Question

An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 10⁴ N C⁻¹ (Millikan’s oil drop experiment). The density of the oil is 1.26 g cm⁻³. Estimate the radius of the drop. (g = 9.81 m s⁻²; e = 1.60 × 10⁻¹⁹ C).

Answer 1

Excess electrons on an oil drop, n = 12

Electric field intensity, E = 2.55 × 10⁴ N C⁻¹

Density of oil, ρ = 1.26 gm/cm³ = 1.26 × 10³ kg/m³

Acceleration due to gravity, g = 9.81 m s⁻²

Charge on an electron, e = 1.6 × 10⁻¹⁹ C

Radius of the oil drop = r

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

F = W

Eq = mg

Ene = ${\displaystyle \frac{4}{3}\pi {\text{r}}^3\times \rho \times \text{g}}$

Where,

q = Net charge on the oil drop = ne

m = Mass of the oil drop

= Volume of the oil drop × Density of oil

= ${\displaystyle \frac{4}{3}\pi {\text{r}}^3\times \rho }$

∴ ${\displaystyle \text{r}=\sqrt[3]{\frac{3\text{Ene}}{4\pi \rho \text{g}}}}$

= ${\displaystyle \sqrt[3]{\frac{3\times 2.55\times {10}^4\times 12\times 1.6\times {10}^{-19}}{4\times 3.14\times 1.26\times {10}^3\times 9.81}}}$

= ${\displaystyle \sqrt[3]{946.09\times {10}^{-21}}}$

= ${\displaystyle 9.82\times {10}^{-7}}$ m

= 9.82 × 10⁻⁴ mm

Therefore, the radius of the oil drop is 9.82 × 10⁻⁴ mm.