When a proton is accelerated through $1\,V$,then its kinetic energy will be.....$eV$.

$A$ particle of mass $m$ and charge $q$ is thrown perpendicular to an electric field of intensity $E$ with an initial velocity $v$. The particle moves a distance $x$ perpendicular to the field and a distance $y$ along the direction of the field. If $y = \alpha x^{2}$,then $\alpha$ is given by:

An electron of mass $m$ and charge $e$ is accelerated from rest through a potential difference $V$ in vacuum. The final speed of the electron will be

What is the magnitude of the upward electric field (in $V/m$) required to keep a coin of mass $10^{-6} \ kg$ and charge $10^{-6} \ C$ in equilibrium? (Take $g = 10 \ m/s^2$)

$A$ 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 $m/s$):

$A$ particle of mass $2 \times 10^{-5} \ kg$ and charge $4 \times 10^{-3} \ C$ starts from rest in a uniform electric field of $5 \ V/m$. What is its kinetic energy after $10 \ s$?