Electric field on Gaussian surface due to external charge

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?

• For Gauss law to tell the electric field things must be symmetrical to pull out E from the cyclic integration. Commented Mar 8, 2018 at 6:06

• 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? Commented Mar 11, 2018 at 9:26
• @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 ? Commented Jan 15, 2020 at 4:43
• @Gokul Yes. It would be like saying $5+2=7$ vs. $5+2+3-3=7$ Commented Jan 15, 2020 at 13:16