Gauss' law states that the net outward normal electric flux through a closed surface is equal to $q_{total, inside}/\epsilon_0$. However, I'm a bit confused of why the presence of an external charge doesn't influence the solution. I always hear as a response – "any field lines entering the enclosed surface, will also leave it", which is fine. However, the field is inversely proportional to $r^2$, therefore, it will "exit" with a different magnitude and thus, with a different flux. So how they possibly cancel each other out? And if possible, I really look forward to a simple explanation, without involving too much math. I just want to get the concept. I'm sorry for such a newbie question.
[Physics] Gauss’ law and an external charge
electrostaticsgauss-law
Best Answer
In an attempt to be brief: The big thing to remember is that the flux is also proportional to the area (technically, the surface integral of the field over the area). Crudely speaking, the side of the enclosed surface with exiting field lines are further away from the external charge than the side with "entering" field lines, and the surface area increases by $r^2$ (remember the formula for the surface area of a sphere). Gauss's Law is just a mathematically precise way of stating this compensation.
(Image source: http://www.ux1.eiu.edu/~cfadd/1360/24Gauss/Gauss.html)