A tiny spherical oil drop of mass 8 mg is kept suspended by applying a potential difference of 100 V between two metallic plates kept seperated by a distance 1 cm. The number of electrons on the oil drop are: (Take $g =10 ms^{-2}$)

$5 \times 10^{10}$

The correct answer is Option (1) → $5 \times 10^{10}$

$F_{gravity}=m×g$

$=8×10^{-6}×9.8$

$=7.84×10^{-5}N$

$F_{electric}=q.E$

$=q\frac{V}{d}=q×\frac{100}{0.01}$

$=q.10^4V/m$

$∴q=\frac{7.84×10^{-5}}{10^4}=7.84×10^{-9}$