$A$ charge $Q$ is placed at the center of a closed cube. The flux passing through any one face of the cube will be .......

If an electric field is given by $\vec{E} = 10 \hat{i} + 3 \hat{j} + 4 \hat{k}$,calculate the electric flux through a surface of area $A = 10 \text{ units}$ lying in the $yz$-plane.

If the electric flux entering and leaving an enclosed surface is $\phi_1$ and $\phi_2$ respectively, then the charge enclosed in the surface is ($\varepsilon_0 =$ permittivity of free space).

An electric field,$\overrightarrow{E} = \frac{2 \hat{i} + 6 \hat{j} + 8 \hat{k}}{\sqrt{6}} \ V/m$,passes through a surface of $4 \ m^2$ area having a unit normal vector $\hat{n} = \left( \frac{2 \hat{i} + \hat{j} + \hat{k}}{\sqrt{6}} \right)$. The electric flux through that surface is:

$A$ point charge of $+12 \,\mu C$ is at a distance $6 \,cm$ vertically above the centre of a square of side $12 \,cm$ as shown in the figure. The magnitude of the electric flux through the square will be ....... $\times 10^{3} \,Nm^{2}/C$.

How do electric field lines depend on the area or the solid angle subtended by the area?