A particle of mass $1\ gm$ and charge $ – 0.1\,\mu C$ is projected from ground with a velocity $10\sqrt 2 $ at an $45^o$ with horizontal in the area having uniform electric field $1\ kV/cm$ in horizontal direction. Acceleration due to gravity is $10\ m/s^2$ in vertical downward direction. Select $INCORRECT$ statement

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 positive charge particle of $100 \,mg$ is thrown in opposite direction to a uniform electric field of strength $1 \times 10^{5} \,NC ^{-1}$. If the charge on the particle is $40 \,\mu C$ and the initial velocity is $200 \,ms ^{-1}$, how much distance (in $m$) it will travel before coming to the rest momentarily

An electron falls from rest through a vertical distance $h$ in a uniform and vertically upward directed electric field $E.$ The direction of electric field is now reversed, keeping its magnitude the same. A proton is allowed to fall from rest in it through the same vertical distance $h.$ The time of fall of the electron, in comparison to the time of fall of the proton is