$A$ uniform electric field $E = (8m/e) \text{ V/m}$ is created between two parallel plates of length $1 \text{ m}$ as shown in the figure (where $m = \text{mass of electron}$ and $e = \text{charge of electron}$). An electron enters the field symmetrically between the plates with a speed of $2 \text{ m/s}$. The angle of the deviation $(\theta)$ of the path of the electron as it comes out of the field will be:

Two balls of charge $q_1$ and $q_2$ initially have a velocity of the same magnitude and direction. After a uniform electric field is applied for a certain time,the direction of the velocity of the first ball changes by $60^{\circ}$,and the velocity magnitude is reduced by half. The direction of the velocity of the second ball changes by $90^{\circ}$. In what proportion will the velocity of the second ball change? Determine the magnitude of the charge-to-mass ratio for the second ball if it is equal to $k_1$ for the first ball. The electrostatic interaction between the balls should be neglected.

$A$ small oil drop of mass $10^{-6} \,kg$ is hanging at rest between two plates separated by $1 \,mm$ having a potential difference of $500 \,V$. The charge on the drop is $(g = 10 \,ms^{-2})$

$A$ 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

$A$ proton is fired at an initial velocity of $150 \, m/s$ at an angle of $60^\circ$ above the horizontal into a uniform electric field of $2 \times 10^{-4} \, N/C$ between two charged parallel plates as shown in the figure. The total time the particle is in motion is:

How many electrons should be removed from a coin of mass $1.6 \,g$,so that it may float in an electric field of intensity $10^9 \,N/C$ directed upward?