Question

In Millikan's experiment, an oil drop of mass $4.9\times {10}^{-14}kg$ is balanced by applying a potential difference of $9.8kV$ between the two plates which are $12.8mm$ apart. Calculate the number of elementary charges on the drop. (Take $g=10{ms}^{-2}$).

Answer 1

$Eq=mg$ (or) $q=\frac{mg}{E}$
$\because E=\frac{v}{d}$ and $q=ne$
$ne=\frac{mgd}{v}$
$\therefore n=\frac{mgd}{ve}=\frac{4.9\times {10}^{-14}\times 10\times 12.8\times {10}^{-3}}{9.8\times {10}^3\times 1.6\times {10}^{-19}}$
$n=4$.