$A$ positive charge $Q$ is placed on a conducting spherical shell with inner radius $R_1$ and outer radius $R_2$. $A$ particle with charge $q$ is placed at the center of the spherical cavity. The magnitude of the electric field at a point in the cavity,at a distance $r$ from the center,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.

The electric field due to a uniformly charged sphere of radius $R$ as a function of the distance $r$ from its centre is represented graphically by

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

The electric field at a distance of $20 \ cm$ from the center of a uniformly charged dielectric sphere of radius $R = 10 \ cm$ is $100 \ V/m$. What will be the electric field $E$ at a distance of $3 \ cm$ from the center of the sphere (in $V/m$)?

Mention applications of Gauss's law.