An electron is moving towards $x$-axis. An electric field is along $y$-direction then path of electron is

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 $cm$?

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

An electron and a proton are in a uniform electric field, the ratio of their accelerations will be

Under the influence of the Coulomb field of charge $+Q$, a charge $-q$ is moving around it in an elliptical orbit. Find out the correct statement$(s)$.

A uniform electric field of $10\,N / C$ is created between two parallel charged plates (as shown in figure). An electron enters the field symmetrically between the plates with a kinetic energy $0.5\,eV$. The length of each plate is $10\,cm$. The angle $(\theta)$ of deviation of the path of electron as it comes out of the field is  $.........$(in degree).

A stream of a positively charged particles having $\frac{ q }{ m }=2 \times 10^{11} \frac{ C }{ kg }$ and velocity $\overrightarrow{ v }_0=3 \times 10^7 \hat{ i ~ m} / s$ is deflected by an electric field $1.8 \hat{ j } kV / m$. The electric field exists in a region of $10 cm$ along $x$ direction. Due to the electric field, the deflection of the charge particles in the $y$ direction is $...........mm$