以下电磁场知识点整理及题目均基于 马西奎 工程电磁场导论 北京 高等教育出版社, 2000. Print. 面向21世纪课程教材 Mian Xiang 21shi Ji Ke Cheng Jiao Cai = Textbook Series for 21st Century Eng. （第一版） 旨在为四川大学电气工程及其自动化专业的电磁场课程提供一个知识纲要性质的文本，以供课程平时学习及期末复习使用。考点及考试范围主要参考王仲老师的讲课内容及2024-2025学年期末考试题目，由于历年考试范围不同，其覆盖范围可能与实际考试范围有所出入。
第一章 静电场
一、公式部分（记记部分）
1. 库仑定律 ：
E ⃗ = q 4 π ε 0 r 2 r ⃗ \vec{E} = \frac{q}{4 \pi \varepsilon_0 r^2} \vec{r}
E = 4 π ε 0 r 2 q r
其中 q q q r r r r ⃗ \vec{r} r
2. 电位的定义 ：
基于静电场环路定律的积分形式 $$\oint \vec{E} \cdot \vec{dl} = 0$$
∇ × E ⃗ = 0 \nabla \times \vec{E} = 0
∇ × E = 0
可说明电场是一个无旋场。
E = − ∇ φ E = -\nabla \varphi
E = − ∇ φ
进而有
U A B = φ A − φ B = ∫ A B E ⃗ ⋅ d l ⃗ U_{AB} = \varphi_A - \varphi_B = \int_{A}^{B} \vec{E} \cdot \vec{dl}
U A B = φ A − φ B = ∫ A B E ⋅ d l
而
φ A = ∫ A R E F E ⃗ ⋅ d l ⃗ \varphi_A = \int_{A}^{REF} \vec{E} \cdot \vec{dl}
φ A = ∫ A REF E ⋅ d l
其中 R E F REF REF
φ ( r ⃗ ) = q 4 π ε 0 ∣ r ⃗ ∣ \varphi(\vec{r}) = \frac{q}{4\pi \varepsilon_0 |\vec{r}|}
φ ( r ) = 4 π ε 0 ∣ r ∣ q
这是叠加法计算电位的基础。
3. 静电场中导体的特点 ：
电荷均匀分布于导体表面，导体内电场为零，整个导体为一等位体，其表面为零位面。
4. 电介质在电场中会发生极化
使得介质内存在连续电偶极子分布，进而形成附加电场。
5. 定义电极化强度 ：
P ⃗ = Σ p ⃗ Δ V \vec{P} = \frac{\Sigma \vec{p}}{\Delta V}
P = Δ V Σ p
其中 Σ p ⃗ \Sigma \vec{p} Σ p
σ p = P ⃗ ⋅ e ⃗ n \sigma_p = \vec{P} \cdot \vec{e}_n
σ p = P ⋅ e n
电介质极化电荷体密度
ρ D = − ∇ ⋅ P ⃗ \rho_D = -\nabla \cdot \vec{P}
ρ D = − ∇ ⋅ P
且满足
∫ V ( − ∇ ⋅ P ⃗ ) d V + ∮ S P ⃗ ⋅ e ⃗ n = 0 \int_V (-\nabla \cdot \vec{P}) dV + \oint_S \vec{P} \cdot \vec{e}_n = 0
∫ V ( − ∇ ⋅ P ) d V + ∮ S P ⋅ e n = 0
对于各向同性的线性介质，满足
P = χ ϵ 0 E ⃗ P = \chi \epsilon_0 \vec{E}
P = χ ϵ 0 E
其中 χ \chi χ
6. Gauss定律 ：
记电位移矢量 $$\vec{D} = \varepsilon_0 \vec{E} + \vec{P}$$，从而有
∮ D ⃗ ⋅ d S ⃗ = ∫ V ρ d V \oint \vec{D} \cdot d\vec{S} = \int_V \rho dV
∮ D ⋅ d S = ∫ V ρ d V
此即一般形式的Gauss定律。其意义为，任意闭合有向曲面 Σ ⃗ \vec{\Sigma} Σ D ⃗ \vec{D} D
∇ ⋅ D ⃗ = ρ \nabla \cdot \vec{D} = \rho
∇ ⋅ D = ρ
为Gauss定律的微分形式，它表明 D ⃗ \vec{D} D
7. 电介质本构方程
D ⃗ = ε 0 E ⃗ + P ⃗ 0 \vec{D} = \varepsilon_0 \vec{E} + \vec{P}_0
D = ε 0 E + P 0
称为电介质本构方程，对于各向同性电介质，则有
D ⃗ = ε 0 E ⃗ + χ ε 0 E ⃗ = ( 1 + χ ) ε 0 E ⃗ \vec{D} = \varepsilon_0 \vec{E} + \chi \varepsilon_0 \vec{E} = (1+\chi) \varepsilon_0 \vec{E}
D = ε 0 E + χ ε 0 E = ( 1 + χ ) ε 0 E
若定义$ 1 + \chi = \epsilon_r $ , $ \epsilon_r \epsilon_0 = \epsilon $ ,则有$$\vec{D}=\epsilon \vec{E}$$ϵ r \epsilon_r ϵ r ϵ \epsilon ϵ
8. 静电场基本方程 ：
积分形式
∮ D ⃗ ⋅ d S ⃗ = ∫ V ρ d V ( Gauss定律 ) \oint \vec{D} \cdot d\vec{S} = \int_V \rho dV \quad (\text{Gauss定律})
∮ D ⋅ d S = ∫ V ρ d V ( Gauss 定律 )
∮ E ⃗ ⋅ d l ⃗ = 0 ( 环路定律 ) \oint \vec{E} \cdot d\vec{l} = 0 \quad (\text{环路定律})
∮ E ⋅ d l = 0 ( 环路定律 )
微分形式
∇ ⋅ D ⃗ = ρ ( Gauss定律 ) \nabla \cdot \vec{D} = \rho \quad (\text{Gauss定律})
∇ ⋅ D = ρ ( Gauss 定律 )
∇ × E ⃗ = 0 ( 环路定律 ) \nabla \times \vec{E} = 0 \quad (\text{环路定律})
∇ × E = 0 ( 环路定律 )
本构方程
D ⃗ = ϵ E ⃗ \vec{D}=\epsilon \vec{E}
D = ϵ E
9. 静电场分界面衔接条件
D 2 n − D 1 n = σ ( 用Gauss定律 ) D_{2n} - D_{1n} = \sigma \quad (\text{用Gauss定律})
D 2 n − D 1 n = σ ( 用 Gauss 定律 )
E 1 t = E 2 t ( 用环路定律 ) E_{1t} = E_{2t} \quad (\text{用环路定律})
E 1 t = E 2 t ( 用环路定律 )
σ = 0 \sigma = 0 σ = 0
tan α 1 tan α 2 = ε 1 ε 2 \frac{\tan \alpha_1}{\tan \alpha_2} = \frac{\varepsilon_1}{\varepsilon_2}
tan α 2 tan α 1 = ε 2 ε 1
此外，利用 φ 2 − φ 1 = E n d \varphi_2 - \varphi_1 = E_n d φ 2 − φ 1 = E n d d → 0 d \to 0 d → 0 φ 2 − φ 1 = 0 \varphi_2 - \varphi_1 = 0 φ 2 − φ 1 = 0 φ 1 = φ 2 = C \varphi_1 = \varphi_2 = C φ 1 = φ 2 = C D 2 n = − ε 2 ∂ φ 2 ∂ n D_{2n} = -\varepsilon_2 \frac{\partial \varphi_2}{\partial n} D 2 n = − ε 2 ∂ n ∂ φ 2 D 1 n = − ε 1 ∂ φ 1 ∂ n D_{1n} = -\varepsilon_1 \frac{\partial \varphi_1}{\partial n} D 1 n = − ε 1 ∂ n ∂ φ 1
ε 1 ∂ φ 1 ∂ n − ε 2 ∂ φ 2 ∂ n = σ \varepsilon_1 \frac{\partial \varphi_1}{\partial n} - \varepsilon_2 \frac{\partial \varphi_2}{\partial n} = \sigma
ε 1 ∂ n ∂ φ 1 − ε 2 ∂ n ∂ φ 2 = σ
此为用电位函数表达的衔接条件（用于求解静电场边值问题）。
10. 静电场边值问题
静电场边值问题是给定空间某一区域内电荷分布（可以为零），同时给定该区域边界上的电位或电场（即边值/边界条件），求解该区域内的电位函数、电场强度分布。
11. Poisson与Laplace方程 ：
∇ 2 φ = − ρ ε 0 \nabla^2 \varphi = -\frac{\rho}{\varepsilon_0}
∇ 2 φ = − ε 0 ρ
为Poisson方程 $$\nabla^2 \varphi = 0$$ 为Laplace方程。
12. 边界条件共有三类 ：
(i) φ ∣ S = f 1 ( s ) \varphi |_{S} = f_1(s) φ ∣ S = f 1 ( s ) ∂ φ ∂ n ∣ S = f 2 ( s ) \frac{\partial \varphi}{\partial n}\bigg|_{S} = f_2(s) ∂ n ∂ φ S = f 2 ( s ) σ P \sigma_P σ P ( φ + β ∂ φ ∂ n ) ∣ S = f 3 ( s ) \left( \varphi + \beta \frac{\partial \varphi}{\partial n} \right) \bigg|_{S} = f_3(s) ( φ + β ∂ n ∂ φ ) S = f 3 ( s )
此外，对于无穷域场，若电荷分布于有限域，应有自然边界条件 lim x → ∞ φ ( r ) = 0 \lim_{x \to \infty}\varphi(r) = 0 lim x → ∞ φ ( r ) = 0
13. 唯一性定理 （特殊方法解静电场的基础）
在静电场中凡满足电位微分方程和给定边界条件的解为给定静电场的唯一解。
14. 静电镜像法与电轴法 ：见二。
15. 电容
对于两个导体组成的静电系统，电容 C C C
C = Q V C = \frac{Q}{V}
C = V Q
孤立导体的电容则指导体与无限远处导体间电容。
16. 部分电容
对于多导体静电系统（由 n > 1 n > 1 n > 1 n + 1 n+1 n + 1
φ 0 = 0 \varphi_0 = 0
φ 0 = 0
则由叠加原理有
[ φ ] = [ α ] [ Q ] [\varphi] = [\alpha][Q]
[ φ ] = [ α ] [ Q ]
α \alpha α α i i \alpha_{ii} α ii α i j ( i ≠ j ) \alpha_{ij} (i \neq j) α ij ( i = j ) α i j > 0 \alpha_{ij} > 0 α ij > 0 α i i > α j j \alpha_{ii} > \alpha_{jj} α ii > α jj α \alpha α α i j = α j i \alpha_{ij} = \alpha_{ji} α ij = α ji
取
[ β ] = [ α ] − 1 [\beta] = [\alpha]^{-1}
[ β ] = [ α ] − 1
则有
[ Q ] = [ β ] [ φ ] [Q] = [\beta][\varphi]
[ Q ] = [ β ] [ φ ]
β \beta β
(i) β i i > 0 , β i j < 0 \beta_{ii} > 0, \beta_{ij} < 0 β ii > 0 , β ij < 0 β i i > ∣ β i j ∣ \beta_{ii} > |\beta_{ij}| β ii > ∣ β ij ∣
再令 φ 1 = 0 = U 10 \varphi_1 = 0 = U_{10} φ 1 = 0 = U 10 φ 2 − φ 1 = U 12 \varphi_2 - \varphi_1 = U_{12} φ 2 − φ 1 = U 12 ⋯ \cdots ⋯ φ k − φ 1 = U 1 k \varphi_k - \varphi_1 = U_{1k} φ k − φ 1 = U 1 k ⋯ \cdots ⋯ Q n − φ 1 = U n 1 Q_n -\varphi_1 = U_{n1} Q n − φ 1 = U n 1
则有
Q 1 = ( β 11 + β 12 + ⋯ + β 1 n ) ( φ 1 − φ 0 ) − β 12 ( φ 1 − φ 2 ) − ⋯ − β 1 k ( φ 1 − φ k ) − ⋯ − β 1 n ( φ 1 − φ n ) Q_1 = (\beta_{11} + \beta_{12} + \cdots + \beta_{1n})(\varphi_1 - \varphi_0) - \beta_{12}(\varphi_1 - \varphi_2) - \cdots - \beta_{1k}(\varphi_1 - \varphi_k) - \cdots - \beta_{1n}(\varphi_1 - \varphi_n)
Q 1 = ( β 11 + β 12 + ⋯ + β 1 n ) ( φ 1 − φ 0 ) − β 12 ( φ 1 − φ 2 ) − ⋯ − β 1 k ( φ 1 − φ k ) − ⋯ − β 1 n ( φ 1 − φ n )
进而可令
C 10 = ∑ i = 1 n β 1 i , C 12 = − β 12 , ⋯ , C 1 k = − β 1 k , ⋯ , C 1 n = − β 1 n C_{10} = \sum_{i=1}^n \beta_{1i}, \quad C_{12} = -\beta_{12}, \cdots, C_{1k} = -\beta_{1k}, \cdots, C_{1n} = -\beta_{1n}
C 10 = i = 1 ∑ n β 1 i , C 12 = − β 12 , ⋯ , C 1 k = − β 1 k , ⋯ , C 1 n = − β 1 n
同理可令 U i j = φ i − φ j U_{ij} = \varphi_i - \varphi_j U ij = φ i − φ j i ≠ j i \neq j i = j C i 0 = ∑ j = 1 n β i j C_{i0} = \sum_{j=1}^n \beta_{ij} C i 0 = ∑ j = 1 n β ij C i j = − β i j C_{ij} = -\beta_{ij} C ij = − β ij [ Q ] = [ β ] [ φ ] [Q] = [\beta][\varphi] [ Q ] = [ β ] [ φ ]
化为 [ Q ] = [ C ] [ U ] [Q] = [C][U] [ Q ] = [ C ] [ U ] C C C C i 0 C_{i0} C i 0 C i j C_{ij} C ij j ≠ i j \neq i j = i
17. 部分电容可说明静电荷屏蔽问题
Q 1 = C 10 U 10 + C 12 U 12 Q_1 = C_{10} U_{10} + C_{12} U_{12}
Q 1 = C 10 U 10 + C 12 U 12
Q 2 = C 20 U 20 + C 21 U 21 Q_2 = C_{20} U_{20} + C_{21} U_{21}
Q 2 = C 20 U 20 + C 21 U 21
令 Q 1 = 0 Q_1 = 0 Q 1 = 0 U 10 = 0 U_{10} = 0 U 10 = 0 U 12 = 0 U_{12} = 0 U 12 = 0 Q 2 = 0 Q_2 = 0 Q 2 = 0
而 U 12 U_{12} U 12 C 12 = 0 C_{12} = 0 C 12 = 0 Q 2 = 0 Q_2 = 0 Q 2 = 0
则有 C 21 ⋅ C 12 U 12 = 0 C_{21} \cdot C_{12} U_{12} = 0 C 21 ⋅ C 12 U 12 = 0 C 21 = 0 C_{21} = 0 C 21 = 0 Q 1 = C 10 U 10 Q_1 = C_{10} U_{10} Q 1 = C 10 U 10 Q 2 = C 20 U 20 Q_2 = C_{20} U_{20} Q 2 = C 20 U 20
这说明 0 号导体切断了 1、2 号导体间的联系。
18. 电荷系统能量
W e = 1 2 ∫ φ ρ d V = 1 2 ∫ D ⃗ ⋅ E ⃗ d V W_e = \frac{1}{2} \int \varphi \rho dV = \frac{1}{2} \int \vec{D} \cdot \vec{E} dV
W e = 2 1 ∫ φρ d V = 2 1 ∫ D ⋅ E d V
从而有空间静电能密度
w e = 1 2 D ⃗ ⋅ E ⃗ = 1 2 ε 0 E ⃗ ⋅ E ⃗ w_e = \frac{1}{2} \vec{D} \cdot \vec{E} = \frac{1}{2} \varepsilon_0 \vec{E} \cdot \vec{E}
w e = 2 1 D ⋅ E = 2 1 ε 0 E ⋅ E
19. 静电力可使用虚位移法求解：且有
f = − ∂ W e ∂ g ∣ Q = 常量 = ∂ W e ∂ g ∣ U = 常量 f = -\frac{\partial W_e}{\partial g} \bigg|_{Q=\text{常量}} = \frac{\partial W_e}{\partial g} \bigg|_{U=\text{常量}}
f = − ∂ g ∂ W e Q = 常量 = ∂ g ∂ W e U = 常量
20. 法拉第对静电力有如下观点：
对于任意电通密度管，其沿轴向受到纵张力，垂直于轴向受到侧向压力。换言之，其沿纵向收缩趋势，沿横向有扩张趋势。张力与压力大小均为 1 2 D ⃗ ⋅ E ⃗ \frac{1}{2} \vec{D} \cdot \vec{E} 2 1 D ⋅ E
二、题型整理
1. 静电场求解
(1) 直接积分法（利用库仑定律）（一般运算较复杂，少用）
例 1.1.1 （P3,7-4）真空中一密度为 2 × 10 6 C/m 2 \times 10^6 \, \text{C/m} 2 × 1 0 6 C/m 0.1 nC/m 2 0.1 \, \text{nC/m}^2 0.1 nC/m 2 − 0.1 nC/m 2 -0.1 \, \text{nC/m}^2 − 0.1 nC/m 2 z = 3 m z=3 \, \text{m} z = 3 m z = − 4 m z=-4 \, \text{m} z = − 4 m E ⃗ \vec{E} E
由题意知 P 距 y 轴 l = 5 m l=5 \, \text{m} l = 5 m r = 3 m r=3 \, \text{m} r = 3 m z = 5 m z=5 \, \text{m} z = 5 m
先求距线电荷距离为 r 的点处电场强度。由对称性易知电场方向沿 x 轴方向向外。
E ⃗ = ∫ − ∞ + ∞ 2 π × 10 − 9 ⋅ r ⃗ 0 4 π ε 0 ( r 2 + x 2 ) d x \vec{E} = \int_{-\infty}^{+\infty} \frac{2 \pi \times 10^{-9} \cdot \vec{r}_0}{4 \pi \varepsilon_0 (r^2 + x^2)} dx
E = ∫ − ∞ + ∞ 4 π ε 0 ( r 2 + x 2 ) 2 π × 1 0 − 9 ⋅ r 0 d x
E = 10 − 9 2 ε r ⋅ 2 = 10 − 9 ε r E = \frac{10^{-9}}{2 \varepsilon r} \cdot 2 = \frac{10^{-9}}{\varepsilon r}
E = 2 ε r 1 0 − 9 ⋅ 2 = ε r 1 0 − 9
从而 E ⃗ ( 5 ) = 10 − 9 ε 5 \vec{E}(\sqrt{5}) = \frac{10^{-9}}{\varepsilon \sqrt{5} } E ( 5 ) = ε 5 1 0 − 9
再求到无限带电平面距离为 r 的点处场强 E ⃗ 2 \vec{E}_2 E 2
E ⃗ 2 = ∫ σ d S 4 π ε 0 ( r 0 2 + r 2 ) ⋅ r r 2 + r 0 2 = ∫ 0 + ∞ σ 2 π r 0 d r 0 4 π ε 0 ( r 0 2 + r 2 ) 3 / 2 ⋅ r d r 0 \vec{E}_2 = \int \frac{\sigma \, dS}{4 \pi \varepsilon_0 (r_0^2 + r^2)} \cdot \frac{r}{\sqrt{r^2 + r_0^2}} = \int_0^{+\infty} \frac{\sigma \, 2 \pi r_0 dr_0}{4 \pi \varepsilon_0 (r_0^2 + r^2)^{3/2}} \cdot r \, dr_0
E 2 = ∫ 4 π ε 0 ( r 0 2 + r 2 ) σ d S ⋅ r 2 + r 0 2 r = ∫ 0 + ∞ 4 π ε 0 ( r 0 2 + r 2 ) 3/2 σ 2 π r 0 d r 0 ⋅ r d r 0
= σ r 2 ε 0 ∫ 0 + ∞ r 0 d r 0 ( r 0 2 + r 2 ) 3 / 2 = \frac{\sigma r}{2 \varepsilon_0} \int_0^{+\infty} \frac{r_0 dr_0}{(r_0^2 + r^2)^{3/2}}
= 2 ε 0 σ r ∫ 0 + ∞ ( r 0 2 + r 2 ) 3/2 r 0 d r 0
= σ r 2 ε 0 [ − 1 r 0 2 + r 2 ] 0 + ∞ = σ 2 ε 0 = \frac{\sigma r}{2 \varepsilon_0} \left[ -\frac{1}{\sqrt{r_0^2 + r^2}} \right]_0^{+\infty} = \frac{\sigma}{2 \varepsilon_0}
= 2 ε 0 σ r [ − r 0 2 + r 2 1 ] 0 + ∞ = 2 ε 0 σ
∴ E ⃗ 2 ( 1 ) = 0.1 × 10 − 9 2 ε 0 e ⃗ z = 0.1 × 10 − 9 2 ε 0 ( − e ⃗ z ) \therefore \vec{E}_2(1) = \frac{0.1 \times 10^{-9}}{2 \varepsilon_0} \vec{e}_z = \frac{0.1 \times 10^{-9}}{2 \varepsilon_0} (-\vec{e}_z)
∴ E 2 ( 1 ) = 2 ε 0 0.1 × 1 0 − 9 e z = 2 ε 0 0.1 × 1 0 − 9 ( − e z )
E ⃗ 3 ( 4 ) = 0.1 × 10 − 9 2 ε 0 e ⃗ z \vec{E}_3(4) = \frac{0.1 \times 10^{-9}}{2 \varepsilon_0} \vec{e}_z
E 3 ( 4 ) = 2 ε 0 0.1 × 1 0 − 9 e z
E ⃗ = E ⃗ 1 + E ⃗ 2 + E ⃗ 3 = 10 − 9 ε 5 ≈ 51 V/m \vec{E} = \vec{E}_1 + \vec{E}_2 + \vec{E}_3= \frac{10^{-9}}{\varepsilon \sqrt{5} } \approx 51 \text{V/m}
E = E 1 + E 2 + E 3 = ε 5 1 0 − 9 ≈ 51 V/m
方向由线电荷指向 P，且于 y 轴垂直。
(2) Gauss定理（常用）
例 1.1.2 （P67 T1-4）用双层电介质制成的同轴电缆如图。介电常数 ε 1 = 4 ε 0 \varepsilon_1 = 4\varepsilon_0 ε 1 = 4 ε 0 ε 2 = 2 ε 0 \varepsilon_2 = 2\varepsilon_0 ε 2 = 2 ε 0 + τ +\tau + τ − τ -\tau − τ
两种电介质中的 D ⃗ \vec{D} D E ⃗ \vec{E} E P ⃗ \vec{P} P
极化电荷分布。
(1) 当 ρ < R 1 \rho < R_1 ρ < R 1
2 π ρ ⋅ D 1 = τ ⋅ ρ 2 R 1 2 ⇒ D 1 = τ ρ 2 π R 1 2 2\pi \rho \cdot D_1 = \tau \cdot \frac{\rho^2}{R^{2}_{1}} \Rightarrow D_1 = \frac{\tau \rho }{2\pi R^2_1 }
2 π ρ ⋅ D 1 = τ ⋅ R 1 2 ρ 2 ⇒ D 1 = 2 π R 1 2 τ ρ
E 1 = D 1 ε 1 = τ ρ 2 π ε 0 R 1 2 E_1 = \frac{D_1}{\varepsilon_1} =\frac{\tau \rho }{2\pi \varepsilon_0 R^2_1 }
E 1 = ε 1 D 1 = 2 π ε 0 R 1 2 τ ρ
① 当 ρ ∈ ( R 1 , R 2 ) \rho \in (R_1, R_2) ρ ∈ ( R 1 , R 2 )
2 π ρ ⋅ D 2 l = τ l 2\pi \rho \cdot D_2 l = \tau l
2 π ρ ⋅ D 2 l = τ l
⇒ D 2 = τ 2 π ρ , E 2 = D 2 ε 1 = τ 8 π ε 0 ρ \Rightarrow D_2 = \frac{\tau}{2\pi \rho}, \quad E_2 = \frac{D_2}{\varepsilon_1} = \frac{\tau}{8\pi \varepsilon_0 \rho}
⇒ D 2 = 2 π ρ τ , E 2 = ε 1 D 2 = 8 π ε 0 ρ τ
② 当 ρ ∈ ( R 2 , R 3 ) \rho \in (R_2, R_3) ρ ∈ ( R 2 , R 3 )
2 π ρ ⋅ D 3 l = τ l 2\pi \rho \cdot D_3 l = \tau l
2 π ρ ⋅ D 3 l = τ l
⇒ D 3 = τ 2 π ρ , E 3 = D 3 ε 2 = τ 4 π ε 0 ρ \Rightarrow D_3 = \frac{\tau}{2\pi \rho}, \quad E_3 = \frac{D_3}{\varepsilon_2} = \frac{\tau}{4\pi \varepsilon_0 \rho}
⇒ D 3 = 2 π ρ τ , E 3 = ε 2 D 3 = 4 π ε 0 ρ τ
③ 当 ρ > R 3 \rho > R_3 ρ > R 3
2 π ρ D 4 l = 0 2\pi \rho D_4 l = 0
2 π ρ D 4 l = 0
⇒ D 4 = 0 , E 4 = 0 \Rightarrow D_4 = 0, \quad E_4 = 0
⇒ D 4 = 0 , E 4 = 0
(2)：
D ⃗ = ε 0 E ⃗ + P ⃗ \vec{D} = \varepsilon_0 \vec{E} + \vec{P}
D = ε 0 E + P
P 1 = D 1 − ε 0 E 1 = τ 2 π ρ − ε 0 ⋅ τ 8 π ε 0 ρ = 3 τ 8 π ρ P_1 = D_1 - \varepsilon_0 E_1 = \frac{\tau}{2\pi \rho} - \varepsilon_0 \cdot \frac{\tau}{8\pi \varepsilon_0 \rho} = \frac{3\tau}{8\pi \rho}
P 1 = D 1 − ε 0 E 1 = 2 π ρ τ − ε 0 ⋅ 8 π ε 0 ρ τ = 8 π ρ 3 τ
P 2 = D 3 − ε 0 E 3 = τ 2 π ρ − ε 0 ⋅ τ 4 π ε 0 ρ = τ 4 π ρ P_2 = D_3 - \varepsilon_0 E_3 = \frac{\tau}{2\pi \rho} - \varepsilon_0 \cdot \frac{\tau}{4\pi \varepsilon_0 \rho} = \frac{\tau}{4\pi \rho}
P 2 = D 3 − ε 0 E 3 = 2 π ρ τ − ε 0 ⋅ 4 π ε 0 ρ τ = 4 π ρ τ
方向均沿矢径向外。
(3) 在两电介质交界面处，由静电场衔接条件有：
D ⃗ ⋅ n ⃗ = 0 \vec{D} \cdot \vec{n} = 0
D ⋅ n = 0
且
σ p = ( P ⃗ 1 − P ⃗ 2 ) ⋅ n ⃗ 12 = − τ 8 π R 2 \sigma_p = (\vec{P}_1 - \vec{P}_2) \cdot \vec{n}_{12} = -\frac{\tau}{8\pi R_2}
σ p = ( P 1 − P 2 ) ⋅ n 12 = − 8 π R 2 τ
(3) 静电场边值问题（偶尔用，公式记得记住）（一般考一维）
例 1.1.3 （P68 T1-9）在平行平板电极上加上一直流电压 U 0 = 2 V U_0 = 2V U 0 = 2 V ρ \rho ρ ρ = − 10 − 6 C/m 3 \rho = -10^{-6} \, \text{C/m}^3 ρ = − 1 0 − 6 C/m 3 ε = ε 0 \varepsilon = \varepsilon_0 ε = ε 0 d = 5 mm d = 5 \, \text{mm} d = 5 mm
由对称性易知电场沿 x 轴方向，取右极板为考察点有：
φ ( 0 ) = 2 V , φ ( 5 mm ) = 0 V \varphi(0) = 2V, \quad \varphi(5\text{mm}) = 0V
φ ( 0 ) = 2 V , φ ( 5 mm ) = 0 V
∇ 2 φ = − ρ ε 0 \nabla^2 \varphi = -\frac{\rho}{\varepsilon_0}
∇ 2 φ = − ε 0 ρ
其中
∇ 2 φ = d 2 φ d x 2 ⟹ φ ( x ) = − ρ 2 ε 0 x 2 + C 1 x + C 2 \nabla^2 \varphi = \frac{d^2 \varphi}{dx^2} \implies \varphi(x) = -\frac{\rho}{2\varepsilon_0} x^2 + C_1 x + C_2
∇ 2 φ = d x 2 d 2 φ ⟹ φ ( x ) = − 2 ε 0 ρ x 2 + C 1 x + C 2
φ ( 0 ) = 2 V ⟹ C 2 = 2 V , φ ( 5 mm ) = 0 V ⟹ C 1 = − 0.005 ρ 2 ε 0 − 2 0.005 \varphi(0) = 2V \implies C_2 = 2V, \quad \varphi(5\text{mm}) = 0V \implies C_1 = -\frac{0.005\rho}{2\varepsilon_0} - \frac{2}{0.005}
φ ( 0 ) = 2 V ⟹ C 2 = 2 V , φ ( 5 mm ) = 0 V ⟹ C 1 = − 2 ε 0 0.005 ρ − 0.005 2
代入数值：ε 0 = 8.85 × 10 − 12 \varepsilon_0 = 8.85 \times 10^{-12} ε 0 = 8.85 × 1 0 − 12 ρ = − 10 − 6 \rho = -10^{-6} ρ = − 1 0 − 6
C 1 = − 0.005 × ( − 10 − 6 ) 2 × 8.85 × 10 − 12 − 400 = 282.5 − 400 = − 117.5 C_1 = -\frac{0.005 \times (-10^{-6})}{2 \times 8.85 \times 10^{-12}} - 400 = 282.5 - 400 = -117.5
C 1 = − 2 × 8.85 × 1 0 − 12 0.005 × ( − 1 0 − 6 ) − 400 = 282.5 − 400 = − 117.5
∴ φ ( x ) = − ρ 2 ε 0 x 2 + C 1 x + C 2 = 5.65 × 10 4 x 2 − 117.5 x + 2 ( V ) \therefore \varphi(x) = -\frac{\rho}{2\varepsilon_0} x^2 + C_1 x + C_2 = 5.65 \times 10^4 x^2 - 117.5x + 2 \quad (\text{V})
∴ φ ( x ) = − 2 ε 0 ρ x 2 + C 1 x + C 2 = 5.65 × 1 0 4 x 2 − 117.5 x + 2 ( V )
E ⃗ = − ∇ φ ( x ) = − d φ d x e ⃗ x = ( − ρ ε 0 x − C 1 ) e ⃗ x = ( 1.13 × 10 5 x + 117.5 ) e ⃗ x ( V/m ) \vec{E} = -\nabla \varphi(x) = -\frac{d\varphi}{dx} \vec{e}_x = (-\frac{\rho}{\varepsilon_0} x - C_1) \vec{e}_x = (1.13 \times 10^5 x + 117.5) \vec{e}_x \quad (\text{V/m})
E = − ∇ φ ( x ) = − d x d φ e x = ( − ε 0 ρ x − C 1 ) e x = ( 1.13 × 1 0 5 x + 117.5 ) e x ( V/m )
方向由正极板指向负极板。
边值问题一般步骤如下 ：
确定场域范围，并利用对称性简化问题。
列出以下方程：① Poisson/Laplace 方程（分域各自列）② 自然边界（如果有）。
强制边界（参考点及电源）④ 分界面衔接条件。
求解表达式，并代入边界条件算出系数。
(4) 镜像法
例 1.1.4 （P68 T1-15）一半径为 a 的接地导体球壳，同心地置于半径为 b 的导体球壳内，两球壳间为真空。一点电荷 q 置于球心 O 处，求两球壳间各点电位。
内球面接地
记点电荷电荷量为 q，易知内球外则分布均匀于表面的感应电荷且电荷量为 − q -q − q
E r = q 4 π ε 0 r 2 E_r = \frac{q}{4\pi\varepsilon_0 r^2}
E r = 4 π ε 0 r 2 q
电位
φ ( r ) = ∫ r b q 4 π ε 0 r 2 d r = q 4 π ε 0 ( 1 r − 1 b ) \varphi(r) = \int_r^b \frac{q}{4\pi\varepsilon_0 r^2} dr = \frac{q}{4\pi\varepsilon_0} \left( \frac{1}{r} - \frac{1}{b} \right)
φ ( r ) = ∫ r b 4 π ε 0 r 2 q d r = 4 π ε 0 q ( r 1 − b 1 )
其中 r 为到球心的距离。下考虑内球内的电场。
设该像电荷电荷量为 q ′ q' q ′ d d d
从而我们得
φ = q 4 π ε 0 r 1 + q ′ 4 π ε 0 r 2 \varphi = \frac{q}{4\pi\varepsilon_0 r_1} + \frac{q'}{4\pi\varepsilon_0 r_2}
φ = 4 π ε 0 r 1 q + 4 π ε 0 r 2 q ′
其中 r 1 , r 2 r_1, r_2 r 1 , r 2 q ′ q' q ′
镜像法结论 ：
① 对于点电荷对接地金属球问题，镜像电荷 $$q’ = -\frac{a}{d} q$$，与球心相距 $$d’ = \frac{a^2}{d}$$
② 对于点电荷对两种无限大电介质平面，镜像电荷则为
q ′ = ε 1 − ε 2 ε 1 + ε 2 q , q ′ ′ = 2 ε 2 ε 1 + ε 2 q q' = \frac{\varepsilon_1 - \varepsilon_2}{\varepsilon_1 + \varepsilon_2} q, \quad q'' = \frac{2\varepsilon_2}{\varepsilon_1 + \varepsilon_2} q
q ′ = ε 1 + ε 2 ε 1 − ε 2 q , q ′′ = ε 1 + ε 2 2 ε 2 q
分别适用于区域 ε 1 \varepsilon_1 ε 1 ε 2 \varepsilon_2 ε 2
镜像法的本质是用点电荷替代感应极化电荷作用。
(5) 电轴法
例 1.1.5 （P69 T1-19）求图中所示带等量异号电荷的偏心圆柱导体间电场，已知介电常数为 ε \varepsilon ε a 1 , a 2 , d a_1, a_2, d a 1 , a 2 , d
电轴法结论 ：
即解如下方程组
{ b 2 = h 1 2 − a 1 2 b 2 = h 2 2 − a 2 2 d = h 2 − h 1 ⟹ { h 1 = d 2 + a 1 2 − a 2 2 2 d h 2 = d 2 + a 2 2 − a 1 2 2 d \begin{cases}
b^2 = h_1^2 - a_1^2 \\
b^2 = h_2^2 - a_2^2 \\
d=h_2-h_1
\end{cases}
\implies
\begin{cases}
h_1 = \frac{d^2 + a_1^2 - a_2^2}{2d} \\
h_2 = \frac{d^2 + a_2^2 - a_1^2}{2d}
\end{cases} ⎩ ⎨ ⎧ b 2 = h 1 2 − a 1 2 b 2 = h 2 2 − a 2 2 d = h 2 − h 1 ⟹ { h 1 = 2 d d 2 + a 1 2 − a 2 2 h 2 = 2 d d 2 + a 2 2 − a 1 2
以上电轴位置确定后，我们便可求解圆柱导体间场强。
电轴法仅适用于带等量异号电荷的平行圆柱导体间场强，并等价于求解方程
h 2 − a 2 = b 2 h^2 - a^2 = b^2
h 2 − a 2 = b 2
其中 h 为两电轴距离一半，b 为电轴至坐标原点距离。
2. 电容及部分电容求解
与推导部分电容的过程类似，设 [ φ ] = [ α ] [ Q ] [\varphi] = [\alpha][Q] [ φ ] = [ α ] [ Q ] [ Q ] = [ β ] [ φ ] [Q] = [\beta][\varphi] [ Q ] = [ β ] [ φ ] C C C β \beta β C C C
例 2.1 （P52 T1-8-3）2 mm 2\,\text{mm} 2 mm
取地面为参考面，列出下述方程：
φ 1 = α 11 Q 1 + α 12 Q 2 , φ 2 = α 21 Q 1 + α 22 Q 2 \varphi_1 = \alpha_{11} Q_1 + \alpha_{12} Q_2, \quad \varphi_2 = \alpha_{21} Q_1 + \alpha_{22} Q_2
φ 1 = α 11 Q 1 + α 12 Q 2 , φ 2 = α 21 Q 1 + α 22 Q 2
令 Q 1 = τ Q_1 = \tau Q 1 = τ Q 2 = 0 Q_2 = 0 Q 2 = 0
φ 1 = 1 2 π ε 0 ( ln 1 r + ln 1 r ′ ) τ = 1 2 π ε 0 ln 8 × 10 3 1 τ \varphi_1 = \frac{1}{2\pi\varepsilon_0} \left( \ln\frac{1}{r} + \ln\frac{1}{r'} \right) \tau = \frac{1}{2\pi\varepsilon_0} \ln\frac{8\times10^3}{1} \tau
φ 1 = 2 π ε 0 1 ( ln r 1 + ln r ′ 1 ) τ = 2 π ε 0 1 ln 1 8 × 1 0 3 τ
φ 2 = 1 2 π ε 0 ( ln 1 3 + ln 1 25 ) τ = 1 2 π ε 0 ln 1 3 τ \varphi_2 = \frac{1}{2\pi\varepsilon_0} \left( \ln\frac{1}{3} + \ln\frac{1}{\sqrt{25}} \right) \tau = \frac{1}{2\pi\varepsilon_0} \ln\frac{1}{3} \tau
φ 2 = 2 π ε 0 1 ( ln 3 1 + ln 25 1 ) τ = 2 π ε 0 1 ln 3 1 τ
∴ α 11 = 1 2 π ε 0 ln 8 × 10 3 , α 21 = 1 2 π ε 0 ln 3 \therefore \alpha_{11} = \frac{1}{2\pi\varepsilon_0} \ln 8\times10^3, \quad \alpha_{21} = \frac{1}{2\pi\varepsilon_0} \ln 3
∴ α 11 = 2 π ε 0 1 ln 8 × 1 0 3 , α 21 = 2 π ε 0 1 ln 3
令 Q 1 = 0 Q_1 = 0 Q 1 = 0 Q 2 = τ Q_2 = \tau Q 2 = τ
φ 1 = 1 2 π ε 0 ln 3 τ , φ 2 = 1 2 π ε 0 ln 7.5 × 10 3 τ \varphi_1 = \frac{1}{2\pi\varepsilon_0} \ln 3 \, \tau, \quad \varphi_2 = \frac{1}{2\pi\varepsilon_0} \ln 7.5\times10^3 \, \tau
φ 1 = 2 π ε 0 1 ln 3 τ , φ 2 = 2 π ε 0 1 ln 7.5 × 1 0 3 τ
∴ α 12 = 1 2 π ε 0 ln 3 , α 22 = 1 2 π ε 0 ln 7.5 × 10 3 \therefore \alpha_{12} = \frac{1}{2\pi\varepsilon_0} \ln 3, \quad \alpha_{22} = \frac{1}{2\pi\varepsilon_0} \ln 7.5\times10^3
∴ α 12 = 2 π ε 0 1 ln 3 , α 22 = 2 π ε 0 1 ln 7.5 × 1 0 3
∴ [ β ] = [ α ] − 1 = [ 0.130 − 0.050 − 0.050 0.131 ] × 2 π ε 0 \therefore [\beta] = [\alpha]^{-1} = \begin{bmatrix} 0.130 & -0.050 \\ -0.050 & 0.131 \end{bmatrix} \times 2\pi\varepsilon_0
∴ [ β ] = [ α ] − 1 = [ 0.130 − 0.050 − 0.050 0.131 ] × 2 π ε 0
C 10 = β 11 + β 12 = 0.080 × 2 π ε 0 F/m , C 12 = − β 12 = 0.050 × 2 π ε 0 F/m C_{10} = \beta_{11} + \beta_{12} = 0.080 \times 2\pi\varepsilon_0 \, \text{F/m}, \quad C_{12} = -\beta_{12} = 0.050 \times 2\pi\varepsilon_0 \, \text{F/m}
C 10 = β 11 + β 12 = 0.080 × 2 π ε 0 F/m , C 12 = − β 12 = 0.050 × 2 π ε 0 F/m
C 21 = − β 21 = 0.050 × 2 π ε 0 F/m , C 20 = β 21 + β 22 = 0.081 × 2 π ε 0 F/m C_{21} = -\beta_{21} = 0.050 \times 2\pi\varepsilon_0 \, \text{F/m}, \quad C_{20} = \beta_{21} + \beta_{22} = 0.081 \times 2\pi\varepsilon_0 \, \text{F/m}
C 21 = − β 21 = 0.050 × 2 π ε 0 F/m , C 20 = β 21 + β 22 = 0.081 × 2 π ε 0 F/m
由电路知识，等效电容 C e q = C 12 + C 10 C 20 C 10 + C 20 = 0.050 + 0.080 × 0.081 0.080 + 0.081 × 2 π ε 0 = 0.090 × 2 π ε 0 F/m C_{eq} = C_{12} + \frac{C_{10}C_{20}}{C_{10}+C_{20}} = 0.050 + \frac{0.080\times0.081}{0.080+0.081} \times 2\pi\varepsilon_0 = 0.090 \times 2\pi\varepsilon_0 \, \text{F/m} C e q = C 12 + C 10 + C 20 C 10 C 20 = 0.050 + 0.080 + 0.081 0.080 × 0.081 × 2 π ε 0 = 0.090 × 2 π ε 0 F/m
3. 静电场能量
例 3.1 （P61 T1-9-3）用 8 mm 8\,\text{mm} 8 mm ε r = 5 \varepsilon_r = 5 ε r = 5 1 pF 1\,\text{pF} 1 pF 1 kV 1\,\text{kV} 1 kV
(1) 移动前，电源已断开
(1)：C 0 = 1 pF C_0 = 1\,\text{pF} C 0 = 1 pF W e 0 = 1 2 C 0 U 0 2 = 5 × 10 − 7 J W_{e0} = \frac{1}{2} C_0 U_0^2 = 5\times10^{-7}\,\text{J} W e 0 = 2 1 C 0 U 0 2 = 5 × 1 0 − 7 J C = Q / U C = Q/U C = Q / U Q Q Q
而 C 0 = 1 pF = ε 0 ε r S d C_0 = 1\,\text{pF} = \frac{\varepsilon_0 \varepsilon_r S}{d} C 0 = 1 pF = d ε 0 ε r S C 1 = ε 0 S d = 1 ε r C 0 = 0.2 pF C_1 = \frac{\varepsilon_0 S}{d} = \frac{1}{\varepsilon_r} C_0 = 0.2\,\text{pF} C 1 = d ε 0 S = ε r 1 C 0 = 0.2 pF
∴ U 1 = Q C 1 = C 0 U 0 C 1 = 5 U 0 = 5 kV \therefore U_1 = \frac{Q}{C_1} = \frac{C_0 U_0}{C_1} = 5U_0 = 5\,\text{kV} ∴ U 1 = C 1 Q = C 1 C 0 U 0 = 5 U 0 = 5 kV W e 1 = 1 2 C 1 U 1 2 = 1 2 × 0.2 × 10 − 12 × ( 5 × 10 3 ) 2 = 2.5 × 10 − 6 J W_{e1} = \frac{1}{2} C_1 U_1^2 = \frac{1}{2} \times 0.2\times10^{-12} \times (5\times10^3)^2 = 2.5\times10^{-6}\,\text{J} W e 1 = 2 1 C 1 U 1 2 = 2 1 × 0.2 × 1 0 − 12 × ( 5 × 1 0 3 ) 2 = 2.5 × 1 0 − 6 J
∴ Δ W e = W e 1 − W e 0 = 2 × 10 − 6 J \therefore \Delta W_e = W_{e1} - W_{e0} = 2\times10^{-6}\,\text{J} ∴ Δ W e = W e 1 − W e 0 = 2 × 1 0 − 6 J A = Δ W e = 2 × 10 − 6 J A = \Delta W_e = 2\times10^{-6}\,\text{J} A = Δ W e = 2 × 1 0 − 6 J
(2) 电源保持连接时，U U U W e 0 = 5 × 10 − 7 J W_{e0} = 5\times10^{-7}\,\text{J} W e 0 = 5 × 1 0 − 7 J C 1 = 0.2 pF C_1 = 0.2\,\text{pF} C 1 = 0.2 pF W e 1 = 1 2 C 1 U 0 2 = 1 2 × 0.2 × 10 − 12 × ( 10 3 ) 2 = 1 × 10 − 7 J W_{e1} = \frac{1}{2} C_1 U_0^2 = \frac{1}{2} \times 0.2\times10^{-12} \times (10^3)^2 = 1\times10^{-7}\,\text{J} W e 1 = 2 1 C 1 U 0 2 = 2 1 × 0.2 × 1 0 − 12 × ( 1 0 3 ) 2 = 1 × 1 0 − 7 J
∴ Δ W e = W e 1 − W e 0 = − 4 × 10 − 7 J \therefore \Delta W_e = W_{e1} - W_{e0} = -4\times10^{-7}\,\text{J} ∴ Δ W e = W e 1 − W e 0 = − 4 × 1 0 − 7 J A = − Δ W e = 4 × 10 − 7 J A = -\Delta W_e = 4\times10^{-7}\,\text{J} A = − Δ W e = 4 × 1 0 − 7 J
仅需记忆以下公式 ：
再代入计算即可。电容器需注意区分常电压系统和常电荷系统。
4. 用虚位移法求静电力
例 4.1 （P6, T1-9-5）极间距离为 d d d U 0 U_0 U 0 ε \varepsilon ε ρ m \rho_m ρ m h h h
设极板宽度为 W W W l l l
C = ε W h d + ε 0 W ( l − h ) d C = \frac{\varepsilon W h}{d} + \frac{\varepsilon_0 W (l-h)}{d}
C = d ε Wh + d ε 0 W ( l − h )
从而
W e = 1 2 C U 0 2 = 1 2 U 0 2 d [ ε W h + ε 0 W ( l − h ) ] W_e = \frac{1}{2} C U_0^2 = \frac{1}{2} \frac{U_0^2}{d} [\varepsilon W h + \varepsilon_0 W (l-h)]
W e = 2 1 C U 0 2 = 2 1 d U 0 2 [ ε Wh + ε 0 W ( l − h )]
∂ W e ∂ h = f e = W U 0 2 2 d ( ε − ε 0 ) \frac{\partial W_e}{\partial h} = f_e = \frac{W U_0^2}{2d} (\varepsilon - \varepsilon_0)
∂ h ∂ W e = f e = 2 d W U 0 2 ( ε − ε 0 )
而液体重力
G = ρ m ⋅ d ⋅ W ⋅ h ⋅ g = f e G = \rho_m \cdot d \cdot W \cdot h \cdot g = f_e
G = ρ m ⋅ d ⋅ W ⋅ h ⋅ g = f e
∴ h = W U 0 2 ( ε − ε 0 ) 2 ρ m g d 2 W = U 0 2 ( ε − ε 0 ) 2 ρ m g d 2 \therefore h = \frac{W U_0^2 (\varepsilon - \varepsilon_0)}{2 \rho_m g d^2 W} = \frac{U_0^2 (\varepsilon - \varepsilon_0)}{2 \rho_m g d^2}
∴ h = 2 ρ m g d 2 W W U 0 2 ( ε − ε 0 ) = 2 ρ m g d 2 U 0 2 ( ε − ε 0 )
一般步骤：① 找出静电能量关于某一变量的表达式 ② 对该变量求偏导
此题亦可使用法拉第观点，详见 P6 例1-2。
