In an electron gun,the electrons are accelerated by the potential $V$. If $e$ is the charge and $m$ is the mass of an electron,then the maximum velocity of these electrons will be

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

$A$ horizontal electric field $E = (mg)/q$ exists as shown in the figure. $A$ mass $m$ with charge $q$ is attached to the end of a light rod of length $l$. If the mass $m$ is released from the position shown in the figure $(\theta = 45^{\circ})$,find the angular velocity of the rod when it passes through the bottom-most position.

An electron falls through a small distance in a uniform electric field of magnitude $2 \times 10^4 \ N C^{-1}$. The direction of the field is reversed keeping the magnitude unchanged and a proton falls through the same distance. The time of fall will be

$A$ body having a specific charge of $8\,\mu \text{C/g}$ is resting on a frictionless plane at a distance of $10\,\text{cm}$ from a wall. It starts moving towards the wall when a uniform electric field of $100\,\text{V/m}$ is applied horizontally toward the wall. If the collision of the body with the wall is perfectly elastic,then the time period of the motion will be (in seconds):

$A$ proton is located at coordinates $(x, y) = (0, 0)$,and an electron is at $(d, h)$,where $d >> h$. At time $t = 0$,a uniform electric field $E$ of unknown magnitude but pointing in the positive $y$ direction is turned on. Assuming that $d$ is large enough that the proton-electron interaction is negligible,at what $y$ coordinate will the two particles have the same $y$ position at the same time?