An electron of mass $m$,charge $e$ falls through a distance $h$ meters in a uniform electric field $E$. Then the time of fall is:

$A$ ball of mass $1 \text{ g}$ having a charge of $20 \mu\text{C}$ is tied to one end of a string of length $0.9 \text{ m}$ and can rotate in a vertical plane in a uniform electric field of $100 \text{ NC}^{-1}$ directed upwards. The minimum horizontal velocity that must be given to the ball at the lowest position so that it completes the vertical circle is (Let $g = 10 \text{ ms}^{-2}$): (in $\text{ ms}^{-1}$)

$A$ sphere carrying charge $Q$ and having weight $w$ falls under gravity between a pair of vertical plates at a distance $d$ from each other. When a potential difference $V$ is applied between the plates,the acceleration of the sphere changes as shown in the figure,along the line $BC$. The value of $Q$ is:

In Millikan's oil drop experiment,a charged particle is held in equilibrium between two plates by an electric field. If the direction of the electric field is reversed,calculate the acceleration of the charged particle. (in $, g$)

In the absence of gravity,a charge $q$ and mass $2m$ is placed stationary in a uniform electric field of intensity $E$. When the charge is released,its speed after $n$ seconds is . . . . . . .

There is a uniform electric field of strength $10^3 \ Vm^{-1}$ along the $Y$-axis. $A$ body of mass $1 \ g$ and charge $10^{-6} \ C$ is projected into the field from the origin along the positive $X$-axis with a velocity of $10 \ ms^{-1}$. Its speed in $ms^{-1}$ after $10 \ s$ is (Neglect gravitation).