对在 $P_0$ 及 $P_1$ 处分别置放 $-q$ 及 $+q$ 的点电荷所形成的电偶极子, 其偶极距 ${\bf m}=q{\bf l}$, ${\bf l}=\vec{P_0P_1}$. 试证明当 $l\to 0$, $q\to+\infty$, 但 $m=ql$ 保持不变时, 此偶极子产生的电场的电势为 $$\bex \phi(P)=-\cfrac{1}{4\pi\ve_0}{\bf m}\cdot\n_P\sex{\cfrac{1}{r_{P_0P}}}, \eex$$ 其中 $\n_P$ 表示关于 $P$ 点的梯度.
证明: $$\beex \bea \phi(P)&=\lim_{l\to0,q\to+\infty\atop m=ql\mbox{ 不变}} \sez{\cfrac{1}{4\pi\ve_0}\cfrac{-q}{r_{P_0P}} +\cfrac{1}{4\pi\ve_0}\cfrac{q}{r_{P_1P}}} =\cfrac{qr_{P_0P_1}}{4\pi\ve_0} \lim_{l\to0,q\to+\infty\atop m=ql\mbox{ 不变}} \cfrac{\cfrac{1}{r_{P_1P}}-\cfrac{1}{r_{P_0P}}}{r_{P_0P_1}}\\ &=\cfrac{qr_{P_0P_1}}{4\pi\ve_0}\cfrac{\p}{\p l}\sex{\cfrac{1}{r_{QP}}}|_{Q=P_0} =\cfrac{qr_{P_0P_1}}{4\pi\ve_0} \cfrac{{\bf l}}{l}\cdot\n_Q\sex{\cfrac{1}{r_{QP}}}|_{Q=P_0}\\ &=-\cfrac{1}{4\pi\ve_0} {\bf m}\cdot\n_P\sex{\cfrac{1}{r_{QP}}}|_{Q=P_0} =-\cfrac{1}{4\pi\ve_0}{\bf m}\cdot\n_P\sex{\cfrac{1}{r_{P_0P}}}. \eea \eeex$$
