In Millikan's oil drop experiment,an oil drop of mass $16 \times 10^{-6} \ kg$ is balanced by an electric field of $10^6 \ V/m$. The charge in coulomb on the drop,assuming $g = 10 \ m/s^2$,is:

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:

$A$ 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?

If a proton is moved against the Coulomb force of an electric field,then

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

Acceleration of a charged particle of charge $q$ and mass $m$ moving in a uniform electric field of strength $E$ is