An electron falls through a small distance in a uniform electric field of magnitude $2 \times {10^4}N{C^{ - 1}}$. The direction of the field is reversed keeping the magnitude unchanged and a proton falls through the same distance. The time of fall will be

An electron of mass ${m_e}$ initially at rest moves through a certain distance in a uniform electric field in time ${t_1}$. A proton of mass ${m_p}$ also initially at rest takes time ${t_2}$ to move through an equal distance in this uniform electric field. Neglecting the effect of gravity, the ratio of ${t_2}/{t_1}$ is nearly equal to

A charged particle of mass $m$ and charge $q$ is released from rest in a uniform electric field $E.$ Neglecting the effect of gravity, the kinetic energy of the charged particle after ‘$t$’ second is

An electron is released from the bottom plate $A$ as shown in the figure $(E = 10^4\, N/C)$. The velocity of the electron when it reaches plate $B$ will be nearly equal to

An electron having charge ‘$e$’ and mass ‘$m$’ is moving in a uniform electric field $E$. Its acceleration will be

Under the influence of the Coulomb field of charge $+Q$, a charge $-q$ is moving around it in an elliptical orbit. Find out the correct statement$(s)$.