An electron falls through a distance of $1.5\; cm$ in a uniform electric field of magnitude $2.0 \times 10^{4} \;N C ^{-1} \text {[Figure (a)]} .$ The direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance [Figure $(b)] .$ Compute the time of fall in each case. Contrast the situation with that of 'free fall under gravity'.

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 $\vec E$ exists between the plates of a charged condenser. A charged particle enters the space between the plates and perpendicular to $\vec E$ . The path of the particle between the plates is a

A particle of mass $m$ and charge $(-q)$ enters the region between the two charged plates initially moving along $x$ -axis with speed $v_{x}=2.0 \times 10^{6} \;m\, s ^{-1} .$ If $E$ between the plates separated by $0.5 \;cm$ is $9.1 \times 10^{2} \;N / C ,$ where will the electron strike the upper plate in $cm$?

$\left(|e|=1.6 \times 10^{-19} \;C , m_{e}=9.1 \times 10^{-31}\; kg .\right)$

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 $'t'$ second is