Q. In Millikans oil drop experiment, a charged drop of mass $1.8 \times 10^{- 14}kg$ is stationary between its plates. The distance between its plates is $0.90\ cm$ and potential difference is $2.0\ kilo\ volts.$ The number of electrons on the drop is
A) 500
B) 50
C) 5
D) 0
Explanation
(c) $QE = mg \Rightarrow Q = \frac{mg}{E} \Rightarrow n = \frac{mgd}{Ve}$
$\Rightarrow n = \frac{1.8 \times 10^{- 14} \times 10 \times 0.9 \times 10^{- 2}}{2 \times 10^{3} \times 1.6 \times 10^{- 19}} = 5$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.