$A$ 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. . . .

$A$ square of side $20 \ cm$ is enclosed by a spherical surface of radius $80 \ cm$. The centers of the square and the sphere are the same. Four charges $2 \times 10^{-6} \ C, -5 \times 10^{-6} \ C, -3 \times 10^{-6} \ C$,and $6 \times 10^{-6} \ C$ are placed at the four corners of the square. The total flux coming out of the spherical surface in $N \cdot m^2/C$ is:

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

The electric field in a region of space is given by $\vec{E} = (5\hat{i} + 2\hat{j}) \text{ N/C}$. Calculate the electric flux through a surface of area $2 \text{ m}^2$ lying in the $YZ$-plane.

$A$ sphere encloses an electric dipole with charges $\pm 3 \times 10^{-6} \; C$. What is the total electric flux across the sphere in $\text{N m}^2 / \text{C}$?

The electric flux from a cube of edge $l$ is $\phi$ for an enclosed charge. If the edge of the cube is made $\frac{2}{3} l$ and the charge enclosed in the cube is doubled,then the electric flux value will be