An alpha particle and a proton are accelerated from rest in a uniform electric field. The ratio of the times taken by proton and alpha particle to attain equal displacements is

An oil drop having charge $2e$ is kept stationary between two parallel horizontal plates $2.0 \, cm$ apart when a potential difference of $12000 \, V$ is applied between them. If the density of oil is $900 \, kg/m^3$,the radius of the drop will be

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 projected from origin with an initial velocity $\vec{v} = (\frac{v_0}{\sqrt{2}}\hat{i} + \frac{v_0}{\sqrt{2}}\hat{j})$. There exists a uniform magnetic field $\vec{B} = B_0\hat{z}$ and a space varying electric field $\vec{E} = E_0 e^{-\lambda x}\hat{x}$ within the region $0 \leq x \leq L$. After travelling a distance such that $x$-coordinate has changed from $x = 0$ to $x = L$,the change in the kinetic energy 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 $(g = 10\, m/s^2)$.

In Millikan's oil drop experiment,a charged particle is held in equilibrium between two plates by an electric field. If the direction of the electric field is reversed,calculate the acceleration of the charged particle. (in $, g$)