Question

In Millinkan's oil drop experiment, a charged oil drop of mass $3.2\times {10}^{-14}kg$ is held stationary between parallel plates $6mm$ apart, by applying a potential difference of $1200V$ between them. How many electrons does the oil drop carry?
(Given $g=10{ms}^{-2}$)

• A. 7
• B. 8
• C. 9
• D. 10

Answer 1

The correct option is D 10

According to figure, it is clear that electrical force is balanced by weight of body, i.e

$qE=mg$

$ne\frac{V}{l}=mg$

$n\times 1.6\times {10}^{-19}\times \frac{1200}{6\times {10}^{-3}}=3.2\times {10}^{-14}\times 10$

$n=\frac{3.2\times {10}^{-14}\times 6\times {10}^{-3}}{1.6\times {10}^{-19}\times 1200}$

$n=10$