In Millikan's oil drop experiment, a charged oil drop of mass 3.2 x 10^-14 kg is held stationary between two parallel plates 6 mm apart by applying a potential difference of 1200 V between them. How many excess electrons does the oil drop carry? Take g = 10 ms^-2 :

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answer is D.

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Detailed Solution

The upward electric force on oil drop is balanced by its weight. I.e.

$\begin{aligned} qE = mg \end{aligned}$

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

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

$\begin{aligned} n = \frac{3.2 \times 10^{- 13} \times 6 \times 10^{- 3}}{1.6 \times 10^{- 19} \times 1200} \end{aligned}$

n = 10

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