When is electric flux said to be positive,negative,or zero?

The electric field ( $\overrightarrow{E}$ in $N C^{-1}$ ) in a region is given by $\overrightarrow{E} = 3 \hat{i} + 5 \hat{j}$. The net electric flux through a square area of side $2 \ m$ parallel to the $y-z$ plane is:

$A$ charge $q$ is placed at the center of one of the faces of a cube. What is the electric flux linked with the cube?

$A$ circular disc of radius $R$ carries surface charge density $\sigma(r) = \sigma_0 \left(1 - \frac{r}{R}\right)$,where $\sigma_0$ is a constant and $r$ is the distance from the center of the disc. Electric flux through a large spherical surface that encloses the charged disc completely is $\phi_0$. Electric flux through another spherical surface of radius $\frac{R}{4}$ and concentric with the disc is $\phi$. Then the ratio $\frac{\phi_0}{\phi}$ is:

Explain the electric field lines and the magnitude of the electric field.

$A$ charged shell of radius $R$ carries a total charge $Q$. Let $\Phi$ be the flux of the electric field through a closed cylindrical surface of height $h$,radius $r$,with its center coinciding with that of the shell. The center of the cylinder is a point on the axis of the cylinder equidistant from its top and bottom surfaces. Which of the following option$(s)$ is/are correct? [$\epsilon_0$ is the permittivity of free space]
$(1)$ If $h > 2R$ and $r > R$,then $\Phi = \frac{Q}{\epsilon_0}$
$(2)$ If $h < \frac{8R}{5}$ and $r = \frac{3R}{5}$,then $\Phi = 0$
$(3)$ If $h > 2R$ and $r = \frac{4R}{5}$,then $\Phi = \frac{2Q}{5\epsilon_0}$
$(4)$ If $h > 2R$ and $r = \frac{3R}{5}$,then $\Phi = \frac{Q}{5\epsilon_0}$