$A$ charged particle of mass $m$ and charge $q$ is released from rest in a uniform electric field $E$. Neglecting the effect of gravity,the kinetic energy of the charged particle after $t$ seconds is

An electron falls through a distance of $1.5 \, cm$ in a uniform electric field of magnitude $2.0 \times 10^4 \, N/C$ as shown in the figure. The time taken by the electron to fall through this distance is ($m_e = 9.1 \times 10^{-31} \, kg$,neglect gravity).

In Millikan's oil drop experiment,a charged particle is held in equilibrium between two plates by an electric field. If the direction of the electric field is reversed,calculate the acceleration of the charged particle. (in $, g$)

$A$ proton is $1840$ times heavier than an electron. When it is accelerated through a potential difference of $1 \ kV$,what will be its kinetic energy in $keV$?

In a certain region,a uniform electric field $E$ and a magnetic field $B$ are present in opposite directions. At the instant $t = 0$,a particle of mass $m$ carrying a charge $q$ is given a velocity $v_0$ at an angle $\theta$ with the $y$-axis,in the $yz$-plane. The time after which the speed of the particle would be minimum is equal to

$A$ particle of mass $0.2 \ g$ and charge $2 \ C$ is released from rest in a uniform electric field of $20 \ N \ C^{-1}$. The kinetic energy of the particle after moving a distance of $20 \ cm$ is (in $J$)