# The value of $E$ in the LHS of Gauss law equation

Suppose there are multiple point charges in a region and I only take the Gaussian surface which encloses only one of the charges $$q$$. I have read that the $$E$$ term on the LHS of the Gauss law equation is the net electric field due to the principle of superposition.

Then my question is this - that if we're to take the net value of 'E' due to all the charges present (ie the charges inside as well as outside of the Gaussian surface ) in the Gauss law equation then wouldn't the q enclosed which we were to calculate after we put the net value of E in the LHS be more than the charge which is actually enclosed by the chosen surface that is 'q' ?

As in the proof of the law when we are considering a Gaussian surface enclosing a single point charge with no other charges in vicinity the value of 'E' at the surface is only due to the enclosed charge and the terms cancel out each other and the flux of E through the surface comes out to be 'q enclosed divided by epsilon'

And if we really have to write the net value of 'E' in the LHS of the equation how do we write the value of E due to the outside charge at the Gaussian surface when the surface encloses only the single charge'q'?

The flux through your entire Gaussian surface due to the charges not contained in your surface will total to $$0$$. The only charge whose field will not give $$0$$ flux is the charge contained in your Gaussian surface.
• @p0803 Just because two integrands are different doesn't mean their integrals over the same region must be different. Ex: $$\int_0^1x\text dx=\int_0^1\frac32x^2\text dx=\frac12$$ – Aaron Stevens Jul 6 at 12:32