A proton is kept at rest. A positively charged particle is released from rest at a distance `d` in its field. Consider two experiments, one ini which the charged particle is also a proton and in another, a position. In the same time `t`, the work done on the two moving charged particles is

Options

A. same as the same force law is involved in the two experiments
B. less for the case of a positron, as the positron moves away more rapidly and the force on it weakens
C. more for the case of a position, as the positron moves away a larger distance
D. same as the work done by charged particle on the stationary proton

Explanation

To solve the problem, we need to analyze the situation of two positively charged particles (a proton and a positron) being influenced by the electric field of a stationary proton. We will calculate the work done on each particle as they move in the electric field created by the stationary proton.

### Step-by-Step Solution:

1. **Understanding the Forces**:
- The force \( F \) between two charged particles is given by Coulomb's law:
\[ F = \frac{k \cdot q_1 \cdot q_2}{r^2} \]
- Here, both the moving particle (proton or positron) and the stationary proton have the same charge \( e = 1.6 \times 10^{-19} \, C \).

2. **Calculating the Force on Each Particle**:
- For both the proton and the positron, the force acting on them due to the stationary proton is:
\[ F_1 = F_2 = \frac{k \cdot e^2}{d^2} \]
- This means the force acting on both particles is equal.

3. **Work Done Calculation**:
- The work done \( W \) on a charged particle when it moves in an electric field is given by:
\[ W = \int F \, dx \]
- Since the force is constant, we can simplify this to:
\[ W = F \cdot x \]
- For both particles, the work done can be expressed as:
\[ W_1 = F_1 \cdot x_1 \quad \text{(for the proton)} \]
\[ W_2 = F_2 \cdot x_2 \quad \text{(for the positron)} \]

4. **Relating Displacement and Acceleration**:
- The acceleration \( a \) of each particle can be expressed using Newton's second law:
\[ a_1 = \frac{F_1}{m_1} \quad \text{(for the proton)} \]
\[ a_2 = \frac{F_2}{m_2} \quad \text{(for the positron)} \]
- Since both forces are equal, we can relate their accelerations:
\[ a_1 = \frac{F}{m_1}, \quad a_2 = \frac{F}{m_2} \]
- Given that the mass of the positron \( m_2 \) is less than that of the proton \( m_1 \), we have \( a_2 > a_1 \).

5. **Displacement in the Same Time**:
- Since both particles are released from rest and experience constant acceleration, the displacement \( x \) after time \( t \) is given by:
\[ x = \frac{1}{2} a t^2 \]
- Therefore, the displacements can be expressed as:
\[ x_1 = \frac{1}{2} a_1 t^2, \quad x_2 = \frac{1}{2} a_2 t^2 \]
- Since \( a_2 > a_1 \), it follows that \( x_2 > x_1 \).

6. **Comparing Work Done**:
- Substituting the displacements into the work done equations:
\[ W_1 = F \cdot x_1, \quad W_2 = F \cdot x_2 \]
- Since \( x_2 > x_1 \) and \( F \) is the same for both, we conclude:
\[ W_2 > W_1 \]
- This means the work done on the positron is greater than the work done on the proton.

### Conclusion:
The work done on the positron is greater than the work done on the proton in the same time \( t \).