$A$ uniform electric field of $10\,N/C$ is created between two parallel charged plates (as shown in the figure). An electron enters the field symmetrically between the plates with a kinetic energy of $0.5\,eV$. The length of each plate is $10\,cm$. The angle $(\theta)$ of deviation of the path of the electron as it comes out of the field is $.........$ (in degrees).

An electron of mass $m_e$ initially at rest moves through a certain distance in a uniform electric field in time $t_1$. $A$ proton of mass $m_p$ also initially at rest takes time $t_2$ to move through an equal distance in this uniform electric field. Neglecting the effect of gravity,the ratio of $t_2/t_1$ is nearly equal to

$A$ small sphere of mass $m = 0.5 \, kg$ carrying a positive charge $q = 110 \, \mu C$ is connected to a light,flexible,and inextensible string of length $r = 60 \, cm$ and whirled in a vertical circle. If a vertically upwards electric field of strength $E = 10^5 \, N/C$ exists in the space,what is the minimum velocity of the sphere required at the highest point so that it may just complete the circle? $(g = 10 \, m/s^2)$

Charge $Q$ is given a displacement $\vec{r} = a\hat{i} + b\hat{j}$ in an electric field $\vec{E} = E_1\hat{i} + E_2\hat{j}$. The work done is

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:

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).