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Gauss' law gives me the electric field at point P due to the charge +q enclosed by the Gaussian surface, but what if there is an external charge +Q? Won't it influence the electric field at point P, as electric field vector at that point due to both the charges will get added according to vector laws? This confuses me, does the Gauss law give me the electric field at point P due to the charge the Gaussian surface encloses or does it give me the net electric field , that is the resultant of electric fields of both +q and +Q?

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  • $\begingroup$ For Gauss law to tell the electric field things must be symmetrical to pull out E from the cyclic integration. $\endgroup$ – SmarthBansal Mar 8 '18 at 6:06
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Gauss's law in general relates the total electric field flux through a closed surface to the charge enclosed in that surface. It does not give you the field at a certain point in space in general. You can, however, exploit symmetries in certain geometries (points, spheres, lines, cylinders, planes, etc.) to determine the field at certain points in space. If you can't exploit some symmetry to pull E out of your flux integral, then you can't use Gauss's law to determine the field at a point in space.

It's like saying "I have 10 numbers that add to 100." If this is all you know then you can't tell me which numbers I have used to get a total of 100. But if I also tell you "each number is the same number", then you can for sure say I used 10 10's to get a total of 100.

As for your diagram, yes the charge Q will influence the field at P. It will not change the total flux through your Gaussian sphere, or any other Gaussian surface not enclosing Q. You will not be able to use Gauss's law now to find the field at P since the symmetry that lets you pull E out of the integral is no longer present. (Although, you could argue that you can use Gauss's law on each charge individually and then use superposition to get the total field, but you can't use Gauss's law for the entire configuration to get the field at P in this case).

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    $\begingroup$ I think I get it now. But I got a question, as in the above situation, does $$\vec{E}$$ in the equation $$\oint \vec{E}.d\vec{A}=q/\epsilon_{0}$$ represent total electric field due to both charges outside and inside or electric field only due to charge +q? I think it represents e field due to both charges +q and +Q but as $$\oint [\vec{E_{+q}} +\vec{E_{+Q}}].d\vec{A} = q/\epsilon_{0} \Rightarrow \oint \vec{E_{+q}}.d\vec{A}+\oint \vec{E_{+Q}}.d\vec{A} = q/\epsilon_{0}$$ The integral $$\oint\vec{E_{+Q}}.d\vec{A}=0$$, so we only consider e field due to charge +q in the Gauss' equation. Right? $\endgroup$ – Noah Cygnus Mar 11 '18 at 9:26
  • $\begingroup$ Yes that is right. Although if you used Gauss's law like this you would only determine the field due to charge q, not the field for the entire configuration. $\endgroup$ – BioPhysicist Mar 11 '18 at 12:38
  • $\begingroup$ @AaronStevens So there is essentially no difference between taking $\vec{E}$ to be field due to only the charge enclosed vs field due to all charges, because the flux term due to external charges will be 0 ? $\endgroup$ – Gokul Jan 15 '20 at 4:43
  • $\begingroup$ @Gokul There is no difference for what? Certainly the field due to one point charge is different for two point charges. However, if you are doing the surface integral over some closed surface, then Gauss's law tells us that fields due to charges outside of that surface does not change the flux across the closed surface. $\endgroup$ – BioPhysicist Jan 15 '20 at 5:26
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    $\begingroup$ @Gokul Yes. It would be like saying $5+2=7$ vs. $5+2+3-3=7$ $\endgroup$ – BioPhysicist Jan 15 '20 at 13:16

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