Answer to Homework-07 of Electromagnetics.A

This page is the ONELINE MIRROR VERSION of the answer to homework 7 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 3.4-3.17

3.4

解：(1) 电子电荷均匀分布在半径为 $a$ 的球面上，球面电势

$$ U = \frac{e}{4\pi\varepsilon_0 a} $$

电子静电势能

$$ W = \frac{1}{2} e U = \frac{e^2}{8\pi\varepsilon_0 a} $$

令静电势能等于电子静止能量 $m c^2$：

$$ \frac{e^2}{8\pi\varepsilon_0 a} = m c^2 $$

解得

$$ a = \frac{e^2}{8\pi\varepsilon_0 m c^2} $$

(2) 电荷均匀分布在半径为 $a$ 的球体内，电荷密度

$$ \rho = \frac{3e}{4\pi a^3} $$

由高斯定理，球内、外电场：

$$ \mathbf{E}(r < a) = \frac{\rho r}{3\varepsilon_0} \mathbf{e}_r = \frac{e r}{4\pi\varepsilon_0 a^3} \mathbf{e}_r,\qquad \mathbf{E}(r > a) = \frac{e}{4\pi\varepsilon_0 r^2} \mathbf{e}_r $$

球内电势分布

$$ U(r) = \int_r^\infty \mathbf{E} \cdot \mathrm{d}\mathbf{r} = \int_r^a \frac{e r}{4\pi\varepsilon_0 a^3} \mathrm{d}r + \int_a^\infty \frac{e}{4\pi\varepsilon_0 r^2} \mathrm{d}r = \frac{e(3a^2 - r^2)}{8\pi\varepsilon_0 a^3} $$

电子静电势能

$$ W = \frac{1}{2} \iiint_V \rho U \, \mathrm{d}V = \frac{1}{2} \int_0^a \frac{3e}{4\pi a^3} \cdot \frac{e(3a^2 - r^2)}{8\pi\varepsilon_0 a^3} \cdot 4\pi r^2 \mathrm{d}r = \frac{3e^2}{20\pi\varepsilon_0 a} $$

令 $W = m c^2$，得

$$ a = \frac{3e^2}{20\pi\varepsilon_0 m c^2} $$

(3) 电子经典半径

$$ r_0 = \frac{e^2}{4\pi\varepsilon_0 m c^2} = \frac{10^{-7} \times (1.6 \times 10^{-19})^2}{9.11 \times 10^{-31}} = 2.8 \times 10^{-15} \, \text{m} $$

3.5

解：带电后，壳内壁感应电荷 $-Q$，外表面带电 $+Q$。系统总储能为球形电容器储能与孤立导体球储能之和：

$$ W = \frac{Q^2}{2C_1} + \frac{Q^2}{2C_2} $$

其中

$$ C_1 = \frac{4\pi\varepsilon_0 R_1 R_2}{R_2 - R_1},\quad C_2 = 4\pi\varepsilon_0 R_3 $$

代入得

$$ W = \frac{(R_2 - R_1)Q^2}{8\pi\varepsilon_0 R_1 R_2} + \frac{Q^2}{8\pi\varepsilon_0 R_3} = \frac{Q^2}{8\pi\varepsilon_0} \left( \frac{1}{R_1} - \frac{1}{R_2} + \frac{1}{R_3} \right) $$

代入数值 $Q = 3.0 \times 10^{-8},\text{C}$，$R_1 = 0.02,\text{m}$，$R_2 = 0.04,\text{m}$，$R_3 = 0.05,\text{m}$，$\varepsilon_0 = 8.85 \times 10^{-12},\text{F/m}$：

$$ W = \frac{(3 \times 10^{-8})^2}{8 \times \pi \times 8.85 \times 10^{-12}} \times \left( \frac{1}{0.02} - \frac{1}{0.04} + \frac{1}{0.05} \right) = 1.82 \times 10^{-4} \, \text{J} $$

用导线连接球与壳后，系统等效为半径 $R_3$ 的孤立导体球，储能：

$$ W' = \frac{Q^2}{8\pi\varepsilon_0 R_3} = \frac{(3 \times 10^{-8})^2}{8 \times \pi \times 8.85 \times 10^{-12} \times 0.05} = 8.09 \times 10^{-5} \, \text{J} $$

3.6

解：由3.4结论，半径为 $r$、电量为 $Q$ 的均匀带电球静电能

$$ W = \frac{3Q^2}{20\pi\varepsilon_0 r} $$

(1) 铀235原子核静电势能

$$ W_1 = \frac{3 \times (92 \times 1.6 \times 10^{-19})^2}{20 \times \pi \times 8.85 \times 10^{-12} \times 9.2 \times 10^{-15}} = 1.27 \times 10^{-10} \, \text{J} = 795 \, \text{MeV} $$

(2) 分裂后每块碎片电量为 $Q/2$，由体积关系：

$$ 2 \cdot \frac{4\pi}{3} r_1^3 = \frac{4\pi}{3} r^3 \quad\Rightarrow\quad r_1 = \frac{r}{2^{1/3}} $$

释放能量

$$ \Delta W = \frac{3Q^2}{20\pi\varepsilon_0 r} - 2 \cdot \frac{3(Q/2)^2}{20\pi\varepsilon_0 r_1} = \frac{3Q^2}{20\pi\varepsilon_0 r} \left(1 - 2^{-2/3}\right) = 4.70 \times 10^{-11} \, \text{J} = 294 \, \text{MeV} $$

(3) 1 kg 铀235原子核数目

$$ N = \frac{1}{235 \times 1.66 \times 10^{-27}} = 2.56 \times 10^{24} $$

总释放能量

$$ \Delta W_{\text{总}} = N \times \Delta W = 2.56 \times 10^{24} \times 4.70 \times 10^{-11} = 1.2 \times 10^{14} \, \text{J} $$

3.7

解：(1) 由对称性与高斯定理，$a < r < b$ 区域：

$$ D = \frac{Q}{2\pi r l},\quad E = \frac{D}{\varepsilon} = \frac{Q}{2\pi\varepsilon r l} $$

电场能量密度

$$ w = \frac{1}{2} D E = \frac{1}{2} \cdot \frac{Q}{2\pi r l} \cdot \frac{Q}{2\pi\varepsilon r l} = \frac{Q^2}{8\pi^2 \varepsilon r^2 l^2} $$

(2) 介质内总电场能

$$ W = \iiint w \, \mathrm{d}V = \int_a^b \frac{Q^2}{8\pi^2 \varepsilon r^2 l^2} \cdot 2\pi r l \, \mathrm{d}r = \frac{Q^2}{4\pi\varepsilon l} \int_a^b \frac{\mathrm{d}r}{r} = \frac{Q^2}{4\pi\varepsilon l} \ln \frac{b}{a} $$

(3) 两极电势差

$$ \Delta U = \int_a^b E \, \mathrm{d}r = \int_a^b \frac{Q}{2\pi\varepsilon r l} \mathrm{d}r = \frac{Q}{2\pi\varepsilon l} \ln \frac{b}{a} $$

电容

$$ C = \frac{Q}{\Delta U} = \frac{2\pi\varepsilon l}{\ln(b/a)} $$

验证：

$$ \frac{Q^2}{2C} = \frac{Q^2}{2} \cdot \frac{\ln(b/a)}{2\pi\varepsilon l} = \frac{Q^2}{4\pi\varepsilon l} \ln \frac{b}{a} = W $$

证毕。

3.8

证：由3.7，半径 $r$ 以内电场能量

$$ W(r) = \frac{Q^2}{4\pi\varepsilon_0 l} \ln \frac{r}{a} $$

电容器总能量

$$ W = \frac{Q^2}{4\pi\varepsilon_0 l} \ln \frac{b}{a} $$

令 $W(r) = \dfrac{1}{2}W$：

$$ \frac{Q^2}{4\pi\varepsilon_0 l} \ln \frac{r}{a} = \frac{1}{2} \cdot \frac{Q^2}{4\pi\varepsilon_0 l} \ln \frac{b}{a} $$

$$ \ln \frac{r}{a} = \frac{1}{2} \ln \frac{b}{a} = \ln \left(\frac{b}{a}\right)^{\frac12} $$

$$ \frac{r}{a} = \sqrt{\frac{b}{a}} \quad\Rightarrow\quad r = \sqrt{ab} $$

证毕。

3.9

解：由高斯定理，圆柱形电容器内电场

$$ E = \frac{\lambda_e}{2\pi\varepsilon_0 r} $$

内导体表面场强最大，取临界击穿场强 $E_{\text{max}} = E_b$：

$$ E_b = \frac{\lambda_e}{2\pi\varepsilon_0 R_1} \quad\Rightarrow\quad \lambda_e = 2\pi\varepsilon_0 E_b R_1 $$

两极电势差

$$ U = \int_{R_1}^{R_2} E \, \mathrm{d}r = \frac{\lambda_e}{2\pi\varepsilon_0} \ln \frac{R_2}{R_1} = E_b R_1 \ln \frac{R_2}{R_1} $$

单位长度储能

$$ W = \frac{\lambda_e^2}{4\pi\varepsilon_0} \ln \frac{R_2}{R_1} = \pi\varepsilon_0 E_b^2 R_1^2 \ln \frac{R_2}{R_1} $$

(1) 求 $U$ 极值，令 $\dfrac{\mathrm{d}U}{\mathrm{d}R_1} = 0$：

$$ \frac{\mathrm{d}}{\mathrm{d}R_1} \left( E_b R_1 \ln \frac{R_2}{R_1} \right) = E_b \left( \ln \frac{R_2}{R_1} - 1 \right) = 0 $$

得 $\ln \dfrac{R_2}{R_1} = 1$，即 $R_1 = \dfrac{R_2}{e}$。最大电势差

$$ U_{\text{max}} = \frac{E_b R_2}{e} $$

(2) 求 $W$ 极值，令 $\dfrac{\mathrm{d}W}{\mathrm{d}R_1} = 0$：

$$ \frac{\mathrm{d}}{\mathrm{d}R_1} \left( \pi\varepsilon_0 E_b^2 R_1^2 \ln \frac{R_2}{R_1} \right) = \pi\varepsilon_0 E_b^2 \left( 2R_1 \ln \frac{R_2}{R_1} - R_1 \right) = 0 $$

得 $\ln \dfrac{R_2}{R_1} = \dfrac12$，即 $R_1 = \dfrac{R_2}{\sqrt{e}}$。最大储能

$$ W_{\text{max}} = \frac{\pi\varepsilon_0 E_b^2 R_2^2}{2e} $$

对应电势差

$$ U_{\text{max}} = \frac{E_b R_2}{2\sqrt{e}} $$

(3) 代入 $E_b = 3 \times 10^6,\text{V/m}$，$R_2 = 0.01,\text{m}$，$e = 2.718$：第一种方案：

$$ U_{\text{max}} = \frac{3 \times 10^6 \times 0.01}{2.718} = 1.1 \times 10^4 \, \text{V} $$

第二种方案：

$$ U_{\text{max}} = \frac{3 \times 10^6 \times 0.01}{2 \times \sqrt{2.718}} = 9.1 \times 10^3 \, \text{V} $$

3.10

解：(1) 内球壳自身电势 $U_1 = \dfrac{Q}{4\pi\varepsilon_0 R_1}$，自能

$$ W_{\text{自}}(+Q) = \frac{1}{2} Q U_1 = \frac{Q^2}{8\pi\varepsilon_0 R_1} $$

外球壳自身电势 $U_2 = -\dfrac{Q}{4\pi\varepsilon_0 R_2}$，自能

$$ W_{\text{自}}(-Q) = \frac{1}{2} (-Q) U_2 = \frac{Q^2}{8\pi\varepsilon_0 R_2} $$

(2) 内电荷在外球壳处电势 $U_{12} = \dfrac{Q}{4\pi\varepsilon_0 R_2}$，外电荷在内球壳处电势 $U_{21} = -\dfrac{Q}{4\pi\varepsilon_0 R_2}$，互能

$$ W_{\text{互}} = Q U_{12} = -\frac{Q^2}{4\pi\varepsilon_0 R_2} $$

(3) 系统总能量

$$ W = W_{\text{自}}(+Q) + W_{\text{自}}(-Q) + W_{\text{互}} = \frac{Q^2}{8\pi\varepsilon_0} \left( \frac{1}{R_1} - \frac{1}{R_2} \right) $$

用球形电容 $C = \dfrac{4\pi\varepsilon_0 R_1 R_2}{R_2 - R_1}$ 验证：

$$ W = \frac{Q^2}{2C} = \frac{Q^2}{8\pi\varepsilon_0} \left( \frac{1}{R_1} - \frac{1}{R_2} \right) $$

结果一致。

3.11

解：设介质拉出长度为 $x$，系统等效为两个圆柱形电容并联，总电容

$$ C = \frac{2\pi\varepsilon_0 x}{\ln(b/a)} + \frac{2\pi\varepsilon (l-x)}{\ln(b/a)} = \frac{2\pi[\varepsilon l - (\varepsilon - \varepsilon_0)x]}{\ln(b/a)} $$

电容器接电源，电压 $V$ 不变，介质受力

$$ F = \left( \frac{\partial W}{\partial x} \right)_V = \frac{V^2}{2} \frac{\mathrm{d}C}{\mathrm{d}x} = -\frac{\pi(\varepsilon - \varepsilon_0)V^2}{\ln(b/a)} $$

负号表示电场力为吸引力。维持介质不动，外力

$$ F_{\text{外}} = \frac{\pi(\varepsilon - \varepsilon_0)V^2}{\ln(b/a)} $$

方向向外。

3.12

解：系统等效为两组平行板电容并联。无介质部分电容：

$$ C_1 = \frac{\varepsilon_0 b (a-x)}{d} $$

含介质部分等效串联电容，记 $t’ = \dfrac{\varepsilon_r - 1}{\varepsilon_r} t$：

$$ C_2 = \frac{\varepsilon_0 b x}{d - t'} $$

总电容

$$ C = C_1 + C_2 = \frac{\varepsilon_0 b \big[(d - t')a + x t'\big]}{d(d - t')} $$

极板总电量 $Q$ 不变，介质受力

$$ F = -\left( \frac{\partial W}{\partial x} \right)_Q = \frac{Q^2}{2C^2} \frac{\mathrm{d}C}{\mathrm{d}x} $$

求导

$$ \frac{\mathrm{d}C}{\mathrm{d}x} = \frac{\varepsilon_0 b t'}{d(d - t')} $$

代入得

$$ F = \frac{Q^2 b (d - t') t' d}{2\varepsilon_0 \big[S(d - t') + x b t'\big]^2} $$

其中 $S=ab$。力沿 $x$ 增大方向，指向介质原位置，得证。

3.13

解：初始电容（介质厚度 $t$，相对介电常数 $\varepsilon_r$），记 $t’ = \dfrac{\varepsilon_r - 1}{\varepsilon_r} t$：

$$ C_0 = \frac{\varepsilon_0 S}{d - t'} $$

抽出介质后电容

$$ C = \frac{\varepsilon_0 S}{d} $$

初始电量 $Q = C_0 V$。

(1) 断开电源，$Q$ 不变，外界做功等于静电能增量：

$$ A_1 = \frac{Q^2}{2C} - \frac{Q^2}{2C_0} = \frac{\varepsilon_0 S t' V^2}{2(d - t')^2} $$

(2) 接通电源，$V$ 不变，外界做功：

$$ A_2 = \frac{C_0 V^2}{2} - \frac{C V^2}{2} = \frac{\varepsilon_0 S t' V^2}{2(d - t')d} $$

(3) 换为导体板即 $\varepsilon_r \to \infty$，$t’\to t$：

$$ A_1 = \frac{\varepsilon_0 S t V^2}{2(d - t)^2},\quad A_2 = \frac{\varepsilon_0 S t V^2}{2(d - t)d} $$

3.14

解：(1) 电像法配置像电荷，原电荷 $(a,0,a)$ 受力：

$$ \mathbf{F} = \frac{q^2}{4\pi\varepsilon_0} \left[ -\frac{1}{(2a)^2} \mathbf{e}_x - \frac{1}{(2a)^2} \mathbf{e}_z + \frac{1}{(2\sqrt{2}a)^2} \left( \frac{1}{\sqrt{2}} \mathbf{e}_x + \frac{1}{\sqrt{2}} \mathbf{e}_z \right) \right] $$

$$ = \frac{q^2}{32\pi\varepsilon_0 a^2} \left[ \left(-2 + \frac{1}{\sqrt{2}}\right) \mathbf{e}_x + \left(-2 + \frac{1}{\sqrt{2}}\right) \mathbf{e}_z \right] $$

力的大小

$$ F = \frac{(2\sqrt{2} - 1)q^2}{32\pi\varepsilon_0 a^2} $$

方向在 $xz$ 平面指向原点。

(2) 沿 $z=x,,y=0$ 移动电荷，外界做功

$$ A = \frac{(4 - \sqrt{2})q^2}{16\pi\varepsilon_0} \int_a^\infty \frac{\mathrm{d}x}{x^2} = -\frac{(4 - \sqrt{2})q^2}{16\pi\varepsilon_0 a} $$

(3) $(a,0,0)$ 处总电场

$$ \mathbf{E} = \frac{q}{2\pi\varepsilon_0 a^2} \left( \frac{1}{5\sqrt{5}} - 1 \right) \mathbf{e}_z $$

面电荷密度

$$ \sigma = \varepsilon_0 E = \frac{q}{2\pi a^2} \left( \frac{1}{5\sqrt{5}} - 1 \right) $$

3.15

解：(1) 均匀外场 $\mathbf{E}_0$ 中导体球，球外电势设为

$$ U = U_0 - E_0 r \cos\theta + \frac{p \cos\theta}{4\pi\varepsilon_0 r^2} $$

球面 $r=a$ 为等势面，$\cos\theta$ 项系数为零：

$$ -E_0 a + \frac{p}{4\pi\varepsilon_0 a^2} = 0 \quad\Rightarrow\quad p = 4\pi\varepsilon_0 a^3 E_0 $$

偶极矩 $\mathbf{p} = 4\pi\varepsilon_0 a^3 \mathbf{E}_0$。

(2) 球外电势

$$ U_{\text{外}} = U_0 - E_0 r \cos\theta + \frac{a^3 E_0 \cos\theta}{r^2} $$

球面带电量 $q$ 时，球内电势

$$ U_{\text{内}} = U_0 + \frac{q}{4\pi\varepsilon_0 a} $$

(3) 球面外侧径向场强

$$ E_{r2} = -\frac{\partial U_{\text{外}}}{\partial r}\bigg|_{r=a} = 3E_0 \cos\theta $$

导体内部场强为0，感应面密度

$$ \sigma = \varepsilon_0 E_{r2} = 3\varepsilon_0 E_0 \cos\theta $$

感应电荷电势

$$ U_{\text{感}} = E_0 a \cos\theta $$

静电自能

$$ W = \frac{1}{2} \iint \sigma U_{\text{感}} \, \mathrm{d}S = 2\pi\varepsilon_0 a^3 E_0^2 $$

3.16

解：无限大介质分界面 $z=0$，点电荷 $q$ 位于 $(0,0,z)$。由边界条件解得像电荷：

$$ q' = \frac{1 - \varepsilon_r}{1 + \varepsilon_r} q,\quad q'' = \frac{2\varepsilon_r}{1 + \varepsilon_r} q $$

原电荷仅受像电荷 $q’$ 的库仑力，间距 $2z$：

$$ F(z) = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q q'}{(2z)^2} = -\frac{q^2}{16\pi\varepsilon_0 z^2} \cdot \frac{\varepsilon_r - 1}{\varepsilon_r + 1} $$

外力与电场力等大反向，将电荷移至无穷远，外力做功

$$ A = \int_{z}^{\infty} -F \, \mathrm{d}z' = \frac{q^2}{16\pi\varepsilon_0 z} \cdot \frac{\varepsilon_r - 1}{\varepsilon_r + 1} $$