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 find the number of electrons on a charged dust particle moving in an electric field, we can follow these steps:

### Step 1: Understand the Forces Acting on the Particle
The charged dust particle experiences two main forces:
1. **Viscous Force (F_viscous)** due to the surrounding medium.
2. **Electric Force (F_electric)** due to the electric field.

When the particle moves with a uniform speed, these two forces are equal in magnitude.

### Step 2: Write the Expression for Viscous Force
The viscous force acting on a sphere moving through a fluid is given by the formula:
\[ F_{\text{viscous}} = 6 \pi \eta r v \]
Where:
- \(\eta\) = coefficient of viscosity
- \(r\) = radius of the particle
- \(v\) = velocity of the particle

### Step 3: Write the Expression for Electric Force
The electric force acting on a charged particle in an electric field is given by:
\[ F_{\text{electric}} = qE \]
Where:
- \(q\) = charge on the particle
- \(E\) = electric field intensity

Since the charge \(q\) can be expressed in terms of the number of electrons \(n\) (where \(e\) is the charge of a single electron):
\[ q = n \cdot e \]
Thus, the electric force can be rewritten as:
\[ F_{\text{electric}} = n \cdot e \cdot E \]

### Step 4: Set the Forces Equal
Since the particle is moving with uniform speed, we can equate the two forces:
\[ 6 \pi \eta r v = n \cdot e \cdot E \]

### Step 5: Solve for the Number of Electrons (n)
Rearranging the equation to solve for \(n\):
\[ n = \frac{6 \pi \eta r v}{e \cdot E} \]

### Step 6: Substitute the Given Values
Now we can substitute the known values into the equation:
- \(\eta = 1.6 \times 10^{-15} \, \text{Ns/m}^2\)
- \(r = 5 \times 10^{-7} \, \text{m}\)
- \(v = 0.01 \, \text{m/s}\)
- \(e = 1.6 \times 10^{-19} \, \text{C}\) (charge of an electron)
- \(E = 6.28 \times 10^{-5} \, \text{V/m}\)

Substituting these values gives:
\[ n = \frac{6 \pi (1.6 \times 10^{-15}) (5 \times 10^{-7}) (0.01)}{(1.6 \times 10^{-19}) (6.28 \times 10^{-5})} \]

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

Calculating the denominator:
\[ (1.6 \times 10^{-19}) (6.28 \times 10^{-5}) \approx 1.0048 \times 10^{-23} \]

Now, divide the numerator by the denominator:
\[ n \approx \frac{5.024 \times 10^{-22}}{1.0048 \times 10^{-23}} \approx 50 \]

### Final Answer
The number of electrons on the charged dust particle is approximately **50**.

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