Consider an electric dipole and its electric field lines.

There will be many field lines that do not extend to or originate from infinity, but rather begin at the positive charge and loop back around to terminate at the negative charge.

However, the field lines exactly on the dipole axis will extend out to infinity (or originate from infinity).

But are these the only field lines that extend to infinity, or are there other off-axial lines that also go to infinity? How can one prove this?

I have tried writing out the differential equations for the field lines, but they seem impossible to solve. No heuristic argument has satisfied me either. One can assert that all off-axial electric field lines must curve away from the axis as one goes to infinity, but does this curvature ultimately cause them to curve all the way back or do they reach some sort of linear asymptote?


1 Answer 1



$\def\vp{{\vec p}}\def\ve{{\vec e}}\def\l{\left}\def\r{\right}\def\vr{{\vec r}}\def\ph{\varphi}\def\eps{\varepsilon}\def\grad{\operatorname{grad}}\def\vE{{\vec E}}$ $\vp:=\ve Ql$ constant $l\rightarrow 0$, $Q\rightarrow\infty$. \begin{align} \ph(\vr,\vr') &= \lim_{l\rightarrow0}\frac{Ql\ve\cdot\ve}{4\pi\eps_0 l}\l(\frac{1}{|\vr-\vr'-\ve\frac l2|}-\frac{1}{|\vr-\vr'+\ve\frac l2|}\r)\\ &=\frac{\vp}{4\pi\eps_0}\cdot\grad_{\vr'} \frac1{|\vr-\vr'|}\\ &=\frac{\vp\cdot(\vr-\vr')}{4\pi\eps_0|\vr-\vr'|^3}\\ \vE(\vr)&=-\grad\ph(\vr)\\ \vE(\vr) &= \frac{1}{4\pi\eps_0}\l(\frac{-\vp}{|\vr-\vr'|^3}+3\frac{\vp\cdot(\vr-\vr')}{|\vr-\vr'|^5}(\vr-\vr')\r) \end{align}

Field Lines

We calculate the field lines with $\vp=\ve_1$ and $\vr'=\vec0$. For calculating the field lines we may multiply the vector field with any scalar field called integrating factor. We choose $4\pi\eps_0|\vr|^3$ as integrating factor and get the differential equation \begin{align} \begin{pmatrix} \dot x\\ \dot y \end{pmatrix} &= \begin{pmatrix} -1+3\frac{x^2}{x^2+y^2}\\ 3\frac{xy}{x^2+y^2} \end{pmatrix} \end{align} for the field lines. In the complex representation $x(t)+i y(t)=r(t)e^{i\phi(t)}$ this reads as \begin{align} \dot x + i\dot y &= -1 + 3\frac{x(x+iy)}{x^2+y^2}\\ &=-1 + 3\cos(\phi)e^{i\phi}\\ \dot r e^{i\phi} + ir e^{i\phi} \dot\phi &= -1 + 3\cos(\phi)e^{i\phi}\\ \dot r + i r\dot\phi &= -e^{-i\phi} + 3\cos(\phi) \end{align} Written in components: \begin{align} \dot r &= -\cos(\phi) + 3\cos(\phi) = 2\cos(\phi)\\ r\dot \phi &= \sin(\phi) \end{align} If we divide these two equations we get \begin{align} \frac{\dot r}{r\dot\phi}=2\cot(\phi) \end{align} We re-parameterize such that $\phi$ becomes the parameter \begin{align} \frac{r'}{r} = 2\cot(\phi) \end{align} Integrate: \begin{align} \ln\l(\frac{r}{r_0}\r) &= \int_{\phi_0}^{\phi} 2\cot(\phi) d\phi = 2\ln\l(\frac{\sin(\phi)}{\sin(\phi_0)}\r)\\ \frac r{r_0} &= \l(\frac{\sin(\phi)}{\sin(\phi_0)}\r)^2 \end{align} As soon as you have some $\phi_0\neq 0$ for $r_0>0$ you get a bounded curve.

Ling-Hsiao Lyu gets the same formula (only that she uses $\phi_0=\frac\pi2$ and $r_0=r_{\rm eq}$) in Section "Dipole Magnetic Field Line" of her Lecture Notes.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.