An electron with mass $9.1 \times 10^{-31} \ kg$ and charge $1.6 \times 10^{-19} \ C$ is placed in an electric field of $1 \times 10^6 \ V/m$. How much time will it take for the electron to reach a velocity equal to $1/10$th of the speed of light?

Calculate the work done by the electrostatic force on an electron moving in a circular orbit around the nucleus.

An electron falls from rest through a vertical distance $h$ in a uniform and vertically upward directed electric field $E$. The direction of the electric field is now reversed,keeping its magnitude the same. $A$ proton is allowed to fall from rest in it through the same vertical distance $h$. The time of fall of the electron,in comparison to the time of fall of the proton,is:

$A$ small block of mass $5 \, g$ and charge $5 \, \mu C$ is placed on an insulated, frictionless, inclined plane of angle $60^{\circ}$. An electric field is applied parallel to the inclined plane. If the block remains at rest, then the magnitude of the electric field is (Take $g = 10 \, m/s^2$):

Assertion : $A$ deuteron and an $\alpha -$ particle are placed in an electric field. If $F_1$ and $F_2$ are the forces acting on them and $a_1$ and $a_2$ are their accelerations respectively,then $a_1 = a_2$.
Reason : Forces will be same in an electric field.

$A$ 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: