If the number of electric lines of force emerging out of a closed surface is $1000$,then the charge enclosed by the surface is .......... $C$.

$A$ charge $Q \; \mu C$ is placed at the centre of a cube. The flux coming out from any one of its faces will be:

$A$ point charge $q$ is placed at the center $O$ of a cube of side length $L$. Another identical charge $q$ is placed at a distance $L$ from $O$ as shown in the figure. The electric flux through the face $ABCD$ is:

$A$ large charged plane having surface charge density $4.9 \times 10^{-6} \text{ C m}^{-2}$ lies in the $x-y$ plane. $A$ circular plane of radius $1 \text{ cm}$ is lying completely in the region where $x, y$ and $z$ coordinates are all positive. When the plane's normal makes an angle $60^{\circ}$ with the $z$-axis,the electric flux through the circular plane is (Given: $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \text{ N m}^2 \text{ C}^{-2}$)

$A$ few electric field lines for a system of two charges $Q_1$ and $Q_2$ fixed at two different points on the $x$-axis are shown in the figure. These lines suggest that:
$(A)$ $|Q_1| > |Q_2|$
$(B)$ $|Q_1| < |Q_2|$
$(C)$ at a finite distance to the left of $Q_1$ the electric field is zero
$(D)$ at a finite distance to the right of $Q_2$ the electric field is zero

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