Acceleration of a charged particle of charge $q$ and mass $m$ moving in a uniform electric field of strength $E$ is

$A$ problem of practical interest is to make a beam of electrons turn at a $90^{\circ}$ corner. This can be done with the electric field present between the parallel plates as shown in the figure. An electron with kinetic energy $8.0 \times 10^{-17} \ J$ enters through a small hole in the bottom plate. The strength of the electric field that is needed if the electron is to emerge from an exit hole $1.0 \ cm$ away from the entrance hole,traveling at right angles to its original direction,is $y \times 10^5 \ N/C$. The value of $y$ is

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

The acceleration of an electron in an electric field of magnitude $50\, V/cm$ is (given that the $e/m$ ratio of the electron is $1.76 \times 10^{11}\, C/kg$):

In an ink-jet printer,an ink droplet of mass $m$ is given a negative charge $q$ by a computer-controlled charging unit,and then enters at speed $v$ in the region between two deflecting parallel plates of length $L$ separated by distance $d$ (see figure below). All over this region exists a downward electric field $E$ which you can assume to be uniform. Neglecting the gravitational force on the droplet,the maximum charge that can be given so that it will not hit a plate is close to :

If a proton is moved against the Coulomb force of an electric field,then