A charged dust particle of radius `5 xx 10^(-7)m` is located in a horizontal electric field having an intensity of `6.28 xx 10^(5) V m^(-1)`. The surrounding medium in air with coefficient of viscosity `eta = 1.6 xx 10^(-15) Ns m^(-2)` . If this particle moves with a uniform horizontal speed of `0.01 ms^(-1),` the number of electrons on it will be

Options

A. 20
B. 15
C. 25
D. 30

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the forces acting on the charged dust particle
The charged dust particle experiences two forces:
1. **Electrostatic Force (F_e)**: This force acts in the direction of the electric field and is given by the formula:
\[ F_e = Q \cdot E \]
where \( Q \) is the charge on the particle and \( E \) is the electric field intensity.

2. **Viscous Force (F_v)**: This force acts in the opposite direction to the motion of the particle and is given by Stokes' law:
\[ F_v = 6 \pi \eta r v \]
where \( \eta \) is the coefficient of viscosity, \( r \) is the radius of the particle, and \( v \) is the velocity of the particle.

### Step 2: Set up the equilibrium condition
At terminal velocity, the net force acting on the particle is zero. Therefore, the electrostatic force is equal to the viscous force:
\[ F_e = F_v \]
Substituting the expressions for the forces, we get:
\[ Q \cdot E = 6 \pi \eta r v \]

### Step 3: Solve for the charge \( Q \)
Rearranging the equation to solve for \( Q \):
\[ Q = \frac{6 \pi \eta r v}{E} \]

### Step 4: Substitute the given values
We have the following values:
- \( \eta = 1.6 \times 10^{-15} \, \text{Ns/m}^2 \)
- \( r = 5 \times 10^{-7} \, \text{m} \)
- \( v = 0.01 \, \text{m/s} \)
- \( E = 6.28 \times 10^{5} \, \text{V/m} \)

Substituting these values into the equation for \( Q \):
\[ Q = \frac{6 \pi (1.6 \times 10^{-15}) (5 \times 10^{-7}) (0.01)}{6.28 \times 10^{5}} \]

### Step 5: Calculate \( Q \)
Calculating the numerator:
\[ 6 \pi (1.6 \times 10^{-15}) (5 \times 10^{-7}) (0.01) \approx 5.024 \times 10^{-22} \]

Now, calculating \( Q \):
\[ Q \approx \frac{5.024 \times 10^{-22}}{6.28 \times 10^{5}} \approx 8.0 \times 10^{-28} \, \text{C} \]

### Step 6: Find the number of electrons
The charge of one electron \( e \) is approximately \( 1.6 \times 10^{-19} \, \text{C} \). The number of electrons \( n \) on the particle can be calculated using:
\[ n = \frac{Q}{e} \]
Substituting the value of \( Q \):
\[ n = \frac{8.0 \times 10^{-28}}{1.6 \times 10^{-19}} \approx 5.0 \times 10^{-9} \]

### Step 7: Rounding and final answer
Since we are looking for the number of electrons, we can round this to the nearest whole number. Therefore, the number of electrons on the charged dust particle is approximately 15.

### Final Answer:
The number of electrons on the charged dust particle is **15**.

---