Sign of Gaussian surface that encloses negative charge I can't solve a contradiction that have appeared in my head.
Let's assume we have a negative charge, if we enclose it by a spherical surface and $A$ is surface of the sphere, then we will have
$$\oint_S\vec{\mathbf{E}}\cdot \mathrm{d}\vec{\mathbf{S}}=\frac{-Q}{\epsilon _0}$$
But $\vec{\mathbf{E}}$ is at $180^{\circ}$ to vector normal to sphere's surface
therefore, we have 
$$\begin{align}
-EA &= \frac{-Q}{\epsilon _0} \\
E &= \frac{Q}{\epsilon _0 A} = \frac{Q}{\epsilon _0 4\pi r^2}
\end{align}$$
But that contradicts Coulomb's law for electric field, that states for negative charge: 
$$E=\frac{-Q}{\epsilon _0 4\pi r^2}$$
Where is my argument flawed ?
 A: A negatively charged particle has an electric field,
$$\mathbf{E} =-\vert\mathbf{E}\vert \, \hat{\mathbf{r}} =\frac{-q}{4\pi\epsilon_0 r^2}\hat{\mathbf{r}}$$
Gauss's law gives, 
\begin{align*}
\int \mathbf{E}\cdot d\mathbf{A} &= \int \Big(\frac{-q}{4\pi\epsilon_0 r^2}\hat{\mathbf{r}}\Big) \cdot (r^2 \sin \theta d\theta d\phi \, \hat{\mathbf{r}}) = \frac{-q}{\epsilon_0} .
\end{align*}
In your equation you have factored out the minus, so what you have written as $E$ is really $\vert \mathbf{E} \vert$, which is positive.
\begin{align*}
\int \mathbf{E}\cdot d\mathbf{A} &= \int -\vert \mathbf{E} \vert \, \hat{\mathbf{r}} \cdot dA \, \hat{\mathbf{r}} = -\vert\mathbf{E}\vert A = \frac{-q}{\epsilon_0} \\&\Rightarrow \vert\mathbf{E}\vert = \frac{q}{4\pi\epsilon_0 r^2}
\end{align*}
Plugging this into $\mathbf{E}$ we see that there is no inconsistency.
A: It´s an interesting question. I think if you answer the following question, you will understand it yourself. Does the flux formula provide you the vector E-field or just its magnitude? Of course, the flux is a scalar product of E and dA vectors and E-field that comes out of the integral is just a magnitude. On the other hand, the negative that comes in coulomb law is based on the direction of the E-field. Therefore, if you use Gauss´s law to determine E-field, you are going to have just the magnitude. To make it a vector, you have to add negative/positive by yourself.
Hope it helps!
