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1.25 An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 $\times 10^{4} N C^{- 1}$ (Millikan’s oil drop experiment). The density of the oil is 1.26 g / ${c m}^{3}$ . Estimate the radius of the drop.

(g = 9.81 m/ $s^{2}$ ; e = 1.60 $\times 10^{- 19}$ C).

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alok kumar singh | Contributor-Level 10

3 months ago

1.25 Excess electrons on an oil drop, n = 12

Electric field intensity, E = 2.55 $\times 10^{4} N C^{- 1}$

Density of oil, $ρ =$ 1.26 g / ${c m}^{3}$ = 1.26 $\times 10^{3}$ g/ $m^{3}$

Acceleration due to gravity, g = 9.81 m/ $s^{2}$

Charge of an electron, e = 1.60 $\times 10^{- 19}$ C

Let the radius of the oil drop be r

Force (F) due to electric field (E) is equal to the weight of the oil drop (W)

F = W

Eq = mg

Ene = $\frac{4}{3} π r^{3} \times ρ \times g$

$r^{3}$ = $\frac{3 \times E \times n \times e}{4 \times π \times ρ \times g}$ = $\frac{3 \times 2.55 \times 10^{4} \times 12 \times 1.60 \times 10^{- 19}}{4 \times π \times 1.26 \times 10^{3} \times 9.81}$

r = 9.815 $\times 10^{- 7}$ m = 9.815 $\times 10^{- 4}$ mm

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1.25 An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 $\times 10^{4} N C^{- 1}$ (Millikan’s oil drop experiment). The density of the oil is 1.26 g / ${c m}^{3}$ . Estimate the radius of the drop.

(g = 9.81 m/ $s^{2}$ ; e = 1.60 $\times 10^{- 19}$ C).

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1.25 An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 ×104NC-1 (Millikan’s oil drop experiment). The density of the oil is 1.26 g /cm3 . Estimate the radius of the drop. (g = 9.81 m/s2 ; e = 1.60 ×10-19 C).

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1.25 An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 $\times 10^{4} N C^{- 1}$ (Millikan’s oil drop experiment). The density of the oil is 1.26 g / ${c m}^{3}$ . Estimate the radius of the drop.

(g = 9.81 m/ $s^{2}$ ; e = 1.60 $\times 10^{- 19}$ C).