$A$ charge of $1 \,C$ is located at the centre of a sphere of radius $10 \,cm$ and a cube of side $20 \,cm$. The ratio of outgoing flux from the sphere and cube will be

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 $x = a$.

It is not convenient to use a spherical Gaussian surface to find the electric field due to an electric dipole using Gauss's theorem because

$A$ charge $q$ is placed at the centre of the open end of a cylindrical vessel. The flux of the electric field through the surface of the vessel is:

Consider a uniform electric field $E = 3 \times 10^{3} \hat{i} \; N/C$.
$(a)$ What is the flux of this field through a square of $10 \; cm$ on a side whose plane is parallel to the $yz$ 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?

$A$ cylinder of radius $R$ and length $L$ is placed in a uniform electric field $E$ parallel to the cylinder axis. The total flux for the surface of the cylinder is given by-