A mass $m = 20\,g$ has a charge $q = 3.0\,mC$. It moves with a velocity of $20\,m/s$ and enters a region of electric field of $80\,N/C$ in the same direction as the velocity of the mass. The velocity of the mass after $3$ seconds in this region is.......$m/s$

A particle of mass $m$ and charge $(-q)$ enters the region between the two charged plates initially moving along $x$ -axis with speed $v_{x}$ (like particle $1$ in Figure). The length of plate is $L$ and an uniform electric field $E$ is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is $q E L^{2} /\left(2 m v_{x}^{2}\right)$

Compare this motion with motion of a projectile in gravitational field

A charged particle (mass $m$ and charge $q$ ) moves along $X$ axis with velocity $V _{0}$. When it passes through the origin it enters a region having uniform electric field $\overrightarrow{ E }=- E \hat{ j }$ which extends upto $x = d$. Equation of path of electron in the region $x > d$ is 

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

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