If the radius of the spherical Gaussian surface is increased,then the electric flux due to a point charge enclosed by the surface:

The inward and outward electric flux for a closed surface in units of $N \cdot m^2/C$ are respectively $8 \times 10^3$ and $4 \times 10^3$. Then the total charge inside the surface is [where $\varepsilon_0$ = permittivity constant].

$A$ cube of side $l$ is placed in a uniform electric field $E$,where $E = E\hat{i}$. The net electric flux through the cube is:

Assertion: Four point charges $q_1, q_2, q_3$ and $q_4$ are as shown in the figure. The flux over the shown Gaussian surface depends only on charges $q_1$ and $q_2$.
Reason: Electric field at all points on the Gaussian surface depends only on charges $q_1$ and $q_2$.

Draw the electric field lines for a system of two identical positive charges placed near each other.

$A$ thin spherical shell of radius $R$ and surface charge density $\sigma$ is placed in a cube of side $5R$ with their centers coinciding. The electric flux through one face of the cube is $(\varepsilon_0 = \text{Permittivity of free space})$