An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55× $10^{4}$ NC^-1 in Millikan’s oil drop experiment . The density of the oil is 1.26 g cm^-3. Estimate the radius of the drop

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$98 .1 \times 10^{- 4} m$

$9 .81 \times 10^{- 3} m$

$9 .81 \times 10^{- 4} mm$

$9 .81 \times 10^{- 6} mm$

answer is C.

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Detailed Solution

$\begin{array}{l} qE = mg \Rightarrow neE = \frac{4}{3} {πr}^{3} ρg \\ r^{3} = \frac{3 neE}{4 πρg} = \frac{3 \times 12 \times 1 .6 \times 10^{- 19} \times 2 .55 \times 10^{4}}{4 π \times 1 .26 \times 10^{3} \times 9 .8} \\ = 0 .945 \times 10^{- 18} \\ r = \sqrt[3]{0 .945 \times 10^{- 18}} = 0 .981 \times 10^{- 6} m \\ = 9 .81 \times 10^{- 4} mm \end{array}$