This page is the ONELINE MIRROR VERSION of the answer to homework 9 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
4.16(4.15)
假设导体带电$Q$，导体之间电压为$U$，电流为$I$：
$$ C=\frac{Q}{U}=\frac{Q}{I}\frac{I}{U}=\frac{\oint_S \boldsymbol{D}\cdot \mathrm{d}\boldsymbol{S}}{\oint_S \boldsymbol{j} \cdot\mathrm{d}\boldsymbol{S}}\frac{1}{R}=\frac{Q}{I}\frac{I}{U}=\frac{\oint_S \varepsilon \boldsymbol{E}\cdot \mathrm{d}\boldsymbol{S}}{ R\oint_S \sigma \boldsymbol{E} \cdot\mathrm{d}\boldsymbol{S}}=\frac{\varepsilon}{\sigma R} $$4.17
类比均匀介质球置于均匀外电场模型，替换对应物理量即可求解。
(1) 电势与场强分布
$$ \begin{cases} \varphi_1 = -\dfrac{3\sigma_0}{\sigma + 2\sigma_0} E_0 r \cos\theta = -\dfrac{3}{\sigma + 2\sigma_0} j_0 r \cos\theta, & r < R \\[10pt] \varphi_2 = -\dfrac{j_0}{\sigma_0} r \cos\theta + \dfrac{\sigma - \sigma_0}{\sigma + 2\sigma_0} \dfrac{R^3}{r^2} \dfrac{j_0}{\sigma_0} \cos\theta, & r > R \end{cases} $$$$ \boldsymbol{E}_1 = \frac{3}{\sigma + 2\sigma_0} \boldsymbol{j}_0, \quad r < R $$$$ \boldsymbol{E}_2 = \left( 1 + \frac{\sigma - \sigma_0}{\sigma + 2\sigma_0} \frac{2R^3}{r^3} \right) \frac{j_0}{\sigma_0} \cos\theta \,\boldsymbol{e}_r + \left( \frac{\sigma - \sigma_0}{\sigma + 2\sigma_0} \frac{R^3}{r^3} - 1 \right) \frac{j_0}{\sigma_0} \sin\theta \,\boldsymbol{e}_\theta, \quad r > R $$(2) 等效电偶极矩
$$ p = \frac{\sigma - \sigma_0}{\sigma + 2\sigma_0} \, 4\pi \varepsilon_0 R^3 \, \frac{j_0}{\sigma_0} $$(3) 焦耳功率与极值 由焦耳定律，导体球焦耳功率：
$$ P=\int_V\sigma E^2\mathrm{d} V=\frac{9\sigma j_0^2}{(\sigma+2\sigma_0)^2}\cdot\frac{4}{3}\pi R^3=\frac{12\pi j_0^2R^3}{\left(\sqrt{\sigma}+\dfrac{2\sigma_0}{\sqrt{\sigma}}\right)^2}\le\frac{3\pi j_0^2 R^3}{2\sigma_0} $$取等条件：$\sigma=2\sigma_0$
4.18
令 4.17 中 $\sigma=0$ 即可得到对应结果。
补充作业
(1) 证明 $\boldsymbol{A} \times \boldsymbol{A} = \mathbf{0}$
$$ (\boldsymbol{A} \times \boldsymbol{A} )_i = \varepsilon_{ijk} A_j A_k $$交换哑指标 $j \leftrightarrow k$：
$$ \varepsilon_{ijk} A_j A_k = \varepsilon_{ikj} A_k A_j = -\varepsilon_{ijk} A_j A_k $$移项得 $2\varepsilon_{ijk} A_j A_k = 0$，即 $(\boldsymbol{A} \times \boldsymbol{A} )_i = 0$，故
$$ \boldsymbol{A} \times \boldsymbol{A} = \mathbf{0} $$(2) 证明 $\boldsymbol{A} \cdot (\boldsymbol{A} \times \boldsymbol{B} ) = 0$
$$ \boldsymbol{A} \cdot (\boldsymbol{A} \times \boldsymbol{B} ) = A_i (\varepsilon_{ijk} A_j B_k) = \varepsilon_{ijk} A_i A_j B_k $$交换哑指标：
$$ \varepsilon_{ijk} A_i A_j B_k = \varepsilon_{jik} A_j A_i B_k = -\varepsilon_{ijk} A_i A_j B_k $$因此 $2\varepsilon_{ijk} A_i A_j B_k = 0$，即
$$ \boldsymbol{A} \cdot (\boldsymbol{A} \times \boldsymbol{B} ) = 0 $$(3) 矢量混合积恒等式
$$ \begin{aligned} (\boldsymbol{A} \times\boldsymbol{B} )\cdot(\boldsymbol{C} \times\boldsymbol{D} ) &= \boldsymbol{A} \cdot\big[\boldsymbol{B} \times(\boldsymbol{C} \times\boldsymbol{D} )\big] \\ &= \boldsymbol{A} \cdot\big[\boldsymbol{C} (\boldsymbol{B} \cdot\boldsymbol{D} ) - \boldsymbol{D} (\boldsymbol{B} \cdot\boldsymbol{C} )\big] \\ &= (\boldsymbol{A} \cdot\boldsymbol{C} )(\boldsymbol{B} \cdot\boldsymbol{D} ) - (\boldsymbol{A} \cdot\boldsymbol{D} )(\boldsymbol{B} \cdot\boldsymbol{C} ) \end{aligned} $$(IV.31) $\nabla(f+g)=\nabla f+\nabla g$
$$ \partial_i(f+g)=\partial_i f+\partial_i g $$(IV.32) $\nabla\cdot(\boldsymbol{A} +\boldsymbol{B} )=\nabla\cdot\boldsymbol{A} +\nabla\cdot\boldsymbol{B} $
$$ \partial_i(A_i+B_i)=\partial_i A_i+\partial_i B_i $$(IV.33) $\nabla\times(\boldsymbol{A} +\boldsymbol{B} )=\nabla\times\boldsymbol{A} +\nabla\times\boldsymbol{B} $
$$ \varepsilon_{ijk}\partial_j(A_k+B_k)=\varepsilon_{ijk}\partial_j A_k+\varepsilon_{ijk}\partial_j B_k $$(IV.34) $\nabla(fg)=(\nabla f)g+f(\nabla g)$
$$ \partial_i(fg)=(\partial_i f)g+f(\partial_i g) $$(IV.35) $\nabla\cdot(f\boldsymbol{A} )=(\nabla f)\cdot\boldsymbol{A} +f(\nabla\cdot\boldsymbol{A} )$
$$ \partial_i(f A_i)=(\partial_i f)A_i+f(\partial_i A_i) $$(IV.36) $\nabla\times(f\boldsymbol{A} )=(\nabla f)\times\boldsymbol{A} +f(\nabla\times\boldsymbol{A} )$
$$ \varepsilon_{ijk}\partial_j(f A_k)=\varepsilon_{ijk}(\partial_j f)A_k+f\varepsilon_{ijk}\partial_j A_k $$(IV.37) $\nabla(\boldsymbol{A} \cdot\boldsymbol{B} )=(\boldsymbol{B} \cdot\nabla)\boldsymbol{A} +\boldsymbol{B} \times(\nabla\times\boldsymbol{A} )+(\boldsymbol{A} \cdot\nabla)\boldsymbol{B} +\boldsymbol{A} \times(\nabla\times\boldsymbol{B} )$
$$ \begin{aligned} \partial_i(A_j B_j) &= (\partial_i A_j)B_j + A_j(\partial_i B_j) \\ &= (\partial_j A_i)B_j + \varepsilon_{ijk}(\nabla\times\boldsymbol{A} )_k B_j + (\partial_j B_i)A_j + \varepsilon_{ijk}(\nabla\times\boldsymbol{B} )_k A_j \\ &= (\boldsymbol{B} \cdot\nabla)A_i + [\boldsymbol{B} \times(\nabla\times\boldsymbol{A} )]_i + (\boldsymbol{A} \cdot\nabla)B_i + [\boldsymbol{A} \times(\nabla\times\boldsymbol{B} )]_i \end{aligned} $$(IV.38) $\nabla\cdot(\boldsymbol{A} \times\boldsymbol{B} )=\boldsymbol{B} \cdot(\nabla\times\boldsymbol{A} )-\boldsymbol{A} \cdot(\nabla\times\boldsymbol{B} )$
$$ \partial_i(\varepsilon_{ijk}A_j B_k)=\varepsilon_{ijk}(\partial_i A_j)B_k+\varepsilon_{ijk}A_j(\partial_i B_k)=B_k(\nabla\times\boldsymbol{A} )_k - A_j(\nabla\times\boldsymbol{B} )_j $$(IV.39) $\nabla\times(\boldsymbol{A} \times\boldsymbol{B} )=(\boldsymbol{B} \cdot\nabla)\boldsymbol{A} +\boldsymbol{A} (\nabla\cdot\boldsymbol{B} )-(\boldsymbol{A} \cdot\nabla)\boldsymbol{B} -\boldsymbol{B} (\nabla\cdot\boldsymbol{A} )$
$$ \begin{aligned} \varepsilon_{ijk}\partial_j(\varepsilon_{klm}A_l B_m) &= (\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})\partial_j(A_l B_m) \\ &= \partial_j(A_i B_j) - \partial_j(A_j B_i) \\ &= (\partial_j A_i)B_j + A_i(\partial_j B_j) - (\partial_j A_j)B_i - A_j(\partial_j B_i) \end{aligned} $$(IV.40) $\nabla\times(\nabla f)=0$
$$ \varepsilon_{ijk}\partial_j\partial_k f = 0 \quad (\varepsilon_{ijk}\text{ 反对称，}\partial_j\partial_k f\text{ 对称}) $$(IV.41) $\nabla\cdot(\nabla\times\boldsymbol{A} )=0$
$$ \partial_i(\varepsilon_{ijk}\partial_j A_k)=\varepsilon_{ijk}\partial_i\partial_j A_k=0 $$(IV.42) $\nabla\cdot(\nabla f)=\nabla^2 f$
$$ \partial_i\partial_i f = \nabla^2 f $$反馈意见
nabla算子相关证明建议理解爱因斯坦求和约定的写法，不必单纯依靠分量相等证明；对于二阶及更高阶张量，分量形式不再适用。若暂时难以理解也无需担心，期末考试一般不会单独考查此类证明题。