$A$ particle,of mass $10^{-3} \ kg$ and charge $1.0 \ C$,is initially at rest. At time $t = 0$,the particle comes under the influence of an electric field $\vec{E}(t) = E_0 \sin(\omega t) \hat{i}$,where $E_0 = 1.0 \ N \ C^{-1}$ and $\omega = 10^3 \ rad \ s^{-1}$. Consider the effect of only the electrical force on the particle. Then the maximum speed,in $m \ s^{-1}$,attained by the particle at subsequent times is . . . . . .

$A$ charged particle of mass $5 \times 10^{-5} \ kg$ is held stationary in space by placing it in an electric field of strength $10^7 \ N C^{-1}$ directed vertically downwards. The charge on the particle is

$A$ 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 released from rest in a uniform electric field. If there is no other force on the particle,the dependence of its speed $v$ on the distance $x$ travelled by it is correctly given by (graphs are schematic and not drawn to scale)

$A$ particle of charge $1\ \mu C$ and mass $1\ g$ moving with a velocity of $4\ m/s$ is subjected to a uniform electric field of magnitude $300\ V/m$ for $10\ s$. Then its final speed cannot be.......$m/s$.

Two large parallel plates are charged such that the potential difference between them is $V_2 - V_1 = 20\ V$. Plate $2$ is at a higher potential. The plates are separated by a distance of $0.1\ m$ and can be considered infinitely large. An electron is released from rest at the inner surface of plate $1$. What is its speed when it strikes plate $2$? $(e = 1.6 \times 10^{-19}\ C, m_0 = 9.11 \times 10^{-31}\ kg)$