In the following figure,two parallel metallic plates are maintained at different potentials. If an electron is released midway between the plates,it will move:

The acceleration of an electron in an electric field of magnitude $50\, V/cm$ is (given that the $e/m$ ratio of the electron is $1.76 \times 10^{11}\, C/kg$):

An electron enters an electric field of intensity $3200 \ V/m$ perpendicular to the field with a velocity of $4 \times 10^7 \ m/s$. If it travels a horizontal distance of $0.10 \ m$,then its deflection from its path is ........ $mm$.

An electron falls through a distance of $1.5 \; cm$ in a uniform electric field of magnitude $2.0 \times 10^{4} \; N C^{-1}$ [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'.

$A$ particle of mass $m$ and charge $q$ is released from rest in a region of uniform electric field of intensity $E$. After traveling a distance $y$,the kinetic energy of the particle is:

Two metal plates having a potential difference of $800 \; V$ are $0.02 \; m$ apart horizontally. $A$ particle of mass $1.96 \times 10^{-15} \; kg$ is suspended in equilibrium between the plates. If $e$ is the elementary charge,then the charge on the particle is: