$A$ square surface of side $L$ metres is in the plane of the paper. $A$ uniform electric field $\vec{E} \text{ (V/m)}$,also in the plane of the paper,is limited only to the lower half of the square surface,(see figure). The electric flux in $SI$ units associated with the surface is

The electric field in a region is given by $\vec{E}=(2 \hat{i}+4 \hat{j}+6 \hat{k}) \times 10^3 \ N/C$. The flux of the field through a rectangular surface parallel to the $x-z$ plane is $6.0 \ N m^2 C^{-1}$. The area of the surface is . . . . . . $cm^2$.

Explain the electric field lines and the magnitude of the electric field.

If the radius of the spherical Gaussian surface is increased,then the electric flux due to a point charge enclosed by the surface:

$A$ 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

$A$ 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 $\,nC$)