# Is Gauss's law wrong, or is it possible that $\int_s{\vec E} \cdot d\vec{s}=0$ does not imply $\vec E = 0$?

You need to be careful here. Gauss’ law is always true, but it is not always possible to use it to infer the electric field. The crucial step is \begin{align} \oint \vec E\cdot d\vec S=\vert\vec E\vert S \tag{1} \end{align} which only holds if the field has constant magnitude on the Gaussian surface and is perpendicular to the surface where it intersects.

Thus for instance, if you place a charge outside a box and compute $\oint \vec E\cdot d\vec S$ on the surface bounding the box, this integral is $0$ because there is no net charge enclosed, but this does NOT mean $\vec E=0$ inside the box as (1) does not hold: by simple geometry the field does not have the same magnitude at every point on the surface of the box.

In other words, yes it is perfectly possible to have $0$ *net* flux $\oint \vec E\cdot d\vec S=0$ but $\vec E\ne 0$.

A similar situation occurs when a charge distribution does not have a particular symmetry: it becomes very difficult to find a surface on which the magnitude of $\vec E$ is constant and thus use (1) to deduce the field.

In such cases one must resort to the superposition principle for practical calculations.

You are absolutely correct in inferring your conclusion that

$$\oint_S \mathbf{E} \cdot d\mathbf{A} = 0$$

does *not* imply that $\mathbf{E}(P) = 0$ at any point. A very simple counterexample to this is to consider a uniform electric field filling all of space:

$$\mathbf{E}(P) := \mathbf{E}_0$$

for a fixed, nonzero electric field vector $\mathbf{E}_0$. It is not hard to see that the total flux through any closed surface here must be zero, since the field lines are just the infinite straight lines in which the vectors $\mathbf{E}_0$ pegged to each point in space point along, and from geometry, any infinite straight line entering a closed and finite surface must exit it.

Indeed, though you may have seen Gauss's law "used" to find an electric field, if you look more closely you will find that in every case, some sort of additional assumption gets made, such as that the charge distribution has some form of symmetry *and* that this symmetry transfers to the field - and that last point is nontrivial: consider the sum of the field of your favorite Gauss's law problem with the field above, i.e. imagine your charge source were in some preexisting ambient electric field environment. This assumption-making ("handwaving") is necessary precisely *because* Gauss's law is insufficient by itself.