A charged particle of radius `5xx10^(-7)m` is located in a horizontal electric field of intensity `6.28xx10^(5)Vm^(-1)`. The surrounding medium has the coeffecient of viscosity `eta=1.6xx10^(-5)Nsm^(-2)`. The particle starts moving under the effect of electric field and finally attains a uniform horizontal speed of `0.02 ms^(-1)`. Find the number of electrons on it. Assume gravity free space.

Explanation

To solve the problem step by step, we will follow the reasoning laid out in the video transcript.

### Step 1: Understand the Forces Acting on the Particle
The charged particle experiences two forces:
1. **Electric Force (F_E)**: This is the force due to the electric field acting on the charge.
2. **Drag Force (F_D)**: This is the viscous drag force acting opposite to the direction of motion.

### Step 2: Write the Expressions for the Forces
1. **Electric Force** can be expressed as:
\[ F_E = Q \cdot E \]
where \( Q \) is the charge on the particle and \( E \) is the electric field intensity.

2. **Drag Force** can be expressed using Stokes' law as:
\[ F_D = 6 \pi r \eta v \]
where \( r \) is the radius of the particle, \( \eta \) is the coefficient of viscosity, and \( v \) is the velocity of the particle.

### Step 3: Set the Forces Equal
Since the particle attains a uniform speed, the net force acting on it is zero. Therefore, we can set the electric force equal to the drag force:
\[ F_E = F_D \]
This gives us:
\[ Q \cdot E = 6 \pi r \eta v \]

### Step 4: Substitute Known Values
Now, we can substitute the known values into the equation:
- Radius \( r = 5 \times 10^{-7} \, \text{m} \)
- Electric field \( E = 6.28 \times 10^{5} \, \text{V/m} \)
- Coefficient of viscosity \( \eta = 1.6 \times 10^{-5} \, \text{Ns/m}^2 \)
- Velocity \( v = 0.02 \, \text{m/s} \)

Substituting these values into the equation:
\[ Q \cdot (6.28 \times 10^{5}) = 6 \pi (5 \times 10^{-7}) (1.6 \times 10^{-5}) (0.02) \]

### Step 5: Calculate the Right Side
Calculating the right side:
\[ 6 \pi (5 \times 10^{-7}) (1.6 \times 10^{-5}) (0.02) = 6 \times 3.14 \times 5 \times 1.6 \times 0.02 \times 10^{-12} \]
Calculating this gives:
\[ = 6 \times 3.14 \times 5 \times 1.6 \times 0.02 \approx 0.0006 \, \text{N} \]

### Step 6: Solve for Charge \( Q \)
Now, we can solve for \( Q \):
\[ Q = \frac{6 \pi (5 \times 10^{-7}) (1.6 \times 10^{-5}) (0.02)}{6.28 \times 10^{5}} \]
Calculating this gives:
\[ Q \approx 48 \times 10^{-19} \, \text{C} \]

### Step 7: Calculate the Number of Electrons
To find the number of electrons \( n \), we use the formula:
\[ n = \frac{Q}{e} \]
where \( e = 1.6 \times 10^{-19} \, \text{C} \) (the charge of a single electron).

Substituting the values:
\[ n = \frac{48 \times 10^{-19}}{1.6 \times 10^{-19}} = 30 \]

### Final Answer
The number of electrons on the charged particle is \( n = 30 \).

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