$A$ charge $q$ is placed at the centre of the open end of a cylindrical vessel. The flux of the electric field through the surface of the vessel is

$A$ thin spherical shell encloses a concentric solid sphere. The radius of the shell is $(0.060)^{1/2} \ m$ and its surface charge density is $-10^{-5} \ C/m^2$. The radius of the solid sphere is $(0.01)^{1/3} \ m$ and its volumetric charge density is $3 \times 10^{-5} \ C/m^3$. $\varepsilon_0$ is the permittivity of free space in $C^2/Nm^2$. The electric flux through a spherical surface concentric with the spherical shell and of radius greater than that of the shell in $V-m$ is:

Draw the electric field lines of a positive point charge.

What will be the total flux through the faces of the cube as shown in the figure with side of length $a$ if a charge $q$ is placed at:
$(a)$ $C$: centre of a face of the cube.
$(b)$ $D$: midpoint of $B$ and $C$.

$A$ 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:

An electrostatic field line leaves at an angle $\alpha$ from a point charge $q_{1}$ and connects with a point charge $-q_{2}$ at an angle $\beta$ ($q_{1}$ and $q_{2}$ are positive). See the figure below. If $q_{2} = \frac{3}{2} q_{1}$ and $\alpha = 30^{\circ}$,then: