试计算由习题 4 给出的电偶极子的所形成的电场的电场强度.
解答: $$\beex \bea {\bf E}(P)&=\cfrac{1}{4\pi\ve_0} \sez{\cfrac{-q}{r_{P_0P}^3}{\bf r}_{P_0P}+\cfrac{q}{r_{P_1P}^3}{\bf r}_{P_1P}}\\ &=\cfrac{q}{4\pi \ve_0} \sez{ \sex{-\cfrac{1}{r_{P_0P}^3}+\cfrac{1}{r_{P_0P}^3}}{\bf r}_{P_0P} +\cfrac{1}{r_{P_1P}^3}\sex{{\bf r}_{P_1P}-{\bf r}_{P_0P}} }\\ &=\cfrac{q}{4\pi\ve_0} \sed{ \sez{\cfrac{\p}{\p l}\sex{\cfrac{1}{r_{QP}^3}}|_{Q\in [P_0,P_1]}\cdot r_{P_0P_1}}\cdot {\bf r}_{P_0P}-\cfrac{{\bf r}_{P_0P_1}}{r_{P_1P}^3} }\\ &=\cfrac{q}{4\pi\ve_0} \sed{ \sez{ \cfrac{{\bf l}}{l}\cdot \sex{\n_Q\cfrac{1}{r_{QP}^3}}|_{Q=P_0}\cdot r_{P_0P_1} }\cdot {\bf r}_{P_0P} -\cfrac{{\bf r}_{P_0P_1}}{r_{P_0P}^3} }\quad\sex{r_{P_0P_1}\ll r_{P_0P}}\\ &=\cfrac{q}{4\pi \ve_0}\sez{ 3\sex{{\bf l}\cdot \cfrac{{\bf r}_{P_0P}}{r_{P_0P}^3}}\cdot {\bf r}_{P_0P} -\cfrac{{\bf r}_{P_0P_1}}{r_{P_0P}^3} }\\ &=\cfrac{1}{4\pi \ve_0r^3} \sez{3({\bf m}\cdot{\bf n}){\bf n}-{\bf m}}\quad\sex{{\bf n}=\cfrac{{\bf r}_{P_0P}}{r_{P_0P}}}. \eea \eeex$$
