An electron is moving in an electric field and a magnetic field. It will gain energy from:

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.

When a proton is accelerated through $1\,V$,then its kinetic energy will be.....$eV$.

An electron of mass $m$,charge $e$ falls through a distance $h$ meters in a uniform electric field $E$. Then the time of fall is:

$A$ charged particle with charge $q$ and mass $m$ starts with an initial kinetic energy $K$ at the center of a uniformly charged spherical region of total charge $Q$ and radius $R$. $q$ and $Q$ have opposite signs. The spherically charged region is not free to move. The value of $K$ is such that the particle will just reach the boundary of the spherically charged region. How much time does it take for the particle to reach the boundary of the region?

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: