The electric flux from a cube of edge $l$ is $\phi$. If an edge of the cube is made $2l$ and the charge enclosed is halved,its value will be

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

$A$ charge $Q$ is situated at the corner of a cube. The electric flux passing through all the six faces of the cube is:

$A$ square of side $20 \ cm$ is enclosed by a spherical surface of radius $80 \ cm$. The centers of the square and the sphere are the same. Four charges $2 \times 10^{-6} \ C, -5 \times 10^{-6} \ C, -3 \times 10^{-6} \ C$,and $6 \times 10^{-6} \ C$ are placed at the four corners of the square. The total flux coming out of the spherical surface in $N \cdot m^2/C$ is:

If there is only one type of charge in the universe,then ($ \vec{E} $ = Electric field,$ \vec{d}s $ = Area vector):

Consider the charges and the Gaussian surface shown in the figure. When calculating the electric flux through the spherical surface,the electric field is due to which of the following?