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

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 positively charged particle moving along $x$-axis with a certain velocity enters a uniform electric field directed along positive $y$-axis. Its

An electron is released from the bottom plate $A$ as shown in the figure $(E = 10^4\, N/C)$. The velocity of the electron when it reaches plate $B$ will be nearly equal to

A proton sits at coordinates $(x, y) = (0, 0)$, and an electron at $(d, h)$, where $d >> h$. At time $t = 0$, $a$ uniform electric field $E$ of unknown magnitude but pointing in the positive $y$ direction is turned on. Assuming that $d$ is large enough that the proton-electron interaction is negligible, the $y$ coordinates of the two particles will be equal (at equal time)