A cone of base radius $R$ and height $h$ is located in a uniform electric field $\vec E$ parallel to its base. The electric flux entering the cone is

Shown below is a distribution of charges. The flux of electric field due to these charges through the surface $S$ is

The electric field in a region is radially outward with magnitude $E = A{\gamma _0}$. The charge contained in a sphere of radius ${\gamma _0}$ centered at the origin is

A positive charge $q$ is kept at the center of a thick shell of inner radius $R_1$ and outer radius $R_2$ which is made up of conducting material. If $\phi_1$ is flux through closed gaussian surface $S_1$ whose radius is just less than $R_1$ and $\phi_2$ is flux through closed gaussian surface $S_2$ whose radius is just greater than $R_1$ then:-

Gauss's law can help in easy calculation of electric field due to

If atmospheric electric field is approximately $150 \,volt / m$ and radius of the earth is $6400 \,km$, then the total charge on the earth's surface is .......... coulomb