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

As shown in the following figure,an electron falls through a distance of $1.5 \ cm$ in a uniform electric field of magnitude $2.0 \times 10^4 \ NC^{-1}$. Find the acceleration of the electron due to the electric field.

$A$ charged bead is sliding freely through a string held vertically under tension. An electric field is applied parallel to the string so that the bead stays at rest at the middle of the string. If the electric field is switched off momentarily and switched on again,then

What is the magnitude of the upward electric field (in $V/m$) required to keep a coin of mass $10^{-6} \ kg$ and charge $10^{-6} \ C$ in equilibrium? (Take $g = 10 \ m/s^2$)

$A$ particle of mass $m$ and charge $q$ is placed at rest in a uniform electric field $E$ and then released. The kinetic energy attained by the particle after moving a distance $y$ 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 $= 1.6 \times 10^{-19} \ C$)