An electron falls from rest through a distance of $1.5 \text{ cm}$ in a uniform electric field of magnitude $2.0 \times 10^4 \text{ N/C}$. The time taken to cover this distance in seconds is . . . . . . . $(e = 1.6 \times 10^{-19} \text{ C}, m_e = 9.11 \times 10^{-31} \text{ kg})$

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

If an $\alpha$-particle and a proton are accelerated from rest by a potential difference of $1 \text{ MV}$,then the ratio of their kinetic energy will be

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

$A$ charged oil drop is suspended in a uniform electric field of $3 \times 10^{4} \; V/m$ so that it neither falls nor rises. The charge on the drop will be $..... \times 10^{-18} \; C$. (Take the mass of the drop $= 9.9 \times 10^{-15} \; kg$ and $g = 10 \; m/s^{2}$)

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.