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

$A$ particle having mass $m$ and charge $q$ is at rest. On applying a uniform electric field $E$ on it,it starts moving. When this particle travels a distance $x$ in the direction of the force,its kinetic energy will be . . . . . . .

The kinetic energy of an electron which is accelerated through a potential of $100 \, V$ is:

$A$ charged particle,initially at rest at $O$,follows the trajectory shown alongside when released. Such a trajectory is possible in the presence of:

$A$ proton is located at coordinates $(x, y) = (0, 0)$,and an electron is at $(d, h)$,where $d >> h$. At time $t = 0$,a uniform electric field $E$ of unknown magnitude but pointing in the positive $y$ direction is turned on. Assuming that $d$ is large enough that the proton-electron interaction is negligible,at what $y$ coordinate will the two particles have the same $y$ position at the same time?

$A$ 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 $q$ will then be