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 ........ $mm$.

$A$ particle of mass $m$ and charge $q$ is released from rest in a uniform electric field. If there is no other force on the particle,the dependence of its speed $v$ on the distance $x$ travelled by it is correctly given by (graphs are schematic and not drawn to scale)

$A$ particle of mass $m$ and charge $(-q)$ enters the region between two charged plates,initially moving along the $x$-axis with a speed $v_{x} = 2.0 \times 10^{6} \; m \, s^{-1}$. If the electric field $E$ between the plates,which are separated by $0.5 \; cm$,is $9.1 \times 10^{2} \; N/C$,at what distance along the $x$-axis will the electron strike the upper plate (in $cm$)?
$(|e| = 1.6 \times 10^{-19} \; C, m_{e} = 9.1 \times 10^{-31} \; kg)$

The kinetic energy of an electron which is accelerated through a potential of $100 \, V$ is:

$A$ charge of magnitude $3e$ and mass $2m$ is moving in an electric field $\overrightarrow{E}$. The acceleration imparted to the charge is:

An electron is made to enter symmetrically between two parallel and equally but oppositely charged metal plates,each of $10 \ cm$ length. The electron emerges out of the field region with a horizontal component of velocity $10^6 \ m/s$. If the magnitude of the electric field between the plates is $9.1 \ V/cm$,then the vertical component of velocity of the electron is (mass of electron $= 9.1 \times 10^{-31} \ kg$ and charge of electron $= 1.6 \times 10^{-19} \ C$)