$A$ 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 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:

$A$ charged particle is suspended in equilibrium in a uniform vertical electric field of intensity $20,000\, V/m$. If the mass of the particle is $9.6 \times 10^{-16}\, kg$,the charge on it and the excess number of electrons on the particle are respectively $(g = 10\, m/s^2)$.

An electron,initially at rest,is accelerated through a potential difference of $200 \ V$,so that it acquires a velocity of $8.4 \times 10^6 \ m/s$. The value of $e/m$ for the electron will be:

$A$ small block of mass $5 \, g$ and charge $5 \, \mu C$ is placed on an insulated, frictionless, inclined plane of angle $60^{\circ}$. An electric field is applied parallel to the inclined plane. If the block remains at rest, then the magnitude of the electric field is (Take $g = 10 \, m/s^2$):

As shown in the figure below,a point charge $q$ moves from point $P$ to a point $S$ traversing a path $PQRS$ in a uniform electric field $\vec{E}$. The electric field is directed along a direction parallel to the $x$-axis. The coordinates of $P$,$Q$,$R$,and $S$ are $(a, b, 0)$,$(2a, 0, 0)$,$(a, -b, 0)$,and $(0, 0, 0)$ respectively. What is the work done by the field in the process?