$A$ charge is uniformly distributed on the surface of a spherical rubber balloon. As it is blown up,the total electric flux coming out of the surface

$A$ circular disc of radius $R$ carries surface charge density $\sigma(r) = \sigma_0 \left(1 - \frac{r}{R}\right)$,where $\sigma_0$ is a constant and $r$ is the distance from the center of the disc. Electric flux through a large spherical surface that encloses the charged disc completely is $\phi_0$. Electric flux through another spherical surface of radius $\frac{R}{4}$ and concentric with the disc is $\phi$. Then the ratio $\frac{\phi_0}{\phi}$ is:

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

The linear charge density of a wire is $8.85\,\mu C/m$. The radius and height of the cylinder are $3\,m$ and $4\,m$ respectively. Find the electric flux passing through the cylinder.

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 $Q$ is situated at the corner of a cube. The electric flux passing through all the six faces of the cube is: