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:

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.

An electron is projected in the direction of an electric field. Just after the projection of the electron:

An electron experiences a force equal to its weight when placed in an electric field. The intensity of the field will be

An electron enters an electric field of intensity $3200 \ V/m$ perpendicular to the field with a velocity of $4 \times 10^7 \ m/s$. If it travels a horizontal distance of $0.10 \ m$,then its deflection from its path is ........ $mm$.

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