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Suppose that there is a charge configuration of 2 point charges, say an electric dipole. Gauss's Law wouldn't be so useful (but it would be possible, as far as I understand) because there is not a special symmetry that can give a certain advantage to find the electric field, since the field line configuration is non-uniform.

However, suppose that we try to apply the Gauss's Law at a weird Gaussian Surface for which at every point of the surface, the magnitude of the electric field is the same (such as in those cases where we have a spherical symmetry situation, just that in this case the surface clearly wouldn't be a sphere). Before going to my question, if we try to enclose both charges in the electric dipole, clearly Gauss's Law tells us that the electric flux would be 0, since the net charge inside is 0.

However, what if we enclose just one charge?, suppose that I enclose only the +q charge without enclosing the -q charge, with a surface such that the magnitude of the electric field is the same everywhere at the surface. If I solve for E, would that give the right answer for the magnitude at that point? Or do we need to take into consideration the fact that there is a charge -q lying around in there?

My question comes from the fact that the electric flux at a closed surface is equal to just the charge inside (divided by epsilon-zero). In fact, what I am specifically asking is: if there is a charge configuration in space (like a sphere or a plane-sheet of charge or any other charge distribution), and I enclose that configuration, according to Gauss's Law, we can get the electric field at the surface of the charge configuration. But what if the space is not a vacuum (such that there are other charges apart from my sphere, or sheet of charge...)?, What if outside my imaginary surface, there is another charge distribution that I am not taking into consideration?, Would the application of Gauss's Law to get the magnitude of the electric field for my previous charge configuration would be correct?

Clearly the other charge distribution outside the Gaussian Surface enclosing another charge distribution should affect the electric field in the vicinity of the charge distribution enclosed. Therefore, enclosing a charge distribution like a sphere to get the electric field would only be reliable if the sphere is strictly in a vacuum where no other charges distort its electric field lines (but the electric flux I think would indeed be right, obviously), or is this statement incorrect?

To explain myself correctly, I drew some diagrams, since I don't think I am explaining myself that well Gauss's Law application for a weird charge distribution

I appreciate any response. Thanks in advance.

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  • $\begingroup$ Gauss law is only sensitive to the enclosed charges. If you have two charges, and you enclose 1, you have to use charge of the enclosed charge in Gauss law. $\endgroup$
    – Mauricio
    Aug 2, 2021 at 18:14

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Gauss' law would still work if you had two charges in your system but only enclosed one in a Gaussian surface.

The "problem" is that Gauss' law in this case would still look like this:

$$ \oint _S \mathbf{E}\cdot\mathrm{d}\mathbf{S} = Q/\epsilon_0,$$ where $Q$ is the charge enclose by the Gaussian surface $S$.

The "problem" being that this is not particularly useful to extract the electric field $\mathbf{E}$, since the its geometrical orientation (where it's pointing) depends on the other charge. That is, if this were an isolated point charge, you know that, by symmetry, the field is only in the radial direction and hence the dot product is Gauss' law is only giving you one term: $E_r$. By symmetry, the electric field magnitude does not depends on the angle, so you can just use a spherical Gaussian surface so that $E_r$ comes out of the integral and the integral only affects the area element, giving you the $4\pi R^2$.
In your case, $\mathbf{E}$ has components in all directions, so the dot product will give you multiple terms, all of which will have to be integrated over. Hence why I am saying that, if you want to extract $\mathbf{E}$, Gauss' law is not very useful, but you might try to just vectorially add the two electric fields or, better yet, add the two potentials and then differentiate: $\mathbf{E} = \nabla(\phi(\mathbf{r})_{+q}+\phi(\mathbf{r})_{-q})$.

Aside

I think the main conceptual question here is: how can Gauss' law give me the correct total electric field, if I am only enclosing one of the two charges?

Well, it's giving you the total flux, not the field.

Look here (pardon the very crude drawing on paint):

enter image description here

The total field here is the sum of the blue ($+q$ charge) and red ($-q$). The flux in Gauss' law is summing the left side (circle 1) where the two fields cancel out, and the right side (circle 2) where the two fields add up. So, very poorly speaking, 2 sides but "only" 2 arrow lengths gives you 1 arrow length, that is the strength of a single charge (the one included in the Gaussian surface).

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  • $\begingroup$ Perfect explanation, thanks!, the "aside" was really what I was having struggle with. So to sum up, if I enclose just one charge it would give me the total flux through S (for which no charge outside would contribute since it cancels out), but in terms of extracting E, it would give me the E for the single charge enclosed, is this correct? $\endgroup$
    – prado5083
    Aug 2, 2021 at 23:28
  • $\begingroup$ @prado5083 It would give you the total E flux caused by the single charge enclosed, and hence the corresponding E, yes. $\endgroup$
    – SuperCiocia
    Aug 2, 2021 at 23:32

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