In a region of uniform electric field of intensity $E$,an electron of mass $m_e$ is released from rest. The distance travelled by the electron in a time $t$ is

$A$ particle of mass $0.5 \ g$ and charge $10 \ \mu C$ is subjected to a uniform electric field of $8 \ NC^{-1}$. If the particle is initially at rest,the velocity of the particle after a time of $5 \ s$ is: (in $ms^{-1}$)

Two charges $-q$ each are fixed and separated by a distance $2d$. $A$ third charge $q$ of mass $m$ placed at the midpoint is displaced slightly by $x$ $(x \ll d)$ perpendicular to the line joining the two fixed charges as shown in the figure. Show that the charge $q$ will perform simple harmonic motion with a time period $T = \left[\frac{8 \pi^{3} \epsilon_{0} m d^{3}}{q^{2}}\right]^{1 / 2}$.

$A$ small sphere of mass $m$ carrying a charge $q$ is hanging between two parallel plates by a string of length $L$ as shown in the figure. The time period of the pendulum is $T_0$. When the parallel plates are charged as shown,the time period changes to $T$. The ratio $T / T_0$ is equal to:

$A$ proton and an $\alpha$-particle are both accelerated from rest in a uniform electric field. The ratio of work done by the electric field on the proton and the $\alpha$-particle in a given time is

$A$ 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 $ms^{-1}$)