$A$ particle of mass $m$ and charge $q$ travelling with a velocity $v$ along the $x$-axis enters a uniform electric field $\overrightarrow{E}$ directed along the $y$-axis. What will be the trajectory of the particle?

$A$ charged cork ball having mass $1 \text{ g}$ and charge $q$ is suspended on a light string in a uniform electric field as shown in the figure. The ball is in equilibrium at $\theta=37^{\circ}$,when the value of the electric field is $E=(3 \hat{i}+5 \hat{j}) \times 10^5 \text{ NC}^{-1}$. (Assume $T$ as tension in the string.) Which of the following options are correct? (Given,$\sin 37^{\circ}=0.60$ and $g=10 \text{ ms}^{-2}$)

Two long parallel plates $A$ and $B$ are separated by a distance of $4 \ cm$ with an electric field of $45.5 \ Vm^{-1}$ between the plates directed normally from plate $A$ to plate $B$,as shown in the figure. An electron is projected from plate $A$ with velocity $v$ at an angle of $30^{\circ}$ with the surface of plate $A$. The maximum value of $v$ so that the electron does not hit plate $B$ is (Assume gravity-free space,charge of electron $= 1.6 \times 10^{-19} \ C$ and mass of electron $= 9.1 \times 10^{-31} \ kg$): (in $km \ s^{-1}$)

$A$ particle of mass $M$ and charge $q$,initially at rest,is accelerated by a uniform electric field $E$ through a distance $D$ and is then allowed to approach a fixed static charge $Q$ of the same sign. The distance of the closest approach of the charge $q$ will then be

$A$ stream of positively charged particles having $\frac{q}{m} = 2 \times 10^{11} \text{ C/kg}$ and velocity $\overrightarrow{v}_0 = 3 \times 10^7 \hat{i} \text{ m/s}$ is deflected by an electric field $1.8 \hat{j} \text{ kV/m}$. The electric field exists in a region of $10 \text{ cm}$ along the $x$-direction. Due to the electric field,the deflection of the charged particles in the $y$-direction is $........... \text{ mm}$.

An electron falls through a distance of $1.5 \, cm$ in a uniform electric field of magnitude $2.0 \times 10^4 \, N/C$ as shown in the figure. The time taken by the electron to fall through this distance is ($m_e = 9.1 \times 10^{-31} \, kg$,neglect gravity).