A particle of mass 2 g and charge `1 muC` is held at rest on a frictionless surface at a distance of 1m from a fixed charge of 1 mC. If the particle is released it will be repelled. The speed of the particle when it is at distance of 10 m from fixed charge is :

Options

A. 100 m/s
B. 90 m/s
C. 60 m/s
D. 45 m/s

Answer

To solve the problem step by step, we will use the principle of conservation of energy, specifically the relationship between kinetic energy and electrostatic potential energy.

### Step 1: Understand the Given Data
- Mass of the particle, \( m = 2 \, \text{g} = 2 \times 10^{-3} \, \text{kg} \)
- Charge of the particle, \( q_1 = 1 \, \mu\text{C} = 1 \times 10^{-6} \, \text{C} \)
- Fixed charge, \( q_2 = 1 \, \text{mC} = 1 \times 10^{-3} \, \text{C} \)
- Initial distance from the fixed charge, \( r_1 = 1 \, \text{m} \)
- Final distance from the fixed charge, \( r_2 = 10 \, \text{m} \)

### Step 2: Write the Energy Conservation Equation
When the particle is released, the change in kinetic energy of the particle equals the change in electrostatic potential energy:
\[ \Delta KE = \Delta PE \]
This can be expressed as:
\[ \frac{1}{2} mv^2 - 0 = U_f - U_i \]
where \( U_f \) is the potential energy at \( r_2 \) and \( U_i \) is the potential energy at \( r_1 \).

### Step 3: Calculate the Initial and Final Potential Energy
The electrostatic potential energy \( U \) between two point charges is given by:
\[ U = k \frac{q_1 q_2}{r} \]
where \( k \) is Coulomb's constant, \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \).

1. **Initial Potential Energy \( U_i \) at \( r_1 = 1 \, \text{m} \)**:
\[ U_i = k \frac{q_1 q_2}{r_1} = 9 \times 10^9 \cdot \frac{(1 \times 10^{-6})(1 \times 10^{-3})}{1} = 9 \times 10^9 \cdot 1 \times 10^{-9} = 9 \, \text{J} \]

2. **Final Potential Energy \( U_f \) at \( r_2 = 10 \, \text{m} \)**:
\[ U_f = k \frac{q_1 q_2}{r_2} = 9 \times 10^9 \cdot \frac{(1 \times 10^{-6})(1 \times 10^{-3})}{10} = 9 \times 10^9 \cdot 1 \times 10^{-10} = 0.9 \, \text{J} \]

### Step 4: Substitute into the Energy Equation
Substituting \( U_i \) and \( U_f \) into the energy equation:
\[ \frac{1}{2} mv^2 = U_f - U_i \]
\[ \frac{1}{2} mv^2 = 0.9 - 9 = -8.1 \, \text{J} \]
This indicates a mistake, as potential energy should decrease, thus:
\[ \Delta PE = U_i - U_f = 9 - 0.9 = 8.1 \, \text{J} \]
So,
\[ \frac{1}{2} mv^2 = 8.1 \]

### Step 5: Solve for Velocity \( v \)
Rearranging the equation to solve for \( v \):
\[ v^2 = \frac{2 \cdot 8.1}{m} \]
Substituting \( m = 2 \times 10^{-3} \, \text{kg} \):
\[ v^2 = \frac{16.2}{2 \times 10^{-3}} = 8100 \]
\[ v = \sqrt{8100} = 90 \, \text{m/s} \]

### Final Answer
The speed of the particle when it is at a distance of 10 m from the fixed charge is \( 90 \, \text{m/s} \).