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. . . . . . .

A rectangular surface of sides $10 \,cm$ and $15 \,cm$ is placed inside acyniform electric field of $25 \,V / m$, such that the surface makes an angle of $30^{\circ}$ with the direction of electric field. Find the flux of the electric field through the rectangular surface .................. $Nm ^2 / C$

Give reason : ''If net flux assocaited with closed surface is zero, then net charge enclosed by that surface is zero''.

Draw electric field lines of simple charge distribution.

In finding the electric field using Gauss Law the formula $|\overrightarrow{\mathrm{E}}|=\frac{q_{\mathrm{enc}}}{\varepsilon_{0}|\mathrm{A}|}$ is applicable. In the formula $\varepsilon_{0}$ is permittivity of free space, $A$ is the area of Gaussian surface and $q_{enc}$ is charge enclosed by the Gaussian surface. The equation can be used in which of the following situation?

In figure $+Q$ charge is located at one of the edge of the cube, then electric flux through cube due to $+Q$ charge is