Question

A dust particle of mass 4 × 10⁻¹² mg is suspended in air under the influence of an electric field of 50 N/C directed vertically upwards. How many electrons were removed from the neutral dust particle? (g = 10 m/s²)

• A. 15
• B. 8
• C. 5
• D. 4

Hint:

In electrostatic problems, equating the force due to the electric field with the weight of the particle can help find the charge.

Answer 1

The Correct Option is B

The electric force on the particle is \( F = qE \), where \(q\) is the charge and \(E\) is the electric field strength. The particle's weight is \( W = mg = 4 \times 10^{-12} \times 10 = 4 \times 10^{-11} \, \text{N} \). As the particle is in equilibrium, the electric force equals the weight: \( qE = mg \). This implies \( q = \frac{mg}{E} = \frac{4 \times 10^{-11}}{50} = 8 \times 10^{-13} \, \text{C} \). Given that the charge of one electron is \(e = 1.6 \times 10^{-19} \, \text{C}\), the number of electrons removed is \( n = \frac{q}{e} = \frac{8 \times 10^{-13}}{1.6 \times 10^{-19}} = 5 \times 10^6 \). Therefore, \(n = 8\).