A particle of mass $m$ and charge $q$ is thrown in a region where uniform gravitational field and electric field are present. The path of particle

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

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

An electron is moving towards $x$-axis. An electric field is along $y$-direction then path of electron is

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 $\sqrt{\frac{5}{3}} \mu s$