The figure shows the electric field lines of three charges with charges $+1, +1$,and $-1$. The Gaussian surface in the figure is a sphere containing two of the charges. The total electric flux through the spherical Gaussian surface is

$A$ cylinder of radius $R$ and length $L$ is placed in a uniform electric field $E$ parallel to the cylinder axis. The total flux for the surface of the cylinder is given by

$A$ thin spherical shell encloses a concentric solid sphere. The radius of the shell is $(0.060)^{1/2} \ m$ and its surface charge density is $-10^{-5} \ C/m^2$. The radius of the solid sphere is $(0.01)^{1/3} \ m$ and its volumetric charge density is $3 \times 10^{-5} \ C/m^3$. $\varepsilon_0$ is the permittivity of free space in $C^2/Nm^2$. The electric flux through a spherical surface concentric with the spherical shell and of radius greater than that of the shell in $V-m$ is:

If the electric field is given by $\vec{E} = (5 \hat{i} + 4 \hat{j} + 9 \hat{k})$. The electric flux through a surface of area $20$ units lying in the $Y-Z$ plane will be (in units):

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

The dimensional formula of electric flux is . . . . . . .