$A$ 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 $........... \text{ mm}$.

The uniform electric field intensity between the two plates of a parallel plate capacitor is $1 \times 10^3 \ Vm^{-1}$ acting vertically upwards as shown in the figure. The plates are sufficiently long and have a separation of $2 \ cm$. $A$ particle of negative charge $1 \ \mu C$ and mass $2 \ g$ is projected at an angle $45^{\circ}$ with the electric field from the lower plate with a velocity '$u$'. The maximum velocity acquired by the particle,if it does not hit the upper plate,is (in $ms^{-1}$)

There is a uniform electric field of strength $10^3 \ Vm^{-1}$ along the $Y$-axis. $A$ body of mass $1 \ g$ and charge $10^{-6} \ C$ is projected into the field from the origin along the positive $X$-axis with a velocity of $10 \ ms^{-1}$. Its speed in $ms^{-1}$ after $10 \ s$ is (Neglect gravitation).

$A$ charged particle is free to move in an electric field. It will travel

In a region of uniform electric field of intensity $E$,an electron of mass $m_e$ is released from rest. The distance travelled by the electron in a time $t$ is

In the following figure,two parallel metallic plates are maintained at different potentials. If an electron is released midway between the plates,it will move: