Electric field lines?

Question

Obviously, electric field lines for 2 charges, one stronger than the other, behave like this: https://www4.uwsp.edu/physastr/kmenning/images/pop4.19.f.19.gif. Is there a way to measure angles for particular field lines? For example, the field line at theta=0 is just a straight line going from the positive charge to the negative one. Similarly, theta= 180 goes from the positive charge to negative infinity. Are there similar rules for some arbitrary value of theta? Do any theta values do something interesting besides theta=0 and 180?

Answer 1

If the system is reduced to two dimension or equivalently infinite long wire, we can use complex potential flow:

$$f(z)=\phi(x,y)+i\psi(x,y)$$

where $\phi(x,y)$ is the "velocity" potential (or electric potential) and $\psi(x,y)$ the stream function.

The complex "velocity" field is given by

$$w=\overline{f'(z)}$$

where $f(z)$ is analytic so that $\phi(x,y)$ and $\psi(x,y)$ satisfy Cauchy-Riemann relation. Hence $\phi$, $\psi$ are harmonic and therefore satisfy Laplace's equation. That is

$$w =\frac{\partial \phi}{\partial x}+i\frac{\partial \psi}{\partial y} =\frac{\partial \psi}{\partial y}-i\frac{\partial \psi}{\partial x}$$

For a single charge at the origin,

\begin{align} f(z) &= \frac{\lambda}{2\pi \epsilon_0} \ln z \\ \phi(x,y) &= \frac{\lambda}{2\pi \epsilon_0} \ln r \\ \psi(x,y) &= \frac{\lambda}{2\pi \epsilon_0} \theta \\ \end{align}

Assume $+2q$ and $-q$ locating at $(-a,0)$ and $(a,0)$ respectively,

\begin{align} f(z) &= \frac{q}{ \pi \epsilon_0} \ln (z+a)- \frac{q}{2\pi \epsilon_0} \ln (z-a) \\ \phi(x,y) &= \frac{q}{2\pi \epsilon_0} \ln \frac{[(x+a)^2+y^2]^2}{(x-a)^2+y^2} \\ \psi(x,y) &= \frac{q}{2\pi \epsilon_0} \tan^{-1} \frac{y(x^2+y^2-2ax-3a^2)}{x^3+xy^2+ax^2+3ay^2-a^2x-a^3} \end{align}

The equipotential and field lines can be expressed as \begin{align} e^{\alpha} &= \frac{[(x+a)^2+y^2]^2}{(x-a)^2+y^2} \\ \tan \beta &= \frac{y(x^2+y^2-2ax-3a^2)}{x^3+xy^2+ax^2+3ay^2-a^2x-a^3} \end{align}

Answer 2

At large enough distance the charge distribution can be well approximated as a monopole, so the open field lines should distribute themselves evenly around the circle.