$\sigma$ is the uniform surface charge density of a thin spherical shell of radius $R$. The electric field at any point on the surface of the spherical shell is:

$A$ solid ball of radius $R$ has a charge density $\rho$ given by $\rho = \rho_0 \left( 1 - \frac{r}{R} \right)$ for $0 \leq r \leq R$. The electric field outside the ball is

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

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 $\frac{\rho r}{3 \epsilon_0}$.

$A$ conducting sphere of radius $0.1 \ m$ has a uniform charge density $1.8 \ \mu C/m^2$ on its surface. The electric field in free space at a radial distance of $0.2 \ m$ from the center of the sphere is $(\varepsilon_0 = \text{permittivity of free space})$

$A$ uniform rod $AB$ of mass $m$ and length $l$ is hinged at its midpoint $C$. The left half $(AC)$ of the rod has linear charge density $-\lambda$ and the right half $(CB)$ has $+\lambda$,where $\lambda$ is constant. $A$ large non-conducting sheet of uniform surface charge density $\sigma$ is also present near the rod. Initially,the rod is kept perpendicular to the sheet. The end $A$ of the rod is initially at a distance $d$. Now,the rod is rotated by a small angle $\theta$ in the plane of the paper and released. The time period of small angular oscillations is: