An electrostatic field in a region is radially outward with magnitude $E = \alpha r$,where $\alpha$ is a constant and $r$ is the radial distance. The charge contained in a sphere of radius $R$ in this region (centered at the origin) is:

$A$ spherical portion has been removed from a solid sphere having a charge distributed uniformly in its volume as shown in the figure. The electric field inside the emptied space is

$(a)$ Show that the normal component of the electrostatic field has a discontinuity from one side of a charged surface to another given by $(E_2 - E_1) \cdot \hat{n} = \frac{\sigma}{\varepsilon_0}$, where $\hat{n}$ is a unit vector normal to the surface at a point and $\sigma$ is the surface charge density at that point. (The direction of $\hat{n}$ is from side $1$ to side $2$.) Hence, show that just outside a conductor, the electric field is $\frac{\sigma \hat{n}}{\varepsilon_0}$. $(b)$ Show that the tangential component of the electrostatic field is continuous from one side of a charged surface to another.

$A$ non-conducting solid sphere of radius $R$ is uniformly charged. The magnitude of the electric field due to the sphere at a distance $r$ from its centre:

Electric field intensity at points in between and outside two thin separated parallel sheets of infinite dimension with like charges of same surface charge density $(\sigma)$ are . . . . . . and . . . . . . respectively.

If the electric field intensity in a fair weather atmosphere is $100 \, V/m$,then the total charge on the earth's surface is ............ $C$ (radius of the earth is $6400 \, km$).