⎨⎪⎪⎧∇2G=∂x2∂2G+∂y2∂2G+∂z2∂2G∇2G=ρ^[∇2Gρ−ρ21Gρ−ρ22∂φ∂Gφ]+φ^[∇2Gφ−ρ21Gφ+ρ22∂φ∂Gρ]+z^∇2Gz∇2G=r21∂r∂(r2∂r∂G)+r2sinθ1∂θ∂(sinθ∂θ∂G)+r2sin2θ1∂φ2∂2G(1.3.5)
第二章——位函数与矢量方程
1、位函数
矢量磁位函数+动态电位：
其中==(2.1.3)==式推导见<跳转到推导1>
B ⃗ = ∇ × A ⃗ (2.1.1) \boldsymbol{\vec{B}}=\nabla \times \boldsymbol{\vec{A}}\tag{2.1.1} B=∇×A(2.1.1)
E ⃗ = − ∇ U − ∂ A ⃗ ∂ t (2.1.2) \boldsymbol{\vec{E}}=-\nabla U-\frac{\partial \boldsymbol{\vec{A}}}{\partial t}\tag{2.1.2} E=−∇U−∂t∂A(2.1.2)
∇ ⋅ A ⃗ = − μ ε ∂ U ∂ t (2.1.3) \colorbox{cyan}{$\nabla \cdot \boldsymbol{\vec{A}}=-\mu \varepsilon \frac{\partial U}{\partial t}$}\tag{2.1.3} ∇⋅A=−με∂t∂U(2.1.3)
对于无电流区域（ ∇ × H ⃗ = 0 \nabla\times\boldsymbol{\vec H}=0 ∇×H=0），标可以定义量磁位：
H ⃗ = − ∇ U m (2.1.4) \boldsymbol{\vec{H}}=-\nabla U_m\tag{2.1.4} H=−∇Um(2.1.4)
2、矢量方程
达朗贝尔方程：
∇ 2 A ⃗ − μ ε ∂ 2 A ⃗ ∂ t 2 = − μ J ⃗ (2.2.1) \nabla ^2\boldsymbol{\vec{A}}-\mu \varepsilon \frac{\partial ^2\boldsymbol{\vec{A}}}{\partial t^2}=-\mu \boldsymbol{\vec{J}}\tag{2.2.1} ∇2A−με∂t2∂2A=−μJ(2.2.1)
∇ 2 U − μ ε ∂ 2 U ∂ t 2 = − ρ ε (2.2.2) \nabla ^2U-\mu \varepsilon \frac{\partial ^2U}{\partial t^2}=-\frac{\rho}{\varepsilon}\tag{2.2.2} ∇2U−με∂t2∂2U=−ερ(2.2.2)
若 k 2 = ω 2 μ ε k^2=\omega^2\mu\varepsilon k2=ω2με：
∇ 2 A ⃗ + k 2 A ⃗ = − μ J ⃗ (PPA) \colorbox{cyan}{$\nabla ^2\boldsymbol{\vec{A}}+k^2 \boldsymbol{\vec{A}}=-\mu \boldsymbol{\vec{J}}$}\tag{PPA} ∇2A+k2A=−μJ(PPA)
∇ 2 U + k 2 U = − ρ ε (PPU) \colorbox{cyan}{$\nabla ^2U+k^2 U=-\frac{\rho}{\varepsilon}$}\tag{PPU} ∇2U+k2U=−ερ(PPU)
亥姆霍兹方程：
∇ 2 E ⃗ = μ σ ∂ E ⃗ ∂ t + μ ε ∂ 2 E ⃗ ∂ t 2 (2.2.3) \nabla ^2\boldsymbol{\vec{E}}=\mu \sigma \frac{\partial \boldsymbol{\vec{E}}}{\partial t}+\mu \varepsilon \frac{\partial ^2\boldsymbol{\vec{E}}}{\partial t^2}\tag{2.2.3} ∇2E=μσ∂t∂E+με∂t2∂2E(2.2.3)
∇ 2 H ⃗ = μ σ ∂ H ⃗ ∂ t + μ ε ∂ 2 H ⃗ ∂ t 2 (2.2.4) \nabla ^2\boldsymbol{\vec{H}}=\mu \sigma \frac{\partial \boldsymbol{\vec{H}}}{\partial t}+\mu \varepsilon \frac{\partial ^2\boldsymbol{\vec{H}}}{\partial t^2}\tag{2.2.4} ∇2H=μσ∂t∂H+με∂t2∂2H(2.2.4)
若区域内没有电荷（ ρ = 0 \rho=0 ρ=0），设 k 2 = ω 2 μ ε − j ω μ σ k^2=\omega ^2\mu \varepsilon -j\omega \mu \sigma k2=ω2με−jωμσ：
其中==(PPE)==式推导见跳转到推导2
∇ 2 E ⃗ + k 2 E ⃗ = 0 (PPE) \colorbox{cyan}{$\nabla ^2\boldsymbol{\vec{E}}+k^2\boldsymbol{\vec{E}}=0$}\tag{PPE} ∇2E+k2E=0(PPE)
∇ 2 H ⃗ + k 2 H ⃗ = 0 (PPH) \colorbox{cyan}{$\nabla ^2\boldsymbol{\vec{H}}+k^2\boldsymbol{\vec{H}}=0$}\tag{PPH} ∇2H+k2H=0(PPH)
第三章——边界条件方程
1、边界条件方程
{ n ^ × E ⃗ = 0 E t = 0 { n ^ ⋅ D ⃗ = ρ S D n = ρ S { n ^ ⋅ B ⃗ = 0 B n = 0 { n ^ × H ⃗ = J ⃗ S H t = J S (3.1.1) \left\{ \begin{array}{r} \boldsymbol{\hat{n}}\times \boldsymbol{\vec{E}}=0\\ E_t=0\\ \end{array} \right. \left\{ \begin{array}{r} \boldsymbol{\hat{n}}\cdot \boldsymbol{\vec{D}}=\rho _S\\ D_n=\rho _S\\ \end{array} \right. \left\{ \begin{array}{r} \boldsymbol{\hat{n}}\cdot \boldsymbol{\vec{B}}=0\\ B_n=0\\ \end{array} \right. \left\{ \begin{array}{r} \boldsymbol{\hat{n}}\times \boldsymbol{\vec{H}}=\boldsymbol{\vec{J}}_S\\ H_t=\,\,J_S\\ \end{array} \right. \tag{3.1.1} {n^×E=0Et=0{n^⋅D=ρSDn=ρS{n^⋅B=0Bn=0{n^×H=JSHt=JS(3.1.1)
因此：
-
理想导体外侧的电场E必垂直于导体表面，且KaTeX parse error: Expected '}', got 'EOF' at end of input: …colorbox{cyan}{D_n=\rho _SKaTeX parse error: Expected 'EOF', got '}' at position 1: }̲
-
理想导体外侧的磁场H必平行于导体表面，且KaTeX parse error: Expected '}', got 'EOF' at end of input: …colorbox{cyan}{H_t=J_SKaTeX parse error: Expected 'EOF', got '}' at position 1: }̲
第四章——静电场
1、对称/非对称场结论
有限/无限长直导线：
{ E / / = λ 4 π ε a ( sin θ 2 − sin θ 1 ) E ⊥ = λ 4 π ε a ( cos θ 1 − cos θ 2 ) \begin{cases} E_{/\!/}=\frac{\lambda}{4\pi \varepsilon a}\left( \sin \theta _2-\sin \theta _1 \right)\\ E_{\bot}=\frac{\lambda}{4\pi \varepsilon a}\left( \cos \theta _1-\cos \theta _2 \right)\\ \end{cases} {E//=4πεaλ(sinθ2−sinθ1)E⊥=4πεaλ(cosθ1−cosθ2)
{ E ⃗ r → ∞ = λ 2 π ε r r ^ U = ln r (4.1.1) \begin{cases} \displaystyle\colorbox{cyan}{$\boldsymbol{\vec{E}}_{r\to \infty}=\frac{\lambda}{2\pi \varepsilon r}\boldsymbol{\hat{r}}$}\\ \displaystyle U=\ln r\\\end{cases}\tag{4.1.1} ⎩⎨⎧Er→∞=2πεrλr^U=lnr(4.1.1)
有限/无限大圆盘：
E ⃗ = λ 2 ε ( 1 − x R 2 + x 2 ) x ^ \boldsymbol{\vec{E}}=\frac{\lambda}{2\varepsilon }\left( 1-\frac{x}{\sqrt{R^2+x^2}} \right)\boldsymbol{\hat{x}} E=2ελ(1−R2+x2x)x^
E ⃗ r → ∞ = λ 2 ε x ^ (4.1.2) \colorbox{cyan}{$\boldsymbol{\vec{E}}_{r\to \infty}=\frac{\lambda}{2\varepsilon }\boldsymbol{\hat{x}}$}\tag{4.1.2} Er→∞=2ελx^(4.1.2)
对称均匀球：
{ E ⃗ = { q r 4 π ε R 3 r ^ , r < R q 4 π ε r 2 , r > R U = { ( 3 q 8 π ε R − q r 2 8 π ε 0 R 3 ) r ^ , r < R q 4 π ε r , r > R (4.1.3) \begin{cases} \boldsymbol{\vec{E}}=\begin{cases} \displaystyle\frac{qr}{4\pi \varepsilon R^3}\boldsymbol{\hat{r}}&,r
半径R的圆环中垂线：
{ E ⃗ = q x 4 π ε ( x 2 + R 2 ) 3 2 x ^ U = q 4 π ε ( x 2 + R 2 ) 1 2 (4.1.4) \begin{cases} \displaystyle \colorbox{cyan}{$\boldsymbol{\vec{E}}=\frac{qx}{4\pi \varepsilon \left( x^2+R^2 \right) ^{\frac{3}{2}}}\boldsymbol{\hat{x}}$}\\ \displaystyle U=\frac{q}{4\pi \varepsilon \left( x^2+R^2 \right) ^{\frac{1}{2}}}\\\end{cases}\tag{4.1.4} ⎩⎪⎪⎨⎪⎪⎧E=4πε(x2+R2)23qxx^U=4πε(x2+R2)21q(4.1.4)
2、电偶极子
电偶极距：
p ⃗ = Q l ⃗ (4.2.1) \boldsymbol{\vec{p}}=Q\boldsymbol{\vec{l}}\tag{4.2.1} p=Ql(4.2.1)
U = p ⃗ ⋅ r ⃗ 4 π ε r 3 = p cos θ 4 π ε r 2 (4.2.2) \colorbox{cyan}{$U$}=\frac{\boldsymbol{\vec{p}}\cdot \boldsymbol{\vec{r}}}{4\pi \varepsilon r^3}=\colorbox{cyan}{$\frac{p\cos \theta}{4\pi \varepsilon r^2}$}\tag{4.2.2} U=4πεr3p⋅r=4πεr2pcosθ(4.2.2)
E ⃗ = − ∇ U = − ( r ^ ∂ U ∂ r + θ ^ 1 r ∂ U ∂ θ ) = p 4 π ε r 3 ( r ^ 2 cos θ + θ ^ sin θ ) (4.2.3) \colorbox{cyan}{$\boldsymbol{\vec{E}}$}=-\nabla U=-\left( \boldsymbol{\hat{r}}\frac{\partial U}{\partial r}+\boldsymbol{\hat{\theta}}\frac{1}{r}\frac{\partial U}{\partial \theta} \right) =\colorbox{cyan}{$\frac{p}{4\pi \varepsilon r^3}\left( \boldsymbol{\hat{r}}2\cos \theta +\boldsymbol{\hat{\theta}}\sin \theta \right)$}\tag{4.2.3} E=−∇U=−(r^∂r∂U+θ^r1∂θ∂U)=4πεr3p(r^2cosθ+θ^sinθ)(4.2.3)
电转矩：
M ⃗ = p ⃗ × E ⃗ (4.2.4) \boldsymbol{\vec{M}}=\boldsymbol{\vec{p}}\times \boldsymbol{\vec{E}}\tag{4.2.4} M=p×E(4.2.4)
3、电极化
体极化电荷：
ρ p ( V ) = − ∇ ⋅ P ⃗ (4.3.1) \rho _{p\left( V \right)}=-\nabla \cdot \boldsymbol{\vec{P}}\tag{4.3.1} ρp(V)=−∇⋅P(4.3.1)
面极化电荷（其中 n ^ \boldsymbol{\hat{n}} n^为自己指向外面）：
ρ p ( S ) = n ^ ⋅ P ⃗ (4.3.2) \rho _{p\left( S \right)}=\boldsymbol{\hat{n}}\cdot \boldsymbol{\vec{P}}\tag{4.3.2} ρp(S)=n^⋅P(4.3.2)
D、E、P关系：
P ⃗ = χ e ε 0 E ⃗ D ⃗ = ε 0 E ⃗ + P ⃗ ε r = 1 + χ e } ⇒ P ⃗ = ε r − 1 ε r D ⃗ (4.3.3) \left.\begin{array}{l}\boldsymbol{\vec{P}}=\chi _e\varepsilon _0\boldsymbol{\vec{E}}\\\boldsymbol{\vec{D}}=\varepsilon_0\boldsymbol{\vec{E}}+\boldsymbol{\vec{P}}\\\varepsilon _r=1+\chi _e\end{array}\right\}\Rightarrow \boldsymbol{\vec{P}}=\frac{\varepsilon _r-1}{\varepsilon _r}\boldsymbol{\vec{D}}\tag{4.3.3} P=χeε0ED=ε0E+Pεr=1+χe⎭⎬⎫⇒P=εrεr−1D(4.3.3)
极化电荷与原场关系：
ρ p = ( 1 − ε r ε r ) ρ + D ⃗ ⋅ ∇ ( 1 ε r ) = ε r = const ( 1 − ε r ε r ) ρ (4.3.4) \colorbox{cyan}{$\rho _p$}=\left( \frac{1-\varepsilon _r}{\varepsilon _r} \right) \rho +\boldsymbol{\vec{D}}\cdot \nabla \left( \frac{1}{\varepsilon _r} \right) \xlongequal{\varepsilon _r=\text{const}}\colorbox{cyan}{$\left( \frac{1-\varepsilon _r}{\varepsilon _r} \right) \rho$}\tag{4.3.4} ρp=(εr1−εr)ρ+D⋅∇(εr1)εr=const(εr1−εr)ρ(4.3.4)
4、静电场的哈密顿算符/边界方程
边界方程：
D的法向分量（方向为2指向1）：
n ^ ⋅ ( D ⃗ 1 − D ⃗ 2 ) = ρ S (4.4.1) \boldsymbol{\hat{n}}\cdot \left( \boldsymbol{\vec{D}}_1-\boldsymbol{\vec{D}}_2 \right) =\rho _S\tag{4.4.1} n^⋅(D1−D2)=ρS(4.4.1)
E的切向分量为0：
n ^ × ( E ⃗ 1 − E ⃗ 2 ) = 0 (4.4.2) \boldsymbol{\hat{n}}\times \left( \boldsymbol{\vec{E}}_1-\boldsymbol{\vec{E}}_2 \right) =0\tag{4.4.2} n^×(E1−E2)=0(4.4.2)
光学性质（假设边界无自由电荷）：
如果2区是理想导体，那么1区电场夹角近似于0度，即与表面垂直，故分界面处处等势。
D 1 cos θ 1 = D 2 cos θ 2 E 1 sin θ 1 = E 2 sin θ 2 D = ε E } ⇒ tan θ 1 tan