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

An electric field $\overrightarrow{\mathrm{E}}=4 \mathrm{x} \hat{\mathrm{i}}-\left(\mathrm{y}^{2}+1\right) \hat{\mathrm{j}}\; \mathrm{N} / \mathrm{C}$ passes through the box shown in figure. The flux of the electric field through surfaces $A B C D$ and $BCGF$ are marked as $\phi_{I}$ and $\phi_{\mathrm{II}}$ respectively. The difference between $\left(\phi_{\mathrm{I}}-\phi_{\mathrm{II}}\right)$ is (in $\left.\mathrm{Nm}^{2} / \mathrm{C}\right)$

$\mathrm{C}_1$ and $\mathrm{C}_2$ are two hollow concentric cubes enclosing charges $2 Q$ and $3 Q$ respectively as shown in figure. The ratio of electric flux passing through $\mathrm{C}_1$ and $\mathrm{C}_2$ is :

A charge $'q'$ is placed at one corner of a cube as shown in figure. The flux of electrostatic field $\overrightarrow{ E }$ through the shaded area is ...... .

A long cylindrical volume contains a uniformly distributed charge of density $\rho$. The radius of cylindrical volume is $R$. A charge particle $(q)$ revolves around the cylinder in a circular path. The kinetic of the particle is

Sketch the electric field lines for a uniformly charged hollow cylinder shown in figure.