In Millikan's oil drop experiment,a charged particle is held in equilibrium between two plates by an electric field. If the direction of the electric field is reversed,calculate the acceleration of the charged particle. (in $, g$)

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

$A$ small oil drop of mass $10^{-6} \,kg$ is hanging at rest between two plates separated by $1 \,mm$ having a potential difference of $500 \,V$. The charge on the drop is $(g = 10 \,ms^{-2})$

$A$ proton is located at coordinates $(x, y) = (0, 0)$,and an electron is at $(d, h)$,where $d >> h$. At time $t = 0$,a uniform electric field $E$ of unknown magnitude but pointing in the positive $y$ direction is turned on. Assuming that $d$ is large enough that the proton-electron interaction is negligible,at what $y$ coordinate will the two particles have the same $y$ position at the same time?

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