Question

A dust particle of mass \(4 \times 10^{-12} \, {mg}\) 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? [Take, \(g = 10 \, {m/s}^2\)]

• A. 15
• B. 8
• C. 5
• D. 4

Hint:

The number of electrons removed can be found by balancing the electric force with the gravitational force on the particle.

Answer 1

The Correct Option is C

Step 1: {Calculating the weight of the dust particle}
\[ m = 4 \times 10^{-12} \, {mg} = 4 \times 10^{-15} \, {kg} \] \[ W = mg = 4 \times 10^{-15} \times 10 = 4 \times 10^{-14} \, {N} \] Step 2: {Balancing forces}
\[ qE = W \Rightarrow q = \frac{W}{E} = \frac{4 \times 10^{-14}}{50} = 8 \times 10^{-16} \, {C} \] Step 3: {Calculating the number of electrons removed}
\[ q = ne \Rightarrow n = \frac{q}{e} = \frac{8 \times 10^{-16}}{1.6 \times 10^{-19}} = 5 \] Thus, 5 electrons were removed from the neutral dust particle.

Answer 2

Step 1: Understand the physical condition
The dust particle is suspended in air, meaning the net force acting on it is zero.
So, the upward electric force equals the downward gravitational force:
\[ F_{\text{electric}} = F_{\text{gravity}} \Rightarrow qE = mg \]

Step 2: Convert given mass to kilograms
\[ m = 4 \times 10^{-12} \, \text{mg} = 4 \times 10^{-12} \times 10^{-6} \, \text{kg} = 4 \times 10^{-18} \, \text{kg} \]

Step 3: Use the balance of forces to find charge
\[ q = \frac{mg}{E} = \frac{4 \times 10^{-18} \times 10}{50} = \frac{4 \times 10^{-17}}{50} = 8 \times 10^{-19} \, \text{C} \]

Step 4: Find number of electrons removed
Charge of one electron: \( e = 1.6 \times 10^{-19} \, \text{C} \)
\[ n = \frac{q}{e} = \frac{8 \times 10^{-19}}{1.6 \times 10^{-19}} = 5 \]

Final Answer: 5 electrons