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 positron, 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 a proton at rest and a positively charged particle (either a proton or a positron) released from rest in its electric field. We will consider the forces acting on both particles and how they affect the work done on each particle over a time interval \( t \).

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a stationary proton and a positively charged particle (which can be either a proton or a positron) released from a distance \( d \).
- The force acting on the moving charged particle due to the stationary proton is given by Coulomb's law.

2. **Coulomb's Law**:
- The force \( F \) between two charges is given by:
\[ F = \frac{k \cdot q_1 \cdot q_2}{r^2} \]
- Here, \( k \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the charges of the particles, and \( r \) is the distance between them.

3. **Identifying the Charges**:
- For both cases (proton and positron), the charge \( q \) is the same (the charge of a proton, \( e \)).
- Thus, the force acting on both particles (proton and positron) due to the stationary proton will be the same:
\[ F = \frac{k \cdot e^2}{d^2} \]

4. **Calculating Acceleration**:
- The acceleration \( a \) of each particle can be calculated using Newton's second law \( F = m \cdot a \):
- For the proton:
\[ a_p = \frac{F}{m_p} = \frac{k \cdot e^2}{m_p \cdot d^2} \]
- For the positron (mass \( m_e \)):
\[ a_e = \frac{F}{m_e} = \frac{k \cdot e^2}{m_e \cdot d^2} \]

5. **Comparing Accelerations**:
- Since the mass of the proton \( m_p \) is much larger than the mass of the positron \( m_e \) (approximately 1836 times), we can conclude:
\[ a_p < a_e \]
- This means the positron will accelerate more than the proton.

6. **Calculating Distance Traveled**:
- The distance traveled \( S \) by each particle in time \( t \) can be calculated using the equation of motion:
\[ S = \frac{1}{2} a t^2 \]
- For the proton:
\[ S_p = \frac{1}{2} a_p t^2 \]
- For the positron:
\[ S_e = \frac{1}{2} a_e t^2 \]

7. **Comparing Distances**:
- Since \( a_e > a_p \), it follows that:
\[ S_e > S_p \]

8. **Calculating Work Done**:
- The work done \( W \) on each particle is given by:
\[ W = F \cdot S \]
- Since the force \( F \) is the same for both particles, the work done will depend on the distance traveled:
\[ W_p = F \cdot S_p \quad \text{and} \quad W_e = F \cdot S_e \]
- Therefore, since \( S_e > S_p \):
\[ W_e > W_p \]

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