The electric field in a region is given by $\vec E = \frac{3}{5}E_0\hat i + \frac{4}{5}E_0\hat j$ and $E_0 = 2 \times 10^3 \, N/C$. Then,the flux of this field through a rectangular surface of area $0.2 \, m^2$ parallel to the $y-z$ plane is ...... $\frac{N \cdot m^2}{C}$.

Two point charges $8 \mu \text{C}$ and $-2 \mu \text{C}$ are located at $x = 2 \text{ cm}$ and $x = 4 \text{ cm}$,respectively on the $x$-axis. The ratio of electric flux due to these charges through two spheres of radii $3 \text{ cm}$ and $5 \text{ cm}$ with their centers at the origin is . . . . . . .

The dimensional formula of electric flux is . . . . . . .

An uncharged metal sphere is placed between two charged plates as shown. The electric field lines look like:

An electrostatic field line leaves at an angle $\alpha$ from a point charge $q_{1}$ and connects with a point charge $-q_{2}$ at an angle $\beta$ ($q_{1}$ and $q_{2}$ are positive). See the figure below. If $q_{2} = \frac{3}{2} q_{1}$ and $\alpha = 30^{\circ}$,then:

In the figure,a $+Q$ charge is located at one of the corners of the cube. What is the electric flux through the cube due to the $+Q$ charge?