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?

$A$ particle of mass $m$ and charge $(-q)$ enters the region between two charged plates,initially moving along the $x$-axis with speed $v_{x}$ (like particle $1$ in the Figure). The length of the plate is $L$ and a uniform electric field $E$ is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is $q E L^{2} / (2 m v_{x}^{2})$. Compare this motion with the motion of a projectile in a gravitational field.

$A$ particle of mass $m$ and charge $q$ is thrown perpendicular to an electric field of intensity $E$ with an initial velocity $v$. The particle moves a distance $x$ perpendicular to the field and a distance $y$ along the direction of the field. If $y = \alpha x^{2}$,then $\alpha$ is given by:

What is the magnitude of the upward electric field (in $V/m$) required to keep a coin of mass $10^{-6} \ kg$ and charge $10^{-6} \ C$ in equilibrium? (Take $g = 10 \ m/s^2$)

An alpha particle and a proton are accelerated from rest in a uniform electric field. The ratio of the times taken by proton and alpha particle to attain equal displacements 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