A positive and negative charge are some distance d away from each other (Electric dipole configuration) Consider the electric field lines in between them.

That is, there is a straight line electric field vector that points from (+) to (-) along the d direction.

There is also an arbitrary electric field line exiting the (+) charge by some angle alpha with respect to the d direction and it enters the (-) charge by some angle beta with respect to the d direction.

Assume alpha does not equal beta. What are the angles?

This was a question posed by my professor to think about. At first I thought the angles would be equal so I didn't really understand his question. I think the case that they would be equal would be when the positive charge magnitude is equal to the negative charge magnitude (I may be wrong). He did not tell us what the magnitude of the (+) and (-) charges so I think this may be the main point. Would there be a way to calculate the angle? Maybe a ratio?

This is mainly for my own curiosity any insight would be appreciated!

dipole angles

  • $\begingroup$ When their charges are different, you can think the electric field between them as a superposition of monopole + dipole field. $\endgroup$
    – user42298
    Commented Oct 8, 2016 at 0:54
  • $\begingroup$ For macroscopic bodies, as long as the charge density is equal and the dimensions of the bodies is also equal, the angles are equal too. $\endgroup$ Commented Oct 8, 2016 at 5:29
  • $\begingroup$ @DarkLink9110 This link may help you? physics.stackexchange.com/a/335511/104696 $\endgroup$
    – Farcher
    Commented May 26, 2017 at 10:26
  • $\begingroup$ Reference : $''$Aptitude Test Problems in Physics$''$ by S.S.Krotov, English translation 1990, problem 3.2 in page 73 and its solution in page 245. $\endgroup$
    – Frobenius
    Commented Sep 29, 2021 at 11:45

3 Answers 3


Your intuition is correct $-$ an asymmetry between $\alpha$ and $\beta$ is only possible if the two charges are different.

If you want a quantitative relationship between the two angles, the correct (read: the only viable) approach is via Gauss's law for the electric flux. From the geometry of the field, which is symmetric about the inter-charge axis, it is relatively easy to see that if you take the surface of revolution generated by the field line about the axis, then you get a cylinder-with-two-conical-ends which goes from one charge to the other:

By definition, the electric field is tangential to this surface at every point, which means that no field lines leave it, and the electric flux is confined inside it. That means, therefore, that the electric flux that leaves charge 1 into the surface must equal the electric flux that arrives at charge 2.

Moreover, we know how to relate these electric fluxes to the angles: close to charge 1, we can ignore the effect of charge 2, and then we just have the flat integral, i.e., the electric flux is the product of the charge times the solid angle spanned by the cone at its apex, $$ \Phi_1 = q_1 \: \Omega_1, $$ where the solid angle can be calculated explicitly as \begin{align} \Omega_1 & = \int_0^\alpha \int_0^2\pi \sin(\theta)\mathrm d\phi \:\mathrm d\theta \\ & = 2\pi \big[-\cos(\theta)\big]_0^\alpha \\ & = 2\pi(1-\cos(\alpha)). \end{align}

Assuming that $q_1>0>q_2$, we can set $\Phi_1+\Phi_2=0$ and therefore \begin{align} 2\pi(1-\cos(\alpha))|q_1| & = 2\pi(1-\cos(\beta))|q_2| \\ \implies \frac{1-\cos(\alpha)}{1-\cos(\beta)} & = \frac{|q_2|}{|q_1|} . \end{align} This relationship then allows you to find any of the relevant quantities in terms of the other three, which is the most that you can do here.

  • $\begingroup$ Close to charge 1 why we can ignore the contribution from 2? $\endgroup$
    – Orion_Pax
    Commented Nov 17, 2021 at 6:23
  • $\begingroup$ Very nice utilisation of the (abstract) concept of field lines. However, you could have stated certain assumptions of the problem setup for completeness of the answer. In particular, you could make it clear whether you consider the charges existing as pointlike particles (with no size?) or macroscopic bodies of finite sizes (out of conductors, nonconductors, and would it still be considered as a dipole anyway) which could affect the way you would find answers. The answer for conductor bodies would be totally different, and answer for nonducting bodies could be found by simple superposition. $\endgroup$
    – Fat32
    Commented Jun 16, 2022 at 1:08

Here is my best attempt, I couldnt solve for the angle as a function of the other angle so I am not sure if it is completely correct or if there is a simpler way to solve.Attempt


enter image description here enter image description here

I hope you understand my handwriting. This question is one of the most interesting one I've seen in a while. Thanks.

  • $\begingroup$ Sin(A/2) = (q2/q1)^1/2 * sin(B/2) $\endgroup$ Commented Dec 4, 2020 at 7:05
  • 1
    $\begingroup$ Hi, please use MathJax to display formulas in questions and answers (and if needed comments). A quick oveview about MathJax you find here: math.meta.stackexchange.com/questions/5020/… Also please do not include pictures of written text in an answer/question. It makes it hard to find the post and it is harder to read. $\endgroup$ Commented Dec 4, 2020 at 8:59

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