电偶极子

电偶极子

Transclude of 电偶极子-2025-09-20-14.01.52.excalidraw

电矩

电偶极子的电矩 $\vec{p}=q\vec{l}$ 反映了电偶极子的特征。

• ~ 电矩是矢量，与 $\vec{l}$ 同向，由负电荷指向正电荷

电偶极子在均匀电场中所受的力矩

$$\vec{M}={\vec{r}}_{+}\times {\vec{F}}_{+}+{\vec{r}}_{-}\times {\vec{F}}_{-}=q({\vec{r}}_{+}-{\vec{r}}_{-})\times \vec{E}=q\vec{l}\times \vec{E}=\vec{p}\times \vec{E}$$

电场分布

电场分布

$$\vec{E}=\frac{1}{4\pi {ϵ}_0{r}^3}(-\vec{p}+3(\hat{r}\cdot \vec{p})\hat{r})$$

Proof

$$\begin{array}{rl}& \overrightarrow{r_{+}}=\vec{r}-\frac{\vec{l}}{2},\overrightarrow{r_{-}}=\vec{r}+\frac{\vec{l}}{2}\\ & \vec{E}=\frac{q}{4\pi {ϵ}_0}\left(\frac{\overrightarrow{r_{+}}}{r_{+}^3}-\frac{\overrightarrow{r_{-}}}{r_{-}^3}\right)\\ & r_{+}^2=r^2+\frac{l^2}{4}-\vec{r}\cdot \vec{l},r_{-}^2=r^2+\frac{l^2}{4}+\vec{r}\cdot \vec{l}\\ & \frac{\overrightarrow{r_{+}}}{r_{+}^3}=r^{-3}{\left(1+\frac{l^2}{4r^2}-\frac{\vec{r}\cdot \vec{l}}{r^2}\right)}^{-3/2}\approx \frac{\overrightarrow{r_{+}}}{r^3}\left(1+\frac{3}{2}\cdot \frac{\vec{r}\cdot \vec{l}}{r^2}\right)\\ & \frac{\overrightarrow{r_{-}}}{\overrightarrow{r_{-}^3}}\approx \frac{\overrightarrow{r_{-}}}{r^3}\left(1-\frac{3}{2}\cdot \frac{\vec{r}\cdot \vec{l}}{r^2}\right)\\ & \vec{E}=\frac{q}{4\pi {ϵ}_0{r}^3}\left(\overrightarrow{r_{+}}-\overrightarrow{r_{-}}+\frac{3}{2}\cdot \frac{\vec{r}\cdot \vec{l}}{r^2}(\overrightarrow{r_{+}}+\overrightarrow{r_{-}})\right)\\ & =\frac{q}{4\pi {ϵ}_0{r}^3}(-\vec{l}+3\hat{r}\cdot \vec{l}\cdot \hat{r})\\ & =\frac{1}{4\pi {ϵ}_0{r}^3}(-\vec{p}+3(\hat{r}\cdot \vec{p})\hat{r})\end{array}$$

中垂线上电场分布

电偶极子中垂线上各点电场强度与该点离电偶极子中心的距离的三次方成反比。 $\vec{E}=\frac{-\vec{p}}{4\pi {ϵ}_0{r}^3}$

点电荷连线上电场分布

电偶极子连线上各点电场强度与该点离电偶极子中心的距离的三次方成反比

$$\vec{E}=\frac{2\vec{p}}{4\pi {ϵ}_0{r}^3}$$

电势分布

电势分布

$$\varphi =\frac{\vec{p}\cdot \vec{r}}{4\pi {ϵ}_0{r}^3}$$

Proof

$$\begin{array}{rl}\varphi & =\frac{q}{4\pi {ϵ}_0}\left(\frac{1}{r_{+}}-\frac{1}{r_{-}}\right)\\ \frac{1}{r_{+}}& =r^{-1}{\left(1+\frac{l^2}{4r^2}-\frac{\vec{r}\cdot \vec{l}}{r^2}\right)}^{-1/2}\\ & \approx \frac{1}{r}\cdot \left(1+\frac{\vec{r}\cdot \vec{l}}{2r^2}\right)\\ \frac{1}{r_{-}}& \approx \frac{1}{r}\cdot \left(1-\frac{\vec{r}\cdot \vec{l}}{2r^2}\right)\\ \therefore \varphi & =\frac{q}{4\pi {ϵ}_0}\cdot \frac{\vec{r}\cdot \vec{l}}{r^3}\\ & =\frac{p\cos \theta }{4\pi {ϵ}_0{r}^2}\\ & =\frac{\vec{p}\cdot \vec{r}}{4\pi {ϵ}_0{r}^3}\end{array}$$

利用电势梯度求电偶极子电场分布

$$\varphi =\frac{px}{4\pi {ϵ}_0(x^2+y^2{)}^{3/2}}$$ $$\begin{array}{rl}{E}_x& =-\frac{\partial \varphi }{\partial x}=\frac{p}{4\pi {ϵ}_0}\frac{2x^2-y^2}{(x^2+y^2{)}^{5/2}}\\ E_y& =-\frac{\partial \varphi }{\partial y}=\frac{p}{4\pi {ϵ}_0}\frac{3xy}{(x^2+y^2{)}^{5/2}}\\ E& =\frac{1}{4\pi {ϵ}_0}\left(\frac{-\vec{p}}{r^3}+\frac{3\vec{p}\cdot \vec{r}}{r^5}\vec{r}\right)\end{array}$$

电势能

$$W=q{\varphi }_A-q{\varphi }_B=-qlE\cos \theta =-pE\cos \theta =-\vec{p}\cdot \vec{E}$$