$A$ particle with charge $e$ and mass $m$, moving along the $X$-axis with a uniform speed $u$, enters a region where a uniform electric field $E$ is acting along the $Y$-axis. The particle starts to move in a parabola. Its focal length (neglecting any effect of gravity) is

$A$ hot filament liberates an electron with zero initial velocity. The anode potential is $1200 \,V$. The speed of the electron when it strikes the anode is

$A$ proton is fired at an initial velocity of $150 \, m/s$ at an angle of $60^\circ$ above the horizontal into a uniform electric field of $2 \times 10^{-4} \, N/C$ between two charged parallel plates as shown in the figure. The total time the particle is in motion is:

$A$ charged particle is free to move in an electric field. It will travel

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:

Two identical positive charges $Q$ each are fixed at a distance of $2a$ apart from each other. Another point charge $q_0$ with mass $m$ is placed at the midpoint between the two fixed charges. For a small displacement $x$ along the line joining the fixed charges,the charge $q_0$ executes $SHM$. The time period of oscillation of charge $q_0$ will be: