01电磁学——静电场

Lingfeng2024-06-10

01电磁学——静电场

数学基础 熟练掌握以下积分

\begin{align*} \int \frac{x}{\sqrt{ x^{2}+a^{2} }} \, dx & = \sqrt{ x^{2}+a^{2} } +C \\ \int \frac{1}{\sqrt{ x^{2}+a^{2} }} \, dx & = \ln(x+\sqrt{ x^{2}+a^{2} }) +C\\ \int \frac{x}{(x^{2}+a^{2})^{3/2}} \, dx & = - \frac{1}{\sqrt{ x^{2}+a^{2} }} +C \\ \int \frac{1}{(x^{2}+a^{2})^{3/2}} \, dx & = \frac{1}{a^{2}} \frac{x}{\sqrt{ x^{2}+a^{2} }} +C \end{align*}

==同时注意电场和磁场都是矢量！！在积分时要同向才能积分！！==

库伦定律 对于空间中的电荷, 其产生的电场强度为

\overrightarrow{ E } = \frac{q\overrightarrow{ e_{r} } }{4\pi\varepsilon_{0}r^{2}}

连续分布的电荷场强为

\overrightarrow{ E } = \frac{1}{4\pi\varepsilon_{0}} \int _{Q} \frac{dq}{r^{2}}\overrightarrow{ e_{r} }

其中

dq

由带电体的电荷分布所决定, 即

dq=\rho dV, \quad dq=\sigma dS,\quad dq=\eta dl

例1 求均匀带电圆环轴线上的场强分布, 设圆环半径为 $a$ , 带电总量为 $Q$

解
取微元

dq=\eta dl

其中

\eta= \frac{Q}{2\pi a}

故

\lvert d\overrightarrow{ E } \rvert = \frac{1}{4\pi\varepsilon_{0}} \frac{dq}{r^{2}}= \frac{1}{4\pi\varepsilon_{0}} \frac{\eta dl}{x^{2}+a^{2}}

注意到

E_{y}=0

, 此时

dE_{x}=dE\cos\theta=\frac{1}{4\pi\varepsilon_{0}} \frac{x\eta}{(x^{2}+a^{2})^{3/2}}dl

故

E=E_{x}= \frac{1}{4\pi\varepsilon_{0}} \frac{x\eta}{(x^{2}+a^{2})^{3/2}} \int_{0}^{2\pi a} \, dl = \frac{1}{4\pi\varepsilon_{0}} \frac{xQ}{(x^{2}+a^{2})^{3/2}}

当

x\gg a

时

E = \frac{1}{4\pi\varepsilon_{0}} \frac{Q}{x^{2}} \frac{1}{\left( 1+\left( \frac{a}{x} \right)^{2} \right)^{3/2}} \approx \frac{1}{4\pi\varepsilon_{0}} \frac{Q}{x^{2}}

与点电荷一致.

例2 求均匀带电棒中垂面上的场强分布, 设棒长为 $2l$ , 带电量为 $q$ .

解
法一取微元

\lvert d\overrightarrow{ E } \rvert = \frac{1}{4\pi\varepsilon_{0}} \frac{\eta dz}{r^{2}+z^{2}}

其中

\eta= \frac{q}{2l}

. 此时

dE_{r}=2dE\cos\alpha=2 \cdot\frac{1}{4\pi\varepsilon_{0}} \frac{r\eta}{(r^{2}+z^{2})^{3/2}}dz

考虑

E_{z}=0

, 因此

E=E_{r}= \frac{r \eta}{2\pi\varepsilon_{0}} \int _{0}^{l} \frac{1}{(r^{2}+z^{2})^{3/2}} \, dz

注意到

\begin{align*} \int \frac{1}{(x^{2}+a^{2})^{3/2}}\, dx & \xlongequal{ x=a\sinh t }\int \frac{1}{a^{3}\cosh^{3}t}\cdot a\cosh t \, dt \\ & = \frac{1}{a^{2}} \int \frac{1}{\cosh^{2} t} \, dt \\ & =\frac{\tanh t}{a^{2}} +C= \frac{x}{a^{2}\sqrt{ a^{2}+x^{2} }}+C \end{align*}

因此

E= \frac{1}{2\pi\varepsilon_{0}} \frac{\eta l}{r\sqrt{ r^{2}+l^{2} }}=\frac{1}{4\pi\varepsilon_{0}} \frac{q}{r\sqrt{ r^{2}+l^{2} }}

当

l\to \infty

时

E= \frac{1}{2\pi\varepsilon_{0}} \frac{\eta }{r \sqrt{ 1+\left( \frac{r}{l} \right)^{2} }}\approx \frac{\eta}{2\pi\varepsilon_{0}r}

法二先计算电势. 考虑

\varphi = 2 \cdot \frac{1}{4\pi\varepsilon_{0}}\int _{0}^{l} \frac{\eta dz}{\sqrt{ z^{2}+r^{2} }}

注意到

\begin{align*} \int \frac{1}{\sqrt{ x^{2}+a^{2} }} \, dx & \xlongequal{ x=a\sinh t } \int \frac{1}{a\cosh t} \cdot a\cosh t \, dt \\ & = t+C= \ln(x+ \sqrt{ x^{2}+a^{2} }) +C \end{align*}

因此

\varphi = \frac{\eta}{2\pi\varepsilon_{0}} (\ln(l+\sqrt{ l^{2}+r^{2} })-\ln r)

故

\begin{align*} E= -\frac{d\varphi}{dr} & = \frac{\eta}{2\pi\varepsilon_{0}} \left(\frac{1}{r}- \frac{r}{(l+\sqrt{ r^{2}+l^{2} })\sqrt{ r^{2}+l^{2} }} \right) \\ & = \frac{\eta}{2\pi\varepsilon_{0}}\left( \frac{1}{r}- \frac{\sqrt{ r^{2}+l^{2} }-l}{r\sqrt{ r^{2}+l^{2} }} \right) \\ & = \frac{\eta}{2\pi\varepsilon_{0}} \frac{l}{r\sqrt{ r^{2}+l^{2} }} \end{align*}

例3 求长度为 $l$ , 电荷线密度为 $\lambda$ 的均匀带电直细棒的电场.

解
同理, 考虑

dE_{x}=\frac{\lambda dx}{4\pi\varepsilon_{0}r^{2}} \cos \theta

注意到

x=-a\cot\theta

, 故

dx= a\csc^{2}\theta

r^{2}=a^{2}\csc^{2}\theta

, 此时

\begin{align*} E_{x} & = \frac{\lambda}{4\pi\varepsilon_{0}a}\int _{\theta_{1}}^{\theta_{2}}\cos\theta \, dx \\ & = \frac{\lambda}{4\pi\varepsilon_{0}a}(\sin\theta_{2}-\sin\theta_{1}) \end{align*}

同理

\begin{align*} E_{y} & =\frac{\lambda}{4\pi\varepsilon_{0}a}\int _{\theta_{1}}^{\theta_{2}}\sin\theta \, dx \\ & =\frac{\lambda}{4\pi\varepsilon_{0}a}(\cos\theta_{1}-\cos\theta_{2}) \end{align*}

故

E=\sqrt{ E_{x}^{2}+E_{y}^{2} }

当在棒延长线上一点时

\begin{align*} E & =\frac{\lambda}{4\pi\varepsilon_{0}}\int _{b}^{b+l} \frac{1}{x^{2}}\, dx \\ & = \frac{\lambda l}{4\pi\varepsilon_{0}b(b+l)} \to \frac{\lambda l}{4\pi\varepsilon_{0}b^{2}} \end{align*}

当靠近直线场时,

\theta_{1}=0

\theta_{2}=\pi

, 此时

E_{x}=0

E=E_{y}= \frac{\lambda}{2\pi\varepsilon_{0}a}

例4 有一均匀带电的薄圆盘, 半径为 $R$ , 面电荷密度为 $\sigma$ , 求轴线上的场强.

解
法一取微元

dE= \frac{2\pi r\sigma dr}{4\pi\varepsilon_{0} (x^{2}+r^{2}) } = \frac{\sigma}{2\varepsilon_{0}} \frac{rdr}{(r^{2}+x^{2})}

此时

dE_{x}=dE\cos\theta = \frac{\sigma x}{2\varepsilon_{0}} \frac{r dr}{(r^{2}+x^{2})^{3/2}}

故

\begin{align*} E &=\frac{\sigma x}{2\varepsilon_{0}}\int _{0}^{R} \frac{r}{\ (r^{2}+x^{2})^{3/2} } \, dr \\ & = \frac{\sigma x}{2\varepsilon_{0}} \left( \frac{1}{x} - \frac{1}{\sqrt{ R^{2}+x^{2} }} \right) \\ & = \frac{\sigma}{2\varepsilon_{0}}\left( 1- \frac{x}{\sqrt{ R^{2}+x^{2} }} \right) \end{align*}

当

x\ll R

, 此时

E= \frac{\sigma}{2\varepsilon_{0}}\left( 1- \frac{1}{\sqrt{ \left( \frac{R}{x} \right)^{2}+1 }} \right)\approx \frac{\sigma}{2\varepsilon_{0}}

当

x\gg R

时

E= \frac{Q}{2\pi \varepsilon_{0}R^{2}} \left( 1- \left( 1+\left( \frac{R}{x} \right)^{2} \right)^{-1/2} \right)\approx \frac{Q}{2\pi\varepsilon_{0}R^{2} } \frac{R^{2}}{2x^{2}}= \frac{Q}{4\pi\varepsilon_{0}x^{2}}

法二计算电势.

\begin{align*} \varphi & = \frac{1}{4\pi\varepsilon_{0}}\int_{0}^{R} \frac{2\pi\sigma rdr}{\sqrt{ r^{2}+x^{2} }} \\ & = \frac{\sigma}{2\varepsilon_{0}} \int _{0}^{R} \frac{r}{\sqrt{ r^{2}+x^{2} }} \, dr \\ & = \frac{\sigma}{2\varepsilon_{0}}(\sqrt{ R^{2}+x^{2} }-x) \end{align*}

故

E=-\frac{d\varphi}{dx}= \frac{\sigma}{2\varepsilon_{0}}\left( 1 -\frac{x}{\sqrt{ R^{2}+x^{2} }}\right)

例5 求均匀带电环面场强, 电荷密度为 $\sigma$ , 半径为 $R_{1}$ , $R_{2}$ .

解
由例4, 由叠加原理即得

E=\frac{\sigma x}{2\varepsilon_{0}}\left( \frac{1}{\sqrt{ x^{2}+R_{1}^{2} }}-\frac{1}{\sqrt{ x^{2}+R_{2}^{2} }} \right)

电势点电荷相对于参考点 $P_{0}$ 的电势能为

E_{P_{0}P}=-\int _{P_{0}}^{P} \overrightarrow{ F }\cdot \, d \vec{l} =- q_{0}\int _{P_{0}}^{P} \overrightarrow{ E } \cdot\, d \vec{l}

电势为

\varphi_{P}=\frac{E_{P_{0}P}}{q_{0}}= \int _{P}^{P_{0}}\overrightarrow{ E } \cdot \, d \vec{l}

两点之间的电势差为

U_{AB}=\varphi_{A}-\varphi_{B}= \int _{A}^{B} \overrightarrow{ E } \cdot \, d \vec{l}

点电荷的电势为

\varphi_{P}= \frac{Q}{4\pi\varepsilon_{0}r}

电荷连续分布

\varphi= \frac{1}{4\pi\varepsilon_{0}}\int \frac{dq}{r}

例1 细圆环半径为 $R$ , 均匀带电 $q$ , 求轴线上的电势.

解
法一直接考虑电势叠加

\varphi= \frac{1}{4\pi\varepsilon_{0}} \int \frac{dq}{r} = \frac{q}{4\pi\varepsilon_{0}\sqrt{ R^{2}+z^{2} }}

法二利用场强, 由前例1有

E= \frac{q}{4\pi\varepsilon_{0}} \frac{z}{(z^{2}+R^{2})^{3/2}}

因此

\begin{align*} \varphi & =\int _{z}^{+\infty} E \, dz \\ & = \frac{q}{4\pi\varepsilon_{0}}\int _{z}^{+\infty} \frac{z}{(z^{2}+R^{2})^{3/2}} \, dz \\ & = \frac{q}{4\pi\varepsilon_{0}\sqrt{ R^{2}+z^{2} }} \end{align*}

高斯定理 闭合面电场强度通量为

\phi=\subset\!\supset \!\!\!\!\!\!\!\!\!\!\iint \vec{E}\cdot d \vec{S} = \frac{1}{\varepsilon_{0}}\sum_{i = 1}^{\infty}q_{i}

例1 半径为 $R$ 的球体均匀带电 $q$ , 求球内外场强和电势的分布.

解
电荷密度为

\rho= \frac{q}{V}= \frac{3q}{4\pi R^{3}}

I区利用高斯面有

E_{1}\cdot 4\pi r^{2}= \frac{1}{\varepsilon_{0}} \rho \frac{4\pi}{3} r^{3}

得到

E_{1}= \frac{qr}{4\pi\varepsilon_{0}R^{3}}

II区有

E_{2}\cdot 4\pi r^{2} = \frac{q}{\varepsilon_{0}}

得到

E_{2}= \frac{q}{4\pi\varepsilon_{0}r^{2}}

例2 均匀带电平面, 电荷面密度为 $\sigma$ .

解
取闭合圆柱面, 此时

E\cdot 2S= \frac{\sigma S}{\varepsilon_{0}}

因此

E= \frac{\sigma}{2\varepsilon_{0}}

例3 无限长直导线, 电荷密度为 $\eta$ .

解
亦作高斯面, 此时

E\cdot 2\pi rh= \frac{\eta h}{\varepsilon_{0}}

故

E= \frac{\eta}{2\pi \varepsilon_{0}r}