$A$ particle of mass $2 \ g$ and charge $6 \ \mu C$ is accelerated from rest through a potential difference of $60 \ V$. The speed acquired by the particle is (in $ms^{-1}$)

The $e/m$ ratio for an electron is $1.8 \times 10^{11} \ C \ kg^{-1}$. What will be its velocity when accelerated through a potential difference of $9 \ V$?

Two charges each of magnitude $q$ are fixed at a distance $2d$ apart. $A$ third charge (proton) placed at the midpoint is displaced slightly by a distance $x$ $(x << d)$ perpendicular to the line joining the two fixed charges. The proton will execute simple harmonic motion having an angular frequency: ($m =$ mass of the charged particle)

$A$ small sphere carrying a charge $q$ is hanging in between two parallel plates by a string of length $L$. The time period of the pendulum is $T_0$. When the parallel plates are charged,the time period changes to $T$. The ratio $T/T_0$ is equal to

Acceleration of a charged particle of charge $q$ and mass $m$ moving in a uniform electric field of strength $E$ is

$A$ uniform electric field of $10^3 \, V/m$ is directed along the $y$-axis. $A$ particle of mass $1 \, g$ and charge $10^{-6} \, C$ is projected from the origin into the field with a velocity of $10 \, m/s$ along the positive $x$-axis. Its speed after $10 \, s$ is (in $m/s$):