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

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

$A$ charged water drop whose radius is $0.1\,\mu m$ is in equilibrium in an electric field. If the charge on it is equal to the charge of an electron,then the intensity of the electric field will be.......$N/C$ $(g = 10\,m/s^2)$

$A$ charged particle of mass $5 \times 10^{-5} \ kg$ is held stationary in space by placing it in an electric field of strength $10^7 \ N C^{-1}$ directed vertically downwards. The charge on the particle is

$A$ particle of charge $q$ and mass $m$ is subjected to an electric field $E = E_{0}(1 - ax^{2})$ in the $x$-direction,where $a$ and $E_{0}$ are constants. Initially,the particle was at rest at $x = 0$. Other than the initial position,the kinetic energy of the particle becomes zero when the distance of the particle from the origin is:

$A$ particle of mass $m$ and charge $(-q)$ enters the region between two charged plates,initially moving along the $x$-axis with speed $v_{x}$ (like particle $1$ in the Figure). The length of the plate is $L$ and a uniform electric field $E$ is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is $q E L^{2} / (2 m v_{x}^{2})$. Compare this motion with the motion of a projectile in a gravitational field.