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

A square surface of side $L$ meter in the plane of the paper is placed in a uniform electric field $E(volt/m)$ acting along the same plane at an angle $\theta$ with the horizontal side of the square as shown in figure.The electric flux linked to the surface, in units of $volt \;m $

Gauss’s law should be invalid if

A charge $+q$ is placed somewhere inside the cavity of a thick conducting spherical shell of inner radius $R_1$ and outer radius $R_2$. A charge $+Q$ is placed at a distance $r > R_2$ from the centre of the shell. Then the electric field in the hollow cavity

Consider a uniform electric field $E =3 \times 10^{3} i\; N / C .$

$(a)$ What is the flux of this field through a square of $10 \;cm$ on a side whose plane is parallel to the $y z$ plane?

$(b)$ What is the flux through the same square if the normal to its plane makes a $60^{\circ}$ angle with the $x -$axis?

A charged body has an electric flux $\phi$ associated with it. The body is now placed inside a metallic container. The flux $\phi$, outside the container will be