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

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 :

If the electric field intensity in a fair weather atmosphere is $100 \,V / m$, then the total charge on the earth's surface is ............ $C$ (radius of the earth is $6400\,km$ )

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

$(A)$ $\left|Q_1\right|>\left|Q_2\right|$

$(B)$ $\left|Q_1\right|<\left|Q_2\right|$

$(C)$ at a finite distance to the left of $\mathrm{Q}_1$ the electric field is zero

$(D)$ at a finite distance to the right of $\mathrm{Q}_2$ the electric field is zero

Electric flux through surface $s_1$