Acceleration of a charged particle of charge $q$ and mass $m$ moving in a uniform electric field of strength $E$ is

Two balls of charge $q_1$ and $q_2$ initially have a velocity of the same magnitude and direction. After a uniform electric field is applied for a certain time,the direction of the velocity of the first ball changes by $60^{\circ}$,and the velocity magnitude is reduced by half. The direction of the velocity of the second ball changes by $90^{\circ}$. In what proportion will the velocity of the second ball change? Determine the magnitude of the charge-to-mass ratio for the second ball if it is equal to $k_1$ for the first ball. The electrostatic interaction between the balls should be neglected.

Three particles are projected in a uniform electric field with the same velocity perpendicular to the field,as shown in the figure. Which particle has the highest charge-to-mass ratio?

In a uniform electric field of $E = 1 \times 10^{4} \text{ N C}^{-1}$,an electron is accelerated from rest. The velocity of the electron when it has travelled a distance of $2 \times 10^{-2} \text{ m}$ is nearly (given $\frac{e}{m} \text{ of electron } \approx 1.8 \times 10^{11} \text{ C kg}^{-1}$):

$A$ small point mass carrying some positive charge is released from the edge of a table. There is a uniform electric field in this region in the horizontal direction. Which of the following options correctly describes the trajectory of the mass? (Curves are drawn schematically and are not to scale).

An oil drop having a mass $4.8 \times 10^{-13} \,kg$ and charge $2.4 \times 10^{-18} \,C$ stands still between two charged horizontal plates separated by a distance of $1 \,cm$. If now the polarity of the plates is changed,the instantaneous acceleration of the drop is: $(g = 10 \,m/s^2)$ (in $\,m/s^2$)