This page is the ONELINE MIRROR VERSION of the answer to homework 5 of the course Electromagnetics.A, CourseID PHYS1004.09, 2026 Spring, 5401, USTC. Last update : 2026/6/1; Status: @ editing, updating, final
题目
电磁学.第三版.习题 2.12 - 2.21
2.12
解：填入介质前，电容器带电量
$$ Q = CV = \frac{\varepsilon_0 S V}{d} = \frac{2\varepsilon_0 \times 10^4}{5 \times 10^{-3}} = 4\varepsilon_0 \times 10^6 \ \text{(C)} $$由高斯定理，电位移矢量大小
$$ D = \sigma_e = \frac{Q}{S} = 2 \varepsilon_0 \times {10}^{6} \text{(C}\cdot\text{m}^{-2}\text{)} $$填入介质后$D$保持不变。
(1) 介质中的电极化强度：
$$ P_1 = \left(1 - \frac{1}{5}\right) \times 2 \times 8.85 \times 10^{-12} \times 10^6 = 1.4 \times 10^{-5} \text{(C}\cdot\text{m}^{-2}\text{)} $$$$ P_2 = \left(1 - \frac{1}{2}\right) \times 2 \times 8.85 \times 10^{-12} \times 10^6 = 8.9 \times 10^{-6} \text{(C}\cdot\text{m}^{-2}\text{)} $$
(2) 电场强度：
$$ E_1 = \frac{D}{\varepsilon_0 \varepsilon_{r1}} = \frac{2 \times 10^6}{5} = 4 \times 10^5 \text{(V}\cdot\text{m}^{-1}\text{)} $$$$ E_2 = \frac{D}{\varepsilon_0 \varepsilon_{r2}} = \frac{2 \times 10^6}{2} = 10^6 \text{(V}\cdot\text{m}^{-1}\text{)} $$
正极板电势：
$$ U = E_1 d_1 + E_2 d_2 = 4 \times 10^5 \times 2 \times 10^{-3} + 10^6 \times 3 \times 10^{-3} = 3.8 \times 10^3 \ \text{(V)} $$分界面电势：
$$ U_{\text{界面}} = E_2 d_2 = 10^6 \times 3 \times 10^{-3} = 3.0 \times 10^3 \ \text{(V)} $$2.13
解：(1) 采用柱坐标，由对称性可知介质中$\boldsymbol{D}$沿径向且仅与$r$有关。由高斯定理：
$$ \boldsymbol{D} = \frac{\lambda_e}{2\pi r} \boldsymbol{e}_r,\quad \boldsymbol{E} = \frac{\boldsymbol{D}}{\varepsilon} = \frac{\lambda_e}{2\pi \varepsilon r} \boldsymbol{e}_r,\quad \boldsymbol{P} = \boldsymbol{D} - \varepsilon_0 \boldsymbol{E} = \frac{\lambda_e (\varepsilon - \varepsilon_0)}{2\pi \varepsilon r} \boldsymbol{e}_r $$极化电荷体密度：
$$ \rho' = \nabla \cdot \boldsymbol{P} = 0 $$极化电荷面密度：
$$ \sigma_1' = \boldsymbol{P}(r = R_1) \cdot (-\boldsymbol{e}_r) = -\frac{\lambda_e (\varepsilon - \varepsilon_0)}{2\pi \varepsilon R_1} $$$$ \sigma_2' = \boldsymbol{P}(r = R_2) \cdot \boldsymbol{e}_r = \frac{\lambda_e (\varepsilon - \varepsilon_0)}{2\pi \varepsilon R_2} $$
(2) 电势差：
$$ \Delta U = \int_{R_1}^{R_2} E \, \mathrm{d}r = \frac{\lambda_e}{2\pi \varepsilon} \ln \frac{R_2}{R_1} $$(3) 电容：
$$ C = \frac{q}{\Delta U} = \frac{2\pi \varepsilon l}{\ln(R_2 / R_1)} $$2.14
解：由高斯定理得：
$$ D = \frac{Q}{4\pi r^2} \boldsymbol{e}_r,\quad E = \frac{Q}{4\pi \varepsilon_0 r^2} (1 + kr) \boldsymbol{e}_r $$电势差：
$$ V = \int_a^b \boldsymbol{E} \cdot \mathrm{d}\boldsymbol{r} = \frac{Q}{4\pi \varepsilon_0} \left( \frac{1}{a} - \frac{1}{b} + k \ln \frac{b}{a} \right) $$电容：
$$ C = \frac{Q}{V} = \frac{4\pi \varepsilon_0 ab}{(b - a) + abk \ln(b/a)} $$极化强度：
$$ \boldsymbol{P} = (\varepsilon - \varepsilon_0)\boldsymbol{E} = -\frac{kQ}{4\pi r} \boldsymbol{e}_r $$体极化电荷密度：
$$ \rho' = -\nabla \cdot \boldsymbol{P} = \frac{kQ}{4\pi r^2} $$面极化电荷密度：
$$ \sigma_a' = \boldsymbol{P}(a) \cdot (-\boldsymbol{e}_r) = \frac{kQ}{4\pi a},\quad \sigma_b' = \boldsymbol{P}(b) \cdot \boldsymbol{e}_r = -\frac{kQ}{4\pi b} $$2.15
解：金属球接地，电势$U=0$。介质内自由电荷密度：
$$ \rho=\frac{3q_0}{4\pi \left[(2R)^3-R^3\right]} $$设接地金属球表面感应电荷量为$q$，电荷沿球面均匀分布。由高斯定理得介质内电场：
$$ E_1 = \frac{1}{8\pi \varepsilon_0} \left( \frac{q}{r^2} + \frac{q_0 r}{7R^3} - \frac{q_0}{7r^2} \right) $$介质外电场：
$$ E_2 = \frac{q + q_0}{4\pi \varepsilon_0 r^2} $$由边界条件解得：
$$ q = -\frac{16q_0}{21} $$外表面电势：
$$ U = \int_{2R}^{\infty} E_2 \, \mathrm{d}r = \frac{q + q_0}{8\pi \varepsilon_0 R} = \frac{5q_0}{168\pi \varepsilon_0 R} $$2.16
解：(1) 全域 $D = 8.85 \times 10^{-10} \ \mathrm{C \cdot m^{-2}}$。
介质外：
$$ P = 0,\quad E = \frac{D}{\varepsilon_0} = 100 \ \mathrm{(V \cdot m^{-1})} $$介质内：
$$ P = \left(1 - \varepsilon_r^{-1}\right)D = 4.43 \times 10^{-10} \ \mathrm{(C \cdot m^{-2})},\quad E = \frac{D}{\varepsilon_0 \varepsilon_r} = 50 \ \mathrm{(V \cdot m^{-1})} $$(2) 电势分布，其中 $x_1 = 0.01,\text{m},\ x_2 = 0.02,\text{m},\ x_3 = 0.03,\text{m}$：
$$ V(0 < x < x_1) = 100x $$$$ V(x_1 < x < x_2) = 50(x + x_1) = 50x + 0.5 $$
$$ V(x_2 < x < x_3) = 100x - 0.5 $$
2.17
解：(1) 介质中$\boldsymbol{D}$沿径向且仅与$R$有关，由高斯定理：
$$ \boldsymbol{D}=\frac{q}{4\pi R^2}\boldsymbol{e}_R $$电场：
$$ \boldsymbol{E}(R_1 \le R \le r) = \frac{q}{4\pi \varepsilon_1 R^2} \boldsymbol{e}_R,\quad \boldsymbol{E}(r \le R \le R_2) = \frac{q}{4\pi \varepsilon_2 R^2} \boldsymbol{e}_R $$电势差：
$$ \Delta U=\int_{R_1}^{R_2}\boldsymbol{E}\cdot \mathrm{d}\boldsymbol{R} = \frac{q}{4\pi} \left( \frac{1}{\varepsilon_1 R_1} - \frac{1}{\varepsilon_1 r} + \frac{1}{\varepsilon_2 r} - \frac{1}{\varepsilon_2 R_2} \right) $$电容：
$$ C = \frac{q}{\Delta U} = \frac{4\pi \varepsilon_1 \varepsilon_2 R_1 R_2 r}{(\varepsilon_1 - \varepsilon_2)R_1 R_2 + (\varepsilon_2 R_2 - \varepsilon_1 R_1)r} $$(2) 极化强度：
$$ \boldsymbol{P}_1 = -\frac{Q(\varepsilon_1 - \varepsilon_0)}{4\pi \varepsilon_1 R^2} \boldsymbol{e}_R,\quad \boldsymbol{P}_2 = -\frac{Q(\varepsilon_2 - \varepsilon_0)}{4\pi \varepsilon_2 R^2} \boldsymbol{e}_R $$极化电荷面密度：
$$ \sigma'\big|_{r} = -(\boldsymbol{P}_2-\boldsymbol{P}_1)\cdot\boldsymbol{e}_R=-\frac{Q}{4\pi r^2} \cdot \frac{\varepsilon_0(\varepsilon_1 - \varepsilon_2)}{\varepsilon_1 \varepsilon_2} $$$$ \sigma_1' =\boldsymbol{P}_1(R_1)\cdot (-\boldsymbol{e}_R)= \frac{Q(\varepsilon_1 - \varepsilon_0)}{4\pi \varepsilon_1 R_1^2},\quad \sigma_2' =\boldsymbol{P}_2(R_2)\cdot \boldsymbol{e}_R= -\frac{Q(\varepsilon_2 - \varepsilon_0)}{4\pi \varepsilon_2 R_2^2} $$
2.18
证：由电磁场边值关系：
$$ E_1 \sin\theta_1 = E_2 \sin\theta_2,\quad \varepsilon_1 E_1 \cos\theta_1 - \varepsilon_2 E_2 \cos\theta_2 = \sigma $$两式相除得：
$$ \varepsilon_2 \cot\theta_2 = \varepsilon_1 \cot\theta_1 - \frac{\sigma}{E_1 \sin\theta_1} = \varepsilon_1 \left(1 - \frac{\sigma}{\varepsilon_1 E_1 \cos\theta_1}\right) \cot\theta_1 $$2.19
解：由高斯定理：
$$ 2\pi r^2 \varepsilon_1 E + 2\pi r^2 \varepsilon_2 E = q \quad \Rightarrow \quad \boldsymbol{E} = \frac{q}{2\pi (\varepsilon_1 + \varepsilon_2) r^2} \boldsymbol{e}_r $$电位移矢量：
$$ \boldsymbol{D}_1 = \varepsilon_1 \boldsymbol{E} = \frac{\varepsilon_1 q}{2\pi (\varepsilon_1 + \varepsilon_2) r^2} \boldsymbol{e}_r,\quad \boldsymbol{D}_2 = \varepsilon_2 \boldsymbol{E} = \frac{\varepsilon_2 q}{2\pi (\varepsilon_1 + \varepsilon_2) r^2} \boldsymbol{e}_r $$自由电荷面密度：
$$ \sigma_1 = D_1(a)=\frac{\varepsilon_1 q}{2\pi (\varepsilon_1 + \varepsilon_2) a^2},\quad \sigma_2 = D_2(a)=\frac{\varepsilon_2 q}{2\pi (\varepsilon_1 + \varepsilon_2) a^2} $$2.20
解：电场线与介质分界面平行，A、B两点场强大小相等。无介质时电容 $C_0=\dfrac{\varepsilon_0S}{d}$，极板带电量 $Q=C_0V$。
(1) 插入介质后等效为两部分电容并联：
$$ C=\frac{\varepsilon S}{2d}+\frac{\varepsilon_0 S}{2d} $$开关保持接通，电压不变，电场与原场强相等：
$$ E = \frac{V}{d} $$(2) 开关断开，极板总电荷不变，电场：
$$ E = \frac{2\varepsilon_0}{\varepsilon + \varepsilon_0} E_0 $$2.21
解：由电像法得像电荷：
$$ q'' = \frac{\varepsilon_0(\varepsilon_2 - \varepsilon_1)q}{\varepsilon_2(\varepsilon_1 + \varepsilon_2)},\quad q' = \frac{\varepsilon_0(\varepsilon_2 - \varepsilon_1)q}{\varepsilon_1(\varepsilon_1 + \varepsilon_2)} $$受力位置处电场：
$$ E = \frac{q}{16\pi \varepsilon_1 a^2} $$电荷所受作用力：
$$ F = -qE = -\frac{q^2}{16\pi \varepsilon_1 a^2} $$