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 $, g$)

$A$ particle of mass '$m$' and charge '$q$',initially at rest, is accelerated by a uniform electric field '$E$' through a distance '$D$' and is then allowed to approach a fixed static charge '$Q$' of the same sign. The distance of the closest approach of the charge '$q$' is [ $\epsilon_0$ = permittivity of free space ]

$A$ 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 $J$)

$A$ charged cork ball having mass $1 \text{ g}$ and charge $q$ is suspended on a light string in a uniform electric field as shown in the figure. The ball is in equilibrium at $\theta=37^{\circ}$,when the value of the electric field is $E=(3 \hat{i}+5 \hat{j}) \times 10^5 \text{ NC}^{-1}$. (Assume $T$ as tension in the string.) Which of the following options are correct? (Given,$\sin 37^{\circ}=0.60$ and $g=10 \text{ ms}^{-2}$)

$A$ proton is fired at an initial velocity of $150 \, m/s$ at an angle of $60^\circ$ above the horizontal into a uniform electric field of $2 \times 10^{-4} \, N/C$ between two charged parallel plates as shown in the figure. The total time the particle is in motion is:

In a uniform electric field,if a charge is fired in a direction different from the line of the electric field,then the trajectory of the charge will be a