Charges are uniformly spread on the surface of a conducting sphere. The electric field from the centre of the sphere to a point outside the sphere varies with distance $r$ from the centre as:

$A$ hollow insulated conducting sphere is given a positive charge of $10\,\mu C$. What will be the electric field at the centre of the sphere if its radius is $2\,m$?

$A$ non-conducting solid sphere of radius $R$ has a uniform volume charge density $\rho$. The electric potential at the center of the sphere is related to the potential at the surface and outside the sphere due to this uniform charge distribution.
Statement-$1$: When a charge $q$ is moved from the surface to the center of the sphere,the change in its potential energy is $q\rho R^2 / 6\varepsilon_0$.
Statement-$2$: The electric field at a distance $r$ $(r < R)$ from the center of the sphere is $\rho r / 3\varepsilon_0$.

If an insulated non-conducting sphere of radius $R$ has charge density $\rho$,the electric field at a distance $r$ from the centre of the sphere $(r < R)$ will be

An infinite nonconducting sheet of charge has a surface charge density of $10^{-7} \ C/m^2$. The separation between two equipotential surfaces near the sheet whose potential differ by $5 \ V$ is:

Consider a uniform spherical charge distribution of radius $R_1$ centred at the origin $O$. In this distribution,a spherical cavity of radius $R_2$,centred at $P$ with distance $OP = a = R_1 - R_2$ (see figure) is made. If the electric field inside the cavity at position $\vec{r}$ is $\vec{E}(\vec{r})$,then the correct statement$(s)$ is(are):