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

Two masses $M_1$ and $M_2$ carry positive charges $Q_1$ and $Q_2$,respectively. They are dropped to the floor in a laboratory setup from the same height,where there is a constant electric field vertically upwards. $M_1$ hits the floor before $M_2$. Then,

An electron is accelerated through a potential difference of $200 \ V$. If $e/m$ for the electron is $1.6 \times 10^{11} \ C/kg$,the velocity acquired by the electron will be:

$A$ uniform electric field of $10^3 \, V/m$ is directed along the $y$-axis. $A$ particle of mass $1 \, g$ and charge $10^{-6} \, C$ is projected from the origin into the field with a velocity of $10 \, m/s$ along the positive $x$-axis. Its speed after $10 \, s$ is (in $m/s$):

In a Millikan oil drop experiment,a charged drop of mass $1.8 \times 10^{-14} \ kg$ is stationary between its plates. The distance between the plates is $0.90 \ cm$ and the potential difference is $2.0 \ kV$. The number of electrons on the drop is:

$A$ 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?