$A$ charge '$Q$' $C$ is placed at the centre of a cube. The electric flux through two opposite faces of the cube is $(\epsilon_0 = \text{permittivity of free space})$

$A$ 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:

The electric flux from a cube of edge $l$ is $\phi$. If an edge of the cube is made $2l$ and the charge enclosed is halved,its value will be

$(a)$ An electrostatic field line is a continuous curve. That is,a field line cannot have sudden breaks. Why not?
$(b)$ Explain why two field lines never cross each other at any point?

The figure shows the electric field lines for a system of two charges. Choose the correct option regarding the electric field intensity at points $A$,$B$,and $C$.

The electric field in a region is given by $\overrightarrow{E} = \frac{2}{5} E_{0} \hat{i} + \frac{3}{5} E_{0} \hat{j}$ with $E_{0} = 4.0 \times 10^{3} \, N/C$. The flux of this field through a rectangular surface area $0.4 \, m^{2}$ parallel to the $Y-Z$ plane is ....... $N m^{2} C^{-1}$.