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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'?

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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.

If you actually wanted to validate Gauss's law you would use the entire field, but you would find that the flux you calculate would end up being the same thing as the flux if you had only considered the enclosed charge. Note that flux and field are not the same thing. I'm not saying the field at points on the surface is the same as if you only considered the enclosed charge. The flux is the surface integral of the field over the entire Gaussian surface. This integral only depends on the enclosed charge, even if the field itself depends on all charges.

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  • $\begingroup$ your first statement is clear but in the second statement you say that value of field at points on the surface, when we add up the field due to the charges outside as well as due to the enclosed charge is not the same as the value of field when we were to consider only the enclosed charge but still the flux of E in both the cases is equal how could this be when the integral when calculating flux only contains E with the dot product of the surface element ,Then how could both the integrals give the same value if the value of E is different in both but the integral of area is equal in both? Thx $\endgroup$ – p0803 Jul 6 at 12:26
  • $\begingroup$ @p0803 Just because two integrands are different doesn't mean their integrals over the same region must be different. Ex: $$\int_0^1x\text dx=\int_0^1\frac32x^2\text dx=\frac12$$ $\endgroup$ – Aaron Stevens Jul 6 at 12:32
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The flux is the surface integral of the field over the entire Gaussian surface. This integral only depends on the enclosed charge, even if the field itself depends on all .field due to charges outside will be resultant to zero,gauss experiment was also derived from the coulombs law ,after gauss perform several experiments

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