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I have come across this doubt, if electric field lines originating from a $+2q$ charge terminate on $-q$ charge (both point charges), will all field lines terminate on the negative charge or only half of them will ? That is, are field lines coming from or coming to a charge proportional to magnitude of charge or not? (ISOLATED point charges) I have found many books saying it's proportional but I can't seem to find a credible/ authentic source or book . So please answer with proof and provide a source too if possible. Thanks.

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Field lines are a visual aid.

Gauss's law is really "counting" the number of field lines which pass through a surface - the flux (flow) of field lines.

So a closed surface around the $+2q$ charge will have twice the flux, "field lines", going out from the charge $+2q$ as the flux going into the $-q$ charge.

Not all the field lines ending on the $-q$ necessarily originate from the $+2q$ charge although if the charges are close together most of them would have come from the $+2q$ charge with the rest coming from other charges.
In the same way nearly half of the field lines originating from the $+2q$ will end up on the $-q$ charge and the rest will go to other charges or go to infinity.

Update in answer to some comments from @Mr.Momo

My answer above was in the "spirit" of the question ie I wrote about electric field lines which I said are visual aids although Faraday did "count" them.

In terms of electric flux I would say that the electric flux through a closed surface around the $+2q$ is double that passing through a closed surface around the $-q$ charge.
The electric flux passing through a closed surface enclosing the two charges is half that of the electric flux passing through a closed surface around the $+2q$.

"…and the rest will go to other charges." I would add here "or go to infinity." Even in the hypothetical situation where the whole universe consists of only these two charges the field lines still make sense If we consider only the two charges to be present that is, isolated 2 charges, what would be the explanation in that case?

I did consider adding "or go to infinity" but did not add the phrase because I do not know what happens at infinity and whether this is only a hypothetical concept.
I do not mind adding it to my answer.
With two isolated charges the flux which does not pass through the surfaces around the two individual charges passes through a closed surface around the two charges.

I think it is worth reading this chapter about how Maxwell applied Faraday's lines of force.

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    $\begingroup$ "…and the rest will go to other charges." I would add here "or go to infinity." Even in the hypothetical situation where the whole universe consists of only these two charges the field lines still make sense $\endgroup$
    – domj33
    Commented May 11, 2017 at 7:08
  • $\begingroup$ If we consider only the two charges to be present that is, isolated 2 charges , what would be the explanation in that case? $\endgroup$
    – m0mosenpai
    Commented May 11, 2017 at 7:11
  • $\begingroup$ Farcher , some of the things you answered were incorrect. Flux and Field Lines are not same . Field lines are proportional to the flux not same. So the FLUX will be half not FIELD LINES. ( Since dot product of Field and Area Vector will be different). Also, can you explain what will happen if the charges are isolated? $\endgroup$
    – m0mosenpai
    Commented May 11, 2017 at 15:33
  • $\begingroup$ @Mr.Momo You do realise the number of field lines is actually a ridiculous thing to talk of? It's the relative number of field lines which has a significance. Else the number of field lines you choose to draw could be 100 times what I chose to draw. And that's practically useless isn't it ? The flux is a relative measure of the number of field lines. $\endgroup$ Commented May 11, 2017 at 16:28
  • $\begingroup$ Yes, I do since it's a hypothetical concept but yes flux does the job of conveying the IDEA of what field lines but we can't say they are the same thing since flux is well defined but field lines are not . So is there a better way to answer my question? $\endgroup$
    – m0mosenpai
    Commented May 11, 2017 at 16:55

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