In MIT's 8.02 course, it is shown in lecture 3 that we can derive Gauss's Law from Coulomb's to get
$ \phi = \oint \vec{E} \cdot \vec{dA} = \frac{Q_{enc}}{\epsilon_{0}} $
However, in the lecture, it was assumed that there were no charges outside the surface. Later, Gauss's Law was used and the charges outside the sphere were ignored.
I've been thinking about it and came across Can someone give an intuitive way of understanding why Gauss's law holds?.
The first answer by user levitopher helped a bit. I think that the lecture's argument still holds when there are charges outside because $ \vec{E} $ is added vectorially.
Suppose we have $ Q_{enc} $ in a sphere and $ Q_{out} $ outside the sphere. Then,
By definition, $ \phi = \oint (\vec{E_{Q_{enc}}} + \vec{E_{Q_{out}}}) \cdot \vec{dA} $
$ = \oint \vec{E_{Q_{enc}}} \cdot \vec{dA} + \oint \vec{E_{Q_{out}}} \cdot \vec{dA}$
$ = \oint \vec{E_{Q_{enc}}} \cdot \vec{dA} + 0 $
$ = \frac{Q_{enc}}{\epsilon_{0}} $
This assumes that $ \oint \vec{E_{Q_{out}}} \cdot \vec{dA} = 0 $ because any field lines that go in must come out (hand wavy but I'm just going to accept it for now).
I'd just like to know if this argument is sound.
Best Answer
The argument is perfectly sound. Hand waving is a diminutive term for this, since this really is the essence of Gauss' Law. Here is a succinct physical answer, without using maths, which you can use to make sense of the situation.
Field lines due to charges located outside will cross the surface first while ''going in'' and then, while ''going out'', and hence, net flux through the closed surface $S$ will be zero. The only way in which they will contribute to non-zero flux is if there is some positive or negative charge(s) located inside, since e.g. electric lines of force originate from a positive charge and terminate on the negative charge. See here for example.
Thus, with a positive charge in the surface, there is non-zero out-flux, and with a negative charge, a net non-zero in-flux. With both positive and negative charge inside, net effect depends on ''which ones are more'', which is why you have that $Q_{encl}$ on the right hand side. An illustration for each of these three cases can be seen here.