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Consider two charges $+q$ and $-Q$ placed at a distance, note- charge q and Q are different In terms of magnitude. like this

My question: is number of flux lines received by $-Q$ proportional to its own charge, or does $+q$ charge have anything to say at all?

As according to gauss law

guass law Source of image: Britannica

The LHS is dependant of field external to Gaussian surface and and RHS of equation depends on charge enclosed within the Gaussian surface.

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    $\begingroup$ The "number of flux lines" is a purely aesthetic choice made by the artist of a diagram. While there are obvious choices of number that give intuition for the physical flux density, the choice of number is not a physical phenomenon that follows any kind of laws. $\endgroup$ – Xerxes Mar 24 at 14:34
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The number of flux lines of each charge is proportional to its own charge. The other charge has nothing to do with that.

See this image with two unequal charges. The right negative charge ($-3Q$) has three times the size of the left positive charge ($+Q$):
image
(image from Chegg Study: physics questions and answers)

This is in accordance with Gauss's law for the electric field:

  • Draw a closed surface around the left charge ($+Q$) only.
    There are 6 field lines coming out of this surface
  • Draw a closed surface around the right charge ($-3Q$) only.
    There are 18 field lines going into this surface.
  • Draw a big closed surface around both charges together ($+Q-3Q = -2Q$).
    There are 12 field lines going into this big surface.
  • Draw a closed surface which does not enclose any of the charges.
    There are $n$ field lines going into and the same $n$ field lines coming out of this surface, thus giving a sum of zero.

In all cases the number of field lines (i.e. the electric flux) through the closed surface is proportional to the charge inside the surface.

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If you draw a Gaussian surface that only includes the $-Q$ charge, the total electric flux through that surface is proportional to the enclosed charged $-Q$ by Gauss' Law. This doesn't depend on the charge $+q$ external to that Gaussian surface. You could move that external charge to infinity and not change that total flux. So, the number of flux lines into $-Q$ is proportional to $-Q$ alone.

update to the address the OP's follow-up question in the comment

But on LHS Of gauss law I.e /E.A , the E is due to all the charges , this is causing the confusion!
... So what does the LHS of the gauss signify then?

Using https://www.glowscript.org/#/user/matterandinteractions/folder/matterandinteractions/program/13-fields (and selecting Measurement type: "Gauss's law"), draw a closed surface then introduce a positive charge inside. Observe the outward flux through each patch.

Glowscript MatterAndInteractions 13-fields empty Glowscript MatterAndInteractions 13-fields centered

Now, as I reposition the charge [which is easier to do with this visualization], note that the flux through each patch changes... but the total remains constant [suggested by Gauss's Law, a physical law that says the total electric flux through a Gaussian surface is equal to the enclosed charge divided by $\epsilon_0$.]

Glowscript MatterAndInteractions 13-fields off-centered

When I move the charge outside the Gaussian surface, the sign of the flux changes for the patches near the charge. The total flux drops to zero.
So, while external charges contribute to the local flux through a patch, their net [total] contribution to the flux is zero through a Gaussian surface that doesn't enclose those external charges.

Glowscript MatterAndInteractions 13-fields external

Note the total electric flux is not just "$EA$",
it's $\sum \vec E_i \cdot \Delta \vec A_i$ summing over all patches.
In integral form, it's $\oint \vec E\cdot d\vec A$.

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  • $\begingroup$ But on LHS Of gauss law I.e /E.A , the E is due to all the charges , this is causing the confusion! $\endgroup$ – user72730 Mar 24 at 11:40
  • $\begingroup$ Consider a single [say, positive] point charge and draw its field lines. Now draw a gaussian spherical surface centered at that point. The flux through that surface is positive. Now reposition that sphere so that it doesn't include the point charge. Note that every field line that entered that repositioned sphere will leave that sphere. This suggests (and can be backed up with a more detailed calculation) the total flux through that sphere is zero [due to an external charge]. While external charges might affect the local flux through a patch of the sphere, it doesn't affect the total flux. $\endgroup$ – robphy Mar 24 at 11:49
  • $\begingroup$ So what does the LHS of the gauss signify then? $\endgroup$ – user72730 Mar 24 at 11:55
  • $\begingroup$ I added some visualizations in my answer to address your follow-up questions. $\endgroup$ – robphy Mar 24 at 12:20
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The answer would be "yes, but...".

In an universe with global neutral electrical charge, the flux through a closed surface is proportional to the charge inside the surface, but it means that it is also proportional to the charge outside the surface, because both are opposite but equal in absolute value.

Your drawing shows only two opposite charges and the rest of the universe is not supposed matter - it is not included in the model. Therefore all the flux received by the negative charge is originated in the positive one - that is, all lines that end in the negative charge start in the positive charge.

However, although you can say that the flux depends on the charge inside the closed surface or the charge outside the closed surface, it doesn't depend on how are those charges distributed. Therefore, the flux that a given negative point charge receives doesn't depend on whether an opposite negative charge is close to it or very far away.

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  • $\begingroup$ @user72730 - If both charges are equal, the amount that goes away to universe equals the amount that comes from universe - you can prove that by applying the Gauss theorem to a closed surface enclosing both charges and nothing else. The distance between charges doesn't matter. $\endgroup$ – Pere Mar 25 at 9:19
  • $\begingroup$ I meant to say I’m not concerned about the distance because you have mentioned it the last line of your answer. $\endgroup$ – user72730 Mar 25 at 9:38

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