Imagine two point charges with equal and opposite charges separated by a sufficiently small distance; such a model is called an electric dipole. We study electric dipoles primarily to understand their mechanical properties. Many things can be approximately described using an electric dipole, such as the magnetic field around a small magnet, the Earth itself (which can be approximated as a dipole to describe its magnetic field), and models of certain diatomic molecules, among others. In the electric dipole model, the distance between the two charges is small enough that we ignore higher-order terms of the distance and retain only the linear part. For physical exploration, however, it is sufficiently accurate and, more importantly, simple enough that we can describe it clearly.
We first study the electric potential generated by an electric dipole. Let their respective charges be q and -q, and the distance between them be \varepsilon. According to Coulomb’s law, the potential generated by a point charge is proportional to the charge and inversely proportional to the distance from the charge. Thus, the potential generated by an electric dipole is: U=C\left(\frac{q}{r}+\frac{-q}{|\vec{r}-\vec{\varepsilon}|}\right) \tag{1} where r=|\vec{r}| and C is a proportionality constant. Directly using equation (1) is somewhat cumbersome. Therefore, we emphasize that \varepsilon is relatively small and expand it using a Taylor series.
\frac{1}{|\vec{r}-\vec{\varepsilon}|}=\frac{1}{\sqrt{r^2+\varepsilon^2-2\vec{r}\cdot \vec{\varepsilon}}} \approx \frac{1}{r\sqrt{1-2\frac{\vec{r}\cdot \vec{\varepsilon}}{r^2}}} \text{ (ignoring terms of order } \varepsilon^2 \text{ and higher)} \begin{aligned} &=r^{-1}\left(1-2\frac{\vec{r}\cdot \vec{\varepsilon}}{r^2}\right)^{-0.5} \\ &=r^{-1}\left(1+\frac{\vec{r}\cdot \vec{\varepsilon}}{r^2}\right) \end{aligned}
Substituting this into equation (1), we get: U=-\frac{Cq\vec{r}\cdot \vec{\varepsilon}}{r^3} \tag{2} This is the vector form. Using polar coordinates, we obtain a concise form: U=-\frac{Cq \varepsilon \cos\theta}{r^2}=-\frac{k \cos\theta}{r^2} \tag{3} This means that the equipotential surfaces (lines) of an electric dipole are given by -\frac{k \cos\theta}{r^2}=\text{const}. The shape of the curves is roughly as follows:
Knowing its potential field, we can obtain its force field. From high school physics, we know that electric field lines are always perpendicular to equipotential surfaces. From a mathematical perspective, that is: E_x=\frac{\partial U}{\partial x}, E_y=\frac{\partial U}{\partial y}
We wish to find the trajectory equation of the electric field lines, but we know the polar coordinate form of the potential energy. One effective but tedious method is to convert the expression for U back to Cartesian coordinates and then use \frac{dy}{dx}=\frac{E_y}{E_x}=\frac{\partial U/\partial y}{\partial U / \partial x} to find the trajectory. Below, we solve for the field line trajectory directly from the polar coordinate perspective.
In polar coordinates, for any curve U(r,\theta)=c, when a point moves from (r,\theta) to (r+dr,\theta+d\theta), the increment of U is: dU=\frac{\partial U}{\partial r}dr+\frac{\partial U}{\partial \theta}d\theta Since U is constant at c, the above differential must be zero. That is: \frac{dr}{d\theta}=-\frac{\partial U}{\partial \theta} \bigg/ \frac{\partial U}{\partial r} \tag{4} We can rewrite this as: \frac{dr}{rd\theta}=-\frac{\partial U}{\partial \theta} \bigg/ \left(r\frac{\partial U}{\partial r}\right) \tag{5} This rewriting is because dr and rd\theta are two mutually perpendicular quantities (basic units of differential), acting like dx and dy in a Cartesian coordinate system. Thus, the left side of equation (5) represents the slope (of the equipotential surface), and the right side represents the ratio of the two perpendicular components of the electric force.
Electric field lines must be perpendicular to equipotential surfaces, so their slope should be -\frac{rd\theta}{dr}. Thus, the differential equation for the electric field lines is: -\frac{rd\theta}{dr}=-\frac{\partial U}{\partial \theta} \bigg/ \left(r\frac{\partial U}{\partial r}\right) \tag{6} \frac{d\theta}{dr}=\frac{\partial U}{\partial \theta} \bigg/ \left(r^2\frac{\partial U}{\partial r}\right) \tag{7} Substituting U=-\frac{k \cos\theta}{r^2} into equation (7), we get: \frac{2\cos\theta}{\sin \theta}d\theta=\frac{1}{r}dr \tag{8} Integrating both sides of equation (8) yields: 2\ln(\sin\theta)+\text{Const}=\ln r Which can also be written as: \frac{\sin^2 \theta}{r}=\text{Const} \tag{9} This is the equation for the electric field lines of a dipole. Its shape is roughly as shown:
At this point, BoJone’s research on electric dipoles is half complete. This article was completed under the inspiration of "The Mathematical Bridge: An Illustrated Guide to Higher Mathematics", but there is a slight lack of rigor in the content of "The Mathematical Bridge" related to this article. On page 278 of the Chinese edition of "The Mathematical Bridge," it states:
Around a dipole (magnet model), any charged particle will move along the curve represented by the following formula: \frac{\sin^2 \theta}{r}=k
However, after my calculations, the trajectory of a charged particle in an electric dipole field can generally only be described by elliptic integrals (which will be discussed in the next article); the above formula is merely the equation for the electric field lines of a dipole. Generally, a charge will not move along the electric field lines (unless the path is a straight line). This was initially quite puzzling to me, and now I would like to "venture a conclusion" that this is a narrative error in the book. Readers of "The Mathematical Bridge" should take note. Of course, I have not read the original English work, so I cannot be certain whether it is a translation issue or an error in the original text.
In the next article, we will explore the motion of charged particles in such an electric dipole field.
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