There exists an electric field of $1 \ N/C$ along the $Y$-direction. The flux passing through a square of $1 \ m^2$ placed in the $XY$-plane inside the electric field is . . . . . . .

Consider a region in free space bounded by the surfaces of an imaginary cube having sides of length $a$ as shown in the figure. $A$ charge $+Q$ is placed at the centre $O$ of the cube. $P$ is a point outside the cube such that the line $OP$ perpendicularly intersects the surface $ABCD$ at $R$,and $OR = RP = a/2$. $A$ charge $+Q$ is also placed at point $P$. What is the total electric flux through the five faces of the cube other than $ABCD$?

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 region,the electric field is given by $\overrightarrow{E}=(3 \hat{i}+5 \hat{j}+7 \hat{k}) \text{ NC}^{-1}$. The electric flux through a surface of area $3 \text{ m}^2$ in the $yz$-plane is (in $SI$ units):

$A$ charge $Q \ C$ is placed at the center of a cube. If $\varepsilon_0$ is the permittivity of vacuum,then the flux through one face and two opposite faces of the cube is respectively:

$A$ charged body has an electric flux $\phi$ associated with it. The body is now placed inside a metallic container. The flux $\phi$ outside the container will be