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 $\Phi=\frac{ Q }{5 \epsilon_0}$
$1,2,3$
$1,2,4$
$1,2$
$1,3$
Assertion : Electric lines of force never cross each other.
Reason : Electric field at a point superimpose to give one resultant electric field.
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