Q. In Millikan’s oil drop experiment, a charged drop of mass $1.8 \times 10^{- 14}$ kg is stationary between the plates. The distance between the plates 0.9 cm and potential difference between the plates is 2000 V. The number of electrons on the oil drop is
Explanation
(b) In equilibrium,
$qE = mg$
Also, $E = \frac{V}{d}\ and\ q = ne$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ne \times \frac{V}{d} = mg$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n = \frac{mgd}{eV}$
Given, $m = 1.8 \times 10^{- 14}\ kg,\ \ \ \ \ \ g = 10{ms}^{- 2},$
$d = 0.9\ cm = 0.9 \times 10^{- 2}\ m,$
$e = 1.6 \times 10^{- 19}$ C,
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ V = 2000$ V
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n = \frac{1.8 \times 10^{- 14} \times 10 \times {0.9 \times 10}^{- 2}}{1.6 \times 10^{- 19} \times 2000}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n = \frac{81}{16} \approx 5$ electrons
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.