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 $x = a$.

Consider an electric field $\vec{E} = E_0 \hat{i}$,where $E_0$ is a constant. The flux through the shaded area (as shown in the figure) due to this field is

In the figure,a point charge $+Q_1$ is at the centre of an imaginary spherical surface and another point charge $+Q_2$ is outside it. Point $P$ is on the surface of the sphere. Let $\Phi _s$ be the net electric flux through the sphere and $\vec E_p$ be the electric field at point $P$ on the sphere. Which of the following statements is $TRUE$?

The expression for an electric field is given by $\vec{E} = 4000 x^2 \hat{i} \text{ V/m}$. The electric flux through the cube of side $20 \text{ cm}$ when placed in the electric field (as shown in the figure) is $......... \text{ V cm}$.

In a region,the intensity of an electric field is given by $\overrightarrow{E}=(2 \hat{i}+3 \hat{j}+\hat{k}) \text{ NC}^{-1}$. The electric flux through a surface of area $10 \hat{i} \text{ m}^2$ in the region is

$A$ few electric field lines for a system of two charges $Q_1$ and $Q_2$ fixed at two different points on the $x$-axis are shown in the figure. These lines suggest that:-