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

Let there be a spherically symmetric charge distribution with charge density varying as $\rho (r) = \rho _0 \left( \frac{5}{4} - \frac{r}{R} \right)$ for $r \le R$,and $\rho (r) = 0$ for $r > R$,where $r$ is the distance from the origin. The electric field at a distance $r (r < R)$ from the origin is given by:

The figure shows the graph of electric field $E(r)$ versus distance $(r)$ from the center of an object. Therefore,...

$A$ solid metal sphere of radius $R$ having charge $q$ is enclosed inside a concentric spherical shell of inner radius $a$ and outer radius $b$ as shown in the figure. The approximate variation of the electric field $\overrightarrow{E}$ as a function of distance $r$ from the centre $O$ is given by:

The electric field near a conducting surface having a uniform surface charge density $\sigma$ is given by:

The electric field in a region is radially outward and at a point is given by $E = 250 r \, V/m$ (where $r$ is the distance of the point from the origin). Calculate the charge contained in a sphere of radius $20 \, cm$ centered at the origin in Coulombs $(C)$.