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 $E_0 = 2\times10^3\, N/C$. Then, the flux of this field through a rectangular surface of area $0.2\, m^2$ parallel to the $y-z$ plane is......$\frac{{N - {m^2}}}{C}$

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 net flux through a closed Gaussian surface containing no net charge, is zero.

$(d)$ When 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

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 $\left.\mathrm{Nm}^{2} / \mathrm{C}\right)$

Which among the curves shown in Figureb cannot possibly represent electrostatic field lines?

Three charges $q_1 = 1\,\mu c, q_2 = 2\,\mu c$ and $q_3 = -3\,\mu c$ and four surfaces $S_1, S_2 ,S_3$ and $S_4$ are shown in figure. The flux emerging through surface $S_2$ in $N-m^2/C$ is

A charge $q$ is surrounded by a closed surface consisting of an inverted cone of height $h$ and base radius $R$, and a hemisphere of radius $R$ as shown in the figure. The electric flux through the conical surface is $\frac{n q}{6 \epsilon_0}$ (in SI units). The value of $n$ is. . . .