An electron having charge '$e$' and mass '$m$' is moving in a uniform electric field $E$. Its acceleration will be

In Millikan's oil drop experiment,a charged particle is held in equilibrium between two plates by an electric field. If the direction of the electric field is reversed,calculate the acceleration of the charged particle. (in $, g$)

$A$ particle of mass $m$ and charge $q$ is fastened to one end $A$ of a massless string having equilibrium length $\ell$,whose other end is fixed at point $O$. The whole system is placed on a frictionless horizontal plane and is initially at rest. If a uniform electric field $E$ is switched on along the $x$-direction as shown in the figure,then the speed of the particle when it crosses the $x$-axis is

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

$A$ charged bead is sliding freely through a string held vertically under tension. An electric field is applied parallel to the string so that the bead stays at rest at the middle of the string. If the electric field is switched off momentarily and switched on again,then

$A$ 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 $T/S$: