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

If an $\alpha$-particle and a proton are accelerated from rest by a potential difference of $1 \text{ MV}$,then the ratio of their kinetic energy will be

In the absence of gravity, a charge $q$ and mass $2m$ is placed stationary in a uniform electric field of intensity $E$. When the charge is released, its speed after $n$ seconds is . . . . . . .

$A$ charged particle of mass $0.003 \, g$ is held stationary in space by placing it in a downward electric field of $6 \times 10^4 \, N/C$. The magnitude of the charge is:

In a certain region,a uniform electric field $E$ and a magnetic field $B$ are present in opposite directions. At the instant $t = 0$,a particle of mass $m$ carrying a charge $q$ is given a velocity $v_0$ at an angle $\theta$ with the $y$-axis,in the $yz$-plane. The time after which the speed of the particle would be minimum is equal to

An electron falls from rest through a vertical distance $h$ in a uniform and vertically upward directed electric field $E$. The direction of the electric field is now reversed,keeping its magnitude the same. $A$ proton is allowed to fall from rest in it through the same vertical distance $h$. The time of fall of the electron,in comparison to the time of fall of the proton,is: