If $\vec E = \frac{E_0 x}{a} \hat i$,then find the electric flux through the shaded area of the cube as shown in the figure,where the shaded face is at $x = a$.

$A$ positive charge $Q$ is situated at the centre of a cube. The electric flux through any face of the cube is (in $SI$ units)

In a region,the intensity of an electric field is given by $\overrightarrow{E}=(2 \hat{i}+3 \hat{j}+\hat{k}) \text{ NC}^{-1}$. The electric flux through a surface of area $10 \hat{i} \text{ m}^2$ in the region is

State Gauss's law in electrostatics and provide its mathematical expression.

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 . . . . . . .

$A$ charge is uniformly distributed on the surface of a spherical rubber balloon. As it is blown up,the total electric flux coming out of the surface