The acceleration of an electron in an electric field of magnitude $50\, V/cm$ is (given that the $e/m$ ratio of the electron is $1.76 \times 10^{11}\, C/kg$):

$A$ 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 . . . . . . .

$A$ moving charge will gain energy due to the application of

An electron enters between two horizontal plates separated by $2\,mm$ and having a potential difference of $1000\,V$. The force on the electron is:

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 $, g$)

In a Millikan oil drop experiment,a charged drop of mass $1.8 \times 10^{-14} \ kg$ is stationary between its plates. The distance between the plates is $0.90 \ cm$ and the potential difference is $2.0 \ kV$. The number of electrons on the drop is: