Two charged particles of masses $m$ and $2m$ have charges $+2q$ and $+q$ respectively. They are kept in a uniform electric field and allowed to move for the same time. The ratio of their kinetic energies will be

$A$ proton and an $\alpha$-particle start from rest in a uniform electric field. The ratio of times taken by them to travel the same distance in the field is

$A$ vertical electric field of magnitude $4.9 \times 10^{5} \, N/C$ just prevents a water droplet of mass $0.1 \, g$ from falling. The value of charge on the droplet will be ........ $\times 10^{-9} \, C$ (Given $g = 9.8 \, m/s^{2}$)

$A$ simple pendulum has a bob with mass $m$ and charge $q$. The pendulum string has negligible mass. When a uniform and horizontal electric field $\vec{E}$ is applied,the final tension in the string changes. The final tension in the string,when the pendulum attains an equilibrium position,is . . . . . . . ($g$: acceleration due to gravity)

As shown in the figure below,a point charge $q$ moves from point $P$ to a point $S$ traversing a path $PQRS$ in a uniform electric field $\vec{E}$. The electric field is directed along a direction parallel to the $x$-axis. The coordinates of $P$,$Q$,$R$,and $S$ are $(a, b, 0)$,$(2a, 0, 0)$,$(a, -b, 0)$,and $(0, 0, 0)$ respectively. What is the work done by the field in the process?

An electron falls through a small distance in a uniform electric field of magnitude $2 \times 10^4 \ N C^{-1}$. The direction of the field is reversed keeping the magnitude unchanged and a proton falls through the same distance. The time of fall will be