Question

In Millikan oil drop experiment a charged drop of mass $1.8\times {10}^{-14}$kg is stationary between the plates. The distance between the plates is 0.9cm and potential difference is 2000 V. The number of electrons in the drop are $(g=10m{s}^{-2})$

• A. 2
• B. 4
• C. 5
• D. 1

Answer 1

Answer: (C)

5

Let the number of electrons in the drop be $n$
According to question, the equilibrium eqn. becomes,
$\Rightarrow 1.8\times {10}^{-14}\times 10=\left(\frac{2000}{9}\right)\times {10}^3\times n\times 1.6\times {10}^{-19}$
$\Rightarrow n=5.06$
$\Rightarrow n\approx 5$

So, the answer is option (C).