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 charge density of a solid sphere of radius $R$ and total charge $Q$ is given by $\rho(r) = \frac{Q}{\pi R^4} r$. The value of the electric field at a point $P$ inside the sphere at a distance $r_1$ from the center 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 $V/m$)?

An electron is projected towards a metal plate with a surface charge density $\sigma = -2 \times 10^{-6} \ C/m^2$. If the electron strikes the plate with zero velocity,calculate the distance $d$ between the electron gun and the plate in terms of its initial kinetic energy $E_k$.

This question has Statement-$1$ and Statement-$2$. Of the four choices given after the statements,choose the one that best describes the two statements.
An insulating solid sphere of radius $R$ has a uniformly positive charge density $\rho$. As a result of this uniform charge distribution,there is a finite value of electric potential at the centre of the sphere,at the surface of the sphere,and also at a point outside the sphere. The electric potential at infinity is zero.
Statement-$1$: When a charge $q$ is taken from the centre to the surface of the sphere,its potential energy changes by $\frac{q \rho R^2}{6 \epsilon_0}$.
Statement-$2$: The electric field at a distance $r (r < R)$ from the centre of the sphere is $\frac{\rho r}{3 \epsilon_0}$.

An infinite non-conducting sheet has a surface charge density $2 \times 10^{-7} \text{ C/m}^2$ on one side. The distance between two equipotential surfaces whose potential difference is $90 \text{ V}$ is (assume $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \text{ Nm}^2/\text{C}^2$):