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
$\frac{{E{q^2}m}}{{2t}}$
$\frac{{2{E^2}{t^2}}}{{mq}}$
$\frac{{{E^2}{q^2}{t^2}}}{{2m}}$
$\frac{{Eqm}}{t}$
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
An electron having charge ‘$e$’ and mass ‘$m$’ is moving in a uniform electric field $E$. Its acceleration will be
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 uniform electric field $E =(8\,m / e ) V / m$ is created between two parallel plates of length $1 m$ as shown in figure, (where $m =$ mass of electron and $e=$ charge of electron). An electron enters the field symmetrically between the plates with a speed of $2\,m / s$. The angle of the deviation $(\theta)$ of the path of the electron as it comes out of the field will be........
A particle of mass $m$ and charge $(-q)$ enters the region between the two charged plates initially moving along $x$ -axis with speed $v_{x}$ (like particle $1$ in Figure). The length of plate is $L$ and an uniform electric field $E$ is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is $q E L^{2} /\left(2 m v_{x}^{2}\right)$
Compare this motion with motion of a projectile in gravitational field