Choose the incorrect statement.

If the electric flux entering and leaving an enclosed surface is $\phi_1$ and $\phi_2$ respectively, then the charge enclosed in the surface is ($\varepsilon_0 =$ permittivity of free space).

Find the electric flux through the shaded face $BCGF$ of the cube,as shown in the figure.

Electric field lines of force of a positive point charge are

$A$ 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 $\Phi = \frac{Q}{5\epsilon_0}$

$A$ metallic sphere of radius $R$ carrying a charge $q$ is kept at a certain distance from another metallic sphere of radius $R/4$ carrying a charge $Q$. What is the electric flux at any point inside the metallic sphere of radius $R$ due to the sphere of radius $R/4$?