静电场

\[ \begin{aligned} F&=\frac{q_0}{4\pi \epsilon_0}\int\frac{\mathrm{d}q}{|r|^2}\frac{z}{|r|}\\ &=\frac{q_0}{4\pi \epsilon_0}\int\frac{z}{(z^2+R^2)^\frac{3}{2}}\frac{q}{2\pi R}R\mathrm{d}\varphi\\ &=\frac{q_0q}{4\pi \epsilon_0}\frac{z}{(z^2+R^2)^\frac{3}{2}}\\ &\approx\frac{q_0q}{4\pi \epsilon_0}\frac{1}{z^2}(when\quad z\gg R)(\vec{k}) \end{aligned} \]

\[ \begin{aligned} F&=\frac{q_0}{4\pi \epsilon_0}\int_0^R\frac{z}{(z^2+r^2)^\frac{3}{2}}\sigma2\pi r\mathrm{d}r\\ &=\frac{q_0}{4\epsilon_0}\frac{q}{\pi R^2}\int_0^R\frac{zd(z^2+r^2)}{(z^2+r^2)^\frac{3}{2}}\\ &=\frac{q_0q}{4\pi \epsilon_0R^2}(-2)\frac{z}{\sqrt{z^2+r^2}}\bigg|^R_0\\ &=\frac{q_0q}{2\pi \epsilon_0R^2}\left(1-\frac{z}{\sqrt{z^2+R^2}}\right)(\vec{k})\\ &=\frac{q_0q}{2\pi \epsilon_0R^2}\left(1-\frac{1}{\sqrt{1+(\frac{R}{z})^2}}\right)\\ &=\frac{q_0q}{2\pi \epsilon_0R^2}\left(1-1+\frac{1}{2}(\frac{R}{z})^2-\frac{3}{8}(\frac{R}{z})^4\cdots\right)\\ &\approx\frac{q_0q}{4\pi \epsilon_0}\frac{1}{z^2}(when\quad z\gg R)(\vec{k})\\ &\approx\frac{q_0q}{2\pi \epsilon_0R^2}(when\quad R\gg z)=\frac{\sigma}{2\epsilon_0}q_0(\vec k) \end{aligned} \]

\[ \begin{aligned} \mathrm{d}F&=\frac{q_0}{4\pi\epsilon_0}\frac{(z-R\cos\theta)\sigma(2\pi R\sin\theta\cdot R\mathrm{d}\theta)}{[(z-R\cos\theta)^2+(R\sin\theta)^2]^\frac{3}{2}}\\ &=\frac{q_0R^2}{2\epsilon_0}\sigma\frac{(z-R\cos\theta)\sin\theta\mathrm{d}\theta}{(z^2+R^2-2zR\cos\theta)^\frac{3}{2}}\\ &=\frac{q_0q}{8\pi\epsilon_0}\frac{(z-R\cos\theta)\sin\theta}{(z^2+R^2-2zR\cos\theta)^\frac{3}{2}}\mathrm{d}\theta(\sigma=\frac{q}{4\pi R^2}) \end{aligned} \]

\[ \begin{aligned} F&=\frac{q_0q}{8\pi\epsilon_0}\int_0^\pi\frac{(z-R\cos\theta)\sin\theta}{(z^2+R^2-2zR\cos\theta)^\frac{3}{2}}\mathrm{d}\theta\\ &\xlongequal{t=-R\cos\theta}\frac{q_0q}{8\pi\epsilon_0R}\int_{-R}^R\frac{z+t}{(z^2+R^2+2zt)^\frac{3}{2}}\mathrm{d}t\\ &=-\frac{q_0q}{8\pi\epsilon_0Rz}\int_{-R}^R(z+t)\mathrm{d}\frac{1}{\sqrt{z^2+R^2+2zt}}\\ &=-\frac{q_0q}{8\pi\epsilon_0Rz}\left(\frac{z+t}{\sqrt{z^2+R^2+2zt}}\bigg|_{-R}^R-\int_{-R}^R\frac{\mathrm{d}t}{\sqrt{z^2+R^2+2zt}}\right)\\ &=-\frac{q_0q}{8\pi\epsilon_0Rz}\left(1-\frac{z-R}{|z-R|}-\frac{1}{z}\sqrt{z^2+R^2+2zt}\bigg|_{-R}^R\right)\\ &=-\frac{q_0q}{8\pi\epsilon_0Rz}\left(1-\frac{z-R}{|z-R|}-\frac{(z+R)-|z-R|}{z}\right)\\ &=\left\{\begin{matrix} \dfrac{q_0q}{4\pi\epsilon_0}\dfrac{1}{z^2},z>R\\ 0,z<R \end{matrix}\right. \end{aligned} \]

\[ \begin{aligned} F&=\int_{-L}^L\frac{q_0\lambda\mathrm{d}l}{4\pi\epsilon_0}\frac{1}{z^2+l^2}\frac{z}{\sqrt{z^2+l^2}}\\ &=\frac{q_0q}{8\pi\epsilon_0L}\int_{-L}^L\frac{z}{(z^2+l^2)^{\frac{3}{2}}}\mathrm{d}l\\ &\xlongequal{l=z\tan\theta}\frac{q_0q}{8\pi\epsilon_0L}\int_{-\arctan\frac{L}{z}}^{\arctan\frac{L}{z}}\frac{\cos^3\theta}{z^2}\cdot z\frac{\mathrm{d}\theta}{\cos^2\theta}\\ &=\frac{q_0q}{8\pi\epsilon_0Lz}\int_{-\arctan\frac{L}{z}}^{\arctan\frac{L}{z}}\cos\theta\mathrm{d}\theta\\ &=\frac{q_0q}{4\pi\epsilon_0}\frac{1}{z\sqrt{L^2+z^2}}(\vec{k})\\ &\approx\left\{\begin{matrix} \dfrac{q_0q}{4\pi\epsilon_0}\dfrac{1}{z^2},z\gg L\\ \dfrac{q_0q}{4\pi\epsilon_0}\dfrac{1}{L}=\dfrac{\lambda}{2\pi\epsilon _0}\dfrac{q_0}{z},z\ll L \end{matrix}\right. \end{aligned} \]

\[ \begin{aligned} F&=\int_{-b}^b\frac{q_0(\sigma\cdot2L\mathrm{d}x)}{4\pi\epsilon_0}\frac{1}{\sqrt{x^2+z^2+L^2}\sqrt{x^2+z^2}}\cdot\frac{z}{\sqrt{x^2+z^2}}\\ &=\int_{-b}^b\frac{q_0}{2\pi\epsilon_0}\frac{q}{4bL}\cdot\frac{z}{x^2+z^2}\mathrm{d}x(L\gg b, z)\\ &=\int_{-b}^b\frac{q_0q}{8\pi b\epsilon_0L}\cdot\frac{1}{(\frac{x}{z})^2+1}\mathrm{d}\frac{x}{z}\\ &=\frac{q_0q}{8\pi b\epsilon_0L}\cdot\arctan\frac{x}{z}\bigg|^b_{-b}\\ &=\frac{q_0q}{4\pi b\epsilon_0L}\arctan\frac{b}{z}(\vec k)\\ &=\frac{q_0\sigma}{\pi \epsilon_0}\arctan\frac{b}{z}\xlongequal{b\gg z}\frac{\sigma}{2\epsilon_0}q_0 \end{aligned} \]

\[ \begin{aligned} &\left\{\begin{matrix} \nabla\times\mathbf{H}=\boldsymbol j+\dfrac{\partial\mathbf{D}}{\partial t}\\ \nabla\times\mathbf{E}=-\dfrac{\partial\mathbf{B}}{\partial t}\\ \nabla\cdot\mathbf{B}=0\\ \nabla\cdot \mathbf{D}=\rho \end{matrix}\right.(\text{微分形式})\\ &\left\{\begin{matrix} \displaystyle\oint_l\mathbf{H}\cdot \mathrm d\boldsymbol{l}=\int_S\boldsymbol j\cdot\mathrm d\mathbf S+\int_S\frac{\partial\mathbf{D}}{\partial t}\cdot\mathrm d\mathbf S\\ \displaystyle\oint_l\mathbf{E}\cdot\mathrm d\boldsymbol{l}=-\int_S\dfrac{\partial\mathbf{B}}{\partial t}\cdot\mathrm d\mathbf S\\ \displaystyle\oint_S\mathbf{B}\cdot\mathrm d\boldsymbol{S}=0\\ \displaystyle\oint_S\mathbf{D}\cdot\mathrm d\boldsymbol{S}=\int_V\rho\cdot\mathrm dV \end{matrix}\right.(\text{积分形式}) \end{aligned} \]

\[ \mathbf{D}=\epsilon \mathbf{E}, \quad \mathbf{B}=\mu \mathbf{H}, \quad \mathbf{j}=\sigma \mathbf{E} \]

\[ \begin{aligned} \vec p&=q\vec d\\ \vec r_\pm&=(r\cos\varphi\mp\frac{d}{2})\hat i+r\sin\varphi \hat j\\ \vec r_\pm^2&=r^2\mp rd\cos\varphi+\frac{d^2}{4}\approx r^2\mp rd\cos\varphi(r\gg d)\\ \vec E_p&=\frac{1}{4\pi\epsilon_0}\left[\frac{q\vec r_+}{r_+^3}+\frac{(-q)\vec r_-}{r_-^3}\right]\\ &=\frac{q}{4\pi\epsilon_0}\left[\frac{(r\cos\varphi-\frac{d}{2})\hat i+r\sin\varphi \hat j}{(r^2-rd\cos\varphi)^{\frac{3}{2}}}-\frac{(r\cos\varphi+\frac{d}{2})\hat i+r\sin\varphi \hat j}{(r^2+rd\cos\varphi)^{\frac{3}{2}}}\right]\\ &=\frac{q}{4\pi\epsilon_0}[\frac{(r\cos\varphi-\frac{d}{2})\hat i+r\sin\varphi \hat j}{r^3}(1-\frac{3d}{2r}\cos\varphi)-\\ &\quad\quad\quad\quad\frac{(r\cos\varphi+\frac{d}{2})\hat i+r\sin\varphi \hat j}{r^3}(1+\frac{3d}{2r}\cos\varphi)]\\ &\left((r^2\pm rd\cos\varphi)^{-\frac{3}{2}}=r^{-3}(1\pm\frac{d}{r}\cos\varphi)^{-\frac{3}{2}}\approx r^{-3}(1\mp\frac{3d}{2r}\cos\varphi)\right)\\ &=\frac{q}{4\pi\epsilon_0r^3}\left[3d\cos\varphi(\cos\varphi\hat i+\sin\varphi \hat j)-d\hat i\right]\\ &=\frac{1}{4\pi\epsilon_0r^3}\left[\frac{3\vec r(\vec r\cdot \vec p)}{r^2}-\vec p\right]\\ &=\left\{\begin{matrix} \dfrac{\vec p}{2\pi\epsilon_0 r^3},x\quad axis\\ -\dfrac{\vec p}{4\pi\epsilon_0 r^3},y\quad axis\\ \end{matrix}\right. \end{aligned} \]

\[ \begin{aligned} E\cdot 2\pi rh&=\frac{\lambda_1h}{\epsilon_0}\Rightarrow E=\frac{\lambda_1}{2\pi\epsilon_0 r}\\ F&=\int_L^{2L}\frac{\lambda_1}{2\pi\epsilon_0 r}\lambda_2\mathrm{d}r\\ &=\frac{\lambda_1\lambda_2}{2\pi\epsilon_0}\int_L^{2L}\frac{\mathrm{d}r}{r}\\ &=\frac{\lambda_1\lambda_2}{2\pi\epsilon_0}\ln 2 \end{aligned} \]

\[ \mathrm{d}\Omega=\frac{\hat r\mathrm{d}\vec S}{r^2} \]

\[ \mathrm d\phi=\vec E\cdot \mathrm d\vec A=\frac{q}{4\pi\epsilon_0r^2}(r^2\mathrm d \Omega)=\frac{q}{4\pi \epsilon_0}\mathrm d\Omega \]

\[ \oint\vec E\cdot\mathrm d\vec A=\phi=\oint\frac{q}{4\pi \epsilon_0}\mathrm d\Omega=\frac{q}{4\pi \epsilon_0}\cdot4\pi=\frac q {\epsilon_0} \]

\[ \oint\vec E\cdot\mathrm d\vec A=\frac {\sum q_i}{\epsilon_0} \]

\[ \begin{aligned} &E(x)(2S)=\frac{\displaystyle2\int_0^x\rho_0xS\mathrm dx}{\epsilon_0}\Rightarrow E(x)=\frac{\rho_0x^2}{2\epsilon_0}(x\in(-b,b))\\ &E(x)=\frac{\sigma}{2\epsilon_0}=\frac{\displaystyle2\int_0^b\rho_0x\mathrm dx}{2\epsilon_0}=\frac{\rho_0b^2}{2\epsilon_0}(|x|>b)\\ \end{aligned} \]

\[ \begin{aligned} &r\in[R_1,R_2],E(2\pi rh)=\frac{-\sigma(2\pi R_1h)}{\epsilon_0}\Rightarrow E=-\frac{\sigma R_1}{\epsilon_0r}\\ &r>R_2,E(2\pi rh)=\frac{-\sigma(2\pi R_1h)+\sigma(2\pi R_2h)}{\epsilon_0}\Rightarrow E=\frac{\sigma (R_2-R_1)}{\epsilon_0r}\\ &\therefore E(r)=\left\{\begin{matrix} 0,r\in(0,R_1)\\ -\dfrac{\sigma R_1}{\epsilon_0r},r\in(R_1,R_2)\\ \dfrac{\sigma (R_2-R_1)}{\epsilon_0r},r\in(R_2, +\infty) \end{matrix}\right. \end{aligned} \]

\[ \begin{aligned} W_{a\to b}&=\int_{a,C}^b(q_0\vec E)\cdot\mathrm d\vec l\\ &=\frac{q_0q}{4\pi\epsilon_0}\int_{a,C}^b\frac{\vec r\cdot\mathrm d\vec l}{r^3} &\text{(库仑定律)}\\ &=\frac{q_0q}{4\pi\epsilon_0}\int_{r_a,C}^{r_b}\frac{\mathrm dr}{r^2} &(\vec r\cdot \mathrm d\vec l=r\mathrm d l\cos \theta\approx r\mathrm dr)\\ &=\frac{q_0q}{4\pi\epsilon_0}\left(\frac{1}{r_a}-\frac{1}{r_b}\right) &\text{(只与初末位置有关)} \end{aligned} \]

\[ W_{a\to b}=\frac{q_0\sum q_i}{4\pi\epsilon_0}\left(\frac{1}{r_a}-\frac{1}{r_b}\right) \]

\[ V_b-V_\infty=-\frac{W_{\infty\to b}}{q_0}=-\int_{\infty}^b\vec E\cdot\mathrm d\vec s=\int_b^\infty\vec E\cdot\mathrm d\vec s \]

\[ \begin{aligned} U&=\frac{1}{2}\frac{1}{4\pi\epsilon_0}\left[q_1(\frac{q_2}{r_{12}}+\frac{q_3}{r_{13}})+q_2(\frac{q_1}{r_{12}}+\frac{q_3}{r_{23}})+q_3(\frac{q_1}{r_{13}}+\frac{q_2}{r_{23}})\right]\\ &=\frac{1}{4\pi\epsilon_0}\left[\frac{q_1q_2}{r_{12}}+\frac{q_1q_3}{r_{13}}+\frac{q_2q_3}{r_{23}}\right]\\ &=\frac{1}{4\pi\epsilon_0}\sum_{i<j,i,j\in\{1,2,3\}}\frac{q_iq_j}{r_{ij}} \end{aligned} \]

\[ \begin{aligned} V&=\frac{1}{4\pi\epsilon_0}\left(\frac{q}{r_+}+\frac{-q}{r_-}\right)\\ &=\frac{q}{4\pi\epsilon_0}\frac{r_--r_+}{r_+r_-}\\ &\approx\frac{q}{4\pi\epsilon_0}\frac{d\cos\theta}{r_+r_-}\\ &(r\gg d，r_--r_+\approx d\cos\theta)\\ &\approx\frac{1}{4\pi\epsilon_0}\frac{qd\cos\theta}{r^2}\\ &\left(r_+r_-\approx (r-\frac d 2\cos\theta)(r+\frac d 2\cos\theta)=r^2-\frac{d^2\cos\theta^2} 4\approx r^2\right)\\ &=\frac{1}{4\pi\epsilon_0}\frac{p\cos\theta}{r^2}\\ &=\frac{1}{4\pi\epsilon_0}\frac{p\cdot r}{r^3} \end{aligned} \]

\[ \begin{gathered} \vec F = q\vec E \\ \tau = \vec r\times\vec F=q\vec r\times\vec F=\vec p\times\vec F \end{gathered} \]

\[ \vec E=\frac{1}{4\pi\epsilon_0}\frac{qz}{(R^2+z^2)^{\frac 3 2}}\hat k\\ \]

\[ \begin{aligned} V&=\int_z^\infty\frac{1}{4\pi\epsilon_0}\frac{qh}{(R^2+h^2)^{\frac 3 2}}\mathrm dh\\ &=\frac{q}{8\pi\epsilon_0}\int_z^\infty\frac{\mathrm d(R^2+h^2)}{(R^2+h^2)^{\frac 3 2}}\\ &=\frac{q}{4\pi\epsilon_0\sqrt{R^2+z^2}}=\frac{\lambda R}{2\epsilon_0\sqrt{R^2+z^2}} \end{aligned} \]

\[ \begin{aligned} V&=\int_0^R\frac{\frac{q}{\pi R^2}(2\pi r\mathrm dr)}{4\pi\epsilon_0\sqrt{r^2+z^2}}\\ &=\frac{q}{4\pi\epsilon_0R^2}\int_0^R\frac{\mathrm d(r^2+z^2)}{\sqrt{r^2+z^2}}\\ &=\frac{q}{2\pi\epsilon_0R^2}(\sqrt{R^2+z^2}-z)\\ &=\frac{\sigma}{2\epsilon_0}(\sqrt{R^2+z^2}-z) \end{aligned} \]

\[ \frac{1}{4\pi\epsilon_0}\frac{q}{R}\text{, 定义积分/对球心计算} \]