A disk of radius $a / 4$ having a uniformly distributed charge $6 \mathrm{C}$ is placed in the $x-y$ plane with its centre at $(-a / 2,0,0)$. A rod of length $a$ carrying a uniformly distributed charge $8 \mathrm{C}$ is placed on the $x$-axis from $x=a / 4$ to $x=5 a / 4$. Two point charges $-7 \mathrm{C}$ and $3 \mathrm{C}$ are placed at $(a / 4,-a / 4,0)$ and $(-3 a / 4,3 a / 4,0)$, respectively. Consider a cubical surface formed by six surfaces $x= \pm a / 2, y= \pm a / 2$, $z= \pm a / 2$. The electric flux through this cubical surface is

Assertion : Electric lines of force never cross each other.

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

Electric flux through a surface of area $100$ $m^2$ lying in the $xy$ plane is (in $V-m$) if $\vec E = \hat i + \sqrt 2 \hat j + \sqrt 3 \hat k$

Electric field in a region is uniform and is given by $\vec{E}=a \hat{i}+b \hat{j}+c \hat{k}$. Electric flux associated with a surface of area $\vec{A}=\pi R^2 \hat{i}$ 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 $\Phi=\frac{ Q }{5 \epsilon_0}$