The $S.I.$ unit of electric flux is

A charged shell of radius $R$ carries a total charge $Q$. Given $\Phi$ as the flux of electric field through a closed cylindrical surface of height $h$, radius $r$ and with its center same as that of the shell. Here, center of the cylinder is a point on the axis of the cylinder which is 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 $h >2 R$ and $r > R$ then $\Phi=\frac{ Q }{\epsilon_0}$

$(2)$ If $h <\frac{8 R }{5}$ and $r =\frac{3 R }{5}$ then $\Phi=0$

$(3)$ If $h >2 R$ and $r =\frac{4 K }{5}$ then $\Phi=\frac{ Q }{5 \epsilon_0}$

$(4)$ If $h >2 R$ and $r =\frac{3 K }{5}$ then $\Phi=\frac{ Q }{5 \epsilon_0}$

The electric field in a region is given by $\overrightarrow{ E }=\frac{2}{5} E _{0} \hat{ i }+\frac{3}{5} E _{0} \hat{ j }$ with $E _{0}=4.0 \times 10^{3}\, \frac{ N }{ C } .$ The flux of this field through a rectangular surface area $0.4 \,m ^{2}$ parallel to the $Y - Z$ plane is ....... $Nm ^{2} C ^{-1}$

The black shapes in the figure below are closed surfaces. The electric field lines are in red. For which case, the net flux through the surfaces is non-zero?

Assertion : Electric lines of force never cross each other.

Reason : Electric field at a point superimpose to give one resultant electric field.

A metallic sphere is kept in between two oppositely charged plates. The most appropriate representation of the field lines is