學院物理/電位

定義

$\begin{array}{c}{U}_r=\frac{kqQ}{r}{V}_r=\frac{U_r}{q}=\frac{kQ}{r}{V}_p=k\int \frac{dq}{r}\mathrm{\Delta }V=-\underset{1}{\overset{2}{\int }}\overrightarrow{E}d\overrightarrow{S}\\ \overrightarrow{E}=\frac{-\partial V}{\partial x}\hat{i}\end{array}$

性質

Template:See also 絕緣體球導體球

r

E

V

$aa$

$\propto r\propto \frac{1}{r^2}\propto \frac{1}{r^2}$

$\begin{array}{cc}& \begin{array}{c}\propto 3a^2-\end{array}{r}^2\\ & \propto \frac{1}{r}\propto \frac{1}{r}\end{array}$

絕緣體球

$\begin{array}{cc}V(r<a)& =\frac{kQ}{2a}(3-\frac{r^2}{a^2})=\frac{kQ}{2a^3}(3a^2-r^2)\\ V(r>a)& =\frac{kQ}{r}\end{array}$

導體球

$\begin{array}{c}V(r<a)=\frac{kQ}{a}\\ V(r>a)=\frac{kQ}{r}\end{array}$

例題

五

Template:See also圖24.14

$\begin{array}{c}V=k\int \frac{dq}{r}=\frac{kQ}{\sqrt{x^2+a^2}}\\ E_x=\frac{-dV}{dx}=\frac{kxQ}{{\sqrt{x^2+a^2}}^3}\end{array}$

六

Template:See also圖24.15

$\begin{array}{c}V=\int dV=\int \frac{kdq}{\sqrt{x^2+r^2}}=\int \frac{k2\pi \sigma rdr}{\sqrt{x^2+r^2}}=k2\pi \sigma \underset{0}{\overset{R}{\int }}\frac{rdr}{\sqrt{x^2+r^2}}\\ =k2\pi \sigma \underset{0}{\overset{R}{\int }}\frac{1}{2r}r{(x^2+r^2)}^{\frac{-1}{2}}d(x^2+r^2)=k\pi \sigma 2{[(x^2+r^2{)}^{\frac{1}{2}}]}_0^R\\ =2k\pi \sigma (\sqrt{x^2+R^2}-x)\\ E_x=\frac{-dV}{dx}=-2k\pi \sigma \frac{d[(x^2+R^2{)}^{\frac{1}{2}}-x]}{dx}=-2k\pi \sigma (\frac{2x}{2}\frac{1}{\sqrt{x^2+R^2}}-1)\\ =2k\pi \sigma (1-\frac{x}{\sqrt{x^2+R^2}})\end{array}$

七

$\begin{array}{c}P\bullet \nwarrow \\ ar\\ \leftarrow x\to \searrow ⊣dx⊢\\ ⟵l⟶\end{array}$

圖24.16

Template:See also $\begin{array}{c}{V}_p=k\int \frac{dq}{r}=k\underset{0}{\overset{l}{\int }}\frac{\lambda dx}{\sqrt{x^2+a^2}}=k\lambda \underset{0}{\overset{l}{\int }}\frac{dx}{\sqrt{x^2+a^2}}\\ \because \int \frac{dx}{\sqrt{x^2+a^2}}=\ln \left(\frac{x+\sqrt{x^2+a^2}}{a}\right)+\text{C}\\ \therefore V_p=k\lambda {[\ln \left(\frac{x+\sqrt{x^2+a^2}}{a}\right)]}_0^l\\ =k\lambda [\ln \left(\frac{l+\sqrt{l^2+a^2}}{a}\right)-\ln \frac{\sqrt{a^2}}{a}]\\ =k\lambda \ln \left(\frac{l+\sqrt{l^2+a^2}}{a}\right)\end{array}$

靜電平衡

導體的靜電平衡有以下性質：

1. 導體內的電場為零
2. 導體的電荷在表面
3. 導體外的電場為$\sigma /ϵ$
4. 導體的電荷分布在曲率半徑越大時密度越大

練習

x

y

${\theta }_1$

${\theta }_2$

r

O

$\begin{array}{cc}& \lambda =a\sin \theta \\ & dE=\frac{kdq}{r^2}\\ & dq=\lambda ds=a\sin \theta rd\theta \\ & d{E}_x=dE\cos \theta =\frac{kdq\cos \theta }{r^2}=\frac{ka\sin \theta \cos \theta d\theta }{r}\\ & d{E}_y=dE\sin \theta =\frac{kdq\sin \theta }{r^2}=\frac{ka{\sin }^2\theta d\theta }{r}\\ & E_x=\frac{ka}{r}\underset{{\theta }_1}{\overset{{\theta }_2}{\int }}\sin \theta \cos \theta d\theta =\frac{ka}{4r}\underset{{\theta }_1}{\overset{{\theta }_2}{\int }}\sin 2\theta d2\theta \\ & =\frac{-ka}{4r}(\cos 2{\theta }_2-\cos 2{\theta }_1)\\ & E_y=\frac{ka}{r}\underset{{\theta }_1}{\overset{{\theta }_2}{\int }}{\sin }^2\theta d\theta =\frac{ka}{4r}\underset{{\theta }_1}{\overset{{\theta }_2}{\int }}(1-\cos 2\theta )d2\theta \\ & =\frac{ka}{4r}[2\theta -\sin 2\theta {]}_{{\theta }_1}^{{\theta }_2}=\frac{ka}{4r}(2{\theta }_2-2{\theta }_1-\sin 2{\theta }_2+\sin 2{\theta }_1)\\ & \overrightarrow{E}=E_x\hat{i}-E_y\hat{j}\\ & =\frac{-ka}{4r}(\cos 2{\theta }_2-\cos 2{\theta }_1)\hat{i}-\frac{ka}{4r}(2{\theta }_2-2{\theta }_1-\sin 2{\theta }_2+\sin 2{\theta }_1)\hat{j}\\ & V_x=-\int E_{x}dS=-\int E_{x}rd\theta \\ & =\frac{ka}{8}\underset{{\theta }_2}{\overset{{\theta }_1}{\int }}\cos 2\theta d2\theta =\frac{ka}{8}[\sin 2\theta {]}_{{\theta }_2}^{{\theta }_1}\\ & =\frac{ka}{8}(\sin 2{\theta }_2-\sin 2{\theta }_1)\\ & V_y=-\int E_{y}dS=-\int E_{y}rd\theta \\ & =\frac{-ka}{4r}\underset{{\theta }_1}{\int }{\theta }_{2}2\theta -\sin 2\theta d\theta \\ & =\frac{-ka}{4r}[{\theta }^2+\frac{\cos 2\theta }{2}{]}_{{\theta }_1}^{{\theta }_2}\end{array}$