Two masses $M_1$ and $M_2$ carry positive charges $Q_1$ and $Q_2$,respectively. They are dropped to the floor in a laboratory setup from the same height,where there is a constant electric field vertically upwards. $M_1$ hits the floor before $M_2$. Then,

$A$ charged bead is sliding freely through a string held vertically under tension. An electric field is applied parallel to the string so that the bead stays at rest at the middle of the string. If the electric field is switched off momentarily and switched on again,then

$A$ charged particle with charge $q$ and mass $m$ starts with an initial kinetic energy $K$ at the centre of a uniformly charged spherical region of total charge $Q$ and radius $R$. Charges $q$ and $Q$ have opposite signs. The spherically charged region is not free to move and kinetic energy $K$ is just sufficient for the charged particle to reach the boundary of the spherical charge. How much time does it take the particle to reach the boundary of the region?

An electron of mass $m_e$ initially at rest moves through a certain distance in a uniform electric field in time $t_1$. $A$ proton of mass $m_p$ also initially at rest takes time $t_2$ to move through an equal distance in this uniform electric field. Neglecting the effect of gravity,the ratio of $t_2/t_1$ is nearly equal to

$A$ charged particle,initially at rest at $O$,follows the trajectory shown alongside when released. Such a trajectory is possible in the presence of:

$A$ particle of charge $q$ and mass $m$ is subjected to an electric field $E = E_{0}(1 - ax^{2})$ in the $x$-direction,where $a$ and $E_{0}$ are constants. Initially,the particle was at rest at $x = 0$. Other than the initial position,the kinetic energy of the particle becomes zero when the distance of the particle from the origin is: