If a particle of mass $10 \ mg$ and charge $2 \ \mu C$ at rest is subjected to a uniform electric field of potential difference $160 \ V$,then the velocity acquired by the particle is (in $ms^{-1}$)

In an electron gun,the electrons are accelerated by the potential $V$. If $e$ is the charge and $m$ is the mass of an electron,then the maximum velocity of these electrons will be

Two metal plates having a potential difference of $800 \; V$ are $0.02 \; m$ apart horizontally. $A$ particle of mass $1.96 \times 10^{-15} \; kg$ is suspended in equilibrium between the plates. If $e$ is the elementary charge,then the charge on the particle is:

There is a uniform electric field of strength $10^3 \ Vm^{-1}$ along the $Y$-axis. $A$ body of mass $1 \ g$ and charge $10^{-6} \ C$ is projected into the field from the origin along the positive $X$-axis with a velocity of $10 \ ms^{-1}$. Its speed in $ms^{-1}$ after $10 \ s$ is (Neglect gravitation).

$A$ 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?

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,