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 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'.

In an ink-jet printer, an ink droplet of mass $m$ is given a negative charge $q$ by a computer-controlled charging unit, and then enters at speed $v$ in the region between two deflecting parallel plates of length $L$ separated by distance $d$ (see figure below). All over this region exists a downward electric field which you can assume to be uniform. Neglecting the gravitational force on the droplet, the maximum charge that can be given so that it will not hit a plate is close to :

A body having specific charge $8\,\mu {C} / {g}$ is resting on a frictionless plane at a distance $10\, {cm}$ from the wall (as shown in the figure). It starts moving towards the wall when a uniform electric field of $100 \,{V} / {m}$ is applied horizontally toward the wall. If the collision of the body with the wall is perfectly elastic, then the time period of the motion will be $....\, S.$

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$(s)$.