When an electron enters an electric field with a velocity in the direction of the electric field lines,then.......

$A$ 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? $(g = 10 \, m/s^2)$

An electron is released from the bottom plate $A$ as shown in the figure $(E = 10^4 \, N/C)$. The velocity of the electron when it reaches plate $B$ will be nearly equal to

$A$ proton, a deuteron, and an $\alpha$-particle having the same momentum enter a region of uniform electric field between the parallel plates of a capacitor. The electric field is perpendicular to the initial path of the particles. Then the ratio of deflections suffered by them is

$A$ 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).

$A$ particle of charge $1\ \mu C$ and mass $1\ g$ moving with a velocity of $4\ m/s$ is subjected to a uniform electric field of magnitude $300\ V/m$ for $10\ s$. Then its final speed cannot be.......$m/s$.