$A$ charge of magnitude $2e$ and mass $4m$ is moving in an electric field $E$. The acceleration imparted to the above charge is

$A$ particle of mass '$m$' and charge '$q$',initially at rest, is accelerated by a uniform electric field '$E$' through a distance '$D$' and is then allowed to approach a fixed static charge '$Q$' of the same sign. The distance of the closest approach of the charge '$q$' is [ $\epsilon_0$ = permittivity of free space ]

If an electron and an $\alpha$-particle are accelerated under a potential difference of $100 \, V$, what is the ratio of their momenta?

An electron of mass $m$,charge $e$ falls through a distance $h$ meters in a uniform electric field $E$. Then the time of fall is:

The kinetic energy of an electron which is accelerated through a potential of $100 \, V$ is:

An oil drop having a mass $4.8 \times 10^{-13} \,kg$ and charge $2.4 \times 10^{-18} \,C$ stands still between two charged horizontal plates separated by a distance of $1 \,cm$. If now the polarity of the plates is changed,the instantaneous acceleration of the drop is: $(g = 10 \,m/s^2)$ (in $\,m/s^2$)