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 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

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)

A uniform electric field, $\vec{E}=-400 \sqrt{3} \hat{y} NC ^{-1}$ is applied in a region. A charged particle of mass $m$ carrying positive charge $q$ is projected in this region with an initial speed of $2 \sqrt{10} \times 10^6 ms ^{-1}$. This particle is aimed to hit a target $T$, which is $5 m$ away from its entry point into the field as shown schematically in the figure. Take $\frac{ q }{ m }=10^{10} Ckg ^{-1}$. Then-

$(A)$ the particle will hit $T$ if projected at an angle $45^{\circ}$ from the horizontal

$(B)$ the particle will hit $T$ if projected either at an angle $30^{\circ}$ or $60^{\circ}$ from the horizontal

$(C)$ time taken by the particle to hit $T$ could be $\sqrt{\frac{5}{6}} \mu s$ as well as $\sqrt{\frac{5}{2}} \mu s$

$(D)$ time taken by the particle to hit $T$ is $\sqrt{\frac{5}{3}} \mu s$

Under the influence of the Coulomb field of charge $+Q$, a charge $-q$ is moving around it in an elliptical orbit. Find out the correct statement$(s)$.