An $\alpha$-particle of mass $6.4 \times 10^{-27} \ kg$ and charge $3.2 \times 10^{-19} \ C$ is situated in a uniform electric field of $1.6 \times 10^{5} \ Vm^{-1}$. The velocity of the particle at the end of $2 \times 10^{-2} \ m$ path when it starts from rest is

$A$ vertical electric field of magnitude $4.9 \times 10^{5} \, N/C$ just prevents a water droplet of mass $0.1 \, g$ from falling. The value of charge on the droplet will be ........ $\times 10^{-9} \, C$ (Given $g = 9.8 \, m/s^{2}$)

$A$ particle of mass $m$ and charge $(-q)$ enters the region between two charged plates,initially moving along the $x$-axis with a speed $v_{x} = 2.0 \times 10^{6} \; m \, s^{-1}$. If the electric field $E$ between the plates,which are separated by $0.5 \; cm$,is $9.1 \times 10^{2} \; N/C$,at what distance along the $x$-axis will the electron strike the upper plate (in $cm$)?
$(|e| = 1.6 \times 10^{-19} \; C, m_{e} = 9.1 \times 10^{-31} \; kg)$

Select the correct statement: (Only force on a particle is due to electric field)

When an electron enters an electric field with a velocity in the direction of the electric field lines,then.......

$A$ charged particle of mass $m$ and charge $q$ is released from rest in a uniform electric field $E$. Neglecting the effect of gravity,the kinetic energy of the charged particle after $t$ seconds is