$A$ charge of magnitude $3e$ and mass $2m$ is moving in an electric field $\overrightarrow{E}$. The acceleration imparted to the charge is:

$A$ particle of mass $m$ and charge $(-q)$ enters the region between two charged plates,initially moving along the $x$-axis with speed $v_{x}$ (like particle $1$ in the Figure). The length of the plate is $L$ and a uniform electric field $E$ is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is $q E L^{2} / (2 m v_{x}^{2})$. Compare this motion with the motion of a projectile in a gravitational field.

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

As shown in the figure, an electron is projected with kinetic energy $K$ at an angle of $\theta = 45^\circ$ between two charged plates. The electron strikes the upper plate. The value of the electric field $E$ must be greater than:

$A$ positive charge particle of $100 \,mg$ is thrown in the opposite direction to a uniform electric field of strength $1 \times 10^{5} \,NC^{-1}$. If the charge on the particle is $40 \,\mu C$ and the initial velocity is $200 \,ms^{-1}$,how much distance (in $m$) will it travel before coming to rest momentarily?

An $\alpha$-particle is accelerated through a potential difference of $200\,V$. The increase in its kinetic energy is.......$eV$.