$A$ metallic sphere of radius $R$ carrying a charge $q$ is kept at a certain distance from another metallic sphere of radius $R/4$ carrying a charge $Q$. What is the electric flux at any point inside the metallic sphere of radius $R$ due to the sphere of radius $R/4$?

$A$ linear charge having linear charge density $\lambda$ penetrates a cube diagonally and then it penetrates a sphere diametrically as shown. What will be the ratio of flux coming out of the cube and the sphere?

Two point charges $2q$ and $q$ are placed at vertex $A$ and the centre of face $CDEF$ of the cube, respectively, as shown in the figure. The electric flux passing through the cube is:

$A$ square loop of sides $a=1 \ m$ is held normally in front of a point charge $q=1 \ C$. The charge is placed at a distance of $a/2$ from the center of the square. The flux of the electric field through the shaded region is $\frac{5}{p} \times \frac{1}{\varepsilon_0} \frac{N m^2}{C}$,where the value of $p$ is . . . . . . .

$A$ square surface of side $L$ meters is in the plane of the paper. $A$ uniform electric field $\vec{E} \text{ (V/m)}$ is also in the plane of the paper,limited only to the lower half of the square surface. The electric flux associated with the surface in $SI$ units is:

$A$ circular disc of radius $R$ carries surface charge density $\sigma(r) = \sigma_0 \left(1 - \frac{r}{R}\right)$,where $\sigma_0$ is a constant and $r$ is the distance from the center of the disc. Electric flux through a large spherical surface that encloses the charged disc completely is $\phi_0$. Electric flux through another spherical surface of radius $\frac{R}{4}$ and concentric with the disc is $\phi$. Then the ratio $\frac{\phi_0}{\phi}$ is: