$A$ particle of mass $M$ and charge $q$ is released from rest in a region of uniform electric field of magnitude $E$. After a time $t$,the distance travelled by the charge is $S$ and the kinetic energy attained by the particle is $T$. Then,the ratio $T/S$:

$A$ particle of mass $m$ and charge $q$ is released from rest in a region of uniform electric field of intensity $E$. After traveling a distance $y$,the kinetic energy of the particle is:

$A$ small sphere of mass $m = 0.5 \, kg$ carrying a positive charge $q = 110 \, \mu C$ is connected to a light,flexible,and inextensible string of length $r = 60 \, cm$ and whirled in a vertical circle. If a vertically upwards electric field of strength $E = 10^5 \, N/C$ exists in the space,what is the minimum velocity of the sphere required at the highest point so that it may just complete the circle? $(g = 10 \, m/s^2)$

$A$ particle of mass $2 \ g$ and charge $6 \ \mu C$ is accelerated from rest through a potential difference of $60 \ V$. The speed acquired by the particle is (in $ms^{-1}$)

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

$A$ uniform electric field,$\vec{E} = -400 \sqrt{3} \hat{y} \text{ NC}^{-1}$ 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 $u = 2 \sqrt{10} \times 10^6 \text{ ms}^{-1}$. This particle is aimed to hit a target $T$,which is $5 \text{ m}$ away from its entry point into the field as shown schematically in the figure. Take $\frac{q}{m} = 10^{10} \text{ Ckg}^{-1}$. 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 $\sqrt{\frac{5}{6}} \mu\text{s}$ as well as $\sqrt{\frac{5}{2}} \mu\text{s}$
$(D)$ time taken by the particle to hit $T$ is $\sqrt{\frac{5}{3}} \mu\text{s}$