Question

An oil drop of radius 2 mm with a density 3 g cm^-3 is held stationary under a constant electric field 3.55 × 10⁵ V m^-1 in the Millikan's oil drop experiment. What is the number of excess electrons that the oil drop will process? 

Consider g = 9.81 m/s²

Options

• 1.73 × 10¹² 

• 1.73 × 10¹⁰ 

• 48.8 × 10¹¹

• 17.3 × 10¹⁰ 

Answer 1

1.73 × 10¹⁰ 

Explanation:

The weight of the oil drop will be balanced by the electrostatic field.

m_oil × g_oil = qE

⇒ `4/3pi"r"^3xxrhoxx"g"` = neE (n = no. of excess electrons)

⇒ n = `(4/3pi"r"^3rho"g")/"eE"`

= `(4/3xxpixx(2xx10^-3)^3xx(3xx10^3)xx9.81)/(1.6xx10^-19xx3.55xx10^5)`

= 173.65 × 10⁸ ≃ 1.73 × 10¹⁰