Two infinite plane parallel sheets separated by a distance $d$ have equal and opposite uniform charge densities $\sigma $. Electric field at a point between the sheets is

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

The flat base of a hemisphere of radius $a$ with no charge inside it lies in a horizontal plane. A uniform electric field $\vec E$ is applied at an angle $\frac {\pi }{4}$ with the vertical direction. The electric flux through the curved surface of the hemisphere is

A charge is kept at the central point $P$ of a cylindrical region. The two edges subtend a half-angle $\theta$ at $P$, as shown in the figure. When $\theta=30^{\circ}$, then the electric flux through the curved surface of the cylinder is $\Phi$ If $\theta=60^{\circ}$, then the electric flux through the curved surface becomes $\Phi / \sqrt{n}$, where the value of $n$ is. . . . . . .

Five charges $+q,+5 q,-2 q,+3 q$ and $-4 q$ are situated as shown in the figure.

The electric flux due to this configuration through the surface $S$ is