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.

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 $= 1.6 \times 10^{-19} \ C$)

$A$ particle of charge $q$ and mass $m$ is subjected to an electric field $E = E_{0}(1 - ax^{2})$ in the $x$-direction,where $a$ and $E_{0}$ are constants. Initially,the particle was at rest at $x = 0$. Other than the initial position,the kinetic energy of the particle becomes zero when the distance of the particle from the origin is:

$A$ charged particle of mass $5 \times 10^{-5} \ kg$ is held stationary in space by placing it in an electric field of strength $10^7 \ N C^{-1}$ directed vertically downwards. The charge on the particle is

An electron enters a parallel plate capacitor with horizontal speed $u$ and is found to deflect by angle $\theta$ on leaving the capacitor as shown below. It is found that $\tan \theta = 0.4$ and gravity is negligible. If the initial horizontal speed is doubled, then the value of $\tan \theta$ will be

Kinetic energy of an electron accelerated in a potential difference of $100 \, V$ is