In a region,the electric field is given by $\overrightarrow{E}=(3 \hat{i}+5 \hat{j}+7 \hat{k}) \text{ NC}^{-1}$. The electric flux through a surface of area $3 \text{ m}^2$ in the $yz$-plane is (in $SI$ units):

$A$ point charge $+q$ is placed at the centre of a cube of side $L$. The electric flux emerging from the cube is

If $\oint_s \vec{E} \cdot \overrightarrow{d S}=0$ over a surface,then:

$A$ charge $q$ is placed at the center of the circular base of an inverted cone of height $h$ and base radius $R$. The cone is capped by a hemisphere of radius $R$ as shown in the figure. The electric flux through the conical surface is $\frac{n q}{6 \epsilon_0}$ (in $SI$ units). The value of $n$ is. . . .

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

Three positive charges of equal value $q$ are placed at the vertices of an equilateral triangle. The resulting lines of force should be sketched as in: