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
$1$
${({m_p}/{m_e})^{1/2}}$
${({m_e}/{m_p})^{1/2}}$
$1836$
A proton and an $\alpha$-particle having equal kinetic energy are projected in a uniform transverse electric field as shown in figure
An electron enters in an electric field with its velocity in the direction of the electric lines of force. Then
A simple pendulum is suspended in a lift which is going up with an acceleration $5\ m/s^2$. An electric field of magnitude $5 \ N/C$ and directed vertically upward is also present in the lift. The charge of the bob is $1\ mC$ and mass is $1\ mg$. Taking $g = \pi^2$ and length of the simple pendulum $1\ m$, the time period of the simple pendulum is ......$s$
An electron falls through a distance of $1.5\, cm$ in a uniform electric field of magnitude $2.0\times10^4\, N/C$ as shown in the figure. The time taken by electron to fall through this distance is ($m_e = 9.1\times10^{-31}\,kg$, Neglect gravity)
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