$A$ cube of side $a$ is placed in an electric field $\vec{E} = E_0 x \hat{i}$. What is the total charge enclosed by the cube?

$A$ particle of charge $-q$ and mass $m$ moves in a circle of radius $r$ around an infinitely long line charge of linear density $+\lambda$. Then the time period will be given as (Consider $k$ as Coulomb's constant).

$A$ spherically symmetric charge distribution is characterised by a charge density having the following variations:
$\rho (r) = \rho_0 \left( 1 - \frac{r}{R} \right)$ for $r < R$
$\rho (r) = 0$ for $r \ge R$
Where $r$ is the distance from the centre of the charge distribution and $\rho_0$ is a constant. The electric field at an internal point $(r < R)$ is:

$A$ hollow spherical shell of radius $r$ has a uniform charge density $\sigma$. It is kept in a cube of edge $3r$ such that the centres of the cube and the shell coincide. Then the electric flux coming out of one face of a cube is ($\varepsilon_0$ - permittivity of free space).

At a distance $l$ from a uniformly charged long wire, a charged particle is thrown radially outward with a velocity $u$ in the direction perpendicular to the wire. When the particle reaches a distance $2l$ from the wire, its speed is found to be $\sqrt{2}u$. The magnitude of the velocity, when it is a distance $4l$ away from the wire is (ignore gravity)

$A$ 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 $\rho r / 3\varepsilon_0$.