$A$ charge $Q \; \mu C$ is placed at the centre of a cube. The flux coming out from any one of its faces will be:

If the electric field is given by $\vec{E} = (5 \hat{i} + 4 \hat{j} + 9 \hat{k})$. The electric flux through a surface of area $20$ units lying in the $Y-Z$ plane will be (in units):

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 ...... $\frac{N \cdot m^2}{C}$.

$A$ spherical rubber balloon carries a charge,uniformly distributed over its surface. As the balloon is blown up and increases in size,the total electric flux coming out of the surface

$A$ linear charge having linear charge density $\lambda$ penetrates a cube diagonally and then it penetrates a sphere diametrically as shown. What will be the ratio of flux coming out of the cube and the sphere?

$A$ cubical Gaussian surface has a side of length $a = 10 \,cm$. Electric field lines are parallel to the $X$-axis as shown in the figure. The magnitudes of the electric fields through surfaces $ABCD$ and $EFGH$ are $6 \,kNC^{-1}$ and $9 \,kNC^{-1}$ respectively. Then,the total charge enclosed by the cube is (Take $\varepsilon_0 = 9 \times 10^{-12} \,Fm^{-1}$): (in $\,nC$)