$A$ particle of mass $m$ and charge $q$ is placed at rest in a uniform electric field $E$ and then released. The kinetic energy attained by the particle after moving a distance $y$ is

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:

Two charges $-q$ each are fixed and separated by a distance $2d$. $A$ third charge $q$ of mass $m$ placed at the midpoint is displaced slightly by $x$ $(x \ll d)$ perpendicular to the line joining the two fixed charges as shown in the figure. Show that the charge $q$ will perform simple harmonic motion with a time period $T = \left[\frac{8 \pi^{3} \epsilon_{0} m d^{3}}{q^{2}}\right]^{1 / 2}$.

$A$ uniform electric field $E = (8m/e) \text{ V/m}$ is created between two parallel plates of length $1 \text{ m}$ as shown in the figure (where $m = \text{mass of electron}$ and $e = \text{charge of electron}$). An electron enters the field symmetrically between the plates with a speed of $2 \text{ m/s}$. The angle of the deviation $(\theta)$ of the path of the electron as it comes out of the field will be:

In Millikan's experiment,an oil drop having charge $q$ becomes stationary upon applying a potential difference $V$ between two plates separated by a distance $d$. The weight of the drop is

The kinetic energy of an electron which is accelerated through a potential of $100 \, V$ is: