There is a uniform electric field of strength ${10^3}\,V/m$ along $y$-axis. A body of mass $1\,g$ and charge $10^{-6}\,C$ is projected into the field from origin along the positive $x$-axis with a velocity $10\,m/s$. Its speed in $m/s$ after $10\,s$ is (Neglect gravitation)

An electron falls through a distance of $1.5\; cm$ in a uniform electric field of magnitude $2.0 \times 10^{4} \;N C ^{-1} \text {[Figure (a)]} .$ The direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance [Figure $(b)] .$ Compute the time of fall in each case. Contrast the situation with that of 'free fall under gravity'.

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 inclined plane making an angle of $30^{\circ}$ with the horizontal is placed in a uniform horizontal electric field $200 \, \frac{ N }{ C }$ as shown in the figure. A body of mass $1\, kg$ and charge $5\, mC$ is allowed to slide down from rest at a height of $1\, m$. If the coefficient of friction is $0.2,$ find the time (in $s$ )taken by the body to reach the bottom. $\left[ g =9.8 \,m / s ^{2}, \sin 30^{\circ}=\frac{1}{2}\right.$; $\left.\cos 30^{\circ}=\frac{\sqrt{3}}{2}\right]$

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 particle of charge $1\  \mu C\  \&\  mass$ $1\  gm$ moving with a velocity of $4\  m/s$ is subjected to a uniform electric field of magnitude $300\  V/m$ for $10\  sec$. Then it's final speed cannot be.......$m/s$