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

An electric line of force in the $X, Y-$ plane is given by the equation $x^2 + y^2 = 1$. $A$ particle with unit positive charge is initially at rest at the point $(1, 0)$ in the $X, Y-$ plane. The particle:

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

Choose the incorrect statement:
$(a)$ The electric lines of force entering into a Gaussian surface provide negative flux.
$(b)$ $A$ charge '$q$' is placed at the centre of a cube. The flux through all the faces will be the same.
$(c)$ In a uniform electric field,the net flux through a closed Gaussian surface containing no net charge is zero.
$(d)$ When the electric field is parallel to a Gaussian surface,it provides a finite non-zero flux.
Choose the most appropriate answer from the options given below:

$A$ charge $q$ is placed at the center of a cubical box of side $a$ with the top open. The flux of the electric field through one of the five closed surfaces of the cubical box is:

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 $\text{Nm}^2/C$):