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

$A$ 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:

$A$ charged particle is suspended in equilibrium in a uniform vertical electric field of intensity $20,000\, V/m$. If the mass of the particle is $9.6 \times 10^{-16}\, kg$,the charge on it and the excess number of electrons on the particle are respectively $(g = 10\, m/s^2)$.

Kinetic energy of an electron accelerated in a potential difference of $100 \, V$ is

An $\alpha$-particle is accelerated through a potential difference of $200\,V$. The increase in its kinetic energy is.......$eV$.

An electric field due to a positively charged long straight wire at a distance $r$ from it is proportional to $r^{-1}$ in magnitude. Two electrons are orbiting such a long straight wire in circular orbits of radii $1 \mathring{A}$ and $2 \mathring{A}$. The ratio of their respective time periods is