$A$ closed surface passes through a spherical conductor as shown in the figure. If a negative charge $-Q$ is placed at point $P$,what will be the nature of the electric flux coming out of the closed surface?

Gauss's law is given by ${\epsilon _0}\oint {\vec E \cdot d\vec s} = q$. If the net charge enclosed by a Gaussian surface is zero,then .......

If the radius of the spherical Gaussian surface is increased, then the electric flux due to a point charge enclosed by the surface

$A$ 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

$A$ 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. . . . . . .

The electric flux through surface $S_1$ is: