For a given surface,the Gauss's law is stated as $\oint {E \cdot ds} = 0$. From this,we can conclude that:

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 $x = a$.

Draw the electric field lines for a system of two identical positive charges placed near each other.

An electric field $\overrightarrow{E} = 4x \hat{i} - (y^2 + 1) \hat{j} \text{ N/C}$ passes through the box shown in the figure. The flux of the electric field through surfaces $ABCD$ and $BCGF$ are marked as $\phi_I$ and $\phi_{II}$ respectively. The difference $(\phi_I - \phi_{II})$ is (in $\text{Nm}^2/C$):

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

Four closed surfaces and corresponding charge distributions are shown below. Let the respective electric fluxes through the surfaces be $\phi_1, \phi_2, \phi_3$ and $\phi_4$. Then: