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For two positive point charges in space separated by some finite distance, the electric field lines we observe look like this:

(for point $P$ being the neutral point, the line shown in red is asymptotic to the field line emerging from a charged point $q_{_1}$ at an angle $\alpha$)

Suppose we have 2 positive charges $q_{_1}$ and $q_{_2}$. If an electric field line emerges from $q_{_1}$ at an angle $\alpha$ from the line joining charges $q_{_1}$ and $q_{_2}$, is there a way we can derive the slope of asymptote to the given field line?

In simple words, can we derive a relationship between $\alpha$ and $\beta$?

Edit

I've been thinking that if somehow we can count the number of field lines emitted in the conical region, having semi vertical angle equal to $\beta$ and with its vertex at $P$, that do not go out of the curved surface area of the cone (hence I mean, measuring the flux through the base of the cone); we can then equate this with the number of field lines emitted from charge $q_{\small{1}}$ within the solid angle $ \Omega= 2\pi\left(1- \cos(\alpha)\right)$. Which is, in magnitude, equal to $\varphi=\frac{q_{\small{1}}}{2\varepsilon_0}\left(1-cos\alpha\right)$.

Hence my question shrinks to:

  • How to calculate the number of field lines emitted inside the solid angle $2\pi\left(1-\cos\beta\right)$ subtended at point $P$?
Emilio Pisanty
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SteelCubes
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    Naïvely, I would expect that asymptotically the field would be that of a point charge with magnitude $q_1 + q_2$, and so you could equate the flux that way. If I didn't have a bunch of grades to submit I'd try to work up an answer along these lines. – Michael Seifert Jan 05 '21 at 15:27
  • You will need to start by defining the line density in lines/coulomb. I like Michael's idea. – R.W. Bird Jan 05 '21 at 16:59
  • In practice, it becomes very difficult to determine the angle between two field lines leaving the surface of a sphere in three dimensions. I'm not even certain that it is possible to uniformly distribute a given number of lines around the surface of a sphere. – R.W. Bird Jan 05 '21 at 18:01
  • @MichaelSeifert I've thought of doing the same, or atleast what I interpret your suggestion that, asymptotic lines to be assumed as the field lines of combined charge $q_1 + q_2$, placed at point P. But I'm concerned that if the point P can be assumed to be at effective charge of $q_1+q_2$, why is the number of field lines zero there, instead of being maximum? – SteelCubes Jan 05 '21 at 22:03
  • The fields from the two positive charges are in opposite directions at point P. – R.W. Bird Jan 06 '21 at 14:53

2 Answers2

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The arguments regarding the conservation of electric flux in this previous answer of mine apply equally well here, with the minor change that the second angle corresponds to the combined charges here: the flux calculation happens, asymptotically, at infinity, where the electric field looks like that of a point charge with the combined charge, located at the center of charge (as defined here), plus a quadrupole correction which is negligible at infinity.

As such, the correct relationship between the two angles, as you have defined them, is $$ (1-\cos(\alpha))q_1 = (1-\cos(\beta))(q_1+q_2). $$

Emilio Pisanty
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  • @SteelCubes The argumentation in this answer is sufficient to deduce that. – Emilio Pisanty Jan 06 '21 at 17:35
  • I finally understand it now. Thanks – SteelCubes Jan 06 '21 at 17:36
  • Nice answer. One minor nitpick, though: I don't think that the center of charge is the same thing as the OP's "neutral point" $P$, which (I assume) is the point where $\vec{E}$ vanishes. Taking into account this difference would add a dipole correction as well. I don't think it really matters in the end, though, since we're measuring the angle $\beta$ at infinity, and the dipole correction can be neglected there as well. – Michael Seifert Jan 07 '21 at 15:00
  • @MichaelSeifert The center of charge is indeed distinct from the 'neutral point'. (At the latter you have ${q_1}/{r_1^2} = {q_1}/{r_2^2}$ in the obvious notation; at the former you have $q_1r_1=q_2r_2$.) The correct starting point for the asymptotes is the COC, which can be easily verified numerically (for future reference: Import["http://halirutan.github.io/Mathematica-SE-Tools/decode.m"]["https://i.stack.imgur.com/o2S2k.png"]). The dipole correction clearly matters in finding the origin for the streamlines (...though I'm not sure I see a fully rigorous argument for this). – Emilio Pisanty Jan 07 '21 at 15:45
  • @SteelCubes The Mathematica notebook in the previous comment (just run the code block on an empty cell) may be of interest to you. – Emilio Pisanty Jan 07 '21 at 15:50
  • @MichaelSeifert ... but yes, I agree $-$ the starting point is irrelevant when determining the angle of the asymptote, as that is only evaluated at infinity. – Emilio Pisanty Jan 07 '21 at 15:53
  • Whats the dipole correction referred here? – Orion_Pax Nov 17 '21 at 09:34
  • @Orion_Pax The dipole term in the multipolar expansion of the field for large $r$. If you want more details, ask separately. – Emilio Pisanty Nov 17 '21 at 10:41
  • I was asking a similar problem but the thread was closed , i was asking for the proof of q1 +q2 and where its placed so as to find flux of asymptote part . Can u tell in there @Emilio_Pisanty using electric field lines ? – Orion_Pax Nov 17 '21 at 12:15
  • And how one can view the code above if one has not a subscription to wolfram mathematica? – Orion_Pax Nov 17 '21 at 12:21
  • And yeah i got the correction part – Orion_Pax Nov 17 '21 at 12:21
  • @Orion_Pax The code linked above can only be accessed using Mathematica. Without it, you can code your own, using your platform of choice; it is fairly straightforward. I'm uninterested in further back-and-forth, though. – Emilio Pisanty Nov 17 '21 at 18:11
  • Yeah its ok @Emilio_Pisanty I was asking a similar problem but the thread was closed , i was asking for the proof of q1 +q2 and where its placed so as to find flux of asymptote part . Can u tell the proof of this if possible? using electric field lines – Orion_Pax Nov 18 '21 at 00:59
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Going with Michael's idea and noting that the field line of interest in your sketch is actually the resultant field, then at very large distances the asymptote becomes a tangent to a radial resultant from a net charge q = $q_1$ + $q_2$. Then the number you are looking will be the total number of lines times the fraction of a sphere occupied by the given solid angle; N = (lines/coulomb)q[Ω/(4π)].

R.W. Bird
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  • I'm not quite able to get your words. Firstly, $\Omega=2\pi(1-\cos\alpha)$. But the field lines emitted at angle $\alpha$ are originated form $q_1$ solely, with or without presence of external influence. So, I can't see why you think of multiplying "q" with $\frac{\Omega}{4\pi}$. Secondly, I still am confused about $\beta$ – SteelCubes Jan 05 '21 at 21:58
  • The field lines shown on the sketch represent the resultant which you get after taking the vector sum of the fields from the two charges. Remember, the field lines from a single spherical charge do not bend. Now that I think about it, β is an independent variable which determines the solid angle. The direction of the field line at a large distance does not depend on α; only on β. – R.W. Bird Jan 06 '21 at 15:16