$A$ charge $Q$ is placed at the centre of a cube of side $a$. The total flux of electric field through the six surfaces of the cube is

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

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].

When is electric flux said to be positive,negative,or zero?

Gauss's law should be invalid if

The figure shows the electric field lines. The spacing between the lines is parallel to the paper at every point. If the magnitude of the field at $A$ is $40 \ N/C$,then the approximate magnitude of the field at $B$ is ....... $N/C$.