Assertion : Electric lines of force never cross each other.
Reason : Electric field at a point superimpose to give one resultant electric field.

Four closed surfaces and corresponding charge distributions are shown below. Let the respective electric fluxes through the surfaces be $\phi_1, \phi_2, \phi_3$ and $\phi_4$. Then:

$A$ line charge of length $\frac{a}{2}$ is kept at the center of an edge $BC$ of a cube $ABCDEFGH$ having edge length $a$ as shown in the figure. If the linear charge density is $\lambda \; C/m$, then the total electric flux through all the faces of the cube will be . . . . . . . (Take $\varepsilon_0$ as the free space permittivity)

The electric field in a region is given by $\overrightarrow{E} = (\frac{3}{5} E_{0} \hat{i} + \frac{4}{5} E_{0} \hat{j}) \, N/C$. The ratio of the flux of this field through a rectangular surface of area $0.2 \, m^{2}$ (parallel to the $y-z$ plane) to that through a surface of area $0.3 \, m^{2}$ (parallel to the $x-z$ plane) is $a : b$,where $a = \dots$ [Here $\hat{i}, \hat{j}$ and $\hat{k}$ are unit vectors along the $x, y$ and $z$-axes respectively].

In a uniform electric field,a cube of side $1 \ cm$ is placed. The total energy stored in the cube is $8.85 \ \mu J$. The electric field is parallel to four of the faces of the cube. The electric flux through any one of the remaining two faces is

$A$ metallic solid sphere is placed in a uniform electric field. The lines of force follow the path$(s)$ shown in the figure as: