An electron of mass $m$ and charge $q$ is accelerated from rest in a uniform electric field of strength $E$. The velocity acquired by the electron when it travels a distance $L$ is

$A$ 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:

$A$ small sphere of mass $m$ carrying a charge $q$ is hanging between two parallel plates by a string of length $L$ as shown in the figure. The time period of the pendulum is $T_0$. When the parallel plates are charged as shown,the time period changes to $T$. The ratio $T / T_0$ is equal to:

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

$A$ simple pendulum has a bob with mass $m$ and charge $q$. The pendulum string has negligible mass. When a uniform and horizontal electric field $\vec{E}$ is applied,the final tension in the string changes. The final tension in the string,when the pendulum attains an equilibrium position,is . . . . . . . ($g$: acceleration due to gravity)

An electron falls from rest through a vertical distance $h$ in a uniform and vertically upward directed electric field $E$. The direction of the electric field is now reversed,keeping its magnitude the same. $A$ proton is allowed to fall from rest in it through the same vertical distance $h$. The time of fall of the electron,in comparison to the time of fall of the proton,is: