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 

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 electric line of force in $X$, $Y-$ plane is given by $x^2+y^2 = 1$. A particle with unit positive charge, initially at rest at the point $x = 1, y = 0$ in the $X, Y-$ plane

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'.

The surface of a planet is found to be uniformly charged. When a particle of mass $m$ and no charge is thrown at an angle from the surface of the planet, it has a parabolic trajectory as in projectile motion with horizontal range $L$. A particle of mass $m$ and charge $q$, with the same initial conditions has a range $L / 2$. The range of particle of mass $m$ and charge $2 q$, with the same initial conditions is