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

$A$ particle of mass $m$ and carrying charge $-q_1$ is moving around a charge $+q_2$ along a circular path of radius $r$. Find the period of revolution of the charge $-q_1$.

$A$ charge of magnitude $3e$ and mass $2m$ is moving in an electric field $\overrightarrow{E}$. The acceleration imparted to the charge is:

$A$ simple pendulum of length $L$ is placed between the plates of a parallel plate capacitor having an electric field $E,$ as shown in the figure. Its bob has mass $m$ and charge $q.$ The time period of the pendulum is given by:

$A$ particle of mass $m$ and charge $(-q)$ enters the region between two charged plates,initially moving along the $x$-axis with a speed $v_{x} = 2.0 \times 10^{6} \; m \, s^{-1}$. If the electric field $E$ between the plates,which are separated by $0.5 \; cm$,is $9.1 \times 10^{2} \; N/C$,at what distance along the $x$-axis will the electron strike the upper plate (in $cm$)?
$(|e| = 1.6 \times 10^{-19} \; C, m_{e} = 9.1 \times 10^{-31} \; kg)$

In a uniform electric field of $E = 1 \times 10^{4} \text{ N C}^{-1}$,an electron is accelerated from rest. The velocity of the electron when it has travelled a distance of $2 \times 10^{-2} \text{ m}$ is nearly (given $\frac{e}{m} \text{ of electron } \approx 1.8 \times 10^{11} \text{ C kg}^{-1}$):